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.
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Tim Folkerts [1] @ur momisugly August 27, 2011 at 9:46 am and [2] @ur momisugly August 27, 2011 at 9:12 pm
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.
Tim Folkerts @ur momisugly August 28, 2011 at 5:02 pm
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
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.
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% @ur momisugly 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”.
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!!!!!!!!!!!
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.
JAE, @ur momisugly 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!
Tim Folkerts @ur momisugly 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)
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”?
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’?
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).
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.
RJ–reduced cooling might limit temperature extremes over a 24-hour period, but it cannot do anything to the average temperature.
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)
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 housethe earth by addinginsulationGHGs that limits how easily energy can escape (assuming the temperatureoutsideof outer space is cooler thaninsidethe temperature of the earth). For a given output frommy furnacethe sun, theinsulationGHGs will make makemy housethe 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)
Double Dang! There are a few mistakes in the strike-throughs above. It should read
I HOPE that fixes it all !
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”?
RJ says: August 29, 2011 at 9:16 am
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?
Tim Folkerts,
Here is as good an answer as you wil get, from the foremost expert in the field of Climatology: click
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.
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?
PPS: The comments I have expressed above have been virtually IGNORED for over 3 years, now. I just gotta ask WHY?
WILLIS?
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??
JAE @ur momisugly August 25, 2011 at 8:26 pm
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.
Jae @ur momisugly August 29, 2011 at 7:18 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.