Radiating the Ocean

Guest Post by Willis Eschenbach

Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.

Figure 1. The question in question.

There are lots of good arguments against the AGW consensus, but this one is just silly. Here are four entirely separate and distinct lines of reasoning showing that DLR does in fact heat the oceans.

Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land. DLR is absorbed in the first mm of rock or soil. Yet the same people who claim that DLR can’t heat the ocean (because it’s absorbed in the first mm) still believe that DLR can heat the land (despite the fact that it’s absorbed in the first mm).

And this is in spite of the fact that the ocean can circulate the heat downwards through turbulence, while there is no such circulation in the land … but still people claim the ocean can’t heat from DLR but the land can. Logical contradiction, no cookies.

Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.

Nor can it be going to evaporation as many claim, because the numbers are way too large. Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount … and some of the evaporation is surely coming from the heating from the visible light.

So if the DLR is not heating the ocean, and we know that a maximum of less than a quarter of the energy of the DLR might be going into evaporation, and the DLR is not heating the air … then where is it going?

Rumor has it that energy can’t be created or destroyed, so where is the energy from the DLR going after it is absorbed by the ocean, and what is it heating?

Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning. At night the top layer of the ocean naturally overturns, driven by the temperature differences between surface and deeper waters (see the diagrams here). DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.

Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.

But the average solar input to the surface is only about 170 watts/square metre.

So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?

Note that each of these arguments against the idea that DLR can’t warm the ocean stands on its own. None of them depends on any of the others to be valid. So if you still think DLR can’t warm the ocean, you have to refute not one, but all four of those arguments.

Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …

w.

4 2 votes
Article Rating
908 Comments
Oldest
Newest Most Voted
Inline Feedbacks
View all comments
Bystander
August 15, 2011 12:53 pm

So the physical properties of say, rock, and water are identical?

jungle
August 15, 2011 1:02 pm

Wouldn’t DLR be absorbed deeper than a mm in water due to the difference between a solid and a liquid?

Theo Goodwin
August 15, 2011 1:06 pm

Great work again, Willis. Would you please take a moment and tell us how Warmista models treat the heat phenomena that you have described as a thermostatic system? With specific regard to infrared radiation, where does your account differ from the Warmista in the way it treats heat and where is there overlap? How do the Warmista models handle what you say about warming from sunlight, if they handle it at all?

David
August 15, 2011 1:08 pm

I think that the arguement is a question of degree and residence time, not a question of does it heat, but how much? “But the average solar input to the surface is only about 170 watts/square metre.” I am not certain average is what is moat important here. SWR at the tropics and even subtropics is far greater, and because it penetrates far deeper it likely has a much greater residence time. Thus any change in SWR flux may accumulate far more energy over time then a similar W/M2 change in SWR caused by a GHG. Yet I have seen none of this quantified.

gnomish
August 15, 2011 1:08 pm

http://en.wikipedia.org/wiki/Solar_pond
if i had back issues of mother earth news i could show that this has been in the green literature almost as much as how to raise goats in your basement.

August 15, 2011 1:11 pm

Bystander said:
“So the physical properties of say, rock, and water are identical?”
And I say . . . Huh? Is there a point to your statement? The physical properties of rock (lets say basalt) and rock (lets say limestone) are not identical, so I would suspect that rock and water do indeed differ. What has that got to do with the article?

Doug Jones
August 15, 2011 1:11 pm

Bystander, no, but they both absorb longwave IR in the 5-15 micron wavelength range in about a millimeter or less. Strawman argument, no cookie.

Gary wilson
August 15, 2011 1:13 pm

Although the end result is the same it is incorrect to state that DLR heats the surface. DLR is radiation that has already left the surface and been absorbed and re-radiated. It is the sun that does the heating. All the DLR does is slow down the heat loss from the surface. Without a heat source such as the sun the surface would gradually cool as the surface flux and DLR decline. So saying DLR heats the surface is not strictly accurate.

Paul Westhaver
August 15, 2011 1:17 pm

In examining surface temperature of night-time compared with day time it can be seen to what portion of the temperature is due to direct IR or remitted IR after absorption.
Imaging if there is a single night with 36 hours rather than 12. The heat loss would be tripled and the temp drop would likely be in the 30 degrees range where I Iive… look at this daily plot:
http://www.cita.utoronto.ca/~rjh/halifax/halifax-weather-ll.html
If night to day has such a huge effect on surface temp…..then why are we torturing ourselves about fractions of a degree? Seems to me the earth has quite a huge capacity to loose heat. We see the effect every night.
I doubt that the earth has chronic heat gain…. heat emission is a forth power of temperature.

August 15, 2011 1:22 pm

So a IR lamp directed at a one square meter tub of water of say one meter in depth to radiate the surface at 170W wouldn’t warm that body of water if it was circulating?
If it only heats the first mm why then are several inches usually warm when going swimming? Or is that an odd question?
170W per square meter, a few billion years . . . it hasn’t done all that much though, except around the equator, has it, I mean the north atlantic, the arctic sea and the antarctic sea is cold as bloody hell year around, but where there’s lots and lots of sun the sea is pretty darn cozy all year around.

Theo Goodwin
August 15, 2011 1:29 pm

gnomish says:
August 15, 2011 at 1:08 pm
Did you get the one where Barry Commoner explained how to heat a 12 story residential building using a V-8 engine in the basement?

Climate Weenie
August 15, 2011 1:31 pm

Dunno about the rest of the notions, but
Evaporation is known to be on the order of 70 w/m2, while average downwelling longwave radiation is more than four times that amount …
is not apt.
it is not the downwelling LWR that is significant, but the net of LW and SW, (the imbalance) that is either met by another process (convection), or results in a temperature change.
Net longwave and shortwave, in the average, is not more than four times 70 W/m^2.

philincalifornia
August 15, 2011 1:31 pm

David says:
August 15, 2011 at 1:08 pm
I think that the arguement is a question of degree and residence time, not a question of does it heat, but how much? “But the average solar input to the surface is only about 170 watts/square metre.” I am not certain average is what is moat important here. SWR at the tropics and even subtropics is far greater, and because it penetrates far deeper it likely has a much greater residence time. Thus any change in SWR flux may accumulate far more energy over time then a similar W/M2 change in SWR caused by a GHG. Yet I have seen none of this quantified.
============================
Yeah, I keep trying to get an answer to this question too (I think it’s part of the same question). What is the change in DLR over the tropical water vapor-containing atmosphere with 390 ppm of CO2, versus the tropical water vapor-containing atmosphere with 280 ppm of CO2, including Doppler Shift effects on the water vapor absorption, of course.

Merrick
August 15, 2011 1:39 pm

And, hate to bring up a long gone painful debate, but if DLR can heat one fluid (water) then it ABSOLUTELY CAN heat another fluid (air) and in EXACTLY the same way. Local thermodynamic equilibrium arguments notwithstanding.

Mac the Knife
August 15, 2011 1:40 pm

Anybody who swims or snorkels the sun warmed ‘top layer’ in a lake and has felt the distinct thermoclines at colder layers beneath, has already confirmed that the surface water absorbs long wave radiation. After a day or two of moderate to brisk winds, the top layer is cooled, confirming cooling from evaporation and turbulent mixing with even cooler water below the nearest thermocline.
Spouting baseless theories and touting irrational models will never refute the simple and direct confirmation that a few days of open water swimming or near surface water temperature measurements will provide.

FerdinandAkin
August 15, 2011 1:43 pm

DLR cannot be absorbed by the first mm of water because it is clear therefore it must be absorbed by the second mm of water.
DLR cannot be absorbed by the second mm of water because it is clear therefore it must be absorbed by the third mm of water.
Actually DLR cannot be absorbed by any of the water because it is mm all the way down.
/sarc

Roy Spencer
August 15, 2011 1:44 pm

Glad to see someone else take the flak for this for a change. Thanks, Willis. 😉
-Roy

tallbloke
August 15, 2011 1:46 pm

Hi Willis,
Argument one asks what the difference is between rock and water. Warm water molecules rise to the top. Warm rock molecules conduct heat to their neighbours, which can’t go anywhere.
Argument two asks where the energy goes. The answer is:
space.
Argument three is not an argument that DLR can warm the ocean, it’s an argument that it can slow its rate of cooling.
Argument four is a numerical misunderstanding. The ocean surface very efficiently absorbs 95% of DLR, and promptly re-emits half of that (the other 5% being reflected). The other half makes it another couple of molecules deeper and then the molecules it warms become more buoyant than their neighbours and rise to the top, losing another half upwards. Now we’re down to ~72W/m^2. Lets remember the net flux is 66W/m^2 upwards at this point. So your ice cube argument fails. The ocean absorbs and re-emits the long wave radiation coming downwards from the atmosphere, the sums balance. In fact it emits 66W/m^2 more long wave radiation than it absorbs. It always has, and the oceans don’t freeze, because solar shortwave warms them to really significant depths of 100 metres and more as internal tides and currents mix its energy downwards. Some of that solar short wave energy is re-emitted as long wave from the surface along with some of the long wave which came from the atmosphere. The rest causes evaporation and thermals or is conducted upwards. The difference is, the solar derived energy can remain deep in the ocean for a long time, controlling it’s bulk temperature.
The question is, do DLR heated water molecules make it downwards far enough for long enough to warm the ocean bulk. I think the answer is no, because warmer water molecules are naturally buoyant, and because the vortices which mix solar energy so efficiently are below the wave troughs, several thousands of times deeper down than the depth DLR penetrates water to. For experimental evidence on this matter I’ve tried putting small soaked pieces of loo paper just under the surface out in the rolling waves away from the shore where they break. They don’t get sucked downwards. So that’s turbulent convection gone, what’s left? Conduction is a non-starter, because water thermally stratifies and anyway is a relatively poor heat conductor unless the heat source is underneath rather than above.
But this isn’t about absolutes. I’m sure the increased DLR warmed the ocean a little bit, or at least slowed its rate of cooling a little bit. I think the increased insolation due to (empirically measured) reduced cloud cover in the tropics 1980-1998 did a lot more to increase ocean heat content. To turn your question back to you, where else could that energy have gone?
Cheers
TB

DirkH
August 15, 2011 1:48 pm

Merrick says:
August 15, 2011 at 1:39 pm
“And, hate to bring up a long gone painful debate, but if DLR can heat one fluid (water) then it ABSOLUTELY CAN heat another fluid (air) and in EXACTLY the same way. ”
Air is mostly occuring in the gaseous phase.

Fred from Canuckistan
August 15, 2011 1:49 pm

You would think that the average ninth grade science student could design an experiment to measure what happens when IR hits water in a tank . . . a great Science Fair experiment.
But if a ninth grader could do it, it will be beyond the tall forehead types who walk posts on the IPCC ramparts.

jimmi_the_dalek
August 15, 2011 1:52 pm

Indeed, the idea that DLR cannot heat the ocean is one of the spurious arguments that should not be used. There are plenty of others. Far too many comments here claim that the Greenhouse Effect cannot be real “because it is not like a real greenhouse”. Or that it contradicts the first or second laws of thermodynamics. Stick to objecting to the infidelities of computer modelling – there is plenty of uncertainty there – the basic physics is much more secure than some people are willing to concede.

Paul Westhaver
August 15, 2011 2:04 pm

The time lag between local apparent noon and max surface temperature is no more than 4 hours. The time constant for the surface heating an cooling rate is VERY short. If every night the surface on the planet cools 2-5 degrees…or more, then what is the big deal about 0.1 degree of warming…notwithstanding the fact that the thermometer sensitivity (globally) is incapable of reporting that level of precision… as Lord Monckton has frequently said.

crosspatch
August 15, 2011 2:04 pm

DLR is radiation that has already left the surface and been absorbed and re-radiated. It is the sun that does the heating. All the DLR does is slow down the heat loss from the surface.

And I believe that is one of the most misunderstood things about greenhouse warming. Greenhouse warming doesn’t have its maximum impact when the sun is shining, it is at night. When the dominant heat flow is, at night, from Earth to space, the DLR would reduce the rate of cooling and it would moderate low temperatures. Greenhouse warming shouldn’t increase daytime highs, it should increase nighttime lows. Daytime high temperatures will be moderated by things like evaporation and clouds but the nighttime temperatures are moderated only by the blanket that lies between the surface and space.
In the daytime, the same greenhouse gasses also absorb IR from the sun and radiate half of that back into space. Roughly half the energy of solar radiation is IR. Greenhouse gasses act to moderate the amount of IR reaching the surface during the day and moderate the amount leaving the atmosphere during the night.
It is quite possible that one might get with increasing greenhouse efficiency is lower high temperatures and higher low temperatures for a reduction in diurnal range and possibly no change at all in average in the tropics. The greatest impact should be seen at the poles in winter where any enhanced “insulation” should be most easily observed.

Robert M
August 15, 2011 2:06 pm

Hmmmm,
So far I have to think that Tallbloke gets the cookie! Sigh! It’s not a chocolate chip cookie is it?
One more thing… What idiot proposed that LR does not heat water?
Well, two more things… Somebody give Bystander a penny so he can buy himself a clue. 🙂

Crispin in Waterloo
August 15, 2011 2:07 pm

The choice of an infrared lamp as a heat is not the best experiment – too much visible light. Put a just-red-hot heating element over the pan of water. The point is to try to heat without visible radiation.
The heater will send out LWR and it will not get nearly as far as 1mm into the water because it is very effective at stopping it penetrating at most IR frequencies. The water will get very much hotter at the surface than a few inches below, unless it is stirred (which of course happens in a real ocean). Water stratifies very well when it is hotter on top. It of course tends to evaporate more even when calm, and when stirred, much more so t here are several considerations.
The 4 arguments seem to mix all frequencies and LWR in together, or shall I say, perhaps do not separate them well enough. Tropical ocean water is definitely heated by the 1 kW/m^2 of incident solar radiation. The question is how much more is it heated by additional LWR (only, not additional sunlight). Well it certainly is not heated deeply by LWR unless there is significant overturning, which in many cases exists, yes?
So, if we are going to live in a real world (and ocean) then we have to consider that there is a great deal of water vapour immediately above the water’s surface absorbing the IR like crazy, remaining heated and not condensing to return, condensed, to the surface. Over a real ocean, the humidity is very high near the surface. How much of the additional LWR reaches the water’s surface, if any? If it does, it will be absorbed in the very top of its surface and promote immediate evaporation in the zone there there is a continuous exchange between the surface and the very bottom of the atmosphere. Any additional LWR that reaches the top of the water will shift the balance to produce slightly more (IR absorbing) water vapour immediately above it. Rapidly too. That thin air-water exchange layer just above the surface is basically in a constant state of saturation or supersaturation and is a really good IR shield.
Please convince me that any of the downwelling IR reaches the surface of the water. I am quite prepared to be convinced and it is essential to the argument that additional downwelling IR can heat the ocean. How much IR hits the water, in Watts? Does this heat flux depend on the local temperature and wind conditions?

Steve from Rockwood
August 15, 2011 2:16 pm

Is the top part of that 1 mm heated up to the same temperature as the bottom part or is there a temperature gradient over that 1 mm?

George
August 15, 2011 2:19 pm

Really? There are people that believe that? How do they have time to post because they must spend all their time rewatching Loose Change.
Without DLR heating the water, there would be no point to Bruce Brown’s Endless Summer. Next.

david
August 15, 2011 2:20 pm

Yes, back in Sweden when I grew up, that first millimeter didn’t absorbed it all summer long, so it was always solide ice year round….. /sarc
In fact with little tidal change ( 1 foot on average), on a good summer (not too many), the first ten feet would be warm. You could see the thermal layer below if you snorkled.
Heat conduction, most likely?
Wave action, not so likely, it would mix out the thin thermal layer?
Heat radiating down that far? I don’t know, can infrared penetrate that far?
Lastly doesn’t heat radiate in all directions? So, that millimeter would radiate up and down right? So the millimeter below would get some heat that way too, and so on?

August 15, 2011 2:26 pm

For those who doubt the greenhouse effect, here is a standard thermodynamics question.
If you know how to solve this, you know why Willis is correct.
Here is the problem (mechanical engineering profs should feel free to include this on the next thermodynamics exam).
1: A continuously heated, flat plate at 15 C is on one side, a parallel, continuously cooled plate at -273C is 20 meters away. What energy flux is required to maintain the heated plate at 15C under the following conditions:
The atmosphere in between is 10% CO2, 72% N2, 18% O2, 1 bar pressure.
The atmosphere in between is 20% CO2, 64% N2, 16% O2, 1 bar pressure.
2: Solve again, assuming that the energy input to the plate is constant, and determine the temperature of the plate at the higher level of CO2 if the temperature at the lower level is 15C and the second plate remains at -273C.
Now comes the part that seems to be missed by our friends in the climate field:
3: At what level of CO2 does the increase in CO2 no longer result in an increase in temperature of the first plate.
If you studied heat transfer at the university level, you may have seen a question like this. If you did, you would know that the plate would be warmer in the second instance (I am procrastinating at work right now, so will leave it as a blog post later to fully answer) and the level of CO2 would be 25%.
Cheers
JE

Rosco
August 15, 2011 2:31 pm

But surely the myth about DLR is that it is the tiny fraction of the atmosphere responsible – the less than 1% of the atmosphere that absorbs infrered is not responsible for more than a fraction of DLR. This is the deception that, like a magician diverting your attention – don’t look at that look at thislook at this -from reality to sell you on their theory
Surely the whole of the atmosphere warms and cools predominantly by conduction and convection ? And surely, like everything else in the Universe, the Nitrogen and Oxygen, being at more than absolute zero, constitute the vast bulk of the DLR – ~98% of it ? It must be impossible to seperately detect the tiny amount from GHGs ?
And surely the latent heat of evaporation of water, being some 2500 times the specific heat capacity of CO2, plays a far more important role in energy distribution on earth than DLR ?
Also surely DLR is a small component of the total radiation in the atmosphere with random emission in any direction but surely mostly upwelling following the rising thermal currents ?
So surely on earth radiation plays a minor role in energy distribution ? Else why do common objects such as a car’s “radiator” really rely on conduction and convection ? Why do they put the fan in the oven if radiation is so effective ?
And why can I sit very close to a radiative heater for long periods (not above it though) but when I touch it I’m immediately off to the emergency room ?
The answer must be that radiation, while it may be the only energy transport mechanism through space, is only a minor player on earth ? Sure it is the only escape but there are well known mechanisms for getting the energy up to the upper levels of the atmosphere and radiation is only a small part of that.

August 15, 2011 2:37 pm

As someone else noted it is the term warming. If DWLR doesn’t reverse the direction of net flux it is not warming. There is also the legitimate question of is conductive heat transfer at the surface more significant than radiative heat transfer. Since the radiative situation is liquid water shooting photons at water vapor which shoots back etc., the average 0.8 degree temperature difference between the water and the air seems pretty reliable. With more CO2 that average may become 0.78 degrees.

R. Gates
August 15, 2011 2:37 pm

Willis,
I’m beginning to see that you are a heretic, and I love it. Yes, this falsehood that DLR can’t heat the oceans has been one of the most absurd things that skeptics have said, this, and the notion that human activity has not been the principal driving cause behind the growth of atmosphere CO2 over the past several centuries.
I realize you are not a “warmist” as I am, though some people insist on lumping all “warmist” thought into the C-AGW category. Regardless, you and I might see eye to eye on more things than might first seem apparent.

John Vetterling
August 15, 2011 2:38 pm

The problem seems to be a lot of folks who just don’t understand thermodynamics.
Light is energy. If the water absorbs light it absorbs that energy, hence its heat increases. It oes not matter how thick the absorbing layer is.
Radiant heat transfer is not the same as conductive heat transfer. Conduction depends on the ratio of the temperatures. Radiance depends only on the temperature of the radiating body not the temperature of any sink.
Convection in the air is much different than convention in the seas. The lowest layers of air are heated then rise as they expand. In the seas it is the upper layers that heat and expand, but there is nowhere to rise because they are already above cooler denser layers.
An awful lot of Zeno’s paradox going on as well.
The basic physics of GW is relatively sound. The weaknesses lie in the guestimates of sensitivity/forcings.

August 15, 2011 2:43 pm

Nice post . You would never see the flip side of a post like this on a AGW blog ( being critical of a AGW principle ). The true believers would not put up with it. Good science requires that weak hypotheses be appropriately dissected, which this post does.
Of course, the real question is how that energy is then distributed / mixed into the ocean,

August 15, 2011 2:52 pm

There is no doubt that some warming takes place somewhere but the issue is whether that warming does in fact get transmitted downwards and/or whether it does in fact slow down the ‘normal’ rate of energy loss from the oceans.
As far as I can tell the DLR warming just affects individual water molecules at the very surface merely microns deep. Other wavelengths get in deeper however.
DLR having warmed those individual molecules there is an increase in all of convection conduction radiation and evaporation. Effectively all that happens is that the moment of evaporation for all the water molecules affected is brought forward by whatever energy is left over after conduction radiation and convection have taken their portions.and of course evaporation is a net cooling process as we all know.
Willis has previously told me that the increase in evaporation alone is not sufficient to deal with ALL the DLR but suppose the increased evaporation only has to mop up what is left over AFTER increased upward radiation convection and conduction. The figures are then much more believable.
Looking for some evidence one way or the other I read about that 1mm deep layer at the top of the oceans that is always (averaged globally) about 0.3C colder than the ocean bulk below.
The thing is that that layer is still there day and night so it isn’t even overcome by the coming and going of sunlight so I don’t see how DLR would make a difference.
That layer represents the net cooling effect of all surface cooling processes.It is on average globally negative.Energy is always leaving the surface layer faster than it is arriving at the surface and is replaced from below by solar shortwave that penetrated deeper. If the energy hitting the surface from DLR were penetrating downward significantly then that layer would become shallower or warmer or dissipate but on average overall it doesn’t.
So to accept what Willis says I need that point addressing. Why does that cooler layer not seem to get warmer, shallower or dissipate (except locally and temporarily) under the influence of more DLR or even under the influence of solar insolation ?
Could it be that it is a permanent feature of the ocean surfaces set by pressure and density differentials such that it provides a buffer between energy arriving in the topmost molecules and that leaving the ocean bulk for a zero effect on both energy into and out of the ocean bulk ?
I like the analogy of a tributary joining a river. The volume of flow downstream of the junction increases but the rate of flow from upstream is not changed. There may be a small area of buffering (akin to that cool layer on the oceans) where there is a slowdown in the rate of flow but nonetheless the rate of flow from higher upsatream is unaffected.
I suspect that that happens at the ocean surface too.

Ian W
August 15, 2011 2:54 pm

1DandyTroll says:
August 15, 2011 at 1:22 pm
So a IR lamp directed at a one square meter tub of water of say one meter in depth to radiate the surface at 170W wouldn’t warm that body of water if it was circulating?
If it only heats the first mm why then are several inches usually warm when going swimming? Or is that an odd question?
170W per square meter, a few billion years . . . it hasn’t done all that much though, except around the equator, has it, I mean the north atlantic, the arctic sea and the antarctic sea is cold as bloody hell year around, but where there’s lots and lots of sun the sea is pretty darn cozy all year around.

Its not an odd question its a rather unthinking question.
the LR in DLR is Longwave Radiation. This does not penetrate far at all into water a millimeter or less. However, you may notice you can see in sunlight. This is because sunlight is largely made up of visible and ultraviolet radiation and this spectrum will penetrate really deeply into the ocean provided that it is at the correct incidence (angle of insolation) and is why the light under water looks greeny blue. If the light is at any angle less than about 40 degrees then most of the light reflects from the surface (see Willis’ last post). This reflection of sunlight at low angles of insolation is another reason that the sunlight on the oceans near the poles does not warm the water much (as you pointed out).
So as you say where there is lots of sun the sea tends to stay warm – but even there – the sea surface water will be considerably cooler by dawn having both radiated and evaporated a lot of the previous day’s heat away.

Rosco
August 15, 2011 2:59 pm

I have just read your steel greenhouse post.
Surely the myth that the earth receives only 235 W/sq m insolation and thus the earth would only be ~minus 18 or 19 C is the big lie of GHG theory.
To model the earth as a disk illuminated by only one quarter of the solar constant is nonsense and we ought to say so. At noon at the equator the insolation is probably at least double the 235 W/sq m (perhaps nearly triple) so often quoted while at one of the poles it is probably close to zero.
The only thing maintaining the poles at the modest temperaturesthey are at compared to absolute zero is the energy transmission of atmospheric and oceanic currents.
I do not believe that at a place like Death Valley or central Australia DLR is responsible for the increase in energy from 235 W/sq m to the ~617 W/sq m necessary to explain a temperature of 50 C as has been observed or the 544 W/sq m required to explain the 40 C temperatures regularly recorded less than 50 km from where I live.
Clearly that energy comes from the sun so why perpetuate this “average” 235 W/sq m myth that the AGW use to distort truth and science ?

Bystander
August 15, 2011 3:03 pm

Hey – I’m not the one with the cartoon and those examples that suggest the behavior is identical.

Dave Springer
August 15, 2011 3:03 pm


“The question is, do DLR heated water molecules make it downwards far enough for long enough to warm the ocean bulk. I think the answer is no, because warmer water molecules are naturally buoyant”
BINGO!
Give the man a cigar.

Chuckarama
August 15, 2011 3:04 pm

Now that I think of it, I have observed sunspots that are shaped much the same as our oceans, like a map cutout, blocking those oceans from being served up as much energy. That would explain why a rock on the beach at the equator is hotter than the ocean waves on that beach, when I touch them. I don’t know how to explain the rabbit and dragon shaped sunspots though…

stevo
August 15, 2011 3:05 pm

No matter how patiently you explain things to some people, they won’t get it. Some people will, no matter how hard you try, still believe that radiation from the atmosphere somehow doesn’t heat the ocean. And yet the physics has been understood for hundreds of years, and tested so thoroughly that to doubt it is simply unreasonable.
Are these people unable to understand? Or are they unwilling? That’s the next interesting question.

MannBitesDog
August 15, 2011 3:11 pm

I notice when I microwave my pasta that the outer layer may become overheated before the center gets warm, that’s why I sometimes use a lower power setting. Sometimes with soup too! Crazy huh?
How did we ever get to the moon?
I wonder how long until “you know who” shows up to comment here?!?

Swift
August 15, 2011 3:17 pm

The Science of Doom has the first of a very good multi post series on DLR and the ocean here http://scienceofdoom.com/2010/10/06/does-back-radiation-heat-the-ocean-part-one/
Most of the questions that have been asked and will be asked are answered there.

Konrad
August 15, 2011 3:18 pm

Willis,
I note with some interest that none of your four arguments match that of the Team. Their offering is that the backscattered LWIR absorbed in the top 1mm cool skin layer of the oceans reduces the thermal gradient across this 1mm, reducing radiative, conductive and evaporative heat losses. Sadly neither their argument or yours appear to have supporting empirical studies.
To much of the AGW doctrine is based on chalk on blackboards and computer models. One thing that many WUWT readers will be aware of is the paucity of empirical testing of the AGW hypothesis. Your suggestion of putting an infra red source over a pan of water was I believe the most sensible part of your post.
This infra red source would need to be emitting only around the 15 micron frequency. The water would need to be salt water at an initial average ocean temperature. An artificial breeze would also be required across the surface of the tank, using air of the appropriate temperature and humidity. A coastal testing location may be appropriate.
I have yet to read a study based on such a test. If you have a link to such a paper Willis, I would greatly appreciate it.

Robin Hewitt
August 15, 2011 3:21 pm

What does it matter? It’s been happening for billions of years without problems, no way is it anthropogenic. Isn’t this a complete red herring in the AGW debate?

August 15, 2011 3:23 pm

Bystander:
In terms of absorbance of radiant energy, the state property of interest is emissivity. Liquid water has an emissivity of about 0.9. Sandstone is closer to 0.7, Basalt 0.7, granite 0.95. So, some land will absorb more radiant energy, including DLR and some will absorb less. Some will indeed be exactly equivalent to water.

August 15, 2011 3:24 pm

Hands – show of hands –
Who HAS swam in Lk Michigan and encountered warm currents on a sunny day?
From whence did they come I wonder?
Anyone – Bueller – Bueller?
.

August 15, 2011 3:25 pm

Get thee back to school Willis 😛
DLR heats the top 10um and the top millimeter has a negative temperature gradient. The very top of the ocean is colder than it is about 1mm down. Think that through and then rephrase your arguments.
Its true the DLR “heats” the ocean but not in the manner you or most others believe. The “heating” is almost entirely related to a reduced rate of cooling. That makes a big difference to how ocean heating should be viewed.

Alexander Duranko
August 15, 2011 3:28 pm

DLR doesn’t exist. It’s the artifact of a mathematical mistake by Sir Arthur Milne in 1922 when he used an infinite atmosphere boundary condition to solve the PDE for IR absorption.
‘Climate science’ made another mistake when it believed the radiometer pointed upwards measured DLR. That signal is real, but exactly counterbalanced at equilibrium by IR in the opposite direction.
Grizzled engineers like me know this because it’s the first law of Radiation – Prevost’s theory of Exchanges [1840].
So the question you pose is irrelevant.

G. Karst
August 15, 2011 3:35 pm

There are a lot of strange and wonderful thing happening in the first mm of water! GK

Dave Springer
August 15, 2011 3:38 pm

Here’s an experiment. Let me know how it turns out.
At normal room temperature fill two identical styrofoam cups with water at 98.6F. Hold your hand about a foot over the top of one of them. Your hand will be providing extra downwelling radiation to that cup. After a few hours measure the temperature in both cups. Let me know how much warmer the cup is that had the extra downwelling longwave radiation. Thanks in advance for actually performing an experiment instead of bloviating about physics you don’t understand, Willis.
.
.

cal
August 15, 2011 3:44 pm

Thanks Willis
I have made the point several times before that people who deny the basic physics undermine the efforts of scientists who challenge the AGW theory. If the theory of DLR and its dependence on CO2 was really as flakey as some would suggest the sceptics would have won the argument 20 years ago. The issue is only about the magnitude of the effect not its existence. The trouble is that climate scientists can point to these silly pseudo scientific arguments as justification for ignoring all the sceptics arguments.

Dave Springer
August 15, 2011 3:48 pm

“Your suggestion of putting an infra red source over a pan of water was I believe the most sensible part of your post.”
Yes, it is. The infrared source should be approximately the same temperature as the water because that’s how it is over the ocean where the air temperature is very near the sea surface temperature. I know what the result will be. Willis is in for a big surprise.

August 15, 2011 3:51 pm

The warming of the surface layer of a lake/ocean isn’t evidence that LWR is absorbed. Solar radiation is clearly the primary source of this heating.
In calm tropical oceans, an hour of cloud versus an hour of sunlight has a noticeable effect on how warm the surface layer of water is.
Otherwise, the important question is,
Does the increase in Downwelling LWR (from increased GHGs) cause ocean warming and if so, by how much?
The answer is almost certainly yes it does, but by (much?) less than the climate models predict (and require for their claim of accurately modeling the climate to stand up).

George E. Smith
August 15, 2011 4:03 pm

“”””” Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things … “””””
Well first off, an “infrared lamp” has a temperature in the range of about 1000 K and is not too bad an imitator of a black body radiator, being incandescent. That said it emits near infra red radiation that peaks at about 1.0 microns wavelength and also at about 4.0 megaWatts per square metre. Compare that to the average atmosphere which has a Temperature around 288 K, and is emitting wavelengths more in the 10 micron range and about 400 W/m^2.
So your heat lamp is 10,000 times the radiance of the atmosphere, and is spectrally peaked where H2O is an extremely good absorber, in fact at 3.0 microns, H2O has its maximum absorption coefficient of around 10^4 cm^-1..
Why not use an ordinary bottle of water at about 15 deg C (288K) as a source to demonstrate how “downwelling” LWIR radiation heats the ocean.
And that 10-15 micron radiation from the atmosphere is absorbed in more like the top 50 microns (99%) of the surface, not the top mm.
On the other hand, the solar energy can go 100s of metres deep in the oceans before full absorption.
So we are being asked to accept that a thermal energy source at a mean Temperature of about 288K (some say it is only 255K) can thermally excite GHG molecules (CO2), causing them to emit LWIR radiation at a wavelength around 15 microns, and that radiation (well maybe half of it) gets absorbed in the ocean which has a much higher specific heat, and will be absorbed in no more than 50 microns of sea water, most of which is already at a much higher temperature than the 255 or even 288 atmospheric source.
Temperature. The Temperature gradient would seem to be in the wrong direction at the surface, to cause much conduction of that surface “heating” into the depths.. I’ve spent enough time out in the deep oceans to believe that most of the time, the deep oceans are rather calm, and deep mixing due to turbulence is far from the norm.
But as to whether “downwelling” LWIR from GHGs can heat the ocean; I don’t know; I’d like to see the results of some actual measurements that demonstrate the phenomenon.

DR
August 15, 2011 4:05 pm

http://www.warwickhughes.com/blog/?p=87
Per Doug Hoyt:

In the laboratory, you can point a 10.6 micron laser at a body of water. Its intensity will be millions of times greater than the intensity increase due to a doubling of carbon dioxide. A thermometer placed just a few centimeters deep in the water will not rise in temperature. It is clear that infrared radiation cannot do bulk heating of water with any efficiency.
At best an increased amount of infrared radiation will slow down any cooling that is occurring. It will not cause a bulk heating.

August 15, 2011 4:07 pm

“You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.”
you know willis I think is one of the major miscommunication problems in radiative physics.
One way to think about it is this.
The shiny surface on a thermos does not warm the coffee inside.
The shiny surface retards the heat loss via radiation.
I sit outside on a freezing winter night with a space blanket.
The blanket doesnt warm me. The blanket slows the heat loss via radiation.
Now go have fun with tallblokes toilet paper experiment

LazyTeenager
August 15, 2011 4:11 pm

I could not have said it any better myself Willis.
But it won’t go away because the desire to believe is stronger than any rational argument. There are even university professors who should know better pushing this particular piece of nonsense.
And if commenters disagree go do the experiment as Willis says.

August 15, 2011 4:18 pm

Of course if that extra downward DLR DOES get into the oceans then we have thousands of years before the ocean temperature would change enough for us to notice any climate effect. The same energy cannot be in two places at once and the heat capacity of the oceans is magnitudes greater than that of air.
Willis, what proportion of DLR do you contend gets into the oceans ?
No, the truth is that it does not warm the oceans but it DOES add to the energy content of the system from those affected molecules upwards through the atmosphere to space.
Thus there IS a climate effect but it is manifested by a change in the surface pressure distribution from more radiation convection conduction and evaporation.
The problem for AGW then is that such changes are miniscule compared to the natural forcings of solar and internal oceanic variability.
Full analysis here:
http://www.irishweatheronline.com/news/environment/wilde-weather/setting-and-maintaining-of-earth%e2%80%99s-equilibrium-temperature/18931.html
This issue is not comparable to denial of the existence of a greenhouse effect. I am here accepting the warming capability of GHGs but simply advancing the description as to how they influence the system and in doing that it is not possible to have such a tiny effect warming up both vast oceans and the atmosphere to a significant degree simultaneously (or even separately) because all they do is accelerate the energy flow through the system to offset that warming effect.
Radiative processes alone need to warm up an entire system to a higher equilibrium temperature in order to regain balance. That is true and the essence of AGW.
In this case radiative processes are not acting alone. Other processes are speeding up the energy flow out to space which reduces the need (or possibly eliminates the need) for any rise in equilibrium temperature for the system as a whole.

DR
August 15, 2011 4:26 pm

It should be very easy to test this per Doug Hoyt’s example.
Sorry, I’m not buying that 100 ppm increase in CO2 can have any measurable effect on ocean temperature.
Water has a shiny surface……

Dave Springer
August 15, 2011 4:29 pm

Swift says:
August 15, 2011 at 3:17 pm
“The Science of Doom has the first of a very good multi post series on DLR and the ocean here http://scienceofdoom.com/2010/10/06/does-back-radiation-heat-the-ocean-part-one/
Most of the questions that have been asked and will be asked are answered there.”
The science of doom article has a huge glaring flaw in part two. It uses a simple application of the Stephen-Boltzman law on perfect black bodies to show the temperature of the ocean with and without DLR and concludes DLR must heat the ocean because otherwise it would be -15C instead of plus 15C.
The flaws are twofold. First of all, unlike the surface of a black body, the ocean is a greenhouse unto itself. Shortwave radiation easily penetrates to a depth of some 100 meters depending on clarity. This is absorbed and warms the water. Water is just about totally opaque to long wave radiation so none of that solar heated water below the surface can cool radiatively until it is somehow mechanically transported to the surface. These are the exact same properties that CO2 has – transparency to visible light, opacity to infrared. You can’t have your cake and eat it too. If CO2 is a greenhouse gas then the ocean is a greenhouse fluid.
The greenhouse properties of liquid water is what raises its temperature far above what it would be if it were a passive black body surface.
The second flaw in the science of doom black body graph is that black bodies don’t give up heat by evaporation. The ocean does. In fact that’s the primary cooling mechanism. 70% of the solar heat in the ocean escapes by evaporation, 20% via radiation, and 10% via conduction. Moreover the ocean retains a lot solar heat absorbed in the summer and releases it in the winter when the air is dryer and evaporation rate is faster. That’s why there’s a much smaller seasonal temperature change in the ocean versus land. The much greater seasonal temperature variation over land is called continentality.
Willis might have learned a lot about the ocean by sailing, surfing, and diving but he evidently needs to learn more about land by doing some driving and digging. The notion that land and water are equivalent in the way they heat and cool is demonstrates utter ignorance of both.

Myrrh
August 15, 2011 4:31 pm

Firstly, who uses “DLR”? Do you mean downwelling thermal infrared? And what do you mean argument against warmistas that it can’t?? The warmistas claim it can’t!
What an odd distortion. The AGWScience argument is that it is “Solar” energy of Visible Light and the two shortwaves either side of UV and Near Infrared which heat land and oceans. They say that longwave infrared, thermal infrared, doesn’t play any part in heating the land and oceans. That’s the picture they give of the Earth as a ‘greenhouse’, that shortwave visible gets through the ‘glass’ of the atmosphere and heats the ground, land and oceans, and thermal long wave infrared doesn’t get through ‘the glass atmosphere’, but then radiates out from the ground and gets trapped by the ‘glass/atmosphere’.
This is ‘standard’ teaching on it, the AGWScience fiction meme gone viral in the general education system – ‘everyone’ takes it so much for granted because it has been so successfully brainwashed.
“When the outer atmosphere or the ozone layer does not trap short wave radiation from the sun, it penetrates the surface of the Earth. This energy is then re-radiated back as energy of a longer wavelength (infrared). This leads to a warming of the Earth’s surface and the lower atmosphere.” http://education.gsfc.nasa.gov/ess/Units/Unit2/U2L5A.html
Light waves, the short electromagnetic of Visible, and the two shortwaves either side are not capable of heating land or oceans. The AGWScience meme is junk science. It claims that blue visible light heats water!!
How???
It’s an argument I’ve been having for quite a while with folks here, those ‘educated’ into believing it, I’m so glad you brought it up.
Just to get this straight. The AGWScience fiction claim is that shortwave converts to heat land and oceans and longwave thermal is radiated out – downwelling SHORTWAVE, Light, upwelling LONGWAVE, heat. By “Solar” they mean these shortwaves and not longwave thermal infrared. See the Kiel/Trenberth 1997 for the AGWScience fiction Basic Energy Balance on which all the has been built. They’ve been denying that longwave thermal infrared heats the Earth!
It’s been the same claim for rather a long time, so long in fact, that even ‘skeptics’ think it is real physics. It’s gobbledegook. Here’s an example: http://wattsupwiththat.com/2011/05/07/visualizing-the-greenhouse-effect-light-and-heat
“Solar “light” radiation in = Earth “heat” radiation to Space out! That’s old news to those of us who understand all energy is fungible (may be converted to different forms of energy)”
Forget about the watts in/watts out and all the aguments about ‘backradiation’ – it’s the basic physics that is unadulterated d cr*p here…
This perverse physics which has given thermal ability to light, shortwave, and denied that longwave thermal infrared heats the Earth has been systematically taken out of references whenever it can be. AGWScience fiction plays around with properties and processes. The example I found earlier of a NASA page for children which taught the real physics that thermal infrared is the heat we feel from the Sun has been dropped – and this nonsense that thermal infrared doesn’t even reach the Earth’s surface has been put in its place – here’s the post:
http://wattsupwiththat.com/2011/07/28/spencer-and-braswell-on-slashdot/#comment-711886
Thanks for trying with the secure connection Anthony, but you go away for the day and it’s back to the same old interference.. It was good while it lasted.

August 15, 2011 4:31 pm

Willis, Gary Wilson, crosspatch, Climate Weenie etc …
I am not sure what the main thrust of this post is (or is ssupposed to be).
The existence of IR radiation downwards is not disputed AFAIK. But as already pointed out, the IR-exchange between to bodies depends on mainly their respective temperatures, and for gases also the spectral properties of the radiating molecules.
And since noone disputes that the net heat flux from the earth’s and ocean surfaces is outgoing, everybody is aware of that in- and outgoing IR-radiation mustbe there at the same time. They are counterparts of each other. The net heating of the surface is by sunlight short wave radiation, making it all the way to the surface. (There is also a part of the sun’s IR being absorbed in the atmosphere and there reradiated in all directions, so that part of the ‘downwelling’ IR also originates from the sun, and only the remainder can be attributed to the ‘greenhouse effect’, but thats a minor detail).
The ‘greenhouse effect’ slowers the IR-radiating cooling mechanism (which would be zero if one had a perfect 100% IR-mirror holding it just above the surface). The net received heat must however be cooled away (in steady state) and the other available cooling mechanisms are mechanical transfer of heat through: Thermal convection and transport of phase-transitioned water vapor, both being heat transported uppwards from the surface.
I don’t think you’ve got this wrong, but I object to the notion that there are ~170 W/m2 (SW-sun)) plus ~330 W/m2 (atmospheric DLR), adding upp to a ~500 W/m2 total. Because the the DLR part is an ongoing inner process, which is only ~half of an ongoing exchange, which by necessity has and must have flows in both directions. between two adjacent bodies each with a temperature. The outgoing counterpart being somewhat larger, ~390 W/m2.
There are only a total of ~342 W/m2 of heating available, and all of that is from the sunlight. Inner processes are going on inside the atmosphere, that’s true. But saying that ~500 W/m2 heats the earth’s surfaces is awkward.

Mark.R
August 15, 2011 4:41 pm

Here in Christchurch N.Z i take ground temperatures at one metre deep.
The ground temp ranges between 8.9c(record, recorded in the last 3 weeks) to a high of 16.8c in summer. What iv noticed when we get a warm sunny day (say 26c) and even if the next 2 days are only 18-20c( cloudy or sunny ) i get a 0.2c increase in the ground temp 3 days after the warm day.The same happens when we get a cold day too 3 days latter a drop in the ground temp of 0.2c.
If we have a prolonged cloudy period say 2weeks temps at one metre underground only change about 0.3c over that period(usually down).If the skys are clearer over the same period the temp can go up or down by up to 1c.
My conclusion is then that DLR even gos deeper into the ground than what most think.

August 15, 2011 4:43 pm

In a discussion elsewhere on this topic, I made a mechanical analogy (for the case of perfect IR-radiation insulation, ie 100% greenhouse effect):

Another highschool example: A balancing scale has a weight of its own, say 2kg. Its function depends on that at zero (external) load the 1kg on either side perfectly balance/cancel each other out, ie equilibrium (mechanical here, thermal with a perfect insulator). Adding 40g on the tray alters that, making the scale tip towards one side.
[Warmist troll] now says: a) You cannot account for the tipping without the 1kg already on that side (true), b) the 1kg contributes much more than these tiny 40g, and c) you are ignoring the 1kg already there!
[Jonas replies: a)moot, b) nonsensical and c) wrong]

The analogy is undefirmed mechanical equlibrium, corresponds to thermal equlibrium (all temperatures being equal), and that deformed state, due to external load (or heat source) correspond to thermal steady state, with net deformation and/or net heat flow, determined uniquely by the the external load, and where it is applied.

Dave Springer
August 15, 2011 4:46 pm

Yet another general misconception is that all the water molecules in any arbitrarily thin layer are at the same temperature. That isn’t how it works. Some of those molecules are boiling hot and some are ice cold. The average of many of them is the temperature. Evaporation occurs because at any one time some water molecules get bumped, literally, over the edge of latent heat of vaporization. If they are surrounded by other water molecules they don’t get to stay vaporized because they are surrounded by cooler molecules which they bump into and give up the heat. However, if they are on the surface, they have a free path to leave as a molecule of water vapor.
Downwelling IR, because it can’t penetrate more than a few microns, is continually bumping surface molecules over the edge of latent heat of fusion and they fly away. For those molecules that don’t get enough of a bump they don’t mix downwards because wave and winds causing mixing well below the surface not actually on the surface and second because warmer water rises above cooler water. At the end of the day downwelling IR does not cause any significant heating (or, for pedants, lowered rate of cooling). All it does is raise the evaporation rate and the energy is carried aloft as latent heat of vaporization and doesn’t get released to the environment as sensible heat until it condenses. The release generally happens at the cloud deck when adiabatic cooling reduces the water vapor temperature below the dewpoint. Downwelling IR warms the cloud deck and has no direct effect on surface temperature. All it does is lowers the temperature gradient of the atmosphere between surface and clouds and raises the temperature gradient between clouds and outer space.

Red Jeff
August 15, 2011 4:57 pm

For what it’s worth, here is a paper that may add to this discussion ‘Induced Emission and Heat Stored by Air, Water and Dry Clay Soil.’ Nasif Nahle http://www.biocab.org/Induced_Emission.html
“Abstract: In this paper, I have resorted to basic formulas obtained from experimentation and observation by several scientists for calculating the heat stored by any substance and the subsequent change of temperature caused on a determined system. I demonstrate that the climate of Earth is driven by the oceans, the ground surface and the subsurface materials of the ground. I explain also how the photon streams from oceans, ground and subsurface materials of ground overwhelm the emission of photons from the atmosphere to the ground during both daytime and nighttime…. Concluding, atmospheric gases do not cause any warming of the surface given that induced emission prevails over spontaneous emission. During daytime, solar irradiance induces air molecules to emit photons towards the surface; however, the load of Short Wave Radiation (SWR) absorbed by molecules in the atmosphere is exceptionally low, while the load of Long Wave Radiation (LWR) emitted from the surface and absorbed by the atmosphere is high and so leads to an upwelling induced emission of photons which follows the outgoing trajectory of the photon stream, from lower atmospheric layers to higher atmospheric layers, and finally towards outer space. The warming effect (misnamed “the greenhouse effect”) of Earth is due to the oceans, the ground surface and subsurface materials. Atmospheric gases act only as conveyors of heat.”

Jose Suro
August 15, 2011 5:00 pm

In peril of sounding like a Warmista, which I’m not, what else could possibly heat the oceans but long wave IR? Reminds me of if not CO2 then what else could it be…. Nothing that I can think of. Well, maybe not the oceans but the near shore waters have another possible explanation.
Back in the 80’s I had a boat and often went to Egmont Key, an island on the mouth of Tampa Bay – beautiful pristine clear waters. One time, and against my better judgement, we went out on Labor Day. Along a 1-mile beachfront I found only one spot where I could back up an 8-foot beam boat onto the beach. While there I did some back of the napkin calculations on the amount of beer consumed by the hundreds of people there, and the resulting urine going into the water at 98-degrees or so. Story short, the waters close to the beach were a lot warmer :). And I stayed out of it!
Best,
J.

Sun Spot
August 15, 2011 5:05 pm

re: Alexander Duranko says: August 15, 2011 at 3:28 pm
Anyone ???, I’m hearing crickets on this item.

Anything is possible
August 15, 2011 5:07 pm

What is the primary source of DLR over the tropical oceans?
Any chance It would happen to be water vapor?

Dave Springer
August 15, 2011 5:15 pm

steven mosher says:
August 15, 2011 at 4:07 pm

One way to think about it is this.
The shiny surface on a thermos does not warm the coffee inside.
The shiny surface retards the heat loss via radiation.

The reflective surface is not the primary mode of insulation. “Thermos” is a brand name that became attached to all of what were orginally known as “vacuum flasks”. The primary means of insulation is the vacuum between the inner vessel and outer walls which stops almost all conductive heat transfer. Most the remaining heat loss is blocked by the reflective coating. That’s why laying tin foil over non-reflective attic insulation doesn’t help a whole lot but does help some.

August 15, 2011 5:20 pm

Addition to the 1:st of my posts just above: The net heat flux (in thermal steady state) in and out from any surface is of cause zero:
I was referring to the net of the IR-radiation. The non-short-wave-radiations from the sun, hitting the earth’s surface of course equals the net heat loss from the very same surface. Due to the three different cooling mechanisms: IR exchange with adjacent bodies, qater evaporation and thermal convections (conduction too, but negligable here)

Rosco
August 15, 2011 5:21 pm

You lost me when you quoted that insolation is reduced to 170 W/sq m at the surface while DLR is more than 4 times the 70 W/sq m you say is accounted for by evaporation – ie ~280 W/sq m.
I guess reality is dead.
How can DLR exceed insolation ? That is impossible !

Bebben
August 15, 2011 5:22 pm

“If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass.”
This beats me. The big ocean with its enourmous heat capacity warms the small atmosphere with its little heat capacity and then the big ocean is heated by the small atmosphere with its little heat capacity, again and again. Like if I should be pounding with my own straight lefts on Mike Tyson.
I kinda sniff that something is seriously wrong here…
Or maybe, if you take the radiation approach, you always have to consider radiation in two directions, like Kiehl and Trenberth. Isn’t this correct? Then the net radiation is outgoing (sum of two vectors) and the DLR is negative so it doesn’t heat the ocean.
Simple as that. From a layman’s perspective, that is.

Dave Springer
August 15, 2011 5:24 pm

Sun Spot says:
August 15, 2011 at 5:05 pm
re: Alexander Duranko says: August 15, 2011 at 3:28 pm
“Anyone ???, I’m hearing crickets on this item.”
It’s so wrong it’s not worth a response would be my guess. That’s certainly why I didn’t bother. A cheap remote IR thermometer aimed upward on a clear night with air temperature exactly the same with the only difference being humidity will register a higher temperature on the more humid night. A rather expensive gadget costing in the five figure range can measure DLR day or night under any conditions. It’s called a pyrgeometer. Doesn’t every grizzled old engineer have one of those puppies on his bench sitting between his ohmmeter and oscilloscope? /sarc

David L. Hagen
August 15, 2011 5:34 pm

Judith Curry posts another round of Slaying the Greenhouse Dragon: Part IV by Vaughan Pratt at Climate Etc.
(Willis is clearer)

August 15, 2011 5:36 pm

As we all know, heat is transferred by conduction, convection and radiation. If the top 50 micrometers of the ocean is warmed by IR, wouldn’t those molecules slough off their extra energy almost instantly via conduction, warming the adjacent molecules? And so on, transferring the heat deeper into the ocean. Am I missing something?

Latitude
August 15, 2011 5:43 pm

Smokey says:
August 15, 2011 at 5:36 pm
=============================================
Me too Smokey…..I know I’m missing something
We don’t plant tomatoes until we can dig down a good foot and a half and the soil has warmed up…

Rosco
August 15, 2011 5:51 pm

I cannot see why it is considered valid to reduce the insolation to Earth by a factor of 3/4 because half the earth is in darkness at any one time and a disk has half the area of a hemisphere.
Almost nowhere on earth therefore receives the 234 W/sq m used to calculate the fictional minus 18/19 C.
Surely the whole of the tropics is subjected to an insolation 2-3 times more than this figure during a significant portion of any day and then it commences cooling at night.
The mean is meaningless and ought not be used. I refuse to believe DLR heats the earth more than the sun – it defies logic.
The real question we ought to consider is given space is ~3 K why is it Earth isn’t even colder ?
Perhaps then we may perform better analyses than being done at present.
Even IPCC documents quote higher insolation figures than the oft quoted 235 W/sq m but they trickily often lead with the Kiehl & Trenberth diagram which defies logic or science.
You can’t argue with someone who is trying to deceive you and using a model which in no way reflects reality is deception.

Rosco
August 15, 2011 5:54 pm

How do they calculate the work of the earth’s climate system or is radiation the only game in town ?

Dave Springer
August 15, 2011 6:02 pm

gnomish says:
August 15, 2011 at 1:08 pm
“http://en.wikipedia.org/wiki/Solar_pond”
Solar ponds are shallow, need a dark bottom, and have to be shielded from turbulence so that saline stratification can occur. The mode of operation is that visible light from the sun is absorbed by the dark bottom and that conductively heats the water from the bottom up. The high salinity of the bottom water changes the temperature/density relationship such that the hot brine can’t rise off the bottom and mix with the low salinity layer on the top.
The ocean is nothing like that. It is effectively bottomless with regard to light from the sun and no signifcant saline stratification occurs in the mixed surface layer which absorbs the solar radiation.
.

August 15, 2011 6:13 pm

First, should it not be an extremely simple task (for a scientist) to measure how much DLR there is? That way it can be determined how much it heats. Say for instance, you cover the ground in a substance that can measure the DLR to some precision out in the open air. Then you move a DLR reflective surface on top and ULR absorbing surface on the bottom over top of the measuring substance, and measure it again without most of the atmosphere shooting the DLR downward.
I hear all the thermodynamics arguments of whether radiative energy can warm an item, but is that really the question we are looking for? I just have not seen any evidence that much radiation ever returns to the surface, if it really increases based on the amount of greenhouse gas present, even with the experiment being done in a laboratory where you can limit the number of contributing factors to come out with a realistic starting point.

gnomish
August 15, 2011 6:14 pm

Dave Springer says:
” Most the remaining heat loss is blocked by the reflective coating. That’s why laying tin foil over non-reflective attic insulation doesn’t help a whole lot but does help some.”
on the floor, where little heat is lost by conduction and absolutely none by convection, a layer of foil, shiny side up, will provide quite excellent insulation – that floor will be the warmest one in the house. next time you tile a floor, try it. it works great.
if you consider co2 functioning as a mirror for IR, then it’s obvious that it is preventing heating from above exactly as much as it is preventing cooling below – except at night, of course.
now, if you could bring that agw ocean.in.a.bottle out of the virtual world inhabited by alarmist witch doctors, the computer models might apply. but not in real life.
no real ocean is anything like that. there is a constant flow of water gas from surface to stratosphere (and back). any co2 gets swept along in a refigeration cycle by the hadley heat pump. it does not sit like a reflective shield over the sea.
you can add all the shiny crap you want to the freon in your air conditioner – it will not alter the efficiency of the refrigeration cycle through radiation physics.
similarly, the heat pump of the hadley cell- IR radiation exchange within the flux has zero effect on the transfer efficiency.
adding heat capacity, however (and co2 is a bit more dense than elemental gasses) improves the efficiency, not the opposite as claimed by the fetishist rent seekers.

DocMartyn
August 15, 2011 6:15 pm

Phytoplankton are packed with chlorophyll and other chromophores. The IR spectrum of different chlorophylls is well known,
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC540707/pdf/plntphys00384-0069.pdf
Plants are only green because we can’t see very far into the red.
The classical way to cool a white light source, for doing photolysis, is to place a round bottomed flask full of water in the beam. This stops sample heating.

richard verney
August 15, 2011 6:18 pm

Willis
First of all and importantly, we are not talking about mm but rather microns. Due to the wavelength, DWLWIR cannot penetrate more than a few microns into water. The problem is that nearly always these first few microns of the ocean are divorced from the ocean being in reality wind swept spume and spray. These few microns are therefore, in the limit, not connected with the ocean and thus even if there was sme physical process (which no one on the warmist side has yet sought to explain) whereby heat can in effect be conducted downwards, it beggars belief that there can be any transport of the warmer micron layer into the bulk ocean. The most likely scenario is that the DWLWIR absorbed by these first few microns are heated and evaporate thereby causing cooling not warming. As you know, the ocean is constantly evaporating and in your earlier article, you discuss the cloud formation which is the result of this evaporation. The DWLWIR cannot overcome the evaporative and convectional forces that are at play (which forces have principally been driven by solar energy). If you like the DWLWIR cannot swim against the tide.
Your dismissal of the first argument is with respect complete and utter rubbish. The DWLWIR absored by the first few microns of the ocean is burnt off and evaporates thereby preventing any transport of heat into the deeper ocean. There is no equivalent process with respect to the land. Rocks or sand or tarmac or whatever does not get burnt off and evaporate away. Further, I suspect that there is a large number of people who consider that backradiation cannot heat the land. Whilst the warmists struggle to describe the basics of their conjecture, many armist acept that backradiation does not heat but rather it acts in some way akin to a blanket and reduces the rate of heat loss, ie., because of backradiation the land cools slower than would be the case if there were no backradiation.
As regards the second argument. DWLWIR is not heat. We all know that the DWLWIR cannot effectively heat anything and cannot do sensible work. That is why no one is seeking to utilise the alleged 333 mw per sqm of backradiation and use this to cure the world’s energy needs. After all, according to Trenberth, the backradiation is nearly twice the solar energy and if it was truly a source of heat or if it could truly do work, man would exploit this valuable resource.
As regards argument 3, I dealt with that at the outset. We are taliking about the first few microns which is wind swept spume and spray and which if anything is evaporated (eventually giving rise to the clouds in your earlier post) and never mixes with the bulk of the ocean.
As regards argument 4, I believe that I have had that argument with you before. One of the biggest problems in the AGW debate is that climate scientists seek to deal with averages. However, this hides what is going on. The reality is that huge amounts of solar energy are being input into the tropical oceans such that these oceans would never freeze. The heat being absorbed by these oceans is then transported towards the poles with the ocean current conveyor belt acting as a huge heat pump distributing the heat absorbed by the tropical oceans. Of course, by late summer/winter there is not quite enough energy being absorbed by the tropical oceans to offset the reduced solar energy being received by say the Artic ocean such that that ocean begins to freeze over.
I had enjoyed your previous recent articles but this latest one is very light weight and nothing more than conjecture.
If you want to suggest that the oceans would be frozen but for the absorption of backradiation, you need to do 2 things. First, prove that DWLWIR is absorbed to a significnt degree by the oceans. Second, you need to carry out a calculation for each kilometre square of ocean based upoj the amount of solar energy received by the ocean at that point and ignoring DWLWIR and detailing at what point in time that part of the ocean would freeze.
Presently, I am extremely sceptical of the points that you raise and consider that they fall well short of discharging the burden of proof that lies on someone claiming that DWLWIR heats the oceans and prevents them from freezing. .

August 15, 2011 6:19 pm

Merrick says:
August 15, 2011 at 1:39 pm
Small issue of heat capacity notwithstanding.

Dave Springer
August 15, 2011 6:29 pm

1DandyTroll says:
August 15, 2011 at 1:22 pm
“So a IR lamp directed at a one square meter tub of water of say one meter in depth to radiate the surface at 170W wouldn’t warm that body of water if it was circulating?”
Not if it’s longwave infrared. An “IR” lamp emits shortwave infrared and a significant amount of visible red light which will penetrate and heat both the water and the walls of the vessel. The infrared coming from the air above the ocean is emitted by a source that’s about the same temperature as the water. DLR cannot possibly warm the water warmer than the air from which the DLR comes from. The argument is over whether it can slow down the rate of cooling. So the experiment to try would be taking two vessels of warm water, say 100F, and suspending a third vessel with water at the same temperature over one of them. The third vessel is your “heat lamp”. If the DLR from the third vessel is capable of slowing down the cooling rate then as all the vessels cool down to room temperature the one without the LWIR “lamp” over it will cool more slowly and you should be able to see this with a therometer in each of the test vessels after some period of time.
Actually duplicating what happens over the ocean makes it sound a lot less likely to the lay person that the “lamp” hung in the air over one of the vessels is going to make any difference in the cooling rate of the vessel beneath it once they realize that the “lamp” is an object the same temperature as the water. Otherwise they imagine the kind of heat lamp commonly used for home heating purposes which has a heated element hot enough to glow cherry red.
“If it only heats the first mm why then are several inches usually warm when going swimming? Or is that an odd question?”
Visible sunlight is heating it far deeper than 1mm with decreasing effectiveness over increasing depth so more heat is added nearer the surface. Without significant turbulence a warm surface layer develops. This discussion isn’t about visible light. It’s about longwave infrared light. There’s very little LWIR in sunlight. Virtually all the LWIR that shines on the surface is LWIR emitted from the surface where a portion of it is reflected back downwards from the atmosphere above it. More or less is reflected back down depending on the exact composition of the gases.

George E. Smith
August 15, 2011 6:31 pm

“”””” Dave Springer says:
August 15, 2011 at 4:46 pm
Yet another general misconception is that all the water molecules in any arbitrarily thin layer are at the same temperature. That isn’t how it works. Some of those molecules are boiling hot and some are ice cold. The average of many of them is the temperature. “””””
Do you want to take a Mulligan on that one Dave ?
At any given Temperature, the average kinetic energy per molecule has some specific value, but the individual molecules have energies which plot as some Maxwell-Boltzmann distribution, which puts the most probable velocity at some value, and tends to cluster the velocities at the low energy side of the peak, giving a long tail on the high energy side; but theoretically having no upper bound on the maximum energy.
For any sample that is at some fixed Temperature, this distribution of velocities (energies) is constant. But any individual molecule can at any time adopt any possible value of energy, as a result of collisions with its neighbors. The thermodynamic Temperature is defined in terms of that distribution and the mean energy per molecule.
I have argued that since the M-B distribution is always fully populated, and any molecule can at any time be anywhere in that distribution, that for any individual molecule, the time averaged distribution of energies , must be the same M-B distribution of the whole population, so one can argue that any individual molecule, has a Temperature that is the same time averaged distribution of kinetic energies as the population as a whole has.
It is not true that near by molecules can have greatly different Temperatures, simply because they have greatly different instantaneous energies.
For example, there is absolutely no tendency for an individual molecule that currently has a high KE (velocity) to seek out and target a neighboring molecule, that has a much lower energy and velocity. The directions of two such molecules are entirely random; and whereas heat would tend to flow in the direction of the highest Temperature gradient; that is only true over time, and nothing would stop the slowest molecule in the bunch from colliding with the fastest; but that is just a random event; not a directed result.

Keith Minto
August 15, 2011 7:01 pm

I have expressed this before, I consider that at least some of the water vapour released from the ocean skin is condensed well before it drifts magically upwards in a pristine state to condense only at a certain altitude.
If the air at the ocean surface is humid that means it has already condensed and released its heat; there are plenty of nuclei available, salt, DMS, dust particles near land.
This could be a source of DLR.

richard verney
August 15, 2011 7:03 pm

When considering the point raised by Willis, it is important to bear in mind that due to the wavelength of solar light, solar light can penetrate well to a 10 a depth of metre or even more ( indeed, it can penetrate to a limiyed extent as far as about 100m). If you can see the bottom of the sea bed, you can see how far solar light can penetrate.
However, the DWLWIR being re-radiated by the GHGs is of a wavelength that can penetrate no more than a few microns into water. Crispin in Waterloo (August 15, 2011 at 2:07 pm) gives a good explanation of the physical processes that are going on and details why the DWLWIR does not make its way to any significant extent into the ocean.
Accordingly, when one goes diving or snorkelling one can feel warm layers of water (sometimes due to currents) and one can feel the warmth of the sunlight. It is solar energy that is absorbed and which heats the water.
The so called greenhouse effect works differently over land than over water. There is a significant diurnal temperature range over land, but very little diurnal range over the deep oceans. It may be that over land, DWLWIR helps the keeping of night tme temperatures up, by reducing the rate of heat loss from the land. However, over the oceans, night time temperatures do not drop off because in the limit (by which I mean for the period of darkness), the ocean itself is an all but limitless heat source having the ability to constantly replenished the air temperature above it. Further, over oceans there is high humidity such that the effects of CO2 if any are substantially dwarfed.by the effects of water vapour and the water vapour being evaporated from the ocean beneath it acts so as to impede DWLWIR coming from the atmosphere high above.

richard verney
August 15, 2011 7:12 pm

@Willis Eschenbach says: August 15, 2011 at 2:04 pm
“…You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.
///////////////////////////////
It is only in climate science that such a statement could be made!!!
It is akin to a bioligist claiming that he has created life by not killing a laboratory rat. Sure, by not killing the rat, the net effect is that you have a living rat, but not killing something is not the same as giving life to something. Likewise slowing the rate of cooling is not the same as warming. They are quite different processes. No wonder climate science is so off the rails.
One obvious reason why Trenberth cannot find his missing heat in the oceans is that it never made its way into the oceans because DWLWIR does not heat the oceans.

Dave Springer
August 15, 2011 7:28 pm

@Willis
Point by point flaws:
“Argument 1. People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land.”
No, it isn’t. Land doesn’t evaporate. Neither does land allow sunlight to warm it uniformly down to any significant depth. That’s why the surface of a lake doesn’t get as hot during the day as a blacktop parking lot and why the lake won’t get as cold at night as parking lot. The solar heating of the parking lot is concentrated on the surface. It gets far hotter during the day and gives up the heat far quicker at night.
“Argument 2. If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.”
The energy is going into latent heat of vaporization. That is the very large amount of energy that water absorbs going from liquid phase to vapor phase with no change in temperature. No change in temperature is why it’s called “latent” heat. The energy needed to turn a pound of water into a pound of water vapor at the same temperature is about 1000 times as much as it takes to raise the temperature of a pound of water by 1F. Water vapor of course rises until adiabatic cooling lowers its temperature below the dewpoint but I’m presuming you already knew how clouds form even if you didn’t know how much energy is transported in the form of latent heat from surface to cloud.
“Argument 3. The claim is often made that warming the top millimetre can’t affect the heat of the bulk ocean. But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers, heating the surface affects the entire upper bulk temperature of the ocean every night when the ocean is overturning.”
Wrong and wrong. Wind driven turbulence doesn’t mix down the top 10 micrometers. That’s a film thinner than the wall of a bubble where viscous forces overwhelm other forces. There’s very little diurnal turning in deep bodies of water and again viscosity is the overwhelming force in the top few micrometers preventing it from mixing deeper. Major turning of deep bodies of water is seasonal not diurnal. If you had much experience with deep freshwater lakes year round (I live on the shore of one) you’d know that from experience – you can smell it when it overturns in the springtime. The ocean is no different in that respect which also explains why so much summer heating is retained and released in the winter causing far less seasonal temperature variation than that of land surfaces at the same latitude.
“Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. ”
Wrong again. Liquid water has the same properties as water vapor when it comes to being transparent to visible light and opaque to infrared. The same properties that make water vapor a greenhouse gas make liquid water a greenhouse fluid. The difference is that liquid water is like water vapor on steroids since there’s more water in the first meter of the ocean than there is water vapor in the column of air above it and sunlight penetrates far beyond the first meter.
You have really failed to think through any of these arguments. You experience sailing and surfing and diving isn’t serving you well at all in understanding the physics of water.

Bob_FJ
August 15, 2011 7:35 pm

Tallbloke @ August 15, 2011 at 1:46 pm seems to have kicked off the concept that DLR is absorbed in what is a nano-skin of the water from where it is rapidly reemitted.

“…Argument four is a numerical misunderstanding. The ocean surface very efficiently absorbs 95% of DLR, and promptly re-emits half of that (the other 5% being reflected). The other half makes it another couple of molecules deeper and then the molecules it warms become more buoyant than their neighbours and rise to the top, losing another half upwards…”

Others have added supportive arguments, to which I’d like to add a point:
I think an important aspect of this concept is that it is an inconceivably rapid process, so fast that wave action and water turn-over etc is not a consideration. Since it all happens so fast in a very thin skin, those naughty photons and evaporating molecules radiate hemispherically back into the air where their free path lengths are very much longer and things are more complicated.

August 15, 2011 7:37 pm

Dave Springer,
Since the ocean doesn’t glow in the infrared, it seems to me that when a water molecule receives an IR photon, it immediately sloughs off the extra energy through conduction to adjacent water molecules. Some of the energy is probably contributing to vaporization, but the air even right above the ocean surface is usually not 100% R.H., so there’s not a lot of vaporization going on. And of course, something is warming the oceans.

Dave Springer
August 15, 2011 7:38 pm

Trenberth et al are looking in the wrong direction for the “missing heat”. After making a quick pitstop in the cloud deck the missing heat from the last 50 years is now unformally distributed radiation in a sphere surrounding the earth with a radius of 50 light years. They aren’t going to find the missing heat looking down into the ocean. It just ain’t there.

Kevin Kilty
August 15, 2011 7:42 pm

Often I find these threads when all others are long since gone, but this one looked like beating up a straw man to me, so I am now glad to have missed it. I have always assumed that when people say “LW does not warm the ocean”, they mean LW does not propagate in the bulk ocean. To have LW warm the bulk ocean requires absorption at the surface and transfer to depth by other means. It does not behave like SW, which I think is the idea most people are trying to convey.
One is not going to illustrate much about heat transfer in a fluid like water using arguments that exclude some modes of heat transfer.

Jer0me
August 15, 2011 7:45 pm

Anyone who has a pool knows full well that the top few inches gets warm first. That goes down quite deep, even in a pool where there is little turbulence (ie not being used) by the end of the day.
If the pool is being used, that warm water is spread pretty much all through.
I would imagine the sea would behave in the same way, and my experience when out surfing makes me believe it is.

Dave Springer
August 15, 2011 7:49 pm

Smokey says:
August 15, 2011 at 7:37 pm
“Dave Springer,
Since the ocean doesn’t glow in the infrared,”
The ocean certainly does glow in the infrared. All matter with a temperature above absolute zero glows in some portion of the spectrum. Discounting glow from rarified ionized gases the glow is pretty much a continuous blackbody spectrum with a center frequency set by its temperature. The difference between land and ocean is that it glows a lot less than land does because most of the energy leaving travels as latent heat of vaporization. The dearth of infrared energy leaving the ocean is made up for by an increase of infrared energy coming out of the cloud deck where that latent heat is being transported to and released.

Dave Springer
August 15, 2011 7:57 pm

Keith Minto says:
August 15, 2011 at 7:01 pm
“If the air at the ocean surface is humid that means it has already condensed and released its heat;”
Wrong. You can see it if it condenses. If it condenses near the surface we call it fog. If it condenses farther up we call it clouds. In either case you can see it after it condenses. If you can’t see it then it hasn’t condensed.

Anonymoose
August 15, 2011 7:58 pm

“believe that DLR can heat the land”
If DLR doesn’t heat the land, the borehole temperature measurements don’t make sense. If heat can penetrate ground without convection, it can penetrate water.

Dave Springer
August 15, 2011 8:05 pm

richard verney says:
August 15, 2011 at 7:12 pm

@Willis Eschenbach says: August 15, 2011 at 2:04 pm
“…You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.
///////////////////////////////
It is only in climate science that such a statement could be made!!!
It is akin to a bioligist claiming that he has created life by not killing a laboratory rat.

I think it’s more like when I told my wife that I earned an extra $20,000 by purchasing a new Corvette instead of a used Ferrari.
The government pulls that one on the taxpayers all the time. They “reduce spending” by increasing it less than they planned. I don’t know who that’s supposed to fool but it won’t work on me and it certainly didn’t fool my wife when I tried it on her.

Tim Folkerts
August 15, 2011 8:22 pm

Smokey says: “Since the ocean doesn’t glow in the infrared, …”
What would lead you to this bizarre conclusion???
For $39 you can buy this “infrared glow meter” from LL Bean to observe water http://www.llbean.com/llb/shop/50247?pi=865287&subrnd=0&qs=3021021_pmd_nextag
If you have a little more money, you can buy a weather satellite and use the “infrared glow meter” to determine global SST. http://en.wikipedia.org/wiki/Sea_surface_temperature

Dave Springer
August 15, 2011 8:29 pm

Smith
“For example, there is absolutely no tendency for an individual molecule that currently has a high KE (velocity) to seek out and target a neighboring molecule, that has a much lower energy and velocity.”
Well sure, it’s not going to seek out slower moving molecules, per se. But it has a greater chance of hitting slower moving targets since they bunch together more.

August 15, 2011 8:38 pm

http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
Complete with experimental evidence. As Willis notes there are much better skeptical arguments. As long as skeptics remain UNSKEPTICAL of bogus skeptical arguments the perception will be that skeptics would not accept the truth if it hit them in the face.
GHGs warm the planet. They do not cool it.
Man’s activities increase GHGs, they do not lower them.
Some general truths that are fully consistent with a skeptical position about catastrophic warming.
When you accept these two facts, then you get to join the science debate. That debate is about
1. How much do GHGs warm the planet
2. Can we and should we do anything about increased GHGs.
But when you try to toss out or deny those basic two facts, you dont get to join the debate.
pretty simple. You wanna know why Willis gets to join the debate? Why spenser gets to join the debate, Lindzen? heck even Monckton.. Because they accept the first two.
If you believe those first two, do a little experiment next time you are on a warmist blog. Announce that you accept those two facts.

August 15, 2011 8:38 pm

Smokey says on August 15, 2011 at 7:37 pm
Dave Springer,
Since the ocean doesn’t glow in the infrared,

Excuse me!? (Smokey or Dave – it’s late an I’m not sure which one wrote the comment)
What do you think you’re seeing on GOES (8–15 µm wavelength) IR satellite image?
I ASSURE YOU lakes/bodies of water *do* show up … and imagery color/shading appears to be proportional to temperature …
Maybe another point is being argued, in which case ‘never mind’ … otherwise take a look for yourselves; pick a region with a body of water for display (like the GREAT LAKES. Remember ground obscured by clouds will read cloud top temperature):
http://weather.rap.ucar.edu/satellite/
.

Dave Springer
August 15, 2011 8:40 pm

astonerii says:
August 15, 2011 at 6:13 pm
“First, should it not be an extremely simple task (for a scientist) to measure how much DLR there is?”
Absolutely. It’s called pyregeometer. Wanna buy one?
http://www.meteorologyshop.eu/Pyrgeometer/ENG_276_EUR_264_1173__.html
A bargain at a mere 5000 euros but if you want to pinch some pence you can wait for the after-Christmas sale in January when they deeply discount the last year’s models to make shelf space for the newer models. /sarc

August 15, 2011 8:45 pm

astonerii . yes there is experimental evidence.
“However, some have insisted that there is a paradox here – how can a forcing driven by longwave absorption and emission impact the ocean below since the infrared radiation does not penetrate more than a few micrometers into the ocean? Resolution of this conundrum is to be found in the recognition that the skin layer temperature gradient not only exists as a result of the ocean-atmosphere temperature difference, but also helps to control the ocean-atmosphere heat flux. (The ‘skin layer‘ is the very thin – up to 1 mm – layer at the top of ocean that is in direct contact with the atmosphere). Reducing the size of the temperature gradient through the skin layer reduces the flux. Thus, if the absorption of the infrared emission from atmospheric greenhouse gases reduces the gradient through the skin layer, the flow of heat from the ocean beneath will be reduced, leaving more of the heat introduced into the bulk of the upper oceanic layer by the absorption of sunlight to remain there to increase water temperature. Experimental evidence for this mechanism can be seen in at-sea measurements of the ocean skin and bulk temperatures.”

August 15, 2011 8:48 pm

“Since the ocean doesn’t glow in the infrared, it seems to me that when a water molecule receives an IR photon, it immediately sloughs off the extra energy through conduction to adjacent water molecules. ”
What? How do you think SST is measured?

Dave Springer
August 15, 2011 8:57 pm

_Jim says:
August 15, 2011 at 8:38 pm
“Dave Springer,
Since the ocean doesn’t glow in the infrared,
Excuse me!? (Smokey or Dave – it’s late an I’m not sure which one wrote the comment)
What do you think you’re seeing on GOES (8–15 µm wavelength) IR satellite image?”
Smokey said it, not me. I quickly corrected him.
What are we seeing on GOES? That depends. You have be looking through some pretty narrow IR windows to see the temperature of the ocean. When viewing a spectrograph from above there are a great many step changes across the spectrum. The tops of each step follow a blackbody curve for a different temperature. The highest steps follow the curve of the ocean temperature. Clear sky only of course. You can’t see through clouds.

anna v
August 15, 2011 8:57 pm

Dave Springer says:
August 15, 2011 at 7:49 pm
The difference between land and ocean is that it glows a lot less than land does because most of the energy leaving travels as latent heat of vaporization.
The ocean starts from a much lower temperature than the land, in sunlight. One can cook eggs on a rock at noon, when the sea is barely 25C in my region of Greece.
The whole package of energy balance as treated by climatology is a mess, in my opinion. They take two meter temperatures when the ground can be up to 60 and 70C and the two meter 36C, the sea 25C. It is not the air that looses energy to space according to T^4. And all this down welling and up welling confusion is like magicians tricks. There is perfectly adequate thermodynamics to describe all thermal situations accurately. The reason they do this sleight of hand is, in my opinion, that if they take the beaten thermodynamic track, the small increase in the heat capacity of the atmosphere due to the anthropogenic CO2 cannot be beaten up into a bogey through bogus feedback arguments that lead to the energy oven .

August 15, 2011 8:58 pm

Dave Springer says on August 15, 2011 at 7:28 pm

Wrong again. Liquid water has the same properties as water vapor when it comes to being transparent to visible light and opaque to infrared.

The gaseous state has some different properties, where molecule resonances/vibration modes are more pronounced in gas state molecules like water vapor vs liquid.
An intro (for some perhaps) – http://en.wikipedia.org/wiki/Infrared_spectroscopy
Note section on “vibrational modes”; this establishes where WV either absorbs or emits energy in certain bands/wavelengths.
WV atmospheric effects as well as liquid water characteristics on the same webpage:
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water#Atmospheric_effects
Of note: “The spectral absorption features of liquid water are shifted to longer wavelengths with respect to the vapor features by approximately 60 nm”
It’s late – maybe I missed something in the discussion prior …
.

Ninderthana
August 15, 2011 9:00 pm

David Springer,
Finally, a victory for solid scientific argument!
On average, the Earth’s surface (both ocean and land) receives 492 W m^-2, 168 W m^-2 from direct sunlight (mostly visible) and 324 W m^-2 from back radiation (mostly infra-red) from the atmosphere. The problem with Will’s argument is that assumes that soil/rock and vegetation act in a similar manner to sea-water.
Both the land and sea are efficient at absorbing visible light, all be it by different mechanisms, however the land is heated far more efficiently by infra-red light than the ocean surface.
On average the Earth surface (both ocean and land) looses 492 W m^-2, 24 W m^-2 by upward convective transport, 78 W m^-2 by evapo-transportation, and 390 W m^-2 by surface radiation.
Most of the 102 W m^-2 that is lost by upward convective transport and evapo-transportation is lost over the oceans and not the land (even if you allow for their differences in area). Will mistakenly compares this 102 W m^-2 with the 324 W m^-2 atmospheric back radiation, when he
should be comparing it to the net infra-red radiation exchange between the ground and the atmosphere/space i.e. 390 W m^-2 (infra-red radiation from the Earth’s surface) – 324 W m^-2 (infra-red back radiation from the atmosphere) = 66 W m^-2.
It is not hard to see that over the oceans the 102 W m^-2 upward infra-red radiation more than over-powers the net 66 W m^-2 downward infra-red radiation. This is where the energy goes!
Sorry, the infra-red back-radiation may warm the oceans by a small amount but it pales in comparison with the far greater warming that occurs by teh absorption of direct visible radiation
that takes place mostly at the tropics.

Tim Folkerts
August 15, 2011 9:01 pm

Some people seem to be under the mistaken impression that the top few microns will be warmer than the bulk water, but in fact that is wrong. The surface, which acts like a black body for IR light will emit IR very efficiently, thereby cooling the surface.
The “DLR” would keep the surface layer from getting quite so cold, but it is mainly limiting the cooling by radiation that would be occurring. This would keep the top microns of water water than before, but still cooler than the without the DLR.
There also seem to be a misconception that since the visible light can penetrate ~ 100 meters, the heating is occurring ~ 100 m down. IN fact, sunlight will heat the top meter the best, the 2nd meter almost as well. By the time we are to the 100th meter, there will be almost no heating. The top meter will be the warmest (ignoring other details like mixing, other materials in the water, etc). .

Ninderthana
August 15, 2011 9:02 pm

Sorry, the second last paragraph should have read:
It is not hard to see that over the oceans the 102 W m^-2 upward energy loss more than over-powers the net 66 W m^-2 downward infra-red radiation. This is where the energy goes!

Dave Springer
August 15, 2011 9:05 pm

@_Jim
Further on spectrograph looking from above. Given the tops of the steps follow a curve for a certain blackbody temperature and given we can see through to the ocean surface we can calculate the temperature difference between the ocean surface and any of the lower steps. Given we also know which gases are causing the lowered steps and given we know the adiabatic lapse rate we can determine the effective emission altitude for any of those gases. Pretty neat, huh? I made a comment in another thread where I did the calculation for CO2 and found it has an effective emission altitude at 15um of about 2000 meters above the surface IIRC.

August 15, 2011 9:07 pm

“jimmi_the_dalek says:
August 15, 2011 at 1:52 pm
Indeed, the idea that DLR cannot heat the ocean is one of the spurious arguments that should not be used. There are plenty of others. Far too many comments here claim that the Greenhouse Effect cannot be real “because it is not like a real greenhouse”. Or that it contradicts the first or second laws of thermodynamics. Stick to objecting to the infidelities of computer modelling – there is plenty of uncertainty there – the basic physics is much more secure than some people are willing to concede.

——————————————————————————–
I think this is something to argue. Why is it that the computer models never match reality? The only true answer is that they use incorrect physics or some mistake somewhere. If they correctly modeled the physics of our planet, they should theoritically spit out a good synopsis of weather patterns but alas, they do not. They in fact never agree with each other on local conditions, and globally, well the IPCC itself split the difference and stated 1.5-4.5C. Why is it that they require such large fudge factors in the first place?
The true scientist would have figured this out and realized that the physics was wrong. Go back to the drawing board, find the mistake and plug in over again. No, instead they fudged it. This is important. The part of science and physics is important. The part of getting this correct so that in the future we do not make the same mistakes. The same mistakes applies to a political and/or ideological movement that looks to make energy scarse and drive our economies into the ground. This movement is nothing but an inquisition all over again.
But alas, the physics might be right if the Earth was actually a black body. But we are a water planet. Water is our dominant feature from the oceans to the atmosphere. Since this is the case and the models like I said have never matched reality, we KNOW there is at LEAST one mistake in their physics calculations. I don’t think closing the book on any part of the physics is the greatest idea in the world. Keep an open mind and realize that yes, there is at least one mistake, and I would guess more since greater minds have attempted this issue. I would point most people towards the lapse rate in general (type does not matter…just look at the changes in pressure and how this effects the atmosphere.)
DLR might have a mistake in it, but I am willing to bet that if it does, its a small one. Focus on where the physics is likely wrong (where Dr. Sagen and Hansen got it wrong on Venus for instance) and explore from there. That is where the mistakes are going to be found.
Falsify the physics, and you falsify every GCM out there. Sure, the theory is still supposedly possible, but I think that blow would be enough to settle the science the other way.

kuhnkat
August 15, 2011 9:09 pm

Willis, when did you start taking acid again???
“We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.”
A man of your intelligence and learning should NOT be throwing BS like this around. The ocean is NOT losing 390 w/m2. It is losing 390-~320w/m2=~60w/m2 which is far less than the 170w/m2 that it is absorbing. Where that heat comes from for the 30+70 is then quite obvious.
This is the kind of BS that you get involved in when you start ignoring real physics. Now, where is the fabled BACKRADIATION in that equation??
You also mention the air has no thermal mass. The air has enough thermal mass that there would be pretty much no convection if the GHG’s did not transfer their energy to it. It may not hold a candle to water or even the earth, but, it is much more substantial than the thermal mass of the lightweight CO2!!!
So, the answer is equivocal, the LW does NOT heat the ocean, the SW does and the GHG’s spread the heat through the atmosphere or cool the atmosphere depending on the balance at that moment!!! Basically all the DLR goes UP one way or another!!!

kuhnkat
August 15, 2011 9:10 pm

Subscribe.

August 15, 2011 9:12 pm

Dave Springer says:
“The ocean certainly does glow in the infrared.”
Yes, of course you’re right. And you’re right that land emits much more IR than the oceans. That was really the comparison I was trying to make. I should not have said the ocean doesn’t emit IR. Most everything does, above absolute zero.

Ninderthana
August 15, 2011 9:18 pm

Just in case some people misunderstand, I had better clarify an important point.
The argument that I have used above involves averages that include both sea and land.
Of course, both the ocean surface and land surface are exposed to ~ 324 W m^-2 infra-red back radiation. The land surfaces absorb virtually all of the 324 W m^-2 and redistributes a large part
to heat subsurface and atmosphere above. Much of this is then re-radiated back to the atmosphere as infra-red light.
The sea surface also absorbs much of the 324 W m^-2 as well. The difference is that this absorption takes place in the top mm or so which is the part of the ocean that is being actively evaporated and transported back into the atmosphere. So little of this absorbed energy actually goes into heating the deeper ocean.

Myrrh
August 15, 2011 9:22 pm

I don’t know where Willis has got the idea that it is the Warmista’s teaching that longwave infrared, (thermal infrared), heats the oceans.. For the last couple of decades it has been the AGWScience fiction meme that only the shortwave Visible (Light) UV & NR, heat the oceans and Thermal Infrared (Heat) doesn’t get through the atmosphere to heat the Earth, it is being taught in schools. It is so viral now that anyone not knowing the real difference between these energies takes it on trust.
“Basic mechanism
The Earth receives energy from the Sun in the form UV, visible, and near IR radiation, most of which passes through the atmosphere without being absorbed. Of the total amount of energy available at the top of the atmosphere (TOA), about 50% is absorbed at the Earth’s surface. Because it is warm, the surface radiates far IR thermal radiation that consists of wavelengths that are predominantly much longer than the wavelengths that were absorbed.”

http://en.wikipedia.org/wiki/Greenhouse_effect
http://en.wikipedia.org/wiki/File:Greenhouse_Effect.svg
THIS is the Warmista’s claim. That “Solar” is only Visible and UV and Nr Infrared (which isn’t thermal) and it is only this which heats the land and oceans.
“Solar” in, Thermal IR out.
Willis is spouting bull, don’t know where he got it from, but: You’re arguing against a strawman set up here…
Wakey, wakey.
It’s the sceptics who point out that this is junk science and that it’s Thermal Infrared (Heat), which heats the Earth’s land and oceans, and us.
Here’s some of the post I linked above, which is when it came to light that NASA is not only making it difficult for real physics to be taught to children, but it is deliberately teaching this AGWScience meme that downwelling thermal infrared doesn’t reach the surface.
“..thankyou for posting that link to the NASA site which shows clearly that it has now stopped teaching traditional well-known and understood differences between Light and Heat energies from the Sun and replacing it with AGWScience fiction memes. This corruption of basic science is deliberate and systematic – dumbing down science education for the masses.
I think this agenda should be brought into the spotlight and a comparison of the NASA pages pre and post corruption is an excellent example as it easily conveys the extent this manipulation has reached. NASA’s reputation is being used to promoted science fiction. I am greatly saddened by it.

I’ll pull a few more quotes into what I posted above, http://wattsupwiththat.com/2011/07/28/spencer-and-braswell-on-slashdot/#comment-711614 for a better look at the difference.
NASA original page teaching previously traditional real world physics to children: http://science.hq.nasa.gov/kids/imagers/ems/infrared.html

From this NASA page:
“Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.
Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!

compare with:
NASA page now teaching that thermal infrared doesn’t even reach us!: http://imagine.gsfc.nasa.gov/docs/science/know_l1/emspectrum.html

Electromagnetic radiation from space is unable to reach the surface of the Earth except at a very few wavelengths, such as the visible spectrum, radio frequencies, and some ultraviolet wavelengths. Astronomers can get above enough of the Earth’s atmosphere to observe at some infrared wavelengths from mountain tops or by flying their telescopes in an aircraft.
http://imagine.gsfc.nasa.gov/Images/introduction/emsurface.gif [Graphic showing downwelling infrared from the Sun stopping short of Earth’s surface, not even reaching mountain tops.]
http://imagine.gsfc.nasa.gov/docs/dict_ei.html#em_waves [link from em spectrum page]:
infrared
Electromagnetic radiation at wavelengths longer than the red end of visible light and shorter than microwaves (roughly between 1 and 100 microns). Almost none of the infrared portion of the electromagnetic spectrum can reach the surface of the Earth, although some portions can be observed by high-altitude aircraft (such as the Kuiper Observatory) or telescopes on high mountaintops (such as the peak of Mauna Kea in Hawaii).

From teaching real physics that the heat we all feel from the Sun is thermal infrared, to the new science fiction paradigm from NASA that no infrared even reaches the mountain tops.. This is one step further than the AGWScience fiction KT97 claim, which says near infrared, (the shortwave not thermal in real physics, not hot), is included in their “Solar” downwelling reaching Earth’s surface, (Visible with the two shortwave either side of UV and Nr IR).
KT97 = Kiehl/Trenberth 1997

To put into science terms, if a new idea contradicting well known and understood and tried and tested real physics as taught traditionally is being promoted, then the promoters must provide proof that the traditional teaching is wrong and the new idea right. Eliminating the traditional teaching from the education system does not constitute proof…
Thank you Willis for bringing it to greater attention.

August 15, 2011 9:30 pm

Since we have not yet quantified exact change in numbers of all “GH” molecules in time, namely hundreds of water vapor molecules, in the atmosphere, it is nonsense to quarrel how much surface warming is caused by one (bad) molecule of CO2 per 10,000 other molecules, which was added to three (good) molecules of CO2 since 1780. How much has changed the cloud cover? Are there 97 or 103 H2O molecules per the bad CO2 molecule on average? These are much more important questions to be answered before.

Dave Springer
August 15, 2011 9:37 pm

_Jim says:
August 15, 2011 at 8:58 pm
“The gaseous state has some different properties, where molecule resonances/vibration modes are more pronounced in gas state molecules like water vapor vs liquid.”
Yes indeed it does. Most significantly there are a number of LWIR infrared windows in the vapor. There are no LWIR windows in the liquid.
“WV atmospheric effects as well as liquid water characteristics on the same webpage:
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water#Atmospheric_effects
One of the poorer technical articles on wickedpedia for sure. Ferinstance:

Water vapor is a greenhouse gas in the Earth’s atmosphere, responsible for 70% of the known absorption of incoming sunlight, particularly in the infrared region

Say what? First of all most of the energy in sunlight is in the shortwave region and this passes through water vapor pretty much unattenuated just like it passes through pure water pretty much unattenuated. “Particularly in the infrared” is quite the understatement. It’s virtually all in the infrared there’s precious little infrared energy in sunlight to begin with. This sentence was probably one of the nocturnal emissions of William Connolley
http://en.wikipedia.org/wiki/William_Connolley
AGW POV warrior extraordinaire who was finally canned as an editor on wikipedia for bias and disinformation and general obnoxiousness so profound that even the librat powers that be at wikipedia couldn’t tolerate it any longer.
“Of note: “The spectral absorption featu
res of liquid water are shifted to longer wavelengths with respect to the vapor features by approximately 60 nm””
Also of note: a shift of 60nm is a 3 degree fahrenheit change in blackbody temperature
Might be worth considering in some of the finer details but given the context here it doesn’t make a bit of difference since liquid water absorbs fully and continuously across the LWIR spectrum. The only way it would be meaningful is if there were some LWIR windows in liquid water where a change in temperature of 3 degrees might move it into or out of a window. No windows, no effect.

Owen
August 15, 2011 9:42 pm

Three great articles from Willis in two days. My head is going to explode !!

ferd berple
August 15, 2011 9:44 pm

Posted on August 15, 2011 by Willis Eschenbach
“So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?”
Willis, no amount of theory can contradict observations. A more difficult question is this: Why are the poles a block of ice if DLR is so high?
If average DLR is so much higher that average solar radiation, then why is there so much of a temperature difference between the equator and the poles? Why is summer in the polar regions with 24 hours of (very weak) sunlight (angle of incidence) so much warmer than the polar winter with 0 hours of sunlight? If solar energy is such a small part of the total energy, then DLR should keep the poles very much the same temperature year round, regardless of the (very weak) amount of sunlight.
If DLR is so high as compared to solar radiation, then the polar winter should be much closer to the polar summer in temperature. Observation contradicts theory, especially in Antarctica, where the moderating effects of water are minimal..

kuhnkat
August 15, 2011 9:51 pm

Willis says,
” DLR heating of the surface slows that overturning.”
Umm, exactly where do all those GHG’s get the energy to radiate all night? Do they have their own tiny personal cold fusion generators?? Me thinks this statement is also very misleading!!!
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHA
Lemme see, the ocean has a huge thermal mass and radiates all night providing energy to the GHG’s, if they didn’t the rest of the atmosphere would be driving the GHG’s radiation, although at MUCH reduced levels!!

Noelene
August 15, 2011 10:01 pm

Dumb question
How does rivers figure in all this?Rivers are always cold are they not?Is that because of their depth?

Brian W
August 15, 2011 10:02 pm

Dave Springer, steven mosher, Jim, smokey
Neither clouds, nor water, fellows DO NOT glow in IR. Since IR is by definition non visible radiation, YOU CANNOT SEE IT. Get it. GOES colors its pictures so you can see something which is incapable of exciting vision. So enough of the crap.

Martin Clauss
August 15, 2011 10:03 pm

Willis,
I look at the semantics of ‘warming’ vs. ‘reducing the rate of cooling’ in this way:
if the ocean is radiating 400 watts/square meter, and there were no CO2 or H2O above radiating back to the ocean (the DLR), it will cool at a certain rate – obviously faster than with CO2/H2O above it, radiating back, say, 300 watts/ square meter. But in both cases, the result is a cooling ocean.
Now if the CO2/H2O were radiating 400 watts/sq meter, then the ocean is neither heating nor cooling, it is staying the same – might one say the two are in equilibrium with each other?
Further, if the CO2/H2O were radiating MORE than 400 watts/sq meter, then that is when I would say the DLR is ‘warming’ the ocean.
That’s just how I look at. One object is warming another when the object doing the warming raises the temperature of the other object ABOVE the temperature state the other object is in.
Now even that could be argued semantically, because if I were out in the cold air, my hands were cold, and I put a pair of gloves on, my hands would become warmer. I might say, ‘my gloves are warming my hands – though I could also say, ” . . my hands are warming AS A RESULT of the gloves” (prevent them from losing heat in the cold air ). . .
So obviously this discussion over ‘warming’ vs’ reduced rate of cooling won’t end, but I still prefer ‘reduced rate of cooling.
I should mention that even with this discussion and my perspective, I agree with your arguments.

kuhnkat
August 15, 2011 10:08 pm

Dave Springer,
“Absolutely. It’s called pyregeometer. Wanna buy one?”
And that is exactly the problem with Climate Science and the way the problem is stated. Stefan Boltzman does not compute 390 w/m2 from the AVERAGE surface of the earth unless it is to a vacum. What would happen to the surface of the ocean if we were able to remove the atmospheric pressure and radiation from above it?? I would love to see a video!!!
These things should be built in pairs to only measure up and down at the same time so Climate Scientists can’t play games!!! Even then we dont’t get the other directions!! 8>)

Dave Springer
August 15, 2011 10:18 pm

Willis Eschenbach says:
August 15, 2011 at 9:35 pm
“I have referred to generally accepted estimates of radiative loss from the surface (usually taken as ~ 390 w/m2), evaporative loss (~ 80 w/m2), convective loss (~ 20 w/m2) and sun hitting the surface (usually taken as 170 w/m2). ”
The same people who generally accept a 3C rise in global average temperature per CO2 doubling. Let’s look at an actual study of ocean heat budget (which I’ve posted before and you either ignored or forgot):
http://www.atmos.umd.edu/~carton/pdfs/foltzetal03.pdf

JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 108, NO. C5, 3146, doi:10.1029/2002JC001584, 2003
Seasonal mixed layer heat budget of the tropical Atlantic Ocean
Gregory R. Foltz, Semyon A. Grodsky, and James A. Carton
Department of Meteorology, University of Maryland, College Park, Maryland, USA
Net surface heat flux is a combination of latent and
sensible heat loss, shortwave radiation absorption, and net
longwave emission. Sensible heat loss is insignificant (<10
W m2) due to small air-sea temperature differences, while
net emission of longwave radiation is a relatively constant
50 W m2 [da Silva et al., 1994].

The math ain’t difficult Willis. The tropical Atlantic absorbs about 200Wm from the sun. What goes in must come out. 10Wm it loses by conduction. 50Wm it loses by radiation. The rest, 140Wm, is lost through latent heat of vaporization.
10/200 = 5% lost by conduction
50/200 = 25% lost by radiation
140/200 = 70% lost by evaporation
What part of that do you and Trenberth and the rest of the bandwagon science brigade not understand?

Dave Springer
August 15, 2011 10:33 pm

@Willis
Can you or someone please explain to me how an ocean that receives only 200Wm of incoming energy can possibly emit more than that? The figure of 390Wm lost by radiative transfer is physically impossible. That would require it be absorbing at least 390Wm which it absolutely does not do. Have you people never heard of the law of conservation of energy? Maybe it’s the thorium dissolved in the ocean that producing all the extra heat, huh? The sun gets focused into a laser beam by water amplification which heats the thorium which causes it to condense and give off more heat than it absorbs through some mechanism that left unexplained except to say it definitely isn’t fission.
God almighty the pseudoscience nonsense pecked out by pikers around here sure gets frustrating at times. It’s so thick in the OP you can cut it with a knife.

August 15, 2011 10:41 pm

Good on you, Willis, for trying to do a little education here. You have great patience.

David Falkner
August 15, 2011 10:56 pm

What happens to the DLR if the ocean doesn’t absorb it? Does it power Santa’s sled back to the North Pole? Go back to space? Why haven’t satellites definitively proved this? Quit wasting your time on BS folks. Willis, you have better things to do, right? I can understand, by the way, the reason for using gross over net. I wouldn’t say acceptable, I would say preferable.

Dave Springer
August 15, 2011 10:58 pm

Brian W says:
August 15, 2011 at 10:02 pm
“Dave Springer, steven mosher, Jim, smokey
Neither clouds, nor water, fellows DO NOT glow in IR. Since IR is by definition non visible radiation, YOU CANNOT SEE IT. Get it. GOES colors its pictures so you can see something which is incapable of exciting vision. So enough of the crap.”
http://www.amazon.com/Sightmark-Night-Raider-2-5×50-Vision/dp/B003UCC68E
I never leave home without my passive night-vision gun scope which brings everything you survey into glowing contrast. It makes you glow with confidence that nothing will escape your notice in even the darkest of dangerous dark alleys.
Seriously, that’s a pretty lame nit to pick especially since we defined the glow as an infrared glow which by definition cannot be seen by the naked eye.

DR
August 15, 2011 11:00 pm

No mention of the Coriolis effect?
How does a rain drop freeze? Did someone figure that out yet?
If this is an exercise in determining if ~100 ppm of CO2 can account for the OHC increases seen for the last ~50 years, I beg the question to be asked what is the heat capacity of CO2 that allows for it to warm the oceans more so than the sun itself.
Just wondering.

Richard111
August 15, 2011 11:06 pm

Having got used to the idea that temperature varies to the fourth power, and recently learning that photon flux varies to the third power, I was hoping I might learn how this fits in the above discussion.

Dave Springer
August 15, 2011 11:08 pm

Brian W says:
August 15, 2011 at 10:02 pm
“Neither clouds, nor water, fellows DO NOT glow in IR. Since IR is by definition non visible radiation, YOU CANNOT SEE IT.”
Yeah, but some animals can. I hope you aren’t some kind of species bigot who thinks human eyeballs are some kind of universal standard for what glows and what doesn’t?
http://en.wikipedia.org/wiki/Infrared_sensing_in_snakes

The ability to sense infrared thermal radiation evolved independently in several different families of snakes. Essentially, it allows these animals to “see” radiant heat at wavelengths between 5 and 30 μm to a degree of accuracy such that a blind rattlesnake can target vulnerable body parts of the prey at which it strikes. It was previously thought that the organs evolved primarily as prey detectors, but recent evidence suggests that it may also be used in thermoregulation and predator detection, making it a more general-purpose sensory organ than was supposed.

Alexander Duranko
August 15, 2011 11:08 pm

Various people above have claimed that DLR is real because you measure it with a radiometer AND it increases when relative humidity increases. Well folks, you really do need a basic course in engineering heat transfer starting with Prevost’s Law then Hoyt C Hottell’s 1954 paper on calculating the emissivity and absorptivity of gases, then moving onto Kirchhoff’s radiation Law, a corollary of the 2nd Law of Thermodynamics.
Prevost’s Law of Exchanges [1840] states that at radiative equilibrium, any body above absolute zero emits radiation and receives exactly the same radiation from the opposite direction. In the case of hot gases, it’s spherical emission and absorption. If there’s an excess in a particular direction, there is no equilibrium. So when you point a radiometer upwards to measure ‘DLR’, you then have to reverse it to measure ‘ULR’ and the difference is what causes heating or cooling.
Hottell showed how you calculate emissivity as a function of pressure and concentration. Increase the concentration of a greenhouse gas and you increase emissivity. So DLR only apparently increases; in reality, the radiative flux in the opposite direction also has to increase.
This is why clouds appear warmer; emissivity tends to unity compared with c. 0.1 for moist air at ambient pressure. This leads onto Kirchhoff’s Law which is that at radiative equilibrium, emissivity = absorptivity.
I find it amazing that any physical scientist should not know these basic scientific facts. Go away until you do know them because this DLR fantasy has to stop.

Dave Springer
August 15, 2011 11:11 pm

@Willis
“Please stop the condescending snarkiness, it just makes you look ugly.”
Then please stop parroting Trenberth, it just makes you look stupid.

August 15, 2011 11:12 pm

The fact is Willis that you’re rightly a stickler for precision. Precision in thinking leads to precision in analysis and thinking the DLR warms the ocean is sloppy and going to lead you astray.
Ooh look clouds overhead, the DLR has increased by 100W/m2, thats got to really be heating the ocean now!
DSR warms the ocean. DLR slows the rate of cooling.

Dave Springer
August 15, 2011 11:30 pm

Hey Willis,
If I sell you a beat up surfboard for $390 and give you an instant rebate of $340 what’s your actual cost for the surfboard?
You and Trenberth sure have some interesting methods of accounting. Maybe you should both be in congress. You’d fit right in. You could collect $390 billion in carbon taxes, refund $340 billion of it in carbon credits, then add to your stump speech that you increased government revenue by $390 billion.
I’m sorry for mocking this but some things just plain deserve mockery.

tallbloke
August 15, 2011 11:33 pm

steven mosher says:
August 15, 2011 at 4:07 pm
I sit outside on a freezing winter night with a space blanket.
The blanket doesnt warm me. The blanket slows the heat loss via radiation.

Actually Mosh, the silvering on the space blanket does very very little. A piece of clear plastic will be very nearly as effective. This is because nearly all of the effectiveness of a space blanket is in reducing convection, and thus, wind-chill. The radiative component is tiny by comparison. Tests have been done on this in the climbing and backpacking magazines years ago. Maybe you can find the info online too.

tallbloke
August 15, 2011 11:36 pm

Smokey says:
August 15, 2011 at 5:36 pm
As we all know, heat is transferred by conduction, convection and radiation. If the top 50 micrometers of the ocean is warmed by IR, wouldn’t those molecules slough off their extra energy almost instantly via conduction, warming the adjacent molecules? And so on, transferring the heat deeper into the ocean. Am I missing something?

Yes, you’re missing the fact that warmed water molecules become more buoyant than their neighbors and head upwards. Rock particles don’t do that.
Smokey says:
August 15, 2011 at 7:37 pm
And of course, something is warming the oceans.

