Guest Post by Willis Eschenbach
A couple of apparently related theories have been making the rounds lately. One is by Nikolov and Zeller (N&Z), expounded here and replied to here on WUWT. The other is by Hans Jelbring, discussed at Tallblokes Talkshop. As I understand their theories, they say that the combination of gravity plus an atmosphere without greenhouse gases (GHGs) is capable of doing what the greenhouse effect does—raise the earth at least 30°C above what we might call the “theoretical Stefan-Boltzmann (S-B) temperature.”
So what is the S-B temperature, theoretical or otherwise?
A curious fact is that almost everything around us is continually radiating energy in the infrared frequencies. You, me, the trees, the ocean, clouds, ice, all the common stuff gives off infrared radiation. That’s how night-vision goggles work, they let you see in the infrared. Here’s another oddity. Ice, despite being brilliant white because it reflects slmost all visible light, absorbs infrared very well (absorptivity > 0.90). It turns out that most things absorb (and thus emit) infrared quite well, including the ocean, and plants (see Note 3 below). Because of this, the planet is often treated as a “blackbody” for IR, a perfect absorber and a perfect emitter of infrared radiation. The error introduced in that way is small for first-cut calculations.
The Stefan-Boltzmann equation specifies how much radiation is emitted at a given temperature. It states that the radiation increases much faster than the temperature. It turns out that radiation is proportional to absolute temperature to the fourth power. The equation, for those math inclined, is
Radiation = Emissivity times SBconstant times Temperature^4
where the Stefan-Boltzmann constant is a tiny number, 0.0000000567 (5.67E-8). For a blackbody, emissivity = 1.
This “fourth-power” dependence means that if you double the absolute temperature (measured in kelvins), you get sixteen (2^4) times the radiation (measured in watts per square metre, “W/m2”). We can also look at it the other way, that temperature varies as the fourth root of radiation. That means if we double the radiation, the temperature only goes up by about 20% (2^0.25)
Let me call the “theoretical S-B temperature” the temperature that an evenly heated stationary blackbody planet in outer space would have for a given level of incoming radiation in W/m2. It is “theoretical”, because a real, revolving airless planet getting heated by a sun with the same average radiation will be cooler than that theoretical S-B temperature. We might imagine that there are thousands of mini-suns in a sphere around the planet, so the surface heating is perfectly even.
Figure 1. Planet lit by multiple suns. Image Source.
On average day and night over the planetary surface, the Earth receives about 240 W/m2 of energy from the sun. The theoretical S-B temperature for this amount of radiation (if it were evenly distributed) is about -18°C, well below freezing. But instead of being frozen, the planet is at about +14°C or so. That’s about thirty degrees above the theoretical S-B temperature. So why isn’t the planet a block of ice?
Let me take a short detour on the way to answering that question in order to introduce the concept of the “elevator speech” to those unfamiliar with the idea.
The “elevator speech” is simply a distillation of an idea down to its very basics. It is how I would explain my idea to you if I only had the length of an elevator ride to explain it. As such it has two extremely important functions:
1. It forces me to clarify my own ideas on whatever I’m discussing. I can’t get into handwaving and hyperbole, I can’t be unclear about what I’m claiming, if I only have a few sentences to work with.
2. It allows me to clearly communicate those ideas to others.
In recent discussions on the subject, I have been asking for that kind of “elevator speech” distillation of Jelbring’s or Nikolov’s ideas, so that a) I can see if whoever is explaining the theory really understands what they are saying and, if so, then b) so that I can gain an understanding of the ideas of Jelbring or Nikolov to see if I am missing something important.
Let me give you an example to show what I mean. Here’s an elevator speech about the greenhouse effect:
The poorly-named “greenhouse effect” works as follows:
• The surface of the earth emits energy in the form of thermal longwave radiation.
• Some of that energy is absorbed by greenhouse gases (GHGs) in the atmosphere.
• In turn, some of that absorbed energy is radiated by the atmosphere back to the surface.
• As a result of absorbing that energy from the atmosphere, the surface is warmer than it would be in the absence of the GHGs.
