Guest Post by Willis Eschenbach.
For all of its faults, the IPCC (Intergovernmental Panel on Climate Change) lays out their idea of the climate paradigm pretty clearly. A fundamental part of this paradigm is that the long-term change in global average surface temperature is a linear function of the long-term change in what is called the “radiative forcing”. Today I found myself contemplating the concept of radiative forcing, usually referred to just as “forcing”.
So … what is radiative forcing when it’s at home? Well, that gets a bit complex … in the history chapter of the Fourth Assessment Report (AR4), the IPCC says of the origination of the concept (emphasis mine):
The concept of radiative forcing (RF) as the radiative imbalance (W m–2) in the climate system at the top of the atmosphere caused by the addition of a greenhouse gas (or other change) was established at the time and summarised in Chapter 2 of the WGI FAR [First Assessment Report].
Figure 1. A graph of temperature versus altitude, showing how the tropopause is higher in the tropics and lower at the poles. The tropopause marks the boundary between the troposphere (the lowest atmospheric layer) and the stratosphere. SOURCE
The concept of radiative forcing was clearly stated in the Third Assessment Report (TAR), which defined radiative forcing as follows:
The radiative forcing of the surface-troposphere system due to the perturbation in or the introduction of an agent (say, a change in greenhouse gas concentrations) is the change in net (down minus up) irradiance (solar plus long-wave; in Wm-2) at the tropopause AFTER allowing for stratospheric temperatures to readjust to radiative equilibrium, but with surface and tropospheric temperatures and state held fixed at the unperturbed values.
In the context of climate change, the term forcing is restricted to changes in the radiation balance of the surface-troposphere system imposed by external factors, with no changes in stratospheric dynamics, without any surface and tropospheric feedbacks in operation (i.e., no secondary effects induced because of changes in tropospheric motions or its thermodynamic state), and with no dynamically-induced changes in the amount and distribution of atmospheric water (vapour, liquid, and solid forms).
So what’s not to like about that definition of forcing?
Well, the main thing that I don’t like about the definition is that it is not a definition of a measurable physical quantity.
We can measure the average surface temperature, or at least estimate it in a consistent fashion from a number of measurements. But we can never measure the change in the radiation balance at the troposphere AFTER the stratosphere has readjusted, but with the surface and tropospheric temperatures held fixed. You can’t hold any part of the climate fixed. It simply can not be done. This means that the IPCC vision of radiative forcing is a purely imaginary value, forever incapable of experimental confirmation or measurement.
The problem is that the surface and tropospheric temperatures respond to changes in radiation with a time scale on the order of seconds. The instant that the sun hits the surface, it starts affecting the surface temperature. Even hourly measurements of radiative imbalances reflect the changing temperatures of the surface and the troposphere during that hour. There is no way that we can have the “surface and tropospheric temperatures and state held fixed at the unperturbed values” as is required by the IPCC formulation.
There is a second difficulty with the IPCC definition of radiative forcing, a practical problem. This is that the forcing is defined by the IPCC as being measured at the tropopause. The tropopause is the boundary between the troposphere (the lowest atmospheric layer, where weather occurs), and the stratosphere above it. Unfortunately, the tropopause varies in height from the tropics to the poles, from day to night, and from summer to winter. The tropopause is a most vaguely located, vagrant, and ill-mannered creature that is neither stratosphere nor troposphere. One authority defines it as:
The boundary between the troposphere and the stratosphere, where an abrupt change in lapse rate usually occurs. It is defined as the lowest level at which the lapse rate decreases to 2 °C/km or less, provided that the average lapse rate between this level and all higher levels within 2 km does not exceed 2 °C/km.
This is an interesting definition. It highlights that there can be two or more layers that look like the tropopause (little temperature change with altitude), and if there is more than one, this definition always chooses the one at the higher altitude.
In any case, the issue arises because under the IPCC definition the radiation balance is measured at the tropopause. But it is very difficult to measure the radiation, either upwelling or downwelling, at the tropopause. You can’t do it from the ground, and you can’t do it from a satellite. You have to do it from a balloon or an airplane, while taking continuous temperature measurements so you can identify the altitude of the tropopause at that particular place and time. As a result, we will never be able to measure it on a global basis.