Yep, it’s called the Sun.

kuhnkat
August 15, 2011 11:37 pm

Willis,
you want serious? OK, dead serious, here is the problem with individual energy flows.
IF there was no atmosphere and no GHG’s above the ocean the instant temperature and energy emissions would have no relation to 390 w/m2. Computing this number is an abortion. My understanding of this from people who understand how to derive the equation is that it simply is not real.
The fact that there IS atmosphere and GHG’s above the ocean means that there has been a relative equilibrium reached where you CAN measure these fluxes, BUT, they are not meaningful without the opposing flux. Saying there is 390 w/m2 is only meaningful if there is ALSO 320 w/m2 down to go back up to create that much energy in the first place. The IR down is the IR up, time shifted by a tiny amount of time, minus the losses that went on out and were transferred to the atmosphere and a few other miscellaneous things. Talking about heating the ocean is poor semantics as the flux is into the ocean with the SW and out of the ocean with IR. IF you wanted to get really picky we would have to include the gravitational energy and electrical energy that disrupts the ocean and atmosphere adding to the wind from the convection. Just like the conduction between dissimilar objects, at a very low level there may be occasional flows against the gradient, but, generally it is all warmer to cooler. I would stick with the SLOWS COOLING and forget the abortion of 390 w/m2 as it only allows people to get confused. If you say 390 you must say 320… (or whatever the correct numbers of the situation are)
“I can hand you a hundred dollar bill, and at the same time you hand me four twenties. That’s one way to describe the transaction, the way I described it above, listing the individual flows. ”
You could also give me a hundred dollar bill without me giving you back anything. The ocean cannot give the atmosphere 390 w/m2 without first getting the 320!!!! At least not with our current set up.
I apologise for misquoting you. Your statment was: “It can’t be heating the air, because the atmosphere has far too little thermal mass.” But some of it is heating the air.
The best idea is to ignore the manner in which Climate Scientists talk about it dividing the fluxes that cannot exist without each other and speak about the net flow. This may make no one happy, but, is most realistic. The surface regime emits about 60 w/m2. The instant fluxes making up this amount are ~390 up and ~320 down. The ocean absorbs plenty of SW to create this balance.
PS: I was addicted to cocaine twice so, if you want to make fun of me for it I deserve it!!! As an absolute and not just because of this. Reading your bio didn’t give me the feeling you had been an addict so I apologize if this was a low blow.

kuhnkat
August 15, 2011 11:52 pm

Tallbloke,
I would add the space blanket also retains the moisture which carries the most energy!! My first experience with a space blanket was one that belonged to a friend. I had a cheap bedroll and we were sleeping on the ground outside a Yosemite campground to save a few bucks. There was a drizzle so my friend loaned me his space blanket. I ended up wetter than if I hadn’t used it!!! All my body moisture condensed in the cheap bedroll soaking me. Of course I didn’t get that cold until I got out of it in the morning in wet clothes. DANG!!!

kuhnkat
August 16, 2011 12:08 am

Tallbloke,
That reminds me of a Greenhouse test with IR treated and NON-IR treated plastic covers of the same material and thickness to prove IR effectiveness. The actual numbers were close to 5 degrees difference between the IR treated cover and the non-IR cover. I was shocked. I then reread the specs and found the IR treated cover ALSO had a treatment to minimize condensation collecting on it. The reduced moisture condensing on the cover made most of the difference I believe!! None of the other Greenhouse covers treated for IR were that effective with the same thickness and type material!!

intrepid_wanders
August 16, 2011 12:12 am

Eschenbach-sensai.
Moderation leads to generosity,
Mercy leads to courage,
Humility leads to leadership.
Please do not take offense with the deriding tones of various “unpleasantries”. As mentioned before, there are issues with working with the discrete and jumping back to the bulk. I see things discussed here very similar to the counter-intuitive concepts of gravity that Galileo discovered with the balls on a ramp. To some, they may be be clear, but to others, the details cloud the picture.
The two concepts that Mosher-sensai mention are not inconceivable, but the discrete level does cause issues. Unless the atmospheric energy transfer model can be conveyed to why the anticyclonic (Red Spot, and the additional coalition of white spots) event on Jupiter has persisted for possibly hundreds of years, there will be questions.
Thank you for contribution to at least my knowledge.
POST SCRIPT (TO ALL THAT ARE DEBATING DLR):
This issue is NOT the the NET Daytime Radiation. The ISSUE is the NIGHTTIME BLANKET EFFECT. When the SUN is out, it will ALWAYS dominate the “energy budget”. Obviously it is the FIREPLACE that warms our world. When the IR saturated H2O, CO2 and other vibrational molecules fall back to sea level at night, some are not ABSORBING new energy. Nighttime tempertures will be “slightly warmer”, which will allow the oceans to retain more heat from the DLR. Vector algebra still works, only in the dark ;).
I believe a successful model of Jupiter would be a huge step in modelling, that is if there are no solids to screw up the equations 😉

Dave Springer
August 16, 2011 12:13 am

Once again, the crazy idea that downwelling longwave radiation (DLR, also called infra-red or IR, or “greenhouse radiation”) can’t heat the ocean has raised its ugly head on one of my threads.
There are lots of good arguments against the AGW consensus, but this one is just silly.
Look, folks, there’s lot’s of good, valid scientific objections against the AGW claims, but the idea that DLR can’t heat the ocean is nonsense. Go buy an infrared lamp, put it over a pan of water, and see what happens. It only hurts the general skeptical arguments when people believe and espouse impossible things …
The craziest, ugliest, silliest, nonsensical impossible thing (you set the tone, Willis, not me) is that an ocean which receives a net input of 200Wm can emit more than that and sustain the loss indefinitely. Conservation of energy is a bitch that isn’t going to budge.
This is ground control to Major Tom.
Your circuit’s dead, there’s something wrong
Can you hear me, Major Tom?
Can you hear me, Major Tom?
Can you hear me, Major Tom?
Can you….

Dave Springer
August 16, 2011 12:23 am

@Willis
“Stop acting like I’m an idiot and like Trenberth is making stupid assumptions, Dave, it’s a foolish error in both cases. We know what we’re talking about. Here’s the part that you seem not to understand:
You (and the cited paper) are talking about net radiation flows. Trenberth and I are talking about individual radiation flows.”
Glad to oblige. Stop acting like an idiot and I’ll treating you like one. The net flow is all that matters. Get that through your thick skull.

tallbloke
August 16, 2011 12:38 am


“The question is, do DLR heated water molecules make it downwards far enough for long enough to warm the ocean bulk. I think the answer is no, because warmer water molecules are naturally buoyant”
Again we are in the realm of semantics, but the energy flow is clear. Since the slowly overturning ocean is warmer than it would be in the absence of nighttime DLR, we say DLR warms the bulk ocean. How? By preventing the bulk ocean from cooling.

I don’t know if Willis is going to respond to my comment directly, so for now I’ll have to pick through the thread to see where he has answered other people by addressing my arguments.
The energy flow is indeed clear. The net effect of long wave radiation at the ocean-air interface is to cool the ocean by some 66W/m^2. Talking about downward flow of long wave radiation without considering upward (and sideways!) flow at the same time doesn’t make sense to me. It’s a flux.
.The bulk ocean is actually warmer because of the DLR … but not because the “DLR heated water molecules make it downwards” as Tallbloke suggests
Willis has this the wrong way round. He said in argument one:
“the ocean can circulate the heat downwards through turbulence”
and I said:
“.do DLR heated water molecules make it downwards far enough for long enough to warm the ocean bulk. I think the answer is no, because warmer water molecules are naturally buoyant, and because the vortices which mix solar energy so efficiently are below the wave troughs, several thousands of times deeper down than the depth DLR penetrates water to.”
But anyway, I’m glad Willis has abandoned argument one.
Willis Eschenbach says:
August 15, 2011 at 10:14 pm
The claim seems to be (e.g. Tallbloke)
Warm water molecules rise to the top. Warm rock molecules conduct heat to their neighbours, which can’t go anywhere.
IR heats the top molecule. It passes some reduced amount of that heat to the molecule below. But what tallbloke forgets is that the top molecule can’t make the second molecule warmer than the top molecule, heat doesn’t flow from cooler to warmer.

Radiative transfer is a quantum operation according to theory, a water molecule can’t pass part of a photon. If a water molecule emits a photon to it’s neighbor below and drops to a lower vibrational state it becomes ‘cooler’ that the molecule below, which will rise to displace it. It’s worth noting at this point that the ocean surface is nearly always warmer than the air above it, and cooler than the water below it, so conduction isn’t going to work to get energy downwards either, as the second law of thermodynamics has to be observed. Any conduction taking place will be from the ocean into the air, because as Willis correctly states:
“heat doesn’t flow from cooler to warmer.”
I haven’t had time to read all the comments. so if I’ve missed another of WIllis’ references to my comment then I hope someone will flag it up. I’d be particularly interested to know if he has addressed my concluding point:
“But this isn’t about absolutes. I’m sure the increased DLR warmed the ocean a little bit, or at least slowed its rate of cooling a little bit. I think the increased insolation due to (empirically measured) reduced cloud cover in the tropics 1980-1998 did a lot more to increase ocean heat content. To turn your question back to you, where else could that energy have gone?”

cal
August 16, 2011 12:49 am

For those who insist that DLR reduces the cooling and does not warm the surface consider the following analogy. Consider two groups of people are facing each other. A machine throws huge rocks over the heads of group A towards group B. Some of the rocks richochet and are lost the rest break into pieces. The people from group B pick up the pieces and throw them back. Some of these pieces fly past group A but others are picked up by group A who throw them back to group B. Now we know that all the rocks originally came from the machine but how do we describe the bombardment of group B? Do we say that they are bombarded by big and small rocks? Or do we say that they are bombarded by big rocks and the presence of group A reduces the number of small rocks that group B throws? I do not think the latter describes reality. One can say that the presence of group A reduces the rate at which small rocks are lost from around the feet of group B but this is not analogeous to warming or reduced cooling it is analogeous to heat accumulation. To recognise the difference one only has to consider the melting of ice which involves heat accumulation but no warming. It would be like group B collecting rocks in piles and waiting before throwing them back.

Dave Springer
August 16, 2011 12:54 am

Another bottom line. Trenberth (and Willis by association) are the ones who can’t find the missing heat from the last 50 years. I told y’all where it is. It’s distributed more or less uniformally in a spherical volume of space 100 light years in diameter surrounding the earth.
Keep spinning your wheels looking downward into the deep ocean if you must. I can lead a horse to water but I can’t make him drink. All I really ask is that you don’t reach into my wallet to cover the cost of your wild goose chase, K?

Mark.R
August 16, 2011 12:55 am

This here is a oiece from a paper by NIWA.
Decadal temperature changes in the Tasman Sea(There was a warming of the Tasman Sea deep to 800 mtre.)
Quote
“The first possible forcing mechanism to consideris air-sea flux. This mechanism is unlikely given the depth penetration of the warming signal, as surfacewarming would increase stratification rather than warm the deep ocean. Air-sea flux can also be eliminated as a possibility by examining European Centre for Medium-range Weather Forecasts(ECMWF) and National Centers for Environmental Prediction (NCEP) heat flux data. The ECMWF andNCEP products indicate that the annually-smoothedheat flux varies between c. 10 W/m2 and c. 40 W/m2 with a mean of c. 25 W/m2. However, the heatflux is always from the ocean to the atmosphere, i.e.,the ocean is losing heat to the atmosphere rather than gaining it”.
publications-journals-nzjm-2005-107-lo.pdf (261.68K)
publications-journals-nzjm-2005-107-lo.pdf

August 16, 2011 1:22 am

“Since the second molecule is not as warm as the top molecule, in contradiction to tallbloke’s claim, it doesn’t rise to the top. And the same for the layers further down. The heat is transmitted down and down, but each layer can’t heat the lower layer more than itself, heat won’t flow uphill. So the water, though warming, doesn’t “rise to the top” as claimed.”
Nice thought but you’re ignoring the fact that the ocean is radiating upwards more than the atmosphere radiates towards it. The energy flow is upwards not downwards. Therefore the molecules at the top are on average losing more energy than they are gaining and so there is no downwards movement of heat as you suggest.
Its my view that the energy the DLR imparts on the topmost ocean surface molecules is very quickly re-radiated back up, so you can think of radiation at the surface of the ocean as bouncing between the atmosphere (CO2 and water vapour) and the water itself. Minus that which is used for evaporation. This is a simplistic explanation so dont bother attacking it on lack of detail. A more detailed explanation was made over at the Science of Doom.

August 16, 2011 1:52 am

Hmm Tallbloke. Rather than trust a magazine about how reflective insulation works or does not work I think I trust the stuff I built for DOD. And I’ll use my thermos to keep my coffee from cooling faster than it would. And If I have to go near a wacking hot fire i’ll also wear a reflective suit
here have some fun. there’s plenty more
http://www.insul.net/howto.php
And yes, if you use a space blanket too long you get hoarfrost in the inside.. cause its working. but eventually get to the fire

eco-geek
August 16, 2011 2:00 am

testing testing the comments under obesity and AGW are not working

August 16, 2011 2:02 am

You note that nobody (except willis) wants to look at the empirical evidence. I posted it up there boys..
Perhaps the D word should be allowed for this topic

Bob_FJ
August 16, 2011 2:04 am

Willis Eschenbach @ August 15, 2011 at 9:21 pm

“…And I still haven’t heard you [TimTheToolMan] or anyone else explain why the ocean is liquid, what mysterious energy source you claim keeps the ocean from freezing soltd. Losing 400 w/m2 as upwelling longwave, gaining only 170 w/m2 from the sun, here comes the ice age …”

Willis, I’m shocked! Putting aside just where you get your numbers from, they are somewhat similar to those in the Trenberth et al 2009 cartoon of energy balance. Do you NOT understand that EMR (radiation) IS A DIFFERENT FORM OF ENERGY TO HEAT? Radiation whizzes around in ALL directions but it is not HEAT. The net global average HEAT loss via radiation from the surface according to Trenberth is a mere 66, not 400 W/m^2, and the incoming averaged solar is 181 W/m^2. But, of course the annual global diurnal/ seasonal/ regional averages etc on that cartoon are a nonsense anyway, if you would like to proceed to a more complicated discussion

kadaka (KD Knoebel)
August 16, 2011 2:52 am

From tallbloke on August 16, 2011 at 12:38 am:

Radiative transfer is a quantum operation according to theory, a water molecule can’t pass part of a photon. If a water molecule emits a photon to it’s neighbor below and drops to a lower vibrational state it becomes ‘cooler’ that the molecule below, which will rise to displace it.

Dang it, tallbloke, I thought you were smarter than that. Guess you’re just not living up to your billing.
You’re examining two individual molecules. The top one passes a photon to the lower one, making the lower one “warmer.” Then what? Why would the lower one rise higher than the other? They are both affected by the same gravity, they have the same mass. Arguably, by acquiring that photon’s worth of energy and thus that much virtual mass, the lower one is infinitesimally more massive. So why would it rise?
Warmer water rises above cooler water because it’s less dense, there is less mass per unit volume. Density is an aggregate quality, takes more than one identical molecule each for two equal volumes to say one is more dense than another.
And why is warmer water less dense that cooler water (assuming both are above 4°C for fresh water and equal pressure)? Collisions. The warmer water has molecules that are more energetic, knocking the fellow molecules they impact farther away, leading to greater spacing than with cooler water. When thermal energy is transferred by collisions that’s known as heat conduction. The possibilities for conductive transfer of energy far outweigh those for radiative transfer in a liquid medium. If one molecule would gain energy by a radiative transfer, chances are excellent it will soon lose that energy to a cooler molecule by conductive transfer.
You’re looking at an unlikely form of energy transfer that effects a highly temporary energy increase, while arguing a molecule will rise above another identical molecule when acted on by gravity. I think you need some more thought about that mechanism you’ve proposed that “disproves” what Willis has said.

Jessie
August 16, 2011 2:59 am

Not quoting any of your words Willis on this post
but after reading ALL of http://wattsupwiththat.com/2011/08/14/its-not-about-feedback/
and visiting the website on buoys data I would suggest speaking with the pilots & or HQ of Mission Aviation Fellowship.
I am sure they would have much experience and comment for your current thesis.
PS the reason for suggesting this is that I flew as a passenger with them over 30 years and they know their industry and environment.

Curious Bystander
August 16, 2011 3:16 am

I’m with Bob_FJ.
I must say thogh I don’t ‘get’ the argument that a blanket around a human body simply slows cooling. That would be so if the body was set at an initial temperature and no longer warmed. But the body generates heat which it loses to the environment. A blanket will help prevent that loss true, but the air it traps and the blanket’s interior surface will be heated by the heat of the body. If the external temperature is ‘cold’ then the interior temperature will be warmer. That is warming, not slowing cooling.

Kelvin Vaughan
August 16, 2011 3:26 am

Willis Eschenbach says:
August 15, 2011 at 2:15 pm
Still unclear on how a so-called “greenhouse” actually works?
The glass stops the hot air rising! Remove a pane from the roof and most of the heat escapes.

tallbloke
August 16, 2011 3:29 am

steven mosher says:
August 16, 2011 at 1:52 am
Hmm Tallbloke. Rather than trust a magazine about how reflective insulation works or does not work I think I trust the stuff I built for DOD. And I’ll use my thermos to keep my coffee from cooling faster than it would. And If I have to go near a wacking hot fire i’ll also wear a reflective suit

Hi Mosh. It’s all about relative scale of effects. Reflectivity becomes useful when the temperature differential between the two surfaces is high, like in the fire situation you mention. In the case of a cragfast climber at 37.5C whose outer layering is radiating at 10C into an ambient temperature of 3C, the benefit the space blanket gives in terms of preventing convective loss is far greater than the benefit of it reflectiing the emitted long wave from a 10C jacket with a 0.7 emissivity back onto the jacket. This is doubly true when the wind is blowing.

JJB MKI
August 16, 2011 3:39 am

@mosher,
Thank you for the thermos analogy, I’d always assumed they worked by ‘trapping heat’ ;). Would a more suitable analogy for greenhouse gasses with regard to reflected DLR and the oceans not be a flask covered with a thin layer of soggy toilet paper? Sure, it would slow down heat loss through IR reflection, but to any remotely significant degree? Not making a point, just askin’. It would be nice to see Tallbloke’s points addressed with something other than snark too..

tallbloke
August 16, 2011 3:42 am

kadaka (KD Knoebel) says:
August 16, 2011 at 2:52 am
From tallbloke on August 16, 2011 at 12:38 am:
Radiative transfer is a quantum operation according to theory, a water molecule can’t pass part of a photon. If a water molecule emits a photon to it’s neighbor below and drops to a lower vibrational state it becomes ‘cooler’ that the molecule below, which will rise to displace it.
You’re examining two individual molecules. The top one passes a photon to the lower one, making the lower one “warmer.” Then what? Why would the lower one rise higher than the other?

It’s true I oversimplified the argument, but in the end I think we agree. Groups of molecules which have warmed each other up by collisions after having radiative energy imparted to them will be less dense per unit volume, and that’s why they are more buoyant against gravity than surrounding groups of cooler molecules and will rise.
The possibilities for conductive transfer of energy far outweigh those for radiative transfer in a liquid medium.
Your conductivity argument shows how ineffective conduction is in heating the ocean bulk, since the ocean surface is cooler than the water below. 2nd law and all that… So if conduction is more likely than radiative transfer, and conduction is limited by the fact that the surface is cooler than the subsurface, I think the general thrust of what I’m saying stands, despite the oversimplification you correctly pointed out.
Thanks for the response.

Konrad
August 16, 2011 3:58 am

Willis your reply to my post was a question relating to chalk on blackboards. I had actually asked if you had any links to empirical evidence of IR radiation around the 15 micron band heating oceans. Sadly I must conclude from the nature of your reply that you do not have links to any such evidence.
After reading over 170 comments I am starting to believe that for all the money spent on AGW promotion, no one has actually done the simple experiment required.
A tank of sea water at a known temperature.
Air at a known temperature, humidity and wind speed.
IR source emitting only between 10 and 20 microns with a spectral peek around 15 microns.
A few accurate thermometers.
Who cares about chalk on blackboards? Has no one done this very simple empirical test?
(And no, Mr. Mosher the RC link is no where close to what is required. Scattered data points from a ship at sea. Measured emission not just backscattering of clouds and only looked at ocean skin temperature. And as to isolated testing of the 15 micron frequency totally useless.)

Spector
August 16, 2011 4:04 am

Note that radiation and absorption are reciprocal. If it cannot absorb, it cannot cool by emission. If you run your recording backwards, most of the basic laws of physics work exactly the same–emission becomes equivalent to absorption. As the concentration of water vapor in the atmosphere near the surface is quite high when compared to that of CO2, I expect that most of the local greenhouse effect near the surface is due to photons emitted from the water vapor in the air. This is often referred to as ‘back-radiation.’ I believe the argument being made is that this radiation is being rejected (perhaps reflected) while the normal LWIR radiation emitted from surface is being allowed to pass. I don’t think nature is that selective.
Yes, something is warming the oceans–sunlight. As far as I could tell; that was not the issue; it was the misconception that LWIR from local greenhouse gas molecules could not return heat to the ocean even though these same molecules had been heated by similar LWIR photons emitted from the surface of the ocean.

Richard S Courtney
August 16, 2011 4:18 am

Willis:
I do not dispute your argument, but I write to point out what I think to be a significant omission from your article: i.e. consideration of how the back radiation inhibits loss of heat from depths of the ocean surface layer. However, some of your responses in the thread suggest to me that you are aware of the point.
I write to explicitly state the point.
The backradiation is IR so only heats the upper ~1 mm thickness of the ocean surface layer. Little of this heat can be moved to lower depths because it adds so much heat to so small a volume that most of the absorbed IR energy is removed by increased evapouration from the surface.
However, solar energy in visible wavelengths penetrates to many meters below the ocean surface and is absorbed at those depths. This heat from visible wavelengths can move up and down. That which reaches the surface will increase evapouration from the surface and, thus, not be available to add heat to lower depths of the ocean.
The heating of the surface layer by back radiation inhibits the heat from visible wavelengths reaching the surface and being removed by evapouration.
In effect, the heating of the ~1 mm surface layer by back radiation acts as an insulator which inhibits heat from below that thin layer reaching the surface. And this ‘insulant’ effect results in more heat absorbed from visible wavelengths being available to add to ocean heat content.
I cannot quantify this ‘insulant’ effect, but I suspect it is greater than the direct ocean heating from the back radiation.
Richard

Alexander Vissers
August 16, 2011 4:27 am

I cannot really see why this has anything to do with climate change. The top mm of the ocean considering the opacity for short wave radiation is relavant for both absorbtion and emission. The temperature of the atmosphere – of which both the weather and climate are attributes – is determined by the ocean absorbtion, emission and evaporation, so whatever happens to the top layer is what determines weather and climate. .

tallbloke
August 16, 2011 4:33 am

Willis Eschenbach says:
August 15, 2011 at 11:46 pm
So the lower molecules, although warmed by the upper molecules, will be cooler than the upper molecules. As a result, despite the fact that they are warming, they won’t “head upwards” as you claim.

Hi Willis,
What I’m saying is generally true. It may get a bit more complex in the ‘skin layer’, see my reply to KD Knoebel above.
These basic points stand.
1) The ocean surface is cooler than the water below. So if your heat conducted from the surface where the water can be excited by back radiation leaves the surface warmer, it doesn’t stay that way for long. Moreover, within a very few iterations of your mechanism, representing only a short distance, the amount of energy available for more downward propagation will be negligible anyway.
2) The second law of thermodynamics says that heat won’t flow from cooler to hotter: So conduction is not going to work to heat the ocean bulk. It will work to cool it though, as heat flows from the warmer subsurface to the cooler surface.
3) The overall radiative flux cools the ocean at a rate of ~66W/m^2
The core point you’re claiming seems to be this:
The additional co2 in the atmosphere causes an increase in the downward component of the radiative flux, therefore the ocean will cool slighty slower than before, thus getting warmer because it can’t get rid of solar energy as fast as before.
Please let me know if you feel I’ve mischaracterised this from your point of view.
I haven’t done the calcs yet, but I’d like to know by how much you believe the downward component of the radiative flux will have theoretically increased from when co2 was at 270ppm.
If the ocean surface is 0.5C warmer than when co2 was at 270ppm, then we also need to know how much the upward component of the flux has increased due to the ocean surface radiating at a higher temperature. We can apply the Stefan- Boltzmann Equation to obtain that.
We need to know this because the key figure is the net figure obtained by subtracting the downward component from the upward component of LW radiation.
Once we have this change in net flux, we can compare that to the estimated forcing caused by the reduction in tropical cloud cover allowing greater insolation to the surface of the ocean 1980-1998 to see which is more likely to have caused most of the warming in the late C20th.
Agreed?

richard verney
August 16, 2011 4:36 am

Willis
Regarding your comment@Willis Eschenbach says:August 15, 2011 at 10:27 pm
commenting upon my comment richard verney says:August 15, 2011 at 6:18 pm
/////////////////////
I have more than 40 years sailing experience. I am well acquainted with the sea first hand.
I have reviewed many thousands of ships logs covering deep ocean voyages, and leaving aside passages through the doldrums,, I can assure you that the times when a ship on a deep ocean voyage experiences weather conditions of BF4 or less is probably less than 10%. Indeed, it is not uncommon to see just a few days of BF4 or less on a deep ocean voyage. The number of days when a ship experiences BF2 or less is probably 2 or 3% (or even less). Of course, this is just an average and I accept that every voyage will be different, and the same voyage will experience different weather conditions from month to month (depending when the voyage takes place) and even if the same voyage is performed in the same month, there will be differences from year to year according to the then prevailing weather.
You have to bear in mind that we are dealing with microns. Even at BF 2, these few microns are wind swept.
When we see spray (white horses), we are observing visible effects. This is much more than the top few microns of the ocean. We are seeing millimetres if not centimetres of turbulence.
Unfortunately, the human eye does not have the resolution to see microns of turbulence. If it did have that resolution, it would see that the top few microns is wind swept even at BF2.
I do not dispute that what goes up, at some stage goes down. Most of this wind swept spray evaporates and eventually leads to the cloud formation of your earlier post and in turn some of the evaporated water is returned to the ocean or over land as rain. I accept that some part of the wind swept spray may not be fully evaporated and may re-marry the ocean. However, by the time that it has remarried, it has probably already re-radiated the photon received from the DWR such that it is no longer in an excited state.
Even if you get through that ‘barrier’ you still have the problem that the top mm of the ocean is cooler than the layer immediately below it and whilst the turning of the ocean may be effective to mix heat contained in the top 1/2m or metre of the ocean but it is difficult to see how it can effectively turn over the top few microns. To what extent do ‘eddy’ currents overlap the top few microns of the medium and drag those few microns downwards.
I would like to see your scientific explanation of the processes involved on a micron level.

August 16, 2011 4:44 am

Any warming by DLWR must be extremely miniscule. There are two beaches near my home both with different types of sand. The sand at Nokomis Beach is cool even at midday in summer. I assume the properties of the sand reflect almost all SWR (the sand looks white) so it does not heat up. The sand a few miles away at Manasota Beach is extremely hot at midday in summer. I assume that the properties of the sand absorb most of the SWR (the sand is grey). This sand gets so hot you can actually burn your feet. However, as soon as the Sun gets lower in the sky, the sand cools off rapidly so that you can walk comfortably on it even before sunset. My point is that if DLWR had any effect the sand at Manasota Beach would not cool so fast. Even if there was a small change in the rate of heat loss due to DLWR it has no discernible effect and is overwhelmed by the changes in other radiations.

August 16, 2011 4:47 am

John Eggert says:
August 15, 2011 at 2:26 pm
“1: A continuously heated, flat plate at 15 C is on one side, a parallel, continuously cooled plate at -273C is 20 meters away.”
If you told us the capacitance in farads we could work out the size of the plates, which in turn would give us an idea of the magnitude of the radiation window.

DocMartyn
August 16, 2011 4:57 am

Here is a paper with the absorbtion vs temperature spectrum of water
http://www.medphys.ucl.ac.uk/research/borl/homepages/veronica/thesis/chapter5.pdf

Alexander Duranko
August 16, 2011 5:11 am

‘Back radiation’ [‘DLR’] is Prevost Exchange Energy’: it can do no work because at equilibrium, it’s exactly offset by radiation from the opposite direction. That it appears to be higher under clouds is simply the result of higher emissivity. This, the oldest of the Radiation Laws, appears not to have been considered by IPCC researchers, nor have they apparently estimated gas emissivity from Hottel Charts, something I used to do when planning industrial heat treatment processes.
This and other major errors [‘cloud albedo effect’ cooling is really heating for thicker clouds, palaeo-data show warming of the ocean deeps started 2100 years before CO2 rose at the end of the last ice age, no check on thermochemical data for CO2-O2 mixtures which would have shown much lower CO2 ‘climate sensitivity’ ] should also have been picked up.
Miskolczi was forced to leave NASA in 2004 when he showed ‘back radiation’ is the artefact of a 1922 mathematical mistake. After commissioning work to find why ‘cloud albedo effect’ cooling wasn’t provable experimentally, NASA published a ‘surface reflection’ justification to claim it’s a small droplet effect. There’s no such physics.
A pattern is emerging. The latest claims, extra heat accumulation in the ocean deeps, double imaginary ‘cloud albedo effect’ cooling, appear to me to be a last ditch attempt to protect the high feedback CO2-AGW hypothesis for which there’s absolutely no experimental evidence.
Reportedly the UK Met. Office trying 50% solar, 50% CO2 in its weather forecasting for the UK; a straw in the wind?

John W.
August 16, 2011 5:19 am

Two words: heat conduction.
As Willis suggests, leave all the silly stuff to the other side.

Dave in Delaware
August 16, 2011 5:33 am

Lets be mindful of our definitions and look carefully at what is happening at different wavelengths of radiant energy. The phrase “Downwelling Longwave Radiation (DLR)” is often used in the context and implied as ‘backradiation from CO2 as a GHG’. The two are NOT identical.
Hence the opening statement – ‘downwelling longwave radiation (DLR also called infra-red or IR, or ‘greenhouse radiation’) – lumps together things which should be considered separately.
First – Does inbound sunlight warm the ocean? Well of course. And that Inbound solar radiant energy includes a mix of UV (very short), visible, and longer wave IR. Yes, inbound sunshine contains IR, but there is virtually NO overlap of those IR wavelengths with the ‘greenhouse radiation’ wavelengths. You can’t just lump them together, which is why one must be careful of ‘heat lamp’ experiments that could include wavelengths other than GHG ‘backradiation from CO2’.
Next – In their paper, Kiehl and Trenberth 1997, the authors include the famous figure with 390 W m-2 upward Surface Radiation, and 324 Wm-2 Back Radiation. That gives (390 UP minus 324 Down), for a NET of 390 – 324 = 66 W m-2 UPward. Very difficult to heat a surface, either land or ocean, with a net upward loss of heat toward space. (we lose a little on every sale, but we make up for it on volume ::chuckle::)
CLOUDS – How did K&T 1997 get to 324 W m-2 backradiation”? From their abstract – “We find that for the clear sky case the contribution due to water vapor to the total longwave radiative forcing is 75 W·m-2, while for carbon dioxide it is 32 W·m-2. Clouds alter these values, and the effects of clouds on both the longwave and shortwave budget are addressed.” The backradiation from CO2 is only 32 W·m-2, or roughly 10% of the 324 total backradiation cited by K&T. The lion’s share of the 324 backradiation is from liquid H2O in the clouds. Once again we see that “Downwelling Longwave Radiation (DLR)” is NOT equivalent to ‘backradiation from CO2 as a GHG’. We must be careful of the wavelengths.
Finally, what about some real measurement data? Measured results cited in Evans and Puckring 2005, show CO2 backradiation of 31 to 35 W m-2 for clear sky winter time results (Ontario, Canada), which is a pretty good match for the K&T value. However at that same location, in summer with higher air temperatures and higher humidity, the H2O values went up, but the CO2 backradiation value DROPS to 10.5 W m-2. If the overlap of H2O and CO2 absorption/emission in air caused the CO2 value to drop with summertime humidity in Canada, I can only surmise that the CO2 backradiation over the ocean must be 10 or lower due to the high humidity. Similar backradiation tests in the very low humidity Antarctic showed that CO2 backradiation results there were always 2 to 2 1/2 times LOWER than H2O.
In conclusion – If the H2O as vapor (humidity) in the air is enough to absorb significantly in the CO2 wavelengths, the liquid surface must be an even better absorber, which makes it easy to believe that absorption of the CO2 wavelengths happens in a ‘thin layer’ at the surface.
If the backradiation of CO2 is on the order of 10 W m-2 to the Ocean, it would be difficult to conclude that CO2 as a GHG is doing much to heat the Oceans. And even more difficult to believe that the man-made part of that 10 W m-2 is doing much of anything significant to change the ocean energy balance.
respectfully,
Dave

kadaka (KD Knoebel)
August 16, 2011 5:39 am

From tallbloke on August 16, 2011 at 4:33 am:

The core point you’re claiming seems to be this:
The additional co2 in the atmosphere causes an increase in the downward component of the radiative flux, therefore the ocean will cool slighty slower than before, thus getting warmer because it can’t get rid of solar energy as fast as before.
Please let me know if you feel I’ve mischaracterised this from your point of view.

Before Willis gets worked up into an even stronger “quote my words” rage, I’ll point out a glaring one right now. I’ve checked the original post and every reply Willis has made at this post. CO2 is mentioned nowhere by Willis. Why are you tagging him with that? The DLR comes from the greenhouse gases, of which water vapor is overwhelmingly the predominant one, which Willis is well aware of. Why muddle up an otherwise reasonable discussion by attributing a CO2 statement to Willis?
The nighttime overturning is driven by the temperature difference between the surface and lower levels, once the surface undergoes its normal nocturnal cooling resulting in cooler water over warmer water. With more DLR the surface doesn’t cool as much, the difference is less, thus less overturning, thus less warmth from deeper down is circulated to the surface. For that much of what was stated, the source of the DLR doesn’t need mentioning. Adding in the conjecture that increased CO2 causes the increased DLR, especially when Willis has said no such thing here, isn’t called for and isn’t helpful.

August 16, 2011 5:46 am

Mosh writes “You note that nobody (except willis) wants to look at the empirical evidence. I posted it up there boys..”
The reference to the Minnett experiment isn’t empirical evidence of anything other than the SST warms relative to the depth of 5cm when the clouds come over. When the DSR induced temperature profile disappears (when the clouds come over) but convection keeps the war water rising for a time what do you think will happen to the SST vs the 5cm depth?
That experiment is far from supporting the argument that temperature gradient determines heat loss and therefore supports ocean warming. AFAIK its not even an actual paper let alone a peer reviewed paper.
Empirical evidence? FAIL.

Bystander
August 16, 2011 5:57 am

Willis says “Please stop the condescending snarkiness, it just makes you look ugly.”
Ah – so condescending snarkiness is only OK when attacking real scientists then….

richard verney
August 16, 2011 5:58 am

Willis
regarding: @Willis Eschenbach says:August 15, 2011 at 10:33 pm commenting on my comment
richard verney says:August 15, 2011 at 7:12 pm
////////////////////////////////////////////
I understand why to a layman one may give a superficial explanation which concentrates on the overall effect rather than the minute detail of the processes involved.
However, when discussing scientific principles with an audience of scientists, precision is paramount. This is a scientific blog and at least those posting what purport to be scientific articles or discussion of scientific effects, should strive themselves to be accurate.
You dismiss this as a question of semantics, it is not. It is a matter of fundamental (or first) principle going to one of the root issues at the foundation of the AGW theory/conjecture.
One of the reasons why it is so fundamental is that CO2 is not the dominant (so called) green house gas. It is well accepted that water vapour is the dominant greenhouse gas. This is material since over the oceans there is on average much more water vapour in the atmosphere when compared with the atmoshere over land. Any increase in CO2 concentrations in the atmosphere above the oceans is completely dwarfed by the naturally occuring high levels of water vapour immediately above the oceans. This is a significant point which is overlooked and is a significant point which is masked when dealing with average conditions. The warmists tend to concentrate on notional average conditions, and when this is done, one fails to see the proper picture since one is not looking at what ia actually occuring in non average conditions. Materially, at times it is the non average condition which leads the way and is the dominamnt factor which needs to be closely and carefully examined.
It is clear from the above that if DWLWIR does not heat the oceans, the reduction in the rate of cooling is predominatly being controlled by the high levels of water vapour above the oceans not by any increase in concentration of CO2 AND the high levels of water vapour are natural (ie., naturally occuring) being the result of solar energy being received by the oceans, ie the effects of sunlight warming the oceans.
It follows from this that the overwhelming probability as to why the oceans have been warming these lasts 30 or 40 years is due to an increase in solar energy received by the oceans very probably due to a reduction in cloudiness (in passing, it appears that the oceans have not been warming these past 8 years or so).
Whilst I am very sceptical of the entire GHG theory/conjecture, I can envisage that changes in CO2 concentrations may have some modest effect over land, however, over the oceans it has all but no effect, simply because the oceans (unlike the land), as a by product of the solar energy they receive, create their own ‘greenhouse gas’ atmosphere immediately above them (ie. WATER VAPOUR) and this water vapour completely dwarfs the effect that might overwise be observed as the result of a 100ppm rise in CO2 levels.
I will revert seperately on the point you raise in the final paragraph. .

tallbloke
August 16, 2011 6:03 am

kadaka (KD Knoebel) says:
August 16, 2011 at 5:39 am
Before Willis gets worked up into an even stronger “quote my words” rage, I’ll point out a glaring one right now. I’ve checked the original post and every reply Willis has made at this post. CO2 is mentioned nowhere by Willis. Why are you tagging him with that? The DLR comes from the greenhouse gases, of which water vapor is overwhelmingly the predominant one, which Willis is well aware of. Why muddle up an otherwise reasonable discussion by attributing a CO2 statement to Willis?

It’s true Willis doesn’t say anything about co2, but the implication is pretty obvious IMO. There is no trend in precipitation so far as anyone knows, so a warming ocean seems to be down to the increase in co2 according to the arguments Willis presented.
I think he’s wrong and that the decrease in tropical cloud cover from 1980-1998 has a lot more to do with the increase in ocean heat content (and thus SST) than increased ‘DLR’ from the atmosphere, whether by co2 or by co2 plus a water vapour feedback.
Anyhow, I’ve invited him to set me straight if he feels I’m mischaracterising the core of the argument, and no doubt he will, in his inimitable style.

Myrrh
August 16, 2011 6:14 am

Dave in Delaware says:
August 16, 2011 at 5:33 am
Lets be mindful of our definitions and look carefully at what is happening at different wavelengths of radiant energy. The phrase “Downwelling Longwave Radiation (DLR)” is often used in the context and implied as ‘backradiation from CO2 as a GHG’. The two are NOT identical.
Hence the opening statement – ‘downwelling longwave radiation (DLR also called infra-red or IR, or ‘greenhouse radiation’) – lumps together things which should be considered separately.

Ah thank you – another apparent sleight of hand meme which has confused many here. Downwelling should only refer to thermal infrared direct from the Sun, and uwelling direct from the Earth, seems a move to sidestep the arguments about ‘backradiation’, of which there are many and already proved nonsense.
AGWScience fiction meme now widespread in the education system has always claimed there is no or no signifcant downwell infrared direct from the Sun, as in KT97.
Which NASA is now promoting contrary to its previous real science fact teaching that the heat we feel from the Sun is thermal infrared. Therefore it must be reaching the surface…
As some mention has already been made by others here, Visible light is transmitted through water – it does not heat it. The KT97 is therefore actually saying that the Sun isn’t heating our Earth at all! Visible, Light, can’t and Thermal IR (Heat) they claim doesn’t reach us..
So what’s the energy budget all about?

richard verney
August 16, 2011 6:25 am

in Delaware says:August 16, 2011 at 5:33 am
/////////////////////////////////////////////////////////////////
ABSOLUTELY
I had not seen your comment (which was no doubt in the pipe;ine) when I posted my above comment (addrssed to Willis).
The warmists need to carefully examine each area and not some conglomerate average.
It is obvious that over the oceans, the role of CO2 is greatly diminished because of the high concentrations of water vapour which dwarf its effect.
Further, the more solar energy received, the more water vapour produced. Now if this increase in water vapour caused the ocean temperature to rise, one would get even more of an increase in water vapour possibly leading to a run away effect. There therefore needs to be some negative feed back in all of this or otherwise the oceans would have evaporated long ago.

kadaka (KD Knoebel)
August 16, 2011 6:34 am

From tallbloke on August 16, 2011 at 6:03 am:

It’s true Willis doesn’t say anything about co2, but the implication is pretty obvious IMO. There is no trend in precipitation so far as anyone knows, so a warming ocean seems to be down to the increase in co2 according to the arguments Willis presented.

Same mistake, different part. On this one, I’m a bit guilty as well. Willis has stuck consistently to how he worded it in Argument 3:

DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer, and it also slows the speed of the overturning once it is established. This reduces the heat flow from the body of the upper ocean, and leaves the entire mass warmer than it would have been had the DLR not slowed the overturning.

The DLR effect is discussed as with or without, on or off. Increases are not mentioned. Actually I see nothing mentioning global warming, anthropogenic or otherwise. The closest he’s come to talking about a “warming ocean” is the ocean is warmer with heating from DLR than if there was no heating from DLR.
So no increasing, no “warming ocean,” no call for invoking CO2 increases.

Matt G
August 16, 2011 6:34 am

Tom in Florida says:
August 16, 2011 at 4:44 am
Exactly, it is extremely small and shows the same result with a volume of water. It is easy to show that DLR can’t warm a volume of water during a day and relies only on solar energy. DLR occurs all the time during day and night so to distinguish this between solar energy, one only has to place identical volumes of water in the sun and in the shade. The shade volume must be outside exposed to the atmosphere and not in the sun so only the DLR is still reaching it. The result which anyone can easily demonstrate (but some at least choose to ignore this observation) shows the water in the shade during one day doesn’t warm, yet the volume in the sun warms greatly. Therefore this experiment provides the scientific evidence that solar energy warms a volume of water and not DLR. The other experiment that also backs this up are based on solar ponds.
http://www.solarponds.com/
All the DLR is mostly (if not all – look later into why) the emissions from solar energy as they escape the Earth’s atmosphere. Thus the 170w/m2 inward and 390w/m2 outward are not really different entities, but from the same source. Both are values measured from the atmosphere only and therefore can’t be compared with a much higher specific heat capacity of water. The water is warmed differently depending on the source of radiation.
Solar energy at 170w/m2 warms the ocean much greater than DLR despite the observed atmospheric value over 2 times greater because it reaches downwards up to 100m. (the experiments above back this with scientific evidence) The warming caused by this is orders greater than DLR just reaching the skin surface. The reason why the oceans don’t freeze is because despite the atmospsheric observed outflux, the energy in the ocean is orders time greater warmed by the sun. If the solar energy 170w/m2 only reached ths skin surface like DLR, then the oceans would freeze.

August 16, 2011 6:36 am

One difference between Dr. Pratt’s post on Climate Etc. and Wilis’ is that at the surface Dr. Pratt emphasizes the conduction plays a larger role than radiation at the skin layer. The change in conduction is still due to reduction in the net radiation flux, but the mechanism limiting the rate of cooling at the surface is primarily the change in conduction due to the increase in air temperature.
In the tropical ocean some one listed the percentage of heat loss was 5% for conduction, 25% for radiation and 70% for evaporation. In a colder ocean the percentages would be closer to 10% conduction, 25% radiation and 65% evaporation. You could find more accurate numbers, but the radiative heat loss would be fairly constant with conduction and evaporation responding more to the temperature difference than radiation.
So the take away is that an increase in DWLR decreases the radiation flux, increasing the air temperature, reducing conductive heat loss. It makes perfectly good sense, since the CO2 and water vapor respond to the DWLR just as they would to the OLR.
It may seem to be picking nits, but Dr. Pratt’s position is more realistic while not violating any laws of physics. Surface warming is a response to a warming atmosphere caused by a reduction in net radiative cooling flux.

tallbloke
August 16, 2011 6:58 am

TimTheToolMan says:
August 16, 2011 at 5:46 am
Mosh writes “You note that nobody (except willis) wants to look at the empirical evidence. I posted it up there boys..”
The reference to the Minnett experiment isn’t empirical evidence of anything other than the SST warms relative to the depth of 5cm when the clouds come over. When the DSR induced temperature profile disappears (when the clouds come over) but convection keeps the war water rising for a time what do you think will happen to the SST vs the 5cm depth?
That experiment is far from supporting the argument that temperature gradient determines heat loss and therefore supports ocean warming. AFAIK its not even an actual paper let alone a peer reviewed paper.
Empirical evidence? FAIL.

Quite so. It was a symposium presentation which never made it into the peer reviewed litereature.

August 16, 2011 7:11 am

Willis,
I did not read all the comments so this may repeat some of what has been said. The DLR can heat water (or the ground) if the water is cooler than the air, and this sometimes occurs (especially at night or in higher latitudes), however on the average, the air temperature drops with increasing altitude in the atmosphere (the lapse rate), and thus, on the average, the DLR does not heat the water or ground. While energy flows down in the DLR, energy is also flowing up (ULR), and heat only flows from warmer to colder. Thus it is the difference in up and down energy flows that cause heating. The fact that DLR can, under special conditions do some heating is not important on the average and is misleading of the cause of the heating due to greenhouse gases. The increase in ground and water temperature and the DLR are a result, not cause, of the atmospheric greenhouse gas warming effect.
Your argument about where does the extra energy go is bases on misunderstanding the insulation effect. Any type of insulation effect over a heat source raises the temperature of the surface. Consider a blanket over a person (internally heated). It is not the blanket that warms, and if the body were for a dead person, the skin would be cool, even under the blanket. Also consider an insulating layer over an electrically heated resistor. The insulator is not doing the heating, but raises the temperature of the resistor.
In the case of Earth, the source of the net heating is absorbed shorter wave Solar energy (and a small amount of radiation decay heating). However, greenhouse gases in the atmosphere are different from passive insulation layers. The air is free to rise from convection so no trapping occurs even though the gases absorb ULR. It is the added feature of the lapse rate that causes the temperature increase, since the temperature at a higher altitude has been raised from the movement up of the location of radiation to space, but the lapse rate is independent of temperature (it only depends on Cp and gravity). The effect of the greenhouse gases is to raise the average altitude of outgoing radiation to space, and it is the lapse rate (due to gravity) that raises the temperature going down from this average location, and thus is the source of the extra heating.
The result is that both the higher surface temperature and significant DLR are the result of, not cause of the surface being warmer.

Richard M
August 16, 2011 7:14 am

The main disagreement here appears to be between net energy flow and gross energy flow. Willis points out the gross radiation flows and others point out that is really meaningless.
OK, if gross flows are important than why hasn’t anyone computed the gross energy flows of kinetic energy? There are trillions and trillions of molecular interactions between the surface and the atmosphere where energy is transferred one way or the other. We always see this discussed by looking ONLY at the net energy flow which is not all that high. So, why isn’t what’s good for the goose also good for the gander? Why don’t we discuss the gross energy flows of conduction? Could it be because the gross flow really isn’t important? So, why would anyone think the gross flows are important for the case of radiation? Maybe Willis or someone can explain to me the difference.

tallbloke
August 16, 2011 7:16 am

kadaka (KD Knoebel) says:
August 16, 2011 at 6:34 am
The closest he’s come to talking about a “warming ocean” is the ocean is warmer with heating from DLR than if there was no heating from DLR. So no increasing, no “warming ocean,” no call for invoking CO2 increases.

Point taken. He’s still wrong though, because the long wave radiative flux cools the ocean rather than warming it, and the DLR component is solar derived energy emitted from the ocean as ULR in the first place. So if there were no DLR, there would be no ULR either, and the ocean would be up in the atmosphere having been boiled by solar shortwave it couldn’t get rid of other than by evaporation.
Warming the ocean is the Sun’s job. Cooling it and losing the heat to space is the atmosphere’s job.

August 16, 2011 7:23 am

Some random thoughts and comments after reading most of the above.
My washing on the line drys much much quicker when there is a breeze; however gentle. No white horses necessary.
Average atmosphere temperature is 14-15DegC. Average ocean temperature is 3DegC.
It’s been well over 10,000 years since the last ice age ended. If the atmosphere could warm the oceans, the oceans should be closer to the 14-15DegC level. 10,000 years not enough?
Regards semantics, if we are saying GHGs “reduce rate of cooling” rather than actually “warming”, then the ocean had to be warmer in the first place for it’s rate of cooling to be reduced. What warmed it in the first place then?
Oceans warm in 3 dimensions but cool in 2 dimensions. So I have my doubts about the figures quoted, i.e. 170Wm2 solar input in 3 dimensions, but 390Wm2 radiant heat loss in 2 dimensions. this doesn’t make sense to me.
According to the AGW theory, the sun warms the surface first, the surface then warms the atmosphere, which in turn further warms the surface.
But if the solar input is only 170Wm2, then that’s the maximum the ocean can radiate back up. The atmosphere cannot ‘back-radiate’ any more than the 170Wm2 it receives from the surface for a total of 340Wm2. Where did the 390Wm2 come from? And I haven’t even allowed for the atmospheric window nor the sunlight that reaches down to about 100 metres, the energy from which doesnt make it back up to the top straight away.
The well mixed portion of the oceans are the top few hundred metres. But the oceans are 4-5 kilometres deep. How does warm water mix “down” thousands of metres? physically impossible I would have thought.
The strongest GHG effect, hence the most DWLWIR happens at the tropics. How come the deep even at the tropics is around 2-3DegC only? Why is the deep much the same temperature at the tropics as it is at the poles?
The “bulk” of the ocean is the deep. The temperature of the tropical deep tells me that GHGs cannot warm the deep.
[Reply] Check your figures, ocean avg surface temp is ~17C, warmer than the atmosphere at sea level. TB-mod

Alexander Duranko
August 16, 2011 7:32 am

Some interesting analysis of ocean heating recently.. Regarding CO2 and H2O specific ‘DLR’, you must realise that because Kirchhoff’s Law requires that emissivity and absorptivity are the same at equilibrium, any change in the ratio of the two partial emissivities/absorptivities simply reflects what is happening on the ground [assuming the radiometer is at ground level].
So, if there’s dew, the H20 emission from the ground will increase and this will lead to more ‘DLR’ from the sky in that wavelength interval. Over the ocean, it’s temperature that counts.
So, as ‘DLR’ isn’t real energy, just a sort of standing wave, and the nearly half the solar energy that is IR is absorbed in the atmosphere, what really heats the oceans and how did it change from the ’80s to the early 00s?
The answer is obvious: Asian industrialisation poured increasing aerosols into the atmosphere and this decreased cloud albedo, particularly in the short wavelengths which penetrate deeply into the sea, possibly up to 150 feet. This was probably the main AGW and because it’s self limiting switched off when aerosol concentration became high [the ‘Asian Brown Cloud’]
Of course, I could be wrong, but at least I hope to have dispelled some of the false assumptions that have driven ‘climate science’ in the wrong direction. Those working in it really do need to readup about Hottel, e.g: 203.158.253.140/media/e-Book/Engineer/…/DK2834_13.pdf
PS cloud IR emission is from the water but it has gettered CO2 so is also a strong CO2 IR emitter because the CO2 is concentrated.

August 16, 2011 7:43 am

From memory the equation for 2 plates (infinite or very large ) radiating at each other.
q/A = (SB(T1^4-T2^4))/ ((1-e1) + (1-e2) -1)
Let T1 be ocean surface and T2 be atmosphere. What emissivity should be used for e2 of a CO2 & H2O atmosphere at 1 atm?

August 16, 2011 7:47 am

@ TB-mod
I beg to differ. Ocean average T is nowhere near that.
oceans

August 16, 2011 7:48 am

hmmm image thingy didn’t work.
Here is the URL
http://www.john-daly.com/deep-sea.gif

tallbloke
August 16, 2011 8:06 am

Baa Humbug says:
August 16, 2011 at 7:47 am
@ TB-mod
I beg to differ. Ocean average T is nowhere near that.

Baa, you’re right. I was thinking average SST, since that’s where the radiative interface is. I’ll correct my note on your comment.

Richard111
August 16, 2011 8:19 am

Baa Humbug says:
August 16, 2011 at 7:48 am
“hmmm image thingy didn’t work.
Here is the URL
http://www.john-daly.com/deep-sea.gif

Many thanks for that link. Never knew sea bottom could get that cold and how shallow is the warm layer is. Learn something new every day! 🙂

DR
August 16, 2011 8:19 am

Thanks Leonard Weinstein.

kadaka (KD Knoebel)
August 16, 2011 8:32 am

From tallbloke on August 16, 2011 at 7:16 am:

Point taken. He’s still wrong though, because the long wave radiative flux cools the ocean rather than warming it, and the DLR component is solar derived energy emitted from the ocean as ULR in the first place. (…)

The net effect may be cooling, but he’s still looking at individual energy flows. When it’s all added up without DLR, there’s a deficit that indicates severe cooling. So either the DLR effect is there, or the upwelling LR figure is very wrong. Nothing else but DLR is in the range where it can cover that deficit.

(…) So if there were no DLR, there would be no ULR either, and the ocean would be up in the atmosphere having been boiled by solar shortwave it couldn’t get rid of other than by evaporation.

That part fails the logic test right at the start. There’s a black surface in vacuum. Sunlight falls on it, the surface warms, the surface emits longwave radiation. No atmosphere, no greenhouse effect, none of that emitted LR goes back to the surface. Thus there can be ULR without DLR. Likewise if there is atmosphere but it lacks GHG’s thus is transparent to LR.

Costard
August 16, 2011 8:56 am

I think I get Willis’ point. Net flow (~390-~320) only matters if you accept that the two are equal currencies and that downwelling LR warms the ocean as much as upwelling LR cools it. And the mechanism for this is a warming of the skin layer, a more even temperature gradient between this layer and the one below it, and less convective cooling.
But this would also suggest a steeper temperature gradient between the (warmer) ocean surface and air, and more heat loss through evaporation and conduction. In the absence of hard numbers, the question becomes like so many, to what extent is this forcing mitigated by the natural processes of the environment?

August 16, 2011 9:30 am

Kadaka,
“Thus there can be ULR without DLR. Likewise if there is atmosphere but it lacks GHG’s thus is transparent to LR.” That is an interesting thought. With a pure nitrogen atmosphere there would be no absorption of ULR of significance so a nitrogen filled double pane glass window with a 100% reflective coating would be a perfect insulator of conductive, convective and radiant heat transfer. We know that is not true, it is a much better insulator, but not perfect. A vacuum with a reflective surface is better. Nitrogen molecules collide so there is some conductive heat transfer.
Generally speaking, the conductive heat transfer is small and can be neglected. There is a point where it is significant enough that it should be considered. Earth with a pure nitrogen atmosphere would be warmer than an Earth with no atmosphere. Earth with a 99.97% nitrogen and .03% CO2 would be warmer than an Earth with no CO2. So how significant would conduction of surface thermal energy to the atmosphere be with respect to radiative retention of the 0.03% CO2?

August 16, 2011 9:33 am

Konrad.
I suppose If I write up the experimental notes like the “famous” “Woods” experiment ( with a green house) that explains I have done the exact experiment you describe that you and everyone else here will accept it with the same eagerness that you accept Woods paragraph.
The point is that no amount of theory, physics, experiment, will convince some people because they do not want to be convinced. We have a word for that. It starts with D

richard verney
August 16, 2011 9:37 am

Willis
Further to my earlier post, I revert on the comment you make in the final paragraph of your post Willis Eschenbach says: August 15, 2011 at 10:33 pm which comments upon my post richard verney says:August 15, 2011 at 7:12 pm
Your final paragraphs reads: “What I don’t understand is what slightest difference this makes. If the ocean is losing 400 w/m2, and it is gaining 170 w/m2, I don’t care in the slightest what you call that. What I want to know is, if DLR isn’t heating the ocean, what makes up the missing energy? Gamma rays? So enough with the semantics, and answer the question—what’s keeping the oceans liquid, call it what you want, if it’s not DLR?”
/////////////////////////////////////////////////////////////////
I have had this debate with you before, and I consider the attitude that you adopt to be somewhat out of character, as a self proclaimed heretic. It is you who make a bald statement that without DLR, the oceans would freeze. I say, I don’t accept this bald statement, and ask you to prove it. Instead of proving it, you seek to reverse the burden of proof and suggest that I must prove that the oceans wouldn’t freeze but for DLR. You have things topsy turvy in the scientific world (albeit this may not be that unusual in climate science where it would appear that climate scientistist do not like dealing with complex issues and seek to parry by passing the akward parcel back to the other side).
I have previously asked you to detail the energy budget for the ocean at 62 N 8 deg 45 E, and at 62N 19 deg 04 42 E. Both of these are at the same latitude (62 deg N) and I envisage that as a consequence, the atmosphere comprises a similar mix of GHGs and they have broadly similar solar and DWLWIR budgets. The ocean at one of these locations freezes each winter, the ocean at the other location does not. I have asked you to explain (with your energy budgets) why that is the case. You have consistently failed/declined to answer. Perhaps this time, ou will not evade/ignore the question and will instead provide the answer.
As I see it, the answer is simple and it lies in the tropics. The fact is that the tropical ocean is a huge heat reservoir. Over the course of a year, the tropical ocean absorbs enormous quanities of solar radiation/energy and in the process it heats up. Some of this heat is distributed/pumped around the globe but not in the same measure in all places. The fact is that the ocean at 62 N 8 deg 45 E receives a greater quantity of the heat absorbed by the tropical ocean and this is sufficient to prevent the ocean at 62 N 8 deg 45 E freezing.
I therefore say to you, that YOU need to do an energy budget for the ocean at the tropics on a daily basis and see whether the ocean at that spot freezes. If it does not (as I am confident is the case), you need to perform successive energy budget calculations (on a daily basis) working your way outwards further north and south until you ascertain the latitude at which the ocean would begin to freeze (based upon the solar/DLR budgets). Once that has been ascertained, you need to introduce the effect of the heat transport/conveyor belt system of the ocean into the equastion. The location where you calculate would first freeze would not in practice freeze due to warm currents being transported by the conveyor belt. Accordingly, you need to carry on with successive energy budget calculations gradually working your way further and further northwards and southwards. You will gradually be working your way to higher latitudes. By the time you get to the Artic you will find that it will freeze over and melt etc with the seasons.
Turning now to your figures, you use Alice in Wonderland figures because you are using those employed in an Alice and Wonderland cartoon (suggested by Trenberth and others). In your comment at 15th August 11:39pm (commenting on a comment by Dave Springer), you state “You (and the cited paper) are talking about net radiation flows. Trenberth and I are talking about individual radiation flows.”
We all know that you can add any figure that you like to both sides of an equation and provided that this is the same figure, it will not alter the result. Hence, if 170 + 60 + 20 is 40 more than 160 + 35 +80 -65, I could add 390 to each of these equations and I would still be left with the result that the first equation is 40 greater than the second. I could subtract 99 from each equation and I would still be left with the fact that the first equation totals 40 more than the second equation. The adding (or the subtraction) of such numbers is of no import since what one is looking at is the net flux between the two equations.
IF WHAT YOU ARE SAYING IS TRUE, IT SHOULD FOLLOW THAT THE OCEANS ARE UNAFFECTED BY THE DIFFERENCE IN CALCULATION WHEN PERFORMED UPON A NET RADIATION FLOW BASIS AND ON A GROSS RADIATION FLOW BASIS. This follows from the fact that whatever additional input they receive from DLR, they give up a corresponding equal and opposite output amount. We are therefore really considering the net flux out.
The problem is that the 390 mw per sqm of DWLWIR can only have come into existence because of and as a factor of the 170 mw per sqm of solar ebergy received. Unless one can truly get something for nothing, this is flase. There is a failure of double accounting.
The reason why all of this is wrong is well explained by Alexander Daranko in his comment of 15 August at 11:08pm and by Kuhnkat in his comment at Aug 15th 11:37pm. It may be that you did not have the benefit of reading those comments before you posted your response of 15th August 11:39pm. There is no such thing in life as a free lunch, and we all know that the 390 mw per sqm of so called back radiation exists as a signal only, and has no energy/ability to do real work because it is cancelled out by equal and oppositite radiative flux in the opposite direction.
If the 390 mw per sqm of so called backradiation had the ability to do work or had the ability to heat up an object which is warmer than it, this would end the worlds energy problems. Rather than wasting time and effort in the pursuit of exploiting solar radiation which is only 170 mw per sqm (on your figures), we would be exploiting the 390 mw per sqm of DLR (on your figures) which is a constant 24/7 energy source come rain, cloud or shine. We are not seeking to exploit this since this ‘imagined’ energy is not sensible energy capable of real work. It is one side of account on which those that propogate this conjecture convienently forget to take into account the opposite cancelling out budget.
The oceans are not radiating away about 490 mw per sq m. This figure is an artificail figure and fails to properly address the net flux out of the oceans. When the net fluxes are properly considered, the oceans do not freeze.
Underpinning this problem is an incorrect assumption that the earth can be considered as if it were a blackbody. It is not. Such an approach is completely unacceptable for a water world which has the ability to absorb and store significant quantities of energy and to release that stored energy not instantaneously but rather at a later date and the position is yet further complicated by the phase changes of water itself.
I consider myself to be a sceptic which means that I do not have absolute views either way and I am open to be persuaded as to the correctness of the AGW conjecture. However, save for a few elements, I am unpersuaded by most of the pillars upon which it is based and I consider it has particular problems with the oceans. Given the heat capacity of the oceans, the oceans and how these behave will determine whether there is any milage in the AGW conjecture.
Finally, I point out the obvious, one explanation as to why there may be less sensitivity to CO2 than the IPCC would have one believe may be due to ocean temperature being less effected by DWLWIR than the warmist would have one believe. This would also explain why Trenberth is having so much trouble finding his missing heat.
I look forward to hearing from you with your detailed calculations expalining why the ocean at 62N 19 deg 04 42 E freezes but not at 62 N 8 deg 45 E.

George
August 16, 2011 9:40 am

Artic vs Tropical and duration (from somewhere above)… The refractive index of ‘water’ is 1.33 yielding a critical angle of 48+ degrees (48.6?) RI = 1/Sin(theta). That assumes that RI of ‘water’ and sea water are equal. With sun angles at lower than 48 degrees, the surface should become, as a sum, reflective. It would bear out if the water temp curve and the air temp curve followed different step functions. The water temp curve would not match the air temp curve increase until the sun angle was above 49 degrees. You could then probably model conductive vs direct heating of the water on the difference using the Critical Angle as the location of the significant phase change.

jae
August 16, 2011 9:42 am

I guess it’s time for me to ask the same question I have asked for years and still have not gotten a decent answer:
A greenhouse made of IR-transparent plastic will get no hotter at noon in
Guam than it gets in Phoenix, despite the much greater DWR in Guam.
Why?

anna
August 16, 2011 9:53 am

In my opinion I think that semantics is a really big and important issue not to be ignored. Much of the confusion and misunderstandings probably arise because of the various interpretations of warming and people waste time arguing things on which they in fact agree upon just because they interpret words differently. For most people warming probably means increasing the temperature rather than “preventing cooling”. Many of the critics to the claim that GHGs can’t warm the planet probably uses this definition of the word warming. To describe the action of GHGs as warming the planet sounds scary because it gives you the impression that even at night, when the heat source (the Sun) is turned off the GHGs will keep on warming the Earth so that when we wake up in the morning the temp has increased from evening +20C to maybe +25C and then when the Sun is turned on again the temp increase will continue and soon the planet boils. So why not try to be scientific and avoid talking about warming? GHG molecules can not create energy, thus they can’t warm, but they can continuously absorb and reemit energy already present and prevent it from dissipating into outer space as quickly as it would otherwise.

Tim Folkerts
August 16, 2011 10:22 am

Matt G says:
August 16, 2011 at 6:34 am

Exactly, it is extremely small and shows the same result with a volume of water. It is easy to show that DLR can’t warm a volume of water during a day and relies only on solar energy. DLR occurs all the time during day and night so to distinguish this between solar energy, one only has to place identical volumes of water in the sun and in the shade. The shade volume must be outside exposed to the atmosphere and not in the sun so only the DLR is still reaching it. The result which anyone can easily demonstrate (but some at least choose to ignore this observation) shows the water in the shade during one day doesn’t warm, yet the volume in the sun warms greatly. Therefore this experiment provides the scientific evidence that solar energy warms a volume of water and not DLR.

Matt, you have missed 1/2 the experiment. Now try allowing ONLY sunlight in, but removing (or at least significantly reducing) the incoming IR. The details would be a bit of a challenge, since you would have to maintain the air temperature over the water, but otherwise have the top of the tank surrounded by something very cold. The surface of the water will still be radiating ~ 400 W/m^2, but the incoming sunlight will only be providing ~ 170 W/m^2. This will ALSO cause the water to cool.
The obvious conclusion is that the SUM of the energy coming in affects the water temperature (along with the sum of the energy out). You can’t logically say that only one source of energy is keeping the water warm.
Come to think of it, a related experiment is performed all the time. On clear nights, the air (and ground and the surface of bodies of water) cool rather effectively. On cloudy nights, every thing stays warmer. The difference is that the clouds are very good emitters of thermal IR across the entire thermal IR spectrum, sending copious amounts of DLR to the surface where it gets absorbed. On clear nights, only GHG’s emit DLR — they only emit in particular bands of the spectrum, so they cannot radiate as much IR downward as the clouds do, so they cannot slow the cooing as effectively.
So the DLR is clearly responsible for slowing the rate of cooling at night. During the day, they ALSO slow the rate of cooling, which is clear if you think about it a little.

kuhnkat
August 16, 2011 10:28 am

Bystander,
“Willis says “Please stop the condescending snarkiness, it just makes you look ugly.”
Ah – so condescending snarkiness is only OK when attacking real scientists then….”
In this case Willis was attempting to point a certain a[***]ole to a better way of interacting. Please do not throw this at him.
[Language. Robt]

Frank
August 16, 2011 10:31 am

Willis: Thanks for increasing the credibility of the skeptical community by taking on this issue.
Many people instinctively assume that when about 330 W/m2 of DLR is deposited in the top 1 mm of the ocean, that massive amount of energy can’t fit in thin layer of water and must be somehow be returned to the atmosphere (possibly by evaporation). After all, that energy can’t penetrate the ocean by convection or conduction. Unfortunately, some of these people don’t recognize that the upward 390 W/m2 of upward LWR emitted by the ocean originates in exactly the same top 1 mm of the ocean that absorbs DLR. Since most DLR originates at altitudes that are colder than the surface of the ocean, the net flux of long wavelength energy must be from the ocean to the atmosphere. Evaporation also removes about 80 W/m2 of energy from the top 1 mm. Only about 80% of solar radiation (SWR), penetrates the top 1 mm of the ocean. Therefore the top 1 mm of the ocean is running an energy deficit despite the large amount of energy that is deposited there by DLR. And measurements show that the top 1 mm is usually colder than the water immediately below.
In the tropics at noon on a sunny day, solar radiation might reach 500-1000 W/m2. If 20% of that much radiation were absorbed by the top 1 mm, the temperature of the top 1 mm might rise – thereby increasing upward radiation and evaporation without “warming” the bulk of the ocean. However, this situation persists for only a small fraction of each day.
One might say that the bulk of the ocean is warmed by the >80% of SWR that penetrates the top 1 mm of the ocean; not the DLR that is absorbed in the top 1 mm. Neither LWR nor evaporation cool the water below the top 1 mm. The surplus energy from SWR that is deposited below the top 1 mm is eventually returned to the cooler top 1 mm by convection and conduction. From the top 1 mm, that energy is eventually returned to the atmosphere and space.

tallbloke
August 16, 2011 10:33 am

Kadaka, point taken again.
richard verney says:
August 16, 2011 at 9:37 am
[…]

Good comment. The LW radiation comprises a flux which isn’t separable into individual ‘up’ and ‘down’ components in any meaningful way. This is what I was trying to get at with the badly thought out and badly worded comment Kadaka rightly criticised.
By the way Richard, this paper looks like good reading in the context of your ocean freezing question to Willis:
http://www.fisica.edu.uy/~barreiro/papers/BarreiroCherchiMasina2011.pdf
It’s under discussion here:
http://judithcurry.com/2011/08/16/climate-sensitivity-to-ocean-heat-transport/

Spector
August 16, 2011 10:38 am

I think a systems approach needs to be taken with respect to the ocean.
From the point of the greenhouse effect it does not matter if a solar photon is absorbed 30ft down and LWIR photons are being absorbed or emitted in the top millimeter. One might say it is the ocean’s business how it arranges its own temperature structure. All that really matters is that heat *energy* is being lost or gained by the ocean as a whole.
Evaporation-condensation is another energy exchange process altogether. At any one moment in time, energy is being exchanged but the temperature is constant. This sort of harkens back to the discussion of Local Thermodynamic Equilibrium (LTE) some months back.

Tim Folkerts
August 16, 2011 10:45 am

“OK, if gross flows are important than why hasn’t anyone computed the gross energy flows of kinetic energy? …. So, why would anyone think the gross flows are important for the case of radiation? Maybe Willis or someone can explain to me the difference.”
Because for radiation, the flow in one direction is due to the object on one side. If I know the temperature and emissivity of one object (such as the surface of the ocean), then I can say how much IR energy it is emitting, INDEPENDENT of the temperature or emissivity or other objects around. So it makes perfect sense to discuss the energy being radiated, not simply the net radiation. (Of course, in the end it is the NET energy that determines the change in temperature, so eventually you need to know about the other objects around).
For conduction you need to know the temperature of BOTH objects and the conditions in between them. This means that knowing the conditions of he ocean surface will not suffice to know the conduction. So in a sense you are already looking at the net flow of energy, and the details about how much goes each direction is not important.
(But such details are indeed considered when looking at microscopic details of the kinetic theory of gas http://en.wikipedia.org/wiki/Kinetic_theory or details of phonons http://en.wikipedia.org/wiki/Phonon)

Tim Folkerts
August 16, 2011 10:53 am

George says:
August 16, 2011 at 9:40 am

Artic vs Tropical and duration (from somewhere above)… The refractive index of ‘water’ is 1.33 yielding a critical angle of 48+ degrees (48.6?) RI = 1/Sin(theta). That assumes that RI of ‘water’ and sea water are equal. With sun angles at lower than 48 degrees, the surface should become, as a sum, reflective. …

You got that backwards, George. The critical angle would apply to light heading UP to the surface at a 48 degree angle and then reflecting back down into the water.
There is, however, a different effect that does come into play for water, making it more reflective as the light hits at a more glancing angle. Look toward the bottom of this link: http://en.wikipedia.org/wiki/Reflectivity

August 16, 2011 10:55 am

How come everybody is talking about AVERAGE radiation levels resulting in AVERAGE temps using the Stefan-Boltzmann formula when there is a fourth power in this formula?
Earth has ONE sun, that radiates on half the earth. Average radiation on this half is 1364/2 = 682 W/m^2 resulting in an average temp of 303K after deduction for albedo.. Directly under the sun the temps could be much higher. These possible temps aren’t seen because the heat is taken up by the oceans and continents. Taking these high temps to the nightside of our earth makes it possible to start talking about reduced cooling by “greenhouse” gasses.http://wattsupwiththat.wordpress.com/wp-admin/edit-comments.php?comment_status=all#comments-form
All we need to do is calculate a temperature budget for earth to see if we’re loosing or gaining 😉
Hasn’t been done afaik

George E. Smith
August 16, 2011 12:17 pm

“”””” Alexander Duranko says:
August 16, 2011 at 7:32 am
Some interesting analysis of ocean heating recently.. Regarding CO2 and H2O specific ‘DLR’, you must realise that because Kirchhoff’s Law requires that emissivity and absorptivity are the same at equilibrium, ….. “””””
Too bad your statement needs those two words “at equilibrium” in order to be correct.
The Earth’s atmosphere is never in (thermal) equilibrium; not even vaguely, so forget Kirchoff’s Law.
While we are considering the validity of “Eschenbach’s Axiom” (Willis to us); that “Downwelling Long Wave Radiation (DLWR) can’t not heat the deep ocean”; we should not forget that fundamental Axiom of Climatism; “Gases cannot radiate thermal (continuum) EM radiation” so therefore Earth’s atmosphere cannot radiate a thermal (Planckian) EM radiation spectrum. And the reason for this fundamental truth, is simply that gases (well at least mono-atomic and homo-diatomic) gases can’t because they have a zero electric dipole moment; and Maxwell’s equations tell us we can’t radiate EM waves without an antenna, so in order to radiate an EM thermal spectrum, based on Temperature of the radiating substance, the atoms/molecules have to have an electric dipole moment that is not zero.
So the molecules must be assymmetrical like H2O for example; they can’t be symmetrical like CO2 or CH4 for example; neither of which has a non-zero electric dipole moment, at any ordinary Temperature. In particular N2, O2, and Ar, the principle gases of earth’s atmosphere all have zero electric dipole moments, at ordinary Temperatures, so they cannot radiate a thermal continuum EM spectrum.
Well that’s the belief anyhow; just like Eschenbach’s Axiom.

richard verney
August 16, 2011 12:37 pm

tallbloke says:
August 16, 2011 at 10:33 am
///////////////////////////////////////////////////////
Thanks for the heads up. When I have a little time, later this week, I shall consider the paper and the post about it on Claimate etc. Looks interesting.

Matt G
August 16, 2011 12:52 pm

Tim Folkerts says:
August 16, 2011 at 10:22 am
It would be better to complete this using only solar sources and no DLW, but would also have to be done outside to keep the same solar radiation 170w/m2 input sources. Changing the percentage radiation source will also change how much penetrates the water. The variable outside atmospheric day temperatures makes very little difference to the volume of water, so leaving it outside would be more accurate. Don’t know how this would cause it to cool compared with a 20c rise in one day, way above the highest atmospheric temperature that day. Also demonstrated with solar ponds, the water can warm much more via the sun, it is latent heat and convection that keeps the ocean surface stable.
“The obvious conclusion is that the SUM of the energy coming in affects the water temperature (along with the sum of the energy out). You can’t logically say that only one source of energy is keeping the water warm.”
I can when this is only demonstrated to do so, but remember this was only for one day. My main point it that for just one day the difference is huge and for over a period a massive change has to occur for this to be even noticeable from the solar source.
On clear nights the atmospheric temperatures can cool quickly, but the ocean SST’s hardly change at all. Only the top 1mm is shown to be a little cooler at night over the ocean, but it makes no difference to SST’s over one full day. Clouds do have an affect on atmospheric temperatures and warm during night and cool during the day. Simply because the clouds cool during the day shows that the suns affect on atmospheric temperatures is greater than any DLR. (despite the 170 and 390 values shown) I agree the DLR is clearly responsible for slowing energy through the atmosphere, but for ocean the changes are very small or make no difference because it always has a high concentration of water vapour above it’s immediate surface.

kuhnkat
August 16, 2011 1:14 pm

Tom Folkert,
“The difference is that the clouds are very good emitters of thermal IR across the entire thermal IR spectrum, sending copious amounts of DLR to the surface where it gets absorbed.”
And where does the energy come from for the clouds to emit energy all night Tom? The amount of thermal mass in them simply is not enough. You have to give up this idea of components of a system functioning the same in isolation.

Stanb999
August 16, 2011 1:31 pm

steven mosher says:
August 16, 2011 at 1:52 am
Hmm Tallbloke. Rather than trust a magazine about how reflective insulation works or does not work I think I trust the stuff I built for DOD. And I’ll use my thermos to keep my coffee from cooling faster than it would. And If I have to go near a wacking hot fire i’ll also wear a reflective suit
here have some fun. there’s plenty more
http://www.insul.net/howto.php
And yes, if you use a space blanket too long you get hoarfrost in the inside.. cause its working. but eventually get to the fire
—————————————————————-
Steven,
Does adding more of this highly reflective aluminium increase the thermal reflectivity?
For instance would you prefer to be in an armour suit like in the days of other dreamy notions of how the world worked… Would it’s greater concentration of molecules even tho reflective of IR enable heat transfer. Or is the real key to not transferring a lot of heat the tiny mass of the aluminium actually being heated. So would adding more co2 make the layer more dense and transfer heat faster?
This post has nothing to do with how the world works…

Martin Lewitt
August 16, 2011 1:35 pm

How did this thread get so long so fast? I haven’t had time to read them all, so hopefully this isn’t repetitive. Willis, do you know what you are talking about? Infrared doesn’t penetrate a whole millimeter, but mere microns. The point is not that it cannot heat the ocean, it can, but the point is it is coupled to the ocean quite differently from solar, while models generally couple them both to the whole mixing layer as if they were equivalent. Solar can penetrate 10s of meters, there has even been Kelp forests at 100 meters depth.
Radiation that penetrates mere microns is more likely to be involved in surface latent heat effects, is more likely to be reradiated quickly and more likely to be less coupled to the ocean by foam or spray or surface biofilms, etc. Your argument amounts to “A watt is a watt”. In a complex nonlinear system a watt isn’t just a watt. It does matter where that watt is. Vertical and horizontal distribution make a difference. The albedo of the ocean is likely to be different in the IR range than in the visible range, just as it is different for snow and other surfaces. You wouldn’t claim a watt absorbed in the desert where it is far more likely to be radiated that night is the same as a watt deposited in a humid climate where it will stick around for longer. What makes you think in a complex nonlinear system you are entitled to the assumption that a watt in the first microns of the ocean surface is the same as one 50 meters down? That is the assumption the modelers are making.

Stanb999
August 16, 2011 1:42 pm

Willis Eschenbach says:
August 16, 2011 at 12:57 pm
Folks, and particularly Tallbloke, you are arguing that radiatively heating a liquid from above the surface causes overturning.
Heating a liquid from above the surface causes stratification, not overturning. You can experience this in the ocean or a lake on a calm sunny day. This is because the heating decreases with depth, no matter what the type of radiation is. As a result, the warmest layers are always on top. In that condition, heat is transferred downwards by conduction.
The air doesn’t heat the water. Period. It doesn’t contain enough energy to do so. Period. Never happens. As you said a cold body never warms a warm one… So the atmosphere except the last few hundred meters are colder than the ocean surface. The top 33 feet of the ocean contains the same mass as the entire atmosphere above it. Being that the last few feet are in actual contact. That is all that can affect change.
If your going with the insulation model…
1. You have diminishing returns with every layer added. So adding more co2 only adds a small portion of actual “benefit”. It’s non-linear and dilative.
2. What gives you the idea that adding heat next to the surface wouldn’t cause other mechanisms to operate faster… IE a higher flame boils water faster. Do you suggest water can only evaporate in a static fashion linearly?

Tim Folkerts
August 16, 2011 1:47 pm

Richard Verney says:

The problem is that the 390 mw per sqm of DWLWIR can only have come into existence because of and as a factor of the 170 mw per sqm of solar ebergy received. Unless one can truly get something for nothing, this is flase. There is a failure of double accounting.

There is no double accounting. If my household receives $170/day and spends $170/day, there is absolutely nothing that prevents my wife and me from handing thousands of dollars back and forth each day. This does not violate and “conservation of money” nor does it require any double accounting to make the budget balance. Same for the earth’s energy budget.
I would challenge you to find any part of the system where the energy in is not equal to the energy out.
A better way to think about this is from one day to the next. At the moment, a column of air 1 m^2 from the ground to the top of the atmosphere has a total thermal energy of about
1000 J/kg*K * 10,000 kg * 250 K = 2.5 billion J.
(These are all only a rough estimate, but it makes the point.) This is energy the atmosphere already has — not energy it needs to get every second from the sun or some place else. This column of atmosphere can easily loose ~ 325 J each second to the earth and 200 J each second to outer space for quite some time without cooling too much. During the day, that column of air will absorb ~ 2*70 = 140 J each second from the sun and a little over 350 + 24 + 78 J from the ground (since evaporation and radiation and convection will be greater during the day then during the night) for a net gain of ~ 75 J each second during the day (ie the atmosphere warms up during the day). At night, it will not get any energy from the sun, but it will still get energy from the ground (via radiation convection and conduction), for a net loss of ~ 75 J each second (ie the atmosphere cools at night). Over the course of a day, the net change is ~ 0.
No double accounting. No violation of conservation of energy. No violation of the 2nd law of thermodynamics.
The details of loses and gains at specific locations and specific times and specific seasons obviously would require much more work, but the “back of the envelope” calculation shows everything is in accordance with the laws of physics.

Mike Rossander
August 16, 2011 1:49 pm

This post is clearly a rebuttal. Less clear is a rebuttal to what?
For those of us who are less well connected to the debate, could we get a little context next time, please? Who believes this and where are they saying it? Then tear into why they’re wrong. Just a little context would go a long way. Thanks.

Tim Folkerts
August 16, 2011 1:59 pm

kuhnkat askes:
August 16, 2011 at 1:14 pm
“And where does the energy come from for the clouds to emit energy all night Tom? The amount of thermal mass in them simply is not enough. ”
See the previous post. There is a surprising amount of thermal mass in the atmosphere.
Besides, the clouds are also ABSORBING energy from the ground very effectively, which means that most of the energy is simply going back and forth continuously. One second, a square meter of cloud emits ~ 325 W thermal IR downward that the ground absorbs. That same second, a square meter of ground emits ~ 390 W thermal IR upward that the clouds absorb ( the exact numbers are not important; I am simply taking representative numbers where the clouds are cooler than the surface). The clouds are actually GAINING energy in the exchange, which would tend to warm them (although there are also many other energy exchanges that will affect the overall energy and temperature of the clouds).

tallbloke
August 16, 2011 2:12 pm

Mike Rossander says:
August 16, 2011 at 1:49 pm
This post is clearly a rebuttal. Less clear is a rebuttal to what?
For those of us who are less well connected to the debate, could we get a little context next time, please? Who believes this and where are they saying it? Then tear into why they’re wrong. Just a little context would go a long way. Thanks.

Hi Mike,
take a look at this post I wrote and Willis’ contributions to it.
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/

tallbloke
August 16, 2011 2:16 pm

Willis Eschenbach says:
August 16, 2011 at 2:12 pm
As a result I wrote the arguments list above, and tallbloke and the folks who think DLR can’t warm the ocean started answering (although none have explained why, if the DLR isn’t warming the ocean, it hasn’t frozen yet.

tallbloke says:
August 15, 2011 at 1:46 pm
Hi Willis,
Argument one asks what the difference is between rock and water. Warm water molecules rise to the top. Warm rock molecules conduct heat to their neighbours, which can’t go anywhere.
Argument two asks where the energy goes. The answer is:
space.
Argument three is not an argument that DLR can warm the ocean, it’s an argument that it can slow its rate of cooling.
Argument four is a numerical misunderstanding. The ocean surface very efficiently absorbs 95% of DLR, and promptly re-emits half of that (the other 5% being reflected). The other half makes it another couple of molecules deeper and then the molecules it warms become more buoyant than their neighbours and rise to the top, losing another half upwards. Now we’re down to ~72W/m^2. Lets remember the net flux is 66W/m^2 upwards at this point. So your ice cube argument fails. The ocean absorbs and re-emits the long wave radiation coming downwards from the atmosphere, the sums balance. In fact it emits 66W/m^2 more long wave radiation than it absorbs. It always has, and the oceans don’t freeze, because solar shortwave warms them to really significant depths of 100 metres and more as internal tides and currents mix its energy downwards. Some of that solar short wave energy is re-emitted as long wave from the surface along with some of the long wave which came from the atmosphere. The rest causes evaporation and thermals or is conducted upwards. The difference is, the solar derived energy can remain deep in the ocean for a long time, controlling it’s bulk temperature.
The question is, do DLR heated water molecules make it downwards far enough for long enough to warm the ocean bulk. I think the answer is no, because warmer water molecules are naturally buoyant, and because the vortices which mix solar energy so efficiently are below the wave troughs, several thousands of times deeper down than the depth DLR penetrates water to. For experimental evidence on this matter I’ve tried putting small soaked pieces of loo paper just under the surface out in the rolling waves away from the shore where they break. They don’t get sucked downwards. So that’s turbulent convection gone, what’s left? Conduction is a non-starter, because water thermally stratifies and anyway is a relatively poor heat conductor unless the heat source is underneath rather than above.
But this isn’t about absolutes. I’m sure the increased DLR warmed the ocean a little bit, or at least slowed its rate of cooling a little bit. I think the increased insolation due to (empirically measured) reduced cloud cover in the tropics 1980-1998 did a lot more to increase ocean heat content. To turn your question back to you, where else could that energy have gone?
Cheers
TB

richard verney
August 16, 2011 2:26 pm

Tim Folkerts says:
August 16, 2011 at 1:47 pm
///////////////////////////////////////////
If the income that you receive from your employer is $170 per day, as you say there is nothing preventing you and your wife handling 1000s of dollar notes per day but this does not mean that your income is any more than $170 per day.
I accept that you can receive $170 and give your wife $165 she then gives you back $164 and you then give her back $163 and she then gives you back $162 etc etc but the fact that so much money is passing through your hands does not mean that your daily income is anyting more than $170 and you could not obatin a mortgage on the basis that you were in receipt of $1000s per day. You are simply counting the same notes more than once.
The 390 w per sqm is a factor of the 170 w per sqm received from the sun. It is created by the solar energy received by the earth. If the sun had never fired up, the 390 w per sq m would not exist. You cannot create something from nothing and whatever is downwelling from the atmosphere it cannot be more than we have received from the sun.
Going back to Trenberth, of course I could go around adding amounts to both sides of the equation without disturbing the balance. But to do so is false. It is net flux that we need to consider when addressing the issues raised by Willis.

tallbloke
August 16, 2011 2:56 pm

Willis Eschenbach says:
August 16, 2011 at 2:41 pm
This means that information has been thrown away to get to the “net flow” number, and I prefer not to throw away information when I don’t have to.

Let’s be super-generous to ourselves and have all the numbers on the table
Downwelling solar absorbed by the ocean ≈ 170 W/m2
Downwelling ‘back radiation’ ≈ 320 W/m^2
Total = 490 W/m^2
Ocean Heat Loss:
Radiation originating from energy very near surface of the ocean ~220 W/m^2
Radiation originating from solar energy in the next 4km depth of ocean ~170 W/m2
Sensible (convection) 30 W/m2
Latent (evaporation) 70 W/m2
Total = 490 W/m^2
Net radiative loss from ocean to atmosphere ~70W/m^2

Stilgar
August 16, 2011 3:07 pm

Willis Eschenbach says:
August 15, 2011 at 2:04 pm
You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.

No, you actually say “it HEATS it”. Which is where it seems most people are getting confused.
Something I have a problem with in any scientific debate is that more often that not, a person arguing a simple idea will use “common parlance” which is actually not as common as as the person assumes.
Stop for a second and think about what you wrote with the non-common parlance in that “it heats it” means that it will increase the temperature above what it currently is (not that is slows cooling but it actually makes it warmer).
If I did not know the common parlance you use, I could refute all of your arguments based on that simple fact alone. Which is what a lot of commenters are trying to do.
If you really want to educate people (instead of generating a mass of comments that simply misunderstand what you are saying), assume they dont know ANY common parlance and acutally say “it slows down the cooling”.

Tim Folkerts
August 16, 2011 3:08 pm

richard verney says:
“I accept that you can receive $170 and give your wife $165 she then gives you back $164 and you then give her back $163 and she then gives you back $162 etc etc …”
Richard, you missed my point, I think. I’m not talking about handing the same bills back and forth, whereby I could only give a maximum of $170 at a time. Suppose I already have $1,000 in my wallet and she has $2,000 in her purse. My boss pays me $170 and I add it to my wallet. I pay my wife $400 for cleaning the house. She pays me $230 for mowing the lawn. Then she goes out and spends $170. We can hand back and forth large sums of money independent of getting $170 per day of new money and spending $170. While the true income is only $170, that does not stop me from giving my wife more than that each day. At the end of the day we both have what we started with.
Similarly, even though one square meter of surface receives only 170 W/m^2 of “new energy” from the sun, that does prevent the surface from “giving away” 390 W/m^2 of IR energy. Even though the atmosphere only receives ~ 70 W/m^2 of “new energy” from the sun, that does not prevent the atmosphere from emitting 325 toward the ground and 200 toward outer space. Looking at the total energy balance shows that the atmosphere gains as much as it looses. The surface gains as much as it looses. Everything stays in balance.

Dave Springer
August 16, 2011 3:33 pm

tallbloke says:
August 16, 2011 at 2:56 pm
“Net radiative loss from ocean to atmosphere ~70W/m^2”
Unless that radiative loss is balanced by a gain in energy then the ocean temperature would be falling like a stone. Clearly it isn’t falling like a stone. What do you propose is the source and mechanism of balancing energy?
Sorry Tallbloke, but I’m calling BS on that net radiative loss. I believe it’s demonstrated there’s a net radiative gain which is balanced by a latent loss. For some reason I thought that’s what you had concluded as well.

August 16, 2011 3:38 pm

Willis,
Firstly, my congratulations on a very sensible article. I have no substantial disagreement,
Just a comment on the nett flow argument. You’re right that it can be analysed either way. But nett flow does clarify something in the boundary layer. Think of the energy balance of a submicron surface layer where conduction is sufficient to keep the temperature fairly uniform. It radiates up, and receives DLR. The imbalance is what has to be made up by other modes of heat transfer. The “inadequacy” of this is the non-problem you began with.
But looking at nett flow, it isn’t even an issue of downward propagation of thermalized DLR. The IR flows are physically added (with sign) at that surface, and the nett IR flow is up. Any issue would be – how can the surface be kept warm enough to maintain nett upward radiation. The answer is of course, turbulent heat transfer from below (ultimately from absorbed insolation). And observation is that it does seem adequate.

August 16, 2011 3:42 pm

Willis wonders “You say the ocean doesn’t warm from DLR … so what keeps it from freezing?”
Boltzmann only requires the ocean to radiate the energy and says nothing about where that energy must come from. As I said in one of my earlier posts, in my view the DLR is absorbed into the top 10um and at about the same rate its absorbed, its re-radiated upwards as part of this requirement. Some is used for evaporation, and there are other factors to be considered but that is the essential process.
At no time does the energy from the DLR make it to the bulk. The ocean isn’t warmed by the DLR, its only ever warmed by DSR. DLR helps keep it warm.
Its not just sematics. Its process. And understanding.

Tim Folkerts
August 16, 2011 3:52 pm

Actually, Stilgar, in thermodynamics “it heats it” is neither necessary nor sufficient to conclude that “it warms it”.
“heats” = “transfers net thermal energy due to a temperature difference”
“warms” = “raised the temperature of”

These are two different concepts.
* I can heat ice without warming it (eg by melting it at 0C)
* I can warm air without heating it (eg by compressing it with a pump)
With these more specific definitions, I think it is perfectly reasonable to say “DLR warms the oceans” since the temperature is higher than it would be without that IR radiation. On the other hand, it is NOT correct to say “DLR heats the oceans” since the DLR comes from a region of cooler temperature, and the net flow of thermal energy can never be from cooler to warmer.
The true “heating” comes from the sun. With no loss of energy from the earth (ie perfect stopping of radiation from the earth), the surface would be heated until it approached the temperature of the sun. With complete loss (ie blackboady radiation loss from the earth), the temperature would be well below 0 C. With some stopping of IR, the temperature is somewhere in between. The exact value depends on how much the IR is reduced. Stop more IR, and the surface warms; stop less IR and the surface cools.

Konrad
August 16, 2011 3:52 pm

steven mosher says:
August 16, 2011 at 9:33 am
“The point is that no amount of theory, physics, experiment, will convince some people because they do not want to be convinced. We have a word for that. It starts with D”
I wholly disagree with your statement. Many WUWT readers are from the engineering professions. A well designed and conducted experiment will indeed convince people. You have failed to acknowledge that study you linked to at RC does not provide adequate empirical data for the question being asked. “Can backscattered IR around the 15 micron frequency heat or slow the cooling of Earth’s oceans to any measurable degree?”
The experiment required is simple
-Enclosed controlled environment preferably cold
-Two tanks of sea water at a known temperature.
-Air sources at a known temperature, humidity and wind speed for both tanks.
-IR source emitting only between 10 and 20 microns with a spectral peak around 15 microns, above one tank only.
-A few accurate thermometers.
This thread now has over 250 comments and no one has pointed to such an empirical study. 100 billion spent on global warming research and no one has done this?
“Do you have any cheese at all?”, he asked, expecting the answer “No”…

Dave Springer
August 16, 2011 3:53 pm

“The 390 w per sqm is a factor of the 170 w per sqm received from the sun. It is created by the solar energy received by the earth. If the sun had never fired up, the 390 w per sq m would not exist. You cannot create something from nothing and whatever is downwelling from the atmosphere it cannot be more than we have received from the sun.”
Evidently the hockey team believes it create something from nothing whenever it’s convenient. They can make ice without water if need be.
Here’s my take. As far as the team goes these instantaneous energy transfers are pure unadulterated obfuscation. When you have two terms on either side of an equation that cancel out you cancel them out. It’s called simplification and is taught in high school mathematics I believe beginning in Algebra 1. If you fail to produce the simplest form of an equation you get dinged for it. Willis evidently has been out of high school too long to remember the basics.
But let’s presume for a moment that Willis could pass a 7th grade algebra test. That raises the question of why on earth he would insist on not following the rules. I don’t think his goal is obfuscation. I think it’s appeasement. Willis is insecure about his position in this debate up against guys with PhD’s so he’s saying to them “Look here, we’re all smart guys and I agree with most of what you say except for clouds”. Willis actually did say to me that he and Trenberth were both smart and understood this. Actually Trenberth is a whole lot smarter than Willis. But that’s not saying much. Trenberth isn’t particularly bright.

Tim Folkerts
August 16, 2011 3:56 pm

Dave Springer says:
“Sorry Tallbloke, but I’m calling BS on that net radiative loss. ”
I think that he specifically meant “net radiative loss IN THE THERMAL IR PART OF THE SPECTRUM”. There is a net overall radiative gain (when solar radiation is included), which allows for a net evaporative/convective transfer upward.

tallbloke
August 16, 2011 3:56 pm

Dave Springer says:
August 16, 2011 at 3:33 pm (Edit)
tallbloke says:
August 16, 2011 at 2:56 pm
“Net radiative loss from ocean to atmosphere ~70W/m^2″
Unless that radiative loss is balanced by a gain in energy then the ocean temperature would be falling like a stone. Clearly it isn’t falling like a stone. What do you propose is the source and mechanism of balancing energy?

The incoming solar (170) less the latent and sensible loss (100) equals 70, which equals the net radiative loss.
These are Trenberth’s figures, which are wrong, but they are the one’s Willis likes to use, so for the sake of keeping the debate on an even keel so far as is possible, I’ll use them to demonstrate the falsity of Willis’ ice cube argument.

tallbloke
August 16, 2011 4:00 pm

Willis Eschenbach says:
August 16, 2011 at 3:40 pm
I’m sorry to have to say this again, but the individual flows are the real, actual, measurable, observable flows. The net flow is a mathematical construct, useful, but without a real-world counterpart.

The real world counterpart to the net flow is the fact that the sum effect of the radiative flux is to cool the ocean by ~66-70W/m^2

tallbloke
August 16, 2011 4:14 pm

NASA have stopped using the K-T energy budget cartoon on their website which shows the separate LW radiation components and replaced it with this one which only shows the net flow:
http://tallbloke.files.wordpress.com/2010/06/energy-budget-new.jpg

Dave Springer
August 16, 2011 4:14 pm

Why does “the team” pursue this obfuscatory complexification of the ocean heat budget one might ask.
The reason is simple. If you toss about all these big numbers about hundreds of watts of LWIR coming out the ocean and hundreds of watts of LWIR flowiing back into the ocean it obscures the fact that primary mechanism of energy loss by the ocean is latent heat, not radiative. The net radiative heat loss at the end of the day is 50Wm, conductive heat loss is 20Wm, latent heat loss weighing in at 140Wm. Energy into the ocean 200Wm which exactly balances out. This accurate correctly simplified equation leaves us arguing over a tiny, hypothetical imbalance on the order 2Wm more input than output. To add insult to injury no one can actually find any sign of the imbalance – it’s all just mathematical creation in a toy computer model with no empirical evidence that it actually exists in nature – hence the infamous “missing heat”.
This would be funny if there weren’t so many imbeciles that believe the output of toy computer models trump reality. It’s sad really and brings to mind something Richard Dawkins said (my modifications for context):
It is absolutely safe to say that, if you meet somebody who claims not to believe in evolution anthropogenic global warming, that person is ignorant, stupid or insane (or wicked, but I’d rather not consider that).
— Richard Dawkins.

August 16, 2011 4:17 pm

Dave Springer says:
Dave you seem to be a smart guy, but Willis has a real talent for explaining things, and I for one appreciate his articles very much. Why don’t you submit an article, and see what it’s like being on the receiving end of criticism?
You could have left your last papragraph @ 3:53 pm out completely. It was petty and juvenile. How old are you, anyway?

Dave Springer
August 16, 2011 4:21 pm

tallbloke says:
August 16, 2011 at 4:14 pm
“NASA have stopped using the K-T energy budget cartoon on their website which shows the separate LW radiation components and replaced it with this one which only shows the net flow:
http://tallbloke.files.wordpress.com/2010/06/energy-budget-new.jpg
Precious. What happened, did Hansen’s high school algebra teacher ring him up on the telly from the nursing home and tell him if he didn’t start following the rules she taught him she’d smack his hand with her ruler?
Pay attention Willis. No more need to appease your foils by agreeing with their errors. You have a new party line to parrot.

Spector
August 16, 2011 4:24 pm

In nonscientific usage, the difference between warming and heating is a matter of degree. Warming is putting a blanket around yourself or standing near the fire. Heating is what you do to make boiling water. Perhaps warming==gentle heating.

tallbloke
August 16, 2011 4:28 pm

Dave Springer says:
August 16, 2011 at 4:14 pm
The net radiative heat loss at the end of the day is 50Wm, conductive heat loss is 20Wm, latent heat loss weighing in at 140Wm. Energy into the ocean 200Wm which exactly balances out.

Dave, where are your figures from please?

Dave Springer
August 16, 2011 4:30 pm

Smokey says:
August 16, 2011 at 4:17 pm
“Dave Springer says:
Dave you seem to be a smart guy, but Willis has a real talent for explaining things, and I for one appreciate his articles very much. Why don’t you submit an article, and see what it’s like being on the receiving end of criticism?
You could have left your last papragraph @ 3:53 pm out completely. It was petty and juvenile. How old are you, anyway?”
I’m 55 and I was writing articles on global warming begining in 2005 and I quit in 2009. Been on the receiving side plenty.
Quite frankly Willis needs the upper hand of being the author and me the commenter to make the playing field a bit more level. This whole damn article was a response to ME for suggesting in a different article of his that LWIR from greenhouse gases doesn’t effect the ocean to any significant degree. I don’t begrudge him the ability to respond to an obscure comment of mine with a headline, large print, pretty pictures, and words directed at me like silly, nonsense, impossible, and ugly. Poor little Willis needs all the advantage like that he can get and at the end of the day he still comes up lacking.

Dave Springer
August 16, 2011 4:34 pm


Dave Springer says:
August 15, 2011 at 10:18 pm
http://www.atmos.umd.edu/~carton/pdfs/foltzetal03.pdf
JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 108, NO. C5, 3146, doi:10.1029/2002JC001584, 2003
Seasonal mixed layer heat budget of the tropical Atlantic Ocean
Gregory R. Foltz, Semyon A. Grodsky, and James A. Carton
Department of Meteorology, University of Maryland, College Park, Maryland, USA
Net surface heat flux is a combination of latent and
sensible heat loss, shortwave radiation absorption, and net
longwave emission. Sensible heat loss is insignificant (<10
W m2) due to small air-sea temperature differences, while
net emission of longwave radiation is a relatively constant
50 W m2 [da Silva et al., 1994].

Bob_FJ
August 16, 2011 4:46 pm

Willis Eschenbach @ August 16, 2011 at 2:12 pm

“…As a result I wrote the arguments list above, and tallbloke and the folks who think DLR can’t warm the ocean started answering (although none have explained why, if the DLR isn’t warming the ocean, it hasn’t frozen yet.”

Willis, please try to understand that EMR (electromagnetic radiation), regardless of wavelength and power, is a different form of energy to HEAT. Put another way, the 400 w/m^2 that you visualise as leaving the surface even though measured in the same units is not rate of HEAT loss. Repeat, EMR is not HEAT.
In understanding science, things are getting desperate when resorting to analogies, but let us compare EMR with DC ELECTRICITY.
1) Electricity is not heat but can be converted to heat by passing it through a resistance, which you might visualise as being like EMR being absorbed in matter. (the photons are akin to electrons)
2) Consider two DC sources of different voltage connected in opposed series, think about what the combined PD is and compare it with opposing EMR’s. Note also that if the voltages or EMR’s are equal, nothing happens. This can be visualised in a typical elemental layer of air where most of the radiation is horizontally opposed.

tallbloke
August 16, 2011 4:52 pm

Dave Springer says:
August 16, 2011 at 4:34 pm
http://www.atmos.umd.edu/~carton/pdfs/foltzetal03.pdf
JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 108, NO. C5, 3146, doi:10.1029/2002JC001584, 2003

Nice, thanks.
Bob_FJ says:
August 16, 2011 at 4:46 pm
In understanding science, things are getting desperate when resorting to analogies, but let us compare EMR with DC ELECTRICITY.
1) Electricity is not heat but can be converted to heat by passing it through a resistance, which you might visualise as being like EMR being absorbed in matter. (the photons are akin to electrons)
2) Consider two DC sources of different voltage connected in opposed series, think about what the combined PD is and compare it with opposing EMR’s. Note also that if the voltages or EMR’s are equal, nothing happens.

As analogies go, it’s a good one I think.

Matt G
August 16, 2011 4:53 pm
Dave Springer
August 16, 2011 4:54 pm


The paragraph from the Geophysical Research Paper (actual experimental science not toy computer model output) is a summary of many individual times and locations. There’s a wealth of information in it. Enough budget tables of individual times and places to make your head spin. Inputs and outputs don’t balance in instant cases. For instance the authors measure how much summertime heating of the mixed layer is entrained until winter when the balance goes the other way which of course it must do and is why there’s little seasonal temperature variation compared to land. Where the energy goes in and where it comes out also moves around to some degree due to winds and ocean currents and precipitation patterns. At the end of the day (or year rather) it all balances out. Energy into the ocean equals energy out of the ocean equals constant average ocean temperature. The argument for any imbalance caused by AGHGs is in the margins which is why this debate is unending. The take home lesson is how small those margins are the utter dearth of empirical evidence saying they fall on side of more energy input than ouput.
.

tallbloke
August 16, 2011 4:58 pm

Dave Springer says:
August 16, 2011 at 4:30 pm
This whole damn article was a response to ME for suggesting in a different article of his that LWIR from greenhouse gases doesn’t effect the ocean to any significant degree. I don’t begrudge him the ability to respond to an obscure comment of mine with a headline, large print, pretty pictures, and words directed at me like silly, nonsense, impossible, and ugly.

There’s a good deal of back history you may not be aware of:
http://tallbloke.wordpress.com/2010/11/16/whatever-happened-to-back-radiation-part-ii/
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/

Myrrh
August 16, 2011 5:02 pm

w.
tallbloke says:
August 16, 2011 at 2:56 pm
Willis Eschenbach says:
August 16, 2011 at 2:41 pm
This means that information has been thrown away to get to the “net flow” number, and I prefer not to throw away information when I don’t have to.
Let’s be super-generous to ourselves and have all the numbers on the table
Downwelling solar absorbed by the ocean ≈ 170 W/m2
Downwelling ‘back radiation’ ≈ 320 W/m^2
Total = 490 W/m^2
#######
I think you’ve all lost the plot.
Downwelling Solar 170 W/m^2 is not heating the Earth. Visible Light, UV and Nr IR are not thermal energies – they cannot be heating the land and oceans to raise the temperature by the amount claimed nor can they therefore create that much Infrared upwelling from the Earth.
You’re in gobbledegook land the lot of you, who are arguing as if this is real.
The ‘missing heat’ is Thermal Infrared direct from the Sun…
Perhaps you can’t hear me because the atmosphere you’re in is empty space, and sound doesn’t travel in empty space..
..now where did I put that popcorn..?

tallbloke
August 16, 2011 5:05 pm

Thanks Matt and Dave – signing off for sleep.
TB

Dave Springer
August 16, 2011 5:15 pm

Bob_FJ says:
August 16, 2011 at 4:46 pm
It can become clumsy at times and difficult for the layperson (including me) to follow but ultimately everything in the universe is composed of energy which can be neither created nor destroyed. Matter itself is a form of energy with the equivalence defined by the equation e=mc2. Thus everything can be described in terms of energy flows measured in any arbitrary term you choose – joules, watts, btus, horsepower, whatever. There are equivalancy equations for converting any of them one to another. EMR and heat are both energy. Following it all its forms is a tough row to hoe but at the end of the day that’s what it boils down to – energy flows – and because energy can be neither created nor destroyed if you can balance the energy accounting from start to finish in any process of interest then you probably got it right. If the books don’t balance then you’ve definitely missed something.