OK, that’s my elevator speech about why the Earth is not a block of ice. Note that it is not just saying what is happening. It is saying how it is happening as well.
I have asked, over and over, on various threads, for people who understand either the N&Z theory or the Jelbring theory, to give me the equivalent elevator speech regarding either or both of those theories. I have gotten nothing scientific so far. Oh, there’s the usual handwaving, vague claims of things like ‘the extra heat at the surface, is just borrowed by the work due to gravity, from the higher up regions of the atmosphere‘ with no mechanism for the “borrowing”, that kind of empty statement. But nothing with any meat, nothing with any substance, nothing with any explanatory value or scientific content.
So to begin with, let me renew my call for the elevator speech on either theory. Both of them make my head hurt, I can’t really follow their vague descriptions. So … is anyone who understands either theory willing to step forward and explain it in four or five sentences?
But that’s not really why I’m writing this. I’m writing this because of the claims of the promoters of the two theories. They say that somehow a combination of gravity and a transparent, GHG-free atmosphere can conspire to push the temperature of a planet well above the theoretical S-B temperature, to a condition similar to that of the Earth.
I hold that with a transparent GHG-free atmosphere, neither the hypothetical “N&Z effect” nor the “Jelbring effect” can possibly raise the planetary temperature above the theoretical S-B temperature. But I also make a much more general claim. I hold it can be proven that there is no possible mechanism involving gravity and the atmosphere that can raise the temperature of a planet with a transparent GHG-free atmosphere above the theoretical S-B temperature.
The proof is by contradiction. This is a proof where you assume that the theorem is right, and then show that if it is right it leads to an impossible situation, so it cannot possibly be right.
So let us assume that we have the airless perfectly evenly heated blackbody planet that I spoke of above, evenly surrounded by a sphere of mini-suns. The temperature of this theoretical planet is, of course, the theoretical S-B temperature.
Now suppose we add an atmosphere to the planet, a transparent GHG-free atmosphere. If the theories of N&K and Jelbring are correct, the temperature of the planet will rise.
But when the temperature of a perfect blackbody planet rises … the surface radiation of that planet must rise as well.
And because the atmosphere is transparent, this means that the planet is radiating to space more energy than it receives. This is an obvious violation of conservation of energy, so any theories proposing such a warming must be incorrect.
Q.E.D.
Now, I’m happy for folks to comment on this proof, or to give us their elevator speech about the Jelbring or the N&Z hypothesis. I’m not happy to be abused for my supposed stupidity, nor attacked for my views, nor pilloried for claimed errors of commission and omission. People are already way too passionate about this stuff. Roger Tattersall, the author of the blog “Tallbloke’s Talkshop”, has banned Joel Shore for saying that the N&Z hypothesis violates conservation of energy. Roger’s exact words to Joel were:
… you’re not posting here unless and until you apologise to Nikolov and Zeller for spreading misinformation about conservation of energy in their theory all over the blogosphere and failing to correct it.
Now, I have done the very same thing that Joel did. I’ve said around the web that the N&Z theory violates conservation of energy. So I went to the Talkshop and asked, even implored, Roger not to do such a foolish and anti-scientific thing as banning someone for their scientific views. Since I hold the same views and I committed the same thought-crimes, it was more than theoretical to me. Roger has remained obdurate, however, so I am no longer able to post there in good conscience. Roger Tallbloke has been a gentleman throughout, as is his style, and I hated to leave. But I did what Joel did, I too said N&Z violated conservation of energy, so in solidarity and fairness I’m not posting at the Talkshop anymore.
And more to the point, even if I hadn’t done what Joel did, my practice is to never post at or even visit sites like RealClimate, Tamino’s, and now Tallbloke’s Talkshop, places that ban and censor scientific views. I don’t want to be responsible for their page views counter to go up by even one. Banning and censorship are anathema to me, and I protest them in the only way I can. I leave them behind to discuss their ideas in their now cleansed, peaceful, sanitized, and intellectually sterile echo chamber, free from those pesky contrary views … and I invite others to vote with their feet as well.
