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.
Willis: Did you forget me again? Waiting for The Link….
Also it might interest you to know Willis that Ernest Rutherford (google him up if you don’t know) was a kiwi born in this very town I live. So what has the world become when a Science Nobel Prize has become as worthless as an Oscar ,Willis. How do I compare Trenberth to Rutherford ? Don’t bother to answer because it was too easy even without the clue that one has (wittingly or unwittingly) denigrated science.
AlecM says:
December 17, 2012 at 12:54 pm
These two statements are true. But, you seemed to be saying that once you had the vector sum of upward and downward PVs, that the resulting PV would represent a beam that was left after summing. That’s not true. You still have two beams (or really, some complicated wavefronts) at every point, including where the photons are emitted: the surface of the Earth, and the atmosphere. The resultant Poynting vectors simply give you the difference between upward and downward waves. This tells you about the net energy flow. This will be large, since it has to equal, on the average, the energy brought in by sunlight. This resultant PV can also be included in calculations about how part of the atmosphere is heated or cooled. But, it still represents energy flowing in different directions; it’s magnitude tells you about imbalances in incoming energy from different directons, and which way the NET energy flow is directed. It doesn’t mean that all the energy ends up actually going in that direction. The individual vectors partially cancel, but the photons, and wavefronts, themselves don’t annihilate.
You are correct that if the camera is colder than the desk, there will be a net flow of energy from the desk to the camera. But this is NOT because of an increase in the energy flow from the desk to the camera. That stays the same, because the temperature of the desk stays the same.
To the contrary, it is because of a decrease in the flow of energy from the camera to the desk. Because the camera is now cold, it radiates less energy.
Nope, the camera which was defined as colder is still radiating toward the desk, it is just that the net flow will aways be in the direction of the warmer to the colder object.
Lets get away from simplistic examples.
You have two stars in a double star system. one star is a red giant with a surface temperature of 5000 degrees K. The diameter of the star is 1 Astronomical unit (140 million kilometers)
The second star is a main sequence star like the sun with a diameter of 0.06 AU with a surface temperature of 10,000 degrees K.
The net energy flow between the two will ALWAYS be from the main sequence star to the red giant star due to the difference in temperature. This is true even though the total energy radiated by the red giant is orders of magnitude higher than from the main sequence star.
Lets go to the satellite world.
I have a flat spinning disk in orbit around the Earth with its flat face pointing toward the Earth (nadir), with its opposite space pointing toward Deep space (zenith). The altitude is 150 statute miles so that the nadir face sees only the Earth. The temperature of the Earth as it radiates energy to space is about 20 degrees C (this is a well known quantity in spacecraft design). The temperature toward Zenith (deep space) is the 4 degrees kelvin of the background of the universe. The disk is black on the nadir and the zenith face and for this example we ignore the edges.
What happens to the temperature of the disk?
Answer that question. The answers will surprise you.
Next consider the case of the disk made from metal that is a perfect thermal conductor and the opposite case of a disk made of aerosol, a perfect insulator.
If you can answer these questions correctly you understand the subject.
D.I. says:
December 17, 2012 at 1:25 pm
First, I’d say that if you’re going to buck the conventional wisdom and tell the physicists who buy into it that they’re wrong, you’d better be absolutely, completely familiar with the current state of the art, and also its history, so you don’t end up misinterpreting something or bringing up an apparent paradox that’s already been hashed out. “Discovering a ‘New Law'” is one thing that a physicist might do by bucking the trend, although invalidating an old law, or established theory, would be enough. It’s a lot easier to get people to buy in to the idea that the old law was wrong if you have a new law with greater explanatory power, e.g., relativity, quantum mechanics, QCD. Or, you could come up with a law explaining things that nobody really thought about before, predicting effects which then show up in observations. Thomas Young did this in the early 1800s, with his wave theories, followed by his famous two-slit experiment that showed the wave nature of light. Still, if one could simply point out a major flaw in accepted theory, and then, armed with a deep and detailed knowledge of that theory, convince everyone that this really was a fundamental flaw and not just some manageable complication, then that would be a way of “telling the physicists that they don’t understand the physics.” Anyone who adds a new law, or convincingly undermines an old one, has made a major contribution, and that would qualify them as a “world class physicist.” But it’s very unlikely that one could do that without having a deep and detailed understanding of the known physics, which would at least qualify as “highly competent,” and would fit a broader definition of “world class” (those who haven’t made that major contribution but could, given the opportunity).
Physicists who are “popular for reciting ‘old laws'” might very well have that level of knowledge, though that might not be necessary, depending on what exactly they “recite.” But if that’s all they do, then, if they “recite” these laws correctly, i.e., the way other physicists understand them, then, by definition, they won’t be telling other physicists that they are wrong.