Dave Springer
August 16, 2011 5:23 pm

tallbloke says:
August 16, 2011 at 4:58 pm
“Dave Springer says:
August 16, 2011 at 4:30 pm
This whole damn article was a response to ME for suggesting in a different article of his that LWIR from greenhouse gases doesn’t effect the ocean to any significant degree. I don’t begrudge him the ability to respond to an obscure comment of mine with a headline, large print, pretty pictures, and words directed at me like silly, nonsense, impossible, and ugly.
There’s a good deal of back history you may not be aware of:
http://tallbloke.wordpress.com/2010/11/16/whatever-happened-to-back-radiation-part-ii/
http://tallbloke.wordpress.com/2011/03/03/tallbloke-back-radiation-oceans-and-energy-exchange/
Yeah but in this particular case the raiser of the silly, impossible, ugly headed nonsense was me. The nerve that I touched was already exposed but it was still me who touched it this time and set Willis off on a rant. Thus I took those harsh words personally and considered them to be fighting words so I fully intend to kick Willis’ sorry intellectual ass for it.

Dave Springer
August 16, 2011 5:33 pm

Truth be told there’s some residual anger in me for the boorish article Willis penned last week with regard to Native Americans. Even Anthony said he wouldn’t have approved it. So I guess the boor in me, which is admittedly legion, is coming out in response. I don’t really mind lowering myself to that level. It’s kind of fun wrestling with pigs if you don’t mind the mud.

Costard
August 16, 2011 5:33 pm

This is absurd. Willis’ argument is that if heat is lost when the ocean emits LR – and it is, indisputably – then it must be gained (or not lost) when the ocean receives DLR. If there is some mechanism by which heat in the ocean is transferred to the surface and radiated – and there is, convection – then a decrease in net radiation means (in the absence of other variables) less heat lost, less convection, and less heat sent to the surface. Sorry but it makes no substantive difference whether you call it less cooling or more warming, when either way the effect of DLR is a higher ocean temperature than would otherwise exist.
What is important is the extent to which less heat radiated = more heat conducted and/or converted into evaporation.

Myrrh
August 16, 2011 5:33 pm

Not all electromagnetic energy heats up organic matter. It’s not simply the case of ‘balancing’ energy in/energy out, this claim is specific to heating land and oceans to raise the temperature.

Dave Springer
August 16, 2011 5:36 pm

tallbloke says:
August 16, 2011 at 4:28 pm
“Dave Springer says:
August 16, 2011 at 4:14 pm
The net radiative heat loss at the end of the day is 50Wm, conductive heat loss is 20Wm, latent heat loss weighing in at 140Wm. Energy into the ocean 200Wm which exactly balances out.
Dave, where are your figures from please?”
A typo in there, by the way. Conductive heat loss, per the paper, is 10Wm.

wobble
August 16, 2011 6:08 pm

Dave Springer says:
This whole damn article was a response to ME for suggesting in a different article of his that LWIR from greenhouse gases doesn’t effect the ocean to any significant degree. I don’t begrudge him the ability to respond to an obscure comment of mine with a headline, large print, pretty pictures, and words directed at me like silly, nonsense, impossible, and ugly.

Dave Springer, I was one of the commenters that asked you about your statement in the last thread.
Personally, I’m happy this issue was raised, and I’ve learned quite a bit reading through this thread. However, I’m still not sure what I believe. Can you clarify your position for me? I’m really just trying to understand, and I’m not trying to trap you.
It seems as if you agree that downward infrared radiation, such as that being caused by green house gases, does retard the cooling of land surfaces. Is this correct?
If so, then it also seems as if you agree that an increase in CO2 might provide some marginal amount (even if such marginal amount is minuscule and undetectable) of increased downward infrared radiation towards the earth. Is this correct?
If so, then it also It seems as if you agree that such marginal amount of increased downward infrared radiation also exists over the oceans. Is this correct?
If so, then it also seems as if there is disagreement regarding the effect that such increased downward infrared radiation has on the oceans. Is this correct?
If so, then it seems as if you assert that increased downward infrared radiation has any effect on ocean temperatures. Is this correct?
If I’m right about everything so far, then please let me know if you believe that: 1) an increased amount of downward infrared radiation is unable to have any effect at all (including theoretical) on ocean temperatures; or 2) an increased amount of downward infrared radiation could theoretically retard ocean cooling, but due to offsetting mechanisms (such as evaporation) ocean cooling isn’t retarded in any measurable amount.
Again, I’m really trying to understand what your specific claims are.

Dave Springer
August 16, 2011 6:15 pm

Matt G says:
August 16, 2011 at 4:53 pm
tallbloke says:
August 16, 2011 at 4:14 pm
Excellent. Actual net surface energy from NASA. A clear picture unsullied by obfuscatory unnecessary opposing flows that cancel out.
I culled it down to global means to make a point:
Average net shortwave radiation at the Earth’s surface: January 1984-1991.
Global mean = 162 W/m2
Average net longwave radiation at the Earth’s surface: January 1984-1991.
Global mean = -48 W/m2
Average net radiation at the Earth’s surface: January 1984-1991.
Global mean = 114 W/m2
So we see here in simple terms that at the surface, which is where we live and why I eschew obfuscatory Top of Atmosphere budget, we have at the end of the day (or rather end of the decade) a net radiative flow of 114 W/m2.
This 114 W/m2 is the amount of energy that doesn’t leave the surface radiatively. Compare this to 48 W/m2 which is the amount of energy that does leave the surface radiatively. LWIR emission only accounts for 48/114 or 42% of all radiative heat loss from the surface, land and ocean combined. For the ocean-only the radiative loss is only 25% (see my previous link to ocean heat budget).
What this means is that the lion’s share of heat loss is not via radiation and that’s especially true over the ocean. Where radiative loss is not a large factor neither can greenhouse gases be a large factor because radiation is only mechanism by which GHGs do their GHG thing.
Given the ocean is 70% of the surface, for the combined total radiative heat loss to reach 42% the land masses must be very much dominated by radiative loss. But over the ocean, not much. It’s mostly about latent heat loss over the ocean.
This is just more in the way of empirical evidence (not toy models or thought experiments with heat lamps and pans of water) that the GHG effect is predominantly a land based phenomenon.
QED
Peace. Out.

Bob_FJ
August 16, 2011 6:27 pm

Myrrh @ August 16, 2011 at 5:02 pm

“…Downwelling Solar 170 W/m^2 is not heating the Earth. Visible Light, UV and Nr IR are not thermal energies – they cannot be heating the land and oceans to raise the temperature by the amount claimed nor can they therefore create that much Infrared upwelling from the Earth…”

Sorry Myrrh, but the term “thermal radiation” is sometimes confused and misused. It is really a misnomer for EMR, (electromagnetic radiation) which includes visible light which does indeed heat matter when its photons are absorbed. Try here at Wikipedia:
http://en.wikipedia.org/wiki/Thermal_radiation

kuhnkat
August 16, 2011 6:49 pm

Willis, you can say it as many times as you want and it still is only a half truth.
I get about 275K for the average with the 170 in. Of course, this close averages may hide stuff, but, 275k isn’t frozen. Also of course, I may have screwed up trying stuff I barely understand even with the BB Calculator.

kuhnkat
August 16, 2011 6:56 pm

Willis,
“This means that information has been thrown away to get to the “net flow” number, and I prefer not to throw away information when I don’t have to.”
Except when you only say 390 you threw away the 324…

wobble
August 16, 2011 7:08 pm

Willis Eschenbach, do you admit that you don’t actually know the amount of warming the “DLR” can cause? Isn’t it possible that the amount of warming that it causes is negligible and even unmeasurable?

Tim Folkerts
August 16, 2011 7:09 pm

Bob_FJ says:
August 16, 2011 at 4:46 pm

In understanding science, things are getting desperate when resorting to analogies, but let us compare EMR with DC ELECTRICITY.
1) Electricity is not heat but can be converted to heat by passing it through a resistance, which you might visualise as being like EMR being absorbed in matter. (the photons are akin to electrons)
2) Consider two DC sources of different voltage connected in opposed series, think about what the combined PD is and compare it with opposing EMR’s. Note also that if the voltages or EMR’s are equal, nothing happens. This can be visualised in a typical elemental layer of air where most of the radiation is horizontally opposed.

Tallbloke seems to think this is a good analogy — I think it is a fair to poor analogy. Electrons are not really all that much like photons. Once a photon is created, it travels pretty freely (until it gets in the neighborhood of an atom). There are no competing fields pushing on it as it flies thru space. A photon heading one way will pass right by a photon traveling the other way.
Electrons, on the other hand are pushed and pulled by fields all thru the circuit. Electrons interact and scatter as they pass each other.
A better analogy would be two separate circuits, one sending electrons counterclockwise around one wire while the other circuit sends electrons clockwise around a parallel wire. The “net current” will be one current minus the other, but electrons really would be flowing both directions.

Stanb999
August 16, 2011 7:10 pm

In a few years… Maybe decades when the current regime is old and feeble. Well they are already feeble in certain regards.
The true case of how the “green house” works will be proven…
The simple fact is water retains heat very well. The atmosphere due to it’s limited mass dissipates heat very poorly. This is what creates the magic that scientists are trying to prove exists. You don’t need hocus-pocus.
You do all that in the dawn of astronomy it was proven that the earth was in the center of the solar system… Worked perfectly on paper. But it was in fact non-sense.

Barry Day
August 16, 2011 7:20 pm

tallbloke says:
August 16, 2011 at 4:14 pm
“NASA have stopped using the K-T energy budget cartoon on their website which shows the separate LW radiation components and replaced it with this one which only shows the net flow:”
http://tallbloke.files.wordpress.com/2010/06/energy-budget-new.jpg
======================================================
Why is the huge 400 percent INCREASE (of three million+) in submarine volcanic and thermal vent activity LEFT OUT OF THEIR EQUATION???It surely would be greater than a miniscule amount of AWG.
Out of sight out of mind,no doubt.

Steve from Rockwood
August 16, 2011 7:20 pm

Bob_FJ says:
August 16, 2011 at 4:46 pm
Willis Eschenbach @ August 16, 2011 at 2:12 pm
“…As a result I wrote the arguments list above, and tallbloke and the folks who think DLR can’t warm the ocean started answering (although none have explained why, if the DLR isn’t warming the ocean, it hasn’t frozen yet.”
Willis, please try to understand that EMR (electromagnetic radiation), regardless of wavelength and power, is a different form of energy to HEAT. Put another way, the 400 w/m^2 that you visualise as leaving the surface even though measured in the same units is not rate of HEAT loss. Repeat, EMR is not HEAT.
=======================================================
When EMR interacts with a conductor currents are formed and dissipated as heat. So an EM field generated in air would transfer heat to the ocean (which is a conductor).
EMR radiating back out into space will only find resistance, but EMR striking the earth (and the oceans) will generate heat.

Tim Folkerts
August 16, 2011 7:22 pm

Here is the simplest argument I can come up with for the “reality” of ~ 390 W/m^2 upward and ~ 320 W/m^2 downward thermal IR, as opposed to the “reality” of ~ 70 W/m upward thermal IR.
Take a 1 m x 1 m sheet of material with an emissivity of 1. Make it thick enough to have a heat capacity of ~ 700 J/K. Create a vacuum around it to eliminate conduction & convection. Cool the sheet close to 0 K and then set it out horizontally someplace above the ground.
CASE 1) If there are indeed two fluxes of 390 W/m^2 up toward the bottom of the sheet and 320 W/m^s down onto the top of the sheet, then the two surfaces will absorb = 710 J/s and warm at rate of ~ 1 K every second.
CASE 2) If there is only a “net flux” of 70 W/m^2 upward, then the top will receive nothing and the bottom will receive 70 W, and the sheet will warm at ~ 0,1 K every second. This is a 10x slower rate and would be easy to measure.
Is there anyone who seriously thinks the object will warm at a rate of 0.1 K/s rather than 1 K/s ???

Tim Folkerts
August 16, 2011 7:30 pm

“Why is the huge 400 percent INCREASE (of three million+) in submarine volcanic and thermal vent activity LEFT OUT OF THEIR EQUATION???It surely would be greater than a miniscule amount of AWG.”
Al the estimates I have seen put geothermal energy flux at ~ 0.1 W/m^2. Even if the energy flow did increase by 400 % to 0.4 W/m, this is still a pretty small affect. A forcing of 0.3 W/m^s would be significant, but hardly “huge”.
And that all assumes that there was a true increase, not just an increase in the known activity. If there had been 0.4 W/m^2 all along, then just discovering new volcanoes would not CHANGE the climate.

gnomish
August 16, 2011 7:34 pm

just a word of support for mr springer.
meritocracy is wonderful; pecking order is not tolerable.
i’m probably not the only one who can distinguish the difference.
i really am fed up with the aroma of domination growing here. willis is now like romm and mosher like gavin. stuff the drama and go back to mama.
i liked it when Anthony ran the place and held to standards which distinguished wuwt from most other sites that quickly become vulgar from ego trippers running down other guests- a clique in quest of a claque. i’m not off topic, either. they make themselves the topic every chance they get.
forum clingers are always a problem that way. they don’t do any of the work to create the venue or get it to become a going concern, they just camp on it and try to take it over and sacrifice it to their flaming vanities.

August 16, 2011 7:37 pm

gnomish,
So you support ad hominem attacks. Got it.

August 16, 2011 7:46 pm

Willis says “Dave, you still haven’t explained how the ocean can lose 400 w/m2 in radiation, gain 170 w/m2 from the sun, and still stay liquid.”
How about inflated measurements of DLR and OLR using supposed gross energy transfers when only the net flow should be used:
http://claesjohnson.blogspot.com/2011/08/how-to-fool-world-by-measuring-masive.html
http://claesjohnson.blogspot.com/2011/08/how-to-fool-yourself-with-pyrgeometer.html

kuhnkat
August 16, 2011 8:02 pm

Myrrh,
“The ‘missing heat’ is Thermal Infrared direct from the Sun…”
In case no one else has answered, that was one of my issues until I found the data. The earth and sun radiation levels cross in the 4-5 micron range. The near IR from 4 microns down is actually included in the visible absorbed numbers I believe. That is where the sun substantially exceeds the earth. From 5 up it is very small and absorbed in the atmosphere. Of course, if they don’t… I still haven’t had anyone explicitly state what the IPCC uses. Probably will have to read the one section to find out for sure. I don’t really believe they are trying to ignore almost 50% of the sun’s output. That is just a bit more than the DLR.

Tim Folkerts
August 16, 2011 8:06 pm

Dave Springer says: August 16, 2011 at 6:15 pm
This 114 W/m2 is the amount of energy that doesn’t leave the surface radiatively. Compare this to 48 W/m2 which is the amount of energy that does leave the surface radiatively. LWIR emission only accounts for 48/114 or 42% of all radiative heat loss from the surface, land and ocean combined. For the ocean-only the radiative loss is only 25% (see my previous link to ocean heat budget).
Actually (using your numbers), LWIR accounts for 100% of all radiative heat loss, but 48/162 = 29% of total heat loss.
What this means is that the lion’s share of heat loss is not via radiation and that’s especially true over the ocean. Where radiative loss is not a large factor neither can greenhouse gases be a large factor because radiation is only mechanism by which GHGs do their GHG thing.
Let me give an analogy as to why I disagree with your conclusion. Suppose i have a large balloon. It is leaking ~ 390 l/min thru a square hole, 80 l/min thru a round hole, and 30 l/min thru a triangular hole. That is a net loss of 500 l/min. But there is a second square hole with a blower that adds 320 l/min to the balloon. And there is a rectangular hole with a blower adding another 180 l/min. With all of this, the balloon is neither inflating nor deflating.
It would be silly to say that the 390 l/min leaking thru the square hole is not very important because there is a different square hole that is simultaneous adding most of this added air back. You would say BOTH holes are important. A 10% change in the square hole leak would be much more important than a 10% change in any of the other leaks.
We can split hairs and discuss if the “sizes of the holes” are quite right, but there is no way to avoid the fact that the huge “square holes” (ie upward & downward IR energy fluxes) are important.
To push the analogy a bit farther than I probably should, we could ask what would happen if we turned the blower up from 390 l/min to 400 l/m. The pressure would increase, causing more air to leak from all three holes. the increase would not necessarily have to increase proportionately for the three leaks, but they would all leak more. And more importantly, the pressure would be raise inside the balloon until the new equilibrium is reached.
For the climate, the analogous argument is obviously that an increase in DLR would increase the surface temperature, causing increases in evaporation, convection and IR radiation. Evaporation will only increase if the surface temperature increases (barring changes to winds, etc), which will by necessity also increase radiation and convection.

August 16, 2011 8:29 pm

Willis, just to remind you I know the DLR plays a role in keeping the oceans warm…but you’re article and understanding at the time of writing were off. I’ll refute your arguments since you keep pleading for someone to do so.
Argument 1. “…”People claim that because the DLR is absorbed in the first mm of water, it can’t heat the mass of the ocean. But the same is true of the land.”
No its not. DLR is NOT absorbed in the first mm of water, its absorbed in the first 10um of the water and the remaining 99990um of that first mm *increase* in temperature with depth. The place where the DLR is absorbed is the coldest place. Conduction cannot happen.
Compare this to land where the top of the earth is the warmest place and conduction can happily warm the earth below.
Argument 2 “If the DLR isn’t heating the water, where is it going?”
As described earlier, it is largely immediately radiated back up. Up and down for basically zero net effect. Except some is lost to space and some to evaporation and the shortfalls are made up from energy from the bulk of the ocean moved to the surface by convection and conduction, supplied in the first place by the sun.
So does this mean DLR heats the ocean? Yes, but not directly. It is the sun that heated the ocean. The DLR ONLY slows that energy from escaping.
Argument 3 “DLR heating of the top mm of the ocean reduces those differences and thus delays the onset of that oceanic overturning by slowing the night-time cooling of the topmost layer”
This is closest you had to correct. However your statement “But in addition to the wind-driven turbulence of the topmost layer mixing the DLR energy downwards into lower layers” betrays the fact you believed the topmost layer to be heated by DLR and warmer than the bulk. The fact is that if the topmost layer were to mix downward, then for the next few mm or so, it would be *cooling* the ocean, not warming it…and its only from 1mm down where the ocean is warmer than below where mixing down warms the ocean below. The important point here is that this warmer water 1mm down wasn’t directly warmed by the DLR. DLR played a role in it being as warm as it is, but not in the sense you meant.
Argument 4 “Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean.”
Is also close to being correct in the sense that DLR plays a role in keeping the ocean warm. Its the specifics that matter here though. Why? Because you’re trying to describe them in your article.

August 16, 2011 9:24 pm

The oceans stay liquid because atmospheric pressure dictates the energy value of the latent heat of evaporation and that sets the rate at which energy can be lost to the air via evaporation by setting the energy cost of the evaporative process.
If atmospheric pressure were reduced to zero the energy cost of the latent heat of evaporation would be so small that not only would the oceans retain no solar energy at all but they would instantly vaporise just as happens to water released to space.
If atmospheric pressure were to be increased then the energy cost of the evaporative process would rise and the oceans would retain more solar energy than they do now for a higher equilibrium temperature.
We all know that water boils at a lower temperature at the top of Everest than at the sea surface.
It is the energy cost of evaporation at a given atmospheric pressure that controls the equilibrium temperature of the ocean bulk and thus the equilibrium temperature of the entire Earth system. In the process it keeps the oceans in a liquid state at current atmospheric pressure.
DLR only affects the rate at which energy is transferred from molecules in the interacting layer in the very top few microns of the sea surface to space via the atmosphere. It does not cool or warm the ocean bulk since it cannot affect the rate at which energy is released by the oceans. That rate is affected only by the energy cost of the evaporative process which is pressure dependent combined with the quantity of solar shortwave input at being retained any given moment.
So more solar input or a rise in atmospheric pressure raises the equilibrium temperature.
Less solar input or a fall in atmospheric pressure reduces the equilibrium temperature.
Differing combinations of such changes alter the equilibrium temperature up or down depending on the relative proportions of changes in those two parameters.
Full explanation here:
http://www.irishweatheronline.com/news/environment/wilde-weather/setting-and-maintaining-of-earth%e2%80%99s-equilibrium-temperature/18931.html

Crispin in Waterloo
August 16, 2011 9:34 pm

Dave Springer says:
August 15, 2011 at 7:57 pm
Keith Minto says:
August 15, 2011 at 7:01 pm
“If the air at the ocean surface is humid that means it has already condensed and released its heat;”
Wrong. You can see it if it condenses. If it condenses near the surface we call it fog. If it condenses farther up we call it clouds. In either case you can see it after it condenses. If you can’t see it then it hasn’t condensed.
++++++++++++
Actually you can only see it if it is larger than 0.1 microns (which reflects visible light). Lots of condensed water vapour is invisible. The droplest form on cloud condensation nuclei (CCN) and are extremely small for some time. Clouds induced in this manner by galactic cosmic rays are particularly small and last longer in the sub-optical state bceause they are slightly charged and avoid each other. This can mislead (a lot) people using visibile light to measure cloud cover by satellite.
Thanks for your many comments above (until you started to get snippy). It was informative and covered many important aspects of the topic.
Willis, I agree with your main argument that the presence of anything above the ocean that impedes direct radiation of LWR to space will slow the cooling, which some are mis-interpreting as ‘warming’. Yes, it may be warmer, but it was not ‘additionally warmed’. This is an extremely important point which all should concede.
There is no doubt that the core of AGW is that the cooling rate will be reduced by additional CO2, and that the interference with the status quo ante will be catastrophic. Obviously the argument is simplistic bceause if any (of many) additional heat loss mechanisms is enhanced or triggered, the heat loss will be pretty much the same or even more. This appears to be the case as natural variation overwhelms significant increases in CO2. End of short story.

August 16, 2011 9:37 pm

“In response to those who say that when the skin layer is warmed by EMR the heat won’t propagate downwards, heating a liquid from the top stratifies it.”
Specifics Willis. You’re ignoring them. This isn’t a “macro” effect, its one that takes place at the very surface of the ocean. Stratification is certainly an effect and an important one…but not one that is relevent to this discussion.
http://en.wikipedia.org/wiki/File:MODIS_and_AIRS_SST_comp_fig2.i.jpg

Spector
August 16, 2011 10:49 pm

I believe many people get led astray when they hear infra-red radiation described as ‘heat radiation.’ This band of electromagnetic energy is the primary heat-transfer radiation band only for objects that are less than red hot. The sun is white hot. Just look at an energy (heat) spectrum of sunlight. The peak energy is around 0.5 microns (500 nm.) The primary CO2 absorption band is at 15 microns, a much longer wavelength.
Sometimes visible light is characterized as ‘cool’ because we are very sensitive to low intensity light levels. “Average indoor lighting ranges from 100 to 1,000 lux, and average outdoor sunlight is about 50,000 lux.”

Keith Minto
August 16, 2011 11:07 pm

Noelene says:
August 15, 2011 at 10:01 pm
Dumb question
How does rivers figure in all this?Rivers are always cold are they not?Is that because of their depth ?

Interesting question,
Guess there is a connection to higher altitude and thus lower air temperature, lower relative relative humidity and the fact that rivers start high and inland. The more or less linear flow plus channel resistance, would result in ‘churning’ turbulence and warm surface/cold bottom water overturning. Just a thought, and it can be tested.

Dixon
August 16, 2011 11:39 pm

Wow, too much discussion! Can we go back to basics with maybe a permanent reference page (Anthony have you considered a ‘pillars of climate science’ primer?) – this stuff is pretty critical. How about an energy balance diagram (probably with citations) to show what we “know” (from actual measurements) and what we theorise (from the gaps). It MUST have been done before.
I can see it wouldn’t be trivial though, EMR is quantised so you’d need all the relevant absorption spectra. You’d need to integrate absorption through the atmosphere and then into the ocean. Then of course you need incoming and outgoing radiation and consequent conduction. Add, aerosols, precipitation, latitude, currents, evaporation, wind, sea-spray, clouds and sediment/algal loading and you’d be pretty complex (and not steady state!) in the real world. But some simple lab experiments would be interesting.

RJ
August 17, 2011 12:33 am

Willis
You are right. To be accurate, DLR means that the surface is warmer than if the DLR weren’t there. So you are technically correct, but in common parlance we don’t usually say “It slows the cooling so it ends up warmer than it would otherwise”. We just say “it warms it”.
But this statement completely contradicts your initial post
The sun heats the surface. At best DLR slightly slows the rate of cooling. To say it warms the surface is incorrect.
DLR can not heat the ocean or the land

August 17, 2011 12:36 am

I was going to ask a ton of questions, but I read through the comments and figured its not worth it in the end. Enough people were being rather feisty that I thought it might be better to just read and learn. I did not really understand the article completely (as in what was the reasoning for these 4 points….?) until I read the back-story, so those of you still posting might want to go back and read that, it helped me understand the reasoning going on here.
Another thing, I think this article was a little lower quality for only that reason. The explanations made me curious enough to read more, so don’t feel bad about this. Just remember that some of us are unaware of previous discussion on the subject and are coming into this subject with just the background information. It helps a ton if we have the counter-points to work with too up front. Why do people dispute this? I think that angle might have helped as well too.
Personally, I have not formed an opinion yet, its late, its time for bed and I want to read through the material one more time. Physics late at night is asking for nightmares!
———————————-
“Dumb question
How does rivers figure in all this?Rivers are always cold are they not?Is that because of their depth ? ”
——————————————-
What river are you talking about? I live in Missouri obviously and I can tell you some rivers are rather warm and tend to match the air temperature to a large extent, while others stay around the same temperature year round…the physics is not really relevant with rivers as they flow quicky enough and are a small enough body that their temperature is mostly depended on the source of the water. The physics is more important in the ocean interaction since that is such a large portion of our Earth, and although rivers react very similar to the oceans (larger rivers that is) the effect is so miniscule that its not even worth noting inside of an energy budget. You are better off just calculating the oceans and estimating the difference between fresh and salt water then anything else.
The factors in the temperature of rivers include snow melt *percentage of river’s source, spring fed or not, and how much of the river is spring fed and where is the source – distance and time taken make a huge difference in River Temperature, and of course ambient temperature. That is quite a mouthful, but its a rather small part of the energy budget being discussed. Its rather not important for the Earth’s climate as a whole. But for individual eco-systems, the river systems can make quite a difference.

RJ
August 17, 2011 12:37 am

Willis
“Nope, not possible, not unless you have an explanation for what keeps the ocean liquid. We know the ocean is losing 400 w/m2 of energy through radiation. We know it’s only getting about 170 w/m2 from the sun … what makes up the difference?”
Maybe your figures are just warmist fiction. LWR radiation does not heat the planet. Its time to move on from this nonsense.

tallbloke
August 17, 2011 12:41 am

Willis Eschenbach says:
August 16, 2011 at 8:05 pm
We know the ocean is losing 400 w/m2 of energy through radiation. We know it’s only getting about 170 w/m2 from the sun … what makes up the difference?

DLR makes up the difference Willis.
>I say DLR, there’s no other energy source around to do it. So no, it’s not possible that the warming is negligible.
This is known as non-sequiteur. The second part doesn’t logically follow from the first. DLR is absorbed in the first few nm of the ocean surface. Because a large number of Watts of energy get concentrated into such a thin layer, significant evaporation takes place. Evaporation is a process which soaks up energy and leaves the surface of the body it has evaporated from cooler rather than warmer.
Absorbing energy doesn’t necessarily result in things getting warmer. Especially if the thing is a body of water. Rocks are different of course.
In the case of the DLR being absorbed in the ocean surface, we have numbers coming out of Trenberth’s models, and numbers coming out of NASA’s real empirical measurements which show convective processes including evaporation to be more significant than they are in the Trenberth model, as Dave Springer has elucidated.
Normally, I’d expect Willis to go with real numbers derived from real empirical tests. In this case however, he prefers the numbers coming out of Trenberth’s model.
Very strange.

RJ
August 17, 2011 12:45 am

Tinthetoolman
So does this mean DLR heats the ocean? Yes, but not directly. It is the sun that heated the ocean. The DLR ONLY slows that energy from escaping.
Not yes. the correct answer is no. LWR does NOT heat either the land or ocean. It may reduce the rate of cooling by a fraction.
The sun and only the sun heats the surface
“Argument 4 “Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean.”
Is also close to being correct in the sense ”
It is not correct. You say the heating is caused by the sun. So if this is correct (and it is) then this statement can not be correct

Bob_FJ
August 17, 2011 12:49 am

Willis Eschenbach @ August 16, 2011 at 6:49 pm

“…I understand that EMR is not heat. It is energy. EMR can heat things that it hits, that’s why we have microwave ovens. Not sure what your point is. You post a quote of mine that doesn’t contain the word “heat” in any form, and claim I don’t understand that EMR is not heat … of course it’s not heat, it’s EMR.

Willis, the problem is that although you may not have mentioned the word HEAT, you have seemingly generally been treating EMR as if it were HEAT, whereas they are actually two different forms of energy with clearly defined different behaviours. Maybe the confusion arises from them both generally being measured in the same units, (they need not be), and from a good few climate scientists apparently not understanding (or “denying“?) the differences.
When you speak of warming, presumably you mean that the temperature of the affected matter has increased as a consequence of heating. Temperature of matter can only be increased by the (non quantum transient) addition of HEAT. (and of course vice versa for cooling). Whilst it is true that EMR can be converted to HEAT, EMR itself cannot be added to matter. What can happen is that certain molecules in matter can be excited by absorption of the photon stream duality in EMR at some wavelengths. When this happens, EMR is converted to HEAT. (it is no longer EMR). There is also other quantum stuff, such as molecular collisions, and other forms of HEAT transfer, but let’s keep it simple.
ALL types of energy transfer require a Potential Difference, including simple stuff like hydraulic energy, where H1 – H2 gives a PD as the difference in height of the liquid, H2 being the gravitationally lower.
In the case of radiative heat transfer, the PD is generally considered as the difference in power of two opposing sources, OR as proportional to T1^4 – T2^4. (in a naive two-body case)
As I pointed out earlier, it doesn’t matter how much EMR is flying around; unless there is a PD, there is ZERO HEAT transfer. This is nicely demonstrated in any elemental layer of air in a discreet parcel, wherein most of the radiation as a consequence of its temperature is horizontally opposed in all directions. Consequence = nothing.
The 400 W/m^2 that you imply as HEAT leaving the surface, is NOT HEAT, but a different form of energy described as EMR, which does not in itself cool the surface.
Do you have any problems with this NASA Energy Budget Diagram?
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif

Ronaldo
August 17, 2011 1:07 am

Perhaps we can just consider this in simple terms – it suits my simple mind!
During the day sunlight warms the sea, which has an enormous capability to store this heat via the specific heat of water and its mass. At night the heating from the sun stops. At this point, the sea will be radiating energy at a rate controlled by its surface temperature and also losing some heat by convection and evaportion. Immediately above its surface is an atmosphere whose vapour content (ie vapour pressure) depends on the temperature of the sea surface and which will be radiating energy towards the sea at a rate determined by its temperature(DLR) and also losing energy by radiation and conduction upwards.
There is no source of heat to provide additional energy, so that the sea will be losing energy at a rate determined by the properties of this atmosphere and will cool. However the sea has a vast store of energy held at a temperature determined by its mass and its specific heat. The cooler water will sink and be replaced by the slightly lower sub-surface temperature in the process, (overturning). In this respect, the DLR from the atmosphere close to the sea surface returns some of the radiated energy and slows the rate of heat loss and thus the rate of sea-surface temperature loss.
Next morning, the sun driven heating starts all over again. The same processes of DLR, evaporation,conduction and convection occur, but in this case the downward HF from the sun dominates theheating process.
This process allows the sea to lose energy at a rate determined by its surface temperature but to maintain a relatively high bulk temperature. It is essentially a “flywheel effect” My slight acquaintance with shallow lakes, where the mass of water is small relative to its surface area suggests that the “flywheel” is much smaller also, i.e. they cool relatively rapidly at night.
In essence the near surface atmosphere is acting to provide the same observable effect as an insulating blanket performs on an initially hot, non heat generating body, after the source of heat has been removed eg a hot water bottle placed in a bed, although the principle method of heat loss from the bottle is by conduction.
If the above explanation is acceptable, it demonstrates, yet again, the remarkable combination of physical properties of H2O, to which we mortals owe a rather large debt of gratitude!

Myrrh
August 17, 2011 1:09 am

kuhnkat says:
August 16, 2011 at 8:02 pm
Myrrh,
“The ‘missing heat’ is Thermal Infrared direct from the Sun…”
In case no one else has answered, that was one of my issues until I found the data. The earth and sun radiation levels cross in the 4-5 micron range. The near IR from 4 microns down is actually included in the visible absorbed numbers I believe. That is where the sun substantially exceeds the earth. From 5 up it is very small and absorbed in the atmosphere. Of course, if they don’t… I still haven’t had anyone explicitly state what the IPCC uses. Probably will have to read the one section to find out for sure. I don’t really believe they are trying to ignore almost 50% of the sun’s output. That is just a bit more than the DLR.

You should have stuck with it being an issue – the data you found is obviously from the AGWScience ficition propaganda department.
Bob_FJ says:
August 16, 2011 at 6:27 pm
Myrrh @ August 16, 2011 at 5:02 pm
“…Downwelling Solar 170 W/m^2 is not heating the Earth. Visible Light, UV and Nr IR are not thermal energies – they cannot be heating the land and oceans to raise the temperature by the amount claimed nor can they therefore create that much Infrared upwelling from the Earth…”
Sorry Myrrh, but the term “thermal radiation” is sometimes confused and misused. It is really a misnomer for EMR, (electromagnetic radiation) which includes visible light which does indeed heat matter when its photons are absorbed. Try here at Wikipedia:
http://en.wikipedia.org/wiki/Thermal_radiation

Do you know how much thermal infrared is given off by the incandescent bulb it gives as an example in the second parargraph? 95%
The 5% visible light doesn’t heat anything.
The first paragraph is what is screwing with your heads in this. “All matter with a temperature greater than absolute zero emits thermal radiation.”
I don’t put my dinner in the fridge to cook. Near infrared is not thermal radiation.
I don’t know how much clearer I can make this – I’ve given the tradition physics and given a page where NASA used to teach this, to children, that the heat we feel from the Sun is thermal infrared and that near infrared isn’t hot at all, it’s not a thermal energy. This is basic primary school level physics. And, I’ve given a page where they are spouting another variation from the AGWScience fiction propaganda department that infrared doesn’t even reach the Earth!
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies, that means they do not heat things, that means they can’t heat land and oceans. That means the ‘energy’ imput from the Sun from these wavelengths has nothing to do with heating the Earth, they have to be taken out of your calculations.
That’s why everything being argued about here is unadulterated gobbledegook! You don’t know how insane you all sound saying Light heats the Earth. You’ve been brainwashed.
The italics should stop end of para 4:
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-720340
I’ve put the comparison out again in http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-720489
There are countless real world applications in use created out of the knowing there is a difference between Light energies and Heat energies. As the NASA page for children used to teach, thermal infrared is used in cooking food. You can’t cook food by shining Blue Visible Light on it!
From the real world example:

http://www.ehow.com/about_6529090_definition-infrared-heating.html
Definition of Infrared Heating
By Jacquelyn Jeanty, eHow Contributor
The sun acts as the earth’s primary source for infrared heat, where energy is transferred from the sun to any material it contacts. Heating methods can vary depending on how heat is transferred to a material and the way it distributes throughout the material. Infrared heating is transferred through thermal energy and uses conduction as its distribution channel.
Infrared Heating
The structure of atoms includes a nucleus an its orbiting electrons. When a force causes these electrons to move faster, energy is generated. According to NASA, infrared heat is made up of energy waves that sit within the far-infrared portion of the electromagnetic spectrum. These waves transfer thermal heat upon making contact with a surface, or material. When heat is transferred, individual atoms take on this energy causing electrons to move faster within their orbits. This results in a chain reaction where energy is transferred, or conducted from atom to atom, which is how heat distributes through the material.
Function
The electromagnetic spectrum is made up of different levels of energy that appear in different forms within the physical world, according to Solaire Infrared Grilling Systems. Energy travels in wavelengths that can vary in length and frequency. The slower frequency waves appear as radio wavelengths, while the fastest waves appear as gamma rays. Wavelengths that sit in the center of the spectrum appear as colors. Infrared wavelengths move just a bit slower than the waves that generate colors and appear as radiant heat energy within the physical world.
Heat Absorption
Since infrared heating works through a radiant effect, much of the heat transfer that takes place inside a material relies on how quickly heat energy is transferred from atom to atom, according to the Casso-Solar Corporation. This transfer of heat energy is known as conductio, and is another method used to heat materials. The type of material heated will affect how quickly heat transfers throughout. A material’s density and thickness can also influence how heat energy is distributed.
Heat Transfer
As with any heating method, the rate in which infrared heat transfers through a material can vary if the material is made up of varying layers and textures, according to Casso-Solar Corporation. A material treated with additives may also heat at a different rate than an untreated material. An example of this would be how large pieces of meat may appear cooked throughout except in areas where the meat attaches to bone. The bone constitutes a different texture and layer that slows the rate in which infrared energy transfers throughout.
Effects
Infrared heating makes contact with a material’s surface then works its way inward. As a result, flat materials or objects receive an even exposure of infrared energy; however, a more three-dimensional object may have some areas more exposed than others, according to Thermovation Equipment. In this case, a rotisserie-type device can increase the amount of surface area exposed. The most efficient heating effects are had within an air-tight environment or oven, so a well-constructed oven can produce the best results.
References
Solaire Infrared Grilling Systems: What Is Infrared?
Casso-Solar Corporation: Introduction to Infrared Heating
Thermovation Equipment: Infrared Ovens
National Aeronautics and Space Administration: Infrared Waves
Read more: Definition of Infrared Heating | eHow.com http://www.ehow.com/about_6529090_definition-infrared-heating.html#ixzz1VGn9UWfn

And, I’ve given the history of that page from NASA she quotes – at the moment someone has managed to keep it up as a live link. There’s a fight going on over there..
All of you posting here who claim that the Solar of Visible and the two non-thermal either side heat the land and oceans NEED to prove it. This is a science blog. I am getting fed up with asking for proof and being ignored.
If you just continue discussing this as if my post doesn’t matter then you’re just a bunch of dreamers arguing about the fictional worlds you’re creating here. You’re not scientists in any bloody shape or form.
I repeat:
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies, that means they do not heat things, that means they can’t heat land and oceans. That means the ‘energy’ imput from the Sun from these wavelengths has nothing to do with heating the Earth, they have to be taken out of your calculations.
If you just continue discussing this as if my post doesn’t matter then you’re just a bunch of dreamers arguing about the fictional worlds you’re creating here. You’re not scientists in any bloody shape or form.

Bob_FJ
August 17, 2011 1:18 am

Tim Folkerts @ August 16, 2011 at 7:09 pm

“…A better analogy would be two separate circuits, one sending electrons counterclockwise around one wire while the other circuit sends electrons clockwise around a parallel wire. The “net current” will be one current minus the other, but electrons really would be flowing both directions.”

Well as I indicated, I don’t like doing analogies because none can be perfect. I’m not comfortable with your “better analogy”. Please explain more what you mean by a ‘net current’ from two separated electrical circuits.

kadaka (KD Knoebel)
August 17, 2011 1:20 am

Noelene said on August 15, 2011 at 10:01 pm:

Dumb question
How does rivers figure in all this?Rivers are always cold are they not?Is that because of their depth?

Answer 1: Air is moving relative to a liquid-containing surface, evaporation happens, wind chill. It’s the water that’s moving but the cooling is still there, with rushing water getting more cooling than a slow-moving river.
Answer 2: The river will feel cool when you get in it, provided it’s cooler than your body temperature, due to the greater heat transfer rate versus air. Your body loses heat faster in water thus it feels cool. (Note: As I was warned long ago in first aid and swim and assorted science classes, you can get hypothermia in 80°F water, as you’ll be loosing heat, it just takes a long while.)
Answer 2a: When standing in moving water (the water is moving around you) the loss of heat is faster than in still water. In still water the water around you warms up a bit, reducing the temperature difference thus slowing the rate of loss. With moving water you don’t get that, rate of loss is higher, water feels cooler than still water.
Sound good to you?

August 17, 2011 1:36 am

Willis said:
“the actual skin surface of the ocean is almost always slightly cooler than the water immediately below. This is because the surface is cooled by evaporation, conduction to the atmosphere, and radiation.
As a result, the skin almost always runs a bit cooler than the water underneath it, with the predictable result—the very surface skin water is always radiating, cooling and sinking a mm or so, then warming from the warm waters below, and rising again. The very surface is constantly being replaced by slightly warmer water from underneath in a very thin skin-based circulation layer. In this way the heat of the ocean makes it to the surface to be evaporated away.
With this constant interchange, with water surfacing, radiating, sinking a mm or so, warming, and rising again to the skin of the ocean, the DLR striking the surface has the same effect it has at night. It slows the thermal circulation, this time the thin vertical circulation at the very surface. Again it slows the motion of the bulk heat to the surface to cool. As a result, the bulk ocean is warmer than it would be without the DLR.”
I don’t agree with that interpretation.
The 1mm deep cooler ocean skin acts as a buffer between the effects of incoming DLR on the topmost molecules and the flow of energy coming up out of the bulk ocean and into the skin from upward convection and conduction (originally from solar input).
All the ‘circulating’ is therefore contained within the skin and does not affect the rate of energy flow from the bulk below and nor does it allow energy to move downward into the bulk. It is an entirely separate process from ‘normal’ ocean mixing.
The interface between the skin and the bulk is therefore the point at which the DLR effect reduces to zero. The depth of that interface and the temperature differential across it is dictated by the flow balance between the layers that is required to negate the DLR effect.
More DLR actually increases the speed of the process within the skin to prevent any effect on the flow rate up from the bulk. The process is self limiting because once all the DLR has been used up there is no more acceleration of the process.
Given that evaporation mops up five times as much energy as is required to provoke it the idea that there is any energy left over to go into the ocean bulk must be wrong.
So DLR does not warm the oceans nor cool the oceans. The energy transfer process from those warmed topmost molecules increases up through the atmosphere to space via more upward radiation convection conduction and evaporation in a self cancelling exercise because of the buffering effect within the ocean skin.
The consequence is a miniscule adjustment in the surface air pressure distribution as the atmosphere adapts to the changed rate of upward energy flow.
GHGs slow down energy loss from atmosphere to space but the whole package of available negative response mechanisms (with the phase change from water to vapour as an essential component) then accelerates it again for a zero or near zero net effect. The ocean bulk remains unaffected.

August 17, 2011 1:38 am

Willis, I’m going to think about it some more but after some thought your steel greenhouse idea is wrong, wrong, wrong. The shell and the surface are at the same temperature in deg K if the difference in radii are small.
BTW we could run this as a lab experiment.
This doesn’t preclude an atmospheric “green house effect” as the effect of infrared absorbing gases is that a layer above the surface will warm and the lapse rate will make the surface warmer as convection occurs.
For those who think the DLR warms water what’s wrong with the incoming short wave to cause the observed surface warming?

tallbloke
August 17, 2011 2:03 am

Willis Eschenbach says:
August 17, 2011 at 12:53 am
You can call it what you want, warming, or slowing the cooling, I don’t understand the difference.

There’s a useful discussion getting underway between Tim and Willis here and I agree with much of what both of them are saying. A lot of confusion has arisen in the past and in this thread because of the conflation of ‘warming’ and ‘slowing the cooling’. Willis thinks that there is no difference and that it’s fine to use ‘common parlance’, I think there is an important difference for the way we conceive of the greenhouse effect, and that ‘common parlance’ should give way to technical definition where it sows confusion otherwise.
The simplest way I can think of to illustrate the difference is to say that ‘warming’ is done by things which are warmer then the things they are warming, whereas ‘slowing the cooling’ is done by things which are cooler than the things they are insulating. In the case of radiative balance we are talking about the surfaces involved rather than the properties of the underlying bulks. The ocean side of the ocean air interface is mostly warmer than the air side.
Earlier in the discussion, Willis attempted to refute my argument that conduction from the surface downwards isn’t effective by saying:
IR heats the top molecule. It passes some reduced amount of that heat to the molecule below. But what tallbloke forgets is that the top molecule can’t make the second molecule warmer than the top molecule, heat doesn’t flow from cooler to warmer.
Since the second molecule is not as warm as the top molecule, in contradiction to tallbloke’s claim, it doesn’t rise to the top. And the same for the layers further down. The heat is transmitted down and down, but each layer can’t heat the lower layer more than itself, heat won’t flow uphill. So the water, though warming, doesn’t “rise to the top” as claimed.

But now Willis says to Tim:
As you point out, the actual skin surface of the ocean is almost always slightly cooler than the water immediately below. This is because the surface is cooled by evaporation, conduction to the atmosphere, and radiation.
As a result, the skin almost always runs a bit cooler than the water underneath it, with the predictable result—the very surface skin water is always radiating, cooling and sinking a mm or so, then warming from the warm waters below, and rising again.

Clearly the discussion is becoming more sophisticated, which is good, because it’s only when we look at the actual physical processes that it becomes clear that while the ocean does indeed absorb the DLR, it doesn’t do much to ‘warm’ the bulk of the ocean. However it does indeed ‘slow the rate of heat loss’ and this means the bulk of the ocean is warmer than it would otherwise be. However, plugging in the empirically derived numbers makes it clear that it doesn’t make it cool a lot slower, and a small increase in DLR relative to ULR caused by increased Co2 wouldn’t make enough difference to account for late C20th warming.
It’s more likely the increased insolation through (empirically measured) reduced tropical low cloud cover which did that.

August 17, 2011 2:34 am

“However it does indeed ‘slow the rate of heat loss’ ”
Not so sure of that. See my above post at
Stephen Wilde says:
August 17, 2011 at 1:36 am
“It’s more likely the increased insolation through empirically measured reduced tropical cloud which did that.”
Yes but I’d go further and suggest global cloud cover variations from the shifting of the jets along with all the other air circulation systems which respond both to top down solar variability and bottom up oceanic variability.
Willis’s thermostat effect is correct but needs to be extended globally.

Bob_FJ
August 17, 2011 2:43 am

Myrrh @ August 17, 2011 at 1:09 am
WRT your comments on:
http://en.wikipedia.org/wiki/Thermal_radiation

Do you [Bob_FJ] know how much thermal infrared is given off by the incandescent bulb it gives as an example in the second parargraph? 95%
The 5% visible light doesn’t heat anything…”

Sorry Myrrh, this is what it actually says, and I‘ve added emphasis to the words ‘visible light‘:

“…Examples of thermal radiation include visible light emitted by an incandescent light bulb, infrared radiation emitted by animals and detectable with an infrared camera, and the cosmic microwave background radiation. Thermal radiation is different from thermal convection and thermal conduction–a person near a raging bonfire feels radiant heating from the fire, even if the surrounding air is very cold…”

You need to go to the original link to access eight embedded links.
Here follows a link to the solar spectrum that falls upon Earth, and which is universally agreed to result in heating of the Earth. You assert a different view. Would you care to draw a vertical line therein that defines each side what does and what does not heat the Earth?
http://en.wikipedia.org/wiki/File:Solar_Spectrum.png

Alexander Duranko
August 17, 2011 2:58 am

Sorry Willis, but DLR is only half local radiative heat transfer. What matters is ‘DLR-ULR’.
DLR on its own is a measure of IR impedance. If an atmosphere had no greenhouse gases, it would have near zero IR impedance; the optical depth would fall to near zero because there would be no optical scattering [absorption, re-emission by other GHG molecules in local thermodynamic equilibrium – remember there may be no thermalisation of IR energy]. So, you’d need low ground temperature to radiate a given amount of energy to space.
This analogy is quite close to what happens in metals when you alloy them. The difference in the way the solid solution atoms’ d-shell electrons interact with the conduction band causes the electrons to scatter thus increasing path length and resistance/impedance so you need a higher potential difference to pass a given electric current for an alloy than a pure metal.
The reason why clouds have much higher DLR than moist air is because they have very high IR impedance. The droplets have dissolved lots of the local CO2. Water and CO2 band IR energy is strongly absorbed and scattered. Because the absorptivity/emissivity is high [c. 1 compared with c. 0.1 for moist air] you get a lot of localised DLR. Because a higher proportion of the IR energy being emitted by the ground/sea is needed to offset that higher DLR, you lose less heat from the ground until its temperature rises. That’s why at night, the sudden appearance of a cloud causes local warming.
But the DLR doesn’t do any actual work. Less heat is lost from the ground until the sum of radiation and convection from the ground and the conduction in the ground change exponentially to a new local equilibrium.
Here’s another view: you’re on the beach and air temperature is 25°C but because the wind speed is high, lots of heat is lost by convection so the sand is only at say 30°C when without convection it would be c.70°C. So, you put up a wind break, sand temperature rises to 45°C and you get pleasantly warm lying on it.
Because [assuming 0.85 emissivity] radiative heat transfer has increased by 86W/m^2 to give constant convection plus radiation, assuming half that extra IR energy is absorbed/scattered by the atmosphere back to the ground and everyone else has put up windbreaks [imagine the Sahara] DLR will rise by 43W/m^2. But that is just the result of the IR impedance causing higher radiative potential for a given power transfer. There is no new energy input to the ground.
DLR can do no work. DLR is a measure of IR impedance. Assess the problems as coupled convection plus radiation.
Answer to George E Smith. local thermodynamic equilibrium exists everywhere and accepting the exponential transients, most of the time emissivity is approximately equal to absorptivity.

August 17, 2011 3:02 am

Willis writes : “Oh, please, I’m not “pleading” for anything”
I could count the number of times you asked people to refute your points. It was more than once…but meh. waste of time.
Willis writes : “You agree, for starters, that DLR a) is absorbed by the ocean”
If by “ocean” you mean the top 10um of ocean’s water molecules, then yes. However you follow up with “What I don’t understand why “conduction cannot happen”.” because you want the conduction to be downward into the bulk and it just doesn’t happen. The location where the DLR is absorbed is colder than the underlying water. It would break the laws of thermodynamics for any “heat” to conduct downwards. Not to mention the fact there is no excess here anyway. It is, as I said, the coldest place.
Willis writes : “As a result, the skin almost always runs a bit cooler than the water underneath it, with the predictable result—the very surface skin water is always radiating, cooling and sinking a mm or so, then warming from the warm waters below, and rising again.”
Get thee back to school Willis 😉
In the skin layer, its conduction not convection that transmits energy.
eg From “Cool-skin simulation by a one-column ocean model – Chia-Ying Tu and Ben-Jei Tsuang”
“Molecular transport is the only mechanism for the vertical diffusion of heat and momentum in the cool skin and viscous layer”
Willis writes : “During the day, sunlight striking the ocean is absorbed most at the surface.”
I’m not quite sure what you’re saying here. There is a temperature profile that DSR creates as its absorbed in water. Most is absorbed not far below the surface and progressively less makes it deeper but its over meters not really “at the surface”. Or at least thats probably not how I’d descibe it in the context of this conversation.
Willis continues : “If the DLR is warming the skin itself, that absorbed solar energy can’t be moved as easily to the skin and radiated/evaporated away.”
Using what physics do you claim this? If there is excess energy being imparted from the sun, the warm water will convect toward the surface and produce/increase the “hook”. The SST increases, Stefan-Boltzmann kicks in and the rate of radiation increases. More is lost to space and to evaporation and the balance is maintained.
Moving along to the second argument (its not worth addressing the warming of the land part. I think you’re just being argumentative)
“I would agree except for the “immediately”, it suggests that it is re-radiated. It is not re-radiated in any sense.”
Re-radiated as a term is wishy washy, I agree. But I think you know what I mean, the DLR is radiated back up pretty much at the same rate its radiated down and I’m not saying anything about individual photons and where they came from or anything like that.
Over at the Science of Doom, SoD calculated that there is about 42J/m2 heat capacity in the topmost 10um layer of the ocean. We know from the Minnett experiment that clouds can increase the DLR by around 100W/m2. We also know from the Minnett experiment that the surface temperature (relative to the 5cm below surface temparature) changed by around 0.5C at most and even though I disagree with their reasoning, it does put an upper limit on any possible “DLR heats the skin” effect.
And using those values that upper limit is reached in a little less than 0.25 seconds to potentially “heat” the surface. If you believe thats what happens then you need to explain where this energy goes after 0.25 seconds and you need to do it within the laws of thermodynamics. Hence its not conducted down. Its not radiated down. Its not convected down (unless you want to try to make proper arguments for any of those) and instead it must be radiated upwards.
Anyway moving right along and this post is already way too long.
Willis writes : “You can call it what you want, warming, or slowing the cooling, I don’t understand the difference.”
I will indeed call it “slowing the cooling” because thats what it is. I think its important to actually understand the processes behind the numbers and I know thats how you feel too. You wouldn’t have bothered with your recent posts on thunderstorms if you’d felt their effect was adequately wrapped up in the coarse averages the AGWers use.

August 17, 2011 4:12 am

As much as I respect and admire you, Willis, I think you’re off base here. No one disputes this: if there was an independently-powered IR heater floating in the air, it would add heat to the water below it. The question is, can water emit heat energy to heat water vapor in the air…then after a time delay, have this energy come back and make the emitting water hotter than it was? You have to be careful not to double-count the energy. If you calculate emission based on a temperature and this temperature already includes the effect of returned energy, then you can’t count it again later…the “when” of emission and reception matters. If I had $100 yesterday and I still have $100 today, at no time can I say I have $200.

tallbloke
August 17, 2011 4:21 am

Stephen Wilde says:
August 17, 2011 at 2:34 am
“However it does indeed ‘slow the rate of heat loss’ ”
Not so sure of that. See my above post at
Stephen Wilde says:
August 17, 2011 at 1:36 am

Thanks Stephen. I’m ready to meet for an in depth discussion. I’m away 14th – 25th sept.

Martin Lewitt
August 17, 2011 4:35 am

@Willis,
“And I still haven’t heard you or anyone else explain why the ocean is liquid, what mysterious energy source you claim keeps the ocean from freezing solid”
At the earth’s surface albedo of about 0.13, the average temperature of the surface is only about 4 degreesC below zero. At the ocean’s albedo of 0.06 or 0.07 the average temperature of the surface is at or above freezing. So in the simplest case the oceans would be liquid except at high latitudes.
This is the calculation at an albedo of 0.07
http://tinyurl.com/3qjxrg2
So as a starting point, before adding clouds, GHGs, diurnal variations and other complexities, most of the surface of the oceans is above the freezing point of water. It is a common error in the climate community to use the earths planetary albedo of about 0.3, which includes clouds when assessing what is needed from the greenhouse gas effect. But since the surface albedo is lower and the ocean albedo lower still, you don’t need the GHGs until you have the clouds. Even in snowball earth scenerios, there is thought to be open ocean in the tropics.

Spector
August 17, 2011 4:36 am

You should all have seen Planck’s law that says the energy of a photon is equal to Planck’s constant times the frequency, usually indicated with the Greek letter nu. This can also be stated as Planck’s constant times the speed of light then divided by the wavelength. This formula indicates the heating potential of a single photon.
It should be obvious that a peak-output, green solar photon with a wavelength of 0.5 microns has 30 times the energy of a 15-micron photon emitted by a CO2 molecule. Anyone can look at a solar radiation spectrum and see that most of the energy is in the visible range. (One micron is equal to 1000 NM or nanometers) That is solar heating power.
The Earth, being so far away from the sun does not have to rise to the same temperature as the sun to reach an equilibrium point. It can export all the heat energy it receives from the sun using the less-energetic, infrared, photons characteristic of normal terrestrial temperatures.

David A
August 17, 2011 5:12 am

tallbloke says:
August 17, 2011 at 2:03 am
“Clearly the discussion is becoming more sophisticated, which is good, because it’s only when we look at the actual physical processes that it becomes clear that while the ocean does indeed absorb the DLR, it doesn’t do much to ‘warm’ the bulk of the ocean. However it does indeed ‘slow the rate of heat loss’ and this means the bulk of the ocean is warmer than it would otherwise be. However, plugging in the empirically derived numbers makes it clear that it doesn’t make it cool a lot slower, and a small increase in DLR relative to ULR caused by increased Co2 wouldn’t make enough difference to account for late C20th warming.”
Indeed, as I said in my post here, David says:
August 15, 2011 at 1:08 pm, I have seen none of this quantified. I lean to the idea that the residence time of SWR entering the oceans is much greater then the residence time of LWR energy entering the oceans, therfore a small change in the radiative balance via SWR flux, due to either the sun and or cloud cover, would, over time, produce a far greater effect on the earths ocean, atmosphere energy budget then a similar flux in DWLR.

Alexander Duranko
August 17, 2011 5:14 am

Martin Lewitt; I’ve done the same calculation. It completely disproves the IPCC claim that present greenhouse gas warming is 33K. That assumes in an IR transparent atmosphere, the -18°C presently in the upper troposphere would move to the earth’s surface.
However, all it does is to estimate the temperature difference between that part of the atmosphere presently in radiative equilibrium with space, a function of emissivity/pressure, and the Earth’s surface; is effect an estimate of lapse rate [5km for -6.5K/km].
In reality, if you took out the greenhouse gases and clouds, the albedo would fall to the no-ice level of 0.07 and the power input would increase. A better estimate would account for aerosols so I believe current net greenhouse warming is c. 10K.
The fact that the IPCC scales to 33K shows how desperately bad is basic physics in climate science. Yet they’ve all got a good education. Mustn’t be too critical though because the DLR argument means the subject is subtle and it takes very good physics to get to the truth.
PS If DLR was an energy source, my beach windbreak argument would mean we could control the temperature of the planet by cutting all vegetation down and by only building underground thereby maximising convective heat transfer from the ground to the air. Oh, sorry, I’ve just invented the reverse UHI effect, make everywhere no tree rural!

Matt G
August 17, 2011 5:37 am

Willis,
The 170w w/m2 and 390 w/m2 are like comparing apples with oranges, they both warm the surface differently and therefore can’t be claimed to be equal energy quantities. This is where you have gone wrong and this is shown by the observed NASA radiation buget. The outgoing radiation buget is lower than claimed by observed compared with the model example represented.
Looking at this at a different way, if the two energy values were equal, then it would be expected that DLWR would warm the surface greater than solar energy. Care to show any real life observation where this shows to be true and is reflected generally all over the planet. Real life observations show when the sun is warming during the day the result are the hottest temperatures possible in this air mass. When it clouds over and DLWR increases the atmospheric temperatures drop during daylight. This would not occur if the claimed energy values were comparing apples with apples. I think you need to consider this and reanalyse a mechanism that doesn’t make sense.

tallbloke
August 17, 2011 5:41 am

Alexander Duranko says:
August 17, 2011 at 5:14 am
In reality, if you took out the greenhouse gases and clouds, the albedo would fall to the no-ice level of 0.07 and the power input would increase. A better estimate would account for aerosols so I believe current net greenhouse warming is c. 10K.

There seem to be a number of people heading to this conclusion at the moment. One being astrophysicist Joseph Postma:
http://www.tech-know.eu/uploads/The_Model_Atmosphere.pdf
http://judithcurry.com/2011/08/16/postma-on-the-greenhouse-effect/
It looks like he’s right that smearing the insolation instantaneously over the day and night side of the globe using the S-B equation is incorrect, unphysical, and leads to a bad result. I’m not sure all his paper is correct, but I agree with him on this bit.

Alexander Duranko
August 17, 2011 6:14 am

One comment on the Curry blog is ‘Gradually the lapse rate should almost vanish in absence of all IR absorption and emission’
This serious lack of basic physics’ knowledge** permeates climate science. I’m not a physicist so I have to check thoroughly. Many do not check so it seems they have been badly educated, or perhaps programmed.
**The Wikipedia article on lapse rate is quite good: http://en.wikipedia.org/wiki/Lapse_rate

August 17, 2011 6:16 am

Alexander Duranko says:
But the DLR doesn’t do any actual work.
Exactly. This is why I have pointed out a number of times that there are no commerical products that use this “heat producing” phenomena.

tallbloke
August 17, 2011 7:02 am

mkelly says:
August 17, 2011 at 6:16 am
Alexander Duranko says:
But the DLR doesn’t do any actual work.
Exactly. This is why I have pointed out a number of times that there are no commerical products that use this “heat producing” phenomena.

Because it makes no sense to consider DLR separately from the ULR and the sideways LR within the LR flux.
You getting the message yet Willis?

Alexander Duranko
August 17, 2011 7:03 am

mkelly: the problem with DLR is that it was predicted by Milne in 1922 from a mathematical mistake so when the poor climate experimentalists measured it with their pyrgeometers, not having good enough physics, they thought is was real.
At the same time Hansen had worked out the amplification of TSI rise at the end of an ice age and concluded it had to be CO2, when there was another possible explanation, reduction of cloud albedo.
35 years’ later, climate science central is still flogging high feedback CO2-AGW and claiming the recent plateau in temperature is because the Chinese have been burning more coal and there’s a magical increase of heat to the ocean deeps. It’s bunkum because DLR isn’t an energy source, simply a signal set by temperature, in turn a function of heat flux, and emissivity.
Lindzen says the IPCC predicts between 2 and 5 times the warming, assuming that came all from CO2-AGW, then uses just enough aerosol cooling, different for every model, so they can calibrate by hind-casting. Kiehl has honestly agreed there’s no proof of any CO2-AGW beyond this uncertainty but hopes that as [CO2] rises, aerosol cooling the same, in time it’ll appear.
We’re dealing with a religious movement, not science. The way to defeat it is to remove all the axioms, one by one. There is no DSL Cloud albedo effect cooling is really heating. That physics’ change means that you have the mechanism explaining the end of an ice age. 33K claimed greenhouse heating is really lapse rate and real greenhouse warming is c. 10K.
The carbon traders are having a last ditch attempt to get their way; Cameron writing to support Gillard, Gore-bull’s day of climate action, but it won’t work. In reality, we face significant global cooling because [1] we’ve just been through a solar grand maximum at one point the highest for 11,000 years and [2] CO2-AGW will be a maximum of 0.2-0.3K by the end of this Century

anna v
August 17, 2011 7:11 am

Ken Coffman says:
August 17, 2011 at 4:12 am
If I had $100 yesterday and I still have $100 today, at no time can I say I have $200.
See the back radiation oven for a demonstration.

wobble
August 17, 2011 8:02 am

tallbloke says:
August 17, 2011 at 2:03 am
Clearly the discussion is becoming more sophisticated, which is good, because it’s only when we look at the actual physical processes that it becomes clear that while the ocean does indeed absorb the DLR, it doesn’t do much to ‘warm’ the bulk of the ocean. However it does indeed ‘slow the rate of heat loss’ and this means the bulk of the ocean is warmer than it would otherwise be.

OK, so it seems as if you agree that DLR does cause the ocean to be warmer than it would otherwise be without any DLR. I’m not sure if Dave Springer agrees with you or not.

However, plugging in the empirically derived numbers makes it clear that it doesn’t make it cool a lot slower, and a small increase in DLR relative to ULR caused by increased Co2 wouldn’t make enough difference to account for late C20th warming.

This is probably the direction that the debate should go.
tallbloke, I have a question for you.
Many have argued that DLR is absorbed within the first few micrometers of the ocean surface which causes evaporation which cools the surface. This has been offered as the reason that DLR can’t materially slow the cooling of the ocean.
However, in the absence of any DLR, the ocean would still evaporate. What would cause this evaporation? In the absence of DLR the evaporation process would extract heat from the air just above the ocean-air boundary and would extract heat from the water just below the ocean-air boundary. In the absence of DLR the evaporation process would therefore be extracting more heat from the water than otherwise gets extracted by a DLR fueled evaporation process.
So it seems as if, in the absence of DLR, it’s possible for the ocean to cool materially more quickly than it otherwise does (with DLR fueling the evaporation process instead of ocean heat).

richard verney
August 17, 2011 8:04 am

Willis,
I note that once again you do not answer my question as to why the ocean at the same latitude (62 degN) freezes at one longitude (19 deg 04 42 E) but not at another longitude (8 deg 45E). Is this because you do not have an answer that fits in with a simple radiative budget?
You quite incorrectly suggest that I have not explained why the oceans do not freeze. I have. The answer lies in the tropics. THE TROPICS IS THE POWER HOUSE OF OUR CLIMATE SYSTEM and understanding what goes on in the tropics is the key to understanding earth’s climate. Fundamental to this, is understanding the ocean in the tropics and the water cycle that takes place there.
You fall into error for a number of reasons such as:
First, you consider only the energy received per square metre. However, what is important is the energy received per metre cube of ocean and this is where the difference in penetrative absorption between solar energy/radiation and DWLWIR is important and comes into play. Solar energy penetrates well to a depth of 1o meteres and heats the ocean to this depth (obviously more heat is absorbed in the first 10s of centimetres and gradually less and less down to 10 metres and relatively trivial amounts below 10 metres). Solar energy can heat a metre cube, in fact it can effectively heat tripple that. This is of utmost importance because it is this volume of water which is being turned over and it is from this volume of water that heat gets transported down into the lower depths of the ocean thereby warming the lower ocean. In constrast, the DWLWIR, at most, is absorbed in the first few microns and boils this off. The energy from the DWLWIR is lost in the evaporation and resulting phase change such that it does not heat the ocean (witness that the very top few micron layer is slightly cooler than the layer below becayse the very top layer has had to give up heat when molecules from it evaporate). The important point is that DWLRIR merely increases the rate of evaporation. Since we know that the very top layer of the surface is slightly cooler than the layer below, even if this very top layer into which the DWLWIR can penetrate could be turned over, there is fact no ‘additional’ heat in this top layer to drag down to lower depths so as to heat the lower depths of the ocean. In fact, I doubt that the top few microns of the ocean could be turned over (but I stand to be corrected on that view).
Second, because as another poster pointed out, a watt is not just a watt. The photonic energy of the 5800k 170 w per sqm is a very different beast to the photonioc energy of the ~250K 390 w per sqm DWLWIR. We know this to be the case since we are able to extract useful work form the former but not from the latter. It is because the 390 w per sq m has no sensible energy in the context of the system in which it is engaged and has no ability to do sensible work that Trenberth and the Team are not multi – billionaires having patented a system to extract energy frrom the DWLWIR thereby solving the world energy problem. It is wrong to consider that Solar radiation and DWLWIR are the same and equally effective, they are not. Further, a watt of energy that has penetrated down to a depth of 2m or 4 or 10m is far more significant in the role of heating the deep oceans than is a watt which never makes it past the first few microns (and which in any event is probably boiled off at that stage). That latter watt does not get to heat the deep oceans.
Third, you consider that because the ocean radiates at the temperature of the surface skin, it is losing a lot of energy. In the overall scheme of things (by which I mean in relation to the total energy contained in the entire volume of the ocean), the amount of energy radiated is trivial and it is important to bear in mind that the so called ~390 – 400 w per sqm being seen as a signal is not heat loss. The ocean is not actually losing a lot of energy (in comparison to that contained in its bulk) and the amount of energy that it loses to the atmosphere is less than it receives from the sun. In this regard, do not forget that unlike land where air temperature and land temperature can be very different, over the deep oceans, ocean temperature and air temperature are very similar.
Fourth, you look at the notional average condition whereas in the context of the workings of the system the dominant condition is that ongoing at the TROPICS. It is here that one has to consider what is going on. As BenAW says:August 16, 2011 at 10:55 am “How come everybody is talking about AVERAGE radiation levels resulting in AVERAGE temps using the Stefan-Boltzmann formula when there is a fourth power in this formula? Earth has ONE sun, that radiates on half the earth. Average radiation on this half is 1364/2 = 682 W/m^2 ….” Ben is quite right (as I have repeatedly been saying to you), you will not understand what is going on if you look only at the notional average condition.
Fifth, the material point (and this follows on from the fourth point), is that at the tropics the ocean is not receiving just 170 w per sqm of solar energy. It is receiving many many times that amount. Even the 682 w per sqm noted by Ben is far less than the ocean is receiving in the tropics, particularly during the 4 hour period between 10 am and 2pm when the sun is near to and/or at its zenith. When one works out how much energy the tropical ocean receives from the sun during the day, it is significantly more than the energy being lost to the atmosphere over a 24 hour period through all ongoing processes (convection, evaporation and radiation) such that the tropical ocean will not freeze. Indeed, not simply will it not freeze, it has surplus energy which energy it is able to distribute around the globe (through the ocean conveyor belts) thereby (for the main part) making up the shortfall of energy which the ocean in higher latitudes receives thereby preventing those oceans from freezing.
Sixth, generally radiative energy yields to and does not overcome convection. This can be seen with a BBQ. The heat given off by the burning embers is radiative heat and is given off in all directions. If you were to point your IR thermoter at the BBQ there would be no difference in the IR temperature reading when looking down at the BBQ or when looking at it from its side. You can cook your food 6 to 12 inches above the charcoal/coals but you cannot cook your food 6 to 12 inces from the side of the BBQ. The reason is that notwithstanding that the source of the heat is radiative, the food is cooked by convection. Convection carries the heat (that was initially radiative heat) upwards towards the food placed above the charcoal/coals. If you place the food say 9 to 12 inches to the side of the BBQ, most of the radiative energy radiating sidewards is carried upawrds by convection such that there is all but no heat energy left 9 to 12 inches from the side of the BBQ. The same process is occuring above the ocean. The DWLWIR is trying its best to get down to the ocean surface and penetrate the water, however water vapour above the ocean impedes its path and the DWLWIR is continually being absorbed by that vapour and then that vapour (which has absorbed the DWLWIR) is being absorbed convected upwards and away from the top layer of the ocean. I envisage that less DWLWIR actually makes its way to the surface of the ocean than is presently assumed.
Seventh, the Trenbeth figures are fantasy. Essentialy they are a fiction of robbing Peter to pay Paul. In the real world the bulk of the energy loss from the oceans is not the radiative loss from the instaneous absoption/emission involved in the receipt of the DWLWIR and its re-radatiation, but rather from vaporation and convection. In practice as I and others have repeatedly observed, it is net flux that is important (but I guess that you are never going to be convinced of that although alarm bells should ring if substantially different results are achived by performing a calculation on so called gross figures and one on net flux figures).
In summary, Willis do your Solar/DLR energy budget calculation for the TROPICAL OCEAN. Whilst I would prefer that you base this on net flux, in practice I do not care whether you use your fantasy figures since even if you do use those figures, you will still find that THE TROPICAL OCEAN DOES NOT FREEZE. Come back to me once you have done that calculation.
PS. I do not understand why so much time in climate science is wasted on peripheral issues. Approximately 70% of the planet is covered by ocean. The ocean contains 99% of the heat capacity of the planet, and it is the oceans that drive the weather. Given that, nearly all study should be directed at studying and understanding the oceans and the water cycle. Of the oceans, it is the tropical ocean that powers everything. Accordingly, the one area to thoroughly understand is the tropical ocean and the water cycle. Study of this would force people away from darn averages that do nothing other than to obscure what is going on. The rest of the globe is of mild interest only, and not of fundamental importance (although as I have mentioned in previous posts I can conceive that DWLWIR may have some relatively modest role to play over land). Enough of that particular gripe.
PPS. Since I last looked in on this article there have nbeen many new posts. I need to do some catch up.

tallbloke
August 17, 2011 8:20 am

wobble says:
August 17, 2011 at 8:02 am
Many have argued that DLR is absorbed within the first few micrometers of the ocean surface which causes evaporation which cools the surface. This has been offered as the reason that DLR can’t materially slow the cooling of the ocean.

No. It’s partly why it can’t directly warm the ocean, along with the inability to conduct heat downwards due to the surface being colder then the subsurface, and the lack of a convective mechanism to do the job, and the fact that the net flux is cooling the ocean anyway.
The reason I provisionally accept the idea that DLR can slow the cooling of the ocean is that if we consider what happens when DLR increases relative to ULR, the net cooling caused by the direction of the 66W/m^2 net flux would be diminished. However, we have yet to see any empirical evidence that the DLR has increased relative to the ULR.
Stephen Wylde seems to be saying that any increasing differential gets automatically compensated for. I’m looking forward to discussing that with him in detail.

Alexander Duranko
August 17, 2011 8:38 am

DWLWIR isn’t an energy source. I agree about the short wave energy.
The dominant organism in the World is phytoplankton because it emits c. 10 times the aerosols of man. It has over 100s of millions of years adapted to control its environment by the peculiar chemistry producing dimethyl sulphide on decay, and because that has very low water solubility it goes into the air to be photo-hydrolysed to sulphuric acid aerosol.
This is what is responsible for the classic smell of the oceans. Without these aerosols, low level thick oceanic cloud albedo would be much greater and the oceans and the World would cool rapidly. Asian industrialisation added more sulphuric acid aerosols thereby causing the late 20th Century increase in ocean heat content which stopped in 2003.
It stopped because the warming due to polluted clouds runs out of steam when albedo gets near 0.5. When Asia reduces emissions, the effect will reverse and Trenberth’s ‘missing heat’ will become negative!
CO2-AGW could well be net zero.

wobble
August 17, 2011 8:53 am

tallbloke says:
August 17, 2011 at 8:20 am
we have yet to see any empirical evidence that the DLR has increased relative to the ULR.

Thanks for addressing my comment, and I understand your claim here.
Can you answer this question that I posted?

Many have argued that DLR is absorbed within the first few micrometers of the ocean surface which causes evaporation which cools the surface. This has been offered as the reason that DLR can’t materially slow the cooling of the ocean.
However, in the absence of any DLR, the ocean would still evaporate. What would cause this evaporation? In the absence of DLR the evaporation process would extract heat from the air just above the ocean-air boundary and would extract heat from the water just below the ocean-air boundary. In the absence of DLR the evaporation process would therefore be extracting more heat from the water than otherwise gets extracted by a DLR fueled evaporation process.
So it seems as if, in the absence of DLR, it’s possible for the ocean to cool materially more quickly than it otherwise does (with DLR fueling the evaporation process instead of ocean heat).

Isn’t it possible that the DLR fueled evaporation process displaces a mechanism that would materially cool the ocean more than the ocean is currently cooled?

richard verney
August 17, 2011 9:16 am

kuhnkat says:
August 16, 2011 at 6:49 pm
Willis, you can say it as many times as you want and it still is only a half truth.
I get about 275K for the average with the 170 in. Of course, this close averages may hide stuff, but, 275k isn’t frozen. Also of course, I may have screwed up trying stuff I barely understand even with the BB Calculator
/////////////////////////////////////////////////
Kuhnkat
Please will you post your calculation.
Your result reflects reasonably well the average temperature of the ocean

kadaka (KD Knoebel)
August 17, 2011 9:28 am

From Ken Coffman on August 17, 2011 at 4:12 am:

If I had $100 yesterday and I still have $100 today, at no time can I say I have $200.

Sure you can. Just be the sole owner of a bank.
1. Deposit the $100 in your bank.
2. Give yourself a $100 no-interest no-fee loan.
3. The loan is an asset on the bank books, so your bank has $100 from that.
4. You have the $100 in your wallet.
5. Add what’s in your wallet and your bank assets together, you have $200.
Alternate 5: $100 in your wallet, $100 deposited in the bank, you have $200.
5 with alternate 5: Actually you have $300.
You should study governmental accounting methods. For example, the US federal government has generated fantastic sums of money over the years by issuing itself US Treasury notes for Social Security tax surpluses so it can direct that money to the general budget. These notes are then easily repaid by the US Treasury printing itself many pretty pieces of paper, which can cost nothing as it only needs to be done “on paper.”

August 17, 2011 9:33 am

Alexander Duranko, richard verney, et al:
Thanks for pointing out that climate science incorrectly assumes gross radiative transfers from cold to hot and vice versa can be assumed in isolation, rather than the correct calculation using only net FLOW of heat.
This is a point repeatedly made by Claes Johnson, who explains very simply how this error came about here:
http://claesjohnson.blogspot.com/2011/08/how-to-fool-world-by-measuring-masive.html
http://claesjohnson.blogspot.com/2011/08/how-to-fool-yourself-with-pyrgeometer.html

tallbloke
August 17, 2011 10:26 am

wobble says:
August 17, 2011 at 8:53 am
tallbloke says:
August 17, 2011 at 8:20 am
we have yet to see any empirical evidence that the DLR has increased relative to the ULR.
Thanks for addressing my comment, and I understand your claim here.
Can you answer this question that I posted?

I don’t know how much of the evaporation is currently caused by DLR and how much of it is caused by solar derived energy, because there seems to be uncertainty about just exactly what the instruments measuring ULR are measuring, the true emission from the ocean surface, or the upward emission of the radiative soup just above it, or a mixture of both, so I can’t answer your question.
This is what Willis affectionately refers to as my “cockamanie ‘dance of the photons’ theory”. 🙂

John Endicott
August 17, 2011 10:27 am

kadaka (KD Knoebel) says:
5. Add what’s in your wallet and your bank assets together, you have $200.
===============
You are forgetting that in addition to assests, you have liabilities, namely a $100 loan that will need to be repaid. You must subtract your liabilities from your assets to come to your actual amount you have, in otherwords $200assets – $100liabilities = $100

richard verney
August 17, 2011 10:56 am

@Willis Eschenbach says: August 16, 2011 at 9:10 pm
///////////////////////////////////////////////////////////
Willis,
The problem with the point you make is that the very very top surface skin layer is cooler than the layer below and hence conduction is not carrying heat from the very very top surface layer to the ocean below. The significance of this is since DWLWIR can only penetrate only a matter of a few microns into the very very top layer of the ocean, ALL the DWLWIR which found its way into the oceans cannot on your account of the process find its way into deeper parts of the ocean.
The principle that you describe is important as regards the Solar energy since this heats the top 10 meteres of the ocean and heat from this part of the ocean (metre by metre) can be conducted downwards in accordance with the principle you describe.
In my earlier comment, may be I did not sufficiently explain the Peter and Paul point. As I see matters, one should visualise the ocean as follows:
1. There is a very very top layer comprising a matter of microns (lets be generous and say up to 10 microns although we are really interested in the top 3 to 5 microns). You may like to think of it as akin to a blanket that protects the deep ocean from losing too much heat.
2. The very very top layer can absorb to limited extent such DWLWIR that reaches the ocean surface.
3. The ocean below the very very top layer cannot be penetrated by DWLWIR due to the wavelength of the DWLWIR. It is effectively opaque to it. However this part of the ocean can be penetrated by the vast majoirty of the incoming solar radiation and this incoming solar radiation (some ~170 w per sqm) heats the top 10 metres of this part of the ocean. Due to over turning and/or the conduction process that you describe in your comment, heat in this part of the ocean gradually makes its way downwards and goes to heat the deep bulk ocean.
4. So what is going on in the very very top layer of the ocean. There are three steps/actions to consider.
5. First, water molecules are constantly being evaporated from the very very top layer with the result that the very very top layer is slightly cooler than the layers of ocean immediately below it. This means that there is no ‘additional’ heat to be overturned nor conducted downwards by the process you describe in your post.
6. Second, we come to the Peter Paul point. Some ~320 to 330 w per sq m of DWLWIR make their way to the first few microns of the very very top layer of the ocean. As they do so, in almost instantaneous manner some ~320 to 330 w per sq m are radiated upwards and away from the very very very top layer of the ocean. Accordingly, the DWLWIR is in effect instaneously extinguished and plays no effective role in the heating/warmth of the ocean. If you like this part of the exchange can be seen to cancel itself out so that there is no need to take it into account when considering what keeps the oceans warm
7. Third, the deep ocean gives up some of its heat. It supplies the very very top layer of the ocean (which is very slightly cooler than the ocean immediately below it) with ~170 w per sqm.
8. Fourth, the very very top layer of the ocean having been supplied with this ~ 170 w per sqm of energy from the ocean below now convects ~20 to 30 w per sq m, evaporates some ~70 to 80 w per sqm and radiates some ~70 w per sqm. Lets say 170w per sqm = 20w + 80w + 70w per sqm.
9. The deep ocean is able to provide the very very top layer with some ~170 w per sqm because it receives some ~170 w per sqm of solar energy in accordance with numbered paragraph 3 above.
10. You will note that when looked upon in this manner, the ocean/atmoshere is at equlibrium. In effect, the ocean (ie., that immediately below the top few microns) is receiving some~170 w per sqm of solar energy and is giving up precisely this quantity of energy. Even without the Peter Paul moment, the ocean would never have given up more. We are concerned with net flux, nothing more.
The behaviour of the oceans is extremely complex. We know that the computer models do not model well either the behaviour of oceans nor clouds. I am sure that we do not properly understand what is going on, and I am therefore quite surprised by the bullishness of your stance.
One of the main problems with climate science (and I have commented upon this problem many times and I note some other people have commented on this in the responses to this article) is that there is no attempt to carry out any physical experimental tests employing empirical observation. Many of the fundamental issues could be clarified and arguments laid to rest, if only some proper physical experimentation was conducted.

Spector
August 17, 2011 11:49 am

RE: Myrrh: (August 17, 2011 at 1:09 am)
“Do you know how much thermal infrared is given off by the incandescent bulb it gives as an example in the second parargraph? 95%
“The 5% visible light doesn’t heat anything.”

I have seen statements like this before and let them go by without thinking. This *can’t* be true. The tungsten filament must be heated to a temperature that is close to the same temperature as the surface of the sun. Not quite, I see from one reference that the peak radiation from an incandescent is around one micron so that light does have a deeper tail into the infrared region. Still, I would be curious how the 5% figure came about. I note that the energy-saving fluorescents typically require about 24% of the equivalent incandescent power. Note that a proper energy integral comparison must be done in the frequency or wave number domain. All these spectra seem to have adopted the wavelength domain for their plots, perhaps to avoid squeezing out the detail in the longer wavelength bands.
Again, the higher the frequency or the shorter the wavelength, the more energetic the photon–Planck’s Law.

tallbloke
August 17, 2011 11:54 am

Willis Eschenbach says:
August 17, 2011 at 10:46 am
What we do know is that however the photons dance, it can’t be driven by more than about a quarter of the DLR. That leaves the rest to warm the top mm of the ocean …
Regarding your “dance of the photons”, you’ve never explained exactly what it is. Somehow your theory (IIRC) had to do with the photons never actually hitting the water surface.
But we know that at least three quarters of the DLR is radiated as thermal radiation. And to be emitted as thermal radiation, first the DLR it has to be converted to thermal energy … making the surface warmer than it would be if there were no DLR
I don’t see any dancing photons in that. To be radiated as thermal radiation at a different frequency than the incoming energy, the DLR must be first converted to thermal energy, which warms/slows the cooling of the surface.
They can’t just dance above the surface.

Hi Willis. I can see a few different possibilies which would account for your observations here. Evaporated molecules form an invisible mist above the ocean surface. The sub visible droplets formed have a much bigger collective surface area than the ocean surface. These droplets become more highly thermalised by radiation emitted both from above and below. The hotter they get, the more buoyant they become, until they rise high enough to cool enough to condense and become visible clouds or a sea fret just above the water. How cloud condensation nuclei grow is a big unknown that Jasper Kirkby and his colleagues are currently working on. We know the gases emitted by decaying plankton which form sulphuric compounds play a part as well as GCR’s.
PS – Truly, I don’t care what you call it. If a room is cold because a door is open to the frozen outdoors, when you close the door the room gets warmer. You can say closing the door warms the room, you can say closing the door slows the cooling of the room, but my point is simple — both the underlying phenomena and the outcome are the same no matter what we name it, the room ends up warmer.
in the same way we can say that DLR warms the ocean or that it slows the cooling of the ocean, it doesn’t matter. The point is that the ocean is warmer with the DLR than without, much warmer. Our name for it doesn’t change the fact that it’s warmer with DLR.

As I said before, when common parlance conflates distinct processes, only confusion can result. Better to be more careful with words IMO.
Nor is it meaningful to say that DLR striking the surface can’t warm the bulk of the ocean. As Tim the Toolman agrees (I think), the bulk of the ocean ends up warmer with DLR than without … so what does “DLR can’t warm the bulk” mean when not just the surface but the bulk ocean undoubtedly ends up warmer with DLR than without DLR?
Tim agrees, as do I, that the presence of the radiative flux slows down the cooling of the ocean. Stephen Wylde thinks the magnitude of this effect is fixed by surface pressure – I think. I’ll be discussing that in person with him in a few weeks time.
I don’t think DLR striking the surface can heat the bulk of the ocean, and neither does Tim, for all the reasons we’ve rehearsed several times on this thread and many others.
LW Radiation only penetrates a few nm.
Conduction can’t go downwards because the surface is cooler than the subsurface.
Turbulent convection isn’t significant because if the eddies aren’t strong enough to pull down fingernail sized pieces of saturated toilet paper then they’re not strong enough to survive destructive interference beyond a few inches. Take it from someone who has designed centrifugal pumps.
Cheers
TB.

August 17, 2011 11:56 am

Willis, you say:
Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. The DLR is about two-thirds of the total downwelling radiation (solar plus DLR). Given the known heat losses of the ocean, it would be an ice-cube if it weren’t being warmed by the DLR. We know the radiative losses of the ocean, which depend only on its temperature, and are about 390 w/m2. In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?”
===============================================
I may be in the wrong discussion here, but I thought the point of it all was that CO2 is adding heat to the system.
My argument against that is – that if, – at the top of the atmosphere, energy in equals’ energy out, – then the “Green House Gas circuit” that has been depicted by our friends Kiehl & Trenberth since the year of our Lord 1997 is totally irrelevant. – It only shows “conservation of energy within the Earth system” – In no way does it show that GHGs add energy to the “Earth System” (ES). – And, -as far as I know, – addition of energy (or use of stored energy) is necessary – anywhere – and on all occasions – if a rise of “Work Done” (WD) which leads to the product we call temperature is to be accomplished.
Therefore, all that DLR can achieve is to help preserve the equilibrium. –And the sucker punch is that the amount of CO2 or any other GHG does not matter. – Only the amount that leaves (the surface) can be returned (to the surface) by GHGs! – According to MIT & the people who measure “Energy in/out” it has not changed (apart from natural fluctuations) since measurements began.

kadaka (KD Knoebel)
August 17, 2011 12:14 pm

From John Endicott on August 17, 2011 at 10:27 am:

You are forgetting that in addition to assests, you have liabilities, namely a $100 loan that will need to be repaid. You must subtract your liabilities from your assets to come to your actual amount you have, in otherwords $200assets – $100liabilities = $100

I forgot nothing, and you obviously will never qualify for a career in government finance. The 2010 Social Security Trustees Report shows a “perpetual projected unfunded liability” of $16.1 trillion. The 2009 Social Security and Medicare Trustees Reports give a combined figure for unfunded liabilities of nearly $107 trillion. Yet as seen in the recent budget battle, right about now the US federal debt is only about $15 trillion.
You have $200, maybe $300.

Editor
August 17, 2011 12:15 pm

Richard Verney.
An excellent post.You said this;
“PS. I do not understand why so much time in climate science is wasted on peripheral issues. Approximately 70% of the planet is covered by ocean. The ocean contains 99% of the heat capacity of the planet, and it is the oceans that drive the weather.”
i agree. Climate science insists on using models that incorporate not only our extremely flawed ‘global’ land temperature calculations
http://wattsupwiththat.com/2011/05/23/little-ice-age-thermometers-%e2%80%93-history-and-reliability-2/
But insist in believing that the temperatures representing 70% of the planet-SST’s- are anything other than nonsensical as any sort of scientific measure prior to 1960;
http://judithcurry.com/2011/06/27/unknown-and-uncertain-sea-surface-temperatures/
(iI am currently in contact with the Met office concerning SST’s)
Modellers seem to believe that the data they use- in all good faith- is of impeccable quality-it isn’t.
tonyb

RJ
August 17, 2011 1:02 pm

@ Willis
“The point is that the ocean is warmer with the DLR than without, much warmer.”
not correct. DLW at best very slightly slows the cooling rate
If there are 100 balls in a box. 20 are thrown upwards and 2 return. It does not increase the number of balls in the box to 102
So DLW does not warmer the surface. It might very slightly slow the cooling rate. As the 2 balls return and increase the number to 82 before they almost immediately leave again
l

Graeme M
August 17, 2011 1:46 pm

As an interested layperson following this discussion I must say I am not sure what is being argued. Everyone seems to have a different take. But the thing that stands out to me is this averaging business. Willis uses an average of 170w/m^2 for solar insolation at the earth’s surface and then explains why that is insufficient on its own to keep the oceans liquid. Say again?
The average figure may be useful for ‘energy budgets’ but why would you consider it a physical reality in this discussion? There would be few places indeed on the earth receiving only that amount every moment of every day.
At the equator, it would vary between zero and 1368 w/m^2 depending on atmospheric conditions, while at the pole far less (I don’t know the figures). Thus the tropical surface cops a lot of heat most of the daylight hours, while the poles get very little most of the time.
Observation shows us that the oceans are liquid in the tropics and ice at the poles. What we see is exactly what Willis asks. If there’s loads of energy at the tropics (probably more than 390 w/m^2 much of the time) the water is liquid, at the poles with little energy it’s frozen. Ocean currents and atmospheric circulation transport heat around which does mean additional heating at the poles and so we see less ice than might otherwise be the case. For example what would happen if there were no currents. More or less sea ice?
So my confusion is, why argue about the effects of a figure that is not real?

Martin Lewitt
August 17, 2011 1:53 pm

@Willis,
“If the DLR isn’t heating the water, where is it going? It can’t be heating the air, because the atmosphere has far too little thermal mass. If DLR were heating the air we’d all be on fire.”
The air had enough thermal mass to emit it. Were we on fire? The downwelling radiation would be coming from a much greater optical depth of air than the few microns of ocean it is directly interacting with. Since a similar flux of radiation is going upwards and the atmosphere is opague over parts of the infrared range, it can be heating the air. Which is heating which, depends on which was warmer.
Water has about 3 times the heat capacity of a similar mass of air. The mass of air over a meter^2 at sea level is over 10,000 kg. A few hundred watts is going to take awhile to reach high temperatures.

RJ
August 17, 2011 2:16 pm

Graeme M
Significant warming due to DLW is based on flawed science that is so obviously flawed that no one should still believe this nonsense
But many do and not just alarmists

richard verney
August 17, 2011 2:49 pm

@Willis Eschenbach says: August 17, 2011 at 10:25 am
richard verney says:
August 17, 2011 at 8:04 am
Willis,
I note that once again you do not answer my question as to why the ocean at the same latitude (62 degN) freezes at one longitude (19 deg 04 42 E) but not at another longitude (8 deg 45E). Is this because you do not have an answer that fits in with a simple radiative budget?
No, I hadn’t seen the question. Nor do I understand it. Is it your clam that the ocean should freeze everywhere at the same latitude? Why would that be?
And having not understood it, I understand even less what it has to do with the subject under discussion. I don’t have time to read every unrelated post. If you start out with something that doesn’t make sense, or that doesn’t catch my interest, I’m likely to just go “Next!” and move on. I said before, to get someone’s attention you need to be brief, clear, and interesting. I have nowhere near enough time to answer every random communication.
//////////////////////////////////////////////////////////////////////
Willis
I have posted many comments in response to your article; only some of which were addressed to you personally. The comments addressed to you personally are: my comments of 15th August at 6:18 pm & 7:12 pm, of 16th August at 4:36am & 5:58 am & 9:37am, and of 17th August at 8:04 am & 10:56 am.
You have responded to some of these comments. See your comments of 15th August at 10:27pm & 10:33pm, 16th August at 2:25pm, and 17th August at 10:25 am. Generally, I have found your response(s) to be inadequate since these simply side step the issues raised.
We have had this debate before, some months ago, when we exchanged with one another a series of comments, and I seem to recall on one other much earlier occasion (but it may be that that exchange was with someone else), I have now asked you several times to explain why using just your Solar/DLR energy budget the ocean freezes at some areas but not at others. You have persistently refused to detail your explanation. I have repeatedly advised you that the answer to your question as to why the oceans do not freeze lies in the tropics. Notwithstanding the fact that I have told you the answer on sever occasions, you keep on saying/suggesting that I have not answered your question as to why the ocean does not freeze.
You now allege: “NO, I HADN’T SEEN THE QUESTION. Nor do I understand it. Is it your clam that the ocean should freeze everywhere at the same latitude? Why would that be?…” (my emphasis).
Obviously, I cannot comment upon the extent of your understanding, but I can state that your excuse that you had not seen the question is plain wrong! You had seen the question which was set out in my comment of 16th August at 9:37 am since you responded to that comment on 16th August at 2:25 pm. See: “Willis Eschenbach says: August 16, 2011 at 2:25 pm richard verney says: August 16, 2011 at 9:37 am…”
I do not wish to wast time in repeating what I have already said. Please re-read my various comments as detailed above and then please revert with your detailed Solar/DLR budget calculation for:
1. The ocean at 62N 8 deg 45E (say for January)
2 The ocean at 62n 19 deg 04 42E (say for January)
3. The ocean at the tropics (say for January).
Alternatively, if you do not wish to produce actual calculations you can advise whether you:
4. Challenge my assertion that the Solar/DLR energy budget would be broadly similar for the ocean at the same latitude. In this regard, I consider that the incident angle of sunlight, the length of the solar day and the albedo of the ocean will be similar when considering the ocean at the same latitude. Likewise there is no reason why the DLR would be markedly different when considering the ocean at the same latitude given the well mixed nature of greenhouse gases (I am not saying that they would be precisely the same because I can see some reasons which may make modest differences). If the Solar/DLR energy budget is broadly similar at two locations one would IF ONLY considering Solar/DLR energy, expect the oceam to behave in similar manner.
5. Challenge my assertion that the reason why the ocean does not freeze at 62degN 8deg 45 E but does freeze at 62N 19 deg 04 42E is due to the former receiving more of the warm ocean currents eminating from the tropics.
6. Dispute my assertion that the tropical ocean does not freeze when you carry out an energy calculation on a net flux basis.
7. Dispute my assertion that the tropical ocean does not freeze even when you carry out a Solar/DLR energy calculation using your gross figures but ignoring the effects of DWR. If you dispute this, please provide the gross figures which you assert are involved for the tropical ocean (ie., not the average figures that you have used in your article) and the figuiures that you are using for latent heat calculations.
Willis, lets make some progress and actually get to the bottom of this. Please respond constructively and not by side stepping the issues raised.
Ps. I will put my hand up and say that some of my terminology has not been as accurate nor as succint as I would have wished but many of my comments were written at circa 3am to 4am in the early hours of the morning when one is not thinking at their best.

Myrrh
August 17, 2011 3:05 pm

Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat.
The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.

To any here that think thermal infrared does not reach the surface of the Earth, and so does not heat the land and oceans: ALL the heat we feel from the Sun is thermal infrared.
This is what heats the land and oceans to raise the temperature.
I repeat:
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies,
That means they do not heat things.
That means they can’t heat land and oceans.
That means the ‘energy’ imput from the Sun from these wavelengths has nothing to do with heating the Earth.
That means they have to be taken out of your calculations.
You are confused between Light and Heat.
Highly energetic does not mean powerful.
Light energies are not powerful enough to create the heat claimed in Your Science Fiction arguments.
It is thermal infrared direct from the Sun which heats the water of the oceans and the land.
Water is a strong absorber of thermal infrared.
In real life applications this is well known.
“The lights use standard HPS and Metal Halide lamps, which are fitted into a water jacket that removes the heat while still allowing all the visible light to reach the plants.”
Read that again.
This is the real world.
You are not in it.
“The project purpose was to develop a mini-farm sufficiently productive to supply food for a manned mission to Mars. It had to be energy efficient, small, light weight, and most importantly, highly productive – all desirable attributes for the home grower as well. The research was done using a high productivity growth chamber (CAAP chamber) fitted with lighting which featured ” ….a recirculating water jacket that absorbs and removes non-photosynthetic energy”. In plain English, they used water cooled lights – the same type of water cooled lights that have been available to home hydroponic growers for several years already.”
http://www.hydroponicsbc.com/watercooledlighting.html
Prove that visible light from the Sun heats water or take it out of your calculations.
You are not scientists if you ignore this challenge.

George E. Smith
August 17, 2011 3:11 pm

“”””” Answer to George E Smith. local thermodynamic equilibrium exists everywhere and accepting the exponential transients, most of the time emissivity is approximately equal to absorptivity. “””””
Well Alexander (Duranko), assuming that what you say (or at least said here) is true; and I don’t have any idea what it means; or why I should accept exponential transients (of what ?), it is somewhat irrelevent, in that it isn’t Kirchoff’s Law.
Kirchoff’s Law requires exact equality; and moreover at every possible wavelength (or wave number if you prefer).
People should stop citing Kirchoff’s Law in support of their otherwise unsupportable claims; in all cases where the conditions are NOT those set forth as necessary under Kirchoff’s Law; and that is spectral point by point exact equality for a medium in thermal equilibrium with a radiation field.
Otherwise such citation is the equivalent of asserting that Oliver Cromwell’s axe, in the British Museum, has only had two new heads, and five new handles since Oliver last used it.

tallbloke
August 17, 2011 3:13 pm

Graeme M says:
August 17, 2011 at 1:46 pm
Sensible assessment Graeme. One minor nitpick. The oceanic transport of heat towards the poles is a lot less significant than atmospheric transport north of 65N (top end of Gulf Stream transport)
Another point which is worth mentioning is that the insolation figure doesn’t just average the solar input over latitude, but also between night and day sides of the planet. This is a serious shortcoming of the models because of the non-linearity of the response of water to thermal input in terms of the microclimate it creates above itself.

Graeme M
August 17, 2011 3:34 pm

Yes Tallbloke, that’s another aspect I don’t follow. The energy budgets all show 340 off w/m^2 coming in at TOA and I understand how that is calculated. But the actual physical environment receives something rather different than an average. At the equator at midday the TOA gets the full benefit of 1368 w/m^2, not 342. There has to be a real measurable difference for that extra 1000 or so w/m^2. The climate is not an average. Nor are the effects of the sun on the earth.

tallbloke
August 17, 2011 3:36 pm

This gives the TOA daily insolation figures throughout the year at all latitudes. Some surprises for those who haven’t thought this through carefully
http://curryja.files.wordpress.com/2011/08/thermo-txt1.pdf

George E. Smith
August 17, 2011 3:56 pm

“”””” Sorry Myrrh, but the term “thermal radiation” is sometimes confused and misused. It is really a misnomer for EMR, (electromagnetic radiation) which includes visible light which does indeed heat matter when its photons are absorbed. Try here at Wikipedia: “””””
Well “Thermal Radiation” has a much more restrictive meaning than is inferred from what is said here (By Myrrh and later commenter (sorry lost the name)).
Thermal Radiation is a continuum Electromagnetic Radiation, of unrestricted wavelength or wave number range, that arises SOLELY because of the Temperature of an assemblage of molecules of any material that has a Temperature higher than zero Kelvins. The spectrum of such radiation is bounded as a limit case, , and at any wavelength or wave number, by the spectrum of a Black Body Radiator, at the same Temperature. Many dense solid bodies can approach quite closely to the thermal emission of the limit black body; but all materials, whether solid, liquid or gas, at temperatures higher than zero Kelvins can and do emit thermal Radiation. Even a single atom or molecule, as a member of an assemblage of molecules, emits Thermal Radiation.
The mechanism of emission is spelled out in Maxwell’s equations, where any varying electric current in any path of greater than zero length (antenna) must radiate EM waves. A varying Electric Current of course is the same thing as an accelerated Electric Charge (Q); and every particle Physicist knows that accelerated Electric Charges MUST Radiate EM waves. There’s a monument to that phenomenon in Silicon Valley, California. It’s known as the Stanford Linear Accelerator Center; a two mile long pin that holds the San Andreas Fault together. It’s purpose is to accelerate electrons (electric charges) to high kinetic energies without suffering the radiation losses that would occur, if they sent the electrons around a race track. But I digress.
Of course, any isolated (neutral) atom or molecule can be in ballistic flight, subject only to the force of gravity. Gravity is so weak, that it for nearly all practical purposes, has a constant velocity. As such, it of course has no Temperature, which is a consequence of collisions occurring within an assemblage of molecules. In free flight, the molecule has neither acceleration, nor in many cases any electric dipole moment, so it does not radiate EM waves.
However, during the time that a molecule is in collision with another, undergoing an elastic scattering process, due to the electrostatic forces between the two molecules, the molecules or atoms are distorted, because the particle momentum is almost entirely contained in the atomic nuclei. The Proton to electron rest mass ratio, is 1837:1, and then the Neutron has a similar mass, so the nuclear momentum is typically about 3675 times that of the electrons; but the electrostatic forces are of the same order. The result is that one charge species has almost 4,000 times the acceleration of the other, so charge neutrality is moot, since only one sign of charge is undergoing significant acceleration, so the other sign doesn’t matter much.
This results in a distortion of the charge distribution, creating a net electric dipole moment, so during the collision interaction time, the atoms/molecules can and do send and receive EM waves. The wavelength range is unbounded; but it is still Thermal Radiation since it arises as a consequence of the Temperature of the assemblage of atoms/molecules.
Whether EM radiation is capable of warming anything, is irrelevant as to whether it is “Thermal Radiation”.

tallbloke
August 17, 2011 4:14 pm

Graeme M says:
August 17, 2011 at 3:34 pm
The energy budgets all show 340 off w/m^2 coming in at TOA and I understand how that is calculated. But the actual physical environment receives something rather different than an average. At the equator at midday the TOA gets the full benefit of 1368 w/m^2, not 342. There has to be a real measurable difference for that extra 1000 or so w/m^2. The climate is not an average. Nor are the effects of the sun on the earth.

This is what really surprises me about Willis’ approach in this thread. When he’s putting forward his own theory, he is acutely aware that averages are masking what is really going on. Climate at the local level is non-linear in its responses to energy input.

Alexander Duranko
August 17, 2011 4:37 pm

George E Smith: I learnt thermodynamics from high temperature physical chemistry and much hard graft in steel and aluminium plants. You are right in that Kirchhoff’s Law of radiation requires exact matching of absorptivity and emissivity of a body at all wavelengths. But this is how radiative equilibrium works: it’s automatic otherwise you wouldn’t get equilibrium..
The problem with climate science is that most people in it haven’t a clue about the way gases behave. This is why I suggested they go back to Hottell’s teaching.
Here’s something to consider: http://www.vermonttiger.com/content/2008/07/nasa-free-energ.html

Graeme M
August 17, 2011 4:42 pm

That is an interesting PDF Tallbloke. Now here’s where not being a scientist makes it hard. I didn’t quite understand the graphic or some of the text. It *seems* to say that figure is demonstrating average daily insolation at TOA. I suspect I don’t understand what TOA is. In my mind I see TOA as a sort of outer shell which is bounded by some point at which our atmosphere is no longer detectable as such. In effect, to me, the TOA simply represents a sphere. Thus, only the points on that sphere which intercept the sunlight can register an incoming energy.
At the exact pole, I’d assume the angle of incidence is enough to ensure no light or energy at any time. At the equator at the zenith however will receive the full 1370 w/m^2. Yet the text and graphic appear to discuss the effect of a horizon and of the seasons (eg “at the pole , sunlight is absent for exactly half the year”) both of which are effects at the surface, not TOA.
What have I missed?

jimmi_the_dalek
August 17, 2011 4:52 pm

Looking at all this, I am puzzled by something. There is a lot of discussion of cooling of the ocean surface by evaporation, but where is the opposite vector included? ie. where is condensation? All that evaporated water must eventually condense, exactly reversing the latent heat transfer, fall as rain, and eventually must get back into the ocean – not necessarily the same part of the ocean, but that does not matter. So when people quote a figure of X watts/sq m for the cooling effect, is that net of the condensation? or just the flow in one direction?

kuhnkat
August 17, 2011 4:55 pm

anna v,
I bought one of those.
How do you turn it on????

tallbloke
August 17, 2011 5:03 pm

Graeme M says:
August 17, 2011 at 4:42 pm
I’d assume the angle of incidence is enough to ensure no light or energy at any time. At the equator at the zenith however will receive the full 1370 w/m^2. Yet the text and graphic appear to discuss the effect of a horizon and of the seasons (eg “at the pole , sunlight is absent for exactly half the year”) both of which are effects at the surface, not TOA.
What have I missed?