But I digress, my point is that passions are running high on this topic, so let’s see if we can keep the discussion at least relatively chill …
TO CONCLUDE: I’m interested in people who can either show that my proof is wrong, or who will give us your elevator speech about the science underlying either N&K or Jelbring’s theory. No new theories need apply, we have enough for this post. And no long complicated explanations, please. I have boiled the greenhouse effect down to four sentences. See if you can match that regarding the N&K or the Jelbring effect.
w.
NOTE 1: Here’s the thing about a planet with a transparent atmosphere. There is only one object that can radiate to space, the surface. As a result, it is constrained to emit the exact amount of radiation it absorbs. So there are no gravity/atmospheric phenomena that can change that. It cannot emit more or less than what it absorbs while staying at the same temperature, conservation of energy ensures that. This means that while the temperature can be lower than the theoretical S-B temperature, as is the case with the moon, it cannot be more than the theoretical S-B temperature. To do that it would have to radiate more than it is receiving, and that breaks the conservation of energy.
Once you have GHGs in the atmosphere, of course, some of the surface radiation can get absorbed in the atmosphere. In that case, the surface radiation is no longer constrained, and the surface is free to take up a higher temperature while the system as a whole emits the same amount of radiation to space that it absorbs.
NOTE 2: An atmosphere, even a GHG-free atmosphere, can reduce the cooling due to uneven insolation. The hottest possible average temperature for a given average level of radiation (W/m2) occurs when the heating is uniform in both time and space. If the total surface radiation remains the same (as it must with a transparent atmosphere), any variations in temperature from that uniform state will lower the average temperature. Variations include day/night temperature differences, and equator/polar differences. Since any atmosphere can reduce the size of e.g. day/night temperature swings, even a transparent GHG-free atmosphere will reduce the amount of cooling caused by the temperature swings. See here for further discussion.
But what such an atmosphere cannot do is raise the temperature beyond the theoretical maximum average temperature for that given level of incoming radiation. That’s against the law … of conservation of energy.
NOTE 3: My bible for many things climatish, including the emissivity (which is equal to the absorptivity) of common substances, is Geiger’s The Climate Near The Ground, first published sometime around the fifties when people still measured things instead of modeling them. He gives the following figures for IR emissivity at 9 to 12 microns:
Water, 0.96 Fresh snow, 0.99 Dry sand, 0.95 Wet sand, 0.96 Forest, deciduous, 0.95 Forest, conifer, 0.97 Leaves Corn, Beans, 0.94
and so on down to things like:
Mouse fur, 0.94 Glass, 0.94
You can see why the error from considering the earth as a blackbody in the IR is quite small.
I must admit, though, that I do greatly enjoy the idea of some boffin at midnight in his laboratory measuring the emissivity of common substances when he hears the snap of the mousetrap he set earlier, and he thinks, hmmm …
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Some people agree that there’s no spectral emission from GHG’s. Thank you.
@Tim Folkerts:
I think your statement is a bit backwards: Convection does not set up a lapse rate. Convection only occurs when the lapse rate exceeds the adiabatic lapse rate and it is what drives a lapse rate exceeding the adiabatic lapse rate back down to the adiabatic lapse rate.
If the lapse rate is already less than the adiabatic lapse rate, the atmosphere is stable and convection is actually suppressed.
A planet (earth, ocean atmosphere) can only cool by radiating to space.
Non GHGs store conductive specific heat, but cannot radiate it to space. (Insulation)
They can however conduct specific heat to radiating GHGs.
Adding more GHGs to an atmosphere accelerates the loss of conducted heat to space, while it slows the loss of radiated heat from the surface. (cooling and warming)
Tim Folkerts says:
January 19, 2012 at 8:58 am
Tim, as long as there is insolation to the surface and conducted heat to the atmosphere,(even sans convection as in a thought experiment) there would be a thermal gradient based on the mass per m2.
I can’t figure out how to subscribe to this thread without posting. So I am posting 🙂 Happy Thursday.
Joe Postma says: “Some people agree that there’s no spectral emission from GHG’s. Thank you.”
And some people are wrong.