So even if we were not already talking about an unmeasurable quantity (radiative change with stratosphere reacting and surface and tropospheric temperatures held fixed), because of practical difficulties we still wouldn’t be able to measure the radiation at the tropopause in any global, regional, or even local sense. All we have is scattered point measurements, far from enough to establish a global average.
This is very unfortunate. It means that “radiative forcing” as defined by the IPCC is not measurable for two separate reasons, one practical, the other that the definition involves an imaginary and physically impossible situation.
In my experience, this is unusual in theories of physical phenomena. I don’t know of other scientific fields that base fundamental concepts on an unmeasurable imaginary variable rather than a measurable physical variable. Climate science is already strange enough, because it studies averages rather than observations. But this definition of forcing pushes the field into unreality.
Here is the main problem. Under the IPCC’s definition, radiative forcing cannot ever be measured. This makes it impossible to falsify the central idea that the change in surface temperature is a linear function of the change in forcing. Since we cannot measure the forcing, how can that be falsified (or proven)?
It is for this reason that I use a slightly different definition of the forcing. This is the net radiative change, not at the troposphere, but at the TOA (top of atmosphere, often taken to mean 20 km for practical purposes).
And rather than some imaginary measurement after some but not all parts of the climate have reacted, I use the forcing AFTER all parts of the climate have readjusted to the change. Any measurement we can take already must include whatever readjustments of the surface and tropospheric temperatures that have taken place since the last measurement. It is this definition of “radiative forcing” that I used in my recent post, An Interim Look at Intermediate Sensitivity.
I don’t have any particular conclusions in this post, other than this is a heck of a way to run a railroad, using imaginary values that can never be measured or verified.
w.
AlecM says:
December 18, 2012 at 3:24 am
Planck didn’t even use photons; he just found that if he assumed that the actual emission and absorption had to take place in little packets, and then the math worked out. This avoided the so-called “ultraviolet catastrophe” wherein the emission spectrum, in theory, increased without bound at shorter wavelengths.
In 1905, Einstein expanded on this idea to explain the photoelectric effect: by postulating that light itself was quantized, with the energy of each photon proportional to its frequency. High frequency photons could eject electrons from a piece of metal, perhaps with energy left over. Below some threshold, lower-frequency photons could not, even if you had a lot of them. Maxwell’s equations couldn’t explain this. Over time, Einstein’s hypothesis gained support, including a Nobel prize in 1921. (Relativity was part ot that too, but soft-pedaled because it still lacked empirical evidence.) Once Arthur Compton bounced photons off of electrons like billiard balls, and demonstrated the corresponding energy and momentum transfer in both, the photon idea was pretty much universally accepted. Compton was awarded a Nobel prize for this, too. At higher energies, it’s relatively easy to see the effects of single photons. In some gamma rayimaging equipment, you can watch photons arrive one at a time; or you can hear them as they’re detected in a Geiger counter.
The Poynting vector of a wavefront is an abstraction, and in constructing it, you lose information about the electric and magnetic fields of that wavefront. The resultant Poynting vector added from several others doesn’t necessarily give the direction of any wavefront. To think otherwise would suggest that you could combine two flashlight beams, or laser beams, at 90 degrees, and get a single beam coming off at 45 degrees from each of the original ones. Or, that you could park two cars facing one another, shine the headlights of each right into the headlights of the other, and make them both go dark, or at least dim. This seems to be like what you are suggesting for IR; if it would work for IR, why not for visible light?
You can stick with a Maxwell/Planck/Poynting interpretation if you like. But modern physicists tend to think that emission and absorption of photons is the best way to describe radiative energy transfer. Pyoynting vectors have their place, but that place is limited. If you have a comprehensive refutation of Einstein and Compton, not to mention Heisenberg and many others, that will convince all or most physicists working with radiation transport, I’ll look forward to seeing it.
in
December 18, 2012 at 3:24 pm
“best way to describe energy transfer” should be “best way to describe radiative energy transfer.”
[FIXED. -w]
Robert Clemenzi says, December 18, 2012 at 2:56 pm: “Greg, assume that there is spherical shell around the Sun with the same radius as the distance between the Sun and the Earth. […] When looking at the Earth from the Sun, the amount of energy available to be absorbed is equal to 1368 W/m2 times the area of a disk with the same radius as the spherical Earth. However, when you consider a period of 24 hours, that energy is distributed over the entire surface of a sphere.”