Basically, you missed the tilt of the Earth with respect to the plane of the ecliptic – 23.5 degrees. The arctic circle is bathed in shallow angle but 24 hour a day sunshine for half the year between equinoxes. As you can see from the figures, the TOA illumination adds to more over the 24 hour period at midsummer than the equator gets with its ~12 hour nights and ~4 hours of shallow angles of incidence near dawn and dusk.

jimmi_the_dalek
August 17, 2011 5:04 pm

Richard Verney
“Prove that visible light from the Sun heats water or take it out of your calculations.”
All electromagnetic radiation of any wavelength, from gamma rays, through ultra violet, visible, infrared, to microwave, and long wave radio, is energy. Different materials have different cross sections to electromagnetic radiation, ie they do not all absorb with the same probability, but none has zero cross section. If something absorbs energy it heats. Below a certain depth in water it is dark. The visible light has therefore been absorbed. Therefore the water has heated. Q.E.D

Graeme M
August 17, 2011 5:05 pm

Oops, forget I spoke. I didn’t take into account the fact that the earth’s axis is tilted from the plane of its orbit…

Graeme M
August 17, 2011 5:06 pm

Snap! 🙂

RACookPE1978
Editor
August 17, 2011 5:19 pm

tallbloke says:
August 17, 2011 at 3:36 pm

This gives the TOA daily insolation figures throughout the year at all latitudes. Some surprises for those who haven’t thought this through carefully
http://curryja.files.wordpress.com/2011/08/thermo-txt1.pdf
Valuable asset … but it is purely theoretical. Accurately finds the average thermal (visible) radiation available – but only at the top of the atmosphere. (Still need to correct for the 3.3% change in received radiation as the earth moves closer and further from the sun each year.)
Otherwise, a good beginning for a perfectly clear, dry day at noon at the tropics where the air mass (or atmospheric thickness) is equal = 1.00. No clouds. No haze. No dust.
Atmospheric losses (not cloud losses!) are equal to a^L where a = 0.85 in the Arctic, and L is the air mass at that latitude. And the simple approximation of L/Sin(latitude) is valid only for latitudes up to around 60 north. (Above that, you need to correct for curvature of the earth, etc.)
Now, follow that calculation down through the atmosphere:
At 80 north, air mass at a 10 degree solar angle (at noon) = 6.12, percent transmitted = 0.353
At 80 north, air mass at a 6 degree solar angle = 11.3, percent transmitted = 0.159

Tim Folkerts
August 17, 2011 5:39 pm

At some level, we will have to live with the limitations of discussing averages in a forum like this. Economists talk about per capita income; actuaries talk about average life expectancy; baseball fans talk about batting averages. Everyone know that people earn different amounts of money; that people live different amounts of time; that a batter will do better in some games than others.
Averages are the the simplest ways to deal with values that vary. Everyone knows that global averages are just that — a simple summary of a much more complex idea. This is just a rough first approximation. For an introduction to an idea, it is great. For an in-depth analysis, it is lousy.
What perhaps amazes me the most is that some people think that experts don’t understand this concept; that climate scientists simply deal with averages rather than taking further steps to look at values at different times of day and different seasons and different latitudes.
So no, a global average won’t tell you …
* if the ocean will be frozen at 62 N
* when and where clouds will form
* the energy balance at the equator.
* how much it will cool overnight.
* how land behaves compared to ocean.
These sorts of details require more calculations and a deeper understanding of the variables involved. It requires computers that can estimate and process data for 1000’s of points and various times and/or locations.
On the other hand, a global average can tell you
* a rough global average temperature.
* a global comparison of how big various energy flows are compared to each other. For example, upward IR carries away much more energy from the surface than upward evaporation.
* what sort of changes would be expected due to hypothetical changes. For example if albedo increases, more sunlight will be reflected, so less IR needs to escape to maintain a balance, so the global temperature must decrease.

Bob_FJ
August 17, 2011 5:40 pm

Myrrh @ August 17, 2011 at 3:05 pm

“…The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
To any here that think thermal infrared does not reach the surface of the Earth, and so does not heat the land and oceans: ALL the heat we feel from the Sun is thermal infrared.
This is what heats the land and oceans to raise the temperature…”

I see that you are apparently ignoring the issues I raised @ August 17, 2011 at 2:43 am including a Wikipedia article on “thermal radiation” as follows:
http://en.wikipedia.org/wiki/Thermal_radiation
If you believe this to be false you could perhaps do a search and be surprised at how many scientists do not support your views.
Meanwhile, here is a thought exercise for you. It is generally agreed that visible light towards the blue end of the spectrum penetrates about 100m in typical seawater. (the visible red and near infrared is “lost” near the surface). Beyond ~100m it gets dark.
Note that a fundamental is that the shorter the wavelength of EMR, the higher the energy level. Also that whilst water is relatively transparent to visible light, it is nevertheless slightly opaque. (in somewhat the same way as GHG’s are partly opaque to IR)
Q: Sunlight is high energy stuff compared with IR. It enters water and then fades away down to ~100m. Where does it go? Energy cannot disappear.
A: It has been converted from EMR (in this case aka as sunlight) to a different form of energy known as HEAT. There is no energy loss. (Re conservation of energy)

RACookPE1978
Editor
August 17, 2011 5:54 pm

Graeme M says:
August 17, 2011 at 4:42 pm
In my mind I see TOA as a sort of outer shell which is bounded by some point at which our atmosphere is no longer detectable as such. In effect, to me, the TOA simply represents a sphere. Thus, only the points on that sphere which intercept the sunlight can register an incoming energy.

You’re right: but remember the actual geometry involved: The earth isn’t a perfect sphere, but conventional wisdom uses a radius of 6371 km. Good enough for all practical purposes.
The atmosphere is a bit more tricky: People debate all day about where the atmosphere ends, and which layer should be counted. It moves up and down with temperature, radiation, tidal pressure from the sun and moon, and the earth’s geode shape as well. I use 51 km, which places the “edge” of the atmosphere right at the top of the stratosphere. Your mileage may vary 8<).
Regardless, the small thickness of the atmosphere (the TOA is "top of atmosphere" in climate terms) compared to the whole radius of the earth's disk, doesn't affect THAT part of the equation. Atmospheric absorption is very, very significant in all other calculations however.

RACookPE1978
Editor
August 17, 2011 6:03 pm

Graeme M says:
August 17, 2011 at 5:05 pm
Oops, forget I spoke. I didn’t take into account the fact that the earth’s axis is tilted from the plane of its orbit…

Well, it depends when you are looking at the world: That 23.5 degree effective angle is true ONLY at mid-summer (June 23 some-odd most years). It declines to 0.0 inclination at the equinox on September 22 – which is the time when the sea ice extents are at a minimum, so you’re earlier assumption is correct. But only for that period right around March 22-23 (the spring equinox – when ice is close to its yearly maximum) and Sept 22-23, the summer minimum.
The Arctic ice melt “season” in the north lies between the two, when the sun is highest. That’s the period he refers to above while talking about diffuse radiation. But at sea ice minimum, the sun is low in the sky even at noon, and disappears each evening in the southwest earlier and earlier each day. Rising to the southeast later and later.

richard verney
August 17, 2011 6:11 pm

jimmi_the_dalek says:
August 17, 2011 at 5:04 pm
/////////////////////////////
I accept that not the entire spectrum of the solar radiation goes to heat the oceans

jimmi_the_dalek
August 17, 2011 6:32 pm

“I accept that not the entire spectrum of the solar radiation goes to heat the oceans”
Really? Why? Heats to different degrees might be plausible – does not heat at all is not.

Myrrh
August 17, 2011 6:40 pm

The claim is that Visible light heats the land and oceans. The claim is the Thermal Infrared direct from the Sun doesn’t contribute to this part of the ‘energy budget’.
================================================================================================================
Thermal Infrared is Heat, it is invisible, it is what we feel as heat from the Sun, it warms us up.
Visible light does not warm us up.
It’s that simple.
Incandescent lightbulbs – the general figure is 95% heat to 5% visible light. This page has 90/10% –

From: http://www.ehow.com/info_8461018_infrared-sunlight-photons.html
“The old Easy-Bake Oven was a testament to the inefficiency of the incandescent bulb. Nearly 90 percent of the energy of an incandescent bulb goes into heat: infrared light. By way of comparison, the same 60-watt light bulb that emits 3 billion billion visible light photons emits 400 billion billion infrared photons. Infrared just refers to electromagnetic radiation with a wavelength between about 0.7 micrometers (millionths of a meter) and up to as long as 100 micrometers.”

The brighter the light given off the more thermal infrared is being given off – it’s heat creating visible light. Visible light is a product, it is not the source of heat. We cannot feel visible light, it is not thermal energy. It cannot raise the temperature of matter as claimed.
The atmosphere is not transparent to visible light, electrons of the nitrogen and oxygen molecules briefly absorb it and send it on its way, this is called reflection/scattering. Water is transparent to visible light, it is transmitted through without being absorbed. Neither of these two scenarios raise the temperature, that is, visible light does not convert to heat.
To call visible light a ‘thermal’ energy is a misnomer, the traditional physics division into Light and Heat as the difference between these has the advantage of keeping common sense in play.
You’ve (generic) have lost all common sense when you argue that visible light heats land and oceans.
I don’t know why you are all arguing against this, your backgrounds of how you came to this will be varied, but I’ve pointed out the NASA used to teach that the heat we feel from the Sun is thermal infrared and now it’s teaching that infrared doesn’t even reach the surface of the Earth.
I’m asking you to think about what I’m saying here.

From : http://en.wikipedia.org/wiki/Radiation
“InfraredMain article: Infrared
Infrared (IR) light is electromagnetic radiation with a wavelength between 0.7 and 300 micrometres, which equates to a frequency range between approximately 1 and 430 THz. IR wavelengths are longer than that of visible light, but shorter than that of terahertz radiation microwaves. Bright sunlight provides an irradiance of just over 1 kilowatt per square meter at sea level. Of this energy, 527 watts is infrared radiation, 445 watts is visible light, and 32 watts is ultraviolet radiation.[1]”

Do you feel heat from the Sun? Are you ‘at sea level’? So why is NASA now teaching that infrared doesn’t even reach the surface of the Earth and as some here have said, that long wave infrared doesn’t heat the surface?
So, there are two things here. That thermal, longwave, infrared has been expunged from the energy budget from the Sun direct to the Earth and that in its place has been put the absurd claim that visible light heats land and oceans..
Since y’all get too easily distracted from the point whenever thermal or heat is mentioned, I’d like you to concentrate on the second part of this. Prove that visible light heats land and oceans. Because until you do, you cannot use these figures. I’m serious. Because all the standard traditional real world physics says it can’t.
Water is transparent to visible light. This means that water does not absorb visible light, it passes through, which is called transmission.

From : http://www.answerbag.com/q_view/134824
“Why is water transparent?
Because water absorbs very little energy at the frequency of visible light.” See Graph.

Gamma rays are not the same as radio waves, there is a qualitative difference between them. There is a qualitative difference between visible and infrared, between light and heat. Heat, thermal infrared, longwave, goes to vibrational states of matter, while visible to electronic transitions.

From: http://en.wikipedia.org/wiki/Transparency_and_translucency
“If the object is transparent, then the light waves are passed on to neighboring atoms through the bulk of the material and re-emitted on the opposite side of the object. Such frequencies of light waves are said to be transmitted.”
(This is visible light in water, it is transmitted, simply passed on.)
“An object may be not transparent either because it reflects the incoming light or because it absorbs the incoming light.”
(This is what happens to visible light in the atmosphere, it is not transparent to visible light.)
“UV-Vis: Electronic transitions ….
When photons (individual packets of light energy) come in contact with the valence electrons of atom, one of several things can and will occur:
An electron absorbs all of the energy of the photon and re-emits it with different color. This gives rise to luminescence, fluorescence and phosphorescence.
An electron absorbs the energy of the photon and sends it back out the way it came in. This results in reflection or scattering.
An electron cannot absorb the energy of the photon and the photon continues on its path. This results in transmission (provided no other absorption mechanisms are active).
An electron selectively absorbs a portion of the photon, and the remaining frequencies are transmitted in the form of spectral color.”

So, the claim that the atmosphere is a transparent medium to visible light is technically inaccurate, because scattering, the blue sky, is from electrons absorbing visible light and reflecting them the way they came in.
Water is actually transparent to visible, the third of the above, the electron can’t absorb the energy and the visible light continues on its path. Visible light is transmitted through water.
There is no heat created.
Your energy budget is junk science.

Tim Folkerts
August 17, 2011 6:47 pm

Alexander Duranko says:
August 17, 2011 at 6:14 am

One comment on the Curry blog is ‘Gradually the lapse rate should almost vanish in absence of all IR absorption and emission’
This serious lack of basic physics’ knowledge** permeates climate science.

Consider this thought experiment. A planet has an atmosphere that is transparent to IR – say pure Argon (no H2O, no CO2). The daytime surface averages 20 C (for 12 hr); the nighttime surface averages 0 C (for 12 hr). The atmosphere is manipulated until it is all at the temperature 10 C (day and night, top to bottom). Then all controls are turned off and the planet is allowed to react.
The night-time atmosphere will cool, but via contact with the surface, since it cannot radiate any appreciable energy to cool directly. The temperature of the first few meters (or even first few km — the scale doesn;t really matter for the sake of this discussion) may cool, but the cooling at the bottom of the atmosphere will not have time to permeate to the top of the atmosphere. The top will still be 10 C.
The daytime side will warm by contact with the surface, Convection will carry this warmed air upward, but only slightly. Once the warmed air (say 20 C) rises ~ 1 kn, it will have cooled to the same temperature as the surrounding 10 C air, so it will stop rising. Once again, vanishingly little energy exchange will occur with the TOA, and the TOA temperarture will remain ~ 10 C.
As the world turns, the air on the warmed side will cool as it goes to night. The air on the cooled side will warm. Changes during the previous 12 hr will be pretty well wiped out.
The question for Alexander Duranko — what physics will ever cause the top of this atmosphere to cool off?
If nothing causes the top to cool, nothing will cause a lapse rate to develop. If nothing causes the top to cool, once the lapse rate disappears, nothing will cause it to come back.
(For those who ask why the lapse rate would disappear to begin with, the answer is conduction, Even though air is a poor conductor, it is indeed a conductor. This will tend to even out the temperatures until they are fairly uniform throughout.)

kuhnkat
August 17, 2011 7:00 pm

Richard Verney,
In your response to Willis you say:
” Even without the Peter Paul moment, the ocean would never have given up more. We are concerned with net flux, nothing more.”
Why would you say that the ocean would not give up more? Doesn’t that require a regulator? (GHG’s) That is, without the GHG’s the ocean really would radiate at 390 decreasing to what ever the relative equilibrium becomes over time?
Just looked up thermal conductivity and find water, glass, red brick all have about the same whereas air is much lower with ice about 3x water and metals 10x and up.
Does this enter into the issue here or is the thermal conductivity fast enough in this scenario to not be a significant regulator?
The overturning at night sounds like it does enter into the issue.

jimmi_the_dalek
August 17, 2011 7:07 pm

“Water is actually transparent to visible, the third of the above, the electron can’t absorb the energy and the visible light continues on its path. Visible light is transmitted through water.”
Rubbish.
Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth.

August 17, 2011 7:08 pm

Tim Folkerts,
That sounds just like Venus.

Bob_FJ
August 17, 2011 7:29 pm

Tim Folkerts @ August 17, 2011 at 5:39 pm

“…On the other hand, a global average can tell you
* a rough global average temperature.
* a global comparison of how big various energy flows are compared to each other. For example, upward IR carries away much more energy from the surface than upward evaporation…”

You have become confused over the difference between EMR (electromagnetic radiation) and HEAT. The following NASA energy diagram removes the distracting up and down IR, and shows only the HEAT transfer, which is what affects global temperatures.
Expressed as percentage of TOA sunlight:
* Thermals = 7%
* Evapotranspiration = 23%
* Radiation absorbed by atmosphere = 15%
* Radiation directly to space = 6%
——————–
Total Heat absorbed by surface = 51%
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif

richard verney
August 17, 2011 9:15 pm

@Tim Folkerts says at 17th August 5:39pm
////////////////////////////////////////////////////////
Tim
I partly agree with the tenor of your comment, but at the same time I bawk at averages.
I do not doubt for one moment that climate scientists are alive to the issues being discussed. I also do not dispute that averages can give you an indication of what might be, if certain average variables are altered. I accept that they have a role to play, but (and this is a big but) they must always be viewed with caution since averages have a tendancy to disguise the detail, and often the devil lies in the detail. One reason why stataticians use mode, medium and standard deviation etc is to get away from the problems that averages can create.
Clouds are a good example of the problems caused by averages. Is there such a thing as average cloudiness? or statistically valid average cloudiness? The problem is the variability in clouds and each component part of this variability leads to different effects and resuls. One needs to consider (amongst other matters) the area of cloud cover, its volume, its composition, the height at which it is formed, the latitude and longitude of its formation, the surface including surface medium and surface albedo over which the cloud is formed, the time of day when the cloud is formed, the calendar date/season in which it is formed etc. Given this variability, I do not accept that there is any such thing as statistically valid average cloudiness. One only needs to be slightly out with one or more of these variables and that in itself can fully explain the warming and cooling trends that appear in the temperature data sets for the past century. For this reason alone, one can never rule out natural variations in clouds as being the explanation (or at any rate the bulk explanation) for the temperature changes that we have seen in the temperature data sets during the past 150 years.
Averages give the false impression that everything is uniform, however, Global Warming (if it is occuring) is not a global event still less a global problem. It is a local event which may or may not give rise to local issues. If the global average temperature is rising, some places will become hotter, but some places may not change at all, and some places may even cool. Becoming warmer may be a good thing for some countries, a neutral thing for others, and a problem for some.
Materially, to find out what will occur it is necessary to consider matters on a local level. For example, will growing seasons simply become longer or will there be a migration of the food belts, and if so to where and with what effect. Will we experiece more or less rain, and will the place where the rainfall occurs alter (or is it largely governed by geographical profiles which profiles are essentially static). Will we get more or less snow, and if so where. A lot of these issues lead to management issues which issues can only be addressed locally. I am not going to list everything, you get the idea.
Even sea level rise (if it is occuring to any significant extent), is not a global problem. Many countries have no sea coast line. Even for countries that do have coast lines, rising sea levels may or may not cause significant issues.
It is for political reasons (not scientific reasons) that we are being forced to look at this issue as if it were a global issue when it is not.

Keith Minto
August 17, 2011 9:19 pm

jimmi_the_dalek says:
August 17, 2011 at 7:07 pm
“Water is actually transparent to visible, the third of the above, the electron can’t absorb the energy and the visible light continues on its path. Visible light is transmitted through water.”
Rubbish.
Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth.

Distilled water is transparent to visible light and has a remarkable diamond like clarity, even compared to tap water.

Keith Minto
August 17, 2011 9:30 pm
Tim Folkerts
August 17, 2011 9:46 pm

Bob_FJ,
1) Myrrh is quite sure of his position and has been for months. Trying to convince him otherwise is futile.
2) I understand your point, but I am not confused about the difference between “heat” and “energy” and “EMR”. I was specifically talking energy flows. Upward IR does indeed carry more energy than evaporation. Downward IR carries quite a bit of energy, too, but in the opposite direction.
Heat is the net flow of energy due to temperature differences. By this definition, there is no reason to only lump the two IR energy flows together, as you are suggesting. ALL the energy transfers due to temperatures differences should be lumped together to calculate the heat. THE heat to the surface would then be + 168 + 324 – 390 – 24 – 78 = 0 (using the old Trenberth numbers).
We could group these in a variety of ways. Your suggestion of grouping by type has much merit — ie combining the +324 – 390 = -66 (presumably the “15% listed above). This highlights types of energy flows.
We could also group them by direction of flow: 492 in and 492 out. This highlights the net balance.
We could subdivide the -390 by where it is going: -350 to the atmosphere and – 40 to space.
My point is that there are many legitimate ways to look at the energy flows. There are many legitimate subsets to look at. Different subsets are interesting and useful in different situations.
PS — I just noticed that the wording in the diagram and your own post use “EMR” and “heat” inconsistently. The third line should be “NET radiation absorbed by atmosphere”. The atmosphere DOES absorb ~ 350 W/m^2 of EMR from the ground, which is a little over 100% of the sun’s rate. The atmosphere does indeed emit large amounts of EMR. It is only the NET EMR (which we could call “EMR heat between ground and atmosphere”) that is ~ 15%.

jimmi_the_dalek
August 17, 2011 9:51 pm

“Distilled water is transparent to visible light and has a remarkable diamond like clarity, even compared to tap water.”
The transparency of water depends on the thickness of the sample – the thicker it is the more light is absorbed. Obviously.
Here
http://en.wikipedia.org/wiki/File:Water_absorption_spectrum.png
is a graph of absorbency versus frequency for pure liquid water. Note that the frequency is given as a log scale. Note the dimensions of the y-axis. The amount of light transmitted falls off exponentially. Thus if the absorption coefficient 1 and the sample is 1 cm thick then the intensity will be reduced by a factor of 1/e where e=2.78 i.e it is reduced to about 36% of its starting intensity. If you look at the values then the absorption coefficient is ~100 in the infra-red. So here it takes a few millimetres to absorb the light. In the visible the coefficient is ~1/1000 to ~1/10000 so it takes a few metres to absorb the light. But the important point is – the coefficient is NEVER zero, and with sufficient depth ALL the light is absorbed, which means ALL the energy is absorbed, and since it is not re-emitted as visible light, ALL that energy is retained as heat (at least until it is emitted as LWIR from the surface)
Now could people please stop talking nonsense and claiming that visible light is not absorbed by water.

RACookPE1978
Editor
August 17, 2011 9:54 pm

Keith Minto says:
August 17, 2011 at 9:19 pm (Edit)
Responding to an earlier comment:
….
Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth.
Distilled water is transparent to visible light and has a remarkable diamond like clarity, even compared to tap water.

OK. You’ve ID’ed part of the problem being discussed.
The IR is absorbed high – at the very top of the (perfectly calm) water column of 100 meters. Up there, the IR-induced heat increases evaporation, and causes a (partial) loss of the heat energy being shined downed from the sun.
A little lower, the visible light penetrates, and some up it turns into plant energy (by the planton and smaller veggies up there) which gets by larger things which get eaten by larger things and …… and eventually they decay on the bottom of the ocean.
The remainder of the visible light turns into ??? (Heat obviously). The wind-swept currents and upper-surface turmoil mix up that mythical first few microns that do absorb the IR and store that IR heat (top 10 cm’s of water ) 1-3 meters down.
It is this moderate-depth absorbed light energy and the real-world mixing of real-world ocean waves and wind that is not (often) discussed above.
My earnest question to Mr. Mosher (or other knowledge persona’s) is what are the latitude by longitude actual heat radiation values “assumed” by the CGM’s to be received at each longitude and latitude on the earth.
If they (the CAGW propagandists who wish to control the world’s life) begin with nothing more accurate than a average world-wide year-wide basis of 100 km x 100 km squares from the equator to the poles, then they are attempting to re-create the pyramids by assuming grains of sand were found randomly spread across the desert. And then proving that the grains of sand cannot move uphill to form the pyramids.

kuhnkat
August 17, 2011 9:57 pm

Myrhh,
Your own quote says “” ….a recirculating water jacket that absorbs and removes non-photosynthetic energy”. Seriously, do you have any idea what energy is and how it is converted between radiation, kinetic energy, heat energy, that it is conserved…???
You might also consider that, although the exterior of a spacecraft has no temperature it is also the biggest vacuum thermos possible?? Handling heat has to be an important issue in the design. You don’t radiate extra heat into your environment that has to be removed by the expenditure of more energy. The plants only need certain wavelengths for photosynthesis. Why give them the whole spectrum?? Why use WATER to remove that radiation if it can’t absorb it???
http://www.botany.uwc.ac.za/ecotree/photosynthesis/spectrum.htm
Shows that the Infrared band is not used by photosynthesis. Guess what you are screaming about as the heat band?? Guess what water absorbs really well?? Not that it doesn’t absorb visible light, just not much or it couldn’t penetrate to 100 meters depth. In other words, most of the infrared will be absorbed by that thin layer of water while very little of the desired visible bands will be absorbed. That water is heated so can be circulated past a transparent window or near the surface of the craft to RADIATE away that heat if they don’t use it elsewhere!!
The reason you FEEL infrared is that it excites the molecules of h2o in your skin similar to a microwave!!! It actually carries less energy but more of it is absorbed!!
Oh, and if you don’t think water absorbs visible light, you need to remember that all those plants and debris in water does. Why it gets really black in the deep oceans. Notice in this spectra that water absorption is LOW in the visible, but, it DOES absorb.
http://en.wikipedia.org/wiki/File:Water_absorption_spectrum.png

Bob_FJ
August 17, 2011 11:50 pm

Tim Folkerts @ August 17, 2011 at 9:46 pm

“…1) Myrrh is quite sure of his position and has been for months. Trying to convince him otherwise is futile…”

Yes Tim, I’ve now come to that same conclusion too, although I didn’t know that he/she has been making such gobsmacking assertions for months. If Myrrh cannot accept that sunlight fades to nothing from around 100m in seawater and what that means, or the principle of conservation of energy, then he would seem to be beyond salvation.
It reminds me of some Christadelphians I encountered in Adelaide (Oz) that were unmovedly convinced that the Grand Canyon was created just a few thousand years ago in the great Biblical flood, and that dinosaurs like T-Rex existed alongside humans, despite that they are not mentioned in the Bible whereas the frequently mentioned king of beasts; the lion, is. Similar stuff I saw on TV docos, like an interview with a Grand Canyon helicopter tour guide, saying it‘s all in the Bible.
Actually, I prefer an hypothesis that was put to me that the Grand Canyon was created when a Scotsman dropped a penny in the area and then futilely attempted to recover it.
Your other points; and thanks for your interest; I’ll respond to separately, possibly in a combined comment to Willis.

Spector
August 18, 2011 12:44 am

IR Heat Radiation Fallacy
As objects become warmer, their peak radiation wavelength becomes ever shorter. We can use the peak radiation energy to measure temperature. Just like the infrared band, light is electromagnetic energy. The equations are the same. As the wavelengths get shorter, we do not get into a new area of behavior until the individual photons become so energetic that they can rip molecules apart. A monochromatic optical laser beam is hot. You can get a nasty burn if you focus an image of the sun on your hand with a large magnifying glass. [Hint: never leave such a glass in a position where it may accidently focus sunlight on paper]
The only reason that you might think of light being cool with respect to infrared is that indoor lighting levels are usually not more than 2 percent of the intensity of natural sunlight. If a stove got white hot, you would not want to be anywhere near; it would be blinding hot–literally.
Yes, infrared light carries heat. Optical light does as well and requires fewer photons for the same amount of heat.
http://en.wikipedia.org/wiki/Black_body

Myrrh
August 18, 2011 2:11 am

Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth
Because energy from the Sun isn’t eternal. Because matter slows down light. Water is matter, the deeper you go the greater the pressure. When you switch off your table lamp, where does the light go?
Get real with your scales, none of you could measure your way out of a paper bag.
You don’t take into consideration the difference in properties. What’s the difference in size between the near infrared and longwave infrared? See the NASA page for children. Short wave works on the scale of electrons. Tiny. Gets bashed about even by the electrons of the molecules of nitrogen and oxygen in the atmosphere – that’s what reflection and scattering is. That’s why we have a blue sky. It’s weak compared with thermal infrared which works on a molecular scale, actually moving molecules to higher energy states, which is what temperature is. But in water, the molecules don’t even let the visible in to their dance. They can’t move the molecules to higher energy states. They do not convert water to heat. They are transmitted through unchanged.
It takes the more powerful heat energy of thermal infrared to move molecules of water to heat. Water is a great absorber of thermal infrared. That’s why the water cooled light to leave only visible which is not hot and which the plants use for photosynthesis which is a chemical change in the use of visible light energy – it does not create heat. It creates sugars which then create heat when they are burned for energy.
Don’t come back with more nitpicking gibberish here. Your energy budget says Visible light raises the temperature of land and oceans to heat. If you can’t appreciate how utterly stupid that concept is, it’s not my fault. You haven’t been paying attention.
Water is transparent to visible light, it is not absorbed, it is transmitted through unchanged. It does not heat water. You have to take the AGWScience fiction claim that you’ve all been brainwashed into believing out of your energy budget. Your arguments are nonsense.
This includes the nonsense that longwave thermal infrared direct from the Sun does not heat the oceans. Either because you’re saying that longwave infrared doesn’t reach the surface of the earth and only back-radiated thermal is in play, or as above, you think visible light is thermal, capable of heating water, and thermal infrared isn’t absorbed by the oceans. You’re in a complete muddle.
Water is a great absorber of thermal infrared, longwave. That’s how your bodies get warmed up internally from the Sun, you’re mostly water. UV light doesn’t make it past the first layer of the epidermis, visible light can penetrate fractionally deeper, then it gets reflected out, that’s why you cast shadows. The shorter near infrared is shorter and so more reflective like visible light, but it penetrates deeper still before being reflected out. That’s how near infrared cameras work in the dark and in getting through clouds to see what is behind them in cosmic photography. This is a different process to thermal imaging which measures the heat radiated from a body. You don’t understand differences in processes because you have zilch understanding of differences in properties. That AGWScience fiction mantra that all electromagnetic energy is the same and all energy creates heat has trapped you in a one dimensional world without scale or volume. You won’t understand what I’ve just written until you step out of it and into the real world this side of the mirror. Until you do, you’re stuck in an impossible world with Alice where you argue among yourselves about your fictional concepts of matter without properties and where you adjust physical laws to suit your make-believe worlds. Water has a great capacity to store heat, it takes longer to warm up and it takes longer to cool down. That’s what’s happening in our oceans. In the real world.
Do not think about the following until you have understood the above.

Heat Capacity and Latent Heat
•If we heat a solid that is at a temperature below its melting point, its temperature will increase until it reaches the melting point. If we continue to add heat, the temperature will not change any further until the solid is melted, and then the temperature will again rise until the liquid has reached its boiling point. If we continue to add heat, the temperature will not change again until all the liquid has become gas, after which adding more heat will again raise its temperature.
•The amount of heat that is required to raise the temperature of a substance by 1oC without changing its phase is called its heat capacity. Heat that changes a substance’s temperature without changing its phase is called sensible heat.
•Due to the hydrogen bond, the heat capacity of liquid water (4.2 J·g–1·oC–1) is higher than any other substance except ammonia (one of a very few compounds other than water that has a hydrogen bond, although it’s hydrogen bond is not as strong water’s). The heat capacity of ice and water vapor are also very high, although they are only about half that of liquid water.
•The amount of heat that must be added to convert a solid to a liquid without changing its temperature, and to convert a liquid to a gas without changing its temperature, are called the latent heat of fusion and the latent heat of vaporization.
•The latent heat of fusion of water (334 J·g-1) is higher than that of any other substance except ammonia, and the latent heat of vaporization of water (2,260 J·g-1) is higher than any other known substance.
•When a gas is allowed to cool to become a liquid and eventually a solid, both the sensible heat and the latent heat of vaporization and fusion are released.
Implications of the High Heat Capacity and Latent Heats of Water
•Water’s extremely high heat capacity, latent heat of fusion, and latent heat of vaporization are all important because they allow water to store and transport heat within the ocean–atmosphere system.
•The high heat capacity of water allows the oceans to store large quantities of the sun’s heat energy with only a small change in the water temperature. Similarly, the oceans can release large amounts of heat to the atmosphere with only a very small temperature change. Thus, ocean water temperature changes much more slowly than land temperature with changes in solar heating or cooling, and this allows the oceans to modify coastal land climates to be much milder than locations in the interior of continents.
•The high latent heat of fusion of water allows large amounts of heat to be added to an ice–water mixture without any change of temperature. In high latitudes, this allows heat to be stored by melting ice in summer and returned by freezing water to ice in the winter without any temperature change occurring in the polar ocean water. Thus, polar ocean water is always at or near the freezing point.
•The high latent heat of vaporization of water is also important to the Earth’s climate (see “Evaporation”).
Evaporation
•Liquids, including water, can be converted to the gaseous phase at temperatures below their boiling point by a process called evaporation. Evaporation occurs because the energy levels of individual molecules within a liquid vary, and some molecules may temporarily possess a high-enough energy level to escape the attractive bonds of other molecules and enter the gaseous phase.
•To evaporate a molecule from liquid below its boiling point, the evaporated molecules require more heat than would be needed to evaporate the molecule if the liquid were at its boiling point. Thus, the latent heat of evaporation of water is higher than its latent heat of vaporization.
•Water evaporated from the oceans carries the very large amount of heat associated with its evaporation with it when it enters the atmosphere. This heat is transported with the water vapor in the atmosphere until the water condenses to become rain or snow when the heat is released to the atmosphere.. This mechanism of heat transport is critical to the distribution of the sun’s heat in the Earth’s atmosphere, especially the transport of heat from low latitudes to higher latitudes and the transport of heat from the oceans onto the continents. As a result, this mechanism is a critical factor in controlling climate and is also the source of energy for weather systems.
http://www.wwnorton.com/college/geo/oceansci/ch/07/welcome.asp

Good bye.

kuhnkat
August 18, 2011 2:46 am

Well Myrrh,
since I am so stupid or deluded you still need to explain why the bottom of the oceans are black.

kuhnkat
August 18, 2011 3:02 am

Oh Myrrh,
While you are straightening me out, how does UV stopped by the skin damage DNA?
Where does all that solar-far infrared that heats us come from at night??
Why don’t plants use far-infrared to drive photosynthesis which is energy intensive?
How does far-infrared absorbed by the surface of the water heat at depth with no reverse convection?
How does that weak visible light when magnified by a magnifying glass burn paper? (oh, sorry, that’s far-infrared isn’t it)
Are you really saying that science has been corrupted as far back as Stefan-Boltzman, Gustav Kirchoff and all the rest of those old guys who gave us those neat equations to compute stuff?
Myrrh, I gotta admit, you do live in a magical world.

tallbloke
August 18, 2011 3:16 am

jimmi_the_dalek says:
August 17, 2011 at 4:52 pm
Looking at all this, I am puzzled by something. There is a lot of discussion of cooling of the ocean surface by evaporation, but where is the opposite vector included? ie. where is condensation? All that evaporated water must eventually condense, exactly reversing the latent heat transfer, fall as rain, and eventually must get back into the ocean – not necessarily the same part of the ocean, but that does not matter. So when people quote a figure of X watts/sq m for the cooling effect, is that net of the condensation? or just the flow in one direction?

Hi Jimmi,
The ‘opposite vector’ would exist if the raindrops were warmer than the ocean by the time they hit it.
In general, they’re not.
Rainfall further cools the ocean.
The Sun warms the ocean, the ocean warms the air, the air loses energy to space. In the middle of this process, the air radiates some energy gained from the ocean back down to the ocean, plus some energy gained directly from the Sun. These are very much second order effects. Heat rises.

kuhnkat
August 18, 2011 3:21 am

Tim Folkerts,
I knew you would have to get around to it. Most of the energy came from the ground which came from the sun.

kuhnkat
August 18, 2011 3:35 am

Got another question for you Myrrh, (do you know Frankencense?)
Since heat generates light why does it generate infrared when it is cooler, but can’t generate visible, UV, and shorter wavelengths until it gets a LOT hotter??

kuhnkat
August 18, 2011 3:49 am

Myrrh,
I just remembered another gentleman who occasionally posts needs your help. Y’see, he uses a heat gun to try and heat this bucket of water. Seems no matter how long he holds that gun above the surface of the water the bucket just doesn’t warm up. He wants to know why it doesn’t warm up but his pool does in the sun, yeah, he could fry an egg on that heat gun, but, it would be messy.

Tim Folkerts
August 18, 2011 10:39 am

Here’s a little info that might move the discussion forward.
The thermal conductivity of water is ~ 0.6 W / m*K
The top of the ocean is about 0.2 K cooler than it is 1 mm down.
This leads to a flow of about 0.6 W/m*K * 0.2 K / 0.001m = 120 W/m^2 by conduction from the “deeper layer” 1 mm down up to the surface.
Note that this is highly uncertain, since the 1 mm and 0.2 K are fairly rough estimates. If the surface was indeed 0.3 K cooler, the value jumps to ~170 W/m^2. Different wind or humidity would play a big role. Also, since the top is cooler, this would lead to some convection in the top mm, increasing the flow of energy. Bottom line — the energy flow up from the ocean thru the surface could easily be pretty much anywhere in the range of roughly 50 – 300 W/m^2.
Note also that this includes the “typical” solar energy into the surface of 170 W/m^2. This is good because it provides a mechanism to remove the solar energy from the ocean so that it is not continually warming. In fact, you could easily argue that the magnitude of this temperature gradient is “self-tuned” so that the upward thermal energy flux roughly matches the average downward solar energy flux, regulating the temperature of the ocean.
And here is an interesting conjecture. Suppose that DLR increases. This would deposit more energy in skin layer of the ocean, raising the temperature slightly. One effect would be to increase the evaporation rate (warm water evaporates faster), which would moderate the amount the surface temperature increases. A second effect would be to decrease the thermal gradient and hence decrease the thermal conduction up from the lower parts of the ocean. And of course, anything that limits the flow of energy from the ocean will necessarily increase the temperature of the ocean.
The logical conclusion from this conjecture — the more DLR, the warmer the oceans will get.

Tim Folkerts
August 18, 2011 10:41 am

Links for my previous comment:
The thermal conductivity of water is ~ 0.6 W / m*K
[http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html]
The top of the ocean is about 0.2 K cooler than it is 1 mm down.
[http://ghrsst-pp.metoffice.com/pages/sst_definitions/]

August 18, 2011 12:16 pm

Willis said:
“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.”
I’m not aware of the ocean skin (as opposed to the interacting zone at the very surface) EVER getting warmer than the bulk globally on average. Locally and temporarily maybe but never globally or long term.
A few surface molecules in the interacting zone warm up but then evaporate earlier than they otherwise would have done which uses up any DLR left over after increased evaporation, convection and radiation from the individual molecules have taken their respective slices.
There are 3 layers not two. The molecules in the interactive zone that do warm up until they evaporate. The ocean skin itself up to 1mm deep which appears NOT to warm up and the ocean bulk which shows no sign of warming up either.
The only logical conclusion is that events in the interacting zone are bufferred from the ocean bulk by the cooler ocean skin.

jae
August 18, 2011 12:20 pm

Still waiting to hear why a IR-transparent greenhouse isn’t any hotter in Atlanta than in Phoenix (same elevation and latitude) on a hot summer day, given that there should be WAY more backradiation in Atlanta (from all the water vapor).

August 18, 2011 12:27 pm

Further to my previous post we need actual evidence that more DLR actually DECREASES the thermal gradient across the interface between the ocean skin and the ocean bulk below.
I have found no evidence that it actually does so on a global basis. The 1mm deep cooler layer seems to be global and permanent.
I am aware of some efforts being made to design experiments and/or sensors that would resolve the issue but we are not there yet.
The cooling power of evaporation is such that it seems unlikely to me that the thermal gradient would decrease.
In fact one would think that in view of the power of evaporative cooling more DLR should actually deepen that cooler layer and INCREASE the gradient. However I don’t think it can because once the DLR has been used up the process is self limiting.
This entire thread boils down to that simple issue.

tallbloke
August 18, 2011 2:11 pm

Willis Eschenbach says:
August 18, 2011 at 10:47 am
Perhaps that works with your friends, Roger, that you wave your hands and say that the dance of the photons is what keeps DLR from warming the ocean, and that the photons are dancing because of an invisible mist above the water … around here we need a bit more than that.

Hi Willis, the tone setting and hand waving you did earlier in this thread seemed to work for you. However I don’t see too many of the back slappers and me-toos left here defending the physically untenable. I’ll come to that.
Perhaps you think sea fogs appear out of thin air?
I think they happen when sub-visible water vapour already present in the air above the sea surface cools and condenses. You can prove this one for yourself in a kitchen experiment. A pan of water which has the heat under it reduced suddenly will get a mist of visible steam suddenly appear on its surface. This isn’t happeneing because more water is evaporated when the heat is reduced WIllis. It’s happeneing because the sub-visible water vapour molecules condense and agglomerate due to the reduction in available energy.
What is the second “distinct process” that the DLR is doing that I’m conflating with DLR hitting something and warming it?
The longwave flux keeps the air warmer than it would otherwise be by thermalising water vapour and co2. This reduces the temperature differential between the sea surface and the air. That slows down the cooling of the ocean. Not by very much, but is is a real physical effect.
TB-“I don’t think DLR striking the surface can heat the bulk of the ocean, and neither does Tim, for all the reasons we’ve rehearsed several times on this thread and many others.”
That’s because you refuse to see that when something “slows down the cooling of the ocean”, the bulk of the ocean ends up warmer. We call that “warming the ocean”, and yes, it occurs simply by slowing the cooling.
That is the effect of the physically distinct process outlined above. Thanks for agreeing that we are talking about ‘slowing the cooling’ rather than ‘heating’.
TB-“LW Radiation only penetrates a few nm.
Conduction can’t go downwards because the surface is cooler than the subsurface.”
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.
I think this is handwaving for which there is plenty of refuting observational evidence in the literature. It’s more imaginitive than your earlier effort though.
Unless your claim is that such a system can’t be heated from the top … but I don’t think that’s your claim.
On the rare occasions the air is warmer than the water, it can be heated from above to a very limited extent by conduction.
TB-“Turbulent convection isn’t significant because if the eddies aren’t strong enough to pull down fingernail sized pieces of saturated toilet paper then they’re not strong enough to survive destructive interference beyond a few inches. Take it from someone who has designed centrifugal pumps.”
Have you ever seen the ocean? There’s turbulence there that will pull down a swimmer, much less fingernail sized pieces of paper. Wind plus ocean equals turbulence that sinks ships, so I haven’t a clue what your claim means, or what designing pumps has to do with it.
I lived on a sea going boat I restored for seven years. I have done a lot of swimming in it, and I’ve never been sucked under in the open sea. The eddies you are talking about take place in tidal rips and down channels between land masses. There’s a lot more open ocean than coastal places where such mayhem can occur. Ship sinking waves on the open sea do occur, but winds that strong and freak waves are rare. I specified the conditions of my experiment and they hold good for the majority of conditions short of breaking wave tops in the open sea. When designing the involute curve of a centrifugal pump casing, you match the velocity of the water flow to the radius it travels round to minimise cavitation and vorticity which forms eddies. This is because eddies interfere with each other and make the flow turbulent, reducing velocity, promoting wear and requiring more power to shift the fluid. I understand the way water moves under various kinds of impelling forces. Turbulent convection at the ocean surface in open water under non-white-top conditions is negligible in terms of forcing warmer packets of water down into cooler.
TB, because the ocean can only lose energy through the top, the top is always slightly cooler than the layer below.
What seems to have escaped you is that this setup, cooler water on top of warmer water, is inherently unstable because the cooler water is denser than the warmer, and it wants to sink, and does so. The cooler surface is kept in existence by water moving to the surface, cooling, sinking a mm or so, rewarming, rising to the surface, cooling, sinking a mm or so, and so on.
I agree that natural convection is continually taking place. It doesn’t help your contention that heat moves downwards from the surface though, because what you have just described is heat moving upwards.
So rather than being thermally isolated from the bulk below as you seem to be claiming, the cooler skin surface is constantly exchanging energy and water molecules with the bulk ocean below. And of course, because of that, any radiation absorbed by the skin surface affects the heat loss (and perforce the temperature) of the bulk of the upper ocean.
I haven’t claimed anything of the sort. Energy is constantly being exchanged, upwards. You are correct that radiation absorbed by the skin surface affects the heat loss, and in combination with the upward long wave radiation, the flux leads to a cooling of the surface by around 66-70W/m^2. However, there are other factors which also maintain the skin temperature and subsurface gradient which are larger in sum than radiation effects.
In other words, the DLR doesn’t need to penetrate to the depths in order to warm the depths.
It can’t penetrate to the depths and it can’t warm the depths through direct interaction with the ocean any more than negligibly. As part of the long wave flux in the air, a change in the amount of DLR relative to ULR can cause the ocean to cool marginally more slowly, though the size of the effect will be small compared to convective and albedo changing processes.
Cheers
TB

Jonathan T Jones
August 18, 2011 2:26 pm

[blockquote] Willis Eschenbach says:
August 18, 2011 at 10:24 am
I have pointed out a number of times how radiation on the top skin layer of the ocean can affect the bulk temperature of the ocean, regardless of the temperature of the skin surface. It does so by slowing the cooling of the ocean, which increases the temperature of the bulk ocean.[/blockquote]
Sorry, but “slowing the cooling” is not “increasing the temperature “. The temperature is still going *down*, whereas an increase in temperature is in the opposite direction.

Spector
August 18, 2011 2:50 pm

RE: Myrrh: (August 18, 2011 at 2:11 am)
“‘Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth’
“Because energy from the Sun isn’t eternal. Because matter slows down light. Water is matter, the deeper you go the greater the pressure. When you switch off your table lamp, where does the light go?”

It appears that you are suggesting that photons after going so far through the water just get tired and wink themselves out of existence. Not quite–conservation of energy applies.
Even though the water is transparent, the chance of photon making it through any given length is always less than 100 percent. Eventually that photon is going to be captured in the water or on the bottom. That will make the capturing molecule warmer by an amount proportional to the frequency of the photon. That ‘lucky’ molecule will then share the energy of its ‘catch’ with all its neighbors.
The only exception to this will be those lucky photons that strike a reflecting surface and make it all the way back up out of the water. In deep water there is very little hope of this.

jimmi_the_dalek
August 18, 2011 4:00 pm

“Sorry, but “slowing the cooling” is not “increasing the temperature “. The temperature is still going *down*, whereas an increase in temperature is in the opposite direction.”
“slowing the cooling” can lead to a temperature increase in cases where there is a continuous input of energy. To use one of those analogies which plague this area, consider a dam which has a steady inflow of water, but there are several leaks in the dam. Now plug one of the leaks (slowing the leaking), and what happens – the water level rises, increasing the amount stored, until it finds a new leak. In terms of heat/temperature , if you have a continuous input of energy, and you slow the rate of departure a bit, then one of the heat sinks in the system will fill up a bit (warm), until a new equilibrium is reached. So the two ideas, “warming” versus “slowing cooling” are not as disjoint as it appears.

richard verney
August 18, 2011 4:40 pm

I am pleased to see that there are some who have more persistence than I do and who are still questioning the dogma being put forward by Willis. I would have liked to have carried on participating in this debate, however, given the number of replies on this article, my computer really slows down and will not permit me to cut and paste etc. In fact, I am having difficulty in scrolling up and down to see who is saying what and what issues have been addressed etc.
I found his recent responses less than impressive. Whilst it is a sign of desperation when left to put forward analogies, I found the analogy he posted in response to my points (opening the door on a warm room), to be very poor. This is particularly so, since I had explained to him how the top few microns, which are cooler than the ocean below, essentially act as a barrier protecting the main bulk of the ocean. One can have an open door on a building if one employs an air blanket to create a barrier.
In fact, he seems to fail to appreciate the significance of the observational evidence that the top few micron layer is cooler than the ocean below. In particularly, since it is cooler than the ocean below, if it is over turned, it actually results in a net cooling of the ocean below. There is therefore an argument that DWLWIR leads to a cooling of the oceans.
Of course, however, the top few micron layer is cool because of the evaporation which occurs from that layer. Irrespective of the contribution to this process by DWLWIR, there would be evaporation in any event and thus it would always be expected that the top few micron layer would be cooler than the ocean below,
Whilst I am unaware of any evidence on the subject, given what is taking place at the air/top of ocean boundary, I would not be surprised if there was in fact a ‘dance of photons’. Leaving that to one side, if the ocean is evaporating at this boundary layer, which we know that it is, it follows that there will always be a mist consisting of evaporated water above this layer. It might be because of temperature profiles, one can not visually see it, but obviously it must be present. It is after all, the first/initial stage of the water vapour that rises to form clouds.
It does appear that very slight progress has been made. It does appear that Willis now agrees that DLR does not warm the oceans, and his argument should therefore be founded on the contention that it slows down the cooling of the oceans. This, of course begs the questions what caused the oceans to be warm in the first place.
Of course, the crux of this debate is net energy flow. Does anyone consider that the classical physicists of the 19th century would have thought that the oceans would freeze had they not seen Trenberth’s energy budget? The classical equations do not involve the effects of so called greenhouse gases, nor back-radiation.
They would simply approach the position, on first principle, that a body at X deg C surrounded by air at Y deg C (where Y is less than X) wants to give up a certain amount of its energy to its surroundings. They would work out how much energy the warmer body wants to give up and then they would look at the processes whereby that energy can be given up (eg., conduction, evaporation, radiation etc and the appropriate combination of these processes).
I for one would love to see the results of an experiment on the cooling profiles of a slab of stone say at 20 deg C which is placed in a giant hermetically sealed and insulated warehouse with air initially set at 15 deg C but with CO2 at 50ppm, 100 ppm, 200ppm. 300ppm and 400 ppm. I would be very surprised if the cooling profiles (ie., rate of cooling) was significantly different (but I am open to be convinced).
Unlike the land, the oceans very much influence their own atmosphere. They control its temperature (witness the small diurnal fluctuation) and the humidity. Given this and the fact that water vapour is a far more potent ‘so called GHG’, it is difficult to see what real difference an increase in CO2 levels from say 280 ppm to 380 ppm would achieve. Any role played by CO2 being wholly dwarfed by the presence of water vapour. Of course, I know that Willis’ article was discussing DLR generally and not CO2 induced DLR specifically and some people have side tracked the issue given that the wider debate concerns the role of CO2 in the atmosphere.

richard verney
August 18, 2011 4:53 pm

I meant to add to my final paragraph:
“If I am right on that, if there has been any warming of the oceans during the last century, it would strongly suggest that this is due to a change in cloudiness, which has allowed more solar radiation to reach and thereby to be absorbed by the oceans.”
That is the fundamental significance of accepting that DWLWIR cannot warm the oceans and at most simply acts so as to reduce the rate of heat loss from the oceans. The additional warmth seen in the oceans must have been created by some other influencing factor. Thus it is important that the significance of Willis’ concession is not overlooked.

Myrrh
August 18, 2011 5:45 pm

Spector says:
August 18, 2011 at 2:50 pm
RE: Myrrh: (August 18, 2011 at 2:11 am)
“‘Visible light is only PARTIALLY transmitted through water, a fact which is obvious if you ask why is is completely dark below a certain depth’
“Because energy from the Sun isn’t eternal. Because matter slows down light. Water is matter, the deeper you go the greater the pressure. When you switch off your table lamp, where does the light go?”
It appears that you are suggesting that photons after going so far through the water just get tired and wink themselves out of existence. Not quite–conservation of energy applies.
What happens when your car runs out of petrol?
Do planes use more or less or the same amount of fuel flying into the wind or flying in calm conditions?
Even though the water is transparent, the chance of photon making it through any given length is always less than 100 percent. Eventually that photon is going to be captured in the water or on the bottom. That will make the capturing molecule warmer by an amount proportional to the frequency of the photon. That ‘lucky’ molecule will then share the energy of its ‘catch’ with all its neighbors.
The bottom of the ocean is dark, light, that is, visible, doesn’t reach it. Water is transparent to visible, it doesn’t capture it.
The only exception to this will be those lucky photons that strike a reflecting surface and make it all the way back up out of the water. In deep water there is very little hope of this.
Ah, but that lucky photon could exist, it would then be reflected all the way out to where it would again be bounced around the sky by all the nitrogen and oxygen molecules and then oops it gets bounced right back into the ocean where it’s still in its lucky streak and gets reflected back out where it gets bounced around the sky again by all the molecules of oxygen and nitrogen and back into the ocean and so on ad infinitum. Oh gosh, just like carbon dioxide backradiation keeps bouncing it back to earth heating the Earth /not allowing the heat to escape making the Earth hotter every time and the more carbon dioxide we put into the air the faster this process will accelerate and we end with runaway global warming! AGWScience is right! You’ve converted me! The light from every fire that has ever been lit must be accumulating too. /s
Sorry, I didn’t mean to be so rude. I’m leaving it in with an explanation, this post, because one of my gripes is that AGWScience Fiction Inc produces these memes that don’t bear scrutiny in the light of real physics, by mixing properties and processes, by denying properties and processes, by giving experiments to do that ignore the real properties and processes, such as above, the hot air gun supposedly proving that radiant heat energy, thermal longwave infrared, can’t heat water because there’s no difference between convection and radiant energy in the heat transfer of AGWScience’s imaginary worlds where all properties and processes are interchangeable as long as they ‘sound’ scientific enough to catch the unwary..
[See – http://en.wikipedia.org/wiki/Heat_gun – “Some heat guns incorporate a built-in rest, so they can be activated and placed on a workbench, which frees the operator’s hand. Heat guns can have nozzles which deflect their air for various purposes, such as concentrating the heat on one area, or thawing a pipe without heating up the wall behind.
Most have a heating element based on electrical resistance but some produce heat by a gas flame. A fan increases and focuses air flow for convection heating.
Other devices used for similar purposes include focused infrared heaters.”
See – http://www.speedheater.us/ ]
Some of the memes are so well ingrained into the education system from a concerted effort to achieve this result in the last decades and taken as if real physics by constant repetition generally, that even scientists normally logical and rational in their own disciples take these fictions for granted because they have no reason to doubt them. So heavier than air molecules can stay up in the atmosphere for hundreds and even thousands of years accumulating and can cross oceans in a matter of hours by Brownian motion… As I’m finding here, even scientists on such a sceptical science blog can’t get their heads around far enough to break out of these memes. What I can’t understand is their unwillingness to check it out for themselves when I have given sufficient information from pukkha sources to show that there is a definite disjunct. I thought I’d cracked the problem with at least one incontrovertable example of this manipulation, in that I had the NASA pages to show that there was manipulation of the basic science.
Which is all that I have been trying to get across. That there is a concerted and deliberate perversion of real basic physics in the dark heart of this AGW scam.
Water is transparent to visible light, it is transmitted through unchanged, unabsorbed, it does not heat water.
It really is a travesty that Trenberth can’t find the missing heat… He’s taken the real heat from the Sun out of his equations.
As have you all. And since visible does not heat the oceans, you not only have to add back in the thermal infrared, you have to take out the visible.
I hope you can at least see the import of what I’m saying here, even if you can’t yet get your heads out of the AGWScience fiction memes.

Tim Folkerts
August 18, 2011 5:50 pm

“I for one would love to see the results of an experiment on the cooling profiles of a slab of stone say at 20 deg C which is placed in a giant hermetically sealed and insulated warehouse with air initially set at 15 deg C but with CO2 at 50ppm, 100 ppm, 200ppm. 300ppm and 400 ppm. I would be very surprised if the cooling profiles (ie., rate of cooling) was significantly different (but I am open to be convinced).”

The factor missing from this experiment is that the walls of said warehouse must be set to 3 K to accurately model the situation, since the earth is surrounded by space @ 3k. (Actually, any temperature WELL below the 15 C would work.) The greenhouse effect works because the CO2 (and other GHGs and clouds for that matter) are much warmer than the background and hence radiate much better than the background.

Matt G
August 18, 2011 6:11 pm

How much of the 390w/m2 are claimed to warm the ocean here? Based on net observed out going DLWR it is ~48 w/m2 + ~30 w/m2 + ~70 w/m2 = 148 w/m2.
The sun 170w/m2 and DLWR 390w/m2 would leave 560w/m2 warming the ocean, so as it’s not burning us to death then it would be causing the oceans to boil instead. This is due to the input is much larger than the output when you claim an equal loss is equal warming, so you can’t have it both ways.
170w/m2 v 148 w/m2 would still leave to the ocean gaining heat, but removing errors could easily leave it balanced and still much more realistic than the barmy combined sun and DLWR claim of 560 w/m2.
560w/m2 would give enough energy to warm the planet to about 373k, which is 100.3c.
560w/m2 v 100w/m2 still nonsense, or if claiming 390w/m2 warm the ocean, then why not the 30w/m2 and 70w/m2 warm them too? This is just as barmy as the claim with only DLWR warms the land and ocean. This is almost applying the same thing, but not to such an extreme way. My conclusion of this is someone trying to move a energy loss into an energy gain and it is utter nonsense. The ocean would be boiling with little energy escaping the planet surface over many thousands of years. (373.3k)

Konrad
August 18, 2011 6:20 pm

Over 400 comments now and still no links to empirical evidence showing that backscattered IR radiation around the 15 micron frequency can warm or slow the cooling of oceans. I find It hard to believe that something so central to AGW theory has no supporting empirical data. This affects 71% of the Earth’s surface. Without empirical evidence, the null hypothesis that there is no “Missing heat’ in the oceans because it was never prevented from leaving still stands.
In all probability the surface area of the earth unaffected by backscattered 15 micron IR is greater than this. Plants that are partially cooled by transpiration should be included the list of real Earth surfaces and materials that do not pay attention to black body equations.
Anyhow, due to the hand-waving, analogising and total lack of empirical data on this thread I have been forced to buy two probe type digital thermometers. Fortunately I already have plenty of computer fans, transformers, poly styrene foam, aluminium foil, cling wrap and sea water. I will test this over the weekend.

Matt G
August 18, 2011 6:29 pm

“a loss is warming”
Sorry, on my last post it should read “a equal loss is equal warming”
Please edit my last post and delete this message.
Thankyou.

richard verney
August 18, 2011 6:58 pm

Tim
You have not thought your comment through.
I am well aware that space is 3K and accordingly, Earth wishes to radiate away some of its heat.
The temperature of the walls of the warehouse play no role in the experiment and would needlessly complicate matters if the air inside wished to give some of its heat to the walls, and the slab some of its heat to the air and the walls. We only want to look at the rate of heat flow from warm slab to air.
All we are interested in is examining the role of CO2 in air (at one atmosphere pressure) at various concentrations on the rate of cooling of an object which is marginally warmer than the air which surrounds the object (obviously, we would require a fully insulated floor so there is no heat loss to the floor). I have suggested a 5 deg C difference in temperature between slab and air, but may be should be looking at less.
To reflect the position over the oceans, perhaps we should in addition repeat the experiment but this time with air with higher levels of humidity to see whether in this scenario any signal from CO2 is dwarfed by the effects of the high humidity.
You may have noted from my last comments that I consider it likely that any effect caused by a change in CO2 from say 280ppm to 380ppm will be wholly dwarfed by the high water vapour contents over oceans such that if the oceans have warmed, it is not due to an increase in CO2 but must be due to some other factor, the prime candidate for which being changes in cloudiness which has allowed more solar radiation to reach the surface of the oceans thereby leading to a warming.

Tim Folkerts
August 18, 2011 7:26 pm

I think it would be instructive to consider three separate parts of the ocean: the TOP 1 mm, the MIDDLE layer from 1 mm to 100 m, and the BOTTOM layer from 100 m on down.
Note: all numbers below are averages and are approximations. I don;t intend to get caught up in those minor details ATM.
MIddle receives ~170 W/m^2 of solar energy. Since Middle is warmer than Bottom, there is basically no energy transfer by convection from Middle to Bottom. And since the temperature gradient is close to zero, there is very little conduction, either. Somehow, Middle needs to release ~ 170 W/m^2 upward to Top, or else Middle would get warmer and warmer.
As estimated in an earlier post in this thread, Middle can transfer this amount of energy by conduction to , as long as a gradient of 0.2 – 0.3 K is maintained across Top. This will keep MIddle in balance.
Of course, Top ALSO needs to keep an energy balance or it will warm. TOP is receiving 170 W/m^2 from MIddle. Top is also receiving ~ 330 W/m^2 of IR energy from GHGs and clouds. Fortunately, it is losing 80 W/m^2 via evaporation, and 30 W/m^2 to convection/conduction, and 390 W/m^2 via upward IR. Give or take a little rounding, we are balanced.
BUT suppose one day I start shining a little extra IR onto the ocean — perhaps some diffuse IR lasers . Top will suddenly be out of balance and start warming! There are a couple of easy solutions to this. A little warming of the surface will increase the evaporation and increase the IR output. Warming the surface by 0.2 C will increase the the IR emissions by a little over 1 W/m^2. A back-of-the-envelope estimate suggests evaporation would also increase by a little over 1 W/m^2.
Whatever the actual numbers are, it appears that only a few extra W/m^2 will erase the temperature gradient across Top. This will erase the 170 W/m^2 of conduction across Top! Middle will start accumulating 170 W/m^2! Of course, MIddle has a large thermal mass and 170 W/m^2 will only slowly change Middle. By the time Middle has warmed 0.2 C, the gradient will have been restored, and Middle will stop warming any further.
SUMMARY — A few extra W/m^2 will lead to a warming of a few tenths of a degree C of the oceans.
NOTE: Yes, these are rough estimates. Yes the details will be horrendous when you start worrying about night/day, latitude, seasons, etc Yes, there are feedbacks (positive and negative) that have been ignored. But overall, the ability of IR to warm the ocean seems inescapable.

philincalifornia
August 18, 2011 7:26 pm

richard verney says:
August 18, 2011 at 6:58 pm
You may have noted from my last comments that I consider it likely that any effect caused by a change in CO2 from say 280ppm to 380ppm will be wholly dwarfed by the high water vapour contents over oceans such that if the oceans have warmed, it is not due to an increase in CO2 but must be due to some other factor, the prime candidate for which being changes in cloudiness which has allowed more solar radiation to reach the surface of the oceans thereby leading to a warming.
===================
I think I asked this same question about 400 comments ago.
The H2O absorption bands never reach zero, even in controlled laboratory conditions, as far as I can tell from the graphs. So, out in the wild, with all kinds of Doppler shifting of the absorption wavelengths going on for 30,000 ppm of water vapor, what’s the contribution of the last 100 ppm of Beer’s Law, logarithmically reduced CO2, even if we charitably assume that this is all anthropogenic ??
Besides f*****g zero.

Tim Folkerts
August 18, 2011 7:36 pm

Sorry, Richard, but if you don’t recognize that the cold surroundings are essential to the experiment, then you don’t understand the greenhouse effect. If the wall and the atmosphere are ~ the same temperature, then the atmosphere will have essentially no effect. The greenhouse effect ONLY works if the surroundings are cooler than the gas. (in fact, of the walls of the room were warmer than 20 C, the CO2 would actually help keep the slab from warming up, rather than helping to keep it from cooling down!

jae
August 18, 2011 8:06 pm

Willis Eschenbach:
You are one of my favorite reads. You OWE me a response to my comment, or you are no longer one of my favorite reads, but just another arrogant poster.

tallbloke
August 18, 2011 11:39 pm

Tim Folkerts says:
August 18, 2011 at 7:26 pm
Whatever the actual numbers are, it appears that only a few extra W/m^2 will erase the temperature gradient across Top. This will erase the 170 W/m^2 of conduction across Top! Middle will start accumulating 170 W/m^2! Of course, MIddle has a large thermal mass and 170 W/m^2 will only slowly change Middle.

Hi Tim, it’s an ingenious argument, but since the Sun also emits IR directly, and the gradient across the top 1mm isn’t erased daily, I think there are many factors other than IR maintaining it which are not so easily displaced by a change in the atmospheric level of a trace gas like co2 from 0.027% of the atmosphere to 0.039%.
Otherwise the world would be a very unstable place indeed.
I did some calcs a while ago which show that a ~2W/m2 solar forcing on the ocean would raise the temperature of the top 700m 0.15C in 10 years. This is in line with the reduction of tropical cloud empirically measured by ISCCP using weather satellites 1980-1998. It also fits the sea level rise caused by thermal expansion of the oceans measured by the satellite altimetry.The timing is right, the magnitude is right, and it’s the most parsimonious explanation for late C20th warming.

Bob_FJ
August 19, 2011 12:04 am

Willis,
Jae has asked the following question at least three times of you on this thread:
Jae @ August 16, 2011 at 9:42 am

“Still waiting to hear why a IR-transparent greenhouse isn’t any hotter in Atlanta than in Phoenix (same elevation and latitude) on a hot summer day, given that there should be WAY more backradiation in Atlanta (from all the water vapor).”

Seems a reasonable question to me Willis. Why don’t you respond?
Oh, and is backradiation/DLR generally higher in the tropics than in hot dry deserts due to water vapour being more plentiful in the tropics? If that is so, why are hot dry deserts typically very much hotter in the day than in the tropics? I remember jae asking this elsewhere long ago, and it seems to remain one of those great gobsmacking mysteries that few want to consider.

Bob_FJ
August 19, 2011 1:08 am

Whoops;
Further to my support just above for jae, might I add that the albedo for desert sand is quite high, commonly quoted at 0.4. Old snow has also been quoted at 0.4 and above, but the topic is confused at high latitudes partly because of low solar zenith angles resulting in increased reflection. (very much so on water, and an unwillingness for ice-melt-feedback alarmists to accept that reality).

Tim Folkerts
August 19, 2011 3:10 am

tallbloke says: August 18, 2011 at 11:39 pm
>Hi Tim, it’s an ingenious argument, but since the Sun also emits IR directly,
But the sun emits different wavelengths of IR. See http://eesc.columbia.edu/courses/ees/slides/climate/absorption.gif for example.
>and the gradient across the top 1mm isn’t erased daily,
The IR from the sun (mostly 0.7- 3 um) penetrates much much better, so it is not stopped in the first few microns like thermal IR is. See http://www.btinternet.com/~martin.chaplin/images/watopt.gif
Thus solar IR doesn’t provide much energy to that top 1 mm. Solar IR would not be effective at heating the top fraction of a mm.
(And even if it was effective at heating the top, that would be part of what set up the current quasi-stable climate conditions. It is only the CHANGES that I am addressing.)
> I think there are many factors other than IR maintaining it which are not so easily
>displaced by a change in the atmospheric level of a trace gas like co2 from
>0.027% of the atmosphere to 0.039%.
Certainly there are many other factors involved. Those will certainly affect the MAGNITUDE of the effect I described. It will not change the SIGN of the effect, however, with some significant factor that I did not include that acts in an OPPOSITE way to increased IR..
The “trace gas” argument doesn’t fly with me, since there MANY things in nature at are very important at low concentrations. Changes in these small concentrations are perfectly capable of creating changes in the forcings. See Willis’ new post about the oddities of MODTRANS, for example, to see estimates about how changes in CO2 affect

Tim Folkerts
August 19, 2011 4:13 am

Tallbloke also says:

I did some calcs a while ago which show that a ~2W/m2 solar forcing on the ocean would raise the temperature of the top 700m 0.15C in 10 years. This is in line with the reduction of tropical cloud empirically measured by ISCCP using weather satellites 1980-1998. It also fits the sea level rise caused by thermal expansion of the oceans measured by the satellite altimetry.The timing is right, the magnitude is right, and it’s the most parsimonious explanation for late C20th warming.

This is interesting and certainly is worth further consideration. The 10 year time frame sounds reasonable (and some quick calculations confirm the magnitude). I don’t know enough about cloud cover to know how much it has changed, so I cannot really confirm nor refute this hypothesis at the moment.
(That is one challenge with climate change — there are so many factors that COULD be important. Sorting it all out takes a LOT of time and knowledge.)

Dave Springer
August 19, 2011 5:16 am

Konrad says:
August 18, 2011 at 6:20 pm
“Over 400 comments now and still no links to empirical evidence showing that backscattered IR radiation around the 15 micron frequency can warm or slow the cooling of oceans. I find It hard to believe that something so central to AGW theory has no supporting empirical data. This affects 71% of the Earth’s surface. Without empirical evidence, the null hypothesis that there is no “Missing heat’ in the oceans because it was never prevented from leaving still stands.”
Amazing, innit? Next up on the hit parade is why the vaunted GCMs have never been run for an earth lacking an ocean. Ya know why? Because the earth would be as cold as the moon in very short order. It isn’t the greenhouses gases that raise the temperature of the planet 33C above blackbody temperature. It’s a liquid global ocean that does it. Non-condensing greenhouse gases serve as “kindling” to ignite the water cycle. They are important only when the global ocean is largely covered by ice.
The modus operandi of greenhouse gases is that they are transparent to short wave solar radiation and opaque to thermal radiation. This allows sunlight to heat the surface instantly and the absorbed energy, when reemitted as thermal radiation, has more tortuous escape path.
The exact same thing holds true for liquid water. Water is transparent to visible light so sunlight can penetrate instantly to some 100 meters depth until impurities in seawater eventually absorb it all. Because water is quite opaque to thermal radiation the absorbed solar energy has a more tortuous path of escape. Specifically the solar-warmed water at depth must be mechanically transported to a surface skin just a few microns thick where the energy can escape to the atmosphere by conduction, radiation, and evaporation. Conduction accounts for little (5%), radiation accounts for a small fraction (25%) and evaporation accounts for most of it (70%).
So we have two important take-home lessons. GHG’s have little effect over the ocean and the ocean itself (not GHGs above it) accounts for most of the greenhouse warming of the planet.
Once you accept these facts all observations fall neatly in place. Yes, Virginia there is some greenhouse warming of the planet by GHGs but the effect is small and it only happens over land. The ocean is running the show, not the atmosphere. The most important role, by far, the atmsphere plays is simply providing enough surface pressure so that liquid water can exist over a temperature range of 0-100C which makes a liquid global ocean possible in the first place. The atmosphere could be pure nitrogen at 14psi and the earth’s climate would be essentially the same so long as the ocean remained liquid. The caveat is that the earth doesn’t receive quite enough energy from the sun to keep the ocean liquid so without some modest amount of non-condensing greenhouse gases you’d get a runaway freeze (which has happened before) that would last until enough greenhouse gases and albedo-lowering volcanic ash accumulated to begin a melt. Then you’d get a runaway melt until the ocean was liquid again.

richard verney
August 19, 2011 5:16 am

Tim
Let me get this straight.
You are saying that CO2 does not in itself have the ability to absorb some of the photons being radiated by the slab and then to re-radiate these in all directions such that some of the re-radiated photons are sent back towards the slab either thereby heating the slab, or alternatively slowing down its cooling? Is that your case? Please revert on clear terms on that point.
Are you saying that where the air is at the same temperature as the surface below, there is no greenhouse effect? Again, please be clear in your response.
If so, if the air above the oceans is the same temperature as the ocean itself (which is commonly the position for the first 50 or 100 or few hundred feet) are you saying that there is no greenhouse effect in the first 50 feet and/or 100 feet and/or few hundred feet?
At what temperature interface does the greenhouse effect suddenly come into existence? Why was it not existing before this, and why and what precisely brought it into existence?
If it depends upon having an interface with colder air above it, to what extent will convection from below overcome in whole or in part the effect of the GHG warming/insulation?
I am quite happy for there to be a further variation in the experiment with the walls of the warehouse cooler than the air inside. In this scenario both the slab and the air will cool and may be a source of uncertainty/error since the temperature of the walls will have to be accurately controlled so that it is the same with all variations of CO2 concentrations. When we are looking for a small change in the profile of cooling iof the slab, it would be unfortunate if slight variations in the control temperature of the walls may mask the results being obtained. In my opinion that will needlessly confuse issues since additionally we will then being having significant convectional forces coming into play.
I look forward to hearing from you.

tallbloke
August 19, 2011 5:54 am

Tim Folkerts says:
August 19, 2011 at 3:10 am
Certainly there are many other factors involved. Those will certainly affect the MAGNITUDE of the effect I described. It will not change the SIGN of the effect, however, with[out] some significant factor that I did not include that acts in an OPPOSITE way to increased IR..

Is there any empirical evidence that the net balance has changed? If we suppose the increase in co2 has increased the DLR by 1.7W/m^2 in the last 100 years, then isn’t that offset largely by the greater upwelling LR coming from a warmer ocean surface? The average SST is thought to be around 17C or so. So how much more LR will that be emitting compared to a surface at 16.3C?

Dave Springer
August 19, 2011 5:57 am

@Willis
“What seems to have escaped you is that this setup, cooler water on top of warmer water, is inherently unstable because the cooler water is denser than the warmer, and it wants to sink, and does so. The cooler surface is kept in existence by water moving to the surface, cooling, sinking a mm or so, rewarming, rising to the surface, cooling, sinking a mm or so, and so on.”
What seems to have either escaped you, or you ignored because I already pointed it out to you, is that viscosity is the dominant force in a surface layer only microns deep. It won’t sink if it’s cooler because viscosity holds it on surface. What actually happens is that this exceedingly thin surface layer evaporates continually exposing the surface below it. Think of DLWR as an ablative process like sand blasting or the heat shield on the space shuttle.

Dave Springer
August 19, 2011 6:06 am

@Willis
“Tallbloke, that’s a lovely theory. Evaporated water molecules form an “invisible mist” above the surface. I’ve never heard of an “invisible mist” above the surface of the ocean … so surely you have some photos or some observations of this “invisible mist”, or at least a claim that someone has actually detected or measured the “invisible mist” …”
First rule of holes, Willis. When you’ve dug yourself into one, stop digging. Now we have this preposterous utterance of demanding pictures of something that’s invisible. Be a sport an take a picture of the oxygen in the air for me, Willis.
Water vapor is as invisible as oxygen. When you see steam, or fog, what you are seeing is water droplets not water vapor. Your deep ignorance is showing. Again. Stop digging.

richard verney
August 19, 2011 6:18 am

Tim
You suggest that the atmosphere because it is much warmer than background space radiates much better. I agree insomuch as background space radiates photons at 3K whereas the atmosphere may be radiating photons at ~250 to 260K.
If I vary my experiment so that I have one slab at say 25deg C and another slab vertically above it at say 10deg C and in between the two I have air. Does the rate of cooling of the warmer slab depend upon the amount of CO2 in the air which separates the two slabs?
If so, why don’t the classical equations detail that, and state that the rate of cooling is dependent upon whether the air has 280ppm of CO2 (which was supposedly the case when the classical equations were formed) or 380ppm of CO2. Indeed, why do modern physic text books not state that the classical equations are no longer valid since the concentration of CO2 in air has increased and this therefore reduces the rate of heat flow from the warmer slab to the cooler slab?
I look forward to your explanations.

Dave Springer
August 19, 2011 6:20 am

Tim Folkerts says:
August 18, 2011 at 10:39 am
“And here is an interesting conjecture. Suppose that DLR increases. This would deposit more energy in skin layer of the ocean, raising the temperature slightly. One effect would be to increase the evaporation rate (warm water evaporates faster), which would moderate the amount the surface temperature increases. A second effect would be to decrease the thermal gradient and hence decrease the thermal conduction up from the lower parts of the ocean. And of course, anything that limits the flow of energy from the ocean will necessarily increase the temperature of the ocean.
The logical conclusion from this conjecture — the more DLR, the warmer the oceans will get.”
It isn’t interesting, it’s wrong. As the water in the top few microns evaporates it exposes a new layer beneath it. It’s ablative. The water beneath is not shielded or insulated by the surface skin because the surface skin is constantly leaving he surface as a vapor exposing new skin a little further down. Sort like sanding a piece of wood. It’s an ablative process.
It’s extremely effective. So effective that 70% of the heat of the ocean escapes via this route while only 5% escapes via conduction and 25% via radiation. It’s a different story over dry land because rocks don’t don’t start evaporating until they reach a temperature of thousands and thousands of degrees.
.

richard verney
August 19, 2011 7:06 am

I note that some commentators have criticised the argument based upon CO2 being a trace gas. I endorse those criticisms.
The mere fact that CO2 may be nothing more than a trace gas does not in itself prevent it from playing a significant role in events such that where there are changes to the concentration of that trace gas, these changes could lead to significant effect. I for one wish that sceptics would not employ the trace gas argument. It is not scientific.
That said, I have for many years been saying that AGW does not work over the oceans: that changes in CO2 concentrations cannot explain the warming of the oceans. My main reasons for this are:
1. Due to the wavelength of DWLWIR, it cannot penetrate the oceans. It is absorbed in the first few microns where if it does anything at all, it merely increases the rate of evaporation which in turn leads to a cooling of the top few micron layer.
2. If the oceans over turns the top few micron layer, this would lead to a cooling of the oceans, not a warming.
3. The oceans are heated solely by solar radiation and it is the amount of solar radiation received which determines whether the oceans heat up.
4. Unlike land, the oceans very much control the atmosphere above them. They control its temperature thus why there is little diurnal range. Further, they control its humidity/water vapour composition. This is a factor of the warming received by the oceans from the input of solar radiation.
5. CO2 is not the dominant GHG. Water vapour is far more significant and given the very high concentration of water vapour over the oceans, AND the ratio between it and CO2, relatively modest changes in CO2 concentrations (up from 280 ppm to 380 ppm) are unlikely to have had much impact on the DWLWIR characteristics which may be in play over oceans. This is subtly different from the trace gas point. It relies upon the ratio between dominant water vapour and less dominant CO2 being overwhelmingly in favour of the dominant water vapour such that the effects of a change in the concentration of the less dominant CO2 are not significant enough to add sufficiently to the effects brought about by the dominant water vapour.
Once one accepts that DWLWIR does not heat the oceans (which Willis now appears to accept) and can at most merely slow down the cooling, if the changes in concentration of CO2 are not sufficient to add anything of consequence to the effects brought about by the dominant water vapour, the warming of the oceans cannot be explained by changes in CO2 concentrations and must be due to other causes. The most likely one of which is changes in cloudiness.
In other words, warming of the oceans is very probably due to natural variation. If the oceans warm, so too does the land and hence most of any warming seen over land is also predominantly accounted for by the natural induced warming of the oceans. I do however accept that changes in CO2 levels may have had some (in my view relatively modest, if any at all) effect on land temperatures, and that other manmade actions such as changes in land use and urbanisation may also have had some effect. However, given the ratio between the heat capacity of the ocean and that of the atmosphere over land, the oceans will always dominate and we are ‘pissing in the wind’ if we think that we can do anything of any significance to control the temperature over land by reducing CO2 emissions. (Please excuse my language) .

August 19, 2011 7:38 am

Willis writes “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.”
No, you’re still missing it or at the very least describing it poorly. The “very surface” is always cooling, not warming. Adding GHGs doesn’t forcibly do anything with DLR warming the surface. The same relationship always exists with DLR < ULR
It is the 1mm depth and below, the "hook", that warms and it does so because the bulk is warming, not because the "very surface" is warming with the bulk following. When the 1mm depth warms from below by convection, it is the SST that has to come along for the ride.
It is the "hook" which moves around from moment to moment trying to set the SST for equilibrium. Its driven by the incoming DSR. In the case of increased GHGs the equilibrium is still set by the incoming DSR and the "hook" except that the SST needs to be a little higher to account for the additional DLR that is (for want of a better term) re-radiated accounting for more of the Steffan-Boltzmann requirment for the ocean to radiate.
So where does any ocean warming ultimately come from? Increased GHGs have caused the SST to be a little higher and if you think of the ocean as having a top temperature of say 15C and a bottom temperature of say 3C then the GHGs will change the top temperature to say 15.5C and given time, the temperatures all the way down to the bottom will change to match. Not by adding 0.5C all the way down but there will be some related increase.
Wow there is a bunch of assumptions in that last statement isn't there. What if the ocean wasn't in equilibrium to start with?

richard verney
August 19, 2011 8:17 am

Dave Springer at August 19 2011 5:16am
////////////////////////////////////////////////////////////
Dave
Very much my take on things.
I have seen you before making your point on model runs without the oceans. I envisage that your predicted result is right. Since a model run in that condition would reveal a lot, it is surprising that such a run has never been made, or is it the case that ‘the Team’ have made such a run but have not published the result.

tallbloke
August 19, 2011 9:28 am

Richard, trace gas point taken. I actually had the prevalence of water vapour in mind when I wrote it, but you are right to pick up on it. Discussions like the one we’ve had are very valuable to help me hone my arguments.
I think we’ve covered some ground here, and hopefully sorted out some of the misunderstandings which lead people to think that the back radiation heating the oceans argument is ‘part of the basic physics’ that sceptics should accept. I’ve been resisting it for years, but it’s amazing how strong this mistaken meme is.

August 19, 2011 9:45 am

tallbloke says:
August 19, 2011 at 9:28 am
…which lead people to think that the back radiation heating the oceans argument is ‘part of the basic physics’ that sceptics should accept.”
If in fact back radiation could heat something or even slow down the rate of cooling then we should exchange the vacuum in thermos bottles for a mixture of CO2 and H2O. Using IPCC’s 5.35(ln C/Co) going from 1 ppm to 1 million yields almost 74 watts/m^2.
But sadly conduction would win out over a phenomena great enough to heat oceans.

Tim Folkerts
August 19, 2011 9:57 am

Richard,
You are still missing at leas one key idea as to how the GH effect works. Let me use your slab analogy and give a few variations. I’ll assume the slab & walls have emissivity = 1 for thermal IR photons. I’ll assume that when warm, they emit 400 W/m^2 of thermal IR (T = ~ 33 C). Let’s also assume that any gas in between is separated from the slab and walls by a thin region of vacuum, so there is no conduction or convection.
CASE 1: Warm Slab; warm walls; vacuum in between.
It should be obvious that the slab will stay the same temperature, receiving and emitting 400 W/m^2, since it is in thermal equilibrium with the walls
NET RADIATION = 0 W/m^2
CASE 2: Warm slab; 0 K walls; vacuum in between.
The slab will cool quickly, since it is radiating 400 W/m and receiving none back.
NET RADIATION = – 400 W/m^2
CASE 3: Warm slab; warm walls; warm CO2 in between. (This is STILL not “the green house effect”)
The CO2 has no net effect. Some of the IR photons leaving the slab get absorbed by the CO2 and some of them will return to the slab. At first glance this might seem to warm the slab, since we have added IR photons to the slab. BUT! we have also blocked some of the IR photons heading from the walls toward the slab. A few photons that should have escaped get returned. A few photons that should have arrived get diverted. The slab will still emit 400 W/m^2. It will still receive 400 W/m^2. The only difference is that the slab receives x W/m^2 from the CO2 (the value of “x” depending on the geometry of the objects involved and the IR spectrum of CO2), and (400-x) W/m^2 from the walls.
NET RADIATION: 0 W/m^2
CASE 4: Warm slab; cold walls; warm CO2 in between. (THIS is “the green house effect”)
The slab loses 400 W/m^2. The CO2 radiates x W/m^2 to the slab (as above). But the walls radiate 0 W/m^2 to the slab because they are at 0 K. So the natural loss from the slab is moderated by the CO2. The slab only looses (400-x) W/m^2, but it gets nothing from the walls. It cools slower than CASE 2.
NET RADIATION – (400-x)
Suppose that x = 100 W/m^2. Then a 300 W/m^2 heater would keep the slab warm with CO2 present (CASE 4), but a 400 W/m^2 heater would be required for CASE 2. The CO2 help “warm” the slab. (For warm walls in CASES 1 & 3, the gas has no effect at all. That is why cold walls are a critical part of the experiment.)

David A
August 19, 2011 10:18 am

If DLWR increases evaporation, then would it not also increase cloud cover? If cloud cover (and or humitity) increases, would it not decrease SWR from the Sun? Would this not reduce a more effective per watt source of energy warming the oceans, (SWR) thus cooling the deeper waters, and increasing the gradient between those waters and the surface?

August 19, 2011 11:28 am

Tim Folkerts says:
August 19, 2011 at 9:57 am
CASE 4: Warm slab; cold walls; warm CO2 in between. (THIS is “the green house effect”)
The slab loses 400 W/m^2. The CO2 radiates x W/m^2 to the slab (as above). But the walls radiate 0 W/m^2 to the slab because they are at 0 K. So the natural loss from the slab is moderated by the CO2. The slab only looses (400-x) W/m^2, but it gets nothing from the walls. It cools slower than CASE 2.
NET RADIATION – (400-x)
Suppose that x = 100 W/m^2. Then a 300 W/m^2 heater would keep the slab warm with CO2 present (CASE 4), but a 400 W/m^2 heater would be required for CASE 2. The CO2 help “warm” the slab. (For warm walls in CASES 1 & 3, the gas has no effect at all. That is why cold walls are a critical part of the experiment.)
How can a 300 W/m^2 heater get the slab to produce 400 W/m^2? Besides you cannot do heat transfer this way. You fail to show a temperature gradient which is required for heat transfer. The only actors in your play that are assigned a T are the slab T=33C and the cold wall of 0 K. So please don’t mix temperature scales and assign a T to the CO2 and so the proper transfer equations. Tell us what will be the emissivity of your CO2 and under what pressure?

Tim Folkerts
August 19, 2011 12:21 pm

Dave Springer says:

It isn’t interesting, it’s wrong. As the water in the top few microns evaporates it exposes a new layer beneath it. It’s ablative. The water beneath is not shielded or insulated by the surface skin because the surface skin is constantly leaving he surface as a vapor exposing new skin a little further down. Sort like sanding a piece of wood. It’s an ablative process.

This is seems like a pretty sure statement from the man who had been bemoaning the lack of empirical evidence related to theories about the ocean. Without experimental backing, I would never endorse one hypothesis over another this adamantly.
You have an interesting and plausible hypothesis (energy absorbed from IR photons goes directly to evaporating the molecules, not to heating the surface). So do I (energy absorbed by IR photons heats the surface layer, which then fuels an increase in upward IR and evaporation). The truth is most likely somewhere in the middle.
But let me give one more argument as to why I am pretty sure that your description is not 100% correct (or even close to 100 % correct). I think we agree that “DLR” is absorbed in the first few μm. But note that this distance is NOT “the first few nm”. If an IR photon is typically absorbed 1 μm = 1000 nm down, and the typical size of a molecule is 1 nm, then the molecule with the extra energy has several hundred or several thousand molecules above it. It cannot “get knocked loose” directly because it will have to hit 100’s of other molecules “shielding” that energetic molecule from escaping. Long before that particular energetic molecule will “evaporate”, it must collide with 100’s (with 100’s be a very conservative estimate) of other molecules, sharing energy during every collision.
“Colliding with 100’s of molecules and sharing the excess energy among them” is pretty much exactly the definition of “raising the temperature”.

kuhnkat
August 19, 2011 12:48 pm

Tim,
Willis is talking about why the oceans don’t freeze. Other than the fact that they won’t radiate at 390w/m2 for very long as they cool and evaporation loss will also decrease, it is the bulk and not just a thin surface layer in question. How does this energy make it down against the gradient to warm the bulk. As we do not have measurements of high enough accuracy to know we are left to speculate.
I would suggest that the highly complex surface area we are talking about has not been defined in its entirety and probably won’t be in the near future (disregarding quantum levels even)
http://www.nytimes.com/2009/07/28/science/28ocea.html
It would seem that these findings would impact the energy flux a bit and the purely mechanical effects also. Would seem to be another reason the surface is distinct from a millimeter down.

tallbloke
August 19, 2011 12:51 pm

Tim says:
“Colliding with 100′s of molecules and sharing the excess energy among them” is pretty much exactly the definition of “raising the temperature”.
And yet the skin surface is cooler and the lower 0.9mm warmer, whereas all the DLR is absorbed in the first 0.05mm. So something other than the radiation is controlling temperature at the skin. Since radiative action only represents 25% of ocean heat loss, I would presume that something else is the latent heat of evaporation, which sucks energy from the surrounding molecules as evaporation occurs.
However, this only accounts for about 80W/m^2. Stephen Wilde has some ideas on this I seem to remember.

Tim Folkerts
August 19, 2011 3:28 pm

Tallbloke,
I think we are not so far apart. I am not arguing that the skin should be warmer that the layers below. When things are in equilibrium, then the skin is indeed cooler by ~ 0.2 – 0.3 K compared with the layers below. This is a consequence of the entire energy balance, especially the loss of energy from the top via evaporation, convection AND IR. As long as things are quasi-stable, this gradient exists. In this situation, ~ 170 W/m^2 can be conducted upward thru the top layer from the layer below. Thus the bulk of the ocean would not be warming or cooling.
What I am saying is that IF there is some extra energy to the skin (for example from extra downward thermal IR, then the balance will get upset. The temperature gradient will decrease (and not merely increase in evaporation rate as Dave Springer is arguing). Less of a gradient –> less conduction of energy from below –> the temperature of the bulk of the ocean must increase.
This simply illustrates one possible mechanism by which extra downward thermal IR will result in higher temperature of the bulk of the ocean without those IR photons ever actually reaching the bulk of the ocean.

Dave Springer
August 19, 2011 4:26 pm

@Myrhh
Visible light can travel thousands of meters through very clear water. Ultimately there is no such thing as absolutely pure water because it’s a universal solvent. Obtaining absolutely pure water is like trying to obtain a complete vacuum or a temperature of absolute zero. They can be approached but never obtained in nature. It is the impurities in seawater which limit how far visible light can penetrate. If it were pure unadulterated H2O you could read a newspaper during the day at the bottom of the Marianas Trench.

Dave Springer
August 19, 2011 4:42 pm

tallbloke says:
August 19, 2011 at 12:51 pm

And yet the skin surface is cooler and the lower 0.9mm warmer, whereas all the DLR is absorbed in the first 0.05mm. So something other than the radiation is controlling temperature at the skin. Since radiative action only represents 25% of ocean heat loss, I would presume that something else is the latent heat of evaporation, which sucks energy from the surrounding molecules as evaporation occurs.
However, this only accounts for about 80W/m^2. Stephen Wilde has some ideas on this I seem to remember.

Where do you get that 80W/m^2 number for latent heat loss? Actual measurements show 200Wm incoming from solar shortwave, 10Wm outgoing via conduction, 50Wm outgoing via radiation, and 140Wm outgoing in latent heat of vaporization. You’re reducing a measured value for latent heat by almost half with the 80Mw figure.

David A
August 19, 2011 4:50 pm

Tim Folkerts says:
August 19, 2011 at 3:28 pm
“What I am saying is that IF there is some extra energy to the skin (for example from extra downward thermal IR, then the balance will get upset. The temperature gradient will decrease (and not merely increase in evaporation rate as Dave Springer is arguing). Less of a gradient –> less conduction of energy from below –> the temperature of the bulk of the ocean must increase.”
Tim if there is extra energy to the skin does it not increase the water vapor in the air just above the skin? Would this increase in water vapor not reduce the SWR entering the oceans below the skin, thus possibly maintaining or increasing the gradient over time? Would it not also increase the absorbtion of LWIR just before it reaches the skin? Also, due to the residence time of SWR being far longer in the oceans, a 2W/m^2 reduction in SWR would have a greater cooling effect, again over time, then a 2 W/m^2 reduction in LWIR. Yet nobody has quantifed these numbers.

August 19, 2011 4:55 pm

“Stephen Wilde has some ideas on this I seem to remember.”
Yes indeed.
I’ve already set out my propositions in this thread and elsewhere but no one seems to have read them because you are all talking around the very same issues without any reference to what I said.

tallbloke
August 19, 2011 5:05 pm

Tim Folkerts says:
August 19, 2011 at 3:28 pm
Tim, I think a small change in the relatively small energy flux at the interface of 66W/m^2 wouldn’t change things because it would be overwhelmed by the other factors maintaining the differential. Not the least being the air temperature itself, which follows the ocean SST fairly closely, about 3 months later on the global average. It is interesting that there seems to be a cycle over ~110 years though.
http://tallbloke.wordpress.com/2011/02/17/roger-andrews-the-solar-sst-relationship-part-ii/

Tim Folkerts
August 19, 2011 5:49 pm

Not surprisingly, there HAVE been studies done on the skin temperature
JOURNAL OF GEOPHYSICAL RESEARCH, VOL. 95, NO. C8, PP. 13,341-13,356, 1990
doi:10.1029/JC095iC08p13341
On the Bulk-Skin Temperature Difference and Its Impact on Satellite Remote Sensing of Sea Surface Temperature
Peter Schluessel, William J. Emery, Hartmut Grassl, Theodor Mammen
Satellite infrared sensors only observe the temperature of the skin of the ocean rather than the bulk sea surface temperature (SST) traditionally measured from ships and buoys. In order to examine the differences and similarities between skin and bulk temperatures, radiometric measurements of skin temperature were made in the North Atlantic Ocean from a research vessel along with coincident measurements of subsurface bulk temperatures, radiative fluxes, and meteorological variables. Over the entire 6-week data set the bulk-skin temperature differences (ΔT) range between −1.0 and 1.0 K with mean differences of 0.1 to 0.2 K depending on wind and surface heat flux conditions. The bulk-skin temperature difference varied between day and night (mean differences 0.11 and 0.30 K, respectively) as well as with different cloud conditions, which can mask the horizontal variability of SST in regions of weak horizontal temperature gradients. A coherency analysis reveals strong correlations between skin and bulk temperatures at longer length scales in regions with relatively weak horizontal temperature gradients. The skin-bulk temperature difference is parameterized in terms of heat and momentum fluxes (or their related variables) with a resulting accuracy of 0.11 K and 0.17 K for night and daytime. A recommendation is made to calibrate satellite derived SST’s during night with buoy measurements and the additional aid of meteorological variables to properly handle ΔT variations.
Measurement show the skin CAN be warmer than the layer below! Only on average is the skin cooler than the surface. If the skin gets extra energy, it apparently can and does decrease (or even reverse) the temperature gradient,

August 19, 2011 6:03 pm

Stephen Wilde says:
August 15, 2011 at 2:52 pm
Stephen Wilde says:
August 15, 2011 at 4:18 pm
Stephen Wilde says:
August 18, 2011 at 12:16 pm
Stephen Wilde says:
August 18, 2011 at 12:27 pm

jae
August 19, 2011 7:00 pm

WTF, Willis?
I think I asked, twice, a very important question that goes to the heart of this (nonsense?) about the putative “atmospheric greenhouse effect.” I have asked this question for about 5 years and have never received a decent response. I had hope HERE, but, alas, I have been completely ignored, and frankly, it bugs me that YOU, of all people, would do that. I have asked the same questions at “science-of-doom” and Surreal-Climate and have received either snips or vague nonsense responses (at least I was not ignored!!!). But YOU are one of my heros, and I am now pissed off. I challenged you, and it appears to me that you have have slithered away into the grass on me?? Colorado cowboys don’t differ much from Montana cowboys, fella. Same sense of FAIRNESS.
THE BASIC QUESTION IS: WHY HAS THE “BACKRADIATION/ATMOSPHERIC GREENHOUSE EFFECT” NEVER BEEN DEMONSTRATED EMPIRICALLY? UNTIL IT IS, IT IS MERELY A THEORY/FIGMENT. CONSULT EINSTEIN AND OTHER SCIENTISTS ABOUT THIS, IF YOU DOUBT IT.
When one asks over and over for a response and is consistently ignored, I guess that one can draw only a few conclusions:
1) He/she is just considered irrelevant/stupid or off-topic and is not worth a response.
2) He/she has a point that the “expert” does not how to address.
3) His/her thought could embarrass the “expert,” so he/she must be ignored (extremely not likely here, considering the integrity of the “expert”).
4) He/she is a butt.
5) He/she has no clue
6) Other possibilities.

August 19, 2011 7:24 pm

“Measurement show the skin CAN be warmer than the layer below! Only on average is the skin cooler than the surface. If the skin gets extra energy, it apparently can and does decrease (or even reverse) the temperature gradient,”
Indeed, but it is the global average that counts as regards the global energy budget.
I suspect that the energy budget between ocean and air is balanced by the average global depth of the interface between skin and bulk and the size of the temperature differential across that interface. That balance is affected by changes in the flow of energy up from the ocean bulk (solar shortwave driven) rather than changes in DLR.
Similarly the energy budget between air and space is balanced at the stratopause which is the point where the balance of ozone destruction/creation changes in response to varying levels of solar activity. The effect of solar variability on ozone quantities apparently reverses at about 45km which is approximately at the stratopause.
The climate zones are controlled by the interaction betwen those two ever shifting points of balance and the visible manifestation of the changing balances is latitudinal shifts in the surface air pressure distribution. Hence climate changes.

August 19, 2011 7:54 pm

Tim Folkerts writes : “Measurement show the skin CAN be warmer than the layer below!”
The paper is behind a paywall. But “bulk” doesn’t mean 1mm below, it means deeper.
You can see that the SST is both warmer and colder than the bulk where the bulk is measured at some number of cm depth (Minnett used 5cm depth in his experiment)
http://en.wikipedia.org/wiki/File:MODIS_and_AIRS_SST_comp_fig2.i.jpg
Having said that there are always exceptions and DSR could be greater than ULR around coastlines for example where warm air comes from the land and onto the water. I’m sure there are other examples. In these cases depending on the rate of evaporation there may be no cool skin.

August 19, 2011 8:16 pm

“depending on the rate of evaporation there may be no cool skin.”
Exactly. Evaporation mops up whatever DLR is left over after upward radiation, convection and conduction have taken their slices.
When evaporation is constrained the coolness and depth of the skin may decline but they will be local variations only. What matters is the global average.

Bob_FJ
August 19, 2011 11:41 pm

Dave Springer @ August 15, 2011 at 7:28 pm
And, attention Willis
I think you described that viscosity is important in the nano-skin of the ocean where most DLR is momentarily absorbed at the quantum level.
(I would add; that is before at astonishing rapidity relative to ocean dynamics, it is reemitted or causes higher energy H20 molecules to fly off, {evaporate}, partly via collisions).
Did you mean the ‘surface tension layer’ perchance? The surface tension layer is also maintained in most of the slow earthly dynamics of the ocean, and can surely only result in stability of the upper surface layer.
It seems to me that there are a lot of unknowns about surface tension, apart from the many observed effects such as meniscus. The following two links are interesting, but simply pose more questions to me.
http://www.engineeringtoolbox.com/surface-tension-d_962.html
http://en.wikipedia.org/wiki/Surface_tension

Septic Matthew
August 19, 2011 11:43 pm

I see that this thread is still alive, so I’ll add a statistical perspective.
The temperature of a region is proportional to the mean kinetic energy of the molecules in that region. As always, there is a distribution of actual energies, and within a region some molecules are reasonably close to the region mean, some are way above the region mean, and some are way below the region mean. For the topic of the thread, the above average molecules in the air radiate energy down to the earth; in the earth’s top layer (solid or liquid) the radiated energy is absorbed by the molecules with below average energy. The above average molecules in the earth radiate to the air, where the energy is absorbed by the below average molecules. in the regions of high temperature, the distribution of the energies in the molecules is shifted upwards compared to the distribution of the energies in the regions of low temperature.
From region to region, there can be no net transfer of heat from the lower temperature regions to the higher temperature regions, but the back radiation from the higher energy molecules in the air to the lower energy molecules in the earth surface can for sure slow the cooling of the earth surface. There are in addition advective and convective processes, but their existence does not contradict the simple scheme I just described. The first major complication is the transfer of energy from the high energy molecules of the air to the low energy molecules of the air by collision. With a few modifications to the details of the total mechanism, the more GHGs there are in the air, other things being equal (which I doubt they are), the slower should be the cooling of the earth surface temperatures from their daytime maxima, hence a gradual increase of those maxima (seasonally adjusted) across time.
That’s the basic narration of radiative heat transfer. If the effect of the increased surface temperature is to increase cloud cover an thereby to reduce surface insolation, that is a major complication to the narration.

Bob_FJ
August 20, 2011 12:26 am

Willis Eschenbach @ August 19, 2011 at 10:34 pm
(Sorry JAE if you think I interfere but but but Grhhhh!)

Willis: “…When the cloud goes away, the clear sky just sucks the warmth out, but when the cloud is over me, I can feel the warmth from the cloud. So I know from my experience that there is DLR…”

Willis, really! You said; I can feel the warmth from the cloud?
Some years ago there was a learned professor that made a claim in the McGraw-Hill scientific encyclopaedia that the moment that cirrus clouds formed (they comprising ice crystals at great very cold altitude) that they rapidly heated the Earth’s surface. I emailed them and that article was subsequently removed. There were similar bizarre claims in Encyclopaedia Britannica, that were similarly addressed.
Willis, your fertile imagination is often very interesting, promoting good debate, but you should not be so assertive about your hypothesises, and listen to what others say to you.

August 20, 2011 12:31 am

“You have an interesting and plausible hypothesis (energy absorbed from IR photons goes directly to evaporating the molecules, not to heating the surface). So do I (energy absorbed by IR photons heats the surface layer, which then fuels an increase in upward IR and evaporation).”
I may as well wade into this discussion with my own view. The energy found at the very surface is made up of energy from the DLR plus energy from the bulk that has conducted through the skin.
I dont believe you can say that the energy used in evaporation exclusively came from the DLR because energy from the bulk is there too. But you cant say its from warmed molecules from the DLR either because on average all the molecules are cooling not warming despite the DLR.
You might be able to look at individual molecules and look at the energy that kicked them into evaporation and I’m sure the DLR will do that often and the energy convected through the skin will also do it.
I do think you can probably say that the energy that makes it to the surface and becomes ULR and subsequently becomes DLR then essentially has two options. Its either going to be radiated again at the surface or its going to be used for evaporation. If its radiated again then its got two options its going to become DLR again or its going to escape to space.
I’m ignoring the exotic options like mixed down by a wave and residence time in the atmosphere…
So in a sense I guess all DLR is used for evaporation or subsequently lost to space.

Bob_FJ
August 20, 2011 1:38 am

Willis,
It disturbs me that you still do not acknowledge advice that it does not matter how much EMR* is whizzing around, unless there is a potential difference (PD) between two sources of it. Otherwise, it amounts to nothing in terms of HEAT transfer. Furthermore, unless there is a change in HEAT level in matter, there is by definition NO change in temperature.
The following illustration shows that by far the most intense EMR fields are towards the horizontal because initially from the surface, radiation is hemispherically equal in all directions, yet mostly nothing happens. Secondarily radiation above the surface is spherical in all directions. The vertical PD is relatively trivial.
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1
There are other implications but let’s see if you can take-in the first step.
* (electromagnetic radiation; including your topical DLR and ULR)

tallbloke
August 20, 2011 2:00 am

Willis Eschenbach says:
August 19, 2011 at 10:13 pm
Now go convince tallbloke that the ocean absorbs DLR

tallbloke says:
August 15, 2011 at 1:46 pm
Hi Willis,
Argument one asks what the difference is between rock and water. Warm water molecules rise to the top. Warm rock molecules conduct heat to their neighbours, which can’t go anywhere.
Argument two asks where the energy goes. The answer is:
space.
Argument three is not an argument that DLR can warm the ocean, it’s an argument that it can slow its rate of cooling.
Argument four is a numerical misunderstanding. The ocean surface very efficiently absorbs 95% of DLR
Stop trying to misrepresent me and instead engage the substantive arguments.

RJ
August 20, 2011 3:01 am

Willis Eschenbach says:
August 19, 2011 at 10:34 pm
“When the cloud goes away, the clear sky just sucks the warmth out, but when the cloud is over me, I can feel the warmth from the cloud.”
And also said in the same post
“Basically, putting in a paragraph of that kind is like waving a big sign saying “I AM A NUTTER! IGNORE ME!”
Were you being serious in the first sentence. Or do you want to withdraw this comment?