GHGs emit very clear bands of IR radiation (due to rotations & vibrations). In addition, ALL gases (including GHGs) emit radiation when electrons jump to/from higher energy orbitals (which is typically visible or UV or near IR, but which does not occur to any measurable extent at room temperature).
People who don’t understand these simple ideas shouldn’t even begin to try explaining the greenhouse effect to others.
Bart,
I think I see the problem. Your solution applies to the radiative temperature, which indeed decreases with r. But the specification is that the atmosphere is perfectly transparent, which means it’s not coupled to the radiative temperature at all. In an atmosphere that isn’t perfectly transparent, the radiative field doesn’t decouple from the atmosphere until the atmosphere becomes optically thin, which is well above the surface.
Before I retired, one of the things I did was argon plasma optical emission spectrometry for elemental analysis. The analytical plasma isn’t at LTE because it’s optically thin and a lot of gas and aerosol is being continually pumped through it. The reason that it’s known not to be in LTE is because there are a variety of ways to measure the temperature of the plasma. There’s the electron number density (it is a plasma), the excitation temperature, the brightness temperature and the gas kinetic temperature. If all those temperatures are the same, LTE can be said to exist. But they aren’t and it doesn’t. But you try to arrange things so you can get as close as possible to LTE in the observation zone because the analysis is more robust to changes that occur when different samples are introduced. The usual way to measure how close you were to LTE was to measure the excitation temperature. That was done by measuring the emission intensities of two lines of the same element with different excitation energies, an intense Mg atomic line and an intense Mg singly ionized line. Since the plasma is optically thin, the intensity is directly proportional to the number density of excited atoms. The number density of excited atoms can be used to calculate an excitation temperature based on the Maxwell-Boltzmann distribution. The emission ratio for a constant temperature can then be calculated. But the observed ratio is always less than the calculated ratio. You adjusted the sample flow, viewing height and RF power to maximize the ratio to achieve what was called in the trade, near LTE conditions.
In the atmosphere, you can do something similar by measuring the emission intensity of a molecule. If the emission intensity matches the calculated emission intensity for the gas kinetic temperature, then LTE must exist. You can do the same thing for a surface except what you look at is the shape of the emission spectrum. That way emissivity, unless it varies strongly with wavelength, doesn’t matter. The best IR thermometers use this technique. The emission intensity is measured at two wavelengths and the ratio calculated. The temperature determined by that ratio from the Planck function is independent of emissivity as long as the emissivity is the same at both wavelengths. Emission spectra from surfaces does, in fact, reasonably closely match the shape of the Planck function for the kinetic temperature as measured by a contact thermometer. The variation in emissivity determined by deviation from the Planck curve is being used in the Diviner moon mission to map the geology of the lunar surface as the emissivity spectrum varies in known ways for different minerals.
Another point. If the atmosphere and surface weren’t at LTE, remote satellite sensing of temperature couldn’t work.
Joe Postma says:
January 19, 2012 at 7:17 am
Bart has a fantastic set of posts here. Excellent scientific analysis, and I need to review it a bit more.
I’d like to point out yet another oddity of standard GHE theory. There’s supposed to be back-radiation from IR-emitting molecules, predominantly CO2, that cause 33C of additional heating. That’s the basic GHE theory.
So tell me then: on a spectrometer plot taken from the surface of the Earth and pointing upwards, or even from one taken from above the Earth and looking down – where is the emission line?
This is embarrassingly bad, your lack of knowledge about spectra in atmospheres is abysmal. Take the advice that I and others have given you about references to read, by scientists who know what they’re talking about, and come back here with some humility. The earth’s atmosphere is not interstellar gas, the physics is different because of the density and the relative rates of collisions to emission rates. A hint to help you, the spectrum of the CO2 band at it’s center has no near surface emissions, that was absorbed and transferred as heat to the rest of the atmosphere. The non-zero signal you see at the band center is the emission from cold CO2 much higher in the atmosphere where the collision frequency is low enough that emission directly to space predominates. Astrophysicists may need to understand many different fields as you claimed before, but you are one whose knowledge in the field of spectroscopy outside interstellar gases is non-existant, I suggest you do what you told me you were going to do last week and talk to one of your professors who understands this area and get educated.
