A Matter of Some Gravity

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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Joe Postma
January 19, 2012 11:45 am

=====================================
Bart says:
January 19, 2012 at 11:34 am
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.
=====================================
Oh yes I agree with that. I didn’t assume my criticism and your analysis were coupled 🙂
Cheers.

Tim Folkerts
January 19, 2012 11:56 am

Joe,
It is easy to see the radiation of GHGs both from below the atmosphere and from above the atmosphere. For example: http://www.skepticalscience.com/images/infrared_spectrum.jpg. (This is only above MOST of the atmosphere, but the point still stands.
GHG’s “take bites out of the spectrum” as seen from space, and “create bites in the spectrum” as seen from the ground. I’m not sure how the evidence could be any clearer than this. Nothing that Phil said implied that you couldn’t see the IR radiation. The fact that you are having trouble understanding how the “bites” as seen from space warm the planet means you are still missing some fundamental understanding of the atmospheric greenhouse effect.

DeWitt Payne
January 19, 2012 12:04 pm

That should be: For a transparent atmosphere… , not: For an isothermal atmosphere…
Bart,

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.

Oh, really. Not according to this: http://teachertech.rice.edu/Participants/louviere/Images/profile.gif
In the Ionosphere, the temperature can reach 4,000K at altitudes above 500km ( http://www.agu.org/pubs/current/si/links/2009JA014485.pdf ). Greenhouse gases (other than ozone) do make the stratosphere cooler than it would be otherwise because the stratosphere is optically thin in the thermal IR range and most greenhouse gases do not absorb significantly in the UV. But the troposphere is optically thick for much of the wavelength range so the reduced emission at higher altitude makes the lower atmosphere warmer than if they weren’t present, not cooler. So the stratosphere has been getting a double whammy: a reduction in ozone causing less heating, and an increase in CO2 causing more cooling.

January 19, 2012 12:06 pm

ZP says:
January 17, 2012 at 9:18 pm
Willis Eschenbach says:
January 17, 2012 at 5:35 pm
“Come back with the absorption lines for argon in the normal IR range.
I say it absorbs none. Zero. It only absorbs or emits in one way, through electron transitions, and that takes a whole bunch of energy to force one of those, and it doesn’t happen in the IR. Because argon is a monoatomic gas, the atoms of argon have no mechanical way to absorb or emit energy through bending or stretching. There is no physical way for the atoms to absorb or emit energy, an atom of argon is a little ball and the energy simple can’t interact with it except in high-energy electron transitions.”
Willis, you are correct that argon cannot have bending or stretching modes of vibration since it is a monoatomic gas. It can only have electronic transitions; the energy of which depends on the orbitals that the electrons are jumping between. However, you are overestimating the energy associated with most transitions, especially those occurring between n > 3 and n = 3. This overestimation leads you to the incorrect conclusion that argon does not absorb/emit IR radiation.

Your over-estimation of the fraction of Argon atoms in the third electronic level under the Earth’s atmospheric conditions leads you to this incorrect conclusion (and perhaps also Anna’s error). Look up the energy of the third excited state and do a Maxwell-Boltzmann distribution calculation at T=300K you’ll find that it’s essentially all in the ground electronic state which does not absorb in the IR.

Editor
January 19, 2012 12:30 pm

The replies to Joe Postma have been too complicated. Surely, the issue is simple.
1 – Sunlight passes through the atmosphere and reaches the surface, heating it.
2 – Consequently, radiation at certain wavelengths is emitted from the surface.
3 – A subset of that radiation – let us call it “wavelength set A” – is substantially absorbed by GHGs, chiefly CO2 and H2O.
4 – Those GHGs then re-radiate at various wavelengths – let us call it “Wavelength set B” – in all directions.
5 – Some (about half) of B reaches the surface.
6 – We should observe less of set A at TOA than we would expect from steps 1 and 2 above on their own.
7 – We should observe more of set B at the surface than we would expect from the sun alone.
So the question that resolves the issue is simply “do we observe less of set A at TOA and more of set B at the surface?“.
NB. There is no assumption that A and B are different sets, and if sunlight contains some A then allowance has to be made for some absorption on the way in.
The same logic can be applied to today’s changing planet, to confrim that the GHG part of AGW is correct. I have assumed that it is, at least wrt CO2, but it would be nice to see it verified this way.