RJ
August 20, 2011 3:07 am

Willis Eschenbach says:
August 19, 2011 at 10:34 pm
“There’s an entire industry out there making instruments that are specifically designed to measure the DLR. There are scientists who make a living measuring the DLR. There are a host of papers about the measurements of DLR”
This point has been discussed before on this site. And also on a Dr. Judith Curry thread
http://www.slayingtheskydragon.com/en/blog/102-climate-follies-encore
The most interesting involved the Holy Grail of Warmist worship, the phantom ‘back-radiation’ force. As the aforementioned “20 Milliseconds” explains there are Warmists, Luke Warmists, and Deniers. This phantom is the defining element of this fable and refuses to fit into classical Physics. The following is the actual exchange, playing the part of the English Lord will be Lord Monckton. The Luke Warmist chorus is played by the who’s who on the CC list and the Slayer is played by your narrator and co-author of Slaying the Sky Dragon.
Lord: “Back radiation can be simply demonstrated by pointing a simple infrared detector at the underside of a cloud. Try it.”
Chorus: “My IR detector only cost $60! Simple! Agreed! Agreed!”
Slayer: “Clouds do not absorb and re-radiate heat back to Earth. Clouds add THERMAL MASS which takes longer to heat and cool. Warmists ‘support’ this false hypothesis with IR thermometer readings, but the IR readings of a hot Barbie is the same from any distance; ENERGY is not. Your $60 REMOTE thermometer is not measuring the radiant energy you are receiving, it is measuring the resonance of the Barbie.”

RJ
August 20, 2011 3:09 am

Jae
“I think I asked, twice, a very important question that goes to the heart of this (nonsense?)”
No need for the question mark

gnomish
August 20, 2011 3:18 am

1- anything which absorbs energy easily is a conductor, not an insulator.
2- additional heat capacity improves the efficiency of a cooling system
3- co2 is not an IR mirror.
4- ‘black body’ is an idealization – there is no such thing IRL.
5- assumptions that an IR detector are measuring temperature of a surface through miles of atmosphere are absolutely false.
6- degrees do not measure joules. temperature is not heat energy
7- attribution of motives to a speaker is always done for the purpose of changing the topic.
8- misattribution of motives is a thinly disguised ad hominem argument.
9- third party dippers who do #8 need to let their sycopants out a notch, smokey.
10- if the numbers don’t match the facts, then the numbers are wrong; not the facts.
11- the facts don’t join anybody’s club and don’t require anybody’s approval.
12- claques are for confidence hustlers.
tallbloke- plz don’t change your ways. you’re about the last one i can still admire – not that you should care; but i do.

gnomish
August 20, 2011 4:36 am

if the question is to be answered, it won’t be by speculation, now, will it?
why not help design an experiment that can take this discussion out of the realm of opinion and provide indisputable facts?
the experiment would be to show what effect IR radiation has on the temperature of a body of water, right?
suppose i have a 30W CO2 laser… (which i do) so there should be no question about it putting out the pertinent LR wavelengths, agreed?
suppose i have an IR ‘thermometer’… (which i do)
suppose i have a cup or water… (which i do)
suppose i point the laser at the surface from a perpendicular direction…
now, where should i put thermometers and where should i take readings with the IR detector to satisfy the requirements of the experiment? what OTHER variables must be controlled for the results to be meaningful?
or would descent from the lofty realms of ‘what i think’ to the level of ‘what really happens’ make this the most boring blog on earth? (gratuitous dig… i acknowledge it without shame)

gnomish
August 20, 2011 4:47 am

btw – when a pyrgeometer is pointed out toward space, it does not show the temperature of space, nor does it show the temperature of the earth when pointed the other way.
and a transparent substance is not in any way properly characterized as a ‘black body’

richard verney
August 20, 2011 5:01 am

Willis
What I am about to say is not intended as a personal criticism of you. I should perhaps observe that I usually very much enjoy your articles and find them stimulating and insightful. Frequently, I find myself agreeing with much of what you have to say. However, this present article, falls well below your usual standard.
The article got off to a very bad start by suggesting that the arguments rejecting the contention that DLR does not HEAT the oceans is “just silly”. Given that the contention that DLR can heat (or slow down the cooling of) the oceans is fundamental to the AGW position, any argument contesting this presumption deserves due consideration. Given that the oceans cover approximately 70% of the globe and they account for 99% of the heat capacity of the earth system (ignoring that in the core etc) and given that the climate and weather is driven by the oceans and ocean conveyor belt and air currents that arise consequential thereto, I would suggest that there is no legs in the AGW conjecture if manmade changes to GHGs cannot result in a warming of the oceans. My point is that the issue raised by your Article is really really fundamental and therefore these issues require addressing in detail not by flimsy rebuttal.
Of the 4 arguments you raise against the validity of the contention that DLR does not heat the oceans, the first is patently bad in that it does not take account of the different nature and properties of matter (ie., the differences between solid land and liquid water) and in particular takes no account of evaporation (and the processes involved in that and the energy involved). The third argument, is also plainly wrong in that it confuses heating with a reduction in the rate of cooling. Thus, the entire thrust of your article is off to a shaky start.
Early on, I picked you up on the point that DLR cannot HEAT. You accepted that this was so and sought to justify your sloppy (my wording not yours) use of language on what is understood in common parlance by the expression to heat. I reverted on that saying that in a scientific article, one has to be precise especially when there is a fundamental difference between the concepts and principles involved since to do otherwise only leads to confusion and promotes misunderstandings. At the time you appeared to accept that.
Now look at your latest response with clouds. You suggest that “when the cloud goes away, the clear sky just sucks the warmth out” as if there was some giant vacumn cleaner up in the sky sucking warmth. There is no such process. This is not a scientific response. You state “I can feel the warmth from clouds”. It is extremely rare for clouds to be warmer than the ground still less warmer than the human body. You cannot feel the warmth from clouds. Again, your terminology is incorrect, and this is really very unfortunate since you were picked up on the incorrect use of these principles at an early stage, and you appeared to accept that you had not been scientifically accurate.
Turning now to the DLR measuring equipment. I have lost count of the number of times that I have had arguments with people who claim because there is DLR measuring equipment on the market DLR is real. There is a fundamental difference between signal and energy. For sure, DLR has a signal but that does not mean that it has sensible energy capable of doing sensible work for example capable of heating something.
If I stand on Mercury and point my DLR meter at the bright burning ball in the sky, I see that I am being bombarded with 5800K photons and I think to myself, no wonder I am ‘bloody’ hot. I then get into my space rocket and fly to Pluto. I get out and I feel rather cold. I point my DLR meter at the burning ball in the sky and observe that I am being bombarded with 5800K photons and wonder why it is so cold that it freezes ‘the balls’ off a brass monkey.
A PVR is essentially a DLR meter tuned to the wavelength of solar radiation being received from the sun. It can convert the solar radiation into energy which can be used to power something. By contrast the DLR meter cannot convert the DLR from the atmosphere into energy or electricity capable of doing work. IF it could we would have solved the world’s energy problems. The DLR being radiated has no sensible energy and one should not confuse signal with energy.
For example, I have a problem with my car. It will not start. I measure the battery and the volt meter suggests that everything is fine. It reads 12 volts. I turn on the ignition, a little red light comes on suggesting that the alternator is not yet charging. I turn the ignition further (to energise the starter) the engine does not turn and even the little red warning light goes out. The battery may be producing 12 volts but is has no power. I am not saying that this is a like for like analogy. I am merely using it as an illustration of the principle that although one can detect a signal, it does not in itself mean that there is any energy in that signal.
As far as I am aware, there is not a single sceptic who denies the existence of DLR. However, there are many who challenge precisely what DLR is capable of doing. Your argument that there are DLR measuring devices on the market does not address that issue. Why people raise it, I do not know.
Finally, I would observe that (in the case of this article) you frequently answer a question with another question or in some other way so as to side step the issue. Now I know that this is difficult, and I know human nature adds to this difficulty but it would be better to respond along the lines: ‘I do not know, but I consider that this is a more telling point etc’ Or, ‘the question you raise is not capable of a quick answer and I do not have the time to properly address it,. but I think that this or that is a more significant point/or you fail to appreciate this or that which is also (or more) relevant’ etc. I can see that many commentators have become frustrated by this side stepping and of course, this side stepping has hindered taking this debate further and has hindered getting to the root of the issues raised.
I emphasise that this comment is not meant as a personal criticism. I do not like to see personal criticisms of the authors of articles posted on this site, and I have on a number of occasions commented that this is disrespectful. Those who take the time to post an article deserve the respect of all those who visit this site. In this you have my respect although I find myself very much in disagreement with much of what you have said in this particular article and your subsequent comments. I consider that the issues raised by your article to be extremely complex ones, and of such complexity compounded by the lack of any empirical data and experiments, to be such that most of us are out of our depths. That said, the burden of proof is firmly upon those who contend that DLR heats the oceans, or slows down the cooling or that the oceans would be frozen but for DLR and to date this burden has not been discharged.

dr.bill
August 20, 2011 5:24 am

Dave Springer: (if you’re still reading)
I have had this thread open since it started, have gradually read the whole thing (in sequence), and have saved most of your comments to a separate file for further reading and rumination. For what it’s worth, I am impressed by your comprehensive grasp of this entire matter, and find your explanations both lucid and cogent. I’ll be plagiarizing/paraphrasing some of your observations next semester for the atmospherics part of my thermodynamics course.
My thanks and compliments,
/dr.bill

August 20, 2011 5:28 am

Sorry, Willis, I’ve read your exposition, Dr. Curry’s exposition, and ScienceOfDoom’s exposition — which are all basically the same — and I simply can’t buy it. No amount of armwaving about turbulence and waves can overcome the enormous differences in magnitudes here. The best picture we have of the ocean’s surface seems to be:
A couple of microns where IR is actually absorbed (and radiated);
About a millimeter of “skin” where the temperature is perhaps a third of a deg C lower than the next layer down, presumably due to evaporation; and
A couple of meters at the “real” ocean surface temperature.
There is some casual talk above confusing microns and millimeters. Please: that is, quite literally, like confusing millimeters and meters. So the claim amounts to:
Turbulent mixing of the topmost membrane has a thermal effect on the skin layer, which is cooled by evaporation and is ~500 time thicker than the membrane; then
Turbulent mixing of the skin layer has a thermal effect on the “real” layer, which is ~1000 times thicker than the skin layer.
This is nearly impossible to believe. As I said, the differences in magnitudes are just too great, no matter how stormy the sea or impressive the surf, and no amount of armwaving and sailing experience (which I also have) can get around that. My conclusion is that nobody really has the faintest idea of what’s actually going on at the water/air interface.
So perhaps it would be better to stick to something like “slows down the cooling” or the impossibly opaqe and convoluted lapse rate explanation. The direct heating by IR simply makes no sense.

August 20, 2011 5:43 am

I’m surprised you don’t get this, Willis. Here’s your definitive statement:
My argument is that the ocean is warmer, both at the surface and in the bulk of the mixed layer, than it would be if it were not absorbing DLR.
– Willis Eschenbach
This is only true if you cherry-pick the time where the temperature reading is taken. Yes, there are times when a small delay in outgoing radiation will create a temperature higher than it otherwise would have been. But it’s also true there are times when the delay causes the temperature to be lower. If you integrate over a whole cycle (like a day or a year). then a small delay in outgoing radiation getting to space contributes nothing to the integrated surface temperature or the integrated ocean temperature. It can’t. It doesn’t.
In addition, a small delay in outgoing radiation cannot lead to an increase in the peak temperature. It contributes nothing toward the warmist claims of higher and higher record temperatures. Re-radiated radiation does not make the emitting surface hotter than it started out…the only thing it can do is modulate (slightly) the rate of cooling. It does not add energy to the system. It can’t. it doesn’t.
I admire you and your work, Willis. I really do. But, in this case, with all due respect, your thinking cap is askew.

David A
August 20, 2011 6:06 am

Any increase in water vapor (Tallblokes invisible layer which I am surprised Willis considers a fairy tale) is in and of itself a spectral modification of incoming TSI reducing SW radiation at the surface, (See A) which we all agree is the PRIMARY mechanism which heats the ocean below the surface. Any increase in cloud cover is an even greater spectral modification of SW TSI at the surface. The question needing to be quantified; Is this surface reduction of SWR entering the oceans, caused by additional GHG, adequate, over time, to reduce the subsurface temperature, despite the change in gradient between the subsurface and the skin? It is amazing, people are throwing around lots of hypothesis, with very little quantifiying of the numbers, this includes Willis. Numbers anyone?
(A) Looking at a solar spectrum chart; “ It show that about 98% of that energy lies between about 250 nm in the UV and 4.0 microns; with the remaining as 1% left over at each end. Such graphs often have superimposed on them the actual ground level (air Mass once) spectrum; that shows the amounts of that energy taken out by primarily O2, O3, and H2O, in the case of H2O which absorbs in the visible and near IR about 20% of the total solar energy is captured by water VAPOR, (CLEAR SKY) clouds are an additional loss over and above that.

Tim Folkerts
August 20, 2011 6:38 am

Gnomish,
I love the idea of trying an experiment! I had contemplated this before, but lack of access to a CO2 laser made me forget the idea
However, I suspect that getting the details right will be devilishly difficult.
Thermometry: Since the temperature gradient between skin and bulk seems to be a key feature, it would be great to confirm that difference even exists, which would require two thermometers (the IR surface thermometer and the bulk thermometer) that could measure accurately to within ~ 0.02 C.
Insulation: We are talking about relatively small differences in energy flow rates. Minimizing unwanted energy flows would be a challenge.
Atmospheric conditions: small changes in temperature or humidity will have noticeable impacts.
With that said, here is one experiment I might propose. It is about as simple as I can imagine to approach the conditions needed. Others are welcome to suggest improvements.
Fill an insulated, uncovered container with water (like a styrofoam cooler). This will ensure that energy primarily enters and leave the system via the surface.
Put a small heater into the water and leave it on until the temperature reaches some equilibrium value. The value should be ~ 170 W per square meter of water surface to approximate the solar energy into the system (although being within a factor of 2 either way would almost certainly work). The water needs to be significantly warmer than the surroundings to ensure appropriate evaporation and to ensure that there is a net flow of IR from the water to the surroundings. Care should be taken not to set up large convection currents.
Aim the CO2 laser down at the water. Run the experiment several times, half the time with the laser on and half the time with the laser off. See if there is a statistically significant difference in water temperature in the two cases. (A single pair of runs may well not be enough, because of other factors like slightly variations in room temperature or humidity.)
For added fun (and rigor), you could intentionally adjust some of the parameters (like room temp, heater power, laser power, humidity). At a minimum, these parameters should be recorded.

Tim Folkerts
August 20, 2011 7:20 am

PS. To me, the most fundamental way to start is looking at the top of the atmosphere. Here there is very little water vapor to worry about because it is so cold ! If more CO2 (or other GHGs are added to the atmosphere, the layer from which the atmosphere radiates will be higher. To maintain energy equilibrium, the temperature of the new, higher radiation layer must adjust to be the same as the older, lower layer had been.
Everything else being equal*, the lapse rate will stay the same, so by the time you get to the surface, it must be warmer than it had been. At this level of analysis, the details at the surface are mostly immaterial, since they are occurring within the system.
I don’t really want to start a new discussion here. There plenty of other places that have discussed this idea, and plenty of other places that will discuss it again.
PPS. Yes, I know that “everything else being equal” sweeps a lot of things under the rug. But Occam’s Razor suggests this as a starting point until some other explanation points otherwise.

August 20, 2011 7:47 am

Tim writes “What I am saying is that IF there is some extra energy to the skin (for example from extra downward thermal IR, then the balance will get upset. The temperature gradient will decrease (and not merely increase in evaporation rate as Dave Springer is arguing). Less of a gradient –> less conduction of energy from below –> the temperature of the bulk of the ocean must increase. ”
This is the argument that RC (and Peter Minnett) used to explain how the DLR warms the ocean. You are quite correct in suggesting the temperature gradient will change with a change in conditions (ie changed DLR or changed DSR) to either increase the rate of conduction through the skin or reduce it.
But the upshot of the change is to alter the “hook” temperature and consequently the actual SST. As the equilibrium is being re-established, the skin’s temperature gradient will assume a value that is optimal to keep the equilibrium and it wont be the same as the value it initially assumed to reduce (or increase) the conduction to get the SST to that value.
So a typical example might be that the ocean starts off in reasonable equilibrium with the DSR balanced by radiative losses to space and evaporation and then thick clouds come over. DSR drops and DLR increases. The immediate effect of DSR dropping is that the energy input into the ocean drops and the temperature profile due to the DSR vanishes as the residual warmth convects towards the surface.
Having the temperature profile drop means an immediate cooling effect throughout the bulk and although the convecting residual warmth will tend to keep the “hook” temperature up for a time, the “hook” is now tending to drop in temperature because its being radiated/evaporated away and not fully replaced by the dwindling warm convected waters from below. The SST itself can tend to remain at its temperature because the ocean’s requirement to radiate is now being “subsidised” by the increased DLR which is absorbed at the “very surface” and (for want of a better term) re-radiated.
So the “very surface” temperature has a tendency to remain while the “hook” temperature has a tendency to drop and that means the temperature gradient of the skin is reducing and therefore energy conducted through the skin is also reducing. The ocean is cooling less as a result of the effect of the DLR.
As the “hook” temperature drops (the energy is being radiated/evaporated away) it necessarily takes the SST with it though. Eventually the SST temperature will reduce to the point where the balance is again restored between the DSR and radiated loss plus evaporation and at this time the temperature gradient of the skin will have assumed the optimum value for the new level of radiation and evaporation.

gnomish
August 20, 2011 7:52 am

Tim Folkerts-
say it was still water – not being mixed …
say no cross-wind – no foam or spray, unrestricted convection only ..
say water is at ambient …
say no plankton or other particulate matter…
(unlensed, laser is @ 5mm diameter beam, delivers @ 25-30W IR)
easy numbers to get:
bulk temp of water – it’s at ambient
alleged reading of ‘surface’ by IR meter, alleged to be a thermometer, with and without laser on…
…at various distances from the spot of the beam
(it will really be registering the heated cloud of water vapor immediately above and around the hot spot, not the surface of the water)
water does not even need to convect in order to rise once it becomes gas because its density is so much lower than the other gases.
is already done some experiment which determined the attenuation of IR as it passes through water vapor?
attenuation, wrt h2o^ (or co2), is synonymous with absorption, right?

gnomish
August 20, 2011 8:09 am

what if –
2 identical cups with same mass of water-
measure the loss over a specific period with IR on one. then stir the water and take its temp.
and the loss over the same period without IR on the other. then stir the water and take its temp
easy math would give the joules absorbed from IR, yes?
it would also need for the heat sourcing beam to be dispersed to ‘dilute’ the power to actual insolation levels…
and also specify the surface area to be completely irradiated…

pochas
August 20, 2011 8:38 am

Longwave absorbs in the ocean surface just like microwaves absorb on the surface of the food you are heating. To heat the food evenly you have to stir it while heating it. The same applies to the surface layer of the ocean. To understand why the surface layer is not at a radically different temperature from the underlying layers, consider an initial state where the ocean, the surface layer and the atmosphere are all in local thermal equilibrium (LTE), which by definition means that they are at the same temperature. In this case there will be no net heat flux from the surface to the ocean sublayers.
For some purposes, LTE is a good assumption, but the deep ocean is quite cold, and there is a slow conductive heat flow downward, enhanced by mixing. The top underlying layer is slowly losing heat to an even deeper layer. To keep things in “near” LTE the top underlying layer must slowly gain heat from the top, IR absorbing layer. The key word is slowly. Not much heat flow downward is necessary to maintain shallow sublayers in near-LTE with the surface, which is why shallow sublayers will be at nearly the same temperature as the surface.

tallbloke
August 20, 2011 9:36 am

TimTheToolMan says:
August 20, 2011 at 7:47 am

Tim, it seems the surface and skin layer and will ‘adjust’ to maximise heat loss from ocean to air at all times.
Is this the constructal law which Willis likes at work? Or simply the entropy law at work?
Tim F: It’s fine to move the action back up 5km into the air, but you thereby concede the ocean surface. As I’ve been saying, the air temperature affects the rate the ocean will cool at, but since DLR isn’t directly heating the ocean bulk, and air temp follows ocean surface temp, it’s clear the effect from changes in DLR on air temp is small in comparison.
Excellent discussion from all by the way.

Tim Folkerts
August 20, 2011 9:41 am

Gnomish says:
August 20, 2011 at 7:52 am
say it was still water…no cross-wind… no foam or spray ..
unrestricted convection only … no plankton or other particulate matter…

Those all sound like good conditions — keeping it simple is always a good starting point.
(unlensed, laser is @ 5mm diameter beam, delivers @ 25-30W IR)
This is problematic. You are delivering ~ 20 W to ~ 0.00002 m^2 = ~ 1 million W/m^2. To be comparable to real conditions, you would want to add only ~ 1-10 W/m^2 extra IR to the surface of the water, and you would want to add it uniformly over the surface. Highly concentrated power over a small bit of the surface would be sooooo far from the conditions we are simulating as to be useless.
easy numbers to get:
This reminds me of a humorous sign I saw once. Paraphrasing slightly … “We offer accurate, easy, inexpensive results. You can pick any two.”
I’m afraid that any “easy” answers will not be sufficient accurate. This is actually a subtle experiment, since we are only looking at small differences that will occur slowly.
Some back-of-the-envelope calculations …
Consider a tank of water 1 m x 1 m that is only 0.1 m deep = 100 kg of water. The net heat capacity is (4200 J/kg*K) x (100 kg) = 4.2E5 J/K. A 420 W heater (similar to daytime solar energy) will raise the temp only ~ 1/1000 K per second (assuming perfect insulation). So it will take ~ 15-20 min to warm 1 K. This heater will raise the temperature quite a bit above ambient temperature before equilibrating, so this might take several hours. (Alternately, we could start with warm water so it will equilibrate sooner).
We want to add just a few W/m^2 of IR to see how it affects the equilibrium temperature. First of all, 10 W/m^2 extra IR will only affect the temperature at a rate of 10 J/s / 4.2E5 = 2E-5 K/s. This would be 1 K in 12 hr. So the experiment will need to run most of the day before the temperature changes are easily noted. (This assumes I am right and there is indeed a change. Dave would presumably expect no change in temperature, only a change in evaporation rate.)
All the while we want to control IR into the water from the room, so the temperature of the walls & ceiling needs to be well maintained. At 300 K, a 1 K shift in the temperature of the walls will change the IR flux about 6 W/m^2. So we would want to control the wall temp within ~ 0.1 K for the control experiment, then raise the wall temperature ~ 2 K to increase the the IR by an appropriate amount.
But the air temperature and humidity must not change, because that would change the evaporation rate and confound the results!
This is not a “throw a thermometer into a cup of water” type of experiment!

tallbloke
August 20, 2011 9:42 am

pochas says:
August 20, 2011 at 8:38 am

I disagree with most of that. Warmer water is more buoyant for a start, so tend upwards. Secondly, water thermally stratifies. This is why the thermocline exists as a relatively sharp boundary. Thirdly, the ocean emits more IR than it absorbs, the net flux is cooling the ocean. Fourthly, the air is colder than the ocean generally, so the heat flow is upwards not down.

pochas
August 20, 2011 12:15 pm

tallbloke says:
August 20, 2011 at 9:42 am
“I disagree with most of that. Warmer water is more buoyant for a start, so tend upwards. Secondly, water thermally stratifies. This is why the thermocline exists as a relatively sharp boundary. Thirdly, the ocean emits more IR than it absorbs, the net flux is cooling the ocean. Fourthly, the air is colder than the ocean generally, so the heat flow is upwards not down.”
If the surface temperature is higher that the sublayer temperature, then heat flow is down.

tallbloke
August 20, 2011 12:50 pm

pochas says:
August 20, 2011 at 12:15 pm
If the surface temperature is higher that the sublayer temperature, then heat flow is down.

I think you probably need to review the discussion of the surface skin layer (warmer underneath than on top), and have a think about buoyancy.

pochas
August 20, 2011 1:46 pm

I’m not going to argue with you, tallbloke. I agree that sunlight is absorbed within the top 100 m of surface and that heat has to be released at the surface, and this involves convection, warm water up and cool water down, yielding a well-mixed surface layer with little temperature variation.
And I’m about to leave it at that.

Myrrh
August 20, 2011 2:35 pm

Willis – please read this.

http://helios.gsfc.nasa.gov/qa_sun.html#sunenergymass
30
How Long for the Sun’s Heat to Reach Earth?
Q. How long does it take heat created on the Sun’s surface to reach Earth? Is it the same as the speed of light?
A. Heat is transmitted through conduction, convection, and radiation. The heat that reaches us from the Sun is infrared radiation, which travels at the speed of light. So, it takes about 8 minutes for it to reach Earth from the Sun.
Dr. Louis Barbier

&

http://sites.bio.indiana.edu/~hangarterlab/courses/b373/lecturenotes/water/waterintro.html
Water
Transparent to visible irradiation (light)

&

http://en.wikipedia.org/wiki/Water_absorption
During the transmission of electromagnetic radiation through a medium containing water molecules, portions of the electromagnetic spectrum are absorbed by water molecules. This water absorption occurs preferentially at certain characteristic wavelengths while the balance of the spectrum is transmitted with minimal effects.

&

http://faculty.weber.edu/nokazaki/Comparative%20Animal%20Physiology/Laboratory/Spectrophotometry04-1.pdf
Spectrophotometers are designed to transmit light of narrow wavelength ranges (see Figure 1 the
electromagnetic spectrum). A given compound will not absorb all wavelengths equally–that’s
why things are different colors (some compounds absorb only wavelengths outside of the visible
light spectrum, and that’s why there are colorless solutions like water).

&
All I’ve given above in other posts.
I don’t know what more I can do to bring this to your attention.
You are talking about “backradiated” infrared. From around 10 microns upwelling from the Earth, which is what we radiate out..
Heat energy direct from the Sun is far more powerful. It’s what you stop feeling on a bright sunlit day when a cloud temporarily hides the Sun and you can feel again when the cloud passes..This is the real, longwave, thermal infrared directly heating the oceans and swimming pools.
Heat, thermal infrared, works on a vibrational level of molecules. It is the heat we feel from the Sun, it reaches the surface in 8 minutes.. Water is a strong absorber of Heat, thermal infrared, and has an extremely high capacity to store it. Water does not absorb Light.
Light, visible, is not thermal energy, it is not hot. It is much smaller than Heat, it works on an electron scale.
Water is transparent to Light, visible. Light is transmitted through unchanged, it is not even absorbed on the electron scale as it is in the atmosphere where it is absorbed by the electrons of the nitrogen and oxygen molecules and bounced back out the way it came.. Visible therefore cannot create heat, it does not heat water because it cannot. It is physically imposssible.
It’s a con.
A deliberate and malign campaign of disinformation. What I have given you is traditional physics. Very well known because tried and tested in real physics and, therefore, real life.
“Many physics teachers traditionally attribute all the heat from the Sun to infrared light.” http://www.newworldencyclopedia.org/entry/Infrared
The above quote mentions this, before disagreeing, most teaching now doesn’t, but erroneously teaches that Heat is Light ..

Transmission
Light is transmitted through substances (such as glass) which we think of as transparent. Transmission occurs when light passes through a substance unchanged, as shown below.
http://water.me.vccs.edu/courses/env211/lesson15_print.htm

Water is transparent to Light, the visible electromagnetic spectrum. Which means it is not absorbed. It cannot heat the oceans.
The backradiated Heat, thermal infrared, is not hotter than the direct Heat from the Sun.
The clouds will also be absorbing thermal infrared from the Sun, however, the main function of the Water Cycle is to take heat away from the Earth; it would be 67°C with our atmosphere but without the water cycle, think deserts.
We really need more traditional teachers of the real basics.

August 20, 2011 3:22 pm

Bob_FJ says: August 20, 2011 at 1:38 am
Willis,
It disturbs me that you still do not acknowledge advice that it does not matter how much EMR* is whizzing around, unless there is a potential difference (PD) between two sources of it. Otherwise, it amounts to nothing in terms of HEAT transfer. Furthermore, unless there is a change in HEAT level in matter, there is by definition NO change in temperature.

Exactly, and that is why the physics literature expresses heat transfer on the potential difference (Ta^4 – Te^4):
E_net = sigma (Ta^4 – Te^4)
rather than using the form assumed by pyrgeometers
E_net = sigma Te^4 – sigma Ta^4
which assumes massive heat transfer is occurring from both cold to hot and vice versa, in violation of the 2nd law.
The whole DLR heating red herring is based on this false assumption of 2 way heat transfer. Doing the calculations this way leads to false answers because it does not consider the fact that if a cold body could heat a warm body, the warm body would simply emit more heat to compensate. Doing the calculation properly using only the potential difference eliminates this error.
http://claesjohnson.blogspot.com/2011/08/how-to-fool-world-by-measuring-masive.html

kuhnkat
August 20, 2011 4:23 pm

Tim Folkerts,
“To maintain energy equilibrium, the temperature of the new, higher radiation layer must adjust to be the same as the older, lower layer had been. ”
Actually, since there would then be more GHG’s at higher altitudes, where there is little delay in IR going to space, the chances are that the extra GHG’s will be as likely to send more IR to space. The idea that an increase of low density will cause a blockage of energy emitted from lower levels simply does not seem reasonable to me.
The real GHE is this idea you are protraying where it is claimed that the extra GHG’s somehow block more radiation from the current levels than they add due to their low temp. My idea is that adding GHG’s does cause a small warming that expands the atmosphere slightly that offsets this blockage. Since the GHG’s added are still minimal as far as actual atmospheric density it simply doesn’t seem believable to me that the conventional explanation will happen.
Over at Curry’s no one has given me any hard data on which way it can be expected to go. As 1998 and 2010 El Nino’s do not appear to have caused a noticeable hot spot, IF it is possible to bottleneck radiation it certainly does not appear to be happening yet.

Spector
August 20, 2011 4:27 pm

Myrrh says: (August 18, 2011 at 5:45 pm)
“It appears that you are suggesting that photons after going so far through the water just get tired and wink themselves out of existence. Not quite–conservation of energy applies.”
“What happens when your car runs out of petrol?”
Cars are not fundamental particles of the universe. They also don’t vanish when they run out of gas. If it is gone when you get back, you are going to assume it was taken away.
Every photon is a packet of energy. That energy can be reconfigured but not destroyed. If photons are observed to be entering the ocean and not coming out, we know that the ocean has acquired their energy. What would you think if you took your car to a garage and later when you came back they told you it had mysteriously vanished?

kuhnkat
August 20, 2011 5:42 pm

Myrrh,
As I and at least two others have pointed out, whether PURE water is TOTALLY transparent to Visible wavelengths or not, the oceans of THIS world Earth contain a rather large amount of organic and inorganic matter that is NOT TRANSPARENT to Visible wavelengths. Our oceans could not provide fish, seaweed, salt, minerals, phytoplankton, mammals, crustaceans and numerous other interesting and tasty things to us if they were pure water. The Arctic and Antarctic would have MUCH more sea ice due to freezing more easily and not absorbing as much energy if they were totally transparent. Sometimes a physical truth is overidden by the messy facts of what we actually are dealing with.
As an article I linked to above shows, there is even a rather busy biozone in the very region under discussion here. Please actually consider this possibility. Denying reality is a lonely job.

Konrad
August 20, 2011 6:05 pm

I have conducted some initial experiments and I believe that backscattered LWIR is not able to slow the cooling of Earth’s oceans to any significant degree. I would encourage others to conduct similar experiments to confirm my results. The equipment required is not overly expensive and the results can be observed in minutes. The results show the dramatic difference between reflecting LWIR back to warm water when its free to evaporatively cool and when it can only cool through conduction and radiation.
What is required –
Two identical probe type digital thermometers with 0.1 degree resolution
Two identical insulated water containers (I used 5 litre Styrofoam cooler boxes)
One IR reflector (I used an A4 sheet of 10mm Styrofoam with aluminium foil attached with spray adhesive.)
One IR window (I built an A4 size “picture frame” of 10mm square balsa wood strips and stretched cling film over it.)
One 10 litre bucket
Extra cling film
Optional extras – computer fans, transformers and an A4 ”dark cool sky” panel of matt black aluminium with ice cooling.
What to do –
Tape probe thermometers in identical positions in both water containers. (I placed the tips 25mm below the water line)
Position IR reflector and IR window 50mm above either water container. You may need to built balsa wood legs or two Styrofoam side walls, but air must be free to move over the surface of the water. (The use of the IR window is to ensure that air flow is similar over each water container.)
Fill bucket with warm water, mix, then fill each water container from the bucket. I used water around 40c as the ceiling was around 18c not a 3k sky.
Observe the temperature change over time for each tank. Less than an hour is required for such a small amount of water. Use a long soft paint brush or similar to keep IR window and reflector free of condensation. You should observe that both tanks cool a the same rate.
Now the important bit – Repeat the experiment, but this time lay a sheet of cling wrap on the surface of the water in each water tank. This allows cooling through radiation and conduction but prevents evaporation. You should be able to observe the water under the IR reflector cools slower.
If results are difficult to observe, try using smaller insulated water containers.
From what I have observed, backscattered LWIR can slow the rate at which substances cool. However in the case of liquid water that is free to cool evaporatively this effect is dramatically reduced. It would appear that including the oceans in the percentage of Earth’s surface that could be affected by backscattered LWIR may be a serious error.
I will conduct further tests using fans to simulate wind, a cooler “Sky” over the IR window and salt in the water. I would encourage others to conduct similar empirical experiments and share their observations. This thread is now over 500 comments long. The time for hand waving and analogising is surely over. Real world observations please.

Myrrh
August 20, 2011 6:30 pm

Spector says:
August 20, 2011 at 4:27 pm
Cars are not fundamental particles of the universe. They also don’t vanish when they run out of gas. If it is gone when you get back, you are going to assume it was taken away.
Let me re-phrase that. What has happened to the petrol in your car when your car stops because it has ran out of gas?
You are postulating that light created by the Sun is eternal.
You are taking laws out of context…
A well-tried and tested and obviously successful way for AGWScience Fiction Inc to spread its disinformation memes.
You should also take into consideration that light slows down dramatically in water, which is, iirc, some 800 times denser than air which is the volume of our fluid gaseous atmosphere, weighing a ton on your shoulders… A plane will use a considerable amount more fuel flying into headwinds because there is resistance, so, similarly, an electromagnetic wave is affected by the density of the medium it travels trhrough. An opaque medium is one which blocks it, so it can’t get through. Sound waves can travel through walls, light waves can’t. So you cast a shadow (though I’m not sure about the people who create these memes).
Back in the finite real world where things get created and destroyed… I gather some sticks on a cold dark night and some time after vigorously drilling one piece with another I get a spark which sets the whole alight until it burns out and only ash is left. There is a finite amount of wood. There is a finite amount of Light and Heat and ash created. The Heat warmed the surroundings, the Light lit up the surroundings, and then there was only a small pile of ash to remind me that I felt warmer and could see better with a fire than without.
There is another well used AGWScience Fiction Inc meme – that all electromagnetic energy is the same. It ain’t. A radio wave is not a gamma ray. This meme is promoted to turn all ‘energy’ into ‘heat’, (‘and here’s the convenient law to enable you to work it out’), and to stop you thinking.
It’s called properties when one looks at the differences between things. They are different sizes, do different stuff, work in different ways on different levels, as the wiki page I gave above describes the differences is a good place to start. One of the ways Light energy is used which does not create heat is in photosynthesis, this is a chemical change, the creation of sugars. Some 90% of the oxygen in our atmosphere is produced by the photosynthesis happening in the oceans. That’s a lot of chemical change that does not produce heat. Even if Light could heat water, which it can’t.
Good luck trying the find the differences explained well, without omissions or downright lies from the AGWScience Fiction Inc’s overbearing influence..
…but it was created.. There’s still hope for us.
http://wattsupwiththat.files.wordpress.com/2011/08/web-ucscalendar-notnuez-attack_sceptical.jpg
(From Marc Hendrick http://wattsupwiththat.com/2011/08/20/saturday-silliness-concerned-cartoon-contest/#more-45511)

jae
August 20, 2011 7:46 pm

Willis seems to be losing it:
“jae, your question doesn’t get answered because it is based on a false assumption. The fallacy is that downwelling longwave radiation can’t be demonstrated empirically.
Willis, I humbly beg you to practice what you preach. Just where the hell did I make this assumption???? Please respond to what I wrote, instead of what you “heard.” DOES THAT SOUND FAMILIAR?
AND I DON’T SHARE YOUR FEAR/DISGUST/?? ABOUT ALL CAPS. IT IS SIMPLY A WAY OF EMPHASIZING SOMETHING. IF YOU DON’T LIKE IT, TOUGH DONUTS, “COWBOY.”

August 20, 2011 7:54 pm

Tallbloke writes “Is this the constructal law which Willis likes at work? Or simply the entropy law at work?”
Certainly at all times the entropy is maximised. There is a general flow of energy to space. I’m not really familiar enough with the nuances of constructal theory to know whether the parts of the ocean interacting to maintain balance represent a system self “constructing” to do that job but it certainly seems possible.

jae
August 20, 2011 8:17 pm

Willis:
Science is ALL about EMPIRICAL EVIDENCE, not diagrams, theoretical prose, “shell models,” etc. Even Einstein was held to that requirement. Perhaps you need to think more about that.
There is absolutely NO empirical evidence for an “atmospheric greenhouse effect,” so all your prose about DLR, etc. has no import. Sorry, man, but you gotta face the facts. In FACT, the empirical evidence is going the other way, if you haven’t noticed (more OCO, less temperature–for about 15 years….No “hot spot” in the upper tropopause, as predicted by your shell nonsense….No warming in the oceans….No increase in sea level…
When do you let go of a flawed theory??

jae
August 20, 2011 8:29 pm

One more thought for Willis:
“I can feel DLR, it’s warmer when a cloud passes over during a winter night. When the cloud goes away, the clear sky just sucks the warmth out, but when the cloud is over me, I can feel the warmth from the cloud. So I know from my experience that there is DLR, regardless of what EINSTEIN AND OTHER SCIENTISTS might say about it.”
BALONEY. This “feeling” is just typical of the guiding principle of all liberals and has no bearing in the scientific literature, AFIK. Prove it with some data, fella!

jae
August 20, 2011 9:06 pm

Oh, and dumb me, I just realized that Willis DID NOT EVEN RESPOND TO THE GIST OF MY POSTS! He just did a RealClimate on me!! I can do a lot of speculating as to why, but I will leave it to his highness to explain this problem.

August 20, 2011 9:27 pm

Jae,
Re this oft repeated observation:
“I can feel DLR, it’s warmer when a cloud passes over during a winter night.”
Clouds come in warm moist air that originates in the tropics traveling polewards.
If there is no cloud and the air is drier. Its been through the wringer somewhere and had its moisture removed. In the process it got colder. Local temperature very much depends upon where the air is coming from.When it gets cold you know the rain is finished.
There is always some direct long wave radiation leaving the surface but as a heat transfer mechanism in the atmosphere radiation is insignificant when you compare it with conduction, evaporation and convection. The exception is when water is either very cold or not present.

kuhnkat
August 20, 2011 9:45 pm

Myrrh,
“You are postulating that light created by the Sun is eternal.”
You do realize that most Cosmology and Physics do come close to that?? Remember the Microwave Background Radiation?? It has alledgedly been literally flying around the universe since the BB!!! That would be about 14 billion years the last I checked. If you disagree with BB, I can understand that. If you think electromagnetic radiation attenuates because it uses up its initial energy just propagating in less than one light year, I think we have a problem. If you think the oceans are so pure that visible light doesn’t get absorbed or reflected by something, you have a serious reality issue.

kuhnkat
August 20, 2011 9:48 pm

Jae,
if a problem is so poorly stated or of no value it should be rather trivial to point this out. Ya gotta wonder with that type of treatment. He seems to be morphing to another Steven Mosher. Going from reality based to conformist.

Bob_FJ
August 20, 2011 10:45 pm

Willis, in my August 17, 2011 at 12:49 am above I concluded with:

“…As I pointed out earlier, it doesn’t matter how much EMR is flying around; unless there is a PD, [potential difference] there is ZERO HEAT transfer. This is nicely demonstrated in any elemental layer of air in a discreet parcel, wherein most of the radiation as a consequence of its temperature is horizontally opposed in all directions. Consequence = nothing.
The 400 W/m^2 that you imply as HEAT leaving the surface, is NOT HEAT, but a different form of energy described as EMR, which does not in itself cool the surface.
Do you have any problems with this NASA Energy Budget Diagram?
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif

I think that many here would appreciate your determination on this.
BTW, when I say NASA, I mean that many divisions of NASA use it. On the other hand, I have not found any of them using the Trenberth/IPCC cartoon at all!

Bob_FJ
August 20, 2011 11:04 pm

Further my post just above, I tested the link I gave and see that it is declared down for a few days.
You could try the following instead, or search around elsewhere; whatever you might feel like:
http://en.wikipedia.org/wiki/Earth's_energy_budget
There are a bunch of similar diagrams in similar vein around the ether.

tallbloke
August 21, 2011 1:39 am

Bob, is this the one you’re trying to link?
http://tallbloke.files.wordpress.com/2010/06/energy-budget-new.jpg

David A
August 21, 2011 1:47 am

Would anyone be so kind as to answer the question I asked in line six of this post? David A says:
August 20, 2011 at 6:06 am

RJ
August 21, 2011 4:06 am

jae says:
August 20, 2011 at 8:17 pm
When do you let go of a flawed theory??
A good question. Yet so many sceptics still believe this flawed GHG backradiation theory.
It means for example if say 100 units of energy are in a box. 25 leave and one (8% times 50%) return (after hitting CO2. 92% pass straight though but 8% do not)
Somehow this returning 1 unit of energy (LWR) replaces not only the energy from the other 24 units but also quite a bit more. It is total and utter nonsense.

richard verney
August 21, 2011 4:13 am

Jae
See my comments of 20th August at 5:01 am.
I have made similar observations.
I can understand your frustration.
Willis has gone very quiet recently, which is disappointing given that he raised this debate. It is likely that Willis is a little out of his depth. This is quite understandable given the lack of empirical data and experimentation carried out on the issues involved. The upshot of this is that inevitably there will be a significant degree of speculation behind the vast majority of arguments being put forward no matter from which side those arguments come. That is the truly amazing fact that comes out of the AGW theory/conjecture. It has been rumbling on for 30 years or so, and no proper experiments have been carried out testing the fundamental issues involved. Quite extraordinary.
Good to see that Konrad has done an experiment. However, I suspect that laboratory conditions are required, before one can draw conclusions. Nonetheless, his experiment is a pointer, and is generally consistent with my view that AGW very probably does not work over water due to the different processes involved. The phase changes of water are fundamental to how the system works and the true consequences and effects of this are in my opinion not fully accounted for.

richard verney
August 21, 2011 4:25 am

David A
People cannot answer your question because the investigatory experiments have not been carried out. Accordingly, the empirical data does not exist so that quantative figures can be put forward.

Konrad
August 21, 2011 5:47 am

richard verney says:
August 21, 2011 at 4:13 am
“Good to see that Konrad has done an experiment. However, I suspect that laboratory conditions are required, before one can draw conclusions”
No Richard, laboratory conditions are not required. While I can achieve those conditions (my new “Dark cool sky “ will be cooled to -50c with a peltier chip), I specifically designed the original experiment to be repeatable by others. I encourage you to attempt it. As to my technical competency, I yell at the screen every time I see the “Mythbusters” try to do controlled experiments. How many years FX experience? Those guys don’t even have movie credits!

Tim Folkerts
August 21, 2011 5:53 am

RJ says:
August 21, 2011 at 4:06 am
It means for example if say 100 units of energy are in a box. 25 leave and one (8% times 50%) return (after hitting CO2. 92% pass straight though but 8% do not)
Somehow this returning 1 unit of energy (LWR) replaces not only the energy from the other 24 units but also quite a bit more. It is total and utter nonsense.

You have the analogy wrong. If your understanding is flawed, it is no wonder you think the theory is flawed. The one key detail you left out of your analogy is the sun! The sun in your analogy would be providing 25 units of energy each second to the “box” (which clearly represents some section of the earth’s surface
CASE 1: No GHG. Every second the “box” gets, say, 25 units of energy from the sun. The “box” adjusts its temperature until it looses 25 units of energy via IR radiation.
CASE 1: With GHG. If I now add a layer of CO2, then two units of energy that leave the box get adsorbed by the CO2. Assuming CO2 is also at its equilibrium, it has also adjusted its net energy flow to be zero. It will radiate one unit to the ground and one unit to space.
The one returning unit up energy PLUS THE 25 NEW UNITS FROM THE SUN, result in the “box” having one extra unit = more thermal energy = higher temperature. Eventually the box (and atmosphere) will equilibrate at some new, higher temperature.
Of course, this is a gross over-simplification. To get closer to the real world, we need multiple layers of CO2. And water vapor. And clouds. And evaporation. But that does not change the basic principle governing thermal equilibrium or IR radiation to/from the atmosphere.

Jonathan T Jones
August 21, 2011 6:27 am

jimmi_the_dalek says:
August 18, 2011 at 4:00 pm
“slowing the cooling” can lead to a temperature increase in cases where there is a continuous input of energy. To use one of those analogies which plague this area, consider a dam which has a steady inflow of water, but there are several leaks in the dam. Now plug one of the leaks (slowing the leaking), and what happens – the water level rises, increasing the amount stored, until it finds a new leak.

Sorry, but you are making the same fundamental mistake that WIllis is, you are attributing the properties of your inflow to your outflow. The only way to increase temperature (or in your analogy, to increase the water level) is to add heat (or add water). slowing the loss of heat (or the loss of water) does not do that. You can see this in your water level analogy by doing a simple experiment – fill a swimming pool that has a leak in it. plug the leak and then turn off the hose that was filling the pool, does the water leave rise after you plug the leak and turn off the hose? no not at all, because plugging the leak (slowing the loss of water) does not add a single drop of water to the pool – only the inflow of water from the hose does that. So “slowing the cooling” is not the same thing as warming because only one of those two things is the actual cause of increasing temperature

A. C. Osborn
August 21, 2011 6:53 am

Konrad says:
August 20, 2011 at 6:05 pm
I like your simple experiment, you appear to have proved the “Greenhouse” effect just like a greenhouse when no evaporation takes place.
But in the real world it has little effect because a lot of evaporation does take place.
Of course it is far too simple for scientists as it doesn’t involve expensive machines and computer models.

A. C. Osborn
August 21, 2011 6:54 am

To say nothing of Convection as well, which is what the evaporation turns in to.

A. C. Osborn
August 21, 2011 6:58 am

Tim Folkerts says:
August 21, 2011 at 5:53 am
RJ says:
August 21, 2011 at 4:06 am
The one key detail you left out of your analogy is the sun!
You seem to have frogotten that the sun does not shine on the same surface 24 hours per day. so half the time RJ is correct.

A. C. Osborn
August 21, 2011 7:00 am

Konrad says:
August 21, 2011 at 5:47 am
Did you run your experiment at night time?

Tim Folkerts
August 21, 2011 7:22 am

Let’s work with this “potential difference” idea.
The sun is at a vastly higher “thermal potential” than the earth. There is a huge flow of thermal energy from the sun to the earth. The amount of energy flow is determined not only by this “thermal potential difference”, but also by the properties of the earth (albedo) and the geometry of he situation (if the earth were closer to the sun, the sun would appear larger in the sky and we would get more energy, and the earth is a sphere).
(This is just like the idea that the energy delivered by an electric potential difference depends of the geometry of resistor and why it is made of.)

E_net = sigma (T_sun^4 – T_e^4)*(geometry) = 342 W/2^2

Adding in the albedo reduces this to about 235 W/m^2
Space is at a vastly lower “thermal potential” than the earth. And space is visible in basically every direction EXCEPT where the sun is. If there was nothing else in the way, the equation would read.

E_net = sigma (T_space^4 – T_e^4) = sigma (3^4 – 288^4) = – 390 W/m^2

Of course, the earth would be cooling dramatically if that were the case. Without GHGs and clouds, we would have to use 3 K as the temperature of the “cold side”, not the temperature of the atmosphere, since the atmosphere would not be involved in radiating.
Fortunately we, have SOME of this “leak” plugged by the atmosphere.

Tim Folkerts
August 21, 2011 7:41 am

Jonathan T Jones says:
August 21, 2011 at 6:27 am

Sorry, but you are making the same fundamental mistake that WIllis is, you are attributing the properties of your inflow to your outflow. The only way to increase temperature (or in your analogy, to increase the water level) is to add heat (or add water). slowing the loss of heat (or the loss of water) does not do that. You can see this in your water level analogy by doing a simple experiment – fill a swimming pool that has a leak in it. plug the leak and then turn off the hose that was filling the pool, does the water leave rise after you plug the leak and turn off the hose? …

But “turning off the hose” = “shutting off the sun” !
The better analogy is something like this …
A swimming pool (assume it is above ground) has a crack along one side. The deeper the water, the faster it will leak. If you keep pumping water into the pool, the water level will settle at some depth — lets say 3 ft deep.
Now we create a rather bizarre solution to our problem. We build a small pool onto the side of the original pool WITH THE SAME CRACK in it. The water has to leak from the original pool thru the small pool and then out onto the ground.
Since the small pool as the same crack, it will fill to a height of 3 ft to maintain a leak rate equal to the rate the hose is filling the original pool.
BUT the original pool has to be ABOVE 3 ft deep to leak into the second pool! If the original pool an the second pool were both at 3 ft. there would be no pressure driving the water from one to the other (no “potential difference” as others have called it.
The second pool added NO WATER the the first, but it did result in the 1st pool being deeper than in had been, because at least for a while it slowed the flow out of the first pool !
(it is still not a perfect analogy by any means. I can think of several improvements of the top of my head. But now it illustrates the ability of the intermediate atmosphere to help raise the temperature of the surface without the atmosphere having to “create” any energy. The atmosphere simply slows the escape of the energy, resulting in a higher temperature of the surface.)

richard verney
August 21, 2011 8:04 am

Willis
Please just answer this one simple question.
If DLR is ‘real’ as you allege (and not something which in effect is just a signal not capable of performing sensible work in the energy system we experience here on Earth, as I allege), please advise why DLR meters/pyrgeometers are not scaled up (on an industrial scale) and used to produce energy/electricity (or similar) to power something useful for mankind and/or otherwise supplement his energy needs.
After all, if they are measuring power, this power could be exploited. Especially given that there are many places on Earth which are rather cloudy (such as the UK and Northern Europe which places would greatly benefit from being able to tap into the warm clouds which according to you supply so much DLR) and which places would benefit from a 24/7 energy source available come rain, cloud or shine.
I look forward to seeing your answer to this straight forward and simple question.

tallbloke
August 21, 2011 8:29 am

richard verney says:
August 21, 2011 at 4:13 am
Willis has gone very quiet recently, which is disappointing given that he raised this debate. It is likely that Willis is a little out of his depth. This is quite understandable given the lack of empirical data and experimentation carried out on the issues involved.

Richard, those of us who like yourself have put some intensive thought into this fundamentally important issue find it disheartening that fellow sceptics with strong voices like Willis have swallowed the AGW proponents arguments on this issue. It’s clear to me, you, Tim the Tool man, Dave Springer and many others that they have it badly wrong.
Where does ‘back radiation comne from? Well, it was first emitted by the ocean, thereby cooling it. Then something less than half of it gets bounced back down again, along with ‘new’ longwave derived from the absorption of solar shortwave in the atmosphere. Yet instead of differentiating between these quantities, the pro AGW camp just lump it all together and claim it ‘warms’ the ocean by direct absorption. – Just insane IMO.

G. Karst
August 21, 2011 9:02 am

Well, it certainly has been a long and contentious discussion with lots of interesting positions. It seems that most are over confident, that the whole process of photon annihilation is quite well explained and understood. This confidence seems to also extend to the surface tension layer of water as well.
I recall the same identical arguments taking place in a room 40yrs ago in a room full of physicists, chemists, and engineers trying to achieve a heat balance account of a nuclear pile. Nothing much has changed since.
The nano-surface of water is not ordinary water at all. Some will claim it is not ordinary space either. Chemistry and even convection, evaporation, exhibits unusual behavior. Even photon conversion to electron flow, within a solar cell creates heated debate. Is it a void migration or a particle flow?
Take a high energy photon, pass it near certain nucleonic field, and some photons, no longer want to be a discrete packet and split into a positron/electron pair, which depart from each other at specified entangled angles (pair production effect). Matter from light. Indeed E=MC2 quantifies the effect, but does not explain it.
To think all these issues will be resolved within a blog discussion is a little beyond, a reasonable expectation. Quantification is the best we can expect at the moment. Experimental data is the only way, for now. So discuss all you want, but do not scoff at each other’s hypothesis, because none of you are completely correct. IMHO GK

A. C. Osborn
August 21, 2011 9:05 am

Tallbloke, what did you think of Konrad’s simple experiment, it appears to show that both camps are correct to a degree (no pun intended).
There is a slight Greenhouse effect, but it disappears when evaporation/convection occurs.

August 21, 2011 9:11 am

richard verney says: August 21, 2011 at 8:04 am
Willis
Please just answer this one simple question.
If DLR is ‘real’ as you allege (and not something which in effect is just a signal not capable of performing sensible work in the energy system we experience here on Earth, as I allege), please advise why DLR meters/pyrgeometers are not scaled up (on an industrial scale) and used to produce energy/electricity (or similar) to power something useful for mankind and/or otherwise supplement his energy needs.

This ‘experiment’ has effectively already been done using concentrating solar cookers – which show that the DLR claimed to have TWICE the power of the Sun when concentrated causes COOLING. On the bright side, mankind can benefit from using concentrated DLR as a refrigerator for the 3rd world. If you actually want to heat something, you have to use the Sun (even though according to AGWers “backradiation” has twice the heating power).
http://solarcooking.org/research/McGuire-Jones.mht
http://chriscolose.wordpress.com/2010/05/08/stoat-taking-science-by-the-throat-latest-posts-archives-about-rss-contact-profile-me-my-family-and-me-more-make-sure-youre-familiar-with-the-comment-polic/#comment-2363
http://hockeyschtick.blogspot.com/2010/06/why-conventional-greenhouse-theory.html

ferd berple
August 21, 2011 9:50 am

tallbloke says:
August 21, 2011 at 8:29 am
Where does ‘back radiation comne from? Well, it was first emitted by the ocean, thereby cooling it.
This is a very good point. Much if the DLR from the clouds is heat that was removed from the oceans. You have to be very careful when calculating any budget (energy or otherwise) to avoid double counting the same item (energy) twice and creating a false picture.
We see this problem all the time in financial accounting. It is one of the great strengths of double entry accounting is that high-lights double postings that would otherwise be hidden by single entry accounting.
The energy budget created by climate science is single entry accounting and subject to undetected double posting errors. If anything, Trenberth’s “missing heat” reminds me of a single entry financial system that is “out of balance”. Something has been overlooked or double posted, creating missing energy.

A. C. Osborn
August 21, 2011 9:58 am

Hockey Schtick says:
August 21, 2011 at 9:11 am
The Solar Cooker tests were fascinating, particularly the night time ones. An absolute proof that down welling IR does no heating work.
I conducted some similar daytime tests about 10 years ago in the UK using a Torch 4″ parabolic Reflector. A piece of 1/2″ diameter steel placed at the bulb’s Focal point reached 350 degrees C after a few minutes when the sun was focused on that point. We wanted to repeat the tests using the Reflector from a 12″ Searchlight but were unable to obtain one at a reasonable price.
The overall intention was to run a Copper Pipe loop with water through the focal point to see how much water we could heat up. But we were distracted by other things.

ferd berple
August 21, 2011 10:39 am

G. Karst says:
Experimental data is the only way, for now. So discuss all you want, but do not scoff at each other’s hypothesis, because none of you are completely correct.
The history of science shows this time and time again. “Why” something happens is never settled in science. As new understanding arises, new theories about “why” something happens will always emerge. This is the great fallacy of science, that by understanding “why” we will somehow be able to predict. Much of the current problem in science is a result of this mistaken belief.
What science teaches us is the art of prediction, without requiring an answer to “why”. A dog chasing a rabbit will never catch the rabbit if it aims for the rabbit. By the time the dog arrives the rabbit will have moved. To be successful the dog must predict where the rabbit will run, and arrive there at the same time.
This is the essence of science. To predict where the rabbit will run. “Why” the rabbit is running is not an interesting question. Maybe the rabbit is running to get away from the dog, or maybe it just likes to run in random patterns. It doesn’t affect our ability to predict.
What matters in science is the ability to predict where the rabbit will run, not why the rabbit is running. As we learn in English. Who, What, When, Where – these are adverbs – they modify action . “Why” does not belong, it isn’t science, it is philosophy.

August 21, 2011 10:57 am

ferd berple says: August 21, 2011 at 9:50 am
You have to be very careful when calculating any budget (energy or otherwise) to avoid double counting the same item (energy) twice and creating a false picture.

Exactly – and here’s why the AGWers go wrong on the accounting – they assume massive heat transfer is occurring from cold to hot and vice versa, failing to consider that if a cold body could (in violation of the 2nd law) cause any heating of a hot body, the hot body would simply emit this additional heat to compensate. In their single entry accounting system they only assume the former, not the latter.
The proper form of the equation for heat transfer shown in the physics literature acts as a dual entry accounting system to eliminate this error. The AGWers fail to understand this.
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-724144

Tim Folkerts
August 21, 2011 11:18 am

A. C. Osborn says:
August 21, 2011 at 6:58 am
You seem to have frogotten that the sun does not shine on the same surface 24 hours per day. so half the time RJ is correct.
My goodness! If this is what you are worried about, let me change the analogy … “The hose runs for 10 minutes, then is shut off for 10 minutes, continuing this cycle indefinitely”. The pool will fill slightly during the 10 minutes the hose is on, an empty slightly during the 10 minutes the hose is off.
This changes absolutely NOTHING about the analogy. Basically you are pointing that the earth cools at night and warms during the day. The atmosphere is STILL acting as a buffer between earth and the 3 K temperature of outer space. The earth is warmer day and night than it would be without that buffer.

ferd berple
August 21, 2011 11:21 am

“Hockey Schtick says:
August 21, 2011 at 9:11 am
The Solar Cooker tests were fascinating, particularly the night time ones. An absolute proof that down welling IR does no heating work.”
The simple fact that you cannot focus the DLR to reduce the rate of cooling of the surface is fairly strong observational evidence that DLR is not affecting the cooling rate.
Otherwise, we could all place large reflectors outside our houses to reduce the rate of cooling at night. We know for example, that glass (greenhouse) reflects IR. So in theory we should be able to construct heat engines by reflecting the DLR to the hot side of the engine, shading the cools side from the DLR.
So, for example, if you placed two trays of water outside at night, one with a sheet of glass reflecting the DLR onto the other, then the tray with the reflected DLR should cool much slower than the tray shielded from the DLR.
Give this a try and report back to us. Don’t place the glass so close to either tray that if affects evaporation, simply place it so that it intercepts the DLR reaching one tray and concentrates it on the other.
I’m pretty sure that there will be no difference between the two trays, otherwise someone would have used this very simple method to optimize net heating and cooling at night. Given the price of energy and the vast amount of DLR as compared to solar energy, the economics are too great for this to have been overlooked.

Tim Folkerts
August 21, 2011 11:41 am

Fred Berple says :
Much if the DLR from the clouds is heat that was removed from the oceans. You have to be very careful when calculating any budget (energy or otherwise) to avoid double counting the same item (energy) twice and creating a false picture.
I think you (and many others) need to rethink your paradigm. I challenge you to think about what happens between today and tomorrow.
The earth and atmosphere ALREADY have a great deal of thermal energy. They have had billions of years to get to where they are now, so the original source of that energy is not important. What IS important is what happens from here forward. What IS important is the energy balance of each part of the system.
During the next 24 hr on average, the surface:
* gains ~ 170 W/m^2 via photons from the sun
* gains ~ 320 W/m^2 via photons from the atmosphere.
* loses ~ 80 W/m^2 via evaporation
* loses ~ 20 W/m^2 via convection
* loses ~ 390 W/m^2 via photons (some of which end up in the atmosphere, and some of which end up in space.)
NET CHANGE = ~ 0 W/m^2
energy is conserved; nothing is double counted.
During the next 24 hr on average, the atmosphere:
* gains ~ 70 W/m^2 via photons from the sun
* gains ~ 350 W/m^2 via photons from the surface.
* gains~ 80 W/m^2 via evaporation
* gains~ 20 W/m^2 via convection
* loses ~ 520 W/m^2 via photons (some of which end up in the surface, and fewer of which end up in space.)
NET CHANGE = ~ 0 W/m^2
energy is conserved; nothing is double counted.
The atmosphere is not “returning energy that belongs to the surface”. The atmosphere is “giving out some of it’s own energy”.

Tim Folkerts
August 21, 2011 11:55 am

Hockey Schtick says:
August 21, 2011 at 10:57 am
The proper form of the equation for heat transfer shown in the physics literature acts as a dual entry accounting system to eliminate this error. The AGWers fail to understand this.
EVEN USING YOUR “PROPER EQUATION”
WITHOUT GHGS and clouds, the equation would be

E_net = sigma (T_space^4 – T_earth^4) = ~ -390 W/m^2

since the earth would be radiating back and forth with space at 3 K.
WITH GHGS and clouds, the equation would be

E_net = sigma (T_atmosphere^4 – T_earth^4) = ~ -70 W/m^2

Your own “proper equations” show that GHGs & clouds reduce energy loss by ~ 320 W/m^2 from the surface. I know that if I reduce the energy loss thru the walls of my home while keeping the input from the furnace the same, my house will be a higher temperature!
(To be a little more accurate, the second equation should actually some combination of the interactions with the atmosphere and with space. The ~ 70 W/m^2 is a weighted average of the interaction with space and the interaction with .)

Tim Folkerts
August 21, 2011 12:36 pm

ferd berple says:
August 21, 2011 at 11:21 am
So, for example, if you placed two trays of water outside at night, one with a sheet of glass reflecting the DLR onto the other, then the tray with the reflected DLR should cool much slower than the tray shielded from the DLR.
Actually, something like that DOES work. Google “night-sky cooling”.
Or more familiarly, the “sheet reflecting DLR” could be a cloud, which is very effective at directing thermal IR down at the ground. Ask yourself then, do things cool more slowly on clear nights without so much DLR, or on cloudy nights.
Or yet again, a car under a carport will often have no dew on the windows when an uncovered car does have dew. Why? The sides of the carport are open, so even a slight breeze will ensure the same temperature air for both cars. The covered car has a “DLR” source above (the roof at ambient temperature). The uncovered car has a DLR source above it at a much cooler temperature.
SUMMARY: Increase the DLR, the objects cool slowly. Decrease the DLR, and the objects cool more quickly.
(PS. the terms “DLR” & “ULR” are potentially misleading. It should be “ILR” (incoming longwave radiation) and OLR (outgoing longwave radiation). For the earth’s surface, LR can only arrive from above heading down. For a container of water, much of the “DLR” would actually be coming from the ground.)

Tim Folkerts
August 21, 2011 12:47 pm

PS Here is a “cool” link to radiative cooling. http://solarcooking.org/plans/funnel.htm
It includes actual construction details and actual data.

“How to Use the Solar Funnel as a Refrigerator/Cooler
A university student (Jamie Winterton) and I were the first to demonstrate that the BYU Solar Funnel Cooker can be used – at night – as a refrigerator. Here is how this is done….
As a result, the cooking vessel now becomes a small refrigerator. We routinely achieve cooling of about 20º F (10º C) below ambient air temperature using this remarkably simple scheme. …
That night, the minimum outdoor air temperature measured was 47.5 degrees – but the water in both jars had ICE.”

RJ
August 21, 2011 1:11 pm

Tim Folkerts says:
August 21, 2011 at 5:53 am
“The one key detail you left out of your analogy is the sun!”
no I have not. where does the initial energy or heat come from
The sun and only the sun warms the planet. Back radiation does not. This GHG backradiation theory is just illogical nonsense.
My analogy is valid. At best LWR very slightly slows the cooling rate by a small fraction. But as radiation travel at the speed of light the reduced cooling is very small. Energy leaves the surface in part (but what %?) as radiation (at the speed of light) and a small fraction of this energy (4% at most) might return due to CO2. But with less heating impact than when it left the surface.
And as cool can not warm hotter etc. The GHG backradiation theory should be kicked into touch. Let warmists / alarmists hold onto it. But sceptics should no longer believe in this theory.

RJ
August 21, 2011 1:22 pm

tallbloke says:
August 21, 2011 at 8:29 am
“Then something less than half of it gets bounced back down again”
Much less than half due to CO2. Around 92% of radiation leaving the surface just passes straight through CO2
So in fact less than half of 8%. (and this is only of the small percentage that does actually leave as radiation).

richard verney
August 21, 2011 1:23 pm

@Hockey schtick at August 21, 2011 at 9:11 am
////////////////////////////////////////////////////////////////////////////////////////////
Unfortunately, your response is typical of the confusion propagated in the AGW debate.
Solar cookers can concentrate ‘hot’ solar 5800K radiation to heat something/to cook. Solar cookers can concentrate ‘cold’ 250K and below DLR radiation to cool something/freeze something.
What solar cookers can not do is concentrate ‘cold’ DLR to in some way raise something above ambient temperature.
Contrast solar energy which we can use to produce electricity which can then be used to heat (your kettle) or cool something (your fridge), or we can use solar energy to run a turbine thence to create electricity etc. DLR cannot be used in this way.

richard verney
August 21, 2011 1:43 pm

@Hockey schtick
I fired off my last response without fully/properly reading your comment. Sorry for that.
I think that we are both agreed, that ‘solar cookers’ can when using DLR only cool, not heat. They do not demonstrate that DLR is somehow warming the ground or even slowing the rate of cooling.

RACookPE1978
Editor
August 21, 2011 1:50 pm

OK. Tim, Willis, and all others who wish to contribute:
Please try use those (average, whole earth) values and equations to address this specific question at this specific location:
Two 1 square meter areas in the Arctic at latitude 85 are hit by energy. The first is smooth but ice-covered, the second is calm water, open ocean.
Each is hit by 50 watts/meter ambient rays from directly above, and by 100 Watts/meter^2 direct rays at 85 degrees incidence angle.
Emissivity of open water = .98; emissivity of ice = .99
Albedo of open water = 0.07; albedo of ice = 0.80
At 85 degrees zenith angle, 60% of received energy is reflected from a smooth surface.
Temperature of open water = 276K,
temperature top of ice = temp air = 274 K,
temp of space = 3K
What are the net radiation fluxes into (and out of) the two meter squares.

G. Karst
August 21, 2011 1:51 pm

ferd berple says:
August 21, 2011 at 10:39 am
…What science teaches us is the art of prediction, without requiring an answer to “why”. A dog chasing a rabbit will never catch the rabbit if it aims for the rabbit. By the time the dog arrives the rabbit will have moved. To be successful the dog must predict where the rabbit will run, and arrive there at the same time.
…As we learn in English. Who, What, When, Where – these are adverbs – they modify action . “Why” does not belong, it isn’t science, it is philosophy.

Why-o-why are you going on about why? Nowhere in my comment did I use that word.
Obviously, you have never hunted rabbit or any other game for that matter. Rabbits are not rubber toy balls thrown for your dog to intercept. They have little rabbit brains and big rabbit eyes which see a dog on a intercept vector, and change course frequently. A successful hunt of any game depends on knowledge of the game and his habits and habitats. Knowing why a (ie) rabbit behaves like a rabbit is essential to predicting where he will be. However I cannot see what any of this has to do with photon annihilation. GK

Phil.
August 21, 2011 2:04 pm

Ian W says:
August 15, 2011 at 2:54 pm
If the light is at any angle less than about 40 degrees then most of the light reflects from the surface (see Willis’ last post). This reflection of sunlight at low angles of insolation is another reason that the sunlight on the oceans near the poles does not warm the water much (as you pointed out).

This is incorrect, 50% reflectivity occurs at angles of less than 10º to the horizontal, the Brewster angle for water is ~53º.

kuhnkat
August 21, 2011 2:29 pm

Tim Folkerts,
“Of course, this is a gross over-simplification.”
Yes it is. You left out the major player, convection, that GHG’s encourage, that takes all that extra heat the GHG’s transfer to their local parcel to the upper trop where it is radiated away.

kuhnkat
August 21, 2011 2:47 pm

Tallbloke,
“Where does ‘back radiation come from? Well, it was first emitted by the ocean, thereby cooling it. Then something less than half of it gets bounced back down again, along with ‘new’ longwave derived from the absorption of solar shortwave in the atmosphere.”
I would suggest that it may be much worse. That is, in the lower trop, GHG’s collide with other particles more often then emitting spontaneously. This would mean a lot of that energy would be used in warming the local parcel and driving convection instead of being radiated up and down. I would love to see some paper on what the breakdown actually is!! Apparently this holds at least until we get to the strat where the pressure is much less.

Tim Folkerts
August 21, 2011 2:48 pm

RACookPE1978 says: August 21, 2011 at 1:50 pm
>Please try use those (average, whole earth) values and equations to address
>this specific question at this specific location:
I’m not sure what your point will be, but I’m game. I will use your numbers, which seem to be in the right ballpark, but which I have not verified independently.
>Each is hit by 50 watts/meter ambient rays from directly above,
I am assuming these are the “DLR” (thermal IR) from the atmosphere.
>Emissivity of open water = .98; emissivity of ice = .99
Well, this allows us to figure out how much IR is absorbed
WATER: 50 W/m^2 * 0.98 = 49 W/m^2
ICE: 50 W/m^2 * 0.99 = 49.5 W/m^2
>and by 100 Watts/meter^2 direct rays at 85 degrees incidence angle.
OK. I take this to mean the solar radiation.
>Albedo of open water = 0.07; albedo of ice = 0.80
>At 85 degrees zenith angle, 60% of received energy is
>reflected from a smooth surface.
Now we can get the incoming solar energy
WATER: 100 W/m^2 * (1-0.07) * (1-.60) = 37.2 W/m^2
ICE: 100 W/m^2 * (1-0.8) = 18.6 W/m^2
[I am assuming that the albedo of the ice is angle-independent, since the ice tends to have a rough surface, not a smooth reflecting surface. If this is not the assumption you were making, then you will need to specify the conditions and/or give the refelction from this particualr patch of ice at 85 degree angle]
>Temperature of open water = 276K,
>temperature top of ice = temp air = 274 K,
>temp of space = 3K
This allows us to get outflowing IR energy
ICE: 0.0000000567*(274^4)*(0.99) = 316 W/m^2
WATER: =0.0000000567*(276^4)*0.99 = 326 W/m^2
[The temperature of space is not directly relevant for two reasons. 1: the number would be so small that it would be negligible 2: You already specified the total incoming flux as 50 W/m^2, which would include radiation from both the atmosphere and from space.]
Both scenarios have a net outflow of energy of about 250 W/m^2. Both the ice and the water would be cooling in these circumstances.
So now that I have done this, what was your point???

Phil.
August 21, 2011 3:24 pm

Myrrh says:
August 17, 2011 at 3:05 pm
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies,
That means they do not heat things.
You are not scientists if you ignore this challenge.

As you have been told many times before you are totally wrong with this assertion.
As proof when I focus my frequency-doubled Nd:YAG laser (532nm, green) on a cloud of soot the soot particles sublime (T ~4500K), they have been heated by the green light. The same experiment can be done using the undoubled light (1064 nm, IR).

August 21, 2011 4:08 pm

Tim Folkerts says: August 21, 2011 at 11:55 am
Your own “proper equations” show that GHGs & clouds reduce energy loss by ~ 320 W/m^2 from the surface.

Wrong, Earth emits to space the exact same amount of energy it receives, as the satellite observations show. The surface is hotter with an atmosphere than without due to pressure – i.e. the adiabatic lapse rate -no GHE necessary. Using the adiabatic lapse rate alone without any GHE fully explains the temperature profiles of Venus, Earth, and Mars. Trying to explain the same with a GHE leads to an underestimate of Venus temp by 419C, and overestimate of Earth by 23C and Mars by 19C. Radiative transfer CANNOT be considered in isolation without the rest of thermodynamics, and pyregometers CANNOT be used to determine HEAT FLOW.

Bob_FJ
August 21, 2011 4:09 pm

Tim Folkerts @ August 21, 2011 at 12:36 pm

“…Or yet again, a car under a carport will often have no dew on the windows when an uncovered car does have dew. Why? The sides of the carport are open, so even a slight breeze will ensure the same temperature air for both cars. The covered car has a “DLR” source above (the roof at ambient temperature). The uncovered car has a DLR source above it at a much cooler temperature.
SUMMARY: Increase the DLR, the objects cool slowly. Decrease the DLR, and the objects cool more quickly…”

Tim, I think something is missing, for instance dew usually forms on still nights, so your carport would mostly not be well ventilated. Also, my car, right now in the open, has dew on its vertical windows, so how do they emit and absorb “ULR” and “DLR“?

“…(PS. the terms “DLR” & “ULR” are potentially misleading. It should be “ILR” (incoming longwave radiation) and OLR (outgoing longwave radiation). For the earth’s surface, LR can only arrive from above heading down. For a container of water, much of the “DLR” would actually be coming from the ground.)”

Radiation from a surface is not just up, but in all directions equally and hemispherically. Radiation from the air is in all directions equally and spherically. Lateral radiation is much greater than up and down.

Bob_FJ
August 21, 2011 4:40 pm

Phil. @ August 21, 2011 at 3:24 pm
You are wasting your time with Myrrh
For instance, when I pointed him to the following in Wikipedia, he retaliated with: an incandescent bulb emits 95% IR which is thermal, and how they use water cooled bulbs to maximise plant growth. (paraphrasing)

“…Examples of thermal radiation include visible light emitted by an incandescent light bulb, infrared radiation emitted by animals and detectable with an infrared camera, and the cosmic microwave background radiation…
…Sunlight is thermal radiation generated by the hot plasma of the Sun. The Earth also emits thermal radiation, but at a much lower intensity and different spectral distribution because it is cooler. The Earth’s absorption of solar radiation, followed by its outgoing thermal radiation are the two most important processes that determine the temperature of the Earth…”

Tim Folkerts
August 21, 2011 4:45 pm

Bob_FJ says: “Tim, I think something is missing, for instance dew usually forms on still nights, so your carport would mostly not be well ventilated. Also, my car, right now in the open, has dew on its vertical windows, so how do they emit and absorb “ULR” and “DLR“?”
I was trying to make the preemptive point that difference was NOT because the air was warmer under the carport, which would ALSO prevent dew. The difference is the “DLR” from the roof of the carport. Since the car-port is warmer than the air higher up, it can “slow the cooling” overnight so that the dew does not form there.
You can also see this when dew forms on the windshield, but not on the side windows. The side windows “see” quite a bit of energy from the ground’s IR to keep warm, while the windshield “sees” mostly the colder sky.
“Radiation from a surface is not just up, but in all directions equally “
Yep — I agree completely.

Konrad
August 21, 2011 5:08 pm

Some further notes to the experiment I described earlier.
I have repeated the experiment using smaller water containers and I would highly recommend this approach to others wanting to check the results. I used two small 200ml plastic Tupperware style tubs, insulated on the sides and base with foam and foil. A hole (force fit) was drilled for the thermometer probe in the side of each container. The lids were cut out in the centre leaving only the clip on rim. This allows the cling wrap to be easily positioned on the water surface and prevents loss of the top layer of water through capillary action.
The use of smaller containers means the results are observable in as little as 15 min. Typical results in a room around 20C were –
For 40C water free to evaporatively cool, both containers cooled 4.5 degrees in 15 min. Both samples continue to cool in sync to room temperature.
For 40C water cooling only through radiation and conduction the containers both cool slower than the evaporatively cooled samples. The container under the foil “Sky” cooled 2.5C in 15 min. The container under the clear “Sky” cooled faster at 2.8C in 15 min. The two samples continue to diverge in temperature as much as 1.2C in the first 45 min. After several hours the samples converge to room temperature.
I believe the results I am getting are consistent and repeatable and possibly even “robust” ;). It should be noted that although my clear “Sky” was actually a ceiling around 20C, the foil “Sky” was reflecting most of the IR emitted by the warm water, not just under 50% of the 10 to 20 micron frequency. Also I have not yet repeated the small scale test with the artificial wind from computer fans. I urge others to repeat this style of experiment to confirm the result.
My conclusion is that backscattered LWIR cannot slow the cooling rate of Earth’s oceans to any measurable degree. Given that the oceans cover 71% of the planets surface, estimates for climate sensitivity should probably be reduced from 1C to per CO2 doubling to 0.29C. This figure is likely to be smaller again, as plants that cool through transpiration may need to be subtracted from equations. I would further speculate that anthropogenic emissions of CO2 cannot cause dangerous, catastrophic or even measurable global warming even if we burnt all known and projected fossil fuel reserves.

Myrrh
August 21, 2011 5:18 pm

kuhnkat says:
August 20, 2011 at 5:42 pm
Myrrh,
As I and at least two others have pointed out, whether PURE water is TOTALLY transparent to Visible wavelengths or not, the oceans of THIS world Earth contain a rather large amount of organic and inorganic matter that is NOT TRANSPARENT to Visible wavelengths. Our oceans could not provide fish, seaweed, salt, minerals, phytoplankton, mammals, crustaceans and numerous other interesting and tasty things to us if they were pure water. The Arctic and Antarctic would have MUCH more sea ice due to freezing more easily and not absorbing as much energy if they were totally transparent. Sometimes a physical truth is overidden by the messy facts of what we actually are dealing with.
The messy fact I’m dealing with is that the claim is that Visible light converts WATER to heat.
That Visible is the main HEATING MECHANISM of the Earth’s surface BECAUSE Thermal Infrared DOESN’T GET HERE. B.S. to such strawish distractions.. ‘That Blue visible travels further into the oceans and therfeore is absorbed creating heat deep in the oceans’ is the claim. Deal with the fact that this claim as it stands is B.S. Which is what I am addressing.
AGWScience Fiction Inc. has REVERSED the properties of Visible and Thermal Infrared, a.k.a. in TRADITIONAL PHYSICS, LIGHT and HEAT, and, it makes various statements to ‘back’ its fiction, including saying that Heat, i.e. long wave thermal infrared, doesn’t even get to the surface! Totally stupid. Heat, thermal infrared is what you feel direct from the Sun. If you can’t feel it, you’re on Earth.
To which, I have given more than sufficient information to show how ludicrous all that AGWSFiction is. I have also shown that it is a deliberate campaign of disinformation. To dumb down the population..
As an article I linked to above shows, there is even a rather busy biozone in the very region under discussion here. Please actually consider this possibility. Denying reality is a lonely job.
What’s lonely here, is defending reality against cr*p science fiction.
Good science fiction is at least internally coherent – in AGWScience fiction we have Blue visible ‘transparent’ in the atmosphere and yet heating the oceans, nothing about it therefore heating the water in the atmosphere..
That’s besides the all the other nonsense such as the atmosphere being transparent to Visible, when reflection/scattering is because of absorption at electron scale, as I’ve already gone through.
So try not to get too distracted, stick with the main claim which is the reversal of properties between Heat and Light. If you understand what AGWScience has done here you’ll realise just how insidiously it has permeated itself into the education system, and so into the ‘general physics knowledge’ of the world. Which will leave the world ignorant again if it isn’t stopped. Or rather, leave the oiks ignorant.
I could give you a page with reference statements about this ‘greenhouse effect’ from various ‘accepted’ sources to show how much they’ve succeeded, but you, now, would just read it and think it was actually real science. I’m asking you to examine these claims, in the very basics.
kuhnkat says:
August 20, 2011 at 9:45 pm
Myrrh,
“You are postulating that light created by the Sun is eternal.”
You do realize that most Cosmology and Physics do come close to that?? Remember the Microwave Background Radiation?? It has alledgedly been literally flying around the universe since the BB!!! That would be about 14 billion years the last I checked. If you disagree with BB, I can understand that. If you think electromagnetic radiation attenuates because it uses up its initial energy just propagating in less than one light year, I think we have a problem. If you think the oceans are so pure that visible light doesn’t get absorbed or reflected by something, you have a serious reality issue.
As I’ve said, stick with what I’m arguing against. You may have noticed that I have given some information on other ways visible energy is actually used in the oceans..

Tim Folkerts
August 21, 2011 5:38 pm

Hockey Schtick says: August 21, 2011 at 4:08 pm
Wrong
Damned if it do; damned if I don’t. YOU were championing the ideas that “that is why the physics literature expresses heat transfer on the potential difference (Ta^4 – Te^4)”
and
“The whole DLR heating red herring is based on this false assumption of 2 way heat transfer. Doing the calculations this way leads to false answers because it does not consider the fact that if a cold body could heat a warm body, the warm body would simply emit more heat to compensate. Doing the calculation properly using only the potential difference eliminates this error.”
YOUR one-way heat transfer equation shows that without GHG & clouds, the net IR transfer away from the surface would be about – 390 W/m^2, while with GHG’s and clouds, it is about -50 W/m^2. This means the surface is not losing ~ about 340 W/m^2 less than it would if there was not the atmosphere to “buffer” the loss. The cooling from current conditions would be dramatic if the atmosphere were not cutting the loss from the surface from -390 to only -40!
“Earth emits to space the exact same amount of energy it receives, as the satellite observations show. “
This much I can agree with. It radiates that much because the solution to YOUR equation suggests and average temperature of ~ 255 K for the earth as seen from outer space. Since the surface is indeed much warmer than 255 K, the atmosphere must blocking some of that surface radiation and re-radiating from a cooler higher altitude (which is, in fact, the greenhouse effect).
“Using the adiabatic lapse rate alone without any GHE fully explains the temperature profiles of Venus, Earth, and Mars. “
True enough. The adiabatic lapse rate is an important idea. Once you know the temperature at the “Top of the atmosphere”, you can use that to predict the surface temperature.
But now, explain to me — without the greenhouse effect or radiation balance from gases in the atmosphere — what sets the top-of-atmosphere temperatures of those planets. Without knowing the starting point, you can’t extrapolate down to the surface, so you need a first-principles estimate of the TOA temperature.

Tim Folkerts
August 21, 2011 6:13 pm

Konrad,
I have enjoyed reading your experiments. It is great to see some actual data.
One thing confuses me — perhaps you can clarify.

For 40C water cooling only through radiation and conduction the containers both cool slower than the evaporatively cooled samples. The container under the foil “Sky” cooled 2.5C in 15 min. The container under the clear “Sky” cooled faster at 2.8C in 15 min. The two samples continue to diverge in temperature as much as 1.2C in the first 45 min. After several hours the samples converge to room temperature.

My conclusion is that backscattered LWIR cannot slow the cooling rate of Earth’s oceans to any measurable degree. </blockquote.
As I understand it, the foil is directly above one of the containers. The foil and the window are horizontal.
The "window" would mostly be allowing 20 C radiation (from the ceiling) to reach the surface.
The "reflector" would be reflecting the 40 C radiation (from the water) back toward the water.
The "reflector" (with more IR radiation toward the water) measurably slowed the cooling rate.
But the conclusion was that extra IR cannot slow the cooling. This seems to be a direct contradiction to your experimental results!
I think you suggested this yourself, but it be interesting to try putting some container (eg a glass baking dish) above the surface, then filling the dish with ice. then you could have a "ceiling" at O C. This should speed the cooling even further.
I have access to some digital thermometers and data loggers, so I may have to try this myself one of these days.

kuhnkat
August 21, 2011 6:31 pm

No Myrrh,
several times you have claimed that the ocean is transparent to VISIBLE light. Make up your mind what you are arguing. You have NOT made it clear other than a garbled message that somehow VISIBLE light carries no energy that can be thermalized.

Myrrh
August 21, 2011 6:34 pm

Phil. says:
August 21, 2011 at 3:24 pm
Myrrh says:
August 17, 2011 at 3:05 pm
Visible light and and the two shortwave energies either side of UV and Nr Infrared are not thermal energies,
That means they do not heat things.
You are not scientists if you ignore this challenge.
As you have been told many times before you are totally wrong with this assertion.
? 🙂 Who are you? Show me the science. And as I’ve said each time also, prove I’m wrong.
Show me how Visible Blue light from the Sun is able to heat the water of the oceans when water is a transparent medium for visible light, that means, it passes through it without being absorbed. It needs to be absorbed to even have a chance of creating heat.
And, the corollary to this, show why the claim that the atmosphere is transparent to visible light is not b.s. because, reflection/scattering is a result of visible light being absorbed by the electrons of the nitrogen and oxygen molecules which comprise practically 100% of our atmosphere. Visible light therefore should be heating up the sky since, the claim is that the oceans absorb it and so get heated..
As proof when I focus my frequency-doubled Nd:YAG laser (532nm, green) on a cloud of soot the soot particles sublime (T ~4500K), they have been heated by the green light. The same experiment can be done using the undoubled light (1064 nm, IR).
What’s a laser? Or is this another one of your ‘instruments I have used’ which further questioning shows you know zilch about? Which page did you get that detail from? How is your green laser different from the green light of the Sun which reaches the surface of Earth? How many such lasers from the Sun would it take to convert land and oceans to heat for them to radiate out the amount of thermal infrared claimed in the KT97? Would there be anything left??

philincalifornia
August 21, 2011 7:20 pm

Myrrh says:
August 21, 2011 at 6:34 pm
Show me how Visible Blue light from the Sun is able to heat the water of the oceans when water is a transparent medium for visible light, that means, it passes through it without being absorbed. It needs to be absorbed to even have a chance of creating heat.
=========================================
Myrrh, a quick tap on the shoulder – visible light doesn’t make it all the way to the bottom (except on SpongeBob SquarePants).

kuhnkat
August 21, 2011 7:56 pm

Myrrh,
“Show me how Visible Blue light from the Sun is able to heat the water of the oceans when water is a transparent medium for visible light, that means, it passes through it without being absorbed. It needs to be absorbed to even have a chance of creating heat. ”
You have apparently changed your claim. You now have narrowed it down to BLUE light. Is this your final claim or are you going to change it again?? It is not easy to respond to changing claims. Blue is only one area of the spectrum. Maybe you can tell us what wavelengths or frequencies you are actually referring to??

kuhnkat
August 21, 2011 8:52 pm

Myrrh, you are very confused about the Greenhouse effect. I don’t necessarily agree with all of it, but, it does NOT claim that atmospheric water vapor converts VISIBLE Light to heat!! It converts your lovely THERMAL RADIATION to rotational energy and more THERMAL RADIATION. These excited water molecules bumping into non-GHG’s then transfer some of the absorbed energy.
If you are disagreeing with the idea that visible energy heats the ocean it would be good to help us understand what does. What actually warms the ocean if not the visible??
Please show your work again. I seem to have skipped over where you showed this.

kuhnkat
August 21, 2011 8:55 pm

Tim Folkerts,
Sorry, just like Willis you ARE double counting. The 324w/m2 down comes from the 390w/m2 up which came from… The SB computation would actually give an ~66w/m2 UP ONLY!!!! This is still only a very rough approximation as SB is only confirmed between objects where we know their temperature and emissivity.
Consider, the equation requires the temperature, or average temperature of the emitting surface. You yourself point out that the surface the ground is emitting against ranges from the 3K space to the 290k or whatever next to the ground. How do we get the average temperature when we do not know it??
When we MEASURE 390 up or 324 down the instrument already has the assumptions of SB built into the circuitry and tells us what we ASSUME!! That is, what is the actual emissivity of the volume the ground is emitting against?? The more expensive instruments are adjustable so we can get more accurate readings, but, how do we determine how it should be set? The cheap ones are set to about .8-.9 I think. How can we get a reliable number when we cannot provide a correct emissivity for the computation??? I believe CO2 is about .2 in the atmosphere and Water vapor may be close to .9. So, what is the emissivity for the whole atmosphere?
http://www.xylenepower.com/Emissivity.htm
uses spectra from the Mars Global Surveyor to provide emissivity of the earth = .7566
Here is a nice link going over the development, implementation and use of pyrometers, pyrgeometers, IR thermocouples, and IR thermometers. Maybe someone can correct me if I’ve garbled the info?
http://www.omega.com/literature/transactions/volume1/thermometers1.html
I am of the opinion that Stefan-Boltzman gives a poor result computing emission between the surface and the atmosphere. Here is another of the arguments:
http://en.wikipedia.org/wiki/Stefan-Boltzmann_law
“Finally, this proof started out only considering a small flat surface. However, any differentiable surface can be approximated by a bunch of small flat surfaces. So long as the geometry of the surface does not cause the blackbody to reabsorb its own radiation, the total energy radiated is just the sum of the energies radiated by each surface; and the total surface area is just the sum of the areas of each surface—so this law holds for all convex blackbodies, too, so long as the surface has the same temperature throughout.”
Notice that it is stated that the geometry cannot allow the black, or grey, body to absorb its own radiation. It also states the temperature should be the same among the pieces. Our atmosphere DOES absorb some of its own radiation and the temperature of the radiating volume varies quite a bit!!! The author also makes a common error. He computes the earth’s temperature using straight SB and says the SURFACE would be 6c ignoring the atmosphere is part of the composite surface that actually is radiatively interacting with the emissions of the sun.
Basically the models and more advanced work may handle these issues, but, these back of the envelope computations using an emissivity of 1 are simply wrong.

Konrad
August 21, 2011 9:07 pm

Tim Folkerts says:
August 21, 2011 at 6:13 pm
“But the conclusion was that extra IR cannot slow the cooling. This seems to be a direct contradiction to your experimental results!”
Tim,
I’m sorry, now I see that my “some further notes” comment skipped over some steps covered in the first description of the experiment.
Two identical containers of water are being tested under two different “Sky” panels (one with aluminium foil, the other a frame with clear cling wrap), which you have correctly surmised are horizontal. These are placed 50mm above each water container. The temperature drop from 40 degrees is then measured for each container. I will call this Test A.
The experiment is then repeated, but this time a small square of clear cling wrap is laid on the surface of the water in each water container. This allows cooling by conduction and radiation, but prevents cooling by evaporation. I will call this Test B.
In Test A the water cools more quickly, however the two water containers temperatures remain very close to each other over time. This indicates that backscattered LWIR has a very limited effect on the rate of cooling for water when it is free to evaporatively cool.
In Test B both water containers cool more slowly than test A, but a divergence in temperature between the two water containers is readily detectable. The container under the foil sky cools more slowly than that under the cling wrap sky. This indicates that backscattered LWIR from a warm material can slow the rate at which that material cools, if radiation and conduction are the only methods for cooling.
Test A represents the evaporative cooling conditions in the real oceans. Test B represents how the climate scientists have modelled the oceans with regard to backscattered LWIR. As Earth’s oceans are not covered in cling wrap, it would appear that the claimed radiative forcing for additional CO2 should be reduced by 71%.

August 21, 2011 9:29 pm

Tim Folkerts:
1. It is not “MY equation.” It is in fact the one and only way RADIATIVE heat transfer is expressed in the physics literature. Here’s one of many examples:
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html
Challenge for you: show a single example in the physics literature where it is expressed in the form suggested by use of pyregometers and as used in climate science (Eq. 2)
2. Your calculations are based upon the fallacy that pyregometers measure net HEAT FLOW. They do not, as explained several times here and on Claes Johnson’s blog (linked above). In addition, it is ridiculous to only use radiative heat transfer calculations when convection plays a much larger role.
3. “Earth emits to space the exact same amount of energy it receives, as the satellite observations show “ YOU: This much I can agree with.”
What you apparently fail to recognize is that the greenhouse effect says that the Earth emits LESS to space than received, due to ‘heat trapping’ gases, accounting for an increase in temperature. This is incompatible with the statement you say you agree with, that INPUT = OUTPUT
4. “Using the adiabatic lapse rate alone without any GHE fully explains the temperature profiles of Venus, Earth, and Mars.“ YOU: True enough”
Thanks for admitting that invoking a ‘GHE’ is unnecessary.
5. “explain to me — without the greenhouse effect or radiation balance from gases in the atmosphere — what sets the top-of-atmosphere temperatures of those planets.”
The TOA temperature is determined by the solar insolation received, as you already agreed to above; INPUT = OUTPUT.

Bob_FJ
August 22, 2011 1:45 am

Kuhnkat and Phil et al,
If you can’t resist responding to Myrrh’s nonsense’s, may I suggest a different approach? Just as Anthony has a ‘Quote of the week‘ and ‘Friday funny‘, why not have a ‘myrth of the day or week‘?
You could deliberately tease him on some laws of physics or with provocative analogies, or whatever turns you on.
One thing that springs to my mind is that plain ornery window glass is I believe fairly opaque to IR and UV light. However modern automotive glass is a step beyond, being deliberately designed to be optimally highly reflective and/or opaque to those EMR frequencies because they are very unwelcome in the cabin, and with absolutely no benefit. The purpose to eliminate UV and IR transmission is to improve passenger comfort in hot weather, especially in modern cars with increased glasshouse‘s, and to reduce air-conditioning workloads. (fuel consumption and performance loss). The EMR “filtering” can be variously achieved in coatings and chemical inclusions in or on the glass. (or in the non-glass laminates in windscreens). The visible light transmissivity is usually also cut down with colour tinting. Nevertheless……. Don‘t park in the summer sun!
It seems hard to find anything authoritative WRT to solar spectrum on this per Google, but the following forum is interesting, if a tad brief:
http://www.infraredtraining.com/community/boards/thread/4343/

Tim Folkerts
August 22, 2011 3:56 am

4. “Using the adiabatic lapse rate alone without any GHE fully explains the temperature profiles of Venus, Earth, and Mars.“ YOU: True enough”
Thanks for admitting that invoking a ‘GHE’ is unnecessary.

Yes, the lapse rate explains the PROFILE = the slope. But determining a line requires not only a slope but also a point.

5. “explain to me — without the greenhouse effect or radiation balance from gases in the atmosphere — what sets the top-of-atmosphere temperatures of those planets.”
The TOA temperature is determined by the solar insolation received, as you already agreed to above; INPUT = OUTPUT.

Thanks for admitting that the temperature of the TOA is determined by INPUT = OUTPUT of radiation, which is a strong function of the GHG’s in the atmosphere. The GHG’s strongly cool the upper atmosphere, thereby requiring the ground to be warmer to compensate.
Once the TOA is determined using radiation to/from GHG’s the internal details of why the surfaces is warmer can me “swept under the rug” using the lapse rate.
(By the way, “adiabatic” means “without energy loss” which is only an approximation since rising warm air can lose energy via conduction; the true lapse rate would have to include the effects of energy loss. Since we agree that GHGs can lose energy by radiation whereas non-GHGs cannot, then the lapse rate itself will be at least partly a function of GHGs.)

Tim Folkerts
August 22, 2011 4:01 am

Konrad,
Thanks for the clarification on the experiment. I will have to think about this a bit more (or to call it a day at the end of a long thread!). I suspect this is tied to the estimates I made earlier that the differences in cooling due to IR would be very small (or non-existent if Dave Springer is correct). I agree that evaporation itself will be the larger effect; the question is MUCH of a correction can we get from IR.

Tim Folkerts
August 22, 2011 4:10 am

kuhnkat says:
August 21, 2011 at 8:55 pm
Sorry, just like Willis you ARE double counting. The 324w/m2 down comes from the 390w/m2 up which came from…
I don’t have much time now. Let me just note that you used the past tense: “which came from”. The energy one second (or microsecond) going one one way is energy that ALREADY had arrived during previous microseconds. There is constantly NEW energy coming and going, originally from the sun and ultimately to outer space. If it is indeed different energy, then there is no double-counting.

RJ
August 22, 2011 4:50 am

“There is constantly NEW energy coming and going, originally from the sun and ultimately to outer space. If it is indeed different energy, then there is no double-counting.”
So the sun not GHG backradiation warms the planet.

Myrrh
August 22, 2011 5:48 am

Bob_FJ says:
August 21, 2011 at 4:40 pm
Phil. @ August 21, 2011 at 3:24 pm
You are wasting your time with Myrrh
For instance, when I pointed him to the following in Wikipedia, he retaliated with: an incandescent bulb emits 95% IR which is thermal, and how they use water cooled bulbs to maximise plant growth. (paraphrasing)
“…Examples of thermal radiation include visible light emitted by an incandescent light bulb, infrared radiation emitted by animals and detectable with an infrared camera, and the cosmic microwave background radiation…
…Sunlight is thermal radiation generated by the hot plasma of the Sun. The Earth also emits thermal radiation, but at a much lower intensity and different spectral distribution because it is cooler. The Earth’s absorption of solar radiation, followed by its outgoing thermal radiation are the two most important processes that determine the temperature of the Earth…”

This is like banging my head against a brick wall. What don’t you understand when I keep pointing out that “Sunlight is thermal radiation” and the like are AGWScience Fiction Inc memes, which, as I keep pointing out, have become ubiqitous in their disinformation through the education system and in general?
YOU NEED TO USE COMMON SENSE HERE.
As an example of thermal radiation an incandescent light bulb is fine. Because practically all of the energy emitted is thermal.
5% is the general figure for that portion which is not thermal infrared, which is Visible Light. In traditional physics these distinct energies having distinct properties were given the names Light and Heat, because that is the distinct difference between them…
So, when wiki rolls onto that ‘incandescent thermal example’ by continuing with the meme from AGWScience that ‘Sunlight is thermal radiation generated by the Sun’ – YOU’ALL read the AGWScience fiction meme that Visible is thermal. And disregard entirely the point I am making.
That an incandescent bulb’s thermal energy is distinct from its light energy which is not thermal, and so, that Visible light (Sunlight) from the Sun is not thermal. ‘Sunlight’ – be careful of that word.
Like ‘absorbed’ it can be used generally, to cover all the electromagnetic spectrum, but Sunlight proper is Visible only. ‘Absorbed’ can be used generally as in descriptions of visible being absorbed at different depths in oceans, but that doesn’t mean it has anything to do with ‘absorbed’ in physics, electron/molecule interactions.
Wiki is ‘lying’ here. It is promoting an AGWScience Fiction Inc meme by disingenuously using a general term in the middle of talking about thermal radiation which makes you associate it with the AGWSF meme. . Now, read that again. What is it actually saying? It is saying that the main energy coming from the Sun is thermal radiation.
Just like the incandescent light bulb..
Got that?
Visible cannot heat because it is not a thermal energy, it doesn’t have the ability to move water molecules into vibrational states, for example, because water is a transparent medium for it. Visible passes through, is transmitted, through water. That means it has no interaction with water. If there is no interaction, nothing happens.
Here: Different properties of electromagnetic waves means that they are different from each other…

How Electromagnetic Waves Behave
When an electromagnetic wave (radiation) hits a material, it can be:
reflected, like light in a mirror;
absorbed, like heat absorbed by a black surface;
transmitted, or passed on, like light passing through glass.
Sometimes two, or even all three processes can occur. Substances behave differently depending on what kind of radiation is falling on them, as the following examples show.
When energy is absorbed by a surface, it heats up. For example microwaves are absorbed by water molecules and warm up. This is how a microwave oven works. Light waves simply pass through water.
A dark painted metal surface absorbs radio and light waves. However X-rays and gamma rays can pass straight through.
Wax can transmit microwaves, but absorbs light waves.
http://www.antonine-education.co.uk/physics_gcse/Unit_1/Topic_5/topic_5_what_are_the_uses_and_ha.htm

When energy is absorbed by a surface, it heats up. For example microwaves are absorbed by water molecules and warm up. This is how a microwave oven works. Light waves simply pass through water.
Traditional physics is internally coherent, that’s why we know how microwave ovens work, that’s how microwave ovens got developed in the first place.
Water absorbs microwaves just as water absorbs thermal infrared, that’s why the Sun’s Heat, thermal infrared, goes into your bodies and warms you up.
Light waves, i.e. Visible, (not thermal infrared, heat waves) simply pass through water.
Light waves, simply pass through water.
Light waves, simply pass through water.
Sunlight simply passes through water.
That’s what they ACTUALLY do!
AGWScience Fiction Inc. has deliberately reversed this, giving the property of thermal infrared heat to the visible light.
This is a deliberate con.
You’all need to thoroughly take this on board before you do any more reading, because until you do you won’t spot the AGWScience meme takeover. And you will thus not be able to understand the basic real world science facts as well established in traditional physics explanations and in real world applications.
And you’ll need to be vigilant, one piece might give you real traditional well understood principles and then slips in an AGWScience meme to confuse you. Remember, you will be passing on your ignorance to the next generation. They won’t be able to create microwave ovens starting from first principles..
Sunheat is absorbed by water. (I’ve just coined that word, maybe it will help.) The HEAT we feel from the Sun is thermal infrared.
See my posts above about NASA and the change to promoting AGWScience Fiction instead of teaching real world science fact.
Here, I’ll fetch one for you..
http://wattsupwiththat.com/2011/08/15/radiating-the-ocean/#comment-720489

NASA original page teaching previously traditional real world physics to children: http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
From this NASA page:
“Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared.
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.
Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared waves are often used in fast food restaurants!

Please, do me a favour, read that several times until you have it quite clear that this is the traditional physics I am defending here against the AGWScience Fiction meme which has altered it, by reversing heat and light energies.
And hopefully then you will see the enormity of the con I’m pointing out.

Myrrh
August 22, 2011 7:15 am

philincalifornia says:
August 21, 2011 at 7:20 pm
Myrrh says:
August 21, 2011 at 6:34 pm
Show me how Visible Blue light from the Sun is able to heat the water of the oceans when water is a transparent medium for visible light, that means, it passes through it without being absorbed. It needs to be absorbed to even have a chance of creating heat.
=========================================
Myrrh, a quick tap on the shoulder – visible light doesn’t make it all the way to the bottom (except on SpongeBob SquarePants).

Where have I said it did?
kuhnkat says:
August 21, 2011 at 7:56 pm
Myrrh,
“Show me how Visible Blue light from the Sun is able to heat the water of the oceans when water is a transparent medium for visible light, that means, it passes through it without being absorbed. It needs to be absorbed to even have a chance of creating heat. ”
You have apparently changed your claim. You now have narrowed it down to BLUE light. Is this your final claim or are you going to change it again?? It is not easy to respond to changing claims. Blue is only one area of the spectrum. Maybe you can tell us what wavelengths or frequencies you are actually referring to??

? I haven’t changed it.. Blue light is visible, the AGWScience Fiction Inc claim is that blue visible heats the oceans further down because it penetrates further.
Come on, stop getting distracted, prove that visible light can heat water or take it out of ‘the’ energy budget. The AGWScience Fiction Inc energy budget.
kuhnkat says:
August 21, 2011 at 8:52 pm
Myrrh, you are very confused about the Greenhouse effect. I don’t necessarily agree with all of it, but, it does NOT claim that atmospheric water vapor converts VISIBLE Light to heat!! It converts your lovely THERMAL RADIATION to rotational energy and more THERMAL RADIATION. These excited water molecules bumping into non-GHG’s then transfer some of the absorbed energy.
Where have I said that?
If you are disagreeing with the idea that visible energy heats the ocean it would be good to help us understand what does. What actually warms the ocean if not the visible??
Please show your work again. I seem to have skipped over where you showed this.

Then don’t skip over…? You could re-read what I’ve posted so far.. I really hope you do, I would really like someone to appreciate the enormity of this con..
The heat we feel from the Sun is thermal infrared, long wave infrared, as the original NASA page teaching traditional science says correctly. We feel it on Earth’s surface, therefore, it is reaching us, therefore, it is reaching the oceans and lands.
It is this heating the oceans.
Thermal infrared, long wave infrared, Heat, is a very powerful energy.

NASA original page teaching previously traditional real world physics to children:
http://science.hq.nasa.gov/kids/imagers/ems/infrared.html
From this NASA page:
“Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to
the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are
about the size of a pin head and the shorter, near infrared ones are the size of cells, or are
microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation
every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm
sidewalk is infrared.
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter
wavelengths are the ones used by your TV’s remote control.
Infrared light is even used to heat food sometimes – special lamps that emit thermal infrared
waves are often used in fast food restaurants!