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Robert, about “W/m2”, please, go back and read my previous comment (December 18, 2012 at 2:35 pm). The argument there is actually sufficient. But I do not mind to talk about other details.
As for 24 hours, the sphere makes one full rotation, hence it is not that only a hemisphere (or only one side of an imaginable disk, for debate’s sake) receives radiation, it is the whole sphere.
Greg House says:
December 18, 2012 at 3:53 pm
OK, the argument there was:
You go on to say:
I’m not sure what the fuss is about here.
The amount of solar energy available at the top of the atmosphere is called the “TSI”, the total solar irradiance. It is expressed in units of watts per square metre perpendicular to the sun’s rays.
It is usually taken to be about 1366 W/m2.
How much of that hits the earth? Well, that would be calculated as the total solar irradiation, TSI, times the area of the sun’s rays intersepted by the earth, which is Pi r^2.
TSI (Watts/m2) * Pi r^2 (m2) = total intercepted energy (watts)
where “r” is the radius of the earth. The units of square metres cancel out, so the answer is in watts. This gives us the total amount of solar energy intercepted by the earth.
Now, what does that work out to be as a per-square-metre average over the entire earth’s surface? To figure that out, we divide total intercepted energy by the area of the earth’s surface, which is 4 Pi r^2
intercepted energy (watts)
—————————————- = average energy (W/m2)
surface area (square metres)
Substituting for the numerator and the denominator, we get
TSI (Watts/m2) * Pi r^2 (m2)
—————————————- = average energy (W/m2)
4 Pi r^2 (square metres)
The Pi r^2 cancels out, leaving
TSI (Watts/m2)
—————————————- = average energy (W/m2) = 1366/4 ≈ 342 W/m2
4 (m2)
So as a global average, it is common to say that the solar irradiation is 342 W/m2. This is not the total sun available at a particular time and place. Heck, at the equator at noon, you get over a kilowatt per square metre at the surface. But it is the global average sun available at the top of the atmosphere.
Is this global average a useful number? Sure, for things that it is useful for … but you don’t want to use it in situations where it is not useful.
Am I missing something? Like I said, not sure what the fuss is about.
Best to all,
w.
Hmmm, Take the average of the lengths of a rectangle 1cm x 9cm and square the average to find the area, then you’ll realise just why playing with averages is not a good idea in non-linear systems.
Robert Clememzi: Use whatever mirror you like Robert. Make it just a polished metal surface if you want. Put a nice absorbent black backing on the other side and allow the heat to conduct out from the bucket. No problem. Entirely up to you. Not sure glass is quite so opaque to IR as you imagine – perhaps you were thinking of UV?
Spacecraft? Perhaps we should wrap them in bubble-wrap filled with CO2? Just a thought? Or maybe they only need 200km of atmosphere around them to keep them warm. Anyway, I’m off to take the radiator out of the car because someone told me that painting the engine white would be all the heat dissipation I need and then I’m going to write to Intel to tell them to take the fans of their processors and put them in clear glass packages to let the IR out.
Willis Eschenbach says, December 18, 2012 at 4:42 pm: “The amount of solar energy available at the top of the atmosphere is called the “TSI”, the total solar irradiance. It is expressed in units of watts per square metre perpendicular to the sun’s rays. It is usually taken to be about 1366 W/m2. […] Am I missing something?”
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Yes, you are: my point.
As I said, a square meter is always a square meter, regardless the form of the surface. Given the premiss of 1366 W/m2, the surface of a hemisphere would have 2 times as much square meter as the surface of a disk of the same radius, hence a hemisphere would receive 2 times as much Watt as the disk of the same radius.
To your “perpendicular”, this is not scientific at all, Willis, because in units like “watts per square metre” or “whatever per square metre” the “square metre” is not a geometrical figure and can not be perpendicular to anything.
Hence, based on he premiss of 1366 W/m2, a hemisphere would receive 2 times as much Watt as a disk of the same radius. Sorry again, but this is a simple math and a simple logic.
If your calculation ignores that, and it does, it is simply wrong.
Greg, I’m sorry, but I couldn’t follow that at all.