Kind of a long answer, I’m afraid, but I liked the question, since it made me think it through.
OK Willis, I’m getting tired of your snappy attitude. If you are so sure that the DLR is what prevents the oceans from freezing here is a simple experiment for you:
Take a mirror about 1sqm. You will accept that a mirror is a higly efficient reflector of photonic radiation, correct? Put a red object in front of it and it will show up just as red in the mirror, right? So it isn’t a big stretch to assume that a fair amount of infra-red gets reflected in much the same way, right? And you’d be right, because the metal coating on a highly polished mirrored surface has a sea of densely packed electrons at its surface that frankly isn’t interested in absorbing photons and simply throws them straight back at you. We use this fact in vacuum flasks, patio heaters, electric fires etc.
OK, so now take a bucket of water and put put it out in the garden on a warm night with the mirror on top, the mirrored surface pointing out to space. According to you, Willis, the bucket of water will freeze, because it is now not getting any heat from the sun or from this DLR which you claim is giving out almost as much heat as the sun during the night, because the mirror will reflect it all back into space with almost 100% efficiency.
I am pretty confident that the bucket of water won’t freeze. No amount of your curt answers will change that. But go ahead, knock yourself out and prove otherwise.
JazzyT: I notice that you imagine photons are part of the initial Poynting Vectors. I suspect this is not the case and here’s my reasoning.
Planck invented the photon as a throwaway concept. He never liked it. It represents the transient when the quantum of energy changes state between the mechanical vibrational and the EM worlds.I don;t believe that the individual PVs have an connection to photons until the net vector forms!
JazzyT: I for one am not claiming the physics is wrong. Willis isn’t actually arguing what the physicists are arguing. He seems to be arguing for the over-simplistic model created by Kevin Trenberth (presumably never reviewed by a physicist).
The physicists make it clear that at the surface the atmosphere is effectively opaque to thermal radiation. Heat transfer is by conduction and convection. It is only above the tropopause where thermal radiation makes a difference. In effect, what the physicists are saying is that because the rate at which the planet tries to cool into the stratosphere is slowed down by adding more CO2 then the stratosphere will get warmer. Then, because the temperature gradient between the warmer stratosphere and the surface would be smaller, the processes of convection in the troposphere are slowed down as well so the troposphere warms up too.
No physicist is claiming that the CO2 causes photons to zap from the nightime warm atmosphere directly into the ground (or ocean) heating it up. Willis has got that completely wrong but it isn’t his fault because Team AGW are passing around so many bad overly simplistic models. There is a fairly good explanation on Wikipedia, although it still has some errors due to lax language.
Ryan,
I’m noticing Trenberth is washing his hands of the latest Earth’s Energy diagrams and leaving younger brainwashed minions to take over. Probably having an edgy feeling about being associated with the latest Train-drivers Guide To Science.
Mack: Talking of Energy diagrams the Wikipedia page on the Greenhouse Effect has quite a good explanation of the physics under “Mechanism” I wouldn’t take issue with too much (I would only take issue with the impact at ground level due to a piddling amount of heating in the stratosphere). However, the energy diagram in the top right of the page is hilarious – it shows twice as much heating energy at ground level as is coming from the sun in the first place! I’m guessing that diagram is a Trenberth special.
Ryan says, December 18, 2012 at 2:02 am: “[…]OK, so now take a bucket of water and put put it out in the garden on a warm night with the mirror on top, the mirrored surface pointing out to space. According to you, Willis, the bucket of water will freeze, because it is now not getting any heat from the sun or from this DLR which you claim is giving out almost as much heat as the sun during the night, because the mirror will reflect it all back into space with almost 100% efficiency.
I am pretty confident that the bucket of water won’t freeze. No amount of your curt answers will change that. But go ahead, knock yourself out and prove otherwise.”
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Ryan, if I was a crazy warmist, I would say: “of course, the bucket of water won’t freeze because of back radiation from the other side of the mirror pointing out to the bucket”. Got you.
Ryan says:
December 18, 2012 at 2:02 am
Instead of a bucket, try a Styrofoam container. And use two of them, one with a mirror and one without. In the morning, see if the temperatures are the same. Be sure to use a clear night. No clouds. The effect will not be very large on a warm night with high humidity. I suggest a night between 35F and 40F.
The main problem with your experiment is that both sides of an optical mirror are very good IR mirrors. Actually, the metal coating will have no effect since the glass itself is IR opaque. (Unless you are using a first surface mirror, of course.)
