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.
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.
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
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?
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?
anna v,
I bought one of those.
How do you turn it on????
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.
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
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…
Snap! 🙂
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
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.
Myrrh @ur momisugly August 17, 2011 at 3:05 pm
I see that you are apparently ignoring the issues I raised @ur momisugly 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)
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.
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.
jimmi_the_dalek says:
August 17, 2011 at 5:04 pm
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I accept that not the entire spectrum of the solar radiation goes to heat the oceans
“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.
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% –
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.
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.
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.
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.
Alexander Duranko says:
August 17, 2011 at 6:14 am
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.)
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.
“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.
Tim Folkerts,
That sounds just like Venus.
Tim Folkerts @ur momisugly August 17, 2011 at 5:39 pm
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
@Tim Folkerts says at 17th August 5:39pm
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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.
Distilled water is transparent to visible light and has a remarkable diamond like clarity, even compared to tap water.
See http://en.wikipedia.org/wiki/File:Water_absorption_spectrum.png