Joel Shore,
That’s true if the surface is isothermal. But it’s not. The meridional temperature gradient forces circulation because of the pressure gradient force. That circulation forms a heat engine that drives what might other wise be an non-radiative isothermal atmosphere back towards the adiabatic rate. For a non-transparent atmosphere, radiation to space cools the atmosphere and also forces the lapse rate towards the adiabatic rate.
Joe Postma,
Those people would be wrong.
Ken Coffman says:
January 19, 2012 at 5:18 am
This is interesting…I didn’t know O2 was magnetic…
By 1891, [James] Dewar could produce liquid oxygen in large quantities, and also showed that it and liquid ozone were strongly attracted by a magnet.
–This Month in Physics History, APS News, January 2012</em.
It's paramagnetic because of the unpaired electrons, this is high school chemistry. Oxygen sensors use this to measure the concentration of O2 because N2 is diamagnetic.
Joel, you are right that I should have said that differently about the lapse rate & convection. The cooling at the top of the atmosphere should have been the main point. There is on the order of 100 W/m^2 of cooling from the TOA. Conduction itself can only provide on the order of 0.001 W/m^2, so conduction will never be able to erase the lapse rate.
(And I might argue that where the downdrafts are from convection, there the convection would help create the temperature gradient. The cool air high in the atmosphere that descends will warm, and so there would be a lapse rate in those areas from convection. In the areas where the air is rising, then the convection would indeed help erase excessive lapses rates. But this is really just a side issue to the point that a lapse rate WILL get set up in any normal sort of atmosphere, even if the “ideal” case is to have no lapse rate.)
David says: “Tim, as long as there is insolation to the surface and conducted heat to the atmosphere,(even sans convection as in a thought experiment) there would be a thermal gradient based on the mass per m2.”
I would simplify this even further. “There is conducted heat if and only if there is a thermal gradient.”
The “mass per m2” could affect the rate of conduction, so in that sense the conduction is “based on” mass per m^2. In my book, the cooling of the upper atmosphere (IR radiation from GHGs) would be the “cause” of conduction of heat and the thermal gradient.
(But as I just argued in a different post, conduction is typically orders of magnitude smaller than convection, evaporation and IR radiation, and so it typically plays an inconsequential role in the dynamics of the troposphere.)
================================
DeWitt Payne says:
January 19, 2012 at 9:58 am
Those people would be wrong.
DeWitt Payne says:
January 19, 2012 at 8:19 am
The fact that there is a valley in the emission spectrum to space at the CO2 band is exactly why CO2 is a greenhouse gas.
================================
Yet the spectral emission from CO2 is supposed to also be why CO2 is a heating GHG. So the presence and lack of CO2 emission is why CO2 is a GHG. Got it, thanks. The spectral emission from CO2 which does and doesn’t exist and why CO2 is a heating and a cooling gas. Thanks.
RE: Bart: (January 17, 2012 at 6:22 pm)
Ref: [The problem with the gravitational heating theory is that it does not provide any mechanism to continuously extract or reflect an average 156 W/m² from a hypothetical average 396 W/m² radiant energy flow of a typical GHG warmed surface so that only the allowed average 240 W/m² actually escapes to outer space.]
“What that means is not that the GHGs heated the surface above what it otherwise would have been. Rather, it means it is maintaining it that much cooler below what it otherwise would have been.”
The Earth is only receiving enough energy from the sun to emit a global average energy flow of about 240 W/m². In that case, the fourth root of the flat area weighted average of the fourth powers of the absolute temperatures around the earth would be on the order of 255 degrees K.
According to the Trenberth diagram, the Earth is actually warm enough to radiate around 396 W/m² from the surface. But the Earth cannot afford to let that radiation get out to space and still maintain a balanced energy budget. Greenhouse gases in the atmosphere solve this problem by continuously capturing 156 W/m² from that flow so that the energy budget remains balanced with our higher surface temperatures. If there were no greenhouse gases in the atmosphere, then there would be nothing to stop the escape of excess radiant energy that the Earth could not afford to lose.