January 19, 2012 12:36 pm

George E. Smith; says:
January 17, 2012 at 2:47 pm
“”””” Leonard Weinstein says:
January 17, 2012 at 8:27 am
Anna & others that think a non-greenhouse gas emits thermal radiation:
Yes there is some radiation even from non-greenhouse gases, but it is so small compared to the ground radiation and greenhouse gas radiation that it can be considered zero for the analysis.
“”””” Anna & others “””””
AKA; Anna and George E. Smith
Well the competing postulates are:-
#1 Stated repeatedly in all sorts of well known Textbooks on Physics. “EVERY object at a Temperature above zero Kelvins, emits “Thermal Radiation”, that being electromagnetic radiation that is emitted solely as a consequence of the Temperature of that object”.
#2 Stated almost solely by “Climate Scientists” in relation to earth’s atmosphere:- “EVERY object EXCEPT GASES at a Temperature above zero Kelvins, emits “Thermal Radiation”, that being electromagnetic radiation that is emitted solely as a consequence of the Temperature of that object”.

This is untrue George, no Physical Chemist would agree with your second postulate, it has nothing to do with Climate science. Homonuclear diatomics do not absorb IR, extremely short lived collisional states, i.e N4, can emit in a continuum but for the purposes of an energy balance this is zero!
Unrelated to the above, but also often cited, with relatively little dispute:- “Some gases having more than two atoms per molecule CAN and DO emit and absorb Electromagnetic Radiation often in the Long Wavelength Infra-red region. These emissions/absorptions are ALWAYS at specific frequencies or bands of frequencies, that correspond to various electron energy levels that are peculiar to each such so-called greenhouse gas species, and are NOT continuum spectra characterized ONLY by the Temperature of the material.”
Not electronic energy levels but rotational and vibrational energy levels, i.e. the nuclei move.
Now when I went to school, EVERY meant EVERY; sans NOTHING if you will.
Gases either emit and absorb thermal EM radiation or they do not. The spectral envelope of such EM radiation is given by the spectrum for a black body at the Same Temperature.
And as Anna said (in effect) nothing is nothing,, not something small.
The radiations from GHGs are a consequence of quantum mechanics, and the electronic structure of the molecule.
Black body Thermal radiation is NOT a consequence of quantum mechanics, it is a resultof ordinary classical physics, and requires no knowledge of any structure of the “particles” that emit and absorb such radiation.
Phil’s very low probability of ionising Argon at 15.8 eV from atmospheric thermal energies, is not challenged; but then you don’t need to ionise argon to get it to emit or absorb radiation, unless it is the specific atomic lines that you want to excite, not thermal.

Anna chose to say that collisional activation in the atmosphere would cause ionization but that this would be ‘quite rare’, not I, I pointed out that this had a probability of 10^-139, a vanishingly small probability of occurring in the Earth’s atmosphere once if you waited for the lifetime of the universe! That is zero by anyone’s definition, Anna made an undefensible comment and was called on it (I notice no mention was made of that gaffe in Anna’s subsequent post).
Even if the probability of the first excited state was the case the chance of that occurring is negligibly small under atmospheric conditions. There’s a reason the spectrum linked by Anna was the result of an electronic discharge!
Well I prefer to have nothing be nothing; not something small.
Sometimes extremely small is the same as nothing, Anna’s example is a case in point, as is the IR spectrum of N2 which consists of about 100 lines 10^10 times weaker than the thousands of lines from CO2 and H2O
Nobody I know of has said that thermal emission from the atmosphere is the major driver of earth’s climate; but either it is there or it is not. Can’t have it both ways. I’m in the same boat as Anna.
Not a good boat to be George in as Anna’s errors show.

Editor
January 19, 2012 12:38 pm

Tim Folkerts – your comment answers the questions. We do as expected see less A at TOA and more B at the surface (I am happy to assume that the graphs do indeed show A and B).

January 19, 2012 12:45 pm

Bart says:
January 17, 2012 at 3:29 pm
So, let me update my elevator one last time:
A) Radiation enters the system
B) it is absorbed by the surface, which fluoresces in the IR
C) heat builds up at the interface between the surface and the atmosphere until its apparent temperature suggests an exceedance of the incoming radiance based on the SB relationship
D) in fact, though, the distribution of surface energy states is non-Planckian and/or not omnidirectional so the entire integrated spatially integrated energy flux is actually not as great as expected
E) the heat from the interface conducts through the atmosphere, establishing a temperature gradient as must exist in a spherically distributed atmosphere
F) the transferred heat accumulates in the atmosphere until:
1) highly energetic emissions are stimulated, which balances the energy fluxes all around in the same way GHGs would in the standard “greenhouse” theory
OR
2) the heat accumulates until the atmosphere achieves escape velocity and vanishes.

This is where you introduce the fundamental error in your analysis.
what it should say is that ‘the heat accumulates in the atmosphere until the layer of gas nearest the surface reaches the surface temperature at which point heat transfer ceases due to ΔT=0. There can be no more heat transfer to the atmosphere unless it cools down and ΔT again exceeds 0. This is the fundamental flaw in all the ‘hotter and hotter’ arguments (it’s independent of the profile with altitude).
(G) if greenhouse gasses which can radiate in the IR are added to the atmosphere, it will result in a net cooler atmosphere than otherwise would be the case, because there is now a lower energy outlet for heat to escape.
Not true if the collision rate is much higher than the rate of emission as it is near the Earth’s surface, this is one of Postma’s errors.