I’ve given it to you in simple math. If you can find a problem with my math, let me know. It’s not complex math, it’s very straightforward.
If you can’t find any problems with my simple math (average solar energy per square metre equals total solar energy available divided by total square metres of Earth surface area, it’s not rocket science), please don’t send further explanations and imaginations of this type about square metres and the like, I can’t make sense of them.
Regarding your curious aversion to the concept of “perpendicular”, I just looked up what NASA says on the subject:
See that part about “perpendicular to the path”? That’s all I was saying. Sorry for the confusion.
w.
It’s amazing how everybody reverts to hypothetical unreal geometrical calculations? Newsflash. The sun is a big ball of fire in the heavens . It is real and the measured flux,power whatever (its irrelevant because it’s timeless) is 1360oddw/sq.m
Mack says:
December 18, 2012 at 5:54 pm
My goodness. Simple arithmetic is now “hypothetical unreal geometrical calculations”.
Mack, these kinds of calculations were developed by the Egyptians for very practical reasons.
Yes, the sun provides 1360-odd watts, as you say.
This is different from the average amount of solar energy received by a square metre of the surface of the Earth.
That is calculated as total energy divided by total surface area.
If you truly think that something as simple as energy divided by surface area is a “hypothetical unreal geometrical calculation”, I fear I don’t know how to assist you, other than to suggest that while the condition persists you stay out of scientific discussions, and that for your own cranial safety you avoid subjects like “normal uniform rational B-spline curves”.
w.
Willis Eschenbach says, December 18, 2012 at 5:45 pm: “If you can find a problem with my math, let me know. […]
Regarding your curious aversion to the concept of “perpendicular”, I just looked up what NASA says on the subject: “In addition, the total solar irradiance is the maximum power the Sun can deliver to a surface that is perpendicular to the path of incoming light.””
=============================================================
Right, theY say that; thus contradicting what they also say on the same page above: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.
Note, by the way: one time the total solar irradiance (TSI) is an average, then it is a maximum. An average can not be a maximum at the same time.
I guess, what they measured was indeed an average, and this average includes the whole hemisphere. Hence, this 1360 W/m2 has nothing to do with a perpendicular. I see a sign of distortion (deliberate or not is another story) there, and their calculation (like yours as well) is therefore wrong.
Now to the problem with your math, it is obvious: your Pi r^2 (m2) (area of a disk) is wrong, 2Pi r^2 (m2) (area of a hemisphere) would be right. OMG, THE “EARTH WITHOUT ATMOSPHERE” IS NOT SO F—ING COLD ANY LONGER! (sarc, sorry)
Of course, there is another general problem with operating with averages, Ryan has already said that (December 18, 2012 at 4:57 pm).
Willis, did I leave you speechless? That WOULD be a FIRST!
I hope the few CAPS did not bug you…
Well, I guess that should read: “Willis, did I leave you RESPONSELESS?” Just where oh where is some physical evidence of a “radiative greenhouse effect?” Nice construct, but no data. No data, no science. Period.
Greg House says:
December 18, 2012 at 6:28 pm
The Earth’s orbit around the Sun is an ellipse. As a result, the amount of energy available varies over the course of a year. Integrated over that year, and then divided by the elapsed time, provides an average TSI for the year.
The Earth rotates on its axis once per day. Integrating over that time, and then dividing by the elapsed time, provides an average value received at the surface which is one fourth the TSI. The surface receives a maximum amount of energy when it is perpendicular to the path of incoming light. At any other angle, the amount of energy “per unit area” is reduced.
Try this. Point a flash light directly at a wall. Compare the area covered with the area covered when the same flash light is pointed at the wall at a large angle. In both cases, the same amount of light hits the wall. However, the area of the two beams is quite different. This is a demonstration of why the watts per unit area are different depending on where it is measured perpendicular to the surface of a sphere (such as the Earth). As long as you measure the power hitting a plane perpendicular to the energy source, the location on the sun lit side of a sphere won’t matter. But that energy is spread out over a larger area depending on the cosine of the angle between a line perpendicular to the surface and the direction of the Sun.
jae says:
///////////////////////////////
The link was bad, but not that hard to debug. Either of these will work.
http://www.warwickhughes.com/papers/idso98.htm
http://www.warwickhughes.com/papers/Idso_CR_1998.pdf
Willis writes “You seem to think that that heat does nothing because at equilibrium, an equivalent amount of heat is being lost … but if we took the DLR away, the ocean would cool rapidly and the mixed layer would end up at a much cooler temperature.”