Visible light isn’t thermal, we do not feel it as heat. What is the difference between UV, which we can’t feel as heat either, and Visible? UV can burn us, Visible does not burn us.
Here’s an example of the AGWScience Fiction Inc. meme takeover of the education system. See if you can spot the flaws, the internal incoherence.

http://www.school-for-champions.com/science/infrared_greenhouse_effect.htm
The Earth is heated to relatively moderate temperatures by the absorption of some of the Sun’s radiated energy.
About 30% of the visible light from the Sun is reflected back into space, while the other 70% is absorbed by the ground and oceans, resulting in heating the Earth. Since the atmosphere is transparent to visible light, it absorbs little energy from the sunlight and does not make much of a contribution in heating the Earth.
The soil, rocks and water that have been heated by the sunlight then give off longer wavelength infrared radiation, some of which is absorbed by the Greenhouse gases in the atmosphere. These gases include water vapor (H2O), carbon dioxide (CO2), methane (CH4) and ozone (O3). Absorption of the infrared radiation heats the atmosphere to acceptable temperatures, provided they are in the proper proportion.

Absorption of infrared
Water vapor (H2O) is the most active molecule in absorbing infrared radiation and thus in heating the atmosphere. It accounts for about 55% of the absorption of thermal radiation in the atmosphere.
</blockquote
Now that you know from traditional science visible does not heat water and thermal infrared is the heat we feel from the Sun and therefore does reach the surface and that water is the great absorber of thermal infrared and has an extremely high heat capacity, you should be able to see the disjunct.
Firstly that they are giving the reversed, through the looking glass with Alice impossible, AGWScience Fiction Inc teaching as if it is real world.
Secondly that in the real world Visible does not heat the land and oceans and thermal infrared, heat, from the Sun does.
Thirdly that in the real world this heat from the Sun will heat the atmosphere on the way down to the Earth's surface as well as the atmosphere receiving the upwelling heat from the Earth's surface.
And, how then can this 'backradiated' thermal infrared heating of the atmosphere be separated from the direct thermal infrared received from the atmosphere which has been directly heated by the Sun?
Remember the blatant lie from AGWScience fiction paradigm is that thermal infrared does not even get through to reach the surface.
One more point, again. Having shown that traditional physics teaches the opposite of AGWScience Fiction Inc, it is encumbent on anyone claiming that AGWScience Fiction Inc's paradigm is real to first show how it has superceded the traditional science.
This is an enormous claim to make. It overturns everything that is already well known in physics and in real life applications. So of course, it cannot be proved.. The only way it can be spread is by repetition and brainwashing of the population and by taking traditional physics out of the general references.
When I ask for proof, I hope that you will actually explore this by trying to find me real proof that visible can heat water. It's in looking for it that you will find that visible can't. Or, you can read what I have put together in exploring this and at least address the fact that these are two completely different explanations of the physical world. They can't both by right.
You'all can't simply continue promoting your version from the memes produced by AGWScience Fiction Inc.
Because I have given you more than sufficient information to show that AGWScience is fiction. Traditional physics falsifies it.

RACookPE1978
Editor
August 22, 2011 8:17 am

Tim Folkerts says:
August 21, 2011 at 2:48 pm (Edit)
RACookPE1978 says: August 21, 2011 at 1:50 pm

>Please try use those (average, whole earth) values and equations to address
>this specific question at this specific location:
I’m not sure what your point will be, but I’m game. I will use your numbers, which seem to be in the right ballpark, but which I have not verified independently. …. Both scenarios have a net outflow of energy of about 250 W/m^2. Both the ice and the water would be cooling in these circumstances.
So now that I have done this, what was your point???

My point? My training. It shows me exactly where I need to look up the derivations an background assumptions about a couple of equations. Thank you for: confirming a few things I thought I knew, exposing a few things that I thought I knew better than I do .

August 22, 2011 9:13 am

From a NASA page on infrared light: “In 1800, William Herschel conducted an experiment measuring the difference in temperature between the colors in the visible spectrum. He placed thermometers within each color of the visible spectrum. The results showed an increase in temperature from blue to red. When he noticed an even warmer temperature measurement just beyond the red end of the visible spectrum, Herschel had discovered infrared light!”
I have no desire to feed Myrrh’s long posts, but blue is of a lower temperature than IR according to this.

Tim Folkerts
August 22, 2011 9:28 am

RACook,
Glad that I could help. I think everything I wrote is correct — and so far no one else has provided any contradictions.
Sorry if I sounded a bit defensive or aggressive. I guess I have gotten a bit cynical, often expecting a “gotcha” from posters who have a specific agenda. It is great when there can be some real learning; some discussions of topics at (moderately) advanced level; some actual data from experiments.

richard verney
August 22, 2011 9:49 am

Myrrh
I must confess that I am somewhat confused, and I need to look into matters further.
I had always assumed that when Trenberth and the Team talk about say 30% of the solar energy being absorbed by the atmosphere and 70% by the surface, they meant 30% and 70% of the total spectrum of solar radiation reaching the TOA. Of course some wavelengths find it more difficult than other wavelengths to penetrate the atmosphere and, hence one reason why some part of the solar radiation is absorbed by the atmosphere, just like DWLWIR finds it more difficult to penetrate the oceans when compared to shorter wavelengths.
I have never taken the reference by the AGW crowd to visible light to mean literally EMR of a wavelength that is visible to the human eye. As I say, I have always assumed this to mean the entire spectrum of solar radiation received at TOA.
Surely, the AGW Team cannot literally mean visible light only, ignoring all other wavelengths in the spectrum of solar radiation received at TOA? If they are limited the energy budget literally to visible light only, it appears that there may be merit in some of what you are saying.

kuhnkat
August 22, 2011 11:22 am

Tim Folkerts,
“If it is indeed different energy, then there is no double-counting.”
As you point out there is constantly NEW energy arriving until the sun goes down. After the sun goes down we have a lengthy unravelling of the double counting that includes energy stored in the atmosphere in non-ghg’s and other areas.
Keeping it simple, the ~320w/m^2 coming down was just counted on the way up. It goes down and then goes up AGAIN!!! This repeats with losses bringing the actual 66 up to 390 up. You are right, it isn’t double counting!!! It is WORSE!! With a laser people easily understand that the radiation is delayed bouncing back and forth between the mirrors only exciting more molecules and atoms in the discharge tube until the discharge at which time it all goes out one end. Do we count ALL the energy reflecting back and forth as NEW energy each time it passes a particular point in the tube or just what was added that cycle??

Tim Folkerts
August 22, 2011 1:10 pm

Kuhnkat,
1) Having the sun’s energy double during the day and cease at night is not “double counting” that needs “unravelling”. Perhaps that is not why you meant, but it is what it sounded like to me.
2) I have rarely heard the principles behind the operation of a laser described as simple! 🙂
3) You seem to be the one double counting. If there are ~320 W/m^2 heading down and ~ 390 W/m^2 heading up, then the NET is ~ 70 W/m^2 upward. No one I know (other than you just now) has ever claimed that any process brings the NET flow from ~ 70 W/m^2 to 390 W.m^2.
3) With a typical He-Ne laser, the exit has a mirror that reflects ~ 99% of the light and emits 1%. So for a laser that emits photons at a rate of 10 mW, the photons hitting the “output coupler” are arriving at a rate of 1000 mW, with 990 mW of photons being reflected. Power would have to added to the lasing medium at 10 mW to generate another. So overall, there is 10 mW being added and 10 mW being extracted. Within the laser, the power levels are much higher.
This is starting to sound like a good reason to BELIEVE the GH effect.
* The mirrors of the laser allow the energy flows inside (990-1000 mW back and forth) to be bigger than the energy input or output (10 mW).
* The greenhouse gases allow the energy flows inside (300-400 W/m^2 of IR back and forth) to be bigger than the individual inputs or outputs (~230 W/m^2 of net incoming solar or outgoing IR).
The nature of the amplification is different, to be sure, but the end results sound pretty similar. The GHE only creates internal signals ~1.5 times the in input, while internal signals 100 times the input are indeed possible. Thanks for the analogy!
PS. I was thinking through an analogy with water flowing from a dam into various pools that I think is rather accurate and informative. But I am not sure people are ready for yet another dam analogy.

t.f.p.
August 22, 2011 2:19 pm

Tim Folkerts
An experiment carried out using a IR pass filter, a vacuum flask, and tap (ie. not salt water) seems to show that IR heats the top few cms of water see:
http://alturl.com/oyxet
Also best to discuss there too!

tallbloke
August 22, 2011 2:29 pm

Tim Folkerts says:
August 22, 2011 at 1:10 pm
3) You seem to be the one double counting. If there are ~320 W/m^2 heading down and ~ 390 W/m^2 heading up, then the NET is ~ 70 W/m^2 upward. No one I know (other than you just now) has ever claimed that any process brings the NET flow from ~ 70 W/m^2 to 390 W.m^2.

The Land/Ocean emits 390W/m^2. 40 goes out of the atmospheric window (maybe 80 but we’ll stick with Trenberth’s numbers for clarity for now). 350 gets absorbed in the water vapour and co2. Apart from whatever gets converted to sensible heat by collisions with Nitrogen and Oxygen, half of that gets radiated back to the surface, along with half the incoming solar absorbed in the atmosphere (34), and half the latent heat and thermals (51).
Now that can’t add up to any more than 260 (assuming no sensible heat conversion), yet the Trenberth cartoon has 324 back radiation.
What have I missed?
And, given that 226 of this 260 already cooled the surface as it left, why is it double counted as warming the surface when it comes back down with the ‘new’ incoming solar derived longwave?
http://tallbloke.files.wordpress.com/2010/06/energy-budget-old.jpg

Myrrh
August 22, 2011 2:34 pm

richard verney says:
August 22, 2011 at 9:49 am
Myrrh
I must confess that I am somewhat confused, and I need to look into matters further.
I had always assumed that when Trenberth and the Team talk about say 30% of the solar energy being absorbed by the atmosphere and 70% by the surface, they meant 30% and 70% of the total spectrum of solar radiation reaching the TOA. Of course some wavelengths find it more difficult than other wavelengths to penetrate the atmosphere and, hence one reason why some part of the solar radiation is absorbed by the atmosphere, just like DWLWIR finds it more difficult to penetrate the oceans when compared to shorter wavelengths.
I have never taken the reference by the AGW crowd to visible light to mean literally EMR of a wavelength that is visible to the human eye. As I say, I have always assumed this to mean the entire spectrum of solar radiation received at TOA.

Please do look into this further, Richard. A great problem here is that (some) people read the 70/30 as if it includes thermal infrared, but when you look into it, you’ll see that actually thermal is taken out completely.
I’ve seen this in traditional scientists arguing against some point or other about AGW, taking it as read that Trenberth included thermal, and unadulterated traditionalists who know that thermal is the heating mechanism for land and oceans wouldn’t have any reason to think Trenberth not a traditionalist in this. .. As the newworldencyclopedia admitted, that thermal infrared is what heats the Earth is still being taught in traditional physics..
So also, I’d be wary of taking anything for granted that you think is ‘real physics’ in explanations from AGWScience influence. For example, the long wave infrared we feel as heat from the Sun penetrates our bodies around 2-4″, water being the great absorber of this, why then does it find it “more difficult to penetrate the oceans compared with shorter wavelengths” when it so easily gets through our skin to warm us up inside?
As I’ve been trying to explain, AGWScience Fiction Inc very cleverly alters real physics, sleight of hand, here it is erroneously comparing it with visible light which is of a completely different electromagnetic character, these are two completely different properties being merged into one view as if they are the same, and they’re not.
Visible ‘penetrates water more deeply’, because it doesn’t interact with water, water being a transparent medium for it visible simply passes through, is transmitted unchanged. Long wave, thermal infrared, immediately interacts to move the molecules of water converting them to heat, the heat is spread by convection.
[You’ll also need to bear in mind that water has strange properties, a very high heat capacity, it can be holding more energy than is readily apparent from its temperature, I posted something on this a few posts up.]
You’ll notice on the KT97 that it says the downwelling ‘Solar’ “converts to heat”. This to anyone traditionally taught immediately means the way thermal infrared heats by moving the molecule. So here a simple blatant lie because visible light isn’t capable of this, see the wiki page above for differences electronic translation and vibrational, a clear manipulation of real physics by changing attributes and processes, swapping them around.
Two things immediately happen, a traditionally trained scientist will just assume because of the wording that what “Solar” means is inclusive of thermal ir which can convert to heat, and doesn’t look further, and those being brainwashed into the AGW paradigm will look at it and associate it with the “Solar” which AGW really teaches, Visible and UV and Nr Ir either side, shortwave.
Near infrared is not hot, it is not thermal infrared, see the NASA page for traditional basics of differences between the two. It has more the character of visible, that is, reflective rather than absorbtive. It penetrates the human body further than visible and UV (which doesn’t even get through the first layer of the epidermis), but like them is reflected back out – hence the near infrared cameras which capture this reflected as visible cameras capture the reflected visible. Thermal infrared cameras which measure the heat being radiated from a body are quite different.
Near infrared penetrates the ocean quite some distance, I’m not sure I can find it again as I was having computer problems at the time, but I found an interesting example re photosynthesis. At deeper levels than red visible reaches, near infrared was being used for photosynthesis by the plants who didn’t have access to red visible, the planty things were violet coloured. So why the different levels of penetration in visible wavelengths? It ain’t because they’re absorbed. Absorbed specifically means interaction as in the atmosphere, where an electron of the molecules nitrogen and oxygen will absorb visible, and then send it out the way it came in, the ‘mechanics’ behind reflection/scattering..
It really isn’t easy to spot these techniques of disinformation, they take on a variety of forms. I’m getting better at spotting them, but as in the wiki on thermal which gave the incandescent light bulb as a like example with the Sun, ‘sunlight’ to anyone educated in AGWScience immediately means Visible, whereas for me alarm bells begin ringing, because I now have learned to read with more care and can see what they’ve done. I had my first practice when I wondered why they were teaching that carbon dioxide accumulated in the atmosphere..
I think what upsets me most about this manipulation of real physics, is that we’ve learned so much about our world, especially in the last hundred years, and this knowledge is being deliberately tampered with because the basics have been changed so much that sensible appreciation will be impossible for those being educated by AGWScience Fiction Inc’s malign influence.
Surely, the AGW Team cannot literally mean visible light only, ignoring all other wavelengths in the spectrum of solar radiation received at TOA? If they are limited the energy budget literally to visible light only, it appears that there may be merit in some of what you are saying.
Well, I’ve been saving the page I found yesterday for someone who could appreciate it… [grin]
http://www.answers.com/topic/greenhouse-effect

“Warming of the Earth’s surface and lower atmosphere caused by water vapour, carbon dioxide, and other trace gases in the atmosphere. Visible light from the Sun heats the Earth’s surface.” Concise Encyclopedia Britannica
“The warming of the atmosphere as some of its gases absorb the heat given out by the earth. Short-wave radiation from the sun warms the earth during daylight hours, but this heat is balanced by outgoing long-wave radiation over the entire 24-hour period.” Oxford Dictionary of Geography
“The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases, and is re-radiated in all directions. Since part of this re-radiation is back towards the surface, energy is transferred to the surface and the lower atmosphere. As a result, the temperature there is higher than it would be if direct heating by solar radiation were the only warming mechanism.[1][2]
Solar radiation at the high frequencies of visible light passes through the atmosphere to warm the planetary surface, which then emits this energy at the lower frequencies of infrared thermal radiation.” Wikipedia

As you can see, not only is this standard AGWScience Fiction Inc’s paradigm, it is in standard reference works which have a reputation for accuracy, the encyclopaedia britannica for goodness sake! But why should I have been shocked, I didn’t think anything could shock me more than what I found NASA had done (see my post on this above).
I’ve investigated this thoroughly to my own satisfaction, through discussions here which is where it first came to my attention – this is the standard teaching now. The science fiction that Visible light heats the Earth. The mantra is, ‘Visible/shortwave heats the Earth, the Earth radiates out Thermal Infrared’ .
“Solar” and “Sunlight” always means this excluding thermal, longwave, infrared in downwelling from the Sun, and it is spread by those I would have thought knew better. I was told in a discussion here by a PhD, ‘the heat you feel comes from the visible light’. But, how can these standard reference books be teaching this fiction as if science fact?
The indoctrination has been going on longer than you might think. It begins with teaching in the elementary schools by teachers who were trained to teach this, a couple of decades down the line and ‘everyone’ takes it for granted. And since concurrently all dissenting voices have been effectively marginalised, it is very hard to explain what traditional science really has to say about this because even those who know often have to toe the party line in giving explanations.. A whole generation with no idea how the world really works.

RACookPE1978
Editor
August 22, 2011 2:56 pm

OK. So take the mythical, easy-to-calculate, “average” 1367 Watt/m^2 number and divide it into the easy-to-calculate perfect black body featureless non-rotating earth-sized disk that CAGW theory begins with. You get a standard, simplified radiation model.
but …
“Radiation shorter wavelengths than 0.3 micro m is absorbed by the upper atmosphere” ((Optical Properties of Snow, Warren, 1982). So, where does this absorbed energy “go” in the standard model? It’s not reflected by clouds or the “mythical earth’s “average” albedo … it never got to the ground to be re-radiated at longer wavelengths as a GHG contributor. But we have not seen it used or “subtracted” in subsequent calculations.
UV intensities have changed by more 8-10% in recent decades … so what changes in the upper atmosphere after the “absorbed” UV flux changes in intensity by that much? How is this vast amount of energy (radiation) affect the “balance” that is being driven by a supposed 3.0 watt/meters “CO2 forcing” into disaster? Or is it?
If 30% of the sun’s energy is reflected by clouds (according to the common models) … where does this energy “go” and who is measuring it with what sensors?
In the high Arctic, what little of the inbound sunlight that is not absorbed by clouds and dust and the atmosphere (and so remains as direct radiation) is reflected by the surfaces (both ice and open ocean); and so it bounces back off the surface and goes back into the atmosphere. If much of the original direct radiation is absorbed by the atmosphere, and most of the reflected radiation (unchanged in wavelength!) is absorbed by the atmosphere, where does the absorbed energy “go” ?

tallbloke
August 22, 2011 3:12 pm

RACookPE1978 says:
August 22, 2011 at 2:56 pm
In the high Arctic, what little of the inbound sunlight that is not absorbed by clouds and dust and the atmosphere (and so remains as direct radiation) is reflected by the surfaces (both ice and open ocean); and so it bounces back off the surface and goes back into the atmosphere. If much of the original direct radiation is absorbed by the atmosphere, and most of the reflected radiation (unchanged in wavelength!) is absorbed by the atmosphere, where does the absorbed energy “go” ?

Shortwave isn’t absorbed by the atmosphere. It goes back out to space if it gets reflected off the ice and snow. See the Trenberth budget I posted above.
By the way, the cartoon shows 67W/m^2 of the incomong solar being directly absorbed by the atmosphere. This is the near IR Myrrh talks about not being thermal. He/she doesn’t seem to be able to hear or read what people are telling him/her about the ocean not consisting of pure water, so the absorption of visible wavelengths in the ocean doesn’t make sense to him/her. Solar Blue and UV carries little energy compared to the other end of the visible spectrum. We have to remember quantities as well as qualities. Herschel didn’t get zero results higher up the spectrum when he refracted incoming sunlight.

kuhnkat
August 22, 2011 4:01 pm

Myrrh,
maybe if you didn’t spend so much time on the kids NASA pages and Wikipedia, just sayin’.

kuhnkat
August 22, 2011 4:04 pm

Myrrh,
show me what heats the oceans. You keep presenting voluminous clips of what you claim are Science Fiction. Show me what heats the oceans.

Bob_FJ
August 22, 2011 4:05 pm

Myrrh, it disturbs me that you feel so passionate about your translation of physics. The effort you put into it is truly impressive. In case you missed something I posted above, I repeat and add to it

One thing that springs to my mind is that plain ornery window glass is I believe fairly opaque to IR and UV light. However modern automotive glass is a step beyond, being deliberately designed to be optimally highly reflective and/or opaque to those EMR frequencies because they are very unwelcome in the cabin, and with absolutely no benefit. The purpose to eliminate UV and IR transmission is to improve passenger comfort in hot weather, especially in modern cars with increased glasshouse‘s, and to reduce air-conditioning workloads. (fuel consumption and performance loss). The EMR “filtering” can be variously achieved in coatings and chemical inclusions in or on the glass. (or in the non-glass laminates in windscreens). The visible light transmissivity is usually also cut down with colour tinting. Nevertheless……. Don‘t park in the summer sun!
It seems hard to find anything authoritative WRT to solar spectrum on this per Google, but the following forum is interesting, if a tad brief:
http://www.infraredtraining.com/community/boards/thread/4343/

Modern automotive glass is deliberately both IR and UV cut, and visible light transmission is also reduced, mainly in order to reduce cabin heat in the summer. (the latter being a compromise with visibility safety). Have you ever noticed that if you park the car in the summer sun, the steering wheel gets uncomfortably hot? (but not if it is a cloudy day?).
Can you explain why?

kuhnkat
August 22, 2011 4:12 pm

MKelly,
thanks. Myrrh appears to be stating that visible light carries NO energy that can be converted to heat. Your post shows he is wrong. Wonder how he will deny it?

Tim Folkerts
August 22, 2011 4:31 pm

tallbloke says:
August 22, 2011 at 2:29 pm
“What have I missed?”
I think what you are missing is something that is not explicitly shown in the simplified diagram. The diagram lumps the entire atmosphere into one system. This is not an error, only a simplification to show the most important features.
The next simplest model of the atmosphere would divide the one average layer into 2 layers. The top layer seems to be emitting 195 W/m^2 upward, so it must be emitting 195 W/m^2 downward. In a two-layer model, this would be absorbed but the lower layer. So the lower layer gets 195 W/m^2 from the top layer and 350 from the surface layer, for a total of 545 incoming IR. The bottom layer is emitting 324 W/m/s down and 324 up, for a total of 648 outgoing IR. This leaves a deficit of roughly 100 W/m^2. This can be made up by the 78 + 24 W/m^2 from evaporation and convection.
In reality, the atmosphere would have to be divided into more and more layers (and divide the earth my zones of latitude and divide the globe by day and night) to get closer & closer to the actual conditions. This would of course require computers, and it beyond my direct experience.
But even the slightly-more-sophisticated 2-layer model removes most of your energy imbalance issues, I think.
“And, given that 226 of this 260 already cooled the surface as it left, why is it double counted as warming the surface when it comes back down with the ‘new’ incoming solar derived longwave?
Suppose that you have $1,000 and I have $1,000. One second you give me $350. The next second I give you $320, and some other friend gives you $170. Is it double counting to say you have $1140?
If we keep doing this, you will quickly gain money, so if you want to stay even. you better find a way to give away some more money. Fortunately there are other “friends” to whom you can give the excess. And I will quickly lose money, especially since I am handing $195 on to someone else every second. Fortunately for me, others are making up the difference.

kuhnkat
August 22, 2011 4:31 pm

Tim Folkerts,
1) Having the Sun’s energy double during the day… HUH????
I said that the energy is being doubled counted causing these large numbers that contribute to no “heating”. Just because energy passes a point in space 2, 3, 4 times or more does NOT mean it has done any work. If the energy did work it could not be reemitted or reflected several times. It would have been lost from the flux.
2) sorry, I don’t know all the quantum details and don’t know the details of all the lasers available. The few I have read about depend on “pumping” the media causing a wave of emissions that then reflect back and forth as more pumping goes on. If you counted the TOTAL energy going past on each wave you would end up with more than is emitted due to “double” counting just like in the atmosphere. The difference that makes the laser so high output, in addition to the pumping, is redirecting almost all the emissions into the same direction by the reflections at both ends. In our atmosphere one end has more than a 50% loss!!
3) Sorry, the 320 heading down wouldn’t be there if you hadn’t had energy going up to excit the GHG’s. That is the problem with the averages, you forget what comes first.
GHE is where the radiation HEATS the surface. if it HEATS the surface it isn’t immediately available to go back up. Since even the IPCC’s charts show the losses at the atmosphere end, doesn’t seem to be much heating going of the ground going on.

kuhnkat
August 22, 2011 4:56 pm

Myrrh,
It does not matter whether Trenberth includes thermal downwelling from the sun or not. As I mentioned before, and you can find any number of TOA solar spectra to substantiate taken outside of the IPCC science era, thermal IR from the sun is NOT measurable at the ground. I chased this myth until I was comfortable that it is a myth. You need to do more research and show me why you think that downwelling SOLAR thermal IR is measurable at the surface. The spectra we have simply do not support this idea.

Tim Folkerts
August 22, 2011 5:07 pm

RACookPE1978 says: August 22, 2011 at 2:56 pm
“Radiation shorter wavelengths than 0.3 micro m is absorbed by the upper atmosphere” ((Optical Properties of Snow, Warren, 1982). So, where does this absorbed energy “go” in the standard model?
That would be part of the “67 W/m^2” of incoming solar energy shown being absorbed by the atmosphere. (See the image Tallbloke posted earlier)
“UV intensities have changed by more 8-10% in recent decades … so what changes in the upper atmosphere after the “absorbed” UV flux changes in intensity by that much? How is this vast amount of energy (radiation) affect the “balance” that is being driven by a supposed 3.0 watt/meters “CO2 forcing” into disaster? Or is it? “
UV is about 10% of the Solar energy, or 0.1 * 342 = 34.2 W/m^2 on average. 9% of that is ~ 3 W/m^2. Assuming your numbers are correct, then both UV & CO2 forcings have changed about 3 W/m^2, suggesting both would be of similar importance.

Myrrh
August 22, 2011 5:22 pm

tallbloke says:
August 22, 2011 at 3:12 pm
Herschel didn’t get zero results higher up the spectrum when he refracted incoming sunlight.
All Herschel discovered was that there was an invisible wavelength other side of Visible which was immensely hotter. Near Infrared isn’t hot. See the NASA page. Neither is Visible. Neither is UV. We cannot feel them. We cannot feel them.
So, what were the supposed temps of these? Where did this come from if these wavelengths aren’t hot? We do not feel any heat from them. What was the thermometer actually measuring?
What you generic are significantly failing to take on board is that I have given you specific details of the change in teaching in the NASA example. There is a disjunct. I’m sticking with the traditional tried and tested physics used in countless real world applications where thermal infrared is Heat, and this is what we feel on the surface of the Earth and this is what is doing the heating. Light does not heat stuff.
Surely you have just tons and tons of specifics on this? This is your basic premise.
Give me real tested proof that Visible is a) hot, b) capable of heating the oceans and land.
BECAUSE the claim is that this is what is raising the temperature of the Earth.
You’re the one’s teaching something different. You’re the ones who have to prove what you’re saying. Deal with it.

Konrad
August 22, 2011 5:39 pm

t.f.p. Says:
August 22, 2011 at 2:19 pm
I had a quick look at your set up, and I would have a couple of points.
The first is that your radiation source, even with the filter, appears to be very short wave. With regard to heating through CO2 back radiation, a spectral peak around 15 microns is appropriate. As George E. Smith pointed out further back on this thread, a suitable source for such LWIR would be warm water itself. In my experiment I placed a foil reflector above warm water to slow its rate of cooling.
The second point would be that your set up restricts evaporative cooling. My results indicate that evaporative cooling is the deciding factor in whether backscattered LWIR can slow the rate at which liquid water cools.
For those interested, this is an photo of the set up used in the experiments I have described earlier on this thread.
http://i56.tinypic.com/2zs2lbs.jpg
The side with the foil “Sky” is shown disassembled. The strange black shapes wired to the gel cell are small centrifugal blower fans for simulating wind.

Myrrh
August 22, 2011 6:36 pm

“What is thermal energy ?
Thermal Energy: A specialized term that refers to the part of the internal energy of a system which is the total present kinetic energy resulting from the random movements of atoms and molecules.
The ultimate source of thermal energy available to mankind is the sun, the huge thermo-nuclear furnace that supplies the earth with the heat and light that are essential to life. The nuclear fusion in the sun increases the sun’s thermal energy. Once the thermal energy leaves the sun (in the form of radiation) it is called heat. Heat is thermal energy in transfer. Thermal energy is part of the overall internal energy of a system.
At a more basic level, thermal energy comes form the movement of atoms and molecules in matter. It is a form of kinetic energy produced from the random movements of those molecules. Thermal energy of a system can be increased or decreased.
When you put your hand over a hot stove you can feel the heat. You are feeling thermal energy in transfer. The atoms and molecules in the metal of the burner are moving very rapidly because the electrical energy from the wall outlet has increased the thermal energy in the burner. We all know what happens when we rub our hands together. Our mechanical energy increases the thermal energy content of the atoms in our hands and skin. We then feel the consequence of this – heat”
http://thermalenergy.org/
=================
Thermal energy, heat on the move, is invisible. It is also known as thermal infrared, to differentiate it from near infrared which is not heat on the move. Light, also known as visible, is not heat on the move. It is not hot. You cannot feel it. It does not heat stuff up.
It could not be what Herschel was measuring..
Do let that sink in.
Enough of your prevaricating. Show me proof from your science that Visible is heating the land and oceans. You can’t. You won’t find this anywhere, because it is not physically possible.
I know that. I’ve given you real explanations from real physics to show that Visible, Light, is not an energy powerful enough to heat up stuff. I’m asking you to look for proof that it can as you claim, so you get to know that it can’t.
Until you can come back with solid proof that it is Visible and not Thermal Infrared heating the land and oceans, what the heck are you talking about? What are all your arguments about ‘energy’ budgets worth? You don’t even know what you’re arguing about because you have no idea what heat from the Sun really is.

“Infrared light lies between the visible and microwave portions of the electromagnetic spectrum. Infrared light has a range of wavelengths, just like visible light has wavelengths that range from red light to violet. “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. The temperature-sensitive nerve endings in our skin can detect the difference between inside body temperature and outside skin temperature
Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control.”

What NASA , or someone at NASA, is still trying to teach children – traditional, well tested, well understood, used in countless applications because well understood.
You are saying something different. It is incumbent on you, who have changed traditional physics, to prove that traditional teaching is wrong and you are right.
So do it.

jae
August 22, 2011 6:38 pm

Sheesh, where did the noble cowboy go? Hiding in the barn?
Willis: this IS still YOUR DOMAIN!

August 22, 2011 7:34 pm

Myrrh,
ALL energy heats molecules which absorb it.
In the process of being absorbed by a molecule energy is stopped in its travels and instead is converted to vibrational movement of the absorbing molecule.
The wavelength determines the amoiunt of light energy available to be transferred to the receiving molecule, the shorter the wavelength the more energy is involved.
Visible wavelengths entering water are slowed down as they pass through water up to about a maximum of 200 metres or so and because they are generally of shorter wavelengths than infra red they impart more energy to deeper molecules than does infra red but in the end they are all absorbed and impart their energy to water molecules that then heat up.
One cannot ‘feel’ the energy in visible light (or of ANY wavelength) directly. Instead one just feels the additional energy acquired by the skin when skin molecules vibrate faster after having absorbed visible light energy. There is no doubt that visible light has a warming effect on skin (and everything else) but there is a level below which one does not significantly notice it. Nevertheless it is present.
The oceans are heated by ALL energy that gets past the evaporative barrier.

kuhnkat
August 22, 2011 8:02 pm

I and others have had questions about the bandwidths that K&T use for their energy balance cartoons. Here is the Abstract from their 1997 budget:
http://www.cgd.ucar.edu/cas/abstracts/files/kevin1997_1.html
This links to their shortwave graph:
http://www.cgd.ucar.edu/cas/papers/bams97/fig5.gif
that appears to show about .125-3.0 micron.
Their longwave graph:
http://www.cgd.ucar.edu/cas/papers/bams97/fig1.gif
which appears to run from about 3.0-50.0 microns.
Here is a link to the PDF of the 1997 paper:
http://atoc.colorado.edu/~dcn/ATOC7500/members/Reading/KiehlTrenberth.pdf
Hope this helps.

Tim Folkerts
August 22, 2011 8:28 pm

kuhnkat says:
3) Sorry, the 320 heading down wouldn’t be there if you hadn’t had energy going up to excit the GHG’s. That is the problem with the averages, you forget what comes first.
You need a new paradigm. You keep claiming double counting, but never show any specific case. Double counting would be something like “there is a net flow of ~ 70 W/m^2 downward, but the atmosphere emits ~ 320 W/m^2 downward. This leads to a total of 390 W/m^2 downward.
The atmosphere and the ground ALREADY have millions of J of energy in each square meter. These joules could have come from the sun 1 second ago or 1 hour ago or 1000 years ago. Or they could have come from when the earth was hot billions of years ago. The could have come from cosmic rays from outer space. They could have come from radioactive Rn decaying in the atmosphere. In fact, they came from all of these sources. They did NOT come only from the sun during the last second.
Now that we know the surface and the atmosphere have lots of energy, the question is “how does that energy change during the next second?” If the energy decreases in the ground, the ground cools; if the energy in the atmosphere increases, the atmosphere warms. Our “energy accountant” needs to track every joule of energy received and emitted by each “division” over the course of a second. Then our “energy accountant will come up with an “annual report” of the trends.
The numbers obviously fluctuate from day to night; from poles to equator; from summer to winter.
But for an average m^2 of the surface during an average second:
* it looses 390 J via IR (from the millions it already had)
(NOTE: 320 of these go to the atmosphere and 40 go to outer space, but that is not the business of the ground. its accounting only deals with how much it gains or loses itself).
* it loses 80 J via evaporation (from the millions it already had)
* it loses 20 J via evaporation (from the millions it already had)
* it gains 170 J via sunlight (from the gazillions of J the sun had)
* it gains 320 J via IR (form the millions the atmosphere had.)
ACCOUNTANT’S REPORT: The average square meter of ground gained 490 J of energy and lost 490 J of energy during an average second for a net change of 0 J, so it is at the same temperature as it was the start of the year.
And for an average m^2 of the atmosphere during an average second:
* it looses 520 J via IR (from the millions it already has)
(NOTE: 320 go to the surface and 200 go to space)
* it gains 80 J via evaporation (from the millions the surface already has)
* it gains 20 J via evaporation (from the millions the surface already has)
* it gains 70 J via sunlight (from the gazillions of J the sun has)
* it gains 350 J via IR (from the millions the land had.)
ACCOUNTANT’S REPORT: The average square meter of atmosphere gained 520 J of energy and lost 520 J of energy during an average second for a net change of 0 J, so it is at the same temperature as it was the start of the year.
Each J of energy is counted exactly once. You could pass dollars or rocks back and forth the same way. There are no magical Enron dollars mysteriously appearing. There are no rocks suddenly appearing or disappearing from the conveyor moving the rocks from one pile to another.
(You could, of you want, combine the IR energies to give only net transfers. The surface and the atmosphere would both have 320 less arriving and 320 less leaving. The annual reports would decrease to 170 J and 200 J respectively both in and out for surface and atmosphere, but the year-end accounting would still have zero net flow)
(Even a small imbalance of 1 J for each square meter during an average second would result in an imbalance of a significant 32 million J during a year)
You have to give up the mindset that a given joule of energy is ricochetting back and forth.
You have to give up the mindset that a given square meter with millions of joules can only transfer 170 J of energy if it only gets 170 J of energy from the sun.

RJ
August 22, 2011 10:38 pm

tallbloke says:
August 22, 2011 at 2:29 pm
At some point in the future people will look back and laugh at science today because of this earth radiation sketch.
For a start it does not add up
Incoming surface radiation = 168
Yet surface outgoing is
78
+24
+390
Total 492
Somehow another 324 units of energy is magically created by backradiation. This is just complete and utter nonsense and I’m surprised that anyone still tries to defend this. If the extra energy comes form the sun then the incoming surface solar radiation should be increased to 492. And total incoming solar radiation adjusted accordingly as well.
And even if the energy leaving the surface was reduced to 66 units. 40% leaving as radiation from the surface still seems far too high. But 390 / 492 = 79%. Surely this is just more science fiction.

RJ
August 22, 2011 10:51 pm

surface.Tim Folkerts says:
August 22, 2011 at 8:28 pm
“You need a new paradigm.”
Like CO2 and water vapor are really energy generators in the atmosphere with LWR acting as their power source. So say 20 units of LWR go in and say 40 units come out and are fired back at the surface as shorter wave radiation. To create a heating impact when they hit the surface.
But this CO2 and water vapor power source is somehow regulated to only ever produce just so much new energy to stop the earth overheating.

kuhnkat
August 22, 2011 11:00 pm

Tim Folkerts,
Try turning off the sun and see what happens to the energy fluxes. Which would you say stays stronger, the atmosphere, the ground, or the oceans?
Now that wasn’t that hard to figure out now was it. When are you going to quit double counting??

Spector
August 22, 2011 11:16 pm

RE: Myrrh: (August 20, 2011 at 6:30 pm)
“You are postulating that light created by the Sun is eternal.”
Not quite. The energy is eternal. It was not created by the Sun; it was released by the Sun’s fusion processes.
Energy can be reconfigured, subdivided or reflected but it cannot be destroyed. This reconfiguration is accomplished by interacting with some other particle or field.
The chemical bonds in the fuel of your car are stored energy. When those bonds are broken, that energy is released as heat and kinetic motive action.

Konrad
August 23, 2011 12:38 am

Is anyone on this thread still interested in the original question about whether backscattered LWIR can heat or slow the cooling of Earth’s oceans? Willis appears to have run away, but it would be difficult to blame him given that the discussion has got very off topic. If backscattered 15 micron IR cannot slow the cooling of the oceans, then the global warming hoax is over.

Myrrh
August 23, 2011 12:57 am

Stephen Wilde says:
August 22, 2011 at 7:34 pm
Myrrh,
ALL energy heats molecules which absorb it.
In the process of being absorbed by a molecule energy is stopped in its travels and instead is converted to vibrational movement of the absorbing molecule.
The wavelength determines the amoiunt of light energy available to be transferred to the receiving molecule, the shorter the wavelength the more energy is involved.
Visible wavelengths entering water are slowed down as they pass through water up to about a maximum of 200 metres or so and because they are generally of shorter wavelengths than infra red they impart more energy to deeper molecules than does infra red but in the end they are all absorbed and impart their energy to water molecules that then heat up.
One cannot ‘feel’ the energy in visible light (or of ANY wavelength) directly. Instead one just feels the additional energy acquired by the skin when skin molecules vibrate faster after having absorbed visible light energy. There is no doubt that visible light has a warming effect on skin (and everything else) but there is a level below which one does not significantly notice it. Nevertheless it is present.
The oceans are heated by ALL energy that gets past the evaporative barrier.

Stephen, water is a transparent medium for Light, the molecules do not absorb Light. That’s a basic real world physics definition. It means that Light does not impart any energy to the molecules of water, it passes straight through, the term here is transmitted. This can be observed in clear water and glass which are transparent mediums.
So, here’s the disjunct from AGWScience Fiction Inc’s manipulation.
It says that the atmosphere is transparent for Visible light, a clear medium in which it passes straight through without being absorbed, it says that this is like the glass in a greenhouse window, which visible light passes straight through without being absorbed.
So, when I point out that in real world physics water is a known transparent medium for visible light just like clear glass, that it is transmitted through without being absorbed, why do you’all keep insisting that it is being absorbed and not transmitted straight through?
Because you’re repeating a ‘meme’ from AGWScience Fiction Inc.
Try to find some real pages from the real world on Light which give the information from traditional science, I gave examples above of this.
Water and glass are transparent mediums for visible light, they are transmitted through without being absorbed.
Find the link I gave above to the wiki page on translucency, and read the complete section I extracted from.
There are several things explained that visible light and UV are capable of doing, transmission means that they pass straight through without being absorbed, so no energy is received by the medium.
I’m under no illusion that what I’m saying about the way AGWScience twists real physics to claim something different is easy to grasp, my mind felt like a yo-yo when I began exploring this. But this is a good example of how subtle the change from real science is which results in a completely different explanation of something – which means that if believed without questioning or having no reason to doubt the explanations the person accepting it has lost a real grasp on the actual physics. AGWScience Fiction Inc creates impossible worlds by such methods.
AGWScience Fiction Inc, likes to create confusion. It doesn’t want you to know real physics. It has insinuated itself into the education system, so now it is very very hard to pick it apart..
There are two parts to just the above which shows the extent of this manipulation, but you really need to concentrate on this, it isn’t easy to see.
That’s the first part, that it says glass is a transparent medium, which is correct, but says water isn’t, which is a lie.
The second part is that it says the atmosphere is a transparent medium, but it isn’t in real physics.
Why isn’t the atmosphere a transparent medium? Because visible is actually absorbed on an electron level by the molecules of nitrogen and oxygen, it is this which produces reflection and scattering, the blue sky.
Now, at this point you need to bring in what absorbed means here in real physics, the wiki extract explains it. If visible light is being scattered it means that the electron is energised and then sends out the light the way it came in. No heat is being created. The energy is being used for motion.
So, you’ll need to look at what ‘absorbed’ means. It doesn’t always mean that heat is created. As in photosynthesis for example, it can mean that the energy of light is used for a chemical conversion, the creation of sugars. This process doesn’t create heat any more than reflection/scattering creates heat.
What is happening here? AGWScience Fiction Inc has given the meme that all absorption is direct creation of heat. That’s not true as these examples show.
Confused? That’s the point of these memes, to confuse. To take real physics out of the general populations’ ken.
So, in real physics the molecules of nitrogen and oxygen reflect/scatter visible light, which means they are absorbed, so my question earlier – how hot is the atmosphere from this?
Convoluted isn’t it?
If you say the atmosphere is a transparent medium and visible doesn’t heat it, and I show you that on an electron level it is absorbed, by AGWScience fiction that means it must be creating heat in the atmosphere because the individual molecules absorb it to reflect and scatter.
The only way out of this AGWScience Fiction Inc’s trap is to get back to real traditional physics, and that is getting harder to find explained..
The reason it is getting harder is because the avenues for it are being closed down. My examples from standard reference sources, Encyclopedia Britannica and the Oxford Dictionary of Geography. Wiki was taken over completely by a virulent Greenie, who kept taking any references to real physics and the real arguments out of the pages. He is now ousted and the pages have to some extent been cleaned up. But not completely, because the fiction that visible is the heating method for the Earth is now ‘establishment’ science, and, there are subtle nuances which confuse as I’ve gone into re the thermal sun’s energy is like incandescent light bulb with the insertion of the word ‘sunlight’.
This is a very clever and now very well established con because the people establishing it have the means to do this, various methods. They introduced it into the education system from elementary level by teaching the would be teachers that this is ‘science’, at the same time instilling fear in the children the lie that carbon dioxide is a toxic, a poison, which in real physics it isn’t, and, by giving experiments and explanations which are nonsense in real life physics, like heating water with a heat gun which is gobbledegook, how many people bother to check and find out that they are calling this heat transfer by radiation when it is heat transfer by convection? But people repeat it and so it is passed on to create more confusion down the line.. That’s why the meme that ‘the science is settled’ was so widely promoted for years and years and concurrent demonisation and ad homs and blocking access to publishing papers to those scientists arguing against it, because it is now in the consciousness of the general public through constant repetition of the lie, who believe it without thinking, because they have no reason to doubt it. It’s a technique of brainwashing on a grand scale.
So here, the simple technique of reversing the properties of Heat and Light, attributing the properties of one to the other, of saying that it is Visible which is thermal heating the Earth, has a whole string of misattributions attached, as memes. Each part has to be looked at separately and examined because sometimes the definitions are used correctly as in transparent meaning that something is passed through without being absorbed, but then misattributed.
Absorbed technically does not always mean that heat is created. Transparent means that something is passed through without being absorbed. The atmosphere is not transparent to visible. Water is transparent to visible. You’ll need to replace the fiction memes with memes from real physics to get internal coherence back.
Until you do, you can’t see the fictional world the AGWScience Fiction Inc has created, impossible.

tallbloke
August 23, 2011 1:08 am

Tim Folkerts says:
Two layer system

Hi Tim, yes, but that’s not what Trenberth’s cartoon shows, and this is why NASA no longer has that cartoon on it’s website, they just show the net flow, because the truth is, all these multilayer models are speculative. Just as speculative as my suggestion that a lot of the downwelling IR may be getting absorbed in evaporated molecules just above the surface and re-emitted from there.
One second you give me $350. The next second I give you $320, and some other friend gives you $170. Is it double counting to say you have $1140?
If we keep doing this, you will quickly gain money, so if you want to stay even. you better find a way to give away some more money.

Lets get the numbers right. 168 from the Sun, which has to end up leaving as IR. 390 IR out of the ocean, plus 102 latent and thermal = 492 energy upwards. Trenberth is simply balancing the 168 in from the Sun with the 492 upward to get his 324 back radiation. I suspect both the outgoing and back radiation averages may be inflated to inflate the LR figure in relative importance to convection. Biomass converts incoming shortwave to biomass which when dead releases sensible heat as it rots, in forest fires etc. Leaves maintain a steady temperature through evapotranspiration, not by emitting IR so much. This is why diurnal temperature variation over the Amazon is steady throughout the year despite a 7% variation in incoming solar due to orbital eccentricity.
kuhnkat says:
August 22, 2011 at 4:31 pm
I said that the energy is being doubled counted causing these large numbers that contribute to no “heating”. Just because energy passes a point in space 2, 3, 4 times or more does NOT mean it has done any work. If the energy did work it could not be re-emitted or reflected several times. It would have been lost from the flux.

Energy in = energy out. How much it contributes to the bulk temperature in the meantime is the question at issue. I suspect a very short ‘residence time’ for downwelling IR, because it is absorbed in the first few microns, can’t conduct downwards, can’t convect downwards and can’t work miracles heating the abyss by magically bypassing these hurdles.

tallbloke
August 23, 2011 1:19 am

Myrrh says:
August 22, 2011 at 5:22 pm
tallbloke says:
August 22, 2011 at 3:12 pm
Herschel didn’t get zero results higher up the spectrum when he refracted incoming sunlight.
All Herschel discovered was that there was an invisible wavelength other side of Visible which was immensely hotter. Near Infrared isn’t hot.

Have a look at this solar spectrum
http://tallbloke.files.wordpress.com/2011/08/spectral-content.gif
The absorption around 1750 and 3500 nm is your incoming solar IR. It amounts to the ~67W/m^2 directly absorbed by the atmosphere. Don’t forget the sum is divided by four to average the radiation incident on the Earth as a whole, half of which is in shadow at any one time.
From what I’ve been told, the point is that there’s lots more of the visible energy coming from the Sun to Earth than IR, as you can see from the plot, so despite the fact it doesn’t excite water as much, there’s more of it to get the job done. UV burns your skin, which is why sunblock blocks UV. To burn your skin, it has to heat it.
Your innards are not warmed by radiation, they are warmed by the chemical processes breaking down your food, and to a small extent by the conduction of heat in blood if evaporation isn’t cooling you enough. That pretty soon leads to heat exhaustion.
I’ll do some more looking into it, thanks for your contribution. Please shorten your comments and point to long previous ones with links by right clicking the timestamp in the previous comment and copying the url so we don’t get swamped. Thanks.

tallbloke
August 23, 2011 1:38 am

G. Karst says:
August 21, 2011 at 9:02 am
To think all these issues will be resolved within a blog discussion is a little beyond, a reasonable expectation. Quantification is the best we can expect at the moment. Experimental data is the only way, for now. So discuss all you want, but do not scoff at each other’s hypothesis, because none of you are completely correct. IMHO GK

Well said.
Konrad says:
August 21, 2011 at 5:08 pm
For 40C water cooling only through radiation and conduction the containers both cool slower than the evaporatively cooled samples. The container under the foil “Sky” cooled 2.5C in 15 min. The container under the clear “Sky” cooled faster at 2.8C in 15 min. The two samples continue to diverge in temperature as much as 1.2C in the first 45 min. After several hours the samples converge to room temperature.
I believe the results I am getting are consistent and repeatable and possibly even “robust” ;). It should be noted that although my clear “Sky” was actually a ceiling around 20C, the foil “Sky” was reflecting most of the IR emitted by the warm water, not just under 50% of the 10 to 20 micron frequency. Also I have not yet repeated the small scale test with the artificial wind from computer fans. I urge others to repeat this style of experiment to confirm the result.
My conclusion is that backscattered LWIR cannot slow the cooling rate of Earth’s oceans to any measurable degree. Given that the oceans cover 71% of the planets surface, estimates for climate sensitivity should probably be reduced from 1C to per CO2 doubling to 0.29C. This figure is likely to be smaller again, as plants that cool through transpiration may need to be subtracted from equations. I would further speculate that anthropogenic emissions of CO2 cannot cause dangerous, catastrophic or even measurable global warming even if we burnt all known and projected fossil fuel reserves.

Very well done Konrad. I’m sure the warmista will want to pick holes in your experimental procedure, at which point we can say:
“OK, using some of the 122 billion dollars you’ve had to investigate global warming over the last 20 years, where is the better controlled experiment you did to test this fundamentally important issue?”
If the best they can do is point to the Minnett – Realclimate experiment which never got published, they are looking pretty bad…
If it’s ok with you, I’ll feature your experiment on my blog, let me know here.
Cheers

Myrrh
August 23, 2011 2:45 am

Konrad says:
August 23, 2011 at 12:38 am
Is anyone on this thread still interested in the original question about whether backscattered LWIR can heat or slow the cooling of Earth’s oceans? Willis appears to have run away, but it would be difficult to blame him given that the discussion has got very off topic. If backscattered 15 micron IR cannot slow the cooling of the oceans, then the global warming hoax is over.
I haven’t gone off topic. If you want to know how much the ocean absorbs thermal infrared you have to begin with the thermal infrared reaching the Earth from the Sun, which is what actually warms the oceans. Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known. It heats up because thermal infrared is capable of heating it as water and organic substances have resonant frequencies with it, which means they vibrate vigorously. There is a general depth of absorption of 2-4 inches. This nonsense that the ‘surface tension’ of water will repel any ‘backradiating’ thermal is ridiculous, thermal is a powerful energy. It packs a punch as you should know every time you step into the Sun and feel its heat warming you up.
The question really is, can heat flow from a cooler to a warmer body, if the ‘backradiating’ longwave is colder than what the earth is radiating up to it, then it can’t.

Konrad
August 23, 2011 3:26 am

tallbloke says:
August 23, 2011 at 1:38 am
I would be happy for you to cover the experiment over at the Talkshop.
An image of the present set up can be seen here –
http://i56.tinypic.com/2zs2lbs.jpg
The black frame on the left hand side is covered with cling wrap, which does not show up well in the image. I have found a peltier chip and some thermal paste in the shed, so I will be able to improve on the “non” backscattered side over the weekend. In the meantime feel free to cut and paste from the comments here.
I would be happy for the moderators to pass my email on to you, so I can provide you further details.

Tim Folkerts
August 23, 2011 3:41 am

tallbloke says:
“Lets get the numbers right. 168 from the Sun, which has to end up leaving as IR. 390 IR out of the ocean, plus 102 latent and thermal = 492 energy upwards. Trenberth is simply balancing the 168 in from the Sun with the 492 upward to get his 324 back radiation.”
I pretty much agree other than the word “simply”.
I have glanced thru there paper, but it was a while ago. I can pretty much guarantee that one way or another they either measured the 324 W/m^2 (eg find the average of value from pyrgeometers around the world) or calculated it (from measured temperatures and compositions of the atmosphere around the world). I suspect the “thermals” component is tougher to get right than the IR, since it requires estimating the magnitude of updrafts and their intensity around the globe.
I suspect they had to work a bit to get the numbers to all balance. I suspect that the uncertainty in the numbers is significant. They may well have used the idea of “simply balancing” the numbers t get them to indeed balance. Some thing like “we got a value of 320 +/- 6 for the back radiation; we will call it 324 in the diagram since we know the balance is very close to zero overall.” Again this is just speculation on my part. I don;t have time to dig thru the article right now, but anyone who is truly interested could go read the paper and report what they did and how they got this number.

tallbloke
August 23, 2011 3:47 am

Question for Tim Folkerts:
O2 and N2 don’t absorb or radiate photons, but they must be excited by collisions with excited co2 and h2o molecules. Presumably they jiggle around more. What will the effect of this be? A priori, it seems likely they will more readily permit convection of more buoyant molecules. Won’t this speed cooling of the atmosphere to space?

Tim Folkerts
August 23, 2011 3:50 am

Two minor points and a not-so-minor point, Tallbloke, and then I must go.
“The absorption around 1750 and 3500 nm is your incoming solar IR. It amounts to the ~67W/m^2 directly absorbed by the atmosphere.”
Presumably the absorbed UV also contributes to this total.
” UV burns your skin, which is why sunblock blocks UV. To burn your skin, it has to heat it. “
UV damages individual molecules. The damage is not due to general warming of the skin.
See http://en.wikipedia.org/wiki/Sunburn
From what I’ve been told, the point is that there’s lots more of the visible energy coming from the Sun to Earth than IR
No, the totals are pretty close to equal. with slightly more IR than visible energy. Depending on just where you draw the somewhat arbitrary lines between UV, visible, and IR, the numbers are ~ 10 UV, 40% visible, and 50 % IR.

Konrad
August 23, 2011 4:56 am

Myrrh says:
August 23, 2011 at 2:45 am
“Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known.”
The fact that Myrrh writes this statement after I have taken the time to design a simple, repeatable empirical experiment and reported the methodology and results to this thread proving that statement false should indicate that responding to Myrrh is not worth the wear on your keyboard.
Myrrh, follow the instructions, do the experiment and learn.

tallbloke
August 23, 2011 4:59 am

Tim Folkerts says:
August 23, 2011 at 3:41 am
they either measured the 324 W/m^2 (eg find the average of value from pyrgeometers around the world) or calculated it (from measured temperatures and compositions of the atmosphere around the world). I suspect the “thermals” component is tougher to get right than the IR, since it requires estimating the magnitude of updrafts and their intensity around the globe.

The worlds foremost expert on pyrgeometry says no change in atmospheric optical depth during co2 increase: http://www.warwickhughes.com/blog/?p=87 see comment 10 particularly.
The thermals also work through thunderstorm vortices as per Willis’ thermostat hypothesis. This means the atmospheric window is probably a good deal bigger then Trenberth thinks, and varies to deal with extra heat from radiation or albedo reduction.
Tim Folkerts says:
August 23, 2011 at 3:50 am
“The absorption around 1750 and 3500 nm is your incoming solar IR. It amounts to the ~67W/m^2 directly absorbed by the atmosphere.”
Presumably the absorbed UV also contributes to this total.

Yes, but the main climatic effect seems to be related to ozone destruction affecting plankton density.
the totals are pretty close to equal. with slightly more IR than visible energy. Depending on just where you draw the somewhat arbitrary lines between UV, visible, and IR, the numbers are ~ 10 UV, 40% visible, and 50 % IR
Well, ok, I was talking about the IR Myrrh is interested in and I lumped in near IR with visible. My bad.
Anyway, ~67W/m^2 absorbed in atmosphere, around 170W/m^2 absorbed in ocean: 2.5 times more.

tallbloke
August 23, 2011 5:37 am

Myrrh says:
August 23, 2011 at 2:45 am
Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known.

Microwaves at 122mm wavelength can warm water. IR at atmospheric and solar wavelengths can’t penetrate it.
From wiki’s page on microwave ovens:
http://en.wikipedia.org/wiki/Microwave_oven#Design
“More recently, some manufacturers have added high power quartz halogen bulbs to their convection microwave models, marketing them under names such as “Speedcook”, “Advantium” and “Optimawave” to emphasize their ability to cook food rapidly and with good browning. The bulbs heat the food’s surface with infrared (IR) radiation, browning surfaces as in a conventional oven. The food browns while also being heated by the microwave radiation and heated through conduction through contact with heated air. The IR energy which is delivered to the outer surface of food by the lamps is sufficient to initiate browning”
IR browns the outside, microwaves heat the inside:
Got that?

Tim Folkerts
August 23, 2011 5:46 am

Tallbloke, I suspect we may mostly be agreeing, but often talking past each other.
“The worlds foremost expert on pyrgeometry says no change in atmospheric optical depth during co2 increase: http://www.warwickhughes.com/blog/?p=87 see comment 10 particularly.”
I was specifically talking about atmospheric IR and pyrgeometers, while this comment is about solar radiation and pyrheliometers, which is a side issue (i think, unless you have a connection in mind that I am missing). I was merely saying you can estimate the 324 W/m^2 IR by measuring the IR directly. Presumably you could also watch for trends in the pyrgeometers to see if there is indeed an increase in downward IR. I’d love to see reports/papers about that!
“Well, ok, I was talking about the IR Myrrh is interested in and I lumped in near IR with visible. My bad.
Anyway, ~67W/m^2 absorbed in atmosphere, around 170W/m^2 absorbed in ocean: 2.5 times more.”

I was talking about the “top” of the curve in the diagram you were referencing. The incoming solar radiation corresponds to ~ 10/40/50 UV/visible/IR.
The yellow part is the part absorbed by the atmosphere — that would presumably be the 67 W/m^2. I can’t easily estimate from the curve how much is due to different bands. For example, almost all of the UV is absorbed by the atmosphere, but it is a small amount to begin with.
The 170 W/m^2 that actually lands would presumably the red part of the graph. There may well me more visible energy hitting the surface than IR.
P.S. I have found from experience that Myrrh is very convinced of his position and on the authority of the websites he chooses to reference for us. No amount of evidence to the contrary is likely to change his view.

Spector
August 23, 2011 5:47 am

RE: Myrrh: (August 23, 2011 at 12:57 am)
“…water is a transparent medium for Light, the molecules do not absorb Light…
Only the empty vacuum of outer space does not absorb light. ‘Transparency’ has limitations. Destroyers and submarines use sonar instead of flashlights because the attenuation of the sounds they are using can be measured as a few dB/kyd (decibels per kiloyard). An attenuation of one dB indicates that the logarithm base 10 of the power of the signal has been reduced by 0.1.
On the Wikipedia ‘Electromagnetic absorption by water’ page, they are using a centimeter scale to measure EM absorption in water. This, I assume, is pure water uncontaminated by algae or plankton. Although undefined here, as far as I could find, the absorption coefficient usually indicates by what value the natural logarithm of the signal is reduced when passing through the specified distance (1 cm). An absorbed photon delivers its packet of energy to some ‘lucky’ molecule.
BTW, this page obviously appears to have been written by someone who has fully bought in to the greenhouse catastrophe theory. That does not mean that this largely unrelated data is incorrect or fudged—it also includes a cute water molecule vibration mode diagram.
http://en.wikipedia.org/wiki/Electromagnetic_absorption_by_water

gnomish
August 23, 2011 6:04 am

Myrrh :

you’ve got a weird notion about electromagnetic energy. best review physics 101 and get the basics right.
tallbloke: think about it = an object at the focal point of a parabolic reflector can NOT radiate more efficiently than it can without any reflector at all.
in fact, the extra chill that makes the ice is due to evaporation in very dry conditions.
back in the day of the raj, the brits had iced drinks in india – made the same way but with no reflectors anywhere.
http://query.nytimes.com/mem/archive-free/pdf?res=F20E1FF63E5F15738DDDA10894D9405B8284F0D3

Tim Folkerts
August 23, 2011 6:12 am

Oops .. in my last reply to Tallbloke I left out the clouds and changes during the day! The red part is the sunlight that hits the surface at noon when it is actually sunny. Some of that Red part would be the reflection from clouds. Part of it would contribute to the atmosphere when it is not noon and more incoming light gets absorbed by the atmosphere.
http://tallbloke.files.wordpress.com/2011/08/spectral-content.gif

August 23, 2011 6:21 am

tallbloke says:
August 23, 2011 at 3:47 am
Question for Tim Folkerts:
O2 and N2 don’t absorb or radiate photons, but they must be excited by collisions with excited co2 and h2o molecules. Presumably they jiggle around more. What will the effect of this be? A priori, it seems likely they will more readily permit convection of more buoyant molecules. Won’t this speed cooling of the atmosphere to space?
All gases dissipate heat. We run our lives with this basic understanding. Hair dryers, radiators in cars, home heating systems, we perspire, dogs pant, etc. There are no examples of gases adding heat to a system on their own.

gnomish
August 23, 2011 6:37 am

tallbloke- every bit of matter will absorb and emit photons.
http://brucegary.net/MTP_tutorial/MTP_ch5.html

tallbloke
August 23, 2011 6:58 am

gnomish says:
August 23, 2011 at 6:37 am
tallbloke- every bit of matter will absorb and emit photons.
http://brucegary.net/MTP_tutorial/MTP_ch5.html

Sure, but how much solar or Earth emitted E/M does O2 and N2 absorb or emit at atmospheric T’s and P’s?
P.S. Love the Raj ice-making story.

Tim Folkerts
August 23, 2011 7:10 am

Gnomish says
“tallbloke: think about it = an object at the focal point of a parabolic reflector can NOT radiate more efficiently than it can without any reflector at all.
in fact, the extra chill that makes the ice is due to evaporation in very dry conditions.
back in the day of the raj, the brits had iced drinks in india – made the same way but with no reflectors anywhere.
http://query.nytimes.com/mem/archive-free/pdf?res=F20E1FF63E5F15738DDDA10894D9405B8284F0D3
This is mostly wrong.
It is true that an object does not radiate better simply by putting it at the focus of a mirror. BUT the ground cannot radiate TO the object. The mirror blocks the warm IR from the ground and reflects in the cooler IR from above in its place. This allows the object to continue radiating the same energy but receive less energy, thereby cooling.
Evaporative cooling is also possible, but the article mentioned above only discusses cooling objects to 65 F by swinging wet objects, not freezing them. This chart http://mrsdlovesscience.com/realtivehumidity/pg12b.jpg shows how much cooling you you could get at various temperatures and humidities. For example, at 24 C and 9 % relative humidity, the max possible cooling would be by 14 C to 10 C. Google “wet bulb temperature” for more info)
Any further cooling was by endothermic reactions (“salts”) or refrigerators (“wonderful pneumatic machines”).

Myrrh
August 23, 2011 10:07 am

Konrad says:
August 23, 2011 at 4:56 am
Myrrh says:
August 23, 2011 at 2:45 am
“Water absorbs longwave infrared and heats up, that is a basic bog standard real world physics known.”
The fact that Myrrh writes this statement after I have taken the time to design a simple, repeatable empirical experiment and reported the methodology and results to this thread proving that statement false should indicate that responding to Myrrh is not worth the wear on your keyboard.
Myrrh, follow the instructions, do the experiment and learn.

Why don’t you re-read what I wrote and the particular point I was making. I thought your experiment was good, but what do I know..
Now do something really useful to help bring down All the AGWScience Fiction Inc scam, design something that will measure just how much blue visible light from the Sun heats water..
gnomish says:
August 23, 2011 at 6:04 am
Myrrh : you’ve got a weird notion about electromagnetic energy. best review physics 101 and get the basics right.
Can’t watch it, but you really should put away your ego when talking to me..
It’s very simple. Visible light is transmitted through water without being absorbed because water is a transparent medium for visible light. That is a technical description of what happens to visible light travelling through water. Because, water is a transparent medium for visible light. Transmitted means specifically it is not absorbed. Absorbed in the specific physics meaning of the word and not in the general ‘disappeared somewhere in the deep and so much have been absorbed’ nonsense.
If the ‘greenhouse glass is transparent to visible light, it passes straight through without being absorbed’ as per the AGWScience Fiction telling of the story, which happens to be a physical fact, then the FACT that water is a transparent medium for visible light must mean that it is also not absorbed in water.
You can’t have it both ways.. Water is a transparent medium for visible light, just as glass is.

Tim Folkerts
August 23, 2011 10:19 am

PS There is mention at the very bottom of Gnomish’s article about freezing the water by evaporation, but only when the air temperature is 42 F or lower. I had missed that earlier. So, yes, when the temperature is quite close to freezing, then evaporation can get you the rest of the way (which actually agrees quite well with the table I linked to).
The link I had posted earlier in the thread (http://solarcooking.org/plans/funnel.htm) was for IR cooling in a sealed container, so evaporative cooling was not an option. Here they could freeze water even when the overnight temperature got no lower than 47 F, so this can work at higher temperatures than the evaporative method.

kuhnkat
August 23, 2011 10:42 am

MKelly,
both O2 and N2 absorb UV. It is part of the atmospheric chemistry so important to our ground life. O2 additionally is a minor absorber in the visible range.

kuhnkat
August 23, 2011 11:22 am

MKelly,
Here is a source: http://www.coe.ou.edu/sserg/web/Results/results.htm
Hey Myrrh,
you might want to take a look at this data also. Pure Water is NOT TOTALLY transparent as it DOES absorb and emit at very low levels. Please note the exponents in the numbers.
Here is another H2O data source: http://www.martin.chaplin.btinternet.co.uk/vibrat.html

kuhnkat
August 23, 2011 11:40 am

tallbloke,
“Energy in = energy out. ”
I can’t argue with that!! 8>)
Nope, I am primarily talking about the flux “measured” above the surface. The fact it is a flux means it is not resident in the water or the atmosphere. I am wondering about the residence at the water end myself which is why I am only willing to accept “slowing the cooling”. Whether it is the same energy or not, the fact is the surface loses more than it receives so CAN’T be “warming” due DLR.

gnomish
August 23, 2011 12:31 pm

Tim Folkert: i’ll buy that the foiil is insulating the contents from external energy except from space.
now i read the article more carefully and see, as you point out, that the jar was sealed. i was guilty of skimming instead of paying proper attention.
heh- so i guess the dlr makes good ice of water…lol
myrrh – the video is a kid burning stuff around the house with a blue laser.
lasers are monochromatic. visible light lasers can be used to burn stuff.
visible light, when absorbed, becomes heat.
visible light does not reach the bottom of the deep ocean because it is absorbed.
when it is absorbed, it becomes heat.
glass does not pass IR . when canadians use a magnifying glass in the sun to burn ants, it’s visible light delivering the heat energy.
dude- everybody has been trying to help you get right. my ego can’t be harmed by your stupid, so forget about that. i don’t suppose i’ll continue to be charitable, though.

Tim Folkerts
August 23, 2011 1:11 pm

“heh- so i guess the dlr makes good ice of water…lol”
It is an interesting concept, isn’t it?
The extension, of course, is that if you could somehow aim the device so that it also “missed” the atmosphere’s IR (for example by removing all the greenhouse gases), the water inside the foil cone would mostly “see” the 3 K of outer space. Then the water would get MUCH colder yet.
Good thing we have all those GHGs and their IR helping to keep us all warm!

Myrrh
August 23, 2011 2:17 pm

kuhnkat says:
August 23, 2011 at 11:22 am
Hey Myrrh,
you might want to take a look at this data also. Pure Water is NOT TOTALLY transparent as it DOES absorb and emit at very low levels. Please note the exponents in the numbers.
Here is another H2O data source: http://www.martin.chaplin.btinternet.co.uk/vibrat.html

You’re a comedian right? You’re defending that Visible is the main heating wave to convert to heat all the land and oceans of our planet, to raise the temperature of all the Earth as claimed by AGWScience Fiction Inc’s takeover of the education system, and you give me a page which says:
“Water is almost perfectly transparent to ‘visible’ light, a property which is made good use of by photosynthesis and allowing production of both biomass and oxygen.”
Even if this work is replicated, all it is doing is confirming that water is transparent to visible. What isn’t being used in photosynthesis which is a chemical conversion and not a heat creating conversion, is transmitted through. What can’t you understand about this? The graph shows that water is even in this work, if this work is correct, for all practical purposes not absorbed by water. Not that I can make sense of it, the UV looks very odd.
Why don’t you understand that you’re confirming what I’ve been saying here?
Spector says:
August 23, 2011 at 5:47 am
On the Wikipedia ‘Electromagnetic absorption by water’ page, they are using a centimeter scale to measure EM absorption in water. This, I assume, is pure water uncontaminated by algae or plankton. Although undefined here, as far as I could find, the absorption coefficient usually indicates by what value the natural logarithm of the signal is reduced when passing through the specified distance (1 cm). An absorbed photon delivers its packet of energy to some ‘lucky’ molecule.
BTW, this page obviously appears to have been written by someone who has fully bought in to the greenhouse catastrophe theory. That does not mean that this largely unrelated data is incorrect or fudged—it also includes a cute water molecule vibration mode diagram.

Yes, obviously written by someone wanting to minimise what the graph is saying.. 🙂 “This water absorption occurs preferentially at certain characteristic wavelengths while the balance of the spectrum is transmitted with minimal effects.” That is, zilch absorption for Visible, it is transmitted with zilch effects, and, all the absorption is by the bulk of thermal infrared; the ‘peaks’ here he points out are for the most part below the bulk ‘peak’ of thermal infrared which is a table top mountain kind of peak..
Visible light does not work on a molecular level, it can’t move molecules as does thermal infrared. Visible is tiny. It gets lucky if it hits an electron.. That’s the best it can manage in the atmosphere, where absorbed briefly by an electron it gets bounced back out the way it came, no heat is being created. That’s what reflection and scattering is.
Visible is not powerful enough to move the molecules of oxygen and nitrogen in the atmosphere to convert them to heat. Water is completley transparent to Visible, not proven in any way that it isn’t, you can see straight through clear water.. Visible doesn’t have the mechanism to interact even on an electron scale with water, is really transparent, it is transmitted through unchanged. Unlike the atmosphere which isn’t really transparent, because of absorption by electrons to produce reflection and scattering.
Here, look at this in the real world where applied scientists who actually do things that have to work: “That sunlight can penetrate water is a well known phenomenon. In fact, it is an essential requirement to sustain the life of aquatic plants like algae that grow in water.” But do read the next sentence. http://almashriq.hiof.no/lebanon/600/610/614/solar-water/unesco/29-34.html
..just fades away…
See though, UV is not absorbed, this is all about targetting nasties in the drinking water supply where it’s important to get UV to them (because UV is more energetic and works on DNA levels).
Here, for UV not being absorbed by water which remains unchanged, question 8 http://www.iuva.org/iuva/faqs
‘Oh, we’ve seen a little blue’ said some ascientists digging around in some ice somewhere, we think water is really blue’ Nanas. Blue is more energetic and smaller so travels slightly differently, it’s more easily scattered (so the blue sky).
OK, enough of this (y’all generally). Water is well known to be a completely transparent medium for Visible. This means that it is not absorbed by the molecules of water, but passed through. Visible cannot heat water. It has no method in place to do so.
It cannot move the molecules into the vibrational states necessary to convert to heat. It’s a bloody physical impossibility for the tiny little tikes.
Visible is used for photosynthesis primarily, that’s its role in life.
Unless you can show that real science, tried and tested and used in countless applications worldwide in a huge variety of fields, doesn’t understand this, then you must accept that the AGWScience fiction you’re promoting ‘that visible heats the Earth’, is just that, science fiction.

August 23, 2011 3:49 pm

Strange how oceans get darker with depth if they are transparent to visible light.

Bob_FJ
August 23, 2011 4:52 pm

Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.
The colour of the car has an effect too. White reflects visible light a great deal more than black, (white cars are cooler) but IR absorption is not affected by visible body colours.

Bob_FJ
August 23, 2011 6:16 pm

Stephen Wilde, Re:

Strange how oceans get darker with depth if they are transparent to visible light.

To be fair, I think Myrrh did say early-on that visible light gets tired, and just slows down in the oceans. He/she did not explain what energy is involved in decelerating those naughty photons from ~300,000,000 M/sec to zero though, and I‘m not convinced.

Tim Folkerts
August 23, 2011 6:58 pm

Just sayin’ ….

The Crackpot Index
John Baez
A simple method for rating potentially revolutionary contributions to physics:
1. A -5 point starting credit.
2. 1 point for every statement that is widely agreed on to be false.
3. 2 points for every statement that is clearly vacuous.
4. 3 points for every statement that is logically inconsistent.
5. 5 points for each such statement that is adhered to despite careful correction.

8. 5 points for each mention of “Einstien”, “Hawkins” or “Feynmann”.

15. 10 points for each statement along the lines of “I’m not good at math, but my theory is conceptually right, so all I need is for someone to express it in terms of equations”.

34. 40 points for claiming that the “scientific establishment” is engaged in a “conspiracy” to prevent your work from gaining its well-deserved fame, or suchlike.

36. 40 points for claiming that when your theory is finally appreciated, present-day science will be seen for the sham it truly is. (30 more points for fantasizing about show trials in which scientists who mocked your theories will be forced to recant.)
37. 50 points for claiming you have a revolutionary theory but giving no concrete testable predictions.

http://math.ucr.edu/home/baez/crackpot.html
REPLY: and 100 points for being Tim Folkerts – Anthony

kuhnkat
August 23, 2011 7:04 pm

Tim and gnomish,
before worrying about how to sneak between those hot DLR’s, you might consider exactly how much IR is radiated to the water by the device itself!

kuhnkat
August 23, 2011 7:26 pm

Myrrh,
“Water is almost perfectly transparent to ‘visible’ light”
OK, I think I am starting to see where your problem comes from. That ALMOST is all that is needed to absorb most of the visible energy even without the contamination in the water. You apparently have even less of a concept of the magnitudes that we are dealing with than I do. It really helps to have a LITTLE math skill. Even though that absorption is positively miniscule, the number of layers of molecules the light has to pass through adds up. Do you have any idea of how many molecules of water it takes to make a column of water 500 ft. deep??? Neither do I. It is a rather large number comparable to the attenuation provided by those large negative exponents on the absorption charts. In other words, even though each layer takes away an amount so small it is difficult to measure, after that many layers it has all been nibbled away. Notice that the least absorptive bandwidth is actually in the UV close to your BLUE light you don’t think absorbs or carries energy.
Repeating that water is TRANSPARENT is a childish tool to reinforce your own misunderstanding of the issues. Until you stop it you will simply continue to look silly.
You still haven’t explained to me how the tiny amount of THERMAL or FAR IR from the sun that makes it to the upper atmosphere manages to warm the earth. Want to take a stab at it now?? About 50% of the sun’s output is in the Infrared. Over 90% of that is in the near infrared you don’t think heats anything. So, you are left with less than 5% of the sun’s output attenuated by distance and a small cross section heating the earth. Please explain how this works??

jae
August 23, 2011 7:43 pm

Very troubling that one of my favorite bloggers decided to abandon all dissenting views and ignore challenges to his “integrity?.” The cowboys that raised me would label this behavior as chickenshit. WTF, Willis?

Tim Folkerts
August 23, 2011 8:03 pm

kuhnkat says:
August 23, 2011 at 7:04 pm
“Tim and gnomish,
before worrying about how to sneak between those hot DLR’s, you might consider exactly how much IR is radiated to the water by the device itself!”

Interesting question. I hadn’t specifically thought about that because I assumed it was small. Let’s see …
Emissivity of aluminum foil is ~ 0.04 (http://www.engineeringtoolbox.com/emissivity-coefficients-d_447.html). That ought be be pretty close the the value for the shiny aluminum they used.
Let’s assume the foil is as hot as 300 K (highly unlikely on a cool desert night when this was being used).
The radiation from the device itself would be

5.67E-8 * 0.04 * (300)^4 = 18 W./m^2

On a hot day, this could get a bit higher, but the whole point was using this during cool nights (especially in the desert). Now the emissivity could be a bit higher (especially if it gets dirty). I suppose the number could be 40 or 50 W/m^2 in real-world situations.
Compared to 300 W/m^2 or more of IR from the ground that is being blocked, the radiation from the foil is pretty minimal.

gnomish
August 23, 2011 9:15 pm

300W/m^2 is like 3 100 Watt bulbs every square yard – that seems unrealistic at night.
btw – thermometers of any kind, including IR detectors – do not measure heat.
temperature is not heat and degrees can not be converted to watts. i don’t think you can properly calculate watts via the IR radiative temperature calculation.

gnomish
August 23, 2011 9:17 pm

http://www.sciencephoto.com/media/356062/enlarge
this thing is incandescent but provides very little heat.

Bob_FJ
August 23, 2011 11:59 pm

Gnomish, Re:

this thing is incandescent but provides very little heat.

That’s very interesting, but I don’t think incandescence is applicable in your example. By definition incandescence relies on a major heat source, whereas many other low energy light sources do not. (E.G. glow-worms or my audio system remote control buttons, providing that there has been exposure to light during the day in the latter case)
Check out for example:
http://en.wikipedia.org/wiki/Luminescence
The sun is rather incandescent, but don’t look directly at it!!!!!!!!!!!!!

Bob_FJ
August 24, 2011 1:52 am

Willis Eschenbach @ August 24, 2011 at 12:32 am
In the time you spent evading JAE’s questions, why not instead respond at least in part, or say something like; you will get back to him soon when you have more spare time?
Why do I ask: Because I think jae’s questions are important and I’d like to know your determinations, as would maybe others.

RJ
August 24, 2011 3:27 am

Bob FJ
LWR further heating the surface is fiction. Supporters believe based on faith not science, evidence or logic. JAE will never receive satisfactory answers.
Slowly more and more with see the GHG backradiation theory for the nonsense it is. But it will take time for all warmists (luke and alarmists) to let this go.

Tim Folkerts
August 24, 2011 4:11 am

gnomish says:
August 23, 2011 at 9:15 pm
“300W/m^2 is like 3 100 Watt bulbs every square yard – that seems unrealistic at night.”
Could you give some numbers to back up this impression? What calculations make 300 W/m^2 seem wrong? What do you think is wrong with my calculations?

Tim Folkerts
August 24, 2011 4:24 am

Bob_FJ says: August 23, 2011 at 11:59 pm

Gnomish, Re:
this thing is incandescent but provides very little heat.
That’s very interesting, but I don’t think incandescence is applicable in your example.

Incandescence is exactly applicable — since it is a hot glowing object.
One reason he can hold it is that it has very poor thermal conductivity, so the corners can get rather cool while the center is still hot. . Another reason is that silica has good transmittance of visible light, but the transmittance gets much worse as you move into the infrared range (http://www.sciner.com/Opticsland/FS.htm). So if the center is still warm, the visible light can get out, while the infrared is blocked, again limiting the energy transmitted to his hand. Both of these properties help make it possible to see the hot glow while not getting burned.

Tim Folkerts
August 24, 2011 6:07 am

RJ says:
August 24, 2011 at 3:27 am
“LWR further heating the surface is fiction. Supporters believe based on faith not science, evidence or logic. JAE will never receive satisfactory answers.”
If you think it is important, why don’t you provide the answers JAE wants? Find the temperatures of greenhouses in Guam and Phoenix. Explain these temperatures in terms conduction, convection, and radiation balance based on the conditions in Guam and Phoenix. Estimate the LWR in both locations and explain what effects, if any, this has on the temperatures of the two greenhouses.

beng
August 24, 2011 6:39 am

Belated thanks, Willis. I never accepted “100% of the energy increase goes into evaporation” argument. The rate of evaporation at a water surface is determined by water & air temp, air humidity & pressure, water & air movement, etc. It doesn’t correlate w/the wavelength of the affecting radiations.

tallbloke
August 24, 2011 7:33 am

beng says:
August 24, 2011 at 6:39 am
The rate of evaporation at a water surface is determined by water & air temp, air humidity & pressure, water & air movement, etc. It doesn’t correlate w/the wavelength of the affecting radiations.

Yes. This is why we can’t know whether a small change in the net LR flux (if it has happened) affects evaporation rates. It will be lost in the noise of changes in the convection, pressure, humidity etc. Lots of feedbacks are in operation, so we don’t know if the non-feedback theoretical calculation for an increase in DLR for co2 has done anything or not.
However, we do know that heat isn’t conducted from cooler to hotter across the ‘skin’ of the ocean. And we know that when wind disrupts the skin, it also causes more evaporation, so DLR plus wind isn’t going to heat the ocean. And we know natural convection doesn’t happen upside down. And we know IR is stopped within 0.05mm of the surface of the water.
Konrad’s empirical experiment further supports the thesis that changes in DLR rates don’t make much difference to the cooling rate, unless evaporation is suppressed. Although his experiment didn’t include a turbulent surface, my own empirical test shows that in average conditions there is insufficient vorticity at the ocean surface to mix heat downwards against the natural buoyancy of warmer water.

tallbloke
August 24, 2011 8:26 am
Tim Folkerts
August 24, 2011 9:40 am

Tallbloke,
I agree with most of what you wrote at 7:33 AM, but I still disagree with your conclusion — at least to some extent.
You said “we do know that heat isn’t conducted from cooler to hotter”
I’ll go a little farther and say “the greater the difference between hot and cold, the more heat (thermal energy) gets transferred.”
The ~ 0.2-0.3 K temperature gradient across the skin is sufficient to allow conduction of ~170 W/m^2 of heat. This happens to be the energy deposited in the bulk of the ocean (below the skin layer) by solar radiation, so this conduction is sufficient to keep the bulk ocean.
Even a small change in the gradient across the surface will result in a change in the ~ 170 W/m^2 of conduction thru the skin. If the top of the skin warms and the gradient across the skin is cut in half, the conduction up thru the skin will also cut in half (give or take a little).
One way to warm the top of the skin would be to add a little IR to the surface. This IR will raise the surface temperature resulting in 1) more evaporation, 2) more upward IR and 3) less energy loss from the lower parts of the ocean. If less energy is lost from the lower levels while still receiving the same solar energy, the result will necessarily be rising temperatures in the lower levels.
==================================
Konrad’s experiments do seem to show that this effect may be small (ie the IR energy goes mostly into increased IR & evaporation and not much into raising the temperature at the top of the skin, so there is little change in conduction out of the bulk). I tried quantified the magnitudes of he energy flows earlier in the discussion and concluded that the effect might be too small to see with a set-up like Konrad’s. I would love for someone else to check my numbers and provide alternate estimates of the energys and cooling rates involved.

August 24, 2011 10:34 am

“So if the DLR isn’t warming the bulk of the upper ocean … then what is?”
Increased solar shortwave input during the late 20th century when global cloudiness and albedo both reduced.
DLR has but a miniscule effect in comparison and is readily dealt with by increased evaporation, convection, conduction and radiation for a faster/larger water cycle and a miniscule adjustment in the surface air pressure distribution.
Insignificant and unmeasurable compared to the natural solar and ocean induced shifts in the surface air pressure distribution.

gnomish
August 24, 2011 11:25 am

willis, can you justify the conversion of thermal IR signature (a temperature proxy reading) to watts? if that is what you are doing, it is a crucial error.

RJ
August 24, 2011 12:22 pm

“So if the DLR isn’t warming the bulk of the upper ocean … then what is?”
The sun of course.
As explained in this Joseph Postma paper.
climaterealists.com/index.php?id=7457

Tim Folkerts
August 24, 2011 12:24 pm

Stephen,
You missed Willis’ point. He is not addressing the question “What raised the average SST from 18 C to 18.5 C over the last 100 years” (or whatever the actual numbers are). He is addressing the question “What in general has raised the to around 18 C to begin with, since 170 W/m^2 of average solar energy alone would predict an average temperature well below freezing”.

tallbloke
August 24, 2011 12:58 pm

Stephen Wilde says:
August 24, 2011 at 10:34 am
“So if the DLR isn’t warming the bulk of the upper ocean … then what is?”
Increased solar shortwave input during the late 20th century when global cloudiness and albedo both reduced.
DLR has but a miniscule effect in comparison and is readily dealt with by increased evaporation, convection, conduction and radiation for a faster/larger water cycle and a miniscule adjustment in the surface air pressure distribution.
Insignificant and unmeasurable compared to the natural solar and ocean induced shifts in the surface air pressure distribution.

I agree with nearly all of this except that it is a small change in DLR that has but a miniscule effect in comparison…
Willis Eschenbach says:
August 24, 2011 at 10:23 am
Various people keep saying over and over that DLR can’t possibly heat the bulk of the upper ocean.
Despite that claim, the bulk of the upper ocean is much warmer than it would be if heated by solar energy alone. Solar energy absorbed by the ocean is only about 170 w/m2. This is only sufficient to heat the ocean to minus 40°C (or F) or so, and that’s frozen very solid.
When someone comes around to answer that, I’m interested. Until then, your explanations don’t explain that, and I’m not interested.

DLR doesn’t “heat” the upper ocean bulk, but the LR flux does slow the rate of cooling of the upper ocean bulk via it’s support of the adiabatic lapse rate of the atmosphere, and this is why the ocean is as warm as it is. This observation is much more than a disagreement over semantics. The substantive arguments put forward in this thread which seek to demonstrate that DLR does not directly “heat” the ocean bulk have merit. The only way changes in DLR can affect the temperature of the upper ocean bulk is via a change in the differential between surface air temperature and ocean surface temperature. This is a very, very slow way to change the temperature of the bulk of the ocean. It’s certainly not going to account for the change in ocean heat content or surface temperature 1980-2003, even though it explains why the ocean is warmer than it would be under an atmosphere devoid of radiatively active gases.
Even then, there is considerable uncertainty over the degree to which additional co2 might have changed that differential, if it has changed it at all. This is because despite the theoretical non-feedback increase in air temperature an extra 120ppmv of co2 could bring about, negative feedbacks empirically observed from satellite data by Roy Spencer and Richard Lindzen and their co-authors have substantially reduced the non-feedback value, and so the Earth’s response to this additional co2 is probably very small indeed.
Read it again real slow if you have a problem
Please stop it with this obnoxious, arrogant numpetry, I much prefer the old Willis who came up with a brilliant hypothesis about the way thunderstorms in the tropics regulate the amount of energy shooting up through the troposphere, bypassing the radiatively limited ‘atmospheric window’.

kuhnkat
August 24, 2011 12:59 pm

Tim Folkerts,
“Compared to 300 W/m^2 or more of IR from the ground that is being blocked, the radiation from the foil is pretty minimal.”
Hmm, by morning it wouldn’t be 300 w/m^2, but, I think you might have caught me there. The tinfoil will be radiating less than the ground which would be an improvement.
I just ran across something I am confused about. In the SpectralCalc calculator it lists Radiance and Radiant Emittance. I looked up the definitions and Radiance seems to be defined as the radiation in one direction and the Radiant Emittance is in all directions. When we are working with a surface, wouldn’t that be Radiance instead of the Radiant Emittance which would be a gas particle or something radiating in all directions? Or, since even a surface will be radiating in a hemisphere would it be in between??

August 24, 2011 1:22 pm

Willis Eschenbach said:
“Solar is about 170 w/m2. The ocean radiates about 390 w/m2, never mind sensible and latent heat loss. The ocean (being at general thermal equilibrium) must receive as much as it radiates.”
Ok, with the help of Tim I’ve got the point. One needs the back radiation to add to the 170 to reach and maintain the ocean temperature.
However one then has to take a step back and look at first principles. Is it really GHGs that dictate the level of back radiation that we observe.
Perhaps not. Atmospheric pressure dictates the energy value of the latent heat of vaporisation so it is atmospheric pressure that dictates the rate at which energy can leave the oceans. The more it costs in terms of energy to achieve evaporation the warmer the oceans must become before equilibrium is reached.
So the oceans will build up to whatever temperature is permitted by atmospheric pressure with or without any GHGs in the air at all.
Once that ocean temperature is achieved the energy for the back radiation is then supplied to the air by energy leaving the oceans and NOT by energy coming in from the sun.
So the upshot is that the oceans accumulate solar energy until they radiate 390 at current atmospheric pressure, at that point 170 continues to be added by solar but to balance the budget the atmosphere by virtue of its density back radiates whatever is necessary to achieve balance.
A feature of GHGs is that they add to back radiation proportionately more than other gases in the atmosphere but in the end it is surface pressure that controls the energy value of the latent heat of vaporisation which is the ultimate arbiter of what rate of energy transfer can be achieved from oceans to air.
So if GHGs add a surplus to the back radiation over and above that dictated by surface pressure the system has to make an adjustment but what it cannot do is alter the energy value of the latent heat of vaporisation. So instead it is the rate of evaporation that must change to balance the budget in the absence of a significant change in surface pressure.
Do you see the point ?
Back radiation is a consequence not of atmospheric scattering of incoming solar shortwave but rather primarily a consequence of atmospheric density slowing down energy loss from sea surface to space.
So if one increases atmospheric pressure at the surface the amount of energy required to provoke evaporation at the sea surface rises and the equilibrium temperature of the whole system rises including the amount of back radiation.
The opposite if one decreases atmospheric pressure at the surface.
We have been looking at back radiation from the wrong point of view.

August 24, 2011 1:33 pm

In effect back radiation is a reflection of the oceans in the sky and sea surface temperatures change the amount of backradiation and NOT vice versa.

tallbloke
August 24, 2011 1:51 pm

Stephen Wilde says:
August 24, 2011 at 1:33 pm
In effect back radiation is a reflection of the oceans in the sky and sea surface temperatures change the amount of backradiation and NOT vice versa.

Good insight, considering the vast majority of GHG’s come from the ocean anyway. However, less heat from the Sun would mean lower ocean temp, less evaporation, and less back radiation. Albedo is thus the important arbiter.
Less evaporation means less cloud, which means more insolation… Voila – self regulating system which co2 is along for the ride with, though maybe, just maybe making the difference between high ice albedo and solar absorptive open ocean.

gnomish
August 24, 2011 1:55 pm

August 24, 2011 at 12:21 pm
willis, can you justify the conversion of thermal IR signature (a temperature proxy reading) to watts?
No, but Stefan and Boltzmann can justify it. Check Wikipedia if the mathematics is too much, they’ll likely have a simpler explanation. Briefly, the radiation (watts per square metre) is proportional to the temperature to the fourth power.
—————————————————-
well, then, that may just be the root of a lot of apparent inconsistencies because degrees can not be converted to watts.
that’s because they are not measuring the same thing at all.
The watt is defined as one joule per second and measures the rate of energy conversion.
temperature is the average kinetic energy of a bulk of particles in motion at equilibrium.
willis- when water changes phase from gas to liquid, it dumps joules by the score with NO CHANGE IN TEMPERATURE. s/b can’t handle that. s/b may show the ideal radiation spectrum for a particular temperature but it can not distinguish between 100 degree iron and 100 degree feathers. in short, it can not measure heat.
Temperature doesn’t measure latent heat, does it? Therefore, temperature does not measure the energy of a system, does it? Therefore temperature is not a valid measure of heat and should not be claimed to represent the heat in a system.
perhaps the tragically missing heat of lore is what temperature measurements miss?
willis: http://www.sciencephoto.com/media/356062/enlarge
the temperature of that thing is 1300C. s/b calcs on the spectrum it emits might show it, too – but guess what s/b calcs do not and can not show – why that man can hold it in his fingers.
reason this out- if that were a chunk of 1300C iron, his fingers would have crisped and stunk up the room with smoke because of heat.
degrees can not be converted into watts, my friend. s/b is not a measure of heat in an object – it only measures temperature. temperature is not heat = they are not the same
but now i understand why the warmists love s/b black body idealization – it lets them think they’ve done the impossible – convert temperature measurement into a measurement of heat energy in a system. but that notion is entirely fallacious.

August 24, 2011 1:59 pm

“However, less heat from the Sun would mean lower ocean temp, less evaporation, and less back radiation. Albedo is thus the important arbiter.”
Yes indeed.
Only two things can affect the equilibrium temperature of the oceans, solar shortwave input and atmospheric pressure. GHGs other than water vapour itself just alter the speed of the water cycle by altering the surface pressure distribution a miniscule fraction.
Additionally the variable rate at which the oceans can release energy to the air will also be a factor but that seems to be limited by the length of trhe thermohaline circulation at 1000 to 1500 years or whatever.
Which brings me neatly to my hypothesis about top down solar effects altering the surface pressure distribution for changes in cloudiness and albedo.
That pretty much squares the circle.

August 24, 2011 2:13 pm

“Less evaporation means less cloud, which means more insolation”
In extreme situations yes but generally insolation is more dependent on the amount of air mass mixing and cloud distribution. By virtue of the solar effect on the polar vortices a weaker sun gives more meridional/equatorward jets, and thus more clouds even though overall the reduced insolation causes less evaporation.
The key is probably the size of the sub tropical high pressure cells. In a warmer world they expand and cloud dissipates where it most matters over the oceans either side of the equator. The opposite when they shrink.
So an active sun leads to a general warm up of the whole system because of less clouds where it most matters and the opposite when the sun is less active.
Thus tiny solar changes are vastly amplified by a chemically induced change to the vertical temperature profile in the atmosphere which affects cloudiness and albedo.
Then the self regulating processes get to work namely a speed up of the water cycle because atmospheric pressure regulates the energy cost of evaporation so more energy in automatically leads to more energy out to maintain energy budget balance.
So changes in the system equilibrium temperature as a whole are pretty minimal but changes in the rate of energy flow through the system move the climate zones around quite naturally to produce what we perceive as climate changes.
However a persistent reduction in solar input would be another game altogether.

kuhnkat
August 24, 2011 2:20 pm

Willis Eschenbach,
“Briefly, the radiation (watts per square metre) is proportional to the temperature to the fourth power.”
What about emissivity?
I will ask you the same question I asked Tom. On the SpectralCalc page it gives two outputs, Radiant Emittance, which appears to be the number you and many others use, and Radiance. Radiant Emittance seems to be defined as emission in all directions and Radiance in one direction. As a surface such as the ocean would seem to emit in a hemisphere wouldn’t the actual number be between these two??

Tim Folkerts
August 24, 2011 4:08 pm

Kuhnkat,
The difference between the two is a bit technical. I would suggest the wikipedia pages for the two as a starting point.
“Irradiance, radiant emittance, and radiant exitance are radiometry terms for the power of electromagnetic radiation per unit area at a surface. ”
http://en.wikipedia.org/wiki/Radiant_exitance
“Radiance and spectral radiance are radiometric measures that describe the amount of light that passes through or is emitted from a particular area, and falls within a given solid angle in a specified direction. ”
http://en.wikipedia.org/wiki/Radiance
Here is one simple simple example of the difference. Suppose a 1000 W spotlight shines thru a piece of warped glass. 1000 W/m^2 would be the radiant emittance. The warped glass would send different amounts of light in different directions. Radiance would measure how much energy is going in each direction from the surface. IF you integrate the radiance, you would get the radiant emittance.
(I *think* I got that all right).

Bob_FJ
August 24, 2011 4:27 pm

Tim Folkerts @ August 24, 2011 at 4:24 am
Re glowing silica insulation tiles:

Incandescence is exactly applicable — since it is a hot glowing object.

Sorry Tim, and Gnomish, I was a bit hasty there, thinking it did not contain much heat because of very low specific heat.
Gnomish @ August 24, 2011 at 1:55 pm
Found your commentary very interesting including the silica tile example.
I’ve often wondered about the S-B thingy, particularly in the fact that EMR radiates hemispherically from a flat surface, and what is intercepted in a simple two-body situation with various dimensions and separation. I’ve not had need to use it, have never researched it, or been brave enough to pose the question.
An analogy: If I stand in front of a small fire, it may be pleasantly warm. However, If the fire is huge, but the same colour temperature, AOTBE, I may feel unbearably hot.
In the Trenberth cartoon, it shows EMR as plain up and down, whereas in reality it is in all directions, which raises some interesting questions as to how S-B was applied. Isn’t the total energy lost from a surface according to S-B hemispherically distributed?

Myrrh
August 24, 2011 4:32 pm

Stephen Wilde says:
August 23, 2011 at 3:49 pm
Strange how oceans get darker with depth if they are transparent to visible light.
Since water is transparent to visible light, why not try working out why?
Bob_FJ says:
August 23, 2011 at 4:52 pm
Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.
Odd that, I didn’t realise you were talking about mobile greenhouses..
The colour of the car has an effect too. White reflects visible light a great deal more than black, (white cars are cooler) but IR absorption is not affected by visible body colours
Visible light isn’t capable of moving molecules to vibrational states. It doesn’t convert to heat. It takes a far more powerful energy, infrared to the size of a pin head, to knock molecules around. So what’s really happening here?
kuhnkat says:
August 23, 2011 at 7:26 pm
Myrrh,
“Water is almost perfectly transparent to ‘visible’ light”
OK, I think I am starting to see where your problem comes from. That ALMOST is all that is needed to absorb most of the visible energy even without the contamination in the water. You apparently have even less of a concept of the magnitudes that we are dealing with than I do. It really helps to have a LITTLE math skill. Even though that absorption is positively miniscule, the number of layers of molecules the light has to pass through adds up. Do you have any idea of how many molecules of water it takes to make a column of water 500 ft. deep??? Neither do I. It is a rather large number comparable to the attenuation provided by those large negative exponents on the absorption charts. In other words, even though each layer takes away an amount so small it is difficult to measure, after that many layers it has all been nibbled away. Notice that the least absorptive bandwidth is actually in the UV close to your BLUE light you don’t think absorbs or carries energy.

But, it’s a fact of basic well known well understood physics that water is transparent to visible light, that’s simply a fact from observation and understanding the processes of its transmission – visible does not get absorbed by water molecules, that’s what “transmission” means. It is standard physics talk. It relates to something specific. It relates to the way light travels through water without being absorbed. So: ““If the object is transparent, then the light waves are passed on to neighboring atoms through the bulk of the material and re-emitted on the opposite side of the object. Such frequencies of light waves are said to be transmitted.”
What is happening there is that the lightwave can’t join in the dance of the molecule and is passed on to the next, and so repeated. What is also happening here, is that the visible light wave ever so slightly gets delayed. Our fluid gaseous atmosphere slows down electromagnetic waves, our oceans much more so, denser. Anyway, light is transmitted through water and other transparent mediums because it can’t do anything else. It’s a property of visible light in the properties of water.
Even if there is some ‘minimal effect’ of absorption, it cannot be because this relationship doesn’t hold good. There must be some other factor involved in changing it, because the basic ‘mechanism’ must hold good through all the depths of water or it is not a basic ‘mechanism’ of the relationship. Deep sea diving proves that it holds good, shining a light at the bottom of the deepest part of the ocean shows clearly that visible light travels unimpeded through the water molecules, for the distance it has enough energy to do so, lighting up the surroundings and by reflecting back from the objects in the water, they become visible. This is what “transmission” and “reflection” means.
Visible light is not transmitted through the atmosphere, because the volume of the fluid gas atmosphere around us is not completely transparent to visible light, it is reflected and scattered off the molecules of nitrogen and oxygen. By the 2nd method on the list I gave from the wiki page above on transparency and translucency: “UV-Vis: Electronic transitions …
“An electron absorbs the energy of the photon and sends it back out the way it came in. This results in reflection or scattering.”
In our atmosphere, visible light is actually absorbed. On an electron level.
Worth repeating here, number 3 on the list of what can and will happens to visible light:
“An electron cannot absorb the energy of the photon and the photon continues on its path. This results in transmission (provided no other absorption mechanisms are active).”
An electron cannot absorb the energy of the photon and the photon continues on its path.
If no other absorption methods are active. So something could be altering the basic mechanism to give an insignificant minimal effect sometimes, but the basic mechanism is unchanged. Visible light is transmitted through water unchanged because water is a transparent medium for it which means it does not absorb it because its electrons cannot.
Now, listen carefully, the atmosphere then, is not a transparent medium for it, because the electrons of oxygen and nitrogen do absorb it. However, in the real world this is but one of the four things that can happen to visible, no heat is created, the light is simply sent out again, reflection/scattering.
Repeating that water is TRANSPARENT is a childish tool to reinforce your own misunderstanding of the issues. Until you stop it you will simply continue to look silly.
Really?
You still haven’t explained to me how the tiny amount of THERMAL or FAR IR from the sun that makes it to the upper atmosphere manages to warm the earth. Want to take a stab at it now?? About 50% of the sun’s output is in the Infrared. Over 90% of that is in the near infrared you don’t think heats anything. So, you are left with less than 5% of the sun’s output attenuated by distance and a small cross section heating the earth. Please explain how this works??
I have already explained it. The heat you feel from the Sun is THERMAL IR, A.K.A, FAR, A.K.A LONGWAVE, A.K.A HEAT. (See the explanation of what heat is I linked above.) Thermal IR is HEAT on the move, it takes 8 minutes to reach you from the Sun. That’s how powerful it is. The Sun is THE great POWERHOUSE of HEAT in our world. It isn’t stuck at TOA.. Or, you’re not on the surface of the Earth.. It is not a tiny amount, it is a GREAT amount of powerful HEAT energy reaching us from the Sun. Ever been in the desert or in the tropics? Ever been in the Sun at all? ALL the HEAT you feel from the Sun is THERMAL INFRARED.
It is HEAT from the Sun which raises the temperature of water, in the oceans and in us, and raises the temperature of the land.
Do yourself a favour, find the post above in which I gave a comparison of NASA teaching on this, traditional and now AGWScience Fiction Inc’s corruption of it. You are parrotting the corrupted science fiction. Until you get that truly grasped you will continue scrabbling around trying to find ‘something’ that ‘appears’ to confirm the fiction, and it can only be nonsense. A ‘minimal’ effect outside of the well understood norm that visible light is transmitted unabsorbed through water to explain the vast temperature rise of oceans and land, really??! And you won’t see how unrealistic that is until you understand that the property of thermal infrared has been given to wavelengths that are incapable of doing what thermal does, and thermal excluded, because this is a creation of science fiction to dumb down the masses.
‘All energy’ is not the same.. There are distinct differences in the properties between them. ‘All energy’ does not create heat..; chemical conversion to sugar in photosynthesis is not the creation of heat, reflection and scattering is not the creation of heat. Thermal infrared’s powerful vibrational moving molecules does create heat.
The two basic differences between LIGHT and HEAT
“Electronic: Transitions in electron energy levels within the atom (e.g. pigments). These transitions are typically in the ultraviolet (UV) and/or visible portions of the spectrum.
Vibrational: Resonance in atomic/molecular vibrational modes. These transitions are typically in the infrared portion of the spectrum.”
ALL the HEAT your feel from the SUN on the SURFACE of the Earth is THERMAL INFRARED.
You’ve been had.
And that’s it, I’m not posting on this explanation again, read the pages I’ve linked to about this. I’m only posting it to show that my view, re this discussion, has real world physics behind it. That thermal infrared is the heating mechanism of the ocean from the Sun and so any arguing that it can’t, can’t. The argument here is about the so-called ‘backradiated’ thermal infrared, and that has to take into account that thermal ir can heat the oceans because water is in resonance with thermal ir, it readily absorbs it.

kuhnkat
August 24, 2011 5:51 pm

Tim,
How does that work with the numbers you get on the SpectraCalc page??
I input 300k, .8 emissivity (I tend to think that is closer for the atmosphere, would be .9 for the ocean) and 0 for recessional velocity. The rest it spits out (well, you can pick the units at the top also)
http://www.spectralcalc.com/blackbody_calculator/blackbody.php

Tim Folkerts
August 24, 2011 6:28 pm

Bob_FJ says:
“I’ve often wondered about the S-B thingy, particularly in the fact that EMR radiates hemispherically from a flat surface, and what is intercepted in a simple two-body situation with various dimensions and separation…
This is really no easy way around the spheres and planes and angles and areas. The wikipedia page on radiometry has over a dozen different quantities relating to how much EM energy is going where in what directions. To really understand, you will need to be able to calculate surface integrals in spherical coordinates. You will have to be comfortable working with with expressions like L = d²Φ / dA dΩ cos(θ).
Otherwise, you may have to be content that other mathematicians/physicists have measured and/or calculated the net fluxes to/from various objects.
http://en.wikipedia.org/wiki/Radiometry

Tim Folkerts
August 24, 2011 6:51 pm

Kuhnkat,
The radiant emittance and the radiance are related by pi (divide them and you get 3.14159). I’m sure that the fact that there are 4 pi steradians in a sphere comes into play somewhere, but I really don’t know the details well enough to even try to explain it without digging deeper.
Also, the emissivity of the atmosphere would not be a simple fixed number for a few reasons. For one, the clouds are close to 1, while clear air would be much lower. Clear air emissivity would further be a function of temperature and humidity and pressure and altitude (and other things, I’m sure. 0.8 seems like a reasonable guesstimate for an over all average.

jae
August 24, 2011 6:52 pm

Willis:
You finally acknowledged my comments with this snark:
“If you were a cowboy you’d know that calling me chickenshit, just because I don’t do things according to your whims, would get you invited off the ranch pronto.”
If you will read backwards a few posts, you will (or SHOULD) notice that YOU put words in MY mouth. This is contrary to one of your primary standards, emphasized over and over and over, eh? (The stuff about responding to what you SAID, NOT WHAT WE THINK YOU SAID! You STARTED THIS MEME, REMEMBER?) That is where I decided that the chickenshit comment was appropriate. You did not live up to your standard! If you prefer, I shall call it hypocrisy, stead of chickenshit from now on. WHATEVER.
You still need to tell me why I can’t add the backradiation to the solar radiation in a greenhouse to calculate the “expected temperature.” It is a plainly observable fact that the “atmospheric greenhouse effect” cannot explain many empirical observations. The comments in this thread explain that clearly, and you still ignore it. Please answer some of Hockeyschticks comments.
You are doing a RealClimate on me, as I said before.
As to your problem about the 400 watts “going out” and only 170 “going in,” you are screwing things up by not looking at the TOA. 235 in and out. You can play endless games trying to explain why this is true, but HEAT STORAGE can explain it all. You don’t need GHGs to explain anything.
And you are totally silent about why other planetoids show the same heating at 1 atm., regardless of their atmospheres.
I recommend you read some Slaying the Dragons. It appears that you have not.

jae
August 24, 2011 7:42 pm

BTW, Willis: I think you can screw around with MODTRAN in any way you want and never demonstrate 400 (390) Wm-2 “looking down,” even in a tropical atmosphere. So we have a disconnect between the radiation diagrams and the Air Force standard, no?”
As I said earlier, empirical evidence is not your friend.