In orbit, when a satellite enters the Earth’s shadow, its outside temperature drops below freezing in about one minute. Something causes it to take a bit longer on the Earth’s surface. Perhaps you can explain why.
Ryan,
I was wondering if I could get through to Willis in a more simple and direct way,
See this picture here………
http://www.ipcc.ch/graphics/ar4-wg1/jpg/faq-1-1-fig-1.jpg
The incoming solar radiation should be about 1360w/sq.m. NOT about 342w/sq.m.
How wrong can numbers be in science to be wronger than this.
Maybe “wronger” should be “more wrong” 🙂
Mack says:
December 18, 2012 at 10:59 am
342 * 4 = 1368
The area of a circle is πR^2
The area of a sphere is 4πR^2
The Earth absorbs radiation as a circle and emits it over the area of a sphere. As a result, when averaged over a period of 24 hours, the average energy is one fourth the peak.
Robert Clemenzi says, December 18, 2012 at 11:37 am: “The Earth absorbs radiation as a circle and emits it over the area of a sphere.”
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A circle is flat, Robert. Last time I checked the Earth was not flat.
Why do you need to reduce our mother Earth to a circle?
Greg House,
You’re nitpicking. I’m sure you know Robert was referring to a disk.
D Böehm says, December 18, 2012 at 1:36 pm: “I’m sure you know Robert was referring to a disk.”
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The same goes for a disk. Last time I checked the Earth was not a disk.
Greg House:
Don’t be silly. The Earth occludes a disc. This thread is too important for you to waste space on it with nonsense.
Richard
Greg, the earth is not a disk but for reception of the sun’s rays it essentially presents a disk. The square metres of the earth’s surface receive diminishing radiation from the sun as you move away in all directions from the point (centre of the disk) where the sun’s rays meet the earth at right angles. Clearly you can see that at the outside edges of the disk, there is very little energy impinging on the square meterage. Indeed, if you go around to the dark side, their is no sun’s energy impinging on that side of the sphere – as far as the sun is concerned the dark side is the other side of the disk. Robert’s observation is an accurate and cogent one that diserves repeting:
“The Earth absorbs radiation as a circle and emits it over the area of a sphere.”
Robert Clemenzi says: December 18, 2012 at 11:37 am
Yes it would be nice to see a little more elucidation and less of the . . . whatever.
We seemed to have sorted the semantic problem with “NET”: perhaps we can now return to the argument.
richardscourtney says, December 18, 2012 at 1:53 pm: “Don’t be silly. … to waste space on it with nonsense”
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I see a sign of desperation on your side, Richard.
Coming back to Robert Clemenzi’s formulation, it was simply wrong. According to reality, the right formulation would be “the Earth absorbs radiation as a sphere and emits it as a sphere”. If we consider the earth to be a sphere, of course, what we do. And if we remember that this sphere is rotating.
This might make some people unhappy, I understand that. But a sphere is a sphere and not a disk.
Gary Pearse says, December 18, 2012 at 2:06 pm: “Greg, the earth is not a disk but for reception of the sun’s rays it essentially presents a disk. The square metres of the earth’s surface receive diminishing radiation from the sun as you move away in all directions from the point (centre of the disk) where the sun’s rays meet the earth at right angles. Clearly you can see that at the outside edges of the disk, there is very little energy impinging on the square meterage. Indeed, if you go around to the dark side, their is no sun’s energy impinging on that side of the sphere…”
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Gary, I am well aware of this sort of argumentation and take it seriously. I hope you and others would take the counter-argumentation equally seriously.
The first problem you encounter is the handling of the premiss “the incoming solar radiation should be about 1360w/sq.m.”. A “sq.m.” is always a “sq.m.”. Hence a hemisphere would receive 2 times as much Watt as the disk of the same radius. Sorry, but this is a simple math and a simple logic.
Greg House:
re you post to me at December 18, 2012 at 2:20 pm.
There is no “desperation” in pointing out that you are wrong. You could have thanked me (and others) for pointing it out.
I will not reply to any more nonsense you address to me because your nonsense is not worth the bother. And if you think me rude for not answering further posts you address to me then I could not care less.
Richard
Greg, assume that there is spherical shell around the Sun with the same radius as the distance between the Sun and the Earth. The inside of that shell receives about 1368 W/m2 of electromagnetic radiation.
It is fairly obvious that the side of the Earth not facing the Sun receives none of that. It is also obvious that when the Sun is low in the sky, the light is spread out over a larger area than when it directly over head. The idea is to come up with a formula that takes all this into account and determines the average amount of energy being absorbed by the planet as a whole.
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. That is what the equations produce, they average the amount of energy available on one side of a sphere over the entire surface of the same sphere.