You seem to think I dont believe that DLR is what causes the oceans to be at the temperature they are. I have told you many times I do believe DLR is responsible for that and I am trying to discuss that process with you because your earlier comments show a lack of understanding what “cooling” really means.
You mention a rock warming in the sun at equilibrium and then point out that adding more DLR by say adding GHGs makes the rock warmer. You think its because more DLR means the rock absorbs more energy from the DLR and heats up. This is incorrect. Or at least a less than optimal way of looking at it because you’re likely to make errors in reasoning like you did with tallbloke.
Here is a simple thought experiment to show you. If the said rock is in equilibrium in the sun and you instantaneously do two things. Turn off the sun. And add as much DLR producing stuff you like. Have it wrapped in shiny tin foil if you like…
What can you say about the DLR. Can you say anything about its increase or its limit?
What immediately starts to happen to the temperature of the rock.?
The answers to those questions I believe are relatively self evident and they show us that it is not the DLR that heats anything. Not even a little bit. The process is one of reduced cooling and the way it works is that the DLR absorbed by the rock (yes its absorbed) is radiated away again. Net energy flow in the IR range is away from the rock.
You can say that mathematically you simply add the radiation and absorbtion to arrive at the net energy flow and note that the absorbtion component is towards the rock. You can even say that for a rock the absorbed energy can mix freely with any other energy and is completely indistinguishable from it.
The issue with the ocean is that unlike a rock, this DLR cannot be absorbed any deeper than 10um (its really now a physical limitation due to the cool skin) and so the mechanism for warming is different to that of a rock. This is the process you’re getting wrong and I’ve been describing.
I’ve said it before and I’ll say it again, the very suface does not and cannot become warmer than the bulk through increases in DLR and transfer that DLR supplied energy down. Warming is from the bottom up.
Robert Clemenzi says, December 18, 2012 at 7:18 pm : […]
=========================================================
Robert, thank you for the explanation. I agree, actually, with one exception, but this exception is not really relevant at the moment, so maybe later.
The point is, however, at least as I see it, if you look closely at what I quoted from that NASA link Willis generously provided, you will see that those different angles/the whole hemisphere must have been included in their average. Look at their formulation again: “At Earth’s average distance from the Sun (about 150 million kilometers), the average intensity of solar energy reaching the top of the atmosphere directly facing the Sun is about 1,360 watts per square meter, according to measurements made by the most recent NASA satellite missions. This amount of power is known as the total solar irradiance”.
What I suspect is this: one department measured/calculated the real average throughout the hemisphere and another (climate) department either deliberately distorted or simply misunderstood that and got as a result a much colder “Earth without atmosphere”.
jae says:
December 18, 2012 at 6:31 pm
Sorry, jae, I didn’t see your post. I’ve fixed the link.
Thanks for persevering,
w.
Greg House says:
December 18, 2012 at 6:28 pm
You might guess that, but you’d be wrong. Actually, what they mean by “average” is that the earth’s orbit is not a perfect circle, so it is closer to the sun during part of the year and further from the sun the other part of the year. As a result, you have to average over a year to get the average TSI …
In other words, the average has nothing to do with what you fantasize it is about. Instead, it’s about the average distance from the sun. The clue is where they say “At Earth’s average distance from the sun, the average intensity …”
w.
Now, that number (1366) is “sort of” valid – after you follow all of the averages properly.
But – that number is true ONLY for top of atmosphere.
At the ground (where the radiation actually has to heat the ground/ice/water/trees/grass/desert/rock/tundra/etc), the CAGW crowd tends to keep playing the “average radiation per year” bit. At which point they ARE wrong. (Well, actually, they have to keep playing that “average” game, because the real numbers fail their approximations and averages.)
At the ground, each watt of sunlight is diffused over a larger and larger area as the latitude increases towards the pole.
Further, the area being irradiated decreases as you go further north: sunlight between the tropic of Cancer and tropics of Capricorn is higher in the sky, penetrates less air mass, and is more high;y concentrated that between 67.5 (north or south) and the pole.
Yes, at the equator, at local solar noon, the air mass = 1.0 (on a clear day, no haze, dust,clouds, or aerosols, etc.) But at the 82 north, air mass may vary from as much 30.0 down to 3.0 … Does that change how much energy is available to heat the arctic ice or water? Certainly.
The sun shines for 0, 3, 12, 15, 18 or 24 hours in a day – depending on latitude and time-of-year. Does that change how much energy is available to heat the surfaces? Certainly, but they use “averages” despite the meaningless “calculation” achieved.
At 82 north latitude, the sun is only 8 degrees above the horizon – at noon! – at time of minimum ice extents. Run that calculation for how much energy is present and tell me that minimum ice extents in the arctic “heat” the exposed water.
The heat rates from solar irradiation mean something further south, but do not heat exposed water in the Arctic at times of minimum sea ice extent, yet using an “average” value fools you into accepting such claims.
” the instruments say that the ocean is radiating IR constantly at somewhere on the order of 400 W/m”
that’s enough energy to boil nearly a gallon of water in an hour…
jeez…. if i lived in a boat, i’d cook!!!!
TimTheToolMan says:
December 18, 2012 at 7:44 pm
OK, my bad. I read your postings to mean that the ocean was not warmed by the DLR.
So … where are our disagreements?
I fear I don’t understand this part. Is it your contention that the DLR doesn’t cause the bulk of the upper ocean to end up warmer than if there were no DLR?
Here’s how I see it. Let us suppose that on average the ocean is losing energy about half of a kilowatt per square metre. Of this loss, about 400 W/m2 is by radiation, and another 20 in sensible heat and 80 in latent heat. So in say 24 hours, it will radiate on the order of 12 kilowatt-hours of energy.
This radiation is balanced by 160 W/m2 of solar energy, and 340 W/m2 of DLR.
Now, if there is no solar energy or DLR being absorbed, all of that 12 kWhrs of energy comes out of the bulk of the mixed layer. The bulk of the mixed layer would end up much cooler than it is now.
If there were only solar energy warming the ocean, then the ocean would be cooler, but not as much. Only about 8 kWhrs of energy would be coming out of the bulk of the mixed layer.
With the DLR and the sun, we have energetic balance. The amount leaving is equal to the amount entering. As a result, there is no heat lost from the bulk of the mixed layer.
And as a result, if there is DLR, the entire bulk of the mixed layer ends up warmer, despite the fact that the DLR is absorbed in the very skin itself.
Two final comments. First, you are totally glossing over mechanical mixing. Wind, wave, and spray are constantly churning the topmost layers of the ocean. I’ve sailed over much of it, and calm times are very uncommon. As a result, your lovely picture of a permanent microscopically thin skin layer that is cooler than the rest of the ocean simply doesn’t obtain over wide areas of the surface—they are being tossed, turned, and churned.
Finally, note that the situation you correctly describe, of a surface layer which is slightly cooler than the underlying layers, is thermally unstable. As a result, that layer is constantly sinking, and being replaced by warmer water from below.
Let me take some numbers, which are not real numbers but will do for purposes of discussion. Let’s suppose that the surface is at 5°C and the underlying layer is at 7°C. I know that we are actually talking about very slight temperature difference, this is for illustration of a mechanism.
The water at 5° is denser than the water at 7°. So it is constantly sinking, and is constantly being replaced by 7° water, which immediately starts cooling by radiation, conduction, and evaporation.
Now, suppose the DLR is switched on. Lets say it warms the skin water to 6°. Note that it is still cooler than the underlying layer, so your condition is satisfied.
What is the fate of the 6° water?
Well, it is cooler than the underlying 7° water, so it will sink … which mixes the DLR energy down into the bulk … which you seem to think is impossible.
People have the mistaken idea that the ocean is thermally stratified, with the warmest on top. And it is … but only during the day. I learned a lot about the thermal behavior of the ocean when I started night diving.
At night, the ocean undergoes a curious transition. After sunset, the surface of the ocean starts cooling. At a certain point, when surface waters have cooled enough, columns of descending water form. These mix the cooler surface waters down into the mixed layer beneath. This circulation, while slow, moves massive amounts of water over the course of each night. At night, if you are swimming along slowly at a depth of say three metres (10′) under the surface, you can feel the descending columns when you encounter them. They are cooler than the surrounding water.