The real problem here is not the greenhouse theory; I think it is the purported hypersensitivity of the atmosphere to the self-masking, narrow band effect of carbon dioxide, by treating it in the public mind as a uniform darkening agent.
=======================================
Tim Folkerts says:
January 19, 2012 at 9:49 am
GHGs emit very clear bands of IR radiation (due to rotations & vibrations). In addition, ALL gases (including GHGs) emit radiation when electrons jump to/from higher energy orbitals (which is typically visible or UV or near IR, but which does not occur to any measurable extent at room temperature).
People who don’t understand these simple ideas shouldn’t even begin to try explaining the greenhouse effect to others.
=======================================
Show me then. The little bit that you can identify of CO2 emission is a mere couple of Watts.
The GHE is described and exposed here:
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
See, in particular, Figure 5.17 on p. 131 which directly shows that what Joe Postma is spreading here are falsehoods.
======================================
Phil. says:
January 19, 2012 at 9:53 am
This is embarrassingly bad, your lack of knowledge about spectra in atmospheres is abysmal. Take the advice that I and others have given you about references to read, by scientists who know what they’re talking about, and come back here with some humility. The earth’s atmosphere is not interstellar gas, the physics is different because of the density and the relative rates of collisions to emission rates. A hint to help you, the spectrum of the CO2 band at it’s center has no near surface emissions, that was absorbed and transferred as heat to the rest of the atmosphere. The non-zero signal you see at the band center is the emission from cold CO2 much higher in the atmosphere where the collision frequency is low enough that emission directly to space predominates. Astrophysicists may need to understand many different fields as you claimed before, but you are one whose knowledge in the field of spectroscopy outside interstellar gases is non-existant, I suggest you do what you told me you were going to do last week and talk to one of your professors who understands this area and get educated.
======================================
Well I know that you all want to backtrack and will take the opportunity to make this a personal assault instead of a scientific one. With every failed logic that’s newly exposed the alarmists will retaliate with innuendo.
So you agree that you can’t see GHG emission near the surface. And you also can’t see it from space. Thank you. So non-existent spectral emission from GHG causes heating. Got it.
Indeed, it is exactly the collision frequency that describes the processes going on. Being collision dominated means the GHG’s don’t have time to radiate. Therefore they can’t re-heat the surface. But that’s what you needed them to do. Indeed, we SEE that they don’t radiate in the spectra. It also means their internal vibratory state is already activated by collision, not by radiation. Most likely outcome is that GHG’s either do nothing, or they cause cooling. I mean we already know that the thermal mass of the air+ ground column keeps the surface from getting as hot as it could.
DeWitt Payne says:
January 19, 2012 at 8:44 am
“There is no reason whatsoever for that to decline to zero as r increases, at least until the number density of molecules is so low that their mean free path is effectively infinite.”
Really, DeWitt. Is it so hard for you to believe that a region near the heat source is hotter than a region far from it? You are grasping at straws.
“As I pointed out, the kinetic temperature of the actual atmosphere at an altitude of 100 km is ~200K for a surface temperature of 288K.”
Please stop talking about the “actual atmosphere” of the Earth. We’re not talking about the Earth. The hypothetical is, what happens on a planet without GHGs? There is no reason whatsoever that it should be anything like your everyday experience of the Earth.
“That would imply a temperature of ~30,000K.”
Orbital velocity is not counted in temperature because the particle is simply following a geodesic in space. Certainly, our satellites in orbit do not radiate like blackbodies at 30,000K!
“But of course, they don’t have to be in a non-intersecting orbit, and the vast majority aren’t.”
At low Earth altitudes where the atmosphere acts as a fluid, this is indeed true. There, the average vector velocity of air molecules is the velocity needed to keep up with the spin of the Earth, which is not terribly large. When you get up into the rarefied regime, it’s a little more haphazard, but your statement is still generally true.
For the Earth and its atmosphere.
But, there’s still a limit to how much the temperature can increase, and your objection of a 2nd law violation is invalid.
Tim Folkerts says:
January 19, 2012 at 8:58 am
“…the equilibrium temperature profile of the atmosphere will be uniform at the temperature of the surface.”
Quite impossible.
“There will be NO lapse rate.”
There will be no adiabatic lapse rate. Such a lapse rate requires a heat sink in the upper atmosphere. Without that heat sink, you do not get a linear lapse rate. You get a convex, monotonically decreasing function of altitude.
DeWitt Payne says:
January 19, 2012 at 9:51 am
“Your solution applies to the radiative temperature”
It applies to conducted temperature. The Laplacian is the expression for conducted heat glow. It results from Fourier’s law and conservation of energy
“If the atmosphere and surface weren’t at LTE, remote satellite sensing of temperature couldn’t work.”
On Earth. There’s not going to be much of a remote sensing industry on our hypothetical non-GHG planet.
Bart says:
January 19, 2012 at 10:53 am
“Without that heat sink, you do not get a linear lapse rate. You get a convex, monotonically decreasing function of altitude.”
In fact, if you look at a plot of temperature versus altitude for the Earth, you find that it is linear in the troposphere and the mesosphere. This suggests massive heat sinks at the top of those regions.
The stratosphere and the thermosphere show the transition to measured decay of temperature which would be expected when there is no proximate outlet for heat to radiate away.
Coincidence?
I will not be able to post much more today. I want to ask those who have learned the standard greenhouse model to try thinking outside the box. You will find that there is no steady state difference between the standard description, and what I am offering.
Thus, you have to start asking for evidence which would resolve which description fits with all the facts. I have offered the main differentiator: the temperature must decrease to zero at infinite radius. Hence, the heat equation says that there will be a gradient downhill in the (upward) radial direction. There can, thus, be no equilibrium until enough heat is dissipated by radiation.
It’s as plain as the noses on your faces: IR absorbing gasses in the Earth’s atmosphere cool the surface, and prevent runaway temperatures.
“Everything not forbidden is compulsory,” said Murray Gell-Mann. There is nothing forbidden in my prescription.
Bart says:
January 19, 2012 at 10:59 am
DeWitt Payne says:
January 19, 2012 at 9:51 am
Erratum:
“The Laplacian is the expression for conducted heat flow…”
Joe Postma says:
January 19, 2012 at 10:35 am
“So you agree that you can’t see GHG emission near the surface.”
With all due respect to Joe, and without taking sides, I want to decouple this line of argument from my own. It is my contention that, in steady state, my description of the process results in an identical situation to the standard greenhouse theory. The heat sinks will be radiating, the backradiation will cancel with radiation from the surface, and everything balances out in exactly the same way, satisfying the SB relationships in the equilibrium condition.
Bart,
My graduate work was in electrochemistry, so I know something about using Laplace transforms to solve the diffusion equation for spherical (mercury drop electrode, e.g.) as well as cartesian coordinates. When solving a semi-infinite diffusion problem, there is no requirement for the concentration to go to zero at infinity. If it did and the concentration was not zero at the electrode surface, there would be diffusion away from the electrode until the concentration at the surface was also zero. A concentration gradient can only be stable if there’s a continuous flux into and out of a volume. For an isothermal atmosphere there is no flux out and the volume isn’t infinite. Admittedly, it would take a very long time to reach equilibrium because diffusion is slow. For radiation emitted by the surface, though, there is flux in and out of any volume element above the surface and the total volume is effectively infinite. If the energy lost by radiation isn’t replaced somehow, the surface cools. But again, the problem is defined as energy into the surface from short wavelength radiation equal energy out of the surface from long wavelength radiation.
The boundary conditions for the radiation problem is a flux of 240 W/m² at r = 6378 km and a flux of zero as r increases without limit. The flux will then drop as 1/r². That would make the flux at 100 km above the surface equal to (6378/6478)^2 * 240 = 232.65W/m² so Teff would drop from 255K to 253.1K. And of course there is no accumulation of energy at the TOA because the emitted photons just keep going on (transparent atmosphere).