January 19, 2012 12:46 pm

Spector says: “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.”
Using Avogadro number you contend that the 156 W/m^2 are produced by 113800 molecules of CO2. That is the number of CO2 in one square meter at 380 ppm.
But the N2 and O2 (that also absorb heat or they would not be gases) do no “capturing” of heat.

Joe Postma
January 19, 2012 12:47 pm

Ah, now that’s a much more useful plot than the one Shore posted, which didn’t show what these do. Thanks a lot! So that’s good, you do see B-B emission in the CO2 band. We all know what the AGHE is supposed to be; I highlighted the various problems with it here:
http://www.tech-know.eu/uploads/Understanding_the_Atmosphere_Effect.pdf
http://principia-scientific.org/publications/The_Model_Atmosphere.pdf
http://principia-scientific.org/publications/Copernicus_Meets_the_Greenhouse_Effect.pdf
So the plots show that indeed, the atmosphere radiates a bunch of energy by itself, and the ground-surface does the majority. A lot of the emission in the atmosphere comes from GHG’s, as I would have expected from astrophysical theory, so at least we’re back on track here, after my journey a bit out of bounds 😉 That emission from the atmosphere indicates cooling, given that non-GHG’s in the atmosphere can’t cool by radiation. I think Bart must be correct when he says that the atmosphere will heat up until it radiates enough to cool and equilibrate, which is the function of GHG’s.

January 19, 2012 12:50 pm

Bart says:
January 19, 2012 at 11:25 am
“Everything not forbidden is compulsory,” said Murray Gell-Mann. There is nothing forbidden in my prescription.

There is in your elevator description, you propose counter gradient conduction at the surface, that is forbidden!

Tim Folkerts
January 19, 2012 1:13 pm

Bart says: “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.”
My only reply is to look at the actual satellite data I posted. http://www.skepticalscience.com/images/infrared_spectrum.jpg
* With no GHG’s there would be no “bites” as seen from space
* With no bites, the earth would be radiating more
* With more outgoing radiation, the earth would cool (specifically the surface where we live would have to cool).
>> Therefore, the GHGs are warming the surface.
THAT is the simplest explanation of the greenhouse effect.

Bart
January 19, 2012 1:22 pm

DeWitt Payne says:
January 19, 2012 at 11:43 am
“A concentration gradient can only be stable if there’s a continuous flux into and out of a volume.”
DeWitt – I am precisely arguing that the situation is unstable.
DeWitt Payne says:
January 19, 2012 at 12:04 pm
“Not according to this:”
Here’s a better, non-cartoon graph. Figure 3.1.

Tim Folkerts
January 19, 2012 1:26 pm

mkelly makes two (maybe three or four) mistakes: “Using Avogadro number you contend that the 156 W/m^2 are produced by 113800 molecules of CO2. That is the number of CO2 in one square meter at 380 ppm. ”
1) There is no way to say how many molecules are in a square meter. You need to have a volume, not an area.
2) Near the surface anyway, a cubic meter of air is just over 1 kg. At ~ 30 g/mole, that would be ~ 30 moles of gas, or ~ 2E25 molecules. The number of CO2 molecules would be well over 1E21, not 1E5
3) The energy does not need to be absorbed in an altitude of 1 m. Pretty much everyone uses a number of at least 100m to absorb IR, so now we are up to 1E 23 molecules. If you really want the “number in 1 square meter”, the only logical interpretation would be in the entire column of air, which would be ~ 100 times more, or ~ 1E25 molecules.
Suddenly his objections seem much less valid.
(all numbers are intended to be orders of magnitude)

Bart
January 19, 2012 1:44 pm

“Greenhouse gases (other than ozone) do make the stratosphere cooler than it would be otherwise because the stratosphere is optically thin in the thermal IR range and most greenhouse gases do not absorb significantly in the UV. But the troposphere is optically thick for much of the wavelength range so the reduced emission at higher altitude makes the lower atmosphere warmer than if they weren’t present, not cooler. So the stratosphere has been getting a double whammy: a reduction in ozone causing less heating, and an increase in CO2 causing more cooling.”
Don’t you see, DeWitt? You are confirming exactly what I have been saying: the steady state result is the same no matter how you interpret it.
DeWitt Payne says:
January 19, 2012 at 12:04 pm
“In the Ionosphere, the temperature can reach 4,000K at altitudes above 500km”
There, you are dealing with an entirely different state of matter. So, you have to specify the boundary conditions and evaluate an entirely new PDE.