Tim Folkerts
August 24, 2011 8:28 pm

JAE,
MODTRAN deals with radiation from the atmosphere; it has nothing to do with the ~ 400 W/m^2 from the surface. The surface has an emissivity close to 1 for IR. The Stefan-Boltzmann equation gives 390 W/m^2 at ~ 15 C. The energy could also be (and I am sure has been) measured directly with IR detectors.
Since 15 C is about the average surface temperature, it seems that both theory and evidence support 390 W/m^2. In this case, I think empirical evidence sides with Willis.

Bob_FJ
August 24, 2011 9:58 pm

Tim Folkerts @ August 24, 2011 at 6:28 pm
Concerning the S-B law

“[There] is really no easy way around the spheres and planes and angles and areas. The wikipedia page on radiometry has over a dozen different quantities relating to how much EM energy is going where in what directions…”

Thanks for that Tim, and I’m not surprised that mathematicians can come up with formulaic solutions for more complex geometries than are explained in the normal treatments of S-B.
My beef with the normal treatment is that for years, I have not been clear on what the “normal geometry” is, although I have recollection that I read somewhere that the single body consideration is for a small flat body.
Willis has now stirred my renewed interest in the authoritative Trenberth/IPCC cartoon in this respect. (BTW, Willis has his own version, based on extending Trenberth’s into a two-layer IR model, a few years ago, which he linked to above somewhere)
I was thinking that in response to my pondering, that maybe someone would assert that the net consequence of hemispherical surface emissions and subsequent atmospheric spherical re-emissions means that the S-B calculation gives the straight up and down result. (That the sum of the vectors is up and down?)
There are a good few issues in that. Is anyone interested?

Bob_FJ
August 24, 2011 10:30 pm

Willis Eschenbach @ August 24, 2011 at 9:45 am
I was intrigued by you calling me a gerbil, a description that I was not familiar with, so I looked it up in my word processor dictionary, and found this:.

small rodent:a small rodent resembling a mouse with long back legs. Native to: hot dry parts of Africa, Asia.

Erh, I’m mostly domicile in Victoria in the SE corner of Australia. Under what basis do you think that I am a gerbil?

tallbloke
August 25, 2011 1:14 am

Willis Eschenbach says:
August 24, 2011 at 10:23 am
[Edited to add] PS – I don’t care if you call it “warming” or “slowing the cooling”. To me, since the ocean is warmer than it would be in the absence of DLR, I say it warms the ocean. But you can call it what you like, THE END RESULT IS THE SAME no matter what name you give it.

The end result is the same but how it gets there isn’t. Ray Pierrehumbert says that if the Earth didn’t have any way of cooling, it would eventually get hotter than the Sun. It’s all about rate of cooling. Let’s do a few numbers to illustrate the point.
‘Local’ Energy Equilibrium of the ocean (W/m^2):
Solar in = evaporation out + conduction/convection out + net radiative out = input to atmosphere
170 = 78 + 24 + ~66-70 = ~170
‘Local’ Energy Equilibrium of the atmosphere (W/m^2):
Input from the ocean below + Direct absorption from the Sun above = radiation out to space
170 + 67 = ~235-239
The upward and downward components of the radiative flux in the atmosphere cause convection within the atmosphere which inflates it and supports the adiabatic lapse rate. This is partly why the surface air temperature and ocean surface are at a temperature ~33C warmer than an atmosphere devoid of radiatively active gases would be (There would still be some adiabatic lapse due to the thermal difference between equator and poles which would drive a reduced atmospheric circulation).
Surface conduction, pressure, humidity, convection, and the adiabatic lapse rate supported by the radiative flux over the course of several billion years, has kept ocean and atmosphere in a thermal equilibrium whereby the ocean surface is slightly warmer than the surface air, to enable the rate of energy loss the ocean needs to maintain in order to remain in local energy equilibrium, via the mechanisms Stephen Wilde outlines. Both the ocean surface and surface air are warmer than the upper atmosphere where radiation to space takes place at the rate demanded by solar input and the S-B BB equation because of the equilibrium demanded by adiabatic lapse. The radiative flux plays an important role in that equilibrium, but small changes in the balance of the components of the flux will not measurably change the bulk temperature of the ocean at the decadal scale because the thermal inertia of the ocean and its evaporation rate lies behind the flux component’s differential of ~66-70W/m^2.
The tail does not wag the dog.

Myrrh
August 25, 2011 3:49 am

Just a quick p.s. to wrap up. This business of thermal infrared not getting through windows, I resorted to sarcasm too obscure to be of any use, I’d taken out the line ‘how do you know that thermal ir doesn’t go through windows?’. Sorry, I’d just got fed up with the fiction memes, I’d already made the point, a very important point in this, that the production of science fiction memes is endemic to AGWScience Inc. and already gone through a couple of examples, like the hot air convection being promoted as ‘proving thermal can’t heat water’. You’ll need to look at all the basic claims in this. Carbon dioxide is heavier than air, that means something in relation to other gases which have weight and volume..
Bob_FJ says:
August 23, 2011 at 4:52 pm
Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.i>
If you take a look at thermal imaging camera, which as I’ve said, is different from the near ir ‘photo’ capture, you can try answering your own question. Thermal imaging, in the mid to far ir, is something that is used widely in real life applications for a variety of reasons, one of which is in analysing heat loss from buildings, but more simply, why do people put double glazing in their homes? Even tripple glazing in Scandanavian countries. To stop, or rather slow down, heat escaping through the glass. http://epogee.co.uk/page.php?16
It isn’t easy to spot the AGWScience sleight of hand which changes the basics, most is done with a little a real fact and mixed in with fiction.
Oh, I’ve only just seen this..

Bob_FJ says:
August 22, 2011 at 1:45 am
Kuhnkat and Phil et al,
If you can’t resist responding to Myrrh’s nonsense’s, may I suggest a different approach? Just as Anthony has a ‘Quote of the week‘ and ‘Friday funny‘, why not have a ‘myrth of the day or week‘?
You could deliberately tease him on some laws of physics or with provocative analogies, or whatever turns you on.
One thing that springs to my mind is that plain ornery window glass is I believe fairly opaque to IR and UV light.

That does it, continue to believe what you want, stay brainwashed, stay believing the world works to contrived science fiction laws, you’ll be the loser.
But meanwhile, y’all, prove that visible light heats water, because it has no physical property capable of doing such a thing. Your science fiction scenario is junk to begin with, again, what are you arguing about?
The ‘energy’ from non thermal sunlight as if it’s converting to heat, and, you’re looking for the ‘missing heat’ when you’ve taken out all of thermal infrared which is the heat energy and which does reach the surface of the Earth!
How has this been sold to you? One way of course is the ‘peak’ energy of Visible, as if this makes it some ‘powerful’ force which is moving molecules to vibrational states..
Your missing heat Willis is Thermal Infrared. This is around where I came in to explore this, iirc the first discussion I entered where you asked this question. I hadn’t realised then that they had excluded thermal infrared and claiming that Light was thermal.. I thought it was simple stupidity not taking life into account, but then found it was all this nonsense, because claiming that Visible light converted to heat land and oceans raising the temp of the Earth. You’re arguing about
fictions.
And, there is so much Heat, thermal infrared, reaching the surface of the Earth, it’s been estimated that with our fluid gaseous atmosphere, which are all greenhouse gases in the real world, but with the Water Cycle taken out, the temperature would be 67°C. Think deserts.
Your energy budget begins with an impossible premise. GIGO.
You all can all believe what you want, but stop promoting this as real science unless you can prove that Visible directly converts to heat land and oceans. Prove that traditional physics is wrong. Real physics on this is still being taught, it’s just becoming more difficult to find.. Until then, you’re just posturing.

Myrrh
August 25, 2011 4:00 am

And lastly, on the meme about white coloured cars.. http://windowoutdoors.com/WindowOutdoors/Color%20and%20Thermal%20Regulation.html
See if you can find it for yourselves.

Myrrh
August 25, 2011 4:05 am
Myrrh
August 25, 2011 4:14 am

Apologies, missed a bracket at the end here:
Bob_FJ says:
August 23, 2011 at 4:52 pm
Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.i>

gnomish
August 25, 2011 5:40 am

myrrh – you’re just wrong.
burning stuff with a RED laser: http://www.youtube.com/watch?v=nbAltD9Zb-M
burning stuff with a GREEN laser: http://www.youtube.com/watch?v=aShgp6LRt98
burning stuff with a BLUE laser: http://www.youtube.com/watch?v=0xsQnJp4NiA
and guess what is used for safety goggles with an IR laser – plain, clear glass (or plastic) because it is opaque to IR.
but a magnifying glass of the same glass or plastic will focus the sun’s rays and burn stuff too.
how can you understand or explain that with the ‘electromagnetic radiation that has no energy’ flaw in your physics?
and if you ever actually did any IR photography, you’d know you can’t get an image thru ordinary glass or water.

Myrrh
August 25, 2011 8:44 am

Bob_FJ says:
August 23, 2011 at 4:52 pm
Myrrh, I see you have not yet explained why car steering wheels get hot after long parking exposure to summer sunlight through the windscreen. Odd really, considering IR cannot penetrate the glass.
? You, generic, are off the wall, even when the process is explained to you, you still parrot the AGWScience fiction memes.. Didn’t you look at the links I’ve posted?
Sitting in front of a largish picture window this morning the Sun’s late summer heat was streaming in, until a cloud passed over the Sun, we’ve got a few rainclouds scattered around. The Sun’s heat is thermal infrared.
And, enough with the lasers, you don’t understand the basics, whatever you think you’re understanding is going to be gigo, because you’ve taken not the slightest bit of real physics into consideration.
As I said, stay brainwashed in your imaginary science fiction world, I’m not explaining it again. Maybe someone reading this will have the nous to benefit.

Myrrh
August 25, 2011 9:25 am

I look forward to your producing blue light central heating for the home, let me know when your patent gets accepted.

gnomish
August 25, 2011 1:06 pm

myrrh – you are just as wrong now as you were an hour or a day ago.
i introduced you to facts that utterly falsify your notion that ‘visible light can’t heat anything’ and i watch you try to evade them using argument from intimidation (by being insulting).
that’s an indication that you are very embarrassed at your inability to explain these natural phenomenon. it further suggests that you imagine bluster will ward off the natural consequences of defending an obvious error. well, that’s just compounding the error.
it’s your right to be wrong but it’s maladaptive and no good consequences will result. hamdulilah, bro. first you lose your friends. then nobody at all will talk with you unless you pay them. then you get ignored or put somewhere. that’s an interesting career path, myrrh. it’s not my fault or my problem, though – it belongs to you; so i’ll respect your defining property.

August 25, 2011 1:37 pm

Myrrh says:
August 25, 2011 at 9:25 am
Myrrh my degree is in mechanical engineering. I only tell you this to illicit a bona fides from you. I may be ignorant but I paid to get that way.

Myrrh
August 25, 2011 2:33 pm

Moi? Intimidating? Shirley some mistake. Hasn’t stopped you from dedicating a whole post to insulting me and still not dealing with my request – prove that blue visible light from the Sun can heat water. Show how much visible actually heats water and so oceans. You know the kind of thing, real science numbers..
I look forward to your producing blue light central heating for the home, let me know when your patent gets accepted.
Why not describe how it would work? Let’s all take a look at how thermal blue light really is.
Let’s keep it compact, say, flat/apartment, two bedrooms, sitting room, kitchen/dining, and bathroom and box room, you design the layout and footage.

Tim Folkerts
August 25, 2011 3:30 pm

I look forward to your producing blue light central heating for the home.
1) Replace your furnace with several thousand of the blue lasers in the video above. (or use blue LEDs since they are very efficient).
2) Aim them thru fiber optic cables around the house (or even just use ducts with mirrors at the corners).
3) Run the end of the fiber optic cables into enclosed boxes in each room. (You could just shoot the lasers around the room, but the bright lights would be annoying and potentially damaging to eyes. Just like you could shoot steam around a room, but it is much safer and more convenient to send it thru “sealed” boxes that we call “radiators”.)
The video clearly shows the blue light heating things to quite high temperatures. The boxes will get warm, but if we design it right, they will not melt or get holes thru them. They will simple get warm, which will in turn warm the rest of the house. The blue light is transfering 1000’s of watts of light energy around the house, moving energy from the “furnace room” to the rest of the house.
Of course, this would be very expensive, so no one would ever do it. But the principle is clear and foolproof.

Myrrh
August 25, 2011 3:45 pm

mkelly says:
August 25, 2011 at 1:37 pm
Myrrh says:
August 25, 2011 at 9:25 am
Myrrh my degree is in mechanical engineering. I only tell you this to illicit a bona fides from you. I may be ignorant but I paid to get that way.
The only bona fides of any value here is an inquiring mind.
All of you who claim that the AGWScience Fiction Inc’s meme that Visible from the Sun converts Earth’s land and oceans to heat, MUST, surely?, be able to produce some actual mechanism for this on a wavelength/molecule basis? Some actual figures of how much of it it take to warm a pint, an ocean, a pebble, anything? Haven’t any of you actually got anything? Doesn’t that strike any of you as a bit odd?
Or have you really confirmed what I’ve said, and are just too shy to let me know?

tallbloke
August 25, 2011 4:00 pm

Myrrh says:
August 25, 2011 at 2:33 pm .
I look forward to your producing blue light central heating for the home, let me know when your patent gets accepted.

Myrrh, if this didn’t work, the manufacturers would be getting their asses sued:
http://www.solar-energy-for-home.com/solar-water-heating.html
The Passive Solar Hot Water Heater
“The lowest cost type is the passive solar heating system. it consist of 2 elements, the flat plate collector and the hot water tank.”
“The collector is a flat plate solar collector. The flat plate collector is an insulated sealed rectangular casing. The bottom plate is covered with a thin absorber sheet. Above this sheet there is a coil of thin pipes running water in a loop to the tank. The top is made either from a glass or a polycarbonate (transparent polymer) sheet.
The glass allows the sun light to enter in and hit the bottom plate absorber. The absorber heats up, and emits heat (in the infrared wavelengths). This heat radiation heats up the water in the water pipes. The top sheet is transparent to visible light but it is opaque to infrared wavelengths therefore the heat stays within the close casing.
The description isn’t perfect (the IR is absorbed by the water pipes and conducted to the water, not absorbed directly by the water) but you should get the general idea. Key point to note is that the top glass allows visible (and near IR) to pass through and heat the plate, which then radiates in the far IR and heats the water pipes.
Similarly, the energy of visible light and near IR entering the ocean gets absorbed by making water molecules vibrate and by various metal ions (salt in solution) and other impurities, which then radiate and conduct heat into the surrounding water.

gnomish
August 25, 2011 4:13 pm

g’day, TB.
got some spare time, I see. 🙂

tallbloke
August 25, 2011 4:45 pm

Hi gnomish.
Just dropped by to see if Willis had answered any of the substantive points made concerning the most fundamentally important issue in the climate debate…
I live in hope that Myrrh might chew on the bones being thrown, rather than burying them and barking.

Myrrh
August 25, 2011 4:46 pm

? Lasers again?? What’s with you lot? Give me figures for blue light as from the Sun, not artificially enhanced, and let me know how it heats water.
TB – using thermal infrared energies to heat water is because water is a great absorber of Heat which is thermal infrared, this does not use visible – photovotaic is visible, it can’t heat anything, it’s used to create electricity, and that pretty much useless if you want to boil more than a kettle..

August 25, 2011 4:51 pm
tallbloke
August 25, 2011 5:18 pm

Myrrh says:
August 25, 2011 at 4:46 pm
TB – using thermal infrared energies to heat water is because….

Read what I wrote again Myrrh. The glass stops IR getting in, as well as getting out. And the IR doesn’t heat the water directly, it heats the pipe the water is running through.
Try to read what people say, not what you reinterpret them as saying.

Phil.
August 25, 2011 5:21 pm

Myrrh says:
August 25, 2011 at 4:46 pm
? Lasers again?? What’s with you lot? Give me figures for blue light as from the Sun, not artificially enhanced, and let me know how it heats water.

Still talking your usual rubbish I see Myrrh, If blue or green light in the form of a laser beam is absorbed so will the same wavelength radiation from the sun! As far as the absorption of the laser beam is concerned it’s no different than filtering sunlight through blue glass and focussing it with a large mirror. Lasers provide a very well controlled beam of light which is capable of being tightly focussed, its properties as far as absorption are concerned aren’t ‘artificially enhanced’.

tallbloke
August 25, 2011 5:25 pm

Myrrh, another quote:
” Absorption is the conversion of light energy to heat energy at the atomic level. When this happens, the light disappears, and it is no longer visible. The ocean is a selective absorber of light. From the components of light: violet-blue-green-yellow-orange-red (V – B – G – Y – O – R) colors, the longer wavelengths are easily absorbed (red, orange) as they contain less energy. The shorter wavelengths (V – B – G) bands travel deeper into the water column before absorption. Hence,
· Red is absorbed around 10m
· Yellow is absorbed around 60m
· Green is absorbed around 100m
Any light beyond 100m is blue.”
http://faculty.scf.edu/rizkf/OCE1001/OCEnotes/chap6.htm

kuhnkat
August 25, 2011 5:57 pm

Tim, gnomish, MKelly etc,
notice that Myrrh conflates anything you tell him that conflicts with his mania with belief in AGW. I am not sure he is rational. He doesn’t seem to be separating the details of the physics from the whole AGW issue.
He apprently does not understand any of this past a very vague general reading (worse than me in other words!!) His insistence on the use of the term transparent as being an absolute rather than a relative is another example. The poor guy really needs some very basic instruction, yet, refuses it for fear he may have to give up his belief system.
He seems to think everything we say makes us deluded believers in the false science of AGW when, I am pretty sure, we are all sceptics, weel I am a denier but what the hey!!
Then again, I wonder if he is Willis’ way of getting back at us for not going along with him!! 8>)

Myrrh
August 25, 2011 6:38 pm

OK, one last attempt to get you to really engage with what I’m saying. You’re on a science blog, one of you, who shall remain nameless, prides himself on being a superior educated scientist, and there’s nothing I’ve said should be beyond any of your capabilities to assess logically.
So, why aren’t you producing the information I’m actually asking for?

Light
Light in Everyday Life
When we go outside on a sunny day, we see the light from the Sun reflecting everywhere, and at the same time, we feel the heat from the Sun. Both light and heat are forms of radiation. …..So most of what we see is from reflected light.
Light can react with matter in four ways:
Emission — actually generating light (this takes a source of energy)
Absorption — most things absorb at least a little light, unless they are perfectly transparent. (this results in the object gaining energy–usually the object gets hotter)
Transmission — glass, water, or other transparent substances transmit light, or allow light to pass through them (pure transmission means passing all of the energy–the object will not get hotter)
Reflection — shiny surfaces reflect light (called specular reflection) but even a dull surface will reflect light (scattered light).

Standard science, as I’ve been through above. Internally coherent and logical, tried and tested and used in countless applications by people who know the difference between properties.
The AGWScience Fiction Inc’s energy budget you’re using goes against that. It says we don’t feel the heat from the Sun because it says thermal infrared doesn’t reach the surface. It says that Visible heats oceans which is impossible for it to do, and it says thermal infrared direct from the Sun doesn’t, which actually does heat the oceans having the power to move the molecules into vibrational states; there is a disjunct here.
If you’re not interested in exploring this, fine. But until you can prove that AGWScience Fiction Inc’s version of how the world works here is fact, then you have not overturned traditional science. Just because it’s harder to retrieve that well tried and tested traditional information does not mean that it isn’t still being taught, and utilised, it just means that you can’t use it, because you weren’t taught real physics about heat and light.
And that, I really am sorry to say, is a loss for all of us.

August 25, 2011 6:40 pm

Whatever, let’s get back on topic.
Neither Willis nor anyone else has tackled the points I made earlier so a few simple thoughts:
i) If we had oceans but no non -condensing greenhouse gases such as CO2 and methane et al at all would there still be a water cycle? Obviously there would. Water would still evaporate at a rate governed by atmospheric pressure. The latent heat of vaporisation would still have an energy value set by the current atmospheric pressure. There would be evaporation, clouds, rainfall, climate zones, Hadley cells and the whole paraphernalia. The oceans would not freeze because solar shortwave would still get past the ocean skin and warm the ocean bulk. As long as the atmospheric pressure is high enough the oceans would remain as liquid water. It is the current atmospheric pressure plus the liquid watery oceans that keeps the system as a whole above the freezing point of water (and below the boiling point too).
ii) Water vapour being a greenhouse gas, AGW theory proposes that the evaporation that would occur even in the absence of non-condensing GHGs then feeds back on itself to reduce energy loss from the oceans and thus the water vapour would accumulate in the air until the entire oceans had heated to boiling point and converted to vapour form with the water cycle coming to a halt. Solar energy would continue getting in, evaporation would continue and as humidity blocked the evaporative energy flow from the oceans to the air all the oceans would heat up to boiling point. That is the logical implication of their assertion that extra water vapour AMPLIFIES the effect of more non- condensing GHGs. That, however is nonsense, it never happened and it could never happen as long as condensation occurs and keeps pace with evaporation and it always would at current atmospheric pressure. To create that scenario with the entire oceans converted to water vapour the atmospheric pressure would have to be sufficient to raise the energy value of the latent heat of evaporation (the energy cost of vaporisation) to a point where the temperature of all the water on the planet were above the boiling point of water at all times. It just isn’t going to happen, not on Earth anyway.
iii) So now with that basic water cycle let’s start adding non -condensing GHGs.
iv) We can see that adding GHGs in the form of water vapour does not slow the rate of energy loss from the oceans otherwise the oceans would all become vapour all the time. So why would we think that non condensing GHGs would slow the rate of energy loss from oceans to air?
v) They don’t. GHGs including water vapour operate in one direction only. Some GHGs are not more powerful than others just because they themselves do not condense. It does not matter that they do not condense because they simply add energy to the air which creates more water vapour and the extra water vapour does the extra condensing (by proxy) to negate the effect of the GHGs that do not themselves condense. There is no conceivable physical reason why non condensing GHGs would have the opposite effect on the system to that of water vapour yet that is what AGW theory logically requires. All types of GHG ALWAYS and unavoidably ACCELERATE the water cycle for a zero net effect on the basic rate of energy transfer from oceans to air which is set by atmospheric pressure acting on the physical properties of water molecules. If non condensing GHGs were to significantly alter surface pressure that would be a different matter but they do not. Alternatively it would be a different matter if non condensing GHGs left any surplus energy remaining after increasing the rate of evaporation (and upward radiation convection and conduction) and subsequent condensation but they do not as far as I can tell.
vi) So, to show me to be wrong, let’s now hear some proper evidence either that water vapour on its own reduces energy loss from oceans to air (the so called amplification process) or that the extra evaporation from non -condensing GHGs fails to produce enough extra evaporation (and condensation) to negate its own warming effect in the air.
vii) The truth must be that far from AMPLIFYING the effects of more non-condensing GHGs an increase in water vapour NEGATES it. To achieve an amplifying effect water vapour must be capable of reducing the rate of energy loss from the oceans yet obviously it is not.
viii) There is a climate effect nonetheless in that the extra energy in the air shifts the surface air pressure distribution a fraction but that is as nothing compared to natural solar and oceanic effects on that same surface air pressure distribution.

Tim Folkerts
August 25, 2011 7:34 pm

Stephen,
I am confused by this statement: “As long as the atmospheric pressure is high enough the oceans would remain as liquid water. ” Atmospheric pressure is the same all over the world (give or take a tiny bit), but that does not currently keep the oceans melted (ie near the poles). Temperature (governed by net energy flows, not pressure) is obviously also a key factor.
Similarly, you state “Water would still evaporate at a rate governed by atmospheric pressure. ” Again you seem to ignore the affects of temperature. Or you treat temperature as governed entirely by atmospheric presssure. Or I am misunderstanding you.
Could you clarify?
===========================================
You also state “AGW theory proposes that the evaporation that would occur even in the absence of non-condensing GHGs then feeds back on itself to reduce energy loss from the oceans and thus the water vapour would accumulate in the air until the entire oceans had heated to boiling point”
I have never seen any such prediction of feedback leading to uncontrolled warming to the boiling point based on some sort of self-feeding process. Could you provide an example of someone (preferably a professional in the field) making such a claim?

jae
August 25, 2011 8:16 pm

Whatever. As I have been saying for YEARS, the empirical evidence does not support the GHE theory. It is no warmer in areas with large amounts of GHEs than in areas with lesser amounts of GHEs, at constant latitude and altitude. It is no warmer now with increased CO2 than it was in 1930. Until the GHE theory can produce some actual empirical evidence, it is just another con game. Sorry folks, but that is the science that even Einstein had to live with.

jae
August 25, 2011 8:26 pm

Folkerts says:
“Since 15 C is about the average surface temperature, it seems that both theory and evidence support 390 W/m^2. In this case, I think empirical evidence sides with Willis.”
Well, I have not seen any EMPIRICAL EVIDENCE for that 390. I’ve only seen SB calculations. But, the 390 is probably about right, since SB works for solids (I don’t know about water, though). The thing that fascinates me, though, is that MODTRAN doesn’t give anywhere near that 390 number, if you enter 0 (zero) for altitude, looking down, -10 offset in temp. One would think that the model would register “tilt” when one enters those parameters–or give me 390. Oh well!
But… whatever….the 390 from the SURFACE does not have to somehow balance the incoming solar radiation at the SURFACE. That’s where Willis gets off base, IMHO. The only place that incoming and outgoing radiation have to balance, over the long term, is TOA. Willis and many others seem to be completely oblivious to other factors that influence the temperature of the atmospheric gases. Like molecular collisions, thermal capacity, and LTE. There is ABSOLUTELY NO EMPIRICAL EVIDENCE FOR AN ATMOSPHERIC GREENHOUSE EFFECT, and this is very hard for most people to realize. Temperatures on the surface of other planetoids prove this conclusively. Many smarter people than I have explained that in several ways in this thread, and Willis has ignored all of them, as far as I can tell (he could try to answer Hockeyschtick’s comments, for example). Hope he has read some of the Dragon (although I admit that there is much BS there, also).
Folkerts says:
“Since 15 C is about the average surface temperature, it seems that both theory and evidence support 390 W/m^2. In this case, I think empirical evidence sides with Willis.”
Well, I have not seen any EMPIRICAL EVIDENCE for that 390. I’ve only seen SB calculations. But, the 390 is probably about right, since SB works for solids (I don’t know about water, though). One thing that fascinates me, though, is that MODTRAN doesn’t give anywhere near that 390 number, if you enter 0 (zero) for altitude, looking down, -10 offset in temp. One would think that the model would register “tilt” when one enters those parameters–or give me 390. Oh well!
But… whatever….the 390 from the SURFACE does not have to somehow balance the incoming solar radiation at the SURFACE. That’s where Willis gets off base, IMHO. The only place that incoming and outgoing radiation have to balance, over the long term, is TOA. Willis and many others seem to be completely oblivious to other factors that influence the temperature of the atmospheric gases. Like molecular collisions, thermal capacity, and LTE. There is ABSOLUTELY NO EMPIRICAL EVIDENCE FOR AN ATMOSPHERIC GREENHOUSE EFFECT, and this is very hard for most people to accept. Temperatures on the surface of other planetoids prove this conclusively. Many smarter people than I have explained that in several ways in this thread, and Willis has ignored all of them, as far as I can tell (he could try to answer Hockeyschtick’s comments, for example). Hope he has read some of the Dragon (although I admit that there is much BS there, also).
What happens in real scientific pursuits is that folks pose questions to test theories. If the theory cannot accomodate the question, then the theory needs to be revised or trashed. What pisses me off here is that I posed such a question to Willis and it was not even acknowledged, let alone answered. In fact all I got was an RC response that completely ignored the question and blasted me with a tirade that says, in effect, that “you don’t believe in backradiation.” Which is pure bullshit and cannot be supported.
Willis, how about an apology? LOL!

philincalifornia
August 25, 2011 9:12 pm

Smokey says:
August 25, 2011 at 4:51 pm
Burning lasers.
========================
As the thread fades, I thought the mods wouldn’t mind something totally unrelated (except for the cigarette incident). I know you’ll like this one.
I had the good fortune to have a friend who drove in the Carrera Panamericana, when the guys from Pink Floyd were driving in it (Nick Mason and David Gilmour). I bought a Maserati from him.

I got a different bootleg tape at the time which showed more of a concert set-up, where a roadie lit his cigarette on a laser they were using.
I couldn’t find it online, but it’s kinda mentioned here too:
http://www.tpimagazine.com/Chronicle/437975/strung_out_on_lasers.html

gnomish
August 25, 2011 9:29 pm

The Boltzmann constant (k or kB) is the physical constant relating energy at the individual particle level with temperature observed at the collective or bulk level. It is the gas constant R divided by the Avogadro constant NA
“The Maxwell–Boltzmann distribution describes particle speeds in gases, where the particles do not constantly interact with each other but move freely between short collisions. It describes the probability of a particle’s speed (the magnitude of its velocity vector) being near a given value as a function of the temperature of the system, the mass of the particle, and that speed value.”
applying the calculations to a reading taken of the surface of a solid is superficial. it can not make any determination about quantity of energy in a bulk without a lot of additional assumptions. the assumptions include that the object is an ideal blackbody, that the surface is the same as the bulk temperature and that there is nothing else coloring the reading. all the assumptions need to be validated.
even so, it can not distinguish quantity of energy at all – which is what the word ‘heat’ is used for when i use it that way…lol

gnomish
August 25, 2011 10:14 pm

oh, myrrh – thanks for straightening me out.
now that you say it, it’s obvious that the blue lasers must, indeed, use artificially enhanced photons. thermal viagra makes them hot.
the green lasers produce female photons – they have thermal implants.
red lasers are red – like IR wishes it could be – so with a bit of makeup they can pass as hot if nobody looks too closely. that’s why they hang out at bars – to take advantage of the beer goggles.
i’ll check the mirror to see if i look a lot smarter now, mmk?

August 25, 2011 10:36 pm

Tim Folkerts asked various reasonable questions:
i) “I am confused by this statement: “As long as the atmospheric pressure is high enough the oceans would remain as liquid water. ” Atmospheric pressure is the same all over the world (give or take a tiny bit), but that does not currently keep the oceans melted (ie near the poles).
I’m talking about the averages globally and the bulk ocean. At the poles energy loss to space is rapid enough in winter to allow a surface skin of ice.
ii) “Temperature (governed by net energy flows, not pressure) is obviously also a key factor.””
Of course but the temperature of the bulk oceon is a consequence of pressure because the surface pressure dictates the energy cost of evaporation. Water evaporates at a lower temperature at the top of Everest or in space. Thus surface pressure averaged globally controls the rate at which absorbed solar energy can be released to the air above. The only other factor affecting the equilibrium temperature of the bulk ocean is the rate of solar shortwave absorption and that is where global albedo becomes relevant.
iii) “AGW theory proposes that the evaporation that would occur even in the absence of non-condensing GHGs then feeds back on itself to reduce energy loss from the oceans and thus the water vapour would accumulate in the air until the entire oceans had heated to boiling point”
I don’t think anyone has said it in such bald terms but it is implied logically so just think it through. The idea is that more non condensing GHGs warm the air and/or surface to increase water vapour in the air. Water vapour being a GHG itself must then either reduce the energy flow from the oceans to produce a runaway positive feedback until the oceans boil (more water vapour, warmer oceans, more water vapour in a runaway scenario that should have started with the first CO2 molecule but never did) or if that does not happen then the effect must be negated by more condensation taking energy away faster at higher levels. The latter is most certainly the case so if non condensing GHGs increase water vapour then again the effect of that increased water vapour must be negated in the same way so no amplification could be caused. In fact it is likely that the increased evaporation negates all of the warming effect in the air of those non condensing GHGs via an increase in the upward flow of latent heat, convection, conduction and radiation from the surface for a zero effect on ocean equilibrium temperature.
The bottom line is this:
At 1 bar atmospheric pressure the oceans will accumulate solar shortwave input until they emit 390 Wm2.
At that point the average global sea surface temperature reaches about 15C.
Solar input continues at 170 Wm2 with 390 still being emitted by the oceans but equilibrium having been reached the difference is made up by the balance being back radiated down from atmosphere to ocean surface. That backradiation is primarily a reflection of the oceans in the sky with scattering of incoming solar energy merely a minor component.
If one adds GHGs to the mix whether condensing or not the properties of the phase changes of water ensure that the balance is maintained by shuffling energy up and down through the atmosphere at variable rates to remove excesses or deficiencies that might arise from solar, oceanic or any other variations that affect the base energy flow imposed by that 1bar atmospheric pressure and the physical properties of water.
The physical manifestation of all that energy shuffling is expansion contracting and latitudinal shifting of the various components of the surface air pressure distribution which in turn affects the sizes and position of the various climate zones.
That is all that climate change is. Simple really.

Bob_FJ
August 25, 2011 11:08 pm

Jae @ August 25, 2011 at 8:26 pm

Folkerts says: “Since 15 C is about the average surface temperature, it seems that both theory and evidence support 390 W/m^2. In this case, I think empirical evidence sides with Willis.”
Well, I have not seen any EMPIRICAL EVIDENCE for that 390. I’ve only seen SB calculations. But, the 390 is probably about right, since SB works for solids (I don’t know about water, though).

Well yes, but more, here is a fairly typical statement on the S-B law from Wikipedia:

The Stefan–Boltzmann law, also known as Stefan’s law, states that the total energy radiated per unit surface area of a black body per unit time (known variously as the black-body irradiance, energy flux density, radiant flux, or the emissive power), j*, is directly proportional to the fourth power of the black body’s thermodynamic temperature T (also called absolute temperature):

Notice that it says total energy, which is radiated equally in all directions hemispherically. Nowhere do I recollect ever seeing: energy radiated normal to the surface, and as Tim confirmed above, in real-life complex two-body situations, the maths is rather complex in translating S-B. In the case of the Trenberth cartoon, we are not even dealing with a two-body situation, because the back radiation is from a gas mixture with lots of stuff having an impact on radiation going on in it at various altitudes.
The following depiction shows that radiation in the atmosphere is mostly lateral and not up and down as depicted in the cartoon. Furthermore, the initial emission from the surface must mean that many of the first absorptions are close to the surface, even if free path lengths are long. Furthermore that subsequent emissions in the air are spherically distributed so that it only partly returns towards the surface.
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1
I could drone-on on other aspects of this, but I hope that keeping it short, you (and others) will not nod-off and take a sniff on it. Any comments?

Bob_FJ
August 26, 2011 12:03 am

Further my comments just above, the following NASA “Earth’s Energy Budget” diagram seems to be more popular than that of the Trenberth/IPCC (and Willis’s own gloriously extended version).
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
The concept is similarly shown with different art around the ether, wherein most rational thinking is more concerned about HEAT transfer, which has a definitive relationship with temperature. (rather than all that EMR potential whizzing around doing mostly nothing until some of it might be absorbed)

Bob_FJ
August 26, 2011 12:09 am

Ouch,
In the above, in the last line, pls read: some of it might be NET absorbed

tallbloke
August 26, 2011 12:23 am

Stephen: good stuff, I’m seeing where you are coming from now. So more co2 just speeds up the hydrological cycle which has a cooling effect greater than the warming effect of making the atmosphere denser?
Bob-FJ: So are you saying a radiometer pointing sideways will register higher readings than pointing it up or down? Any results online anywhere?

RJ
August 26, 2011 12:30 am

Bob FJ
This is a big improvement on the Keihl and Trenberth radiation budget
So 51% is absorbed by the surface and
7% leaves the surface by conduction
23% by latent heat
21% by radiation (15% of this is absorbed by the atmosphere)
The radiation % though still seems high. As does the % that is absorbed by the atmosphere
And there is no backradiation

tallbloke
August 26, 2011 12:39 am

jae says:
August 25, 2011 at 8:26 pm
the 390 from the SURFACE does not have to somehow balance the incoming solar radiation at the SURFACE. That’s where Willis gets off base, IMHO. The only place that incoming and outgoing radiation have to balance, over the long term, is TOA.

My earlier post shows that the oceanic balance and the atmospheric balance work out fine with the net radiation balance. The radiation flux helps maintain the atmospheric depth and lapse rate which causes the surface at the ocean/air interface to be warmer than it would be without radiative gases in the atmosphere, and that causes the bulk temperature to rise from solar input until the bulk at the top is warmer than the surface. But this has nothing to do with any mythical requirement for absorption of energy from back radiation into the ocean being mixed into the bulk as Willis believes.

Bob_FJ
August 26, 2011 1:16 am

tallbloke @ August 26, 2011 at 12:23 am

Bob-FJ: So are you saying a radiometer pointing sideways will register higher readings than pointing it up or down? Any results online anywhere?

No, I did not say that, but bear in mind that your radiometer[s] needs to be located on the full circumference of a circle for each elemental parcel, and you need to integrate the results of multiple parcels. I simply pointed out that radiation is equal in ALL directions, not just up and down, as a statement of fact. And no, I have no online results for you.
Do you disagree with the following?
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1

Bob_FJ
August 26, 2011 1:41 am

Tallbloke, Further to my above, perhaps I should add that your instrumentals should not point in just one lateral dimension but 360 degrees from east to west cumulatively rather than just north and south.

Myrrh
August 26, 2011 2:42 am

tallbloke says:
August 25, 2011 at 5:25 pm
Myrrh, another quote:
” Absorption is the conversion of light energy to heat energy at the atomic level. When this happens, the light disappears, and it is no longer visible. The ocean is a selective absorber of light. From the components of light: violet-blue-green-yellow-orange-red (V – B – G – Y – O – R) colors, the longer wavelengths are easily absorbed (red, orange) as they contain less energy. The shorter wavelengths (V – B – G) bands travel deeper into the water column before absorption. Hence,
· Red is absorbed around 10m
· Yellow is absorbed around 60m
· Green is absorbed around 100m
Any light beyond 100m is blue.”

I can only suggest that you use the search facility to find all mentions of transparent which explain that water being that for visible light means “conversion of light energy to heat energy at the atomic level” doesn’t apply.
It’s probably because this thread has been so busy, you appear not to have thought through what I’ve been saying:
“Transmission — glass, water, or other transparent substances transmit light, or allow light to pass through them (pure transmission means passing all of the energy–the object will not get hotter)”
Water is transparent to visible light, it does not convert to heat.
It does not convert to heat because water is a transparent medium for it.
The problem here, as I have been pointing out, is that you (generic) are taking AGWScience fiction memes and simply using them without examination and so failing to see they are illogical in real physics.
Early on at the beginning of this Dave Springer pointed out the disjunct in reasoning re visible light and water in Willis’ statements:
“Argument 4. Without the heating from the DLR, there’s not enough heating to explain the current liquid state of the ocean. ”
“Wrong again. Liquid water has the same properties as water vapor when it comes to being transparent to visible light and opaque to infrared. The same properties that make water vapor a greenhouse gas make liquid water a greenhouse fluid. The difference is that liquid water is like water vapor on steroids since there’s more water in the first meter of the ocean than there is water vapor in the column of air above it and sunlight penetrates far beyond the first meter.”

Water is transparent to visible, light. I have given several explanations of this, it cannot heat water because it doesn’t get absorbed on the atomic level – transparent means that it passes through without interacting with the molecules of water. That’s what transparent means.
Thermal infrared, heat, does interact, it moves the molecules of water into vibrational states and converts to heat.
‘Absorbed’ here as a description of process is a technical term in physics. ‘Absorbed’ as a general description should not be confused with it, which is how you have used it in “The shorter wavelengths (V – B – G) bands travel deeper into the water column before absorption”.
Which, because visible is not actually absorbed in the physics meaning of interaction, means only that it has disappeared, ‘absorbed’ in a general sense.
It’s confusion of meaning which is at the heart of this problem. I should be really grateful if you would give this some thought, because then you’ll also be able to see what I meant by the irrationality of confusing ‘transparent’ with regard to the meme that visible light is transparent in the atmosphere, when it isn’t – [reflection/scattering, why we have a blue sky is because the electrons of oxygen and nitrogen absorb and then send visible back out, scattering the blue in the sky. However, this is not a conversion to heat, because it creates none].
Let’s stick with water and the difference between visible and thermal infrared travelling through it. For visible, water is transparent, it passes through without interacting so no heat is created.
“Transmission — glass, water, or other transparent substances transmit light, or allow light to pass through them (pure transmission means passing all of the energy–the object will not get hotter)”
Therefore, ” Absorption is the conversion of light energy to heat energy at the atomic level. When this happens, the light disappears, and it is no longer visible.”
Can’t be a description of visible light, light in water, it can only refer to the invisible thermal infrared, heat.
Water is a great absorber of heat, thermal infrared, which as Dave Springer pointed out, makes liquid water a greenhouse fluid, water vapour on steroids.
So, there are first of all two basic aspects which have to be sorted to see the different processes at work here, the first is to appreciate the difference between light and heat waves in that there is a difference between electronic transitions and vibrational resonance:
“Electronic: Transitions in electron energy levels within the atom (e.g. pigments). These transitions are typically in the ultraviolet (UV) and/or visible portions of the spectrum.
Vibrational: Resonance in atomic/molecular vibrational modes. These transitions are typically in the infrared portion of the spectrum.”
The second aspect is that not all electronic transitions create heat, photosynthesis for example a use of light energy for a chemical conversion which is not a conversion to heat but sugars, and, reflection/scattering as in the blue sky, a use of the light’s energy to reflect it back out, again no heat is created.
Visible light from the Sun passes straight through water, it does not convert water to heat. Therefore, it cannot heat the oceans.
Thermal Infrared, the real heat energy, is in resonance with water, it does not pass through without interacting but moves water into vibrational states raising the temperature.
Water is a great absorber of heat, and has an extremely high capacity to hold it, which is why we put water in the radiators of our central heating systems.
In water vapour this is the mechanism which takes the great thermal heat from the Sun from the surface back up into the atmosphere to release it by condensing out in the colder heights before coming down as rain.
Water is a great absorber of thermal infrared, water passes visible light through unchanged.
The only thing that happens with visible in water is that of a slight delay as visible tries to engage in the dance of the molecule of water but cannot get in, this has the effect of slowing down the light wave.
Here’s a more technical description of the process:

“A transparent material is one in which the charged particles
can’t permanently absorb any photons of visible light. While these
charged particles all try to absorb the visible light photons, they
find that there are no permanent quantum states available to them when
they do. Instead, they play with the photons briefly and then let them
continue on their way. This playing process slows the light down.” http://www.madsci.org/posts/archives/may98/893732585.Ph.r.html

That’s what happens to visible light in water. It cannot heat water. It cannot heat the oceans. The ‘energy budget from the Sun’ as you, generic, are describing it, is impossible in the real physics of our real world.
Now, since visible cannot be heating the oceans, and the meme is that thermal infrared does not reach the surface – what is heating the Earth??
?

Tim Folkerts
August 26, 2011 3:42 am

So many thought-provoking posts lately!
I only have time for one response atm, so let’s try this simple statement: “And there is no backradiation” (in response to the energy diagram from NASA).
Consider this analogy. You have two rooms separated by a thin, gas-tight divider. Both are at the same pressure. One is full of N2 and the other is full of O2. Now, I create an opening between the two rooms. Does any gas flow from one room to the other?
Well, since there is no pressure difference, there will be no net flow. You will not feel any “wind” pushing from one room to the other. You will not see anything happening. No work could be done by this flow.
But at a molecular level, we know that there are indeed molecules on both sides, and there will indeed be billions of molecules passing back and forth every second. There is a real and measurable flow of O2 in one direction and N2 in the other direction as the gases diffuse and mix.
I would argue that the flow of IR photons in both directions is just as real. Talking about the net flow is certainly one legitimate way to talk about the situation, where all we are interested in is the macroscopic scale. But at a microscopic level, there is a very real flow of photons in both directions. This comes to ~ 390 W/m^2 in one direction and ~ 325 W/m^2 in the other.

August 26, 2011 4:08 am

“So more co2 just speeds up the hydrological cycle which has a cooling effect greater than the warming effect of making the atmosphere denser?”
Not quite. To the extent that the atmosphere gets denser there would be a warming but CO2 is such a small part of the atmosphere that a density change would be infinitesimal.
Also, once all the DLR is in the process of being converted to latent heat, conduction, convection and upward radiation the process of increasing speed or size for the hydrological cycle stops so there is a zero effect on the rate of energy flow from the bulk ocean. The process is therefore self limiting so no net cooling of the ocean bulk either. DLR is therefore neutral as regards its effect on the bulk ocean. Events in the top 1mm which acts as a buffer ensure that it be so.
There is also a refinement to cover the fact that the sun is about one third stronger than it was 4 billion years ago but the rise in temperature of the bulk ocean is very small. Various factors result in the system being pretty resistant to solar changes too. One just gets a change in the speed or size of the water cycle to accommodate most of the extra energy passing through.
The big thing is surface pressure because that underpins the whole process. If surface pressure does not change then the temperature of the bulk ocean won’t change except perhaps for a small change if solar input gets much greater.
The water cycle takes the strain whatever forces seek to disrupt the equilibrium temperature of the bulk ocean. All climate changes are a consequence of water cycle changes via a redistribution of surface pressure.

August 26, 2011 4:11 am

“This comes to ~ 390 W/m^2 in one direction and ~ 325 W/m^2 in the other.”
With the difference pretty damned close to solar input at about 170 Wm2

tallbloke
August 26, 2011 4:23 am

Stephen Wilde says:
August 26, 2011 at 4:11 am (Edit)
“This comes to ~ 390 W/m^2 in one direction and ~ 325 W/m^2 in the other.”
With the difference pretty damned close to solar input at about 170 Wm2

390-325 = 65 not 170.
However, 65 is the net flux figure and coincidentally(??) energetically equivalent to the amount of solar radiation directly absorbed by the atmosphere.

tallbloke
August 26, 2011 4:28 am

Myrrh says:
August 26, 2011 at 2:42 am
tallbloke says:
August 25, 2011 at 5:25 pm
Myrrh, another quote:
” Absorption is the conversion of light energy to heat energy at the atomic level. When this happens, the light disappears, and it is no longer visible. The ocean is a selective absorber of light. From the components of light: violet-blue-green-yellow-orange-red (V – B – G – Y – O – R) colors, the longer wavelengths are easily absorbed (red, orange) as they contain less energy. The shorter wavelengths (V – B – G) bands travel deeper into the water column before absorption. Hence,
· Red is absorbed around 10m
· Yellow is absorbed around 60m
· Green is absorbed around 100m
Any light beyond 100m is blue.”
Myrrh’s response:
I can only suggest that you use the search facility to find all mentions of transparent which explain that water being that for visible light means “conversion of light energy to heat energy at the atomic level” doesn’t apply.
Water is transparent to visible light, it does not convert to heat.
It does not convert to heat because water is a transparent medium for it.

Myrrh, my last reply to you, I will not entertain a dialogue of the deaf.
The figures I quoted are actual real empirical measurements, not equivalent to your assumption which is based on pure water not sea water
This thread is about the oceans, not pure lab water in vials.
You need to read up on the physics of radiative transfer of energy between photons and molecules, excitation states and collisional vibrations. So do I.
Bye for now.

tallbloke
August 26, 2011 4:40 am

Bob_FJ says:
August 26, 2011 at 1:16 am
tallbloke @ August 26, 2011 at 12:23 am
Bob-FJ: So are you saying a radiometer pointing sideways will register higher readings than pointing it up or down? Any results online anywhere?
No, I did not say that, but bear in mind that your radiometer[s] needs to be located on the full circumference of a circle for each elemental parcel, and you need to integrate the results of multiple parcels.

Fair point, I see what you mean now.
I simply pointed out that radiation is equal in ALL directions, not just up and down, as a statement of fact. And no, I have no online results for you.
Do you disagree with the following?
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1

I don’t understand the diagram. What is an “elemental parcel of air”?
Why wouldn’t it radiate at all outward angles from all points on its surface?
If it did that, wouldn’t the radiation in any direction be equal from the visible hemisphere?

tallbloke
August 26, 2011 4:51 am

Stephen Wilde says:
August 26, 2011 at 4:08 am
“So more co2 just speeds up the hydrological cycle which has a cooling effect greater than the warming effect of making the atmosphere denser?”
Not quite. To the extent that the atmosphere gets denser there would be a warming but CO2 is such a small part of the atmosphere that a density change would be infinitesimal.

Yes, and combustion uses up oxygen anyway, so we’re losing atmospheric bulk, albeit non-radiative bulk.
There is also a refinement to cover the fact that the sun is about one third stronger than it was 4 billion years ago but the rise in temperature of the bulk ocean is very small. Various factors result in the system being pretty resistant to solar changes too. One just gets a change in the speed or size of the water cycle to accommodate most of the extra energy passing through.
I’ve always thought that if the surface T can stay fairly constant while the Sun gets 25-33% stronger, an extra 100ppm of co2 isn’t going to do a lot. 🙂
The big thing is surface pressure because that underpins the whole process. If surface pressure does not change then the temperature of the bulk ocean won’t change except perhaps for a small change if solar input gets much greater.
This is the part I really want to get into detail on so I can properly ‘realise it’ for myself. For a given mass of atmosphere, the surface pressure will be a constant, whether that mass is bulkier and less dense, or compact and more dense. But the AGW proponents seem to be saying that raising the altitude radiation to space takes place at is what causes the surface T to be what it is, and that more longwave radiative flux raises that altitude by causing thermal expansion of the atmosphere due to slowing down the escape of solar input to space.
Or at least, that’s one of their many arguments, I think.

Myrrh
August 26, 2011 5:08 am

Myrrh, my last reply to you, I will not entertain a dialogue of the deaf.
The figures I quoted are actual real empirical measurements, not equivalent to your assumption which is based on pure water not sea water
This thread is about the oceans, not pure lab water in vials.
You need to read up on the physics of radiative transfer of energy between photons and molecules, excitation states and collisional vibrations. So do I.

Yes you do. This is a property of water, all water, whatever is suspended in the water doesn’t change the fact that visible light cannot join in to the properties of the molecules of water. Do begin by re-reading was transmission, transparent and absorption actually mean. It means that water does not absorb visible light, it cannot therefore heat water.
Then, if you’re making some kind of case that it’s the suspended particles and life in the oceans which is heating up which is heating the water which is heating the Earth, damn well explain exactly how. I’m getting really tired of all the supercilious comments from everyone to cover up that you don’t know what you’re talking about.
So do, go and find out, all of you who are promoting this ridiculous fiction that shortwave converts to heat the land and oceans of the whole planet.

Tim Folkerts
August 26, 2011 6:05 am

“Do you disagree with the following?”
http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1

Of course I disagree with it! I could turn the image 90 degrees and and make exactly the same argument that most of the energy is going up and down instead of to the sides!
By symmetry, it is clear that since any orientation of the image is equivalent, then the radiation in any direction is equivalent.

tallbloke
August 26, 2011 6:07 am

Myrrh:
Nothing is completely transparent, and although the coefficient of absorption of water is low for water in the visible and near IR, it’s enough to convert the proportion of the 1.22 × 10^17 joules of energy from the Sun the Earth absorbs every second which penetrates many metres into the ocean (unlike IR) into the amount of heat we have in the oceans. You need to remember the Earth is well insulated by space.
These coefficients are determined from empirical measurements Myrrh, they don’t depend on atomic theory or anything fancy like that.
The shallow water near the beach that warms up during a sunny day is mostly being warmed by conduction from the sun warmed bottom anyway, abit like the solar hot water system in my earlier comment. Maybe you are being decieved by that.

August 26, 2011 6:21 am

“But the AGW proponents seem to be saying that raising the altitude radiation to space takes place at is what causes the surface T to be what it is, and that more longwave radiative flux raises that altitude by causing thermal expansion of the atmosphere due to slowing down the escape of solar input to space.”
The problem with that is that it is only one possibility. The atmospheric heights will rise from EITHER warming from below or cooling from above. Thus the tropopause will rise both if the oceans release more energy for low level warming (or from more energy in the air from more GHGs) AND if the stratosphere cools from a more active sun (as I contend but the consensus suggests such cooling from human GHGs).
Now if the tropopause rises as a result of a stratosphere cooling from above what does that do to the upward energy flow ? Presumably that involves a faster rate of energy loss upward to space ?
Would the surface really become warmer or would the lapse rate slacken off ?
I’m aware of the basic principles but a bit vague as to how they work out in different real world situations.
Anyway regardless of the detail what we have is a naturally very variable tropospheric height which responds to the interaction between changes in energy flow from both above and below.
And it is the tropospheric heights (or rather the way they vary between equator and pole) that controls the entire tropospheric energy budget by shifting the surface air pressure systems around.
Now I do accept the reality of more energy in the air (but not the oceans) from GHGs. It is just that with such an efficient two way ‘valve’ at the tropopause responding rapidly to oceans and sun the effect from GHGs is going to be irrelevant in comparison.
By way of illustration the mid latitude jets may have shifted up to 1000 miles latitudinally between MWP and LIA. Would our extra CO2 shift them even 1 mile ? Could we ever notice ?

tallbloke
August 26, 2011 7:02 am

Thanks Stephen, lots more food for thought.
I think maybe some of this stuff is getting through to the high priests of AGW because the latest incarnation of their argument doesn’t mention direct heating of the ocean by back radiation anyway.
Have a look and see if your arguments deal with this from Raymond Pierrehumbug:
http://climateclash.com/2011/01/15/g6-infrared-radiation-and-planetary-temperature/
Comments #1 at the top and #45 at the bottom are worth a read I think…as are the rest if you have the time.

August 26, 2011 8:46 am

Whoops. Thanks, tallbloke,for pointing out my erroneous post. Too much rushing about at work at the time.

Spector
August 26, 2011 11:08 am

RE: Myrrh: (August 23, 2011 at 2:17 pm)
Water is completley transparent to Visible, not proven in any way that it isn’t, you can see straight through clear water.. Visible doesn’t have the mechanism to interact even on an electron scale with water, is really transparent, it is transmitted through unchanged. Unlike the atmosphere which isn’t really transparent, because of absorption by electrons to produce reflection and scattering.
You may be confusing statements that only apply to free water molecules in the atmosphere where the energy states are limited to the natural vibration modes of the molecule in free fall. Other vibrations may be possible for a short time when two molecules are bouncing together. In a gas this is happening so rarely that sunlight can reach the surface of the earth with *almost* no attenuation due to this cause. But in liquids and solids, the molecules are all so tightly bound that this is happening much more frequently and a wide range of possible vibration modes are allowed. This is why a solid can be black with a uniform spectrum of resonances.
Molecules in solids and liquids are bound together by electric and magnetic forces. Unlike the neutrino, photons, by nature are electromagnetic and they easily interact with these bonds.
If you looked at that graph at the top of the Wikipedia page, you would see that the absorption coefficient for light in water is in the range of about 0.0001 to 0.01 per centimeter. This means that over distances between 100 cm (red) or 10,000 cm (blue), the light energy will be reduced to about 36.8 percent of its original level (one divided by the base of the natural logarithms) and thus 63.2 percent is absorbed, or converted to heat along the way. When anyone says water is ‘perfectly transparent,’ they usually are referring to distances that are less than 10 cm.
Light does not ‘fade away.’ Photons are packets of energy. Their motto is “We Deliver.” Light intensity is measured in watts per square meter for a continuous flow. That is potential heating power. A standard measure of heat energy is the BTU. A heat flow of one BTU/sec is equivalent to about 1055 watts.
The confusion about infrared being heat energy is common. It is terrestrial temperature range heat radiation. Light is solar temperature range heat radiation. If you look at a plot of the solar spectrum, you will see the vertical axis labeled in watts per square meter per frequency or wavelength interval peaks out in the middle of the optical range.

Tim Folkerts
August 26, 2011 1:01 pm

One very simple question for Myrrh.
Give an definition of “thermal infrared”. When you say “thermal infrared” what exactly are you including and what exactly are you excluding? You use this phrase repeatedly and it is a central point in many of your posts, so I would like to know precisely what your definition is.

Bob_FJ
August 26, 2011 3:37 pm

RJ @ August 26, 2011 at 12:30 am

[The NASA diagram] is a big improvement on the Keihl and Trenberth radiation budget
So 51% is absorbed by the surface and
7% leaves the surface by conduction
23% by latent heat
21% by radiation (15% of this is absorbed by the atmosphere)

Yes, I don’t know about the numbers, but it doesn’t have the annoying distractions in the Trenberth. It’s interesting that the Wikipedia gatekeepers prefer it too. As far as I’m aware, they have never shown the Trenberth cartoon.

Bob_FJ
August 26, 2011 3:45 pm

Tallbloke @ August 26, 2011 at 4:40 am

http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1
I don’t understand the diagram. What is an “elemental parcel of air”?
Why wouldn’t it radiate at all outward angles from all points on its surface?
If it did that, wouldn’t the radiation in any direction be equal from the visible hemisphere?

An elemental parcel of air means a very small quantity approaching zero, as employed in calculus. Molecules within the parcel would radiate in all directions, and at its boundaries too.

Bob_FJ
August 26, 2011 3:52 pm

Tim Folkerts @ August 26, 2011 at 6:05 am

http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1
Of course I disagree with it! I could turn the image 90 degrees and and make exactly the same argument that most of the energy is going up and down instead of to the sides!…”

Sorry, but most of the Earth’s surface can be considered to be flat and horizontal. It follows that the initial radiation from that surface is hemispherical with a horizontal base. Atmospheric layering also varies vertically. It would not be rational to turn this all through 90 degrees.
BTW, my laptop is becoming painfully slow in some operations on this long thread. Anyone else?

Bob_FJ
August 26, 2011 5:16 pm

Tim Folkerts, Re: spherical radiation;
Another way of looking at it is to visualize the spines on a sea urchin as being radiation. (or maybe the dimples on a golf ball). The largest proportion of them tend towards the lateral. Nothing changes if you rotate the sphere through 90 degrees.

Tim Folkerts
August 26, 2011 6:07 pm

Let me clarify about the radiation direction Re: http://farm3.static.flickr.com/2522/3837627461_4fc91e7a03_z.jpg?zz=1 . I was a little hasty. I was concluding from symmetry that any given direction has as much radiation as any other given direction, which would be true.
An “elemental” parcel of air is presumably on that is small compared to other dimensions of interest — say a cubic meter of air (or a cubic cm of air). In this approximation,. there is no variation within the parcel. This will naturally lead to the radiation being isotropic. The same amount of radiation goes up as goes down as goes east as goes west as goes north as goes south.
You could indeed lump 4 of these together and call them “sideways”, which would make exactly 2x as much radiation leaving from the “4 sides” as from the “top & bottom”. The flip side is that there would be twice as much area on the sides to receive radiation.
The sides will be radiating to/from other parcels at the same altitude and same (roughly) temperature, so the net radiation would be zero sideways. In this sense, the sides are of no interest. Only the top and bottom are “aimed” are parcels of a different temperature and hence only they can lead to any net radiation.

Tim Folkerts
August 26, 2011 6:59 pm

To me, the logic comes down to this. Please point out any specific idea that you take issue with.
1) If some parcel of the ocean (say 1m x 1m x 100m deep but NOT counting the skin layer) is balanced in terms of energy, then the temperature will stay constant.
2) The balance for an “average parcel” is approximately + 170 W/m^2 net solar radiation and -170 W/m^2 conduction to the skin layer (with very close to 0 W/m^2 of energy flow sideways or downward).
3) To maintain the -170 W/m^2, the skin layer must maintain a temperature gradient of 0.2 – 0.3 K (as is indeed observed).
4) If the IR energy into the top layer increases, the top layer must respond.
5) The first change will be an increase in temperature right near the top of the skin (since the IR is absorbed there).
6) in response to this increased temperature, the IR up from the surface will increase, as will the evaporation. BUT NOTE: the surface temperature has to increase for either of these to happen. Evaporation will speed up IN RESPONSE TO the elevated temperatures. Evaporation can and does limit the temperature rise, but cannot 100% prevent it.)
7. Even a small rise in the surface temperature will lead to a decrease in the conduction from below.
8. Even a small decrease in conduction out of the bulk of the ocean will lead to warming of the bulk of the ocean over time.
At this point, we could discuss the EXTENT of the warming of the bulk due to an increase in the downward IR. IR might be inefficient at warming the oceans, but IR logically has SOME effect unless one or more steps above are incorrect. Perhaps 10 W/m^2 extra only warms the surface 0.01 K, leading to a 0.01 K increase in the bulk temperature to restore the gradient.
Or someone could point out a flaw in my logic.

jae
August 26, 2011 7:01 pm

Bob_FJ::
“The following depiction shows that radiation in the atmosphere is mostly lateral and not up and down as depicted in the cartoon. Furthermore, the initial emission from the surface must mean that many of the first absorptions are close to the surface, even if free path lengths are long. Furthermore that subsequent emissions in the air are spherically distributed so that it only partly returns towards the surface.”
Very interesting. Am studying. Thanks!

jae
August 26, 2011 7:04 pm

All I KNOW ALMOST FOR CERTAIN NOW is that the GHE doesn’t make any sense from an empirical evidence standpoint. Other planets prove it. Our planet proves it. Wish I was smart enough to explain it.

kuhnkat
August 26, 2011 7:52 pm

Bob_FJ,
please stop with the most of the radiation is lateral. ALL of the radiation is emitted spherically unless there are physical reasons that would distort it (such as being emitted from a flat surface making it hemispherical). Trying to say most is lateral or most is up and down or… is all equally incorrect.

kuhnkat
August 26, 2011 8:00 pm

Tim Folkerts,
“The sides will be radiating to/from other parcels at the same altitude and same (roughly) temperature, so the net radiation would be zero sideways. In this sense, the sides are of no interest”
Yes, this is the NET that is SUPPOSED to be computed when things emit at each other. Of course, for some reason, the Climate Types throw out what is real physics for pseudo physics in computing 390 and 324 when they look at up and down.
I would point out that we have convection in the atmosphere so we have substantial areas where we have substantial differences in the net!! 8>) We also have irregularities where the SB is reduced in accuracy as the object has a strutcture where it can absorb its own black body radiation. I would suggest that the ocean surface is this way most of the time as it is not perfectly smooth and waves radiate against each other and the trough… Brings a whole new aspect to this why the ocean doesn’t freeze bit. It heats itself?? And in reality, this cannot be ignored any more than “backradiation!”

Bob_FJ
August 26, 2011 10:58 pm

kuhnkat @ August 26, 2011 at 7:52 pm

Bob_FJ,
please stop with the most of the radiation is lateral. ALL of the radiation is emitted spherically unless there are physical reasons that would distort it (such as being emitted from a flat surface making it hemispherical). Trying to say most is lateral or most is up and down or… is all equally incorrect.

So in the analogy I made about visualising the spines on a sea urchin as radiation from an elemental sphere of air, does not work for you? Can you not see that there are a lot more spines pointing laterally than up and down? (or if you prefer, the dimples on a golf ball). You appear to be very assertive in your opinion, but should explain why, and your next comment seems to support my point anyway. I did not discuss absorption because neither does S-B which is at the core of the issue.
kuhnkat @ August 26, 2011 at 8:00 pm Re Tim Folkerts @ August 26, 2011 at 6:07 pm

Tim Folkerts,
“The sides will be radiating to/from other parcels at the same altitude and same (roughly) temperature, so the net radiation would be zero sideways. In this sense, the sides are of no interest”
Yes, [says Kuhnkat] this is the NET that is SUPPOSED to be computed when things emit at each other. Of course, for some reason, the Climate Types throw out what is real physics for pseudo physics in computing 390 and 324 when they look at up and down.

Further to my August 25, 2011 at 11:08 pm the issue is that S-B gives the TOTAL radiative energy from a flat surface, which is hemispherically distributed. As I understand it from others, (I’ve not had the fortitude to read in detail the Trenberth et al papers of 1997 and 2009), the 390 W/m^2 up is based on an S-B calculation on global average T. But, a great deal of the total emission, (which is not HEAT), is lateral, and the NET lateral is zero, and not part of upward. Thus the 390 up is seriously overstated

Tim Folkerts
August 26, 2011 11:04 pm

“Climate Types throw out what is real physics for pseudo physics in computing 390 and 324 when they look at up and down.”
NO! If I heat a 1 m^2 piece of ground to ~ 288 K it will emit ~ 390 W/m^2. If the surrounding area is 0K, then there will be no radiation coming back. If the surrounding are 288 K, then there will be the 390 W/m^2 coming back. If the surrounding are warmer, there could by 780 W/m^2 coming back. The temperature of the surrounds does not change the rate at with the photons are emitting.
It is not “pseudo physics” that at the opening between two chambers, molecules are moving both directions.
It is not “pseudo physics” that in a wire, electrons are moving both directions.
It is not “pseudo physics” that in the atmosphere, photons are moving both directions.
The net flow is certainly something interesting to calculate (in any of the cases above). But the actual movement of actual particles in interesting too, and certainly not “pseudo physics”!

Bob_FJ
August 26, 2011 11:11 pm

Jae @ August 26, 2011 at 7:04 pm

All I KNOW ALMOST FOR CERTAIN NOW is that the GHE doesn’t make any sense from an empirical evidence standpoint. Other planets prove it. Our planet proves it. Wish I was smart enough to explain it.

Yep, but I don’t think we can win the war by questioning if there is a GHE though. I prefer to go along for example with atmospheric physicist and elitist alarmist Andrew Dessler who has admitted that if sceptics like Roy Spencer can show that there are major negative feedbacks, then AGW is no problem.

August 27, 2011 5:46 am

Tim Folkerts said:
“7. Even a small rise in the surface temperature will lead to a decrease in the conduction from below.”
That is the exact issue that needs to be resolved. I have described and considered that point at length elsewhere.
That statement represents the application of Fourier’s Law which provides (I think) that the flow of energy across a boundary is proportionate to the temperature differential either side of it.
However I have come to the conclusion that the combined effect of increased evaporation plus upward convection radiation and conduction from the warmed molecules at the very TOP of the ocean skin removes the energy added by DLR fast enough to prevent any change in the temperature differential at the interface between the ocean bulk and the BOTTOM of the skin layer.
Thus the 1mm deep cooler ocrean skin acts as a buffer between the DLR and the ocean bulk.
Reasons for that conclusion:
i) That little ‘hook’ in the temperature profile from warm to cool of 1mm depth below the actual surface and above the bulk remains present day and night. If it is not removed by solar input on a diurnal basis I don’t see how it could be affected by a little more DLR.
ii) That coolness of 0.3C 1mm deep appears to be a fixture of the oceans averaged globally. There seems to be no evidence of significant variation on a globally averaged basis.
iii) If DLR were to reduce the rate of energy flow from the ocean bulk to the skin layer then the depth and intensity of the temperature change at that bulk/skin interface must change but there is no evidence that it does if averaged globally.
iv) There is a nice analogy available namely the concept of a tributary joining a river. The downstream VOLUME of flow may increase but there is no need for any change in the RATE of flow from UPSTREAM in either river or tributary except maybe in a small buffer zone around the actual junction. So it appears to be similar where the energy flow from the bulk meets the energy flow from the DLR within the ocean skin which is most likely a similar type of buffer zone. In the case of river and tributary gravity determines the flow rates. In the case of ocean and skin atmospheric pressure determines the flow rates.
v) Evaporation being an energy hungry process (5 times more energy taken up than is required to provoke it) and since the energy shortfall is taken from where it is most readily available it seems most likely that the shortfall will be taken from any remaining DLR once increased radiation conduction and convection have taken their slices.
So it seems to me to be unlikely that Tim’s point 7 is correct.
I saw something that led me to believe that those ‘in the know’ are aware of this issue and that experiments are in the design stage but not yet ready for deployment.
So it is not safe to say that DLR has ANY effect on the upward flow rate of energy from the oceans. Pronouncing to that effect is not adviseable. Confidence in AGW is misplaced until the issue has been resolved.

RJ
August 27, 2011 6:22 am

Bob_FJ says:
August 26, 2011 at 11:11 pm
“Yep, but I don’t think we can win the war by questioning if there is a GHE though. I prefer to go along for example with atmospheric physicist and elitist alarmist Andrew Dessler who has admitted that if sceptics like Roy Spencer can show that there are major negative feedbacks, then AGW is no problem.”
Battles can be won by employing more than one weapon.
And are sceptics to also support pseudo science because it might help to win this battle. Or promote real science
GHG backradiation is not based on real science. We can acknowledge this but also challenge the alarmists positive feedback theory and the hopelessness of solutions like ET schemes.

kuhnkat
August 27, 2011 8:29 am

Tim Folkerts,
1) if it is balanced it simply means there is the same energy coming and going from that parcel. Does not say much about what is happening outside of that parcel, only that the flux is balanced within that parcel.
2) you are guessing. just because the input and output for the surface (not just the ocean surface) is 170 has little to do with the balance of any particular parcel which isn’t in balance anyway. Warming in the day, cooling at night in general.
3) again this is an average and will be changing due to instant flows.
4) OK, except what would increase this input?? Generally it is increased insolation that cycles through the ocean first meaning the surface temp goes up, evap and OLR is going up a fraction before the DLR that stabilizes it.
5) speculation
6) probably
7) maybe
8) not necessarily. convection is up although I believe Willis is trying to claim that a large enough delay in the flux up and out would allow conduction to move energy down. While certainly possible, I would need to see a lot of measurements to agree this is significant below the absorption layer without gross physical effects like currents carrying the energy.

kuhnkat
August 27, 2011 8:41 am

Bob_FJ,
I look at half of a sea urchin and see ALL of the spines going AWAY from the surface. Since the bottom layer of the atmosphere is the most dense, the LATERAL will be just as likely to end up back at the ocean surface than what goes mostly up. It has more atmosphere to penetrate to leave the system.
You are misinterpreting the importance of the direction here. If you move up, say 100ft, this geometry will have slightly more than half of the energy going away from the earth due to the geometry. Of course, the energy ends up heating the rest of the atmosphere at least as much as it is being reradiated or I don’t think we would get nearly as much convection.

Tim Folkerts
August 27, 2011 9:46 am

Bob_FJ,
You make some good points about radiation. Yes, I agree that there is more power radiating laterally than vertically from any chuck of air. The same could be be said for any small chunk of the surface . I don’t agree that this implies that the 390 W/m^2 is overstated.
A very similar (and more familiar to many people who have taken university physics) situation occurs in electrostatics. The electric field lines spreads outward from a point charge in all directions (like those spines on the sea urchin). Integrating the electric field gives the flux Φ = ∫ E·dA (mathematically identical to integrating the photons heading in various directions from the radiation surface). This is known as Gauss’ Law. Any calculus-based university physics text will show have a chapter discussing exactly the sort of question about flux passing thru spheres and boxes.
The key point for this discussion is that while the flux from a point charge spreads out radially (like the radiation from a point on the ground radially), the flux from an infinite plane does not spread out at all, heading out constantly and perpendicularly from the surface. Analogously, if the radiant flux is 390 W/m^2 at the surface of an infinite plane, the flux would be 390 W/m^2 at any distance from the plane.
And yes, the earth is not an infinite plane, but from an elevation of 1 meter or 100 m or even 10 km, the “infinite plane approximation” will be quite close (if you look out of the windows of a jet, the earth’s surface still covers pretty close to 50% of your entire field of view). The details would be easy to work out. My intuition says that even going up 1% of the earth’s radius (~ 64 km), the flux would only decrease about 2% (~ 8 W/m^2) due to the spherical shape of the earth.

kuhnkat
August 27, 2011 9:59 am

Tim Folkerts,
IF we had been measuring this flux across the oceans to a high degree of accuracy for years and saw the DLR slowly increasing without the 390 increasing THAT MIGHT be a problem as that would seem to show the oceans were absorbing energy without warming the surface. Impossible eh?
If the 324 number were decreasing it would seem that the atmosphere was decreasing in temperature right? Of course, without this energy the ocean surface would seem to have to cool and the 390 drop. right?
We do not have enough measurements of the surface, upper 700 meters, atmosphere, the sun, or any other part of the system to have the vaguest idea of whether what we are seeing is good bad or indifferent, AND the accuracy of our measurements aren’t good enough to tell us what we are seeing within what we THINK we know. We have a range of measurements for the diurnal and annual changes that would seem to be much more informative, yet, we seem to get stuck talking about these silly averages that show us nothing about what happens when the fluxes change, which they are doing all the time.
I have no problem with the idea that the IR being emitted from the surface is absorbed and reemitted, partially downward, by things called GHG’s (and that they warm the local parcel through collisions or cool it if it is warmer). I have no problem with the idea that a surface warmer than the GHG’s appear to be absorbing the IR and in its turn, reemitting the IR (unless it is being reflected by a surface that can’t absorb it??). What I have a problem with is the idea that a surface emitting at a higher rate would be raised to an even higher level by this lower density energy. It simply turns the lower density energy around as it is already emitting at a higher density. This would appear to slow the FLUX within the material making up that surface.
What is interesting is that this is NOT taken back the next steps. The sun is irradiating the ocean in depth with an energy density of a particular level. If the flux is slowed in the ocean then the flux into the ocean from the sun would be slowed also as the ocean warms. Everyone seems to agree with the idea that two bodies will eventually come to an equilibrium radiating against each other. This happens throughout the system, not just at the ground air interface. The NET is what is actually being moved between and through the systems. As the system warms the rate of warming slows due to the decreased delta between.
None of this would necessarily preclude conduction warming the next layer down. What makes this idea a non-sequitur is that we have continuous downwelling in the arctic and antarctic replenishing the cold abyssal waters maintaining the general temperature profile. Even with all the undersea volcanoes, what our limited measurements appear to show is a very stable COLD mass of ocean that is NOT changing due to these small surface anomalies.
Then again, without better measurements I could be wrong!!
I simply am not interested in alarmists who called wolf dozens of times before they even knew what was happening and persist as their scam is almost daily shown to have been based on shoddy science and STILL have no solid evidence of a problem. Willis appeared to be worried about why the oceans don’t freeze. If he is worried, he should REALLY worry about what might happen with an effective method of reducing GHG’s in the atmosphere and overshoot!!!!! Frozen oceans and dead plants!!!
HAHAHAHAHAHAHAHAHAHAHAHAHAHAHAHA

Tim Folkerts
August 27, 2011 10:42 am

kuhnkat says:
“What I have a problem with is the idea that a surface emitting at a higher rate would be raised to an even higher level by this lower density energy. ….
If he is worried, he should REALLY worry about what might happen with an effective method of reducing GHG’s in the atmosphere and overshoot!!!!! Frozen oceans and dead plants!!!”

These two statements of yours seem to be completely at odds with each other. First you are saying more GHGs couldn’t warm the earth by increasing their “low density energy”. Then you are saying that less GHGs would cool the earth, which would be due to the loss of that same “low density energy”.
Why should the earth be at such a unique point — where only changes in one direction have an affect?

Tim Folkerts
August 27, 2011 1:34 pm

>>Tim Folkerts said:
>>7. Even a small rise in the surface temperature will lead to a decrease
>>in the conduction from below.”
> Stephen Wilde said:
>That is the exact issue that needs to be resolved.
I am glad we are getting close to agreement. We seem to be focused in on the same issue. The question seems to be:

If 1 W/m^2 of addition “DLR” occurs, the result will be an X °C increase in the ocean temperature.

The “mainstream consensus seem to be that, with no feedback, a doubling of CO2 will lead to 3.7 W/m^2 increase in DLR and subsequently a 1 C increase in temperature. If this were the case, then X = 1/3.7 = 0.27 °C/(W/m^2)
You seem to to be saying that feedback reduces this value to X ≡ 0 °C

However I have come to the conclusion that the combined effect of increased evaporation plus upward convection radiation and conduction from the warmed molecules at the very TOP of the ocean skin removes the energy added by DLR fast enough to prevent any change in the temperature differential at the interface between the ocean bulk and the BOTTOM of the skin layer.

To me, this sounds like “I have a spring so stiff that it prevents any motion when I push on it”. The spring WILL more a bit no matter how stiff (even if that motion is incredibly small). The extra radiation would have to have SOME effect on the water before the water would react with more IR and more evaporation (even if that increase in temperature is incredibly small). I can accept that it may be the case that X << 0.27 W/m^2, but not exactly 0 W/m^2
====================================
I suspect we will not get any closer to an answer in this thread. Stephen presented legitimate arguments why X should be very close to zero — so close that it would represent an insignificant, immeasurable change in temperature. I presented very crude estimate earlier of changes in IR and evaporation that led me to conclude X ~ 0.1 C/(W/m^2). This is smaller than the "no feedback" number = 0.27, which would put this estimate in the "negative feedback" regime. On the other hand, the estimate is mostly just an order-of-magnitude estimate. If I am a factor of 5 high, then X = 0.02, which would be in line with Stephen's estimate. If I am a factor of 5 low, I in in line with "mainstream" estimates of 2C per doubling.
I don't have the time nor expertise nor funding to pursue this to get a more accurate estimate. So as I said a little earlier, we may have to leave it at that …

August 27, 2011 4:52 pm

“To me, this sounds like “I have a spring so stiff that it prevents any motion when I push on it”.”
More like a spring that is bouncy at one end and which becomes progressively stiffer towards the other end.
Note that I am not denying a climate effect. Merely pointing out that the effect is limited to the surface molecules,a portion of the skin and the air.
The climate effect is a slight adjustment of surface pressure distribution but miniscule compared to natural variations forced by solar and oceanic variability. We could never be able to measure it.

Bob_FJ
August 27, 2011 5:35 pm

Kuhnkat & Tim Folkerts, concerning your recent comments on spherical radiation.
I refer you back to your earlier exchange:

Kuhnkat @ August 26, 2011 at 8:00 pm
Tim Folkerts,
“The sides will be radiating to/from other parcels at the same altitude and same (roughly) temperature, so the net radiation would be zero sideways. In this sense, the sides are of no interest”
Yes, [said Kuhnkat] this is the NET that is SUPPOSED to be computed when things emit at each other…”

And, repeating, the S-B law gives the TOTAL emitted radiant energy for a single flat body, which of course is hemispherically distributed. It does NOT give radiation only normal to the surface.
As can be visualized in the sea urchin illustration, a very large proportion of its spines point laterally through 360 degrees, not upwards. (In conceptual terms, treat 0 to 45 degrees as lateral and 45 to 90 degrees as vertical). Translating these spines as representing radiation, it follows that adjacent elemental parcels in an elemental layer give the net result as you describe above. Also, adjacent ELEMENTAL layers contain lateral radiation that deviates slightly from the horizontal but still meet and result in a similar outcome between such layers. (up & down) Thus we have a lot of the original S-B calculated emission that does not comprise part of the vertical effect you (Tim) are looking for, or in other words, the vertical components are less than the total S-B emission. It might be interesting to someone to pursue the maths, and integrate the vectors etc, but in my old age, I’ve lost pleasure in such activity long ago.
Kuhnkat, I’m sorry but your most recent comment; me no understand
Tim, generally speaking, I don’t like analogies. The more complicated they are, the greater the tendency is for them to become problematic. I think your latest is inappropriate for this forum, and we don’t want to wander off into new territory do we?

Tim Folkerts
August 27, 2011 9:12 pm

Bob_FJ,
Gauss’s Law is not rally an analogy — it is exactly the same math applied to a different situation. In that sense it is much better than analogies about urchins or golf balls.
When you say :

“It might be interesting to someone to pursue the maths, and integrate the vectors etc, but in my old age, I’ve lost pleasure in such activity long ago.”

quess what — this IS the integral I was pointing you to with gauss’s law. And the result is very clearly not what you suggest when you say

“Thus we have a lot of the original S-B calculated emission that does not comprise part of the vertical effect you (Tim) are looking for, or in other words, the vertical components are less than the total S-B emission. “

If 390.00 W/m^2 leave from everywhere on the surface, then 390.00 W/m^s WILL be going up thru a 1m x 1m square mounted 1 meter above the surface.

Bob_FJ
August 28, 2011 12:41 am

Dear moderator/webmaster, this is driving me absolutely nuts! My mobile (Oz-wireless) laptop (notebook) computer has almost become stationary on this very long myrrhed thread. Please can we have a page #2 continuation to speed things up?
Regards, Bob-FJ
REPLY: Sorry, no facility for page 2 on a single thread, I’d have to change the whole blog settings. More laptop memory or a bigger swap file or different browser might help – Anthony

kuhnkat
August 28, 2011 7:33 am

Tim Folkerts,
Slow the cooling Tim. Yes that means it is warmer that it would be otherwise but that is confusing terminology and should not be used.

Bob_FJ
August 28, 2011 3:33 pm

Tim Folkerts @ August 27, 2011 at 9:12 pm
Quickly, returning to S-B and first principles:
The hemispherical emissions from the surface can be replicated by perpendicular and horizontal vectors. The integration of the latter will give a greater quantity than that normal to the surface. The horizontal is PART OF your 390 W/m^2 radiative energy leaving the surface, but does NOT go up. Thus from first principles, the Trenberth cartoon shows a vertical number that is far too high
A further difficulty with S-B is that it only describes single body emissions and does not embrace consequent absorptive effects for a large body immersed in air. On Earth, most of the initial emissions will be absorbed close to the surface, some very close, and this has consequences as the emission mode switches to spherical. And of course the atmosphere has myriad layering effects. Lots of stuff going on that disturbs the radiative considerations quite markedly.

Tim Folkerts
August 28, 2011 5:02 pm

“The horizontal is PART OF your 390 W/m^2 radiative energy leaving the surface, but does NOT go up. Thus from first principles, the Trenberth cartoon shows a vertical number that is far too high.
From first principles, consider a large grid of 1m x 1m squares on the ground, each radiating 390.0 W. (Or a sea urchin with 390 quills placed in every grid. Or 390 1W lasers pointing out randomly from each square). Then consider a 1m x 1m grid somewhere above the ground — perhaps 1 m up or 1000 m up.
I agree that the 390.0 W from one square will not all pass thru the 1m x 1m square in the grid directly above it. Some will go sideways to varying degrees and pass thru some OTHER part of the upper grid. (None will go absolutely horizontally because then it would not be emitted from the surface to begin with).
But the upper square that we are looking at will also GAIN some photons from OTHER parts of the lower grid that are radiating sideways (this is the part you seem to have neglected). The sideways radiation can’t just disappear. By symmetry, in a transparent atmosphere, the total thru the upper grid must be 390.0 W for each square meter.
Now the atmosphere is not transparent. This is where the “350 W/m^2” and “40 W/m^2” come into play. The calculations would get a bit involved to know how the “sideways” component factors into the absorption, but I will assume that the clever people who put together MODTRAN and similar calculations were aware of this and built it into the program. But if you think they are wrong, then you are welcome to perform a detailed analysis of there methods looking for errors.
And the earth is not flat. The 390 W from 1 square meter will indeed spread out SLIGHTLY as you go up. But this will result in a correction of no more than ~ 2% up to 60 km high and 0.02% up to 6 km high. Most of the action for absorption occurs in the first 6 km, so the errors from the curvature of the earth should be negligible. (And again, there is a good chance that this is indeed built into the models. If not, it could be a potential source of systematic error.)
TO WRAP UP: If people truly think they have a better grasp on radiation than Trenberth and/or the people who put together MODTRAN, then do the math in detail and find the errors.

Bob_FJ
August 28, 2011 6:09 pm

Tim Folkerts [1] @ August 27, 2011 at 9:46 am and [2] @ August 27, 2011 at 9:12 pm

“…[1] If 390.00 W/m^2 leave from everywhere on the surface, then 390.00 W/m^2 WILL be going up thru a 1m x 1m square mounted 1 meter above the surface…”
“…[2]Analogously, if the radiant flux is 390 W/m^2 at the surface of an infinite plane, the flux would be 390 W/m^2 at any distance from the plane…”

The TOTAL S-B radiation from the surface is hemispherically distributed, and it is clear that the total radiation that would pass through your square metre window frame above the surface would reduce with increasing height of the frame. That would still be true with a transparent atmosphere, but more so with GHG’s present. At your 1-metre high target, there would be interception of radiation from other unit areas on the surface, let’s say 10 metres or so away, but these rays would be ever more approaching the horizontal, and aiming at a diminishing target (in perspective). There becomes a point when this is negligible, especially when the shorter photon free path lengths are involved.

Bob_FJ
August 28, 2011 7:24 pm

Tim Folkerts @ August 28, 2011 at 5:02 pm

“…but I will assume that the clever people who put together MODTRAN and similar calculations were aware of this and built it into the program. But if you think they are wrong, then you are welcome to perform a detailed analysis of there methods looking for errors…
…TO WRAP UP: If people truly think they have a better grasp on radiation than Trenberth and/or the people who put together MODTRAN, then do the math in detail and find the errors.”

Ah, I see the anticipated appeal to authority had to come out eventually!
Coming back to those first principles and vectors, since you don’t discuss it, are you claiming that in vector maths as in the following simple graphic, that F does not equal Fx + Fy?
http://0.tqn.com/d/physics/1/G/8/-/-/-/vectorcomponents.jpg

“…Then consider a 1m x 1m grid somewhere above the ground — perhaps 1 m up or 1000 m up…
…The sideways radiation can’t just disappear. By symmetry, in a transparent atmosphere, the total thru the upper grid must be 390.0 W for each square meter…”

So you are saying that the hemispherical radiation from the surface is somehow transformed entirely into polarised radiation normal to the surface? Or, put another way, in the vector diagram above that Fx is transformed to the vertical and is added to Fy? Or F goes vertical? What ARE you saying?
BTW, I think that JAE said somewhere above that MODTRAN does NOT support your 390 W/m^2
Oh, and we have already agreed that The sideways radiation can’t just disappear…. it just goes sideways with greater intensity than vertically.

Tim Folkerts
August 28, 2011 7:25 pm

Radius of earth = 6371 km = 6371000 m (from the internet, but the exact value doesn’t really matter)
Area of earth = 4 pi r^2 = 5.100645 E+014 m^2 (assuming a perfect sphere)
power radiated = 390 W/m^2 * 5.100645 E+014 m^2 = 1.989251E+017 W (using 390 W/m^s as the average for the surface)
By conservation of energy (in a transparent atmosphere where no photons get absorbed), if I draw a sphere around the earth, all of the photons that leave the surface (ie 1.989251E+017 W J worth of IR photons every second), must pass thru the upper sphere.
The following table will probably lose formatting when pasted here, but I’ll give it s shot…
LOCATION RADIUS AREA (m^2) POWER (W) POWER/AREA (W/m^2)
+1 m 6371001 5.100646E+014 4.973129E+016 389.9998776
+10 m 6371010 5.100661E+014 4.973129E+016 389.9987757
+100 m 6371100 5.100805E+014 4.973129E+016 389.9877573
+1000 m 6372000 5.102246E+014 4.973129E+016 389.8775991
The power stays the same (since energy cannot disappear as the photons travel along thru a transparent atmopshere). The area increase SLIGHTLY. So the Power/Area decreases only slowly with altitude. At 1000 m up, the difference is only 0.03 % — very much in agreement with my estimate earlier of 0.02% @ 600 m.
There is no need to look at a “switch from flat to spherical”. The total power is a much simpler approach. (But the “Gauss’ Law” approach will give the correct answer when applied to flux from the surface heading out in spherical direction.)
In a real atmosphere, some of the photons will get absorbed, which will indeed reduce the flux. But that is accounted for in the “Trenberth cartoon”.

jae
August 28, 2011 7:36 pm

BOB_FJ stated:
“Yep, but I don’t think we can win the war by questioning if there is a GHE though. I prefer to go along for example with atmospheric physicist and elitist alarmist Andrew Dessler who has admitted that if sceptics like Roy Spencer can show that there are major negative feedbacks, then AGW is no problem.”
I fail to see what your point is. There IS no PROOF OF A GHE, period! In fact the current information suggests that there is absolutely NO effect from GHEs!
It is time that Willis and all of the other “WARMISTAS” PROVE their conjecture, not simply draw “shell diagrams” and radiation cartoons. Science DEMANDS EMPIRICAL EVIDENCE and it is up to the proponents of this nonsense to prove, EMPIRICALLY, that a GHE exists. THAT IS THE VERY HEART OF SCIENCE, which seems to have been abandonded by the current disgusting “intelligencia elitista,” I am surprised (but not too shocked) that Willis may be a part of that group. There is absolutely NO PROOF/DEMONSTRATION OF A GREENHOUSE EFFECT ON PLANET EARTH OR ANY OTHER PLANET. IT IS TIME TO ADDRESS THIS PROBLEM, WILLIS/AGW FOLKS!!!!!!!!!!!

Tim Folkerts
August 28, 2011 8:42 pm

Bob_FJ says: August 28, 2011 at 7:24 pm
Ah, I see the anticipated appeal to authority had to come out eventually!
No, there was no appeal to authority. I never said they were right simply because they were authorities. In fact, I said quite the opposite. I challenged you to show that they, in fact, NOT authorities. Show that they are wrong and I will believe you instead.
Coming back to those first principles and vectors, since you don’t discuss it, are you claiming that in vector maths as in the following simple graphic, that F does not equal Fx + Fy?
http://0.tqn.com/d/physics/1/G/8/-/-/-/vectorcomponents.jpg

Assuming you mean vector addition, then yes, F = Fx + Fy
So you are saying that the hemispherical radiation from the surface is somehow transformed entirely into polarised radiation normal to the surface? Or, put another way, in the vector diagram above that Fx is transformed to the vertical and is added to Fy? Or F goes vertical? What ARE you saying?
I am saying that if 390 W/m^2 pass up from every square meter of the the surface, then 389.88 W/m^2 will pass up thru every square meter of horizontal area 1000 m above the surface. The vertical component remains ~ 390 W.m^2 (My comment with the calculations was probably being moderated when you wrote this).
BTW, I think that JAE said somewhere above that MODTRAN does NOT support your 390 W/m^2
Rather than appealing to the authority of JAE, why not go to the source? Here is a link:
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.html
Set the sensor to look down from 0 km and you will get ~ 390 W/m^2 (depending on where on earth you are — more power in the tropics; less at the poles)
Oh, and we have already agreed that The sideways radiation can’t just disappear….
See my previous post that does indeed calculate the total energy, not letting any energy disappear or appear.

Bob_FJ
August 28, 2011 10:16 pm

JAE, @ August 28, 2011 at 7:36 pm
I agree scientifically with you that there is no empirical evidence of a greenhouse effect, and everything you said, but my point was that the war is largely a political one, and I don’t think that arguing about it will influence it in any way. It is proven in the lab that CO2 does absorb IR, but that has little relevance in the real atmosphere and all the other stuff that’s going on in a much bigger environment. Unfortunately, the church conflates that point together with speculations and modelling, rather successfully, supported by “thousands of scientists“ HaHa at the IPCC. I think that the only way that that part of the war can be won, is if eventually there is agreement that significant negative feedbacks are proven and exaggeration in the models in positive feedbacks are removed. That is not to say that I disagree with patiently reminding EVERYONE that there is NO empirical evidence, or that it is not scientifically interesting to pursue the matter with alarmists.
Similarly, with the Aspley/Apsley (?) pre-paper release that increasing use of fossil fuels does not result in the CO2 increase observed at ML, even if the final paper is overwhelmingly plausible, will probably not have any benefit politically. Simple minded politicians and MSM etc will be totally unable to believe that the “vast amount” of human pollution we pump into the atmosphere is not the culprit.
Sad really!

Bob_FJ
August 28, 2011 11:51 pm

Tim Folkerts @ August 28, 2011 at 7:25 pm AND August 28, 2011 at 8:42 pm
Thanks Tim, I appreciate that you have put some good effort into your two recent posts, but I’m struggling to understand the relevancies, and I‘ll comment rather briefly WRT the main issue that you have NOT responded to.
The fundamentals that I come back to that you seem to evade is that in vector maths, whilst you seem to understand the basics, (Like you agree that Fx + Fy = F), you don’t seem to be able to apply that to S-B hemispherical radiation from a flat surface.
In any vertical plane relative to plan view, there is a range of radiation relative to the surface that can be described in terms of inclination. These can be transcribed to horizontal and normal (vertical) vectors for each ray of radiation. In turn, these can be integrated into normal and horizontal vector sums. Thus there are integrations describing both normal and 90 degree opposed horizontal radiation.
Please explain why you propound that the major horizontal vector integration is somehow transformed to a normal presentation? (rather than being TOTALLY separate and irrelevant to the normal)

Tim Folkerts
August 29, 2011 5:43 am

JAE:
Wikipedia says “The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases, and is re-radiated in all directions. Since part of this re-radiation is back towards the surface, energy is transferred to the surface and the lower atmosphere. As a result, the temperature there is higher than it would be if direct heating by solar radiation were the only warming mechanism.”
Is this what you mean by “greenhouse effect”?
* If so, what specifically do you disagree with in this statement?
* If not, what is your definition of “greenhouse effect”?

Myrrh
August 29, 2011 5:47 am

Spector says:
August 26, 2011 at 11:08 am
RE: Myrrh: (August 23, 2011 at 2:17 pm)
Water is completley transparent to Visible, not proven in any way that it isn’t, you can see straight through clear water.. Visible doesn’t have the mechanism to interact even on an electron scale with water, is really transparent, it is transmitted through unchanged. Unlike the atmosphere which isn’t really transparent, because of absorption by electrons to produce reflection and scattering.
You may be confusing statements that only apply to free water molecules in the atmosphere where the energy states are limited to the natural vibration modes of the molecule in free fall. Other vibrations may be possible for a short time when two molecules are bouncing together. In a gas this is happening so rarely that sunlight can reach the surface of the earth with *almost* no attenuation due to this cause. But in liquids and solids, the molecules are all so tightly bound that this is happening much more frequently and a wide range of possible vibration modes are allowed. This is why a solid can be black with a uniform spectrum of resonances.
? What is it with you lot? What is so difficult to understand that this subject has been thoroughly investigated and it is known scientific fact that water is transparent to visible light, it does not absorb visible light, it remains unaffected by visible light. The science of visible light is optics.
In fact, all that you claim for visible light in the atmosphere. Which is not transparent to visible light, in real physics. The atmosphere is not transparent to visible light because the molecules of nitrogen and oxygen reflect/scatter visible light, that’s why we a have a blue sky.. Visible light works on an electon transition level, in the atmosphere visible light hits electrons of nitrogen and oxygen and is briefly absorbed and sent out again.
These principles are bog standard science facts. I have given you explanations of them. I am at a loss to understand why you’re not showing any comprehension. For pity’s sake, read the effin section I posted on the difference between electronic transitions of Visible/UV and the vibrational moving of whole molecules by thermal infrared. You (generic) appear to have no concept of scale, among other things..
Molecules in solids and liquids are bound together by electric and magnetic forces. Unlike the neutrino, photons, by nature are electromagnetic and they easily interact with these bonds.
If you looked at that graph at the top of the Wikipedia page, you would see that the absorption coefficient for light in water is in the range of about 0.0001 to 0.01 per centimeter. This means that over distances between 100 cm (red) or 10,000 cm (blue), the light energy will be reduced to about 36.8 percent of its original level (one divided by the base of the natural logarithms) and thus 63.2 percent is absorbed, or converted to heat along the way. When anyone says water is ‘perfectly transparent,’ they usually are referring to distances that are less than 10 cm.

As water is actually transparent to visible light, that might just raise a question of ‘what is really happening here?’ – it is not any proof that water is not transparent to visible light. Visible light is passed through water. That’s what transparent means. It is transmitted unchanged. That’s what transmitted means. It is not absorbed, that’s what not absorbed means.
The atmosphere is not transparent to visible light, reflection/scattering proves it is not transparent to visible light, the electrons of nitrogen and oxygen absorb visible light – so how much is visible light heating the atmosphere?
You can’t say it doesn’t, because you claim that absorption creates heat.
You are in such a muddle about this because because, you have no concept of differences.
So, where’s the amount of heat shown in the ‘energy budget’?
And damn well give me a proper answer. I have shown that reflection/scattering is an outcome of absorption.
Light does not ‘fade away.’ Photons are packets of energy. Their motto is “We Deliver.” Light intensity is measured in watts per square meter for a continuous flow. That is potential heating power. A standard measure of heat energy is the BTU. A heat flow of one BTU/sec is equivalent to about 1055 watts.
Yeah right, you see nothing wrong in creating perpetual motion machines in ‘backradiating’ creating runaway heating effects, so of course you see nothing untoward of creation of an effect producing eternal light, as you see nothing untoward in ‘net flow from hot to cold in radiation by colder also heating the warmer’ means that a volume of colder can raise the temperature of a warmer..
Nuts. All these ideas you keep repeating are just nuts, bearing no relation whatsoever to reality, to real physical properties and processes.
The confusion about infrared being heat energy is common. It is terrestrial temperature range heat radiation. Light is solar temperature range heat radiation. If you look at a plot of the solar spectrum, you will see the vertical axis labeled in watts per square meter per frequency or wavelength interval peaks out in the middle of the optical range.
There’s only confusion in you and your ilk. You’re confusing peak with power to do work. The visible has the power to move electrons, to enable chemical conversion, to sugars as in photosynthesis. It doesn’t have the power to move molecules into vibration in resonance. Thermal infrared, heat on the move, heat energy, does. The great heat that the Sun sends to the surface of the Earth is the invisible thermal infrared. Thermal infrared heats water and land. It goes through windows. You can feel it..
You’re all so wrapped up, I almost typoed warped.., in your irrational memes from AGWScience Inc that you don’t see how much nonsense you’re spouting.
So, I insist: I’ve given you all the method by which visible light is scattered by absorption in the atmosphere, which falsifies your claim that the atmosphere is transparent to visible. How much heat does this create? Why isn’t it noted on your ‘energy budget’?

RJ
August 29, 2011 9:06 am

Tim Folkerts says:
August 29, 2011 at 5:43 am
“if direct heating by solar radiation were the only warming mechanism.”
It is NOT a warming mechanism. At best it might very slightly slow the cooling rate. The sun does the warming nothing else (except in an unusual event when the atmosphere is warmer than the surface).

RJ
August 29, 2011 9:16 am

And something like this would be clearer and more accurate
GHG backradiation causes the average surface temperature over time to be slightly higher than it would be due to reduced cooling at the surface. The reduced cooling and the resulting increased average surface temperature however is very minimal.

August 29, 2011 9:56 am

RJ–reduced cooling might limit temperature extremes over a 24-hour period, but it cannot do anything to the average temperature.

RJ
August 29, 2011 11:35 am

Ken Coffman says:
August 29, 2011 at 9:56 am
Wouldn’t limiting temperature extremes do just that. (impact on the average temperature over a period eg one day or week)

Tim Folkerts
August 29, 2011 12:06 pm

Ken Coffman says:
August 29, 2011 at 9:56 am
RJ–reduced cooling might limit temperature extremes over a 24-hour period, but it cannot do anything to the average temperature.
I can reduce the rate of cooling of my house by adding insulation that limits how easily energy can escape (assuming the temperature outside is cooler than inside). For a given output from my furnace, the insulation will make make my house warmer than it would be without the insulation.
I can reduce the rate of cooling of my house the earth by adding insulation GHGs that limits how easily energy can escape (assuming the temperature outside of outer space is cooler than inside the temperature of the earth). For a given output from my furnace the sun, the insulation GHGs will make make my house the earth warmer.
Once you have concluded that GHGs do indeed limit cooling, you have logically concluded they make he earth warmer than it would be with no greenhouse gases. (You can then ask HOW MUCH it will warm, but that is a different question. A much more complex question. A question with many additional variables that come into play)

Tim Folkerts
August 29, 2011 12:15 pm

Double Dang! There are a few mistakes in the strike-throughs above. It should read

I can reduce the rate of cooling of my house the earth by adding insulation GHGs that limits how easily energy can escape (assuming the temperature outside of outer space is cooler than inside the temperature of the earth). For a given output from my furnace the sun, the insulation GHGs will make make my house the earth warmer than it would be without the insulation GHGs.
Once you have concluded that GHGs do indeed limit cooling, you have logically concluded they make he earth warmer than it would be with no greenhouse gases. (You can then ask HOW MUCH it will warm, but that is a different question. A much more complex question. A question with many additional variables that come into play)

I HOPE that fixes it all !

August 29, 2011 12:51 pm

Tim Folkerts says:
August 29, 2011 at 12:15 pm
Double Dang! There are a few mistakes in the strike-throughs above. It should read
I can reduce the rate of cooling of my house the earth by adding insulation GHGs that limits how easily energy can escape (assuming the temperature outside of outer space is cooler than inside the temperature of the earth).
So what is the R value of the “insulating GHG’s”?

Tim Folkerts
August 29, 2011 5:31 pm

RJ says: August 29, 2011 at 9:16 am

And something like this would be clearer and more accurate
GHG backradiation causes the average surface temperature over time to be slightly higher than it would be due to reduced cooling at the surface. The reduced cooling and the resulting increased average surface temperature however is very minimal.

Since you are making a claim that the effect is “minimal”, can you provide an estimate of how much impart GHGs make? 0.1 K? 1K? 10 K. And what calculations do you base this on? (Be specific). The estimates I have seen range from ~ 15 K to 30 K — what errors do you know of in these estimates?

August 29, 2011 5:59 pm

Tim Folkerts,
Here is as good an answer as you wil get, from the foremost expert in the field of Climatology: click

jae
August 29, 2011 6:38 pm

Tim Folkerts says:
“Wikipedia says “The greenhouse effect is a process by which thermal radiation from a planetary surface is absorbed by atmospheric greenhouse gases, and is re-radiated in all directions. Since part of this re-radiation is back towards the surface, energy is transferred to the surface and the lower atmosphere. As a result, the temperature there is higher than it would be if direct heating by solar radiation were the only warming mechanism.”
Is this what you mean by “greenhouse effect”?
* If so, what specifically do you disagree with in this statement?
* If not, what is your definition of “greenhouse effect”?”
Well, here’s my take: I should be more specific by saying that I don’t accept the “radiative atmospheric greenhouse effect.” There probably is a “greenhouse effect” that involves heat storage by the oceans and atmosphere. It is the radiative concept that bugs me, because I can find absolutely no empirical evidence for it. As I have noted a hundred times, if there were a “radiative atmospheric greenhouse effect,” we should be able to observe it. My favorite example is a comparison of the temperature between Phoenix and Atlanta (same latitude and elevation) in August. Phoenix is MUCH hotter, despite the fact that Atlanta has 3 times as much “greenhouse gases.” WHY? So let’s chalk that up to energy lost by evaporation of water. BUT, then, why is a greenhouse constructed of IR-transparent material in Atlanta colder than a similar one in Phoenix? Just WHERE is the evidence for this GHE?
Now, contrary to what Willis said, I DO NOT SAY THERE IS NO BACKRADIATION. I simply do not know WHY there is no “radiative atmospheric greenhouse effect.” It might be that it is “canceled” by convection, or it may have something to do with the mechanisms involved with LTE. But it has not been demonstrated, so it is not yet scientific! Period.

jae
August 29, 2011 7:10 pm

Oh, and Tim:
I did not EVEN get into the FACTS about other planets with atmospheres. They also demonstrate that there is no “magic” associated with GHGs. Have you read the Dragon?

jae
August 29, 2011 7:18 pm

PPS: The comments I have expressed above have been virtually IGNORED for over 3 years, now. I just gotta ask WHY?
WILLIS?

kuhnkat
August 29, 2011 8:56 pm

jae,
“I did not EVEN get into the FACTS about other planets with atmospheres.”
Don’t need no stinkin’ atmosphere. Pretty much the whole solar system warmed over the period the Alarmists were ranting about CO2, GHG’s, and H2O feedbacks!!!! How do GHG’s warm rocks with no atmosphere??

Bob_FJ
August 29, 2011 9:28 pm

JAE @ August 25, 2011 at 8:26 pm

“…One thing that fascinates me, though, is that MODTRAN doesn’t give anywhere near that 390 number, if you enter 0 (zero) for altitude, looking down, -10 offset in temp. One would think that the model would register “tilt” when one enters those parameters–or give me 390. Oh well!…”

Could you please elaborate on this, whilst I remain puzzled at Tim Folkerts’ evasion of the question I raised of horizontal vectors in hemispherical (S-B) surface radiation. Why/what is your 10 offset in T….. it shouldn‘t make a huge difference in absolute T^4 calculation anyway though? Are you suggesting the calculator is faulty?
Are you using the same model calculator that he has referred to:
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.html
Incidentally, in that thingy, one is allowed to enter different CO2 levels, but there is no difference in results between 280 and 392 PPM….. Maybe it is an invalid entry even if allowable of course, but why do they have that alterable field? Ho hum.

Bob_FJ
August 29, 2011 9:42 pm

Jae @ August 29, 2011 at 7:18 pm

“PPS: The comments I have expressed above have been virtually IGNORED for over 3 years, now. I just gotta ask WHY?…”

Yep, I know how you feel.
(We have no right to argue against authority. Although for but one example, the Japanese in particular might recently agree with Wegener on his theory of tectonics and continental drift, whereas he was dismissed by the then geological church)

kuhnkat
August 29, 2011 9:45 pm

Bob_FJ,
try using your calculator with .1 and .01. I get slightly LESS radiation for the .01 than the .1. What Tim has been trying to tell you about the horizontal stuff. They take it into account. Close to the ground there will be less area radiating through the space horizontally.

Tim Folkerts
August 29, 2011 9:46 pm

I have glanced at Slaying the Skydragon — well the free chapters. I was no especially impressed, and I have seen rather mixed reviews from others as well. It seems other long on philosphpty and short on math.
One particular statement that jumped out as I was jsut glancing again was:
“Backradiation would correspond to audio feedback ”blowup” between
a microphone and loudspeaker without amplifier energy input, which cannot
happen because inevitable losses prevents blowup without energy input.”
This seems like a complete mis-interpretation of “backradiation” to me.
PS I agree 100% that there is no “magic” associated with the GHE.

Tim Folkerts
August 29, 2011 10:51 pm

Bob,
I’m not being evasive about horizontal components — mostly running out of time.
In the end, we are looking at radiative TRANSFER of energy. To calculate the transfer from one region to another, we need a dividing line between the two regions and a calculation of the total flux passing thru that surface. That is exactly what I gave you. And exactly what Gauss’ Law gives you. One region is the ground (and ocean); the other is the atmosphere. The simplest case is to draw the line right at ground level. Then it is painfully clear that 390 W/m^2 is indeed leaving the ground and entering the air . The energy flying around within a region (eg horizontally) may be interesting in some ways, but is not part of the TRANSFER between regions.

tallbloke
August 30, 2011 12:34 am

Tim Folkerts says:
August 26, 2011 at 6:59 pm
At this point, we could discuss the EXTENT of the warming of the bulk due to an increase in the downward IR. IR might be inefficient at warming the oceans, but IR logically has SOME effect unless one or more steps above are incorrect. Perhaps 10 W/m^2 extra only warms the surface 0.01 K, leading to a 0.01 K increase in the bulk temperature to restore the gradient.

Hi Tim,
Other processes are swamping increases in DLR from extra co2. For example, it is estimated that OLR has increased some 6W/m^2 since 1948 (NCEP). That has been counteracted by increased insolation due to reduced cloud cover 1983-1998 of 4Wm^2 and who knoes what before. (ISCCP)
What is the evidence that DLR has increased at all?
If it has, then as you say, there could conceivably be a small increase in skin temp. I suggest this is a function of increased air temp, because rising ocean T is going to increase evap anyway. This is a very, very slow way to heat the bulk of the ocean. It’s not going to account for the rise in T of the top 700m that occured ~1980-2003 before the oceans started cooling again.
It’s the Sun shining through less clouds that did that mate.
Enjoy what’s left of global warming.

tallbloke
August 30, 2011 12:38 am

Myrrh says:
? What is it with you lot? What is so difficult to understand that this subject has been thoroughly investigated and it is known scientific fact that water is transparent to visible light, it does not absorb visible light, it remains unaffected by visible light. The science of visible light is optics

What is it with people who deny empirically measured quantities?
The coefficients of absorption were provided. They show how visible light is absorbed in water and changed to heat. That’s why deep sea divers take torches with them.

Green Sand
August 30, 2011 1:28 am

tallbloke says:
August 30, 2011 at 12:34 am
It’s the Sun shining through less clouds that did that mate.

————————————————————————————
Interesting sunshine hours and temperature charts at the Met Office re the last 80+ years in the UK.
“Rainfall, sunshine and temperature time-series”
http://www.metoffice.gov.uk/climate/uk/actualmonthly/
The UK – Annual – “Mean Temperature” and the “Sunshine Hours” appear to correlate very closely, sort of makes sense, more sunshine = warming, less = cooling? Admittedly a gross over simplification, but it is actual observational data.

David A
August 30, 2011 4:10 am

Tim Folkerts says:
August 26, 2011 at 6:59 pm
“At this point, we could discuss the EXTENT of the warming of the bulk due to an increase in the downward IR. IR might be inefficient at warming the oceans, but IR logically has SOME effect unless one or more steps above are incorrect. Perhaps 10 W/m^2 extra only warms the surface 0.01 K, leading to a 0.01 K increase in the bulk temperature to restore the gradient.”
Tim, the ocean can cool even if the gradient decreases. Simply decrease the amount of SWR entering below the surface and you both cool the ocean and decrease the gradient. Any increase in water vapor reduces SWR reaching below the surface, (via spectral modification of TSI) despite the GHG effect on the atmosphere. Thus water vapor temporairly increases the energy in the atmosphere, while reducing the longer residence time SW energy entering the oceans, thus over time the overall feedback may be negative.

Myrrh
August 30, 2011 4:15 am

tallbloke says:
August 30, 2011 at 12:38 am
Myrrh says:
? What is it with you lot? What is so difficult to understand that this subject has been thoroughly investigated and it is known scientific fact that water is transparent to visible light, it does not absorb visible light, it remains unaffected by visible light. The science of visible light is optics
What is it with people who deny empirically measured quantities?
The coefficients of absorption were provided. They show how visible light is absorbed in water and changed to heat. That’s why deep sea divers take torches with them.

It’s pretty well meaningless when water is transparent to visible and doesn’t absorb it. Since water is transparent to visible and transmits it unchanged then whatever that is saying it can’t be saying that it is heating up the water when it says ‘absorbed’, or, there is something else happening.
That water is a transparent medium for visible light is a fact of real physics. It doesn’t get absorbed, it doesn’t get absorbed because water doesn’t absorb it. This is a physical known. It is what is. The process is understood.
That water ‘disappears’ at certain depths does not mean that it is being ‘absorbed’ in the technical meaning of this, that the molecules of water are actually taking in the energy. Because, water doesn’t. Water is a transparent medium for visible light. That divers at depth need to take a torch with them is supposed to disprove that??!!
For goodness sake.
What I am also saying is that the claim that the atmosphere is transparent to visible light is nonsense. Read the wiki on the difference between UV/Visible and Thermal – they are different from each other. They work on different levels. Light scattering in the atmosphere is a result of absorption – the molecules of nitrogen and oxygen absorb visible on an electron scale and send it back out again.
So, there are two things here. Prove that visible light from the Sun can actually heat water. And yes, you could try Tim’s LED… See how long it takes to get your bath up to a temp you can have a good hot soak in.. And, give me the amount of heat that is being generated in the atmosphere from the known physics of absorption of visible light, electronic transition, which results in reflection/scattering, since you claim that ‘absorption’ means ‘creation of heat’.
You really won’t understand what I’m saying until you think about this by actual properties and processes.
That you have bought into the AGWScience Fiction Inc.’s meme that Visible light is a thermal energy capable of converting to heat land and oceans, is not my fault. That you persist in reading every mention of ‘absorbed’ through that paradigm, is not my fault. That you keep rejecting what I’m trying to explain to you, is not my fault. It is my fault that I keep nagging at y’all to look at this objectively, I’m sorry you find it irritating.
I care about this aspect of the AGW con. It means that a whole generation of children now don’t understand the difference between Light and Heat. This is a controlled dumbing down of the population, just the same as the perversion of science by demonising carbon dioxide and the general postmodernism of ‘science is just another social construct’, where all scientific objectivity is denied in its postmodern reality and absorbed can mean whatever you want it to mean.
That water is transparent to visible light is an objective scientific fact.
That the atmosphere is not transparent to visible light is an object scientific fact.
The physics of this are very well known, in traditional science. Whatever else is happening must be related to the tried and tested basics.

tallbloke
August 30, 2011 6:17 am

Good link, thanks.
I’ll put a comparison up for discussion on my site.
I ran this last year:
http://tallbloke.wordpress.com/2010/06/21/willie-soon-brings-sunshine-to-the-debate-on-solar-climate-link/

ferd berple
August 30, 2011 7:29 am

My question is a simple one. If the DLR is so much greater than solar radiation, then why are the poles so much colder than the equator? The small difference in total radiation (solar + DLR) between summer and winter should mean that we don’t experience any large difference in temperature between winter and summer.
Especially Antarctica? The DLR should heat the Antarctic substantially during the winter, when their is no sunshine, preventing the extreme cold we see at the poles. During summer, since the angle of incidence is low, the solar radiation should have almost no effect. Antarctica should be about the same temperature summer and winter.
Since we don’t see this, this would appear to be direct, first order evidence that DLR is not significantly heating the planet. If it was, if DLR was so much greater than solar radiation, then the Antarctic in winter should be about the same temperature as the Antarctic in summer. Europe in the winter should be about the same temperature as Europe in the summer.

Tim Folkerts
August 30, 2011 9:52 am

ferd berple says:
August 30, 2011 at 7:29 am
“The DLR should heat the Antarctic substantially during the winter, when their is no sunshine, preventing the extreme cold …”
Outer space is 3 K. With no sunshine and only outward radiation, the surface should cool toward 3 K (-270 C). I would say the radiation from the atmosphere is QUITE effective at preventing the extreme cold (-270 C), keeping it at a much “balmier” ~ -55 C during the winter at the South Pole (http://en.wikipedia.org/wiki/South_Pole#Climate.2C_and_day_and_night). During the summer, the sunlight can push this all the way up to ~ -25 C (Yes, there are other effects at play, but “DLR” is certainly one significant factor limiting the cooling at the poles during the winter.)

tallbloke
August 30, 2011 10:22 am

Tim Folkerts says:
August 30, 2011 at 9:52 am
Outer space is 3 K. With no sunshine and only outward radiation, the surface should cool toward 3 K (-270 C). I would say the radiation from the atmosphere is QUITE effective at preventing the extreme cold (-270 C), keeping it at a much “balmier” ~ -55 C during the winter at the South Pole

Tim, please. Convective transport from the equator is what supports the temperatures at the poles, not radiation, of which there is comparitively little.

tallbloke
August 30, 2011 10:29 am

Myrrh says:
August 30, 2011 at 4:15 am
It’s pretty well meaningless when water is transparent to visible and doesn’t absorb it.

Empirically determined coefficients of absorption are strong evidence that water does indeed absorb visible light. Better evidence than anything you’ve come up with to say it doesn’t.
That the atmosphere is not transparent to visible light is an object scientific fact.
Yeah? How come I can see for 50 miles or more from a mountain top on a clear day, and yet it’s completely dark a lot less than a mile under the surface of the sea?
??

A. C. Osborn
August 30, 2011 11:04 am

jae says:
August 29, 2011 at 6:38 pm
Now you have really done it, confusing people with observed facts, that is not how science is done, you must have equations and models. Sarc off

kuhnkat
August 30, 2011 1:05 pm

Tim Folkerts,
“This seems like a complete mis-interpretation of “backradiation” to me. ”
The Slayers mostly address the mythology that has built up in place of the science. The IPCC types can scream about this approach all they want. If they had communicated the science better initially there wouldn’t be all those poorly understood myths out there!! Of course, if they had tried to explain the REAL science they wouldn’t have gotten anywhere in the first place!! They apparently made the decision to allow the myths to propagate in place of the science and are now paying for that decision by losing support.
You can call it knocking down strawmen, but, when the majority of the population believes those strawmen…

Tim Folkerts
August 30, 2011 1:13 pm

JAE says:
“My favorite example is a comparison of the temperature between Phoenix and Atlanta (same latitude and elevation) in August. Phoenix is MUCH hotter, despite the fact that Atlanta has 3 times as much “greenhouse gases.” WHY? So let’s chalk that up to energy lost by evaporation of water. ”
I’m confused. You seem to be saying that your favorite comparison is not really a good example because there is another factor (evaporation) that should be important, making the comparison ineffective to begin with? (Not to mention differences in cloud cover and vegetation.)
And please provide evidence of your other assertion about greenhouses. Do you have data on similar greenhouses showing that the one in Atlanta is indeed cooler on a day with similar sunshine?
In fact, this map pretty well explains both assertions: http://www.azsolarcenter.org/images/articles/az/solmap.gif
AZ gets much more sunshine, that GA, which would be the primary factor in the higher average temperatures of both the cities and the greenhouses. Do note that Phoenix cools down significantly more at night than Atlanta. Hmmm — now there indeed is an observed effect of the extra GHGs (water vapor) around Atlantla, providing DLR to limit the cooling of Atlanta overnight.

August 30, 2011 1:19 pm

Tim F says:
“AZ gets much more sunshine, than GA, which would be the primary factor in the higher average temperatures of both the cities and the greenhouses.”
So it’s clouds that regulate the temperature? That makes way more sense than “carbon”.

kuhnkat
August 30, 2011 2:13 pm

Green Sand,
“The UK – Annual – “Mean Temperature” and the “Sunshine Hours” appear to correlate very closely, sort of makes sense, more sunshine = warming, less = cooling? Admittedly a gross over simplification, but it is actual observational data.”
You didn’t use the terms forcing or feedback or unprecedented. Couldn’t have anything to do with Climate Science!! 8>)

Tim Folkerts
August 30, 2011 3:39 pm

Smokey asks: “So it’s clouds that regulate the temperature? ”
Of course clouds regulate temperature. Clouds are one of the key factors. But you are way too well-informed to feign ignorance that there are MANY factors that influence climate – changes in the sun, changes in the earth’s orbit, changes in clouds. And, yes, changes in “DLR” also influence the climate.

David A
August 30, 2011 3:40 pm

David A says:
August 30, 2011 at 4:10 am
Tim Folkerts, do you have a comment on this??

Bob_FJ
August 30, 2011 4:11 pm

Tim Folkerts @ August 29, 2011 at 10:51 pm

“Bob, I’m not being evasive about horizontal components — mostly running out of time.
In the end, we are looking at radiative TRANSFER of energy. To calculate the transfer from one region to another, we need a dividing line between the two regions and a calculation of the total flux passing thru that surface. That is exactly what I gave you. And exactly what Gauss’ Law gives you. One region is the ground (and ocean); the other is the atmosphere. The simplest case is to draw the line right at ground level. Then it is painfully clear that 390 W/m^2 is indeed leaving the ground and entering the air. The energy flying around within a region (eg horizontally) may be interesting in some ways, but is not part of the TRANSFER between regions.”

I think you may still be missing my point. The S-B calculation gives the total radiant energy distributed hemispherically from the surface. (per Stefan’s experiment with a small flat body pointing into a hemispherical body). It is apparent that your 390W/m^2 contains both horizontal and vertical vectors. In other words, the horizontal vectors are part of the 390, but by definition, they do not go up. Thus the vertical (normal) vector integration should be less than 390.
I agree that it seems incredible that Trenberth and others could have made a mistake, such as in thinking that S-B gives the normal radiation rather than hemispherical, but some strange things have happened in climate science. It could be something silly like thinking that “total” means the total of all (Planck distribution) wavelengths in the normal, but if that is not the case, there is clearly a paradox that you seem unable to address.

jae
August 30, 2011 6:40 pm

Tim Folkerts says:
“I’m confused. You seem to be saying that your favorite comparison is not really a good example because there is another factor (evaporation) that should be important, making the comparison ineffective to begin with? (Not to mention differences in cloud cover and vegetation.)”
Look here for my data: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/state.html
Look at maximums (clear days in Atlanta and Phoenix). Phoenix does have less solar radiation (10 kwh/m2/day than Phoenix (12 kwh/m2/day), but Atlanta has over three times as much GHGs, so the direct solar radiation should be MORE than made up by the backradiation from all those GHGs, eh?
You got me on data from actual greenhouses, but I’ll bet you a six-pack that I’m correct.
But you also note:
“Do note that Phoenix cools down significantly more at night than Atlanta. Hmmm — now there indeed is an observed effect of the extra GHGs (water vapor) around Atlantla, providing DLR to limit the cooling of Atlanta overnight”.”
Do note from the data source I provide that you are completely wrong. You suffer from the “cold desert night myth.”
Your turn.

jae
August 30, 2011 7:20 pm

Tim:
More on the possible source of your “confusion:”
IF the GHE content of the atmosphere is a significant “actor” in our climate, then WHY OH WHY aren’t the areas with maximum GHEs (tropics) warmer than they are? Where is that “amplification” provided by water vapor? Why are the tropics not getting warmer, with additional OCO? IF clouds and evaporation do, indeed, “compensate” for additional GHEs, then the additional GHEs don’t have any real influence, right?
SO, even if the radiation cartoons make any sense, the postulated radiative exchanges don’t affect the temperature/climate. Maybe THAT is why the empirical evidence doesn’t match the hype?
Willis’ “thermostatic control systems” theories probably explain very well what is going on (Mother Earth has a governor on heat content), regardless of the effects of the GHGs.

Tim Folkerts
August 30, 2011 7:21 pm

Bob_FJ says: “It is apparent that your 390W/m^2 contains both horizontal and vertical vectors. In other words, the horizontal vectors are part of the 390, but by definition, they do not go up. ”
No — I would say every photon emitted has heads off in some direction that contains both a horizontal and vertical component. And in fact, the horizontal component cannot be exactly zero, since then it would not leave the ground. So, a purely horizontal vector is NOT part of the 390 W/m^2. Every photon leaving the ground has at least SOME vertical component. Every photon will eventually leave thru any arbitrary sphere I draw around the earth.
OK — one last new explanation I just thought of.
Suppose 390 W of IR energy leaves a 1m x 1m square of the ground. I build a 100 m tall square hollow column of perfect mirrors that channels the light upward. This will reflect the 390 W worth of photons so that all the IR energy passes thru the 1m x 1m square at the top of the column. Now I build a whole grid of these all over the world. Every 1m x 1m square that is 100 m above the ground will have a flux of 390 W.
Now the fun part. I remove all of the mirrors. The energy from my first square will now spread out and pass thru thousands of other upper squares. But conversely, light from thousands of other lower squares will spread out and pass thru the original upper square. The mirrors didn’t create or destroy any energy. We must still have 390 W passing thru any upper square. And the 100 m was arbitrary; the same would be true at any height.

Tim Folkerts
August 30, 2011 8:29 pm

JAE:
“Do note from the data source I provide that you are completely wrong. You suffer from the “cold desert night myth.””
Well, “from the data you provided”, Atlanta has an average change from day to night of 11.1 C, while Phoenix has a change of 14.8 C. So apparently you are completely wrong about me being completely wrong.
Look at maximums (clear days in Atlanta and Phoenix). Phoenix does have less solar radiation (10 kwh/m2/day than Phoenix (12 kwh/m2/day)
First, you need to use “flat plat, 0 degree tilt” data. Your numbers inflate the true isolation.
Looking specifically at July, Pho gets a max of 7.1 kWh/m^2 (~ 300 W/m^2 average over 24 hr) and Atl gets 5.7 kHh/m^2 (~ 240 W/m^2). That is an extra ~ 60 W/m^2 averaged over 24 hr, or more than 100 W/m^2 extra during the daylight hours during the best day of July. The average difference is a little less, but comparable. That is a considerable bonus for Pho.
the direct solar radiation should be MORE than made up by the back radiation from all those GHGs, eh?
I would not necessarily assume that. How much difference do you think the difference in conditions from Alt to Pho will make on the DLR? Will it be more than 60 W/m^2 to make the temperatures similar in the two locations? Oh, and it has to make up for the difference in evaporative cooling edge Atlanta has, too.

Bob_FJ
August 30, 2011 11:58 pm

Tim Folkerts @ August 30, 2011 at 7:21 pm

“Bob_FJ says: “It is apparent that your 390W/m^2 contains both horizontal and vertical vectors. In other words, the horizontal vectors are part of the 390, but by definition, they do not go up. ”
No — I [Tim] would say every photon emitted has heads off in some direction that contains both a horizontal and vertical component. And in fact, the horizontal component cannot be exactly zero, since then it would not leave the ground. So, a purely horizontal vector is NOT part of the 390 W/m^2. Every photon leaving the ground has at least SOME vertical component. Every photon will eventually leave thru any arbitrary sphere I draw around the earth…”

No Tim, the horizontal components ARE part of the 390, but they go “sideways”, not up, and as you have severally agreed, (E.G. on August 29, 2011 at 10:51 pm The energy flying around within a region (eg horizontally) may be interesting in some ways, but is not part of the TRANSFER between regions) This followed your lecture on radiative HEAT transfer. In fact, in any typical elemental homogenous layer in the atmosphere, from simple logic/observation, there is no HEAT transfer because T is uniform because there is no potential difference in the EMR level, despite that the lateral emissions are far more intense than in the vertical. Note that in the horizontal plane that this occurs, whilst directions of radiation vary through 360 degrees, somehow those naughty photons and molecules sort it all out with net zero HEAT transfer. You raise an objection that the initial surface emission in itself may not be perfectly horizontal, but this is not an issue. The initial surface emission is not the end of the story, because it drives the secondary spherical absorption/emissions, and the surface emissions are balanced by those in the elemental layers above
When you said; Every photon will eventually leave thru any arbitrary sphere I draw around the earth I wonder if that is what you meant to say, and why. Only a minority of photons are emitted from the surface out to space, most are absorbed and cease to exist, to be replaced by others, not necessarily of the same value or molecule source, quite close to the surface. And of course there are collisions and a lot of other stuff going on.
I thought your additional thought exercise was a bit fanciful, and my concluding points above would seem to be pertinent in part.

tallbloke
August 31, 2011 1:14 am

Hi Bob,
since energy is finally leaving Earth at around the same rate it’s arriving, and since as you correctly state the horizontal flux doesn’t do anything much in terms of heat transfer, I’m led to ask;
So what?
Seems to me that if (directional) IR meters are measuring 390 up from or near the surface and 320 down, then all the horizontally emitted radiation that builds up is just buzzing around doing no work until it just happens to get re-emitted up or down and become part of the up/down flux.
Or have I misunderstood something you are saying? (quite possible).

tallbloke
August 31, 2011 1:30 am

Tim Folkerts says:
August 30, 2011 at 3:39 pm
And, yes, changes in “DLR” also influence the climate.

Leaving aside the question of whether “changes in “DLR” also influence the climate” for now, can you point me to any empirical evidence that the DLR has changed? If Miscolzi is correct that Tau remains constant according to the radiosonde data and his line by line study, then it hasn’t, and we can all stop worrying.
So where is the empirical data to show he is wrong?
Thanks.

Bob_FJ
August 31, 2011 3:31 am

Tallbloke @ August 31, 2011 at 1:14 am
Quickly: I don’t know if there is any good IR instrumental data that can give a global average result, but my understanding is that the Trenberth cartoon derives the 390 from an S-B calculation based on nominal global average T. To me there appears to be a paradox in that the 390 is radiated hemispherically from the surface and that a good part of that 390 is radiated horizontally (or close to), such that in terms of nominal global average T, it is continuous, and not in itself part of the vertical flux.
It seems that Trenberth and others claim that the 390 is polarised normal to the surface, and that is what I’m having difficulty with.

tallbloke
August 31, 2011 3:50 am

Belmiloud, Djedjiga, Roland Schermaul, Kevin M. Smith, Nikolai F. Zobov, James W. Brault, Richard C. M. Learner, David A. Newnham, and Jonathan Tennyson, 2000. New Studies of the Visible and Near-Infrared Absorption by Water Vapour and Some Problems with the HITRAN Database. Geophysical Research Letters, Vol. 27, No 22, pp. 3703-3706, November 15, 2000
http://www.tampa.phys.ucl.ac.uk/djedjiga/GL11096W01.pdf
Abstract. New laboratory measurements and theoretical
calculations of integrated line intensities for water vapour
bands in the near-infrared and visible (8500-15800 cm−1) are
summarised. Band intensities derived from the new measured
data show a systematic 6 to 26% increase compared
to calculations using the HITRAN-96 database.
The recent
corrections to the HITRAN database [Giver et al., J. Quant.
Spectrosc. Radiat. Transfer, 66, 101-105, 2000] do not remove
these discrepancies and the differences change to 6 to
38 %. The new data is expected to substantially increase the
calculated absorption of solar energy due to water vapour in
climate models based on the HITRAN database.

Conclusions
Table 1 (Final column) also sets out values for the comparison
of our “best” total intensities of the water polyads
with those given in HITRAN-COR. It should be stressed
that the line intensities of our observations and the HITRAN
database differ from line to line and that the given values
are only valid for room temperature. The measurements
at 252K yielded slightly different ratios, but are omitted
here for clarity. For a major fraction of the lines, the principal
part of the change takes the form of a re-scaling of
the data on a polyad-by-polyad basis, we recommend the
use of the factors set out in Table 1 as an interim solution.
Other databases, such as GEISA [Jacquinet-Husson et
al., 1999], are based on the same laboratory data and will
therefore require the same correction. Detailed line-by-line
data including both experiment and theory will be published
[Schermaul et al., 2000].
There is another lesson to be learned. Making sure the
database is valid is necessary foundation for all modelling
of atmospheric radiation transfer, especially so when theory
and observation fail to agree.

David A
August 31, 2011 4:47 am

tallbloke says:
August 31, 2011 at 3:50 am
It is apparent to me that there is a natural fluctuation to the oceans between SWR and DLWR. In other words same tropical area, less water vapor = more sun hitting the surface (in the near-infrared and visible)= oceans warm and release more water vapor which equates to greater spectral modification of TSI (in the near-infrared and visible) which of course then cools the oceans. So there is a natural flux whereby TSI is first absorbed more in the oceans, then due to water vapor spectral modification, (noted in your comment that it is greater then the models show) TSI is then absorbed more in the atmosphere. The oceans are alternately heated and then cooled by this natural process of changes in spectral modification of TSI.
Of course there are many other processes going on here which explains why none of them have a perfect correlaton to temperature, but this cycle of spectral modification is happening within them.

tallbloke
August 31, 2011 6:11 am

David A:
Yes, there are negative feedbacks operating at all levels and timescales of climate activity.

jae
August 31, 2011 10:18 am

Folkerts:
You initially SAID the following, remember:?
““Do note that Phoenix cools down significantly more at night than Atlanta. Hmmm — now there indeed is an observed effect of the extra GHGs (water vapor) around Atlantla, providing DLR to limit the cooling of Atlanta overnight”.
Now you are changing your story to discuss diurnal variation, instead of “cooling down more?” The data show very clearly that Phoenix is much hotter than Atlanta, EVEN AT NIGHT.
Then you say:
“I would not necessarily assume that. How much difference do you think the difference in conditions from Alt to Pho will make on the DLR? Will it be more than 60 W/m^2 to make the temperatures similar in the two locations? Oh, and it has to make up for the difference in evaporative cooling edge Atlanta has, too.”
Well…if a few ppm CO2 can make a measureable difference, then a factor of three for GHGs should have way more effect than a mere 60wm-2 over 24 hours.
But let’s look at some other evidence that suggests no GHE is needed to explain Earth’s temperature:
http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
Don’t get hung up on the first part of the article. Juest look at the graph at the end. It seems that atmospheric temperatures on other planetoids have NO dependence upon the molecular makeup of those atmospheres. WTF??

Tim Folkerts
August 31, 2011 1:21 pm

jae says:
August 31, 2011 at 10:18 am
>>Folkerts: note that Phoenix cools down significantly more at night than Atlanta
>JAE: Now you are changing your story to discuss diurnal variation,
No, I said what I meant — you need to read more carefully. Phoenix cools down more at night (~ 15 C) than Atlanta cools down at night (~ 11 C). I did not say “Phoenix gets cooler than Atlanta …” It was always about diurnal variation.
>Well…if a few ppm CO2 can make a measureable difference,
>then a factor of three for GHGs should have way more effect
>than a mere 60wm-2 over 24 hours.
The typical estimate is that doubling CO2 leads to an increase of ~ 3.7 W/m^2. So tripling CO2 would be ~ 6 W/m^2. I have also seen estimates that H2O contributes ~ 5 times as much to the GHE than CO2. From this estimate, the increases water vapor will only add ~ 30 W/m^2, or half of difference from sunlight. So I conclude water vapor changes are less important than sunlight changes. Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”?
>But let’s look at some other evidence that suggests no GHE is needed
>to explain Earth’s temperature:
> http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
Actually, as I read the article, they discuss that (quite rightly) that other effects besides GHGs do indeed affect temperatures, which is not a surprise to anyone. But concluding that other factors are important does not imply (as you and they conclude) “no GHE is needed”. (Ironically, the article they quote to get the figure is all about how well radiative transfer & GHE predicts the profiles of atmospheres.)
Do you really think it is a surprise that heat capacity affects temperatures? Do you really think that people working in this field don’t know about lapse rates?
Its logically the same as saying “I know two people who eat the same amount of sugar each day but only one exercises; one is thin and the other is fat. Therefore sugar is not important, only exercise.” This is not proof that sugar is not important, any more than showing that heat capacity is important negates that GHE might be important. To positively declare that GHE is not needed, you have to not only show that other things are important, but that they can explain all the values without the GHE.

Myrrh
August 31, 2011 2:19 pm

tallbloke says:
August 30, 2011 at 10:29 am
Myrrh says:
August 30, 2011 at 4:15 am
“It’s pretty well meaningless when water is transparent to visible and doesn’t absorb it.”
Empirically determined coefficients of absorption are strong evidence that water does indeed absorb visible light. Better evidence than anything you’ve come up with to say it doesn’t.
And how does this compare with thermal infrared absorption?
“That the atmosphere is not transparent to visible light is an object scientific fact.”
Yeah? How come I can see for 50 miles or more from a mountain top on a clear day, and yet it’s completely dark a lot less than a mile under the surface of the sea?
??

OK, I’m convinced you’re not able to see the disjunct, in what amused me about these claims. However, I am in awe of you being able to see the whole of the starry vastness around us through our transparent atmosphere during the day..
http://www.madsci.org/posts/archives/2000-03/953560966.Ph.r.html
This AGWScience fiction version of reality claims that the atmosphere is transparent to visible light, clearly, pun intended, it ain’t; we can see immediately for ourselves that on a clear day the sky is blue, it is not transparent, and on investigation we can discover that the sky is blue from light being scattered by the molecules of nitrogen and oxygen in electronic transitions, which means the light is absorbed by the electrons.
So, how much does this absorption visible light heat the atmosphere? Where is this heat reflected in the numbers of this fictional ‘energy budget’ which claims that absorption always creates heat?

jae
August 31, 2011 2:21 pm

Tim:
“Do you really think it is a surprise that heat capacity affects temperatures? Do you really think that people working in this field don’t know about lapse rates?”
Are you somehow “reading betweeen the lines?” Or something?
You don’t seem to comprehend the point of that graph. Heat capacity, lapse rate, and atmospheric pressure explain EVERYTHING! NO GHGs need be added!

Tim Folkerts
August 31, 2011 3:25 pm

Bob,
You ask some good question about radiation, but unfortunately I think this is a poor medium for trying to converse about such advanced topics. Without being able to sit and chat with a piece of paper between us, this medium of typed words in a blog (about a somewhat different topic to begin with) is just to unwieldy for discussions of surface integrals and radiation and idealizations and real-world conditions. So I think I will have to drop out of this conversation. I do hope that somewhere, somehow you succeed in your quest for knowledge about SB radiation and the paradox you think is there.

tallbloke
August 31, 2011 3:33 pm

Myrrh says:
August 31, 2011 at 2:19 pm
tallbloke says:
August 30, 2011 at 10:29 am
Empirically determined coefficients of absorption are strong evidence that water does indeed absorb visible light. Better evidence than anything you’ve come up with to say it doesn’t.
And how does this compare with thermal infrared absorption?

So far as heat generated in the ocean is concerned, about 50/50 on average.
on investigation we can discover that the sky is blue from light being scattered by the molecules of nitrogen and oxygen in electronic transitions, which means the light is absorbed by the electrons.
Very little visible light is absorbed in the atmosphere. Which is why I can see 50 miles from mountain tops. Unlike in the ocean, where it gets pretty dark pretty fast as you go down.
Are you ready to concede this?

jae
August 31, 2011 3:53 pm

Tim:
You say: “The typical estimate is that doubling CO2 leads to an increase of ~ 3.7 W/m^2. So tripling CO2 would be ~ 6 W/m^2. I have also seen estimates that H2O contributes ~ 5 times as much to the GHE than CO2. From this estimate, the increases water vapor will only add ~ 30 W/m^2, or half of difference from sunlight. So I conclude water vapor changes are less important than sunlight changes. Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”? ”
There is something really, really wrong with your calculations and logic! How are you gonna get to even the 324 W/m2 for the “global average downwelling radiation,” using that kind of analysis? And we are not talking about radiation at some 15 C average global average here! The water vapor levels in Phoenix are typically in the neighborhood of 5 g/m^3 in Phoenix and around 20 g/m^3 in Atlanta. How can the difference be only 30 wm-2?

Myrrh
August 31, 2011 4:20 pm

Please read the link I posted.
All the energy that is ‘reflected out’ in the fiction budget is being absorbed.

Bob_FJ
August 31, 2011 4:37 pm

Kuhnkat @ August 29, 2011 at 9:45 pm

“Bob_FJ, try using your calculator with .1 and .01. I get slightly LESS radiation for the .01 than the .1. What Tim has been trying to tell you about the horizontal stuff. They take it into account. Close to the ground there will be less area radiating through the space horizontally.”

Sorry, but I’ve only just got around to looking at your suggestion. I don’t think that Tim has suggested that, but what he does claim is that the vertical integration of the surface emission (altitude = 0) a la Trenberth average global T, is 390 W/m^2 normal to the surface, and that it remains normal regardless of altitude. Presumably he is ignoring other effects as inferred in the MODTRAN calculator.
If the horizontal components increase as you claim between 0.01 and 0.1 altitude, and at the same time the normal does also, isn’t the total energy greater? Why would that be so?

Bob_FJ
August 31, 2011 4:39 pm

Tim Folkerts @ August 31, 2011 at 3:25 pm
Thank you for your patience and good manners in the S-B debate. I agree that to continue it would be a bit like sawing sawdust. (I only like simple analogies)

Myrrh
August 31, 2011 4:51 pm

Very little visible light is absorbed in the atmosphere. Which is why I can see 50 miles from mountain tops. Unlike in the ocean, where it gets pretty dark pretty fast as you go down.
Are you ready to concede this?

Just in case.. The point I was making is that you cannot see the stars in sunlight, even from the top of your mountain, therefore…
The ocean is around 800 times denser than our atmosphere, iirc, maybe someone has the maths to work it out, the relationship between the attenuation of light in the oceans and in the atmosphere, to reach the dark – I think that would have to be calculated by reflection of light back into atmosphere from the Earth to be comparable? Thinking of the thin blue line boundary as seen from space.

philincalifornia
August 31, 2011 6:12 pm

tallbloke says:
August 31, 2011 at 3:50 am
Belmiloud, Djedjiga, Roland Schermaul, Kevin M. Smith, Nikolai F. Zobov, James W. Brault, Richard C. M. Learner, David A. Newnham, and Jonathan Tennyson, 2000. New Studies of the Visible and Near-Infrared Absorption by Water Vapour and Some Problems with the HITRAN Database. Geophysical Research Letters, Vol. 27, No 22, pp. 3703-3706, November 15, 2000
http://www.tampa.phys.ucl.ac.uk/djedjiga/GL11096W01.pdf
Abstract. New laboratory measurements and theoretical
calculations of integrated line intensities for water vapour
bands in the near-infrared and visible (8500-15800 cm−1) are
summarised. Band intensities derived from the new measured
data show a systematic 6 to 26% increase compared
to calculations using the HITRAN-96 database. The recent
corrections to the HITRAN database [Giver et al., J. Quant.
Spectrosc. Radiat. Transfer, 66, 101-105, 2000] do not remove
these discrepancies and the differences change to 6 to
38 %. The new data is expected to substantially increase the
calculated absorption of solar energy due to water vapour in
climate models based on the HITRAN database.
========================================================
This might be a dumb question, but it won’t be the first time I’ve asked one of those. I think it’s related to some posts further up this humongous thread, or it may be on another. Could water vapor absorption band broadening be increased even further in the horizontal and horizontal-ish directions by motion due to wind ?? I’m not sure it would make much of a difference to boatloads of water vapor plus 280 ppm CO2 vs. boatloads of water vapor plus 390 ppm CO2, but just wondering.

Tim Folkerts
August 31, 2011 6:42 pm

One quick comment — when discussion the principles of radiation, I was assuming a transparent atmosphere. In MODTRAN, this would mean setting CO2 = CH4 = O3 = H2O to zero. With these settings, the radiation DOES remain pretty constant from the ground up, exactly as I predicted. There are some effects from 1.5 km to 20 km, bit these I suspect are other parameters in the model, like aerosols.
Once the principles of radiation are understood, then the details of absorption by GHGs can start to be added.
(I *am* a little surprised the output of the model does no change from 20 km to 100 km. Over this range, the curvature of the earth should definitely be noticeable, reducing the radiation by a few percent)

jimmi_the_dalek
August 31, 2011 7:19 pm

Re. this stuff about the Stefan-Boltzmann equation and hemispherical sections, would it help if it were pointed out that the derivation of said equation includes an integration over solid angles so that the quantity computed is the emission normal to the surface?

jae
August 31, 2011 8:09 pm

Folkerts?? Your lack of reply is noticed. You morons have no science on your side and you know it.

Tim Folkerts
August 31, 2011 9:15 pm

Hey JAE, the whole world does not revolve around you — I have other things to do with my life. It gets old having you DEMAND that everyone answer your ill-posed questions, yet you provide no calculations yourself. So you fall back on ad hominem attacks..
Present with your OWN analysis and then I will be more impressed.
“How are you gonna get to even the 324 W/m2 for the “global average downwelling radiation,” using that kind of analysis?”
I don’t expect much, but here are two hints to get you started: 1) logarithms, 2) clouds. This will easily get you to the right rough answer using “that kind of analysis”.
PS. Using MODTRAN looking up,mid-latitude, summer settings,
* with the relative humidity = 0.25 –> I = 266.272 W/m^2 (~ phoenix)
* with the relative humidity = 0.75 –> I = 297.232 W/m^2 (~ Atlanta)
Frankly I am amazed my rough estimate of 30 W/m^2 extra due to the higher humidity using “that kind of analysis” was so close to MODTRAN’s result of 31 W/m^2.

gnomish
August 31, 2011 9:27 pm

@ Folkerts:
“Even if I am off a bit, the effect of water is still not “way more”.
What calculations support your claim of “way more”?”
http://www.engineeringtoolbox.com/spesific-heat-capacity-gases-d_159.html
Gas or Vapor kJ/kg
Air 0.287
Carbon dioxide 0.189
Water Vapor 0.462
Steam 1 psia.
120 – 600 oF
That’s what it takes to change the temperature 1 degree K.
When CO2 changes from 1 to -1 C, a change of 2 degrees C, it radiates 2(0.189 kJ/kg) = 0.378 kJ/kg.
http://en.wikipedia.org/wiki/Enthalpy_of_vaporization
When water vapor changes from 1 to -1 (and condenses) it radiates 2257 kj/kg + 2(0.462 kJ/kg) = 2257.853776 kJ/kg.
It does this every single time you see a cloud.
But CO2 has no phase change so it carries no heat – the numbers:
All gases at the same temperature have the same number of molecules per unit volume.
Water, being light, masses 18g/mole and CO2 masses 44 g/mole
Using 1 mole of air, just to make math easy:
We lowball the water in the atmosphere at 1% of the molecules
So, in a mole of atmosphere, we have 0.01 moles of water = 0.18g
now we highball the CO2 at 500ppm which is 0.0005, or 1/2000 of a mole of CO2.
1/2000 * 44g/mole = 0.000484 moles of CO2 = 0.021296g
So in our mole of air with but 1% H2O and a generous 500ppm CO2-
the water condensing radiates 0.18g * 2257.853776 kJ/kg = 406.41367968 J
while the CO2 radiates 0.021296g * 0.378 kJ/kg = 0.008049888 J
the ratio of 0.008049888/406.41367968 = .00001980712855516645290496438242332
or as much to say that water vapor in the example carries 50486.873814890343815963650674393 times more heat than the CO2 does.
And that’s just condensation to liquid phase. If it turns to snow- multiply by 5-6.
Conclusion – the CO2 is insignificant retainer of heat in our atmosphere. Water vapor does 50,000 times more work.
(And that was even done with shorting the estimate on water while boosting the estimate on CO2)

gnomish
August 31, 2011 9:42 pm

oh, yah- please note that when water gas changes to water liquid, there is NO temperature change. Your boltzman forumla calculator measures NO change, eh.
it only calculates temperature and temperature is not heat.

kuhnkat
August 31, 2011 10:04 pm

Bob_FJ,
“If the horizontal components increase as you claim between 0.01 and 0.1 altitude, and at the same time the normal does also, isn’t the total energy greater? Why would that be so?”
The earth is curved. On a small ball you will see that even if you had radiation on a plane from the point of emission you would see that it quickly diverges from the surface. On earth the curve is much slower so the effect is very small comparatively. At the surface you will only have horizontal energy from a very small area to the point that where the emission happens there is no other area emitting to that point. As you move up in altitude you will have energy from an increasingly large area that can emit through the same space.
Think of overlapping hemispheres on the surface of the globe. The center of each hemisphere will not be able to emit directly at the center of the next hemisphere. You will see a wedge that it also cannot emit through being blocked by the curvature. This averages out from there being an almost infinite number of hemispheres. The end result is that the further apart the centers are the lowest area they can both irradiate will be higher above the surface.
I have to figure out how to draw pictures and post them some day. Any suggestions?

tallbloke
August 31, 2011 11:23 pm

Myrrh says:
August 31, 2011 at 4:20 pm
Please read the link I posted.
All the energy that is ‘reflected out’ in the fiction budget is being absorbed.

Nothing at that link says anything about what quantity is absorbed and what quantity is reflected.
I can tell you that according to a paper I read, about 30W/m^2 more is absorbed in clouds than theory can account for. This is likely due to an imperfect understanding of scattering and an underestimate on water’s absorption potential (see the paper I linked above for the latter, and this post on my blog for the former). So I agree Trenberth’s cartoon is far from perfect, but not so far as you think.
Myrrh says:
August 31, 2011 at 4:51 pm
Just in case.. The point I was making is that you cannot see the stars in sunlight, even from the top of your mountain, therefore…

Just in case what Myrrh? Just in case I don’t understand your self delusions? A small amount of the visible light incoming from the sun is scattered to produce the blue sky and…. What? You think this proves that lots of light is absorbed in the atmosphere?
I might not be able to see stars during the day (I can’t see candlelight from a mile away during the day either), but I can see the sunlight reflected by the moon, and I can see the sunlight reflected by Venus and Mercury, and I do see stars starting to appear while there is still blue in the sky in the evening. And when I’m in a jetliner, I see the spectrum of colours in the sky at sunset, with the darkness and stars above, and still I can see streetlights on the ground and the sunset reflected off lakes from 8 miles up. Doesn’t look like much energy is going into making the atmosphere opaque to me.
Tallbloke says:
Very little visible light is absorbed in the atmosphere. Which is why I can see 50 miles from mountain tops. Unlike in the ocean, where it gets pretty dark pretty fast as you go down.
Are you ready to concede this?
Myrrh replies:
The ocean is around 800 times denser than our atmosphere, iirc, maybe someone has the maths to work it out, the relationship between the attenuation of light in the oceans and in the atmosphere, to reach the dark – I think that would have to be calculated by reflection of light back into atmosphere from the Earth to be comparable? Thinking of the thin blue line boundary as seen from space.

So we’re back to light being ‘attenuated’ instead of absorbed in the ocean eh?
Here’s the message: Some light is ‘attenuated’ or scattered, some light is reflected (off aerosols in the atmosphere, off suspended particles in the ocean). In the final analysis, nearly all sunlight that misses clouds makes it through the atmosphere into the ocean, where nearly all of it not reflected off wave flanks is absorbed and converted to heat.
If you are interested in the proportions, either trust the scientists (not just climatologists) who have done the measurements, or do your own. My approach is to hunt around for important papers who’s results have not been included in the models relying on trenberth and work back from there.
I’ll leave you to it.

tallbloke
September 1, 2011 12:13 am

philincalifornia says:
August 31, 2011 at 6:12 pm
tallbloke says:
August 31, 2011 at 3:50 am
New Studies of the Visible and Near-Infrared Absorption by Water Vapour and Some Problems with the HITRAN Database. Geophysical Research Letters, Vol. 27, No 22, pp. 3703-3706, November 15, 2000
http://www.tampa.phys.ucl.ac.uk/djedjiga/GL11096W01.pdf
Abstract. New laboratory measurements and theoretical
calculations of integrated line intensities for water vapour
bands in the near-infrared and visible (8500-15800 cm−1) are
summarised. Band intensities derived from the new measured
data show a systematic 6 to 26% increase compared
to calculations using the HITRAN-96 database. The recent
corrections to the HITRAN database [Giver et al., J. Quant.
Spectrosc. Radiat. Transfer, 66, 101-105, 2000] do not remove
these discrepancies and the differences change to 6 to
38 %. The new data is expected to substantially increase the
calculated absorption of solar energy due to water vapour in
climate models based on the HITRAN database.
========================================================
This might be a dumb question, but it won’t be the first time I’ve asked one of those. I think it’s related to some posts further up this humongous thread, or it may be on another. Could water vapor absorption band broadening be increased even further in the horizontal and horizontal-ish directions by motion due to wind ?? I’m not sure it would make much of a difference to boatloads of water vapor plus 280 ppm CO2 vs. boatloads of water vapor plus 390 ppm CO2, but just wondering.

Dunno – sorry.

tallbloke
September 1, 2011 12:17 am

gnomish says:
September 1, 2011 at 12:00 am
hey TB-
seawater absorbs just about everything if it’s deep enuff:

Don’t tell me, tell Myrrh! 😉
And if it’s not deep enuff, the bottom absorbs it and conducts the heat to the water.

Bob_FJ
September 1, 2011 1:50 am

Tim Folkerts @ August 31, 2011 at 6:42 pm

“One quick comment — when discussion the principles of radiation, I was assuming a transparent atmosphere…”

Tim, you have been very naughty indeed*! Earlier you said that you did not want to continue the discussion, and I concurred. (over two hours before your 6:42 pm) OK, so you want to get the last word in, but no, it will not be the last, for I will respond, if but in brief:
If an integrated hemispherical S-B emission normal to the surface can be established, it is absolutely no surprise to me that in any model it would remain normal in a transparent atmosphere, (or vacuum), unchanged in intensity to great height, apart from a trivial effect arising from the curvature of the oblate spheroid that we live on.
HOWEVER, that is not the issue that I have raised. To re-elaborate in brief; YOU have admitted above: I would say every photon emitted [delete ‘has’, sic?] heads off in some direction that contains both a horizontal and vertical component.
My issue is: what happens to the horizontal stuff, or perhaps ‘nominally tangential’ might be a slightly more precise definition. (Let alone the intermediate lateral, and perhaps below any curvature)
I see that there are a couple of other commenters that I should respond to in order to expand on this, but meanwhile and briefly, your consideration of a transparent atmosphere, seems to make your position look more precarious. What is to prevent the non-normal stuff from freely escaping directly to space? (up to a nominal 90 degrees opposed to any normal viewing mechanism, and which would be very different in our real-life partially opaque atmosphere.)
* Tim; as penance: Please write out 100 lines; of my selection of:
Travesty Trenberth, although having declared overbearing expertise, knows a lot less about hurricanes than does Chris Landsea and others!
No cheating please, no copy and repetitive paste, and Email your truly completed penance to Anthony Watts, swearing upon your IPCC Bible.

Myrrh
September 1, 2011 2:47 am

Well, I don’t know what else to say, you’re all clearly convinced that light from the Sun is heat and heat from the Sun doesn’t reach the Earth’s surface.
I can understand your confusion, I’ve read countless pages now which confuse the two, claiming that visible heats the Earth and thermal heat doesn’t, even pages that give information which show the difference in process will go on to claim visible converts to heat. It looks like the memes deliberately introduced by AGWScience Inc to confuse this issue have become so ingrained in the population that never had earlier teaching, and in older populations who never took any notice of it before AGWscaremongering, that the simple difference between Light and Heat energies is so confused in their minds that they live in an alternate universe and can’t make it back. Even posting that the new paradigm has been deliberately introduced, such as in the NASA example I gave above and the encyclopedia admission that traditional teaching is still that heat on the move from Sun is thermal infrared and it reaches us and does all the heating, isn’t enough to get anyone here to stop and think.
Visible and UV are not thermal energies, nor near IR, we can’t feel them as heat, they’re not heat on the move, they don’t have thermal infrared’s power to move molecules into vibration, because shortwave’s work on the smaller electron level, etc. etc. Yet, all it takes is for this fiction producing propaganda to keep repeating lies such as thermal infrared doesn’t pass through glass and that’s enough to those accepting it as the truth to reject even real world examples to the contrary, such as a real industrial understanding of the differences which produces window coatings to stop thermal infrared, Heat, while allowing visible, Light, through.
(Here, for anyone who might be bothered: http://www.archsky.com/technical-binder/guide/coatings.php Low-E Coatings
Low-E coatings are metallic-oxide coatings that are applied to the interior surface of the glass, which is invisible to the naked eye. Low-E coatings reflect the “long” infrared (IR) portion of sunlight, which is the radiant heat portion. Depending upon which surface of the glass it is applied, Low-E coatings can keep radiant heat from entering, or escaping a building. )
There’s a whole industry now involved in producing thermal infrared energy in heating buildings, for example, actually able to heat walls and the objects in the room including the people with this real Heat energy, and, such applications as saunas, deeply penetrating thermal infrared heats the insides without the scalding effects on the skin, and, in countless variety of heat penetration in healing. Where are all the visible light heating industries if it is such a powerful force which physically converts to heat all the oceans and lands of Earth?? Where the f*** are they?
Instead, we see countless real world applications of Visible from those who still understand the differences of the properties of electromagnetic waves and the differences in the matter through which they travel, such as: http://www.optech.ca/aboutlaser.htm and, there are pages and pages of information about the use of Light in enhancing photosynthesis by eliminating Heat, thermal infrared, from lamps used for this in real world plant growing industries.
I’m sorry, you really don’t know how insane this sounds to someone with traditional tried and tested background knowledge, that I can see clearly how AGWScience has manipulated this by promoting the lie that Light is Heat by the simple expedient of reversing properties is of scant use if I can’t even get any of you to even think about it. Even showing the difference in size of effects, that thermal infrared works on molecule vibrational level and visible on the finer smaller electron isn’t getting through, and all kinds of excuses come out to defend the now ubiquitous in education lie.
A typical page, giving actual facts re the difference, but with the added lie from AGW which then makes nonsense of it all: http://hyperphysics.phy-astr.gsu.edu/hbase/ems3.html
Ask yourselves, please, why would NASA have taught the traditional view for so long only to change it now? Why has the traditional teaching that the heat we feel from the Sun is thermal infrared and therefore is reaching us at the surface, must be reaching us at the surface, been systematically obliterated from education, and the lie that it doesn’t reach us promoted? What proof can you offer that the new fictional paradigm is correct and traditional teaching wrong? None, you only defend the new paradigm by more fictional arguments, but you can’t see that this is what you’re doing. What I find really surprising, and disheartening, is that you think yourselves scientists and instead of being interested that this disjunct has been brought to your attention, you simply dismiss it and all explanations because you’re not willing to explore the contradiction.
Anyway, this will definitely be my last post here so I’ll just leave you with a couple of reminders from what I’ve been saying:
The Traditional, well-examined and understood differences between Visible Light and Thermal Infrared Heat:

” “Near infrared” light is closest in wavelength to visible light and “far infrared” is closer to the microwave region of the electromagnetic spectrum. The longer, far infrared wavelengths are about the size of a pin head and the shorter, near infrared ones are the size of cells, or are microscopic.
Far infrared waves are thermal. In other words, we experience this type of infrared radiation every day in the form of heat! The heat that we feel from sunlight, a fire, a radiator or a warm sidewalk is infrared. ..Shorter, near infrared waves are not hot at all – in fact you cannot even feel them. These shorter wavelengths are the ones used by your TV’s remote control. “

At least, read this until you’re familiar with what it is saying, and bring this to mind every time you read something that contradicts it..
Hopeful that will spark again the scientific curiosity which brought you into whatever field you’re in. No need for more examples of the lies contradicting these traditional facts…
Anyway, one little piece of information from our history of discovery of the properties of the world around us which I hope you find interesting:

http://docs.lib.noaa.gov/rescue/mwr/056/mwr-056-08-0322.pdf
MONTHLY WEATHER REVIEW AUGUST, 1928
FURTHER STUDIES IN TERRESTRIAL RADIATION ‘
By G. C. SIMPSON, C. B. F. H. S.,F. R. Met. Soc.
The new results affect previous work materially. Emden
found that the stratosphere sends no radiation downwards,
and of course the same result came out of my
previous work. The new investigation shows that the
stratosphere sends on the average a downward flux of
longwave radiation of more than .120 cal./cm.2/min., which
is more than 43 per cent of the effective solar radiation.
This agrees with the observations made by Angstrom on
mountain peaks and in balloons, which revealed a downward
radiation of between .13 and .16 cal./cm.2/min.
at helghts between 4,000 and 5,000 metres, where, according
to Emden, there should have been less than .05
cal./cm.2/min.

The beginnings of our real understanding about all this. And, an example of a real scientist, who records his failures, because he had to think again after Angstrom’s work.
And, when and if any of what I have been saying inspires you to actually investigate further, do please find or devise experiments to show the difference by proving or disproving visible light from the Sun capable of heating land and oceans of our wonderful Earth. Convince yourselves first that the memes you’re promoting are physically possible.
And then put thermal infrared, the heat energy on the move from the Sun to reaching us in 8 minutes, back where it belongs in the energy budget..

gnomish
September 1, 2011 4:22 am

heh – i can’t talk to myrrh – i don’t know his frequency.
besides, he only listens to the black dog.

Myrrh
September 1, 2011 5:35 am

OK, one p.s. to get you started: http://www.iquefilms.com/SpectrallySelective.asp and http://www.iquefilms.com/FAQ.asp
Says of it 73FG product: “Originally developed for the US space and military program, 73FG is now being adapted for commercial use. Rejecting as high as 94% of infra-red and ultra-violet of 99% while allowing 70% of visible light to pass through, 73FG is the ideal choice for homes and buildings that would want the high amount of clarity but least amount of heat into the interiors.”
The 43FG will stop even more infra-red from entering – 98%
Why don’t you try telling them that they should be making windows rejecting visible light because that is the mechanism which heats oceans and lands, even at the bottom of the oceans..
From its FAQ page: “Just like V-KOOL, IQuie Selective is a spectrally selective film that is so clear and yet able to cut of 96% of infra-red heat.”
Oh gosh, infra-red is heat coming through windows?? My, who’d have thought it.
I repeat, in the real world the difference between HEAT and LIGHT energies is very well known, and where it is still taught in traditional physics people will still be able to find new ways of utilising the differences, while those who are taught that Light is Heat are at a disadvantage, intellectually and practically. They don’t understand the difference and so, question 4.
Do yourselves and everyone else a favour, quit talking about absorption and ‘energy budgets’ until you understand the basic differences of properties and processes, you are only adding to the disinformation campaign on this.
And, when you do step back into the real world from the other side of your reversed imaginary science fiction universe, I bet you’ll be as appalled and impressed as I at the sheer audacity of the con that has promoted the teaching that solar gets through the greenhouse glass and heats the ground while the real heat, thermal infrared, doesn’t..
..if.

tallbloke
September 1, 2011 5:51 am
tallbloke
September 1, 2011 7:11 am

gnomish says:
September 1, 2011 at 4:22 am
heh – i can’t talk to myrrh – i don’t know his frequency.
besides, he only listens to the black dog.

Yeah. I just saw the latest begin with this:
Myrrh says:
September 1, 2011 at 2:47 am
Well, I don’t know what else to say, you’re all clearly convinced that light from the Sun is heat and heat from the Sun doesn’t reach the Earth’s surface.

So I won’t be bothering to wade through the rest, or the PS…

tallbloke
September 1, 2011 7:58 am

Ok, I couldn’t resist, I did read through Myrrh’s final offerings. There’s no point in making a reply, as I already covered the points raised, from a few different angles, but I just don’t seem to be getting through.
Oh well. Maybe Myrrh will buy some of these fancy IR blocking window film and then take some diurnal temperature readings in the dark and the full sun. In the end, the only way this person is going to be convinced of anything is by his/herself. Which is fine.
For everyone else, when advertising blurb says:
“ideal choice for homes and buildings that would want the high amount of clarity but least amount of heat into the interiors.”
You can be sure that ‘least amount of heat’ means less than before (by blocking 30% of visible light and 95% of IR) but a lot more than none.
Here are the specs:
Visible Light Transmission
70.0%
Visible Light Reflectance
8.0%
Ultra-Violet Rejection
99.0%
Infra-Red Rejection
94.0%
Total Solar Energy Rejection
55.0%
Luminous Efficacy
1.40
Solar Heat Gain Coefficient
0.44
Shading Coefficient
0.50
Emissivity
0.60
U-Value (btu/h.ft2.F)
0.94
If it’s going to ‘reject’ 94% of the Solar IR, but the heat gain coefficient is still going to be 0.44, then Myrrh is going to have to accept that visible light heats stuff after all…
Like I told Myrrh several comments ago, for the ocean, about half the heat is coming from visible and half from solar Near IR.

Tim Folkerts
September 1, 2011 8:17 am

Sorry, Bob! I wasn’t intending to keep active in the discussion, but someone else made a couple points about what I had said and I wanted to make a quick clarifications/suggestions to those who were continuing the discussion.
I don’t know much about Trenberth’s expertise regarding hurricanes (and even less about Chris Landsea’s expertise), but I am more than willing to admit that many experts overstep their fields of expertise and end up say silly things. And many experts get too focused on a narrow area of expertise and forget/downplay/ignore important factors that should be included (eg focusing too much on just CO2, or analyzing temperature data without really considering how poor the siting of the stations might be). A good scientist will keep an open mind (but as the saying goes “not so open that his/er brain falls out), but scientists are human too, and certainly are not perfect.
I have enjoyed my interactions in this thread with you, and Tallbloke and Willis and Stephen Wilde — and a few others as well. It has clarified issues and given me an opportunity to teach a little and to learn a little.
Now back to work!

G. Karst
September 1, 2011 10:36 am

Part of the problem here is we are using old optical terms and mixing them with quantum terms.
There really isn’t any reflection, refraction, or transmission taking place from one medium to the next. All photons are annihilated.
Some molecules/atoms absorb the photon and quickly re-emit (the photon) at the exactly the same angle received. We call this reflection in optics, however the process is not 100% efficient and there will always be some increase in vibrational energy until a ir photon emission or conductance balances the kinetics of the particle. If the emitted photon leaves at a different angle, this of course is refraction.
Another molecule/atom re-emits a photon in the direction of the photons projected path. We regard this optically as transparent, even though it is not. Nor is this process 100% efficient, and some losses occur (as kinetic heat), as stated above. So there really is no such thing as transparent, only photons which are efficient at re-emitting photons at close to their original energy levels.
All of the above contribute to scattering, as the molecule/atom is not perfectly still and have velocity. Blue light is excellent at the straight line re-emission of light (in H2O), therefore appears to penetrate farther. Its energy content is degraded over a greater distance, is all.
I hope this clears some of the misconceptions involved. GK

G. Karst
September 1, 2011 10:55 am

Correction:

So there really is no such thing as transparent, only photons which are efficient at re-emitting photons at close to their original energy levels.

Bolded photon should read molecule/atom as in:
So there really is no such thing as transparent, only molecules/atoms which are efficient at re-emitting photons at close to their original energy levels.
sorry GK

gnomish
September 1, 2011 2:17 pm

“Why don’t you try telling them that they should be making windows rejecting visible light ”
what a concept. windows that nobody can see thru…. hmm…
but i don’t want to be an early adopter cuz i’m heavily invested in solar powered shade generators.
‘sayyeda wegoddago now. etco!’ – anonymous barfly at Louie’s.

jae
September 1, 2011 8:05 pm

Folkerts:
I admit I went over the top with my anger at your obfuscation, but you appear to be the typical slippery AGW liberal, offering only vague undocumented “facts” and “consensus” crap to support your position. To wit, you say:
“Present with your OWN analysis and then I will be more impressed.
“How are you gonna get to even the 324 W/m2 for the “global average downwelling radiation,” using that kind of analysis?”
I don’t expect much, but here are two hints to get you started: 1) logarithms, 2) clouds. This will easily get you to the right rough answer using “that kind of analysis”.
Notice, everyone, how the challenge is directed back to the challenger, ala true believers. Mr. Folkerts, you are violating your own admonition: I challenged you to show just how the hell your logic can lead to a backradiaition of 324 w/m-2? You must realize that you said NOTHING in that comment to answer that challenge or to even address it! You didn’t show your work: please show how you get from 60 Wm-2 to 324 Wm-2 using “clouds” or “logarithms.” Please don’t try to appear that you have the math worked out, when you don’t have a clue.
Oh, and MODTRAN shows just how stupid your numbers are!

kuhnkat
September 1, 2011 8:12 pm

Myrrh,
thanks for the link to the 1928 paper on the Strat.
The really sad thing is that you have so little knowledge of what modern science actually says that you don’t realize that the findings in that paper are what is now understood (within the errors of the instrumentation).
Good luck with the Windmills.

Tim Folkerts
September 2, 2011 9:53 am

JAE,
The challenge is to find some sort of common ground, some common understanding.
You see “true believer” when you read my comments”. Well, I see “true disbeliever” when I read yours. I don’t see evidence in your writing that you under stand Plank’s Law of Radiation or surface integrals. I don’t see evidence that you understand the the quantum mechanics of why CO2 absorbs IR, or why Doppler broadening is important.
Yet without seeming to understand what scientists are saying, you are firmly convinced that they are wrong. This is NOT skepticism.
That is why I challenged you to present your side. For you to be convinced that MODTRAN’s numbers are “stupid”, you really need to understand the numbers, and be able to point out specific mistakes.
“please show how you get from 60 Wm-2 to 324 Wm-2 using “clouds” or “logarithms.”
OK.
1) Clouds are very close perfect blackbody radiators of IR. On a day with low clouds, the clouds will be very close to the temperature of the ground. In these conditions, the ground will radiate its 390 W/m^2 or energy upward, while the clouds will radiate a bit less than 390 W/m^2 downward. In these conditions, there can easily be MORE THAN 325 W/m^2 radiating down.
2) On clear days, the downward IR radiation will be from GHGs (and a bit from aerosols). There is a “rule of thumb” that the effect of more GHG’s is logarithmic — ie DOUBLING the concentration produces a LINEAR increase in IR (its not amazingly accurate, but it works OK). If IR radiation drops by 30 W/m^2 when the humidity drops from 75% to 25%, should drop by another 30W/m^2 @ 8% humidify. looking the other way around, increasing the RH from 0.1% to 75% RH should be ~ 180 W/m^2. The rule of thumb will perforce breakdown at low humidities. But from these numbers, it is easy to see that humidity itself could easily account for a few hundred W/m^2 downward. So as an order-of-magnitude estimate, it is clear GHG’s could provide a few hundred W/m^2 even on a cloudless day. (MODTRAN suggest that these numbers are not exactly accurate, but are in the right ballpark, providing some confirmation.)
Since we are averaging numbers above 325 W/m^s and below 325 W/m^2, we are perfectly on track to get a number like 325 W/m^2 for the average thermal IR heading down toward the surface.
Now, if you want to give specific critiques of these numbers, I’d be happy to listen. If you want to present some numbers of your own, I’d be happy to listen. If you simply want to call the numbers “stupid” or claim the people calculating them are “morons” or suggest the numbers don’t agree with your intuition or insist that only one side of the discussion needs to do any work, well then you are showing yourself to be a “true disbeliever” convinced of your position without any support.

jae
September 2, 2011 6:37 pm

Sorry, Tim, but you lost me again with this “rule of thumb” crap:
“The rule of thumb will perforce breakdown at low humidities. But from these numbers, it is easy to see that humidity itself could easily account for a few hundred W/m^2 downward. So as an order-of-magnitude estimate, it is clear GHG’s could provide a few hundred W/m^2 even on a cloudless day.’
As far as I can tell, you have not added one iota of data or evidence to your argument. Just POINT TO THE FRIGGING EMPIRICAL DATA that shows that “humidity itself could easily account for a few hundred….” (After all, THAT is what this discussion is about: EMPIRICAL DATA). Your “calculations” certainly don’t support that “humidity could easily account for…” To the contrary, you claimed something like 60 watts! Tim, WTF? Hell we are back to where I first challenged you–no data–same point, which you refuse to address.
BTW, “Rules of thumbs” don’t count largely in science, last I knew (but I’m an old fart, so maybe the current climatology neoscience has different concepts???). Can you cite some source for the importance of “rules of thumb?”
I got a different “rule of thumbj, Tim.” It is that ALL THE OTHER PLANETOIDS SHOW THE SAME TEMPERATURE PHENOMENON AS THIS ONE, DESPITE THE COMPOSITION OF GASES IN THE ATMOSPHERE OF THE PLANETIOD. You have still never addressed this “issue,” nor have any of your warmista friends, AFIK ( I quit reading SurrealClimate a long time ago, since it is a political blog, not really a scientific one (as evidenced by their disgusting deletion of “unwanted” comments).

Bob_FJ
September 2, 2011 6:47 pm

Tim Folkerts you said to JAE

“…You see “true believer” when you read my comments”. Well, I see “true disbeliever” when I read yours. I don’t see evidence in your writing that you under stand Plank’s Law of Radiation or surface integrals. I don’t see evidence that you understand the the quantum mechanics of why CO2 absorbs IR, or why Doppler broadening is important…”

Tim, unless I’ve missed something on this long thread, I think that is unfair, and irrelevant to JAE’s issues. I don’t recall any reason why he should mention any of the four topics you raise. I recall him being a true sceptic over the years, especially over at Steve McIntyre’s site and he had a good few exchanges with a Dutch (?) physicist, “the one who knew everything“; Tom Vonk. He remains a true sceptic as far as I’m concerned, and I’m sure he knows a lot more than you credit him for.
For instance, I could repeat to YOU: I don’t see evidence in your writing that you understand Plank’s Law of Radiation. But so what? What has that got to do with the price of cheese? BTW, you mean Planck, not Plank eh?

Bob_FJ
September 2, 2011 7:17 pm

G. Karst @ September 1, 2011 at 10:36 am
I found your quantum theory to be interesting, and new to me. Can you explain how it works WRT radiation pressure and “space sails” etc. Here is the topic introduction in Wikipedia:

Radiation pressure is the pressure exerted upon any surface exposed to electromagnetic radiation. If absorbed, the pressure is the power flux density divided by the speed of light. If the radiation is totally reflected, the radiation pressure is doubled. For example, the radiation of the Sun at the Earth has a power flux density of 1,370 W/m2, so the radiation pressure is 4.6 µPa (absorbed).

http://en.wikipedia.org/wiki/Radiation_pressure
I always thought, (without exploring) that photons simply pass through “transparent” molecules. What are the theoretical objections to that? I take it that theses processes you describe have never been observed.
And yes, I know, I should not rely on Wiki, but it should be OK since it is not an emotive topic like AGW

jae
September 2, 2011 7:17 pm

Folkerts:
You went over the top, again, Timbo. You really FINALLY blew your cover when you made this ridiculous comment:
“I don’t see evidence in your writing that you under stand (sic) Plank’s Law of Radiation or surface integrals. I don’t see evidence that you understand the the quantum mechanics of why CO2 absorbs IR, or why Doppler broadening is important.”
Well, I have a PhD in organic chemistry and actually DO understand radiation quite well. Especially IR, since I’ve used it directly hundreds of times to identify organic compounds. I probably know at least as much about “quantum mechanics” as you do (if you disagree, please state your qualifications on this topic for all the three bloggers still following this thread to see). However, maybe I’m dumb about “surface integrals,” and I hope you can help me understand THAT concept!
The “bottom line” is that THERE IS ABSOLUTELY NO EMPIRICAL EVIDENCE OF ANY RADIATIVE GREENHOUSE EFFECT. And until there is, it is just a theory, not some “law,” “consensus,” “rule-of-thumb,” etc. Period!
Thanks for putting up with my obstinance, but I enjoy the argument.

G. Karst
September 3, 2011 2:07 pm

Bob_FJ says:
September 2, 2011 at 7:17 pm
Wow, you don’t beat around the bush and go right to the meat of the matter (pun intended). I assume you are not looking for the mathematical formula quantifying the forces involved (wiki will do that).
These most excellent question are presently being investigated at the Cern LHC experiment. How exactly does energy packets acquire mass like properties? Or does the force exerted by photons derive from elongated photons whose force is a result of its own ass snapping into its leading frontal lead head. Almost anything I say, would be instantly attacked, as these phenomena are hot topics of contention and are being actively pursued and theorized.
Personally, I am skeptical of present mass explanations (Higgs field) and am more inclined to simply give the entire universe, a velocity (angular or linear or both), and avoid the whole issue. Of course, this creates the conundrum, of what medium is the universe moving in!
You may feel my answer is a “cop out” and you would be correct, it is. There are some excellent minds here, less timid than me, who may care to go further.
Actually, I had this conversation with John Conway, chief scientist at the fermilab (CDF) collider, recently. He poo-pooed, tsk tsked, me very politely:

G:
A particle moving along at some velocity relative to an observer acquires
an apparent mass for that observer, due to the effects of special relativity.
In its own rest frame (and it is always possible to imagine “catching up”
with the particle and traveling with it) however it has a “rest mass”. It is
the question of where the rest mass of particles comes from that we attempt
to answer with the Higgs boson, though the standard model gives us no hint
as to why different particles have different masses.
To speak of the universe as a whole as having a velocity is fine, but then you
have to specify: with respect to what? And anyway then you can still “boost”
in the rest frame of the universe and you are back pondering where the rest
masses of particles come from.

He is a great scientist and I defer to him. For now! GK

Bob_FJ
September 3, 2011 6:14 pm

G. Karst @ September 3, 2011 at 2:07 pm and Myrrh please note particularly 3)
Concerning “radiation pressure”…. Thanks for that G…. very interesting
1) The theory that perfect reflection is the consequence of absorption and simultaneous emission in the reverse direction seems plausible in terms of “photon mass” deceleration and acceleration. (resulting in a doubling of force compared with simple absorption). However, I have difficulty in understanding how the individual or sibling molecules can have the “intelligence,” and “skill,” and have “awareness” that the surface of the body is highly polished. On the other hand, if the photons simply bounce off, it does make sense that a polished surface would promote a pure bounce.
2) If we visualize a perfectly transparent “Space sail”, the theory that transparent matter absorbs incoming radiation and simultaneously emits it with no change of direction, would sensibly result in a net force of zero. However, again, I have difficulty understanding how the molecule figures out what to do. On the other hand, photons might arguably simply pass through a “transparent” molecule, giving the same result.
3) Myrrh, please note that radiation pressure has been proven by experiment*, and that it is related to the flux power REGARDLESS of wavelength. Also be reminded that ocean water is not perfectly transparent. May I suggest that you do some careful unbiased reading, starting at Wikipedia, and follow the links and hints:
http://en.wikipedia.org/wiki/Radiation_pressure
Note also that this has nothing to do with AGW.
* whilst the precise cause at the molecular level cannot be observed, and is the subject of debate.

G. Karst
September 4, 2011 3:21 pm

For those who may be interested, “The Cassiopeia Project – making Science simple” put together an excellent 4 part (10min each) video which does summarize quantum theory. Some of the visuals are a little lame but easily forgiven. The link provided is for part 3, which describes photon annihilation (adsorption) and electon cloud shaping. It is worth watching the other 3 parts. Enjoy! GK
[youtube http://www.youtube.com/watch?v=19EFUyhhch8?version=3&rel=1&fs=1&showsearch=0&showinfo=1&iv_load_policy=1&wmode=window%5D

Tim Folkerts
September 5, 2011 9:00 pm

JAE states: The “bottom line” is that THERE IS ABSOLUTELY NO EMPIRICAL EVIDENCE OF ANY RADIATIVE GREENHOUSE EFFECT.”
How do you define “radiative greenhouse effect”? It is difficult for me to try to defend something without knowing precisely what you mean.
And what would constitute “empirical evidence” to you?
I am fully convinced that there is readily available empirical evidence (and theoretical support) that
1) IR radiation from the atmosphere shines down toward the surface
2) this IR is due to “greenhouse gases” (and clouds and aerosols to a lesser extent)
3) the surface absorbs much of that IR
4) conservation of energy requires that additional energy will result in a higher temperature
5) the observed average surface temperature is higher than would be with only sunlight heating it.
Which of these do you disagree with? Then I could try providing specific support and/or empirical evidence. Or is this not what you mean by the “radiative greenhouse effect” ?
PS I can be a bit obstinate myself , and I can love a good discussion too. 🙂

Bob_FJ
September 6, 2011 1:26 am

Tim Folkerts @ September 5, 2011 at 9:00 pm
Putting aside your obfuscations to JAE, and I hope as one old fart to another that JAE does not mind me interjecting, it seems to me that you totally ignore that he asked you to explain to him your view on “surface integration”.
My understanding of what you describe is that it is mathematically applicable to integrate the normal (vertical) components of the surface hemispherical emissions. I have no problem with that, but the point is that it is only part of the total hemispherical surface emission. There is sideways stuff (horizontal vectors) that is part of the whole that you neglect. What adds to the confusion is that the sideways stuff should initially conceptually be considered in an environment of a transparent atmosphere, in which case it would just escape directly to space. However, the real-world situation is that it is less simple when the atmosphere is partially opaque. Would you care to discuss this trifling complexity?

RJ
September 6, 2011 2:34 am

4) conservation of energy requires that additional energy will result in a higher temperature
This is not correct. The radiation is from a colder body so at best can only very slightly slow the cooling rate. So saying
4) conservation of energy requires that the additional energy might result in slower surface cooling
would be more accurate
In that if 100 balls are in a box and 20 leave and one return. The box now contains 81 balls not 80. But I’m unsure if the one ball does anything more than returns (if it does in fact return) and immediately leaves again without having any impact.

Tim Folkerts
September 6, 2011 7:31 am

Bob,
Trying to type – let alone explain – surface integrals is not really effective in this forum. You will have to pursue that on your own, I fear. (Besides, I am a bit rusty and don’t really want to take the hours to get back up to speed on it all before spending even more hours explaining it. I have done surface integrals before as an undergrad and as a grad student, and I know it will work here)
But the net result in this case is not that tough to understand. One last analogy. You have a very large warehouse with fire sprinklers on the ceiling. The sprinklers are spaced 1 m apart on a grid, and each sprinkler sprays 390 liters/minute. Once the sprinklers have been on for more than a few seconds, then an average of 390 liters/min will arrive at each square meter of the floor of the warehouse, INDEPENDENT OF the direction the water is spraying. The water could be in a stream straight down; the water could be spraying uniformly in all directions; the water could be spraying in a cone @ 45 degrees. The “vector components” of the water shooting out of the sprinklers is not important; only the net flow really matters.
Similarly, the vector components of the photons don’t really matter (for net transfer of energy in a transparent atmosphere). If 390 W/m^2 leave a 1mx1m square of the surface, then 390 W/m^2 will reach a 1mx1m square of a surface drawn around the earth at any height (again ignoring the absorption for the time being). We don’t need to worry the the 390 W are spreading out over some “sphere” or that some of the photons are going mostly sideways.
[Since the atmosphere is not perfectly transparent, then the directions would matter a bit, but that is a next level deeper than what I think we are discussing here.
Gravity is a slight inconvenience since gravity is not significant for photons, but the analogy works just fine in the absence of gravity, since the water is spraying generally toward the floor and will eventually get to the floor with or without gravity.
One final challenge in the analogy is that water drops splatter off one another, while photons tend not to interact, but again that is a minor distraction that doesn’t affect the result.]

Tim Folkerts
September 6, 2011 10:17 am

RJ says:

4) conservation of energy requires that the additional energy might result in slower surface cooling
would be more accurate
In that if 100 balls are in a box and 20 leave and one return. The box now contains 81 balls not 80. But I’m unsure if the one ball does anything more than returns (if it does in fact return) and immediately leaves again without having any impact.

I like your analogy. The box clearly represents a section of ground. The balls represent energy. Since no balls are arriving, it must be night. (During the day, “balls” of sunlight would be arriving.)
So we have a section of ground with some energy (and hence some temperature). During the night it loses 20 “balls of energy” — which must represent IR radiation leaving the ground. The returning “balls” would be IR coming from elsewhere (in this case the atmosphere) down to the ground.
We now consider two difference locations. In one location (call it “Phoenix), less energy heads down to the ground, so the ground has less energy (80 “balls” worth) at the end of the night and is cooler at dawn. In another location (call it “Atlanta”) more energy returns, so it has more energy (81 “balls” worth) and is a little warmer at dawn.
Your model has perfectly shown how “back-radiation” affects diurnal ranges. Areas where more energy shines down (humid, cloudy areas like Atlanta), the diurnal variation is less. Areas where less energy shines down (dry, clear areas like Phoenix), the diurnal variation is greater.
[The next step would be to note that the NEXT day when 20 “balls of sunshine” arrive, the area with more back-radiation will end up with more energy and hence a higher temperature (101 balls vs 100 balls). At this rate, the area with back-radiation would gain 1 “ball of energy” each day, warming up a bit more every day. At some point the area with back-radiation will warm up enough to emit 21 balls of energy (perhaps 110 balls at the start of the night, losing 21, gaining 20 sunshine and 1 “back-radiation” so it holds steady at 110 every morning, as opposed to the other area which will only be at 100 units of energy. This models the “radiative GHE” quite, although the relative values are not quite right. ]

RJ
September 6, 2011 11:25 am

“[The next step would be to note that the NEXT day when 20 “balls of sunshine” arrive, the area with more back-radiation will end up with more energy and hence a higher temperature (101 balls vs 100 balls).”
But it does not work this way. The highest maximum temperature in a day will never increase due to backradiation (unless the atmosphere is warmer than the surface)
All backradiation might do is increase the average surface temperature over a period due to a reduce cooling rate. The sun and only the sun warms (except where the atmosphere is warmer than the surface)
Its like a jacket can not increase the body’s heat above the heat from the body. But if the jersey had its own heat source then it might.

Tim Folkerts
September 6, 2011 1:05 pm

RJ says: “All backradiation might do is increase the average surface temperature over a period due to a reduce cooling rate. ”
YES! The cooling rate of the surface would be greater if it were surrounded by the 3K temperature of outer space (ie well below 0 C on average).
The current GHG’s and clouds “reduce the cooling” (ie prevent the earth from getting so close to 3K) by surrounding the earth with a ~270 K jacket. This allows the sun to warm the earth well above the temperature it could warm the earth without the help of the “reduced cooling”.
If we add more GHGs and reduce the cooling even more, the obvious conclusion is that the surface could be warmed even more by the sun. (At THIS point people can start arguing about how feedbacks might affect the amount of “reduced cooling”.)

Bob_FJ
September 6, 2011 4:03 pm

Tim Folkerts @ September 6, 2011 at 7:31 am
Tim, re your warehouse sprinkler analogy:
If you remove the walls, the outer (Omni directional) sprinklers would wet the ground ballistically outside the warehouse. The inner ones would not do this because of interference. Remove gravity and air drag and the outer sprinklers would spray all the way out to space.
A better analogy would be to replace the sprinklers with lamps that shine equally in all directions. (with no interference, and remove the walls). If you stand some distance from the warehouse you will see light coming at you horizontally. The more lamps there are, or the greater the power, the brighter will be this non-downward light. You could consider just having an opening in the wall at eye level, and painting everything inside black if you are worried about reflection.
I have no problem with integrating the downward light, but it is not all of the light.

jimmi_the_dalek
September 6, 2011 5:35 pm

I like Tim’s analogies with sprinklers and invisible mirrors to try and explain the integration over angles, but at the end of the day they are just analogies and no substitute for the maths. What you need to do is to look at the derivation of the Stefan-Boltzmann equation. The total power emitted from a ‘back body’ is obtained by taking Planck’s formula for the radiation density and integrating it
1) over all frequencies
2) over all solid angles in a hemisphere
3) over the area of the surface.
Now if you do 1) and 2) but not 3) you get the Stefan-Boltzmann formula i.e it is the power emitted by an infinitesimal area (a point) integrated over all angles. So the angles have already been taken care of. All that remains is to integrate over the area, which is simple – it just means take the power per unit area and multiply by the area. In other words ,you do not have to worry about angles , that has already been included.
Now you may not believe that, but if so you should get a good physics book and consult it. Or, this happens to be one of the cases where wikipedia gives the right answer, so you could just search for ‘Stefan-Boltzmann derivation’ and see what you get.

Bob_FJ
September 7, 2011 12:11 am

jimmi_the_dalek @ September 6, 2011 at 5:35 pm
Thanks for that Jimmi,

“…Now you may not believe that, but if so you should get a good physics book and consult it. Or, this happens to be one of the cases where wikipedia gives the right answer, so you could just search for ‘Stefan-Boltzmann derivation’ and see what you get.”

I’ve had a quick look at Wiki’s S-B derivation, and have some difficulty because I’ve not done maths like that for over 50years. (I’m an ol’ fart just like JAE and maybe Tim whom seems to admit recently the same difficulty). However, from a quick look it seems to me that Wiki’ describes an integration of hemispherical emission from a small flat surface emitting into a female hemispherical receptor. (= the original Stefan experiment?). Could you please describe how that integration becomes normal to the emitter (as you imply) rather than the total that it emits in all directions as received by the female hemisphere?
Oh and BTW, it has been mentioned somewhere that in various engineering problems such as in heat exchangers, that field of view calculations in two-body situations are not easy. However, if S-B is normal to the transmitter, it would be fairly straightforward, yes, no?
If for instance two bodies are inclined in modestly different planes and S-B emission is normal to those different planes…… Uh? ……. One of the bodies might receive nothing from the other! (which is obviously not the case)
Oh, and convex or spherical surfaces?

RJ
September 7, 2011 1:13 am

Tim Folkerts says:
September 6, 2011 at 1:05 pm
“the obvious conclusion is that the surface could be warmed even more by the sun.”
NO. It is not the obvious conclusion. In fact it is a flawed conclusion
CO2 and backradiation just at best slows the rate of cooling. (and therefore increases the AVERAGE temperature). It never warms (unless the atmosphere is warmer than the surface).
If the highest surface temperature possible on a given day in a given location is 20 degrees. Backradiation will never rise this to above 20 degrees. This is the mistake many so called climate scientists make.
The sun warms. Backradiation might slow the cooling rate after the sun has done the warming.

Bob_FJ
September 7, 2011 6:17 pm

Further to my September 7, 2011 at 12:11 am to Jimmi, I’ve had another thought on Jimmi’s recommendation to review the Wiki’ article on S-B law, and particularly its derivation. Here is an extract:

Finally, this proof started out only considering a small flat surface. However, any differentiable surface can be approximated by a bunch of small flat surfaces. So long as the geometry of the surface does not cause the blackbody to reabsorb its own radiation, the total energy radiated is just the sum of the energies radiated by each surface; and the total surface area is just the sum of the areas of each surface—so this law holds for all convex blackbodies, too, so long as the surface has the same temperature throughout.

http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Clearly, this is still talking of total energy radiated as having hemispherical distribution, and the sort of integration of surface emission talked of by Tim is not a simple summation of individual small flat surfaces emitting hemispherically. If we look at the Trenberth et al cartoon the field of view of the 390 W/m^2 is normal to the surface. However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.

kuhnkat
September 8, 2011 10:14 am

G Karst,
Don’t you love the voodoo of modern particle physics?? If you are in the FRAME of the particel it will have one mass, BUT, if you are in another frame observing the particle it will have another mass. In my ignorant opinion this is the area where relativity falls apart!!!
On another note, I have read where even photons have mass. This of course ties in with solar sails etc. I am just wondering when this will be built into the GCM’s. If the earth is emitting 390w/m2 maybe someone can compute what the “lift” for the atmosphere will be from all this energy going UP!!! Then there are the issues of the individual particles being struck by all these photons. Is it inertia or heat or both that is being transferred?? I am still waiting for someone to explain all this to us!!

kuhnkat
September 8, 2011 1:18 pm

Tim Folkerts,
“The current GHG’s and clouds “reduce the cooling””
Unfortunately you have stated only part of the issue. The water vapor and other specie absorb incoming solar radiation lowering the maximum the sun can raise the surface temp.
GHG’s are MODERATORS!!! They prevent the hot peaks AND the cold valleys!!!!! Both are necessary for the liveability of the earth!!

Bob_FJ
September 8, 2011 4:58 pm

The following diagram shows that even in a transparent atmosphere the surface emission vectors are not constant at all altitudes as claimed by Tim
http://bobfjones.files.wordpress.com/2011/09/surface-integration.png?w=588&h=568

Bob_FJ
September 8, 2011 5:31 pm

Kuhnkat,
I think you asked above somewhere how to post drawings.
I use MS Paint ,which is now fairly easy and more versatile to use in Windows 7. Older versions are cunningly designed to drive you mad and are harder to learn.
I used to store drawings/photos at the
Flickr website to access URL’s, but Yahoo have taken over, and they have foiled my attempts so far to access my previous work.
Instead, I have a FREE WordPress website for posting stuff on the web, such as several articles here at WUWT and Jo Nova‘s. So in the example above, I just posted it onto my website and grabbed the URL of the image.
It is easy to get a free WordPress website

kuhnkat
September 8, 2011 7:18 pm

Bob_FJ,
thanks for the suggestions.

Tim Folkerts
September 9, 2011 10:33 am

Bob_FJ says: September 6, 2011 at 4:03 pm

Tim, re your warehouse sprinkler analogy:
If you remove the walls, the outer (Omni directional) sprinklers would wet the ground ballistically outside the warehouse. The inner ones would not do this because of interference. Remove gravity and air drag and the outer sprinklers would spray all the way out to space.
A better analogy would be to replace the sprinklers with lamps that shine equally in all directions. (with no interference, and remove the walls). If you stand some distance from the warehouse you will see light coming at you horizontally. The more lamps there are, or the greater the power, the brighter will be this non-downward light. You could consider just having an opening in the wall at eye level, and painting everything inside black if you are worried about reflection.

Bob, the trouble here (and with your diagram in the later post) is that there is no “outside” the warehouse”. You never get to the “end of the earth” so there are always more “sprinklers (or lights) on the other side”.
The lights themselves ARE a good analogy. Let me leave you with this riddle. Suppose you have a large building (large enough that the walls are far away from you compared to the height of the ceiling — or you could paint the walls white and get pretty much the same result in a smaller room). The ceiling is covered with banks of lights (eg lights like this everywhere http://www.kulekat.com/articles/wp-content/themes/tuned-100/images/LED/led-grow-light-panel.jpg). Now suppose you make the roof twice as high. The lights, being farther away, will look dimmer. But being farther away, they will look closer together. Will the floor be illuminated better, worse, or the same as before? (I submit the room will look just a bright — again, assuming the wall are very far away from you, like a large warehouse or big-box store.)

Tim Folkerts
September 9, 2011 11:04 am

kuhnkat says:
Blockquote>Tim Folkerts says, “The current GHG’s and clouds “reduce the cooling””
Unfortunately you have stated only part of the issue. The water vapor and other specie absorb incoming solar radiation lowering the maximum the sun can raise the surface temp.
GHG’s are MODERATORS!!! They prevent the hot peaks AND the cold valleys!!!!! Both are necessary for the liveability of the earth!!
I don’t really disagree with what you said. I was pointing out that the main affect of GHGs on surface temperature is to emit “thermal” IR (> 4 um) toward the ground. Since sunlight has very little energy in the thermal IR spectrum, GHGs themselves absorb very little incoming solar radiation. This was the issue that was being discussed. The net effect of gaseous GHGs (which is a bit redundant since the second “G” = “gas”) like CO2, CH4, and H2O vapor is to keep the earth from being a much colder place. (H2O vapor DOES absorb a bit of near IR, so it would keep a BIT of sunlight out, but the effect on outgoing IR would be greater it seems, so it should still be a net warming effect.)
Clouds (liquid H2O), as you rightly point out, have several competing effects. They reflect away incoming sunlight, leading to a cooling affect (especially during the day). They emit thermal IR, leading to a warming affect (especially at night). This is a very handy moderating effect. They have a second moderating effect too: warmer weather -> more evaporation ->; more clouds ->; less sunlight –>; cooler weather.
* GHGs are “WARMERS” (but have minimal moderating effects that I can imagine).
* water droplets are “MODERATORS” (in a variety of ways)
It is the dual nature of water (gas and liquid) that makes it unique (and challenging) with its affects on climate.

kuhnkat
September 9, 2011 1:48 pm

Tim_Folkerts,
“I don’t really disagree with what you said. I was pointing out that the main affect of GHGs on surface temperature is to emit “thermal” IR (> 4 um) toward the ground.”
Even that is debatable. The tired cartoon shows half up and half down when some part of that energy is being transferred to the local atmosphere speeding convection. We know the conductivity of the ground and the atmosphere is very poor so the transfer speeding convection each morning would not be very fast based on conductivity only.
It would be really nice if they someday gave us a better cartoon to play with. My WAG is that in the lower atmosphere the transfer of energy through collision is actually as, if not more, important than the DLR.

kuhnkat
September 9, 2011 1:51 pm

Tim Folkerts,
Doesn’t water vapor also have very wide absorption characteristics similar to water droplets?? I’ve looked it up before but may have only looked at liquid water spectra.

Bob_FJ
September 9, 2011 4:29 pm

Tim Folkerts @ September 9, 2011 at 10:33 am
Nice try Tim, but you overlook the fundamental issue. Unlike your water sprinkler analogy, there is no interference between light waves, so that even those in the middle of a very large array will keep going until there is some other interference, such as your warehouse walls. In a transparent atmosphere, the horizontal stuff will head out to space, but cannot be seen in the normal field of view. (as depicted in the Trenberth cartoon). That “missing” stuff is part of the Trenberth S-B 390 W/m^2, that he illustrates as entirely normal to the surface, so thus there is a paradox
BTW, did you notice that your hero Travesty Trenberth has been very naughty again, this time in the fracas over the Spencer and Braswell paper.

Bob_FJ
September 9, 2011 4:40 pm

Dear Moderator,
Sorry, but in my September 8, 2011 at 5:31 pm , I made a wee mistake causing later posts to be all in italics. In line 2, I inserted the italics closure slash the wrong side of the I. Sorry for the chaos. Would it help if I Email the whole thing?

Bob_FJ
September 9, 2011 10:31 pm

jimmi_the_dalek,
On a day of awful weather, I’ve just been browsing up through this humungous thread, and found yours of August 31, 2011 at 7:19 pm

Re. this stuff about the Stefan-Boltzmann equation and hemispherical sections, would it help if it were pointed out that the derivation of said equation includes an integration over solid angles so that the quantity computed is the emission normal to the surface?

Nope! Unless you can validly contradict what Wicki’ says about it, as I repeat next:
Further to my September 7, 2011 at 12:11 am to Jimmi, I’ve had another thought on Jimmi’s recommendation to review the Wiki’ article on S-B law, and particularly its derivation. Here is an extract:

Finally, this proof started out only considering a small flat surface. However, any differentiable surface can be approximated by a bunch of small flat surfaces. So long as the geometry of the surface does not cause the blackbody to reabsorb its own radiation, the total energy radiated is just the sum of the energies radiated by each surface; and the total surface area is just the sum of the areas of each surface—so this law holds for all convex blackbodies, too, so long as the surface has the same temperature throughout.

http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Clearly, this is still talking of total energy radiated as having hemispherical distribution, and the sort of integration of surface emission talked of by Tim is not a simple summation of individual small flat surfaces emitting hemispherically. If we look at the Trenberth et al cartoon the field of view of the 390 W/m^2 is normal to the surface. However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.

Tim Folkerts
September 10, 2011 8:30 pm


kuhnkat says:
>Doesn’t water vapor also have very wide absorption characteristics similar
>to water droplets?? I’ve looked it up before but may have only looked at
>liquid water spectra.
here are copies of spectra that represent what I understand of the spectra for liquid and gaseous water.
http://www.randombio.com/spectra.png
http://www.physics.umd.edu/grt/taj/104a/watopt.gif
The scales are rather different, but they show distinct sorts of behavior. H2O gas has several moderately broad bands of absorption in the IR range. H2O liquid is clearest near 400 nm, absorbing progressively more effectively on either side of that minimum.

kuhnkat
September 11, 2011 11:51 am

Tim_Folkerts,
So, water vapor DOES absorb in the near infrared where there is a rather large amount of insolation. (about 51% of solar is IR with most of it in the near infrared) Water vapor, the most important GHG, then is a moderator of insolation.
CO2 will cool or warm depending on the temperatures of what is providing the energy. CO2, at 390ppm, simply has too little mass in the atmosphere to make a difference as far as a storage media while its ability to absorb and either emit and/or transfer energy through collision is significant. Basically it is an energy distributor similar to water vapor, more efficient in a couple of bands but not covering as wide a spectra.

Tim Folkerts
September 11, 2011 3:11 pm

KuhnKat,
Water vapor and water droplets do still have one major difference that springs to mind in relation to “moderation”. The addition of droplets (ie clouds) reflects energy away from the earth, so that energy is completely removed, leading to a definite cooling effect . The addition of H2O gas absorbs the energy — less does indeed get to the surface, but more gets in the atmosphere. I suspect the net result for more H2O gas would would lead ti small changes in the surface temperature, so it is not nearly the same sort of “moderator” that clouds would be..

Tim Folkerts
September 11, 2011 3:47 pm

Bob, I am not sure what the purpose of this statement was: “BTW, did you notice that your hero Travesty Trenberth has been very naughty again, this time in the fracas over the Spencer and Braswell paper.”
Science is not about “heroes” or “villians”. When did I ever claim Trenberth or anyone else was a “hero”? The “fracas” seems to have a lot of losers and no winners — the biggest loser probably being the reputation of “science” in the eyes of the public. But eventually the science will sort itself out, independent of the egos and hot air on both sides.
And speaking of science sorting itself out — if there really was a glaring error in the magnitudes of the IR energy values in the Trenberth paper, don’t you think people like Spencer or Braswell or Lindzen would have pointed it out long ago? If you are going to make an “extraordinary claim” that numbers that are widely accepted even by “skeptical” scientists are wrong, then you will need “extraordinary evidence”, not simply gut feelings or analogies. Do the calculations and show us what the correct numbers should be, and submit the results to be published.

Bob_FJ
September 11, 2011 7:15 pm

Tim Folkerts @ September 11, 2011 at 3:47 pm
I thought my sarcasm over Travesty Trenberth would get you going, and I agree with your response. He does seem to excel in that sort of thing though, what?
Coming back to the paradox, there are some first principles that need to be established before dreaming about mathematical considerations. You did say earlier something like; first understand how it works in a transparent atmosphere before figuring out the effects of GHG’s etc. Now thence is where the initial surface emission gets modified and I guess will approach towards radiation normal to the surface, but very, very complicated, and that is a different argument to the claim that S-B radiation from the surface IS normal.
1) Do you agree that EMR waves pass through each other, without interference?
2) Do you agree that in a transparent atmosphere, the horizontal components of hemispherical radiation will escape to space approaching the tangential rather than normal?
3) Do you agree that an observer looking down, with a normal field of view, will not be able to see the tangential stuff?
4) Do you agree that if the observer shifts to a location where the first mentioned tangential stuff can be seen normally, there will be new tangential stuff from other locations that will then not be viewable?
5) Do you agree that the tangential stuff IS part of the S-B hemispherical emission, but is not normal as depicted by Trenberth et al?

Tim Folkerts
September 12, 2011 4:06 pm

Bob, I really am not quite sure what you are saying …
>1) Do you agree that EMR waves pass through each other, without interference?
Sure. (Well, there are some photon-photon interactions that are possible, but they are not important in this setting).
>2) Do you agree that in a transparent atmosphere, the horizontal components
>of hemispherical radiation will escape to space approaching the tangential
>rather than normal?’
What do you mean by “horizontal components of hemispherical radiation”? There are no purely horizontal photons. Certainly the more horizontal they are , the more tangentially they will approach some arbitrarily drawn sphere that we call the “top of atmosphere”. But all that is really important is if they do indeed escape somewhere, no the angle they are heading.
>3) Do you agree that an observer looking down, with a normal field of view, >
>will not be able to see the tangential stuff?
Do you mean normal = “perpendicular”? And what sort of “observer” do you have in mind. The only truly logical “observer” would be a meter that detects all radiation passing thru a given part of the surface in question. For example, i could put a small, horizontal black body there or a pyrgeometer pointing downward. Either of these would “see” radiation from any direction that happens to hit the “observer”.
It would be odd to only look at radiation that is heading straight up (ie normal = perpendicular). In fact, there would only be an infinitesimal amount of energy heading precisely vertically. So I guess, if you “put on blinders” so that only vertical photons were counted, you would get a much smaller number for the photon energy — way less than 390 W/m^2
>4) Do you agree that if the observer shifts to a location where the
>first mentioned tangential stuff can be seen normally, there will be
>new tangential stuff from other locations that will then not be viewable?
If I simply shifted my ANGLE (not position) and looked only at the photons heading straight in along that angle, then certainly you would see a different set of photons.
>5) Do you agree that the tangential stuff IS part of the S-B hemispherical emission,
>but is not normal as depicted by Trenberth et al?
I am quite sure Trenberth is not trying to depict the upward or downward IR as perpendicular (normal) to the surface.
I agree that the PARTIALLY-tangential stuff that is seen when the angle is shifted is indeed part of the SB emission. And when this is added up over all angles, it will indeed give the 390 W/m of SB radiation. (depending of course on the exact temperature of the ground/oceans. And then we would get to the details of the absorption by GHGs)
Maybe this is the crux: “However, with hemispherical radiation, the horizontal stuff is part of the 390 and in a transparent atmosphere would escape to space outside the field of view.”
There is no “outside the field of view”. The 390 from 1 m^2 of the surface would not all travel vertically thru the corresponding 1 m^2 directly above it. Some would “escape” at an angle thru other square meters. But energy from other square meters of the ground would head slightly sideways to pass thru the given upper square meter.
In the end, these must all balance (other than the slight decrease due to the curvature of the earth.)

Bob_FJ
September 12, 2011 5:30 pm

Tim Folkerts @ September 11, 2011 at 3:47 pm
Tim, you do raise a good point with your;

And speaking of science sorting itself out — if there really was a glaring error in the magnitudes of the IR energy values in the Trenberth paper, don’t you think people like Spencer or Braswell or Lindzen would have pointed it out long ago?

I think it is no secret that there has been a war going on between Dessler et al and Spencer et al concerning water vapour/cloud feedbacks*, what with their conflicting papers in GRL a few years ago etc. More recently, I was puzzled why so much emphasis in this war was concerning the feedback radiative effects, when, if we are to believe the Trenberth cartoon, by far the greatest surface cooling comes from thermals and evapotranspiration (~60%). I had several exchanges with Roy Spencer, suggesting that a small change in “convective factors” would intuitively have more negative feedback effect than was being researched in the feedback radiative effects. The outcome was that he agreed it was important, and that it was taken into account in the GCM’s, but, by reading between the lines, everyone was too busy trying to find the radiative stuff to worry about actually quantifying “convection”.
Perhaps the sceptics do not take the cartoon too seriously, and anyway the 390 W/m^2 is not all that important, it is the net heat loss that is, some of which goes directly to space BTW.
Here is a much more popular diagram, that is all over the NASA divisions etc, that does not even try to portray back radiation:
http://science-edu.larc.nasa.gov/EDDOCS/images/Erb/components2.gif
*That Dessler is desperate not to lose!

Bob_FJ
September 13, 2011 3:49 pm

Tim Folkerts @ September 12, 2011 at 4:06 pm
Groan, since what has been discussed only applies to a transparent atmosphere, let’s move on to what happens, per quantum theory, with opacity.
The S-B emission from the surface is hemispherical, but is instantaneously halted by the GHG’s etc in a rather complicated way. Photon free path length (FPL) will vary considerably according to the Planck energy distribution, the water vapour level, (0 to ~4%), lapse rate, and other stuff, together with heap big thermalization of O2 and N2. For consideration purposes, let’s hazard that the FPL varies between 100m and 1,000m. At the moment of initial absorption, the photons cease to exist, but others, not necessarily of the same energy level will be emitted, some as a consequence of collisions. Some might be from relatively high altitude, but the majority will be close to the surface, because as has been demonstrated earlier, most of the radiation is “sideways” through 360 degrees. Clearly at this point, S-B is no longer relevant, and “reemission” becomes spherical. All of this is a continuous process of course.
Because EMR is weird, (and not HEAT), all that S-B sideways stuff is part of the 390 W/m^2 but doesn’t transfer HEAT. A similar situation exists with the spherical reemissions; a significant proportion whizzes around sideways, doing nothing. So, let’s guess that from about 1m up from the surface, up (and down) flux has nothing to do with S-B. (which applies nominally to black bodies)

Tim Folkerts
September 14, 2011 8:36 am

Bob,
I think we have gotten to (or perhaps gotten well beyond) the point where we can be productive in this setting. I agree with many things you said, but sorting out the precise details would be ineffective here.
Just to name a few details that are tough to hammer out:
* what does it mean to radiate energy from an area (eg “390 W/m^2”) when dealing with energy coming from a volume (eg radiation from GHGs)?
* in thermodynamics, “heat” is formally defined as a process of energy transfer, akin to “work”. You can’t “transfer work” and you can’t “transfer heat”. Often, heat is understood to mean “thermal energy”. Without more precise agreement on definitions, we will easily talk past each other.
* from outer space, you can detect some of the IR photons that left the ground, so SB has something to do with the radiation anywhere in or above the atmosphere. But deciding how much it has to do wit the radiation is more subtle.
* I don’t think we have really agreed on the simplified situation of a sphere at uniform temperature radiating through a transparent atmosphere. Trying to discuss more complex situations from a shaky foundation is a recipe for failure.