Here’s the thing it took me a while to understand. Yes, during the day the ocean picks up a whole lot of energy. This comes from the sun, which penetrates tens of metres deep into the ocean. But at night, every night, it loses just about that same amount of energy. And unlike during the day, it has to lose it all at the surface. That means that each nocturnal overturning of the ocean has to move lots and lots of water up to the surface every night. At night, if you are swimming slowly say three metres (10′) under the surface, you can feel the descending water. it’s cooler than the surroundings, with a distinct thermocline dividing it from the slowly upwelling surroundings. If you hang in it with neutral buoyancy, you slowly drift downwards.
Now, if there is DLR, it does several things to this nightly overturning. First, it slows the heat loss of the surface water. As a result, it delays the onset of nightly oceanic overturning. Finally, it slows the overturning once it is started.
The net result of each of these changes is that the bulk of the mixed layer ends up warmer with the DLR than it would be without DLR. Note that curiously, the energy from the DLR doesn’t need to mix downward in order to affect the bulk temperature of the upper ocean. All it needs to do is slow the overturning and the concomitant heat loss, and the bulk of the upper mixed layer ends up warmer.
So I agree with you that DLR is absorbed by the ocean. I agree that it is absorbed in the very skin. I think where we part company is that I say if there were no DLR the entire ocean would freeze, top to bottom, whereas you seem to think that DLR somehow only affects the very skin layer.
But then, I may be totally misunderstanding your position once again. Anyhow, give me your thoughts.
Regards,
w.
gnomish says:
December 18, 2012 at 11:17 pm
Quick reality check. Four hundred watts per square metre is about the amount of energy radiated constantly from any object at around 20°C (70°F). For example, that’s the rate at which my desk radiates at room temperature. If you have discovered some way to boil water by exposing it to my desk, please let me know.
w.
sure- let’s do have a reality check:
Volume: 1 gallon
Energy: 400W
Start Temp: 0C
End Temp: 100C
Efficiency 100%
Time to Temp: 66 minutes
http://www.phpdoc.info/brew/boilcalc.html
‘those instruments’ are saying if i put a 4 ft diameter parabolic mirror over the water, it could direct enough energy to do my cooking, heat my house and power a steam turbine 24/7.
forget nuclear, coal, solar and wind! energy supply is so solved by conundrum power!!
This 400W/sqm concept is what is causing the problem. The 2nd Law of Thermodynamics states that heat can only flow from a hot object to a colder object. This is equally true of IR as it is of conduction/convection, otherwise the fundamental principle of entropy would be violated.
So 400W/sqm cannot be true of the NET energy flow from two objects at the same temperature. It could only be true if the 2nd object is at a temperature of absolute zero (-273Celsius). At absolute zero a body emits no heat energy at all so we don’t have to subtract that power from the power emitted by the desk, for instance.
Now for radiation from the sun the radiated power is relevant because the earth is, frankly, quite close to absolute zero compared to the temperature of the sun (5500Celsius). So when we talk about radiated IR power from the sun we can effectively use the measured radiated power DIRECTLY because the actual temperature of the Earth is sufficiently close to being absolute zero that the small error introduced is more or less negligible compared to other quantities related to climate. We cannot do the same for the radiated power from a desk to the floor because the desk and the floor are at much the same temperature so no net energy can flow between the two, i.e we would need to subtract the power emitted from the floor from the power emitted by the desk to leave us with a net power transfer of 0W/sqm.
So you see this gives Team AGW an ideal opportunity to misdirect the layman. They can use the radiated power from the sun more-or-less directly, but then do the same with IR to suggest they can add together. This is not in fact the case. In both cases you need to subtract the Earth’s IR emitted power at the point of equilibrium to find the net power transfer. The situation is further complicated because from the Earth to the atmosphere you don’t actually have only IR to deal with – most of the cooling of the planet’s surface is done by conduction/convection which is why you have the rapid cooling with altitude that you see in the graphs above until you reach the tropopause.
It is perhaps worth bearing in mind that the Sun to Earth temperature difference is far greater than the Earth to space temperature difference, with a subsequent impact on the point at which equilibrium is reached with respect to heat outflow being equal to heat inflow.