Bart
January 19, 2012 1:52 pm

Look at it this way, DeWitt. Starting from some temperature at the base, in any finite time, the temperature is going to be less out farther than it is nearer, and it is going to go to zero at infinity. So, there must be a gradient over any finite time.
What is the form of that gradient? Well, through separation of variables, the temperature function is the product of a time varying part, and a spatial varying part. The spatial varying part is a function of spatial coordinates only, so it does not change with time. Therefore, since for any finite time, the spatial solution is of the form A + f(r), where A is a constant and f(r) is a monotonically decreasing function of r, and A must be zero at that time, then it must always be zero, and f(r) must go to zero as r approaches infinity.
QED.

ZP
January 19, 2012 2:02 pm

Phil. says:
January 19, 2012 at 12:06 pm
Your over-estimation of the fraction of Argon atoms in the third electronic level under the Earth’s atmospheric conditions leads you to this incorrect conclusion (and perhaps also Anna’s error). Look up the energy of the third excited state and do a Maxwell-Boltzmann distribution calculation at T=300K you’ll find that it’s essentially all in the ground electronic state which does not absorb in the IR.

Argon is in the third period, therefore the ground state electron configuration is [Ne] 3s2 3p6. The valence electrons are going to be in the third electronic level unless excited.

jjthom
January 19, 2012 4:39 pm

Mike Jonas says: January 19, 2012 at 12:30 pm

6 – We should observe less of set A at TOA than we would expect from steps 1 and 2 above on their own.
7 – We should observe more of set B at the surface than we would expect from the sun alone.
=========
go here
http://www.patarnott.com/atms749/
in the drop down box (presentation by chapter…)
choose chapt 6 thermal emission
slide 9 in the power point presentation gives satellite downward looking spectrum and ground based upward looking spectrum
You will note that where there is a hole in one there is a peak in the other.
This mans website contains a wealth of information – well worth a trawl!

Spector
January 19, 2012 6:01 pm

RE: mkelly: (January 19, 2012 at 12:46 pm)
“Spector says: ‘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.’
“Using Avogadro number you contend that the 156 W/m^2 are produced by 113800 molecules of CO2. That is the number of CO2 in one square meter at 380 ppm.
“But the N2 and O2 (that also absorb heat or they would not be gases) do no ‘capturing’ of heat.”

My comment only applies to radiant energy. There is no reference to Avogadro’s number. If the surface of the Earth is so warm that the average radiant power emitted by that surface must be 396 W/m² according to Trenberth and the Stefan-Boltzmann formula, then the atmosphere *must* extract and return an average power of 156 W/m² from that *radiated* surface power before it gets out to space because the Earth as a whole is only receiving enough power from the sun to emit a nominal average of 240 W/m². It is as simple as that. The Earth cannot afford to continuously emit an average of 396 W/m² to outer space and remain as warm as it is. A greenhouse-gas free, transparent atmosphere cannot, by definition, prevent the escape of this radiation from the surface.
Power in watts is energy flow in joules per second. Neglecting the miniscule amount of geothermal power being generated, the principle of conservation of energy requires that the average power emitted by the Earth be the same as the average power being received in a steady state equilibrium condition. All other internal heat transfer issues are irrelevant.
It appears that water vapor may be twice as effective as carbon dioxide in this role.

Peter Offenhartz
January 19, 2012 6:05 pm

Ah, Willis Eschenbach, you are a breath of fresh air! Three cheers for the First Law of Thermodynamics! And Three cheers for the Elevator Speech. May I suggest an addition to The Speech? Many of your (ignorant) critics seem to think that all substances, including nitrogen and oxygen, can radiate in the infrared, that the laws of blackbody radiation apply without regard to the nature of the radiating substance. So some addition along the lines of “at any given wavelength, only substances that absorb radiation can radiate” might help.
Human ignorance and stupidity always amaze me. You have put a stake through the heart of some of it. Thank you!

Tim Folkerts
January 19, 2012 6:57 pm

Bart,
If you impose boundary conditions that T = 0 infinity, then I agree with your conclusions.
But for the non-radiating atmosphere, the condition is more like “no energy can flow past the top of the atmosphere”. There is no requirement that the edge of the atmosphere be 0 K. Furthermore, the system is not infinite. The condition is that the TOA cannot lose any energy outward and must approach the temperature of the material closer in. This will eventually lead to equilibration at the surface temperature.
As to the “it will take infinitely long”, well that is true of any such problem. It would take infinitely long for my coffee to cool to the temperature of the room. But there will be a characteristic time describes the system. For a cup of coffee, it might be 10 minutes. For the atmosphere I suspect it is more like a few months. But in any case, after you have waited 10x that time or 20x that time, then the change will be complete for all practical purposes. So after a couple hours I consider the coffee to be room temp. After a few years or decades, the atmosphere would be uniform for all practical purposes.
Now if the conditions are change even slightly (for example a small amount of GHGs are present at the TOA), then the boundary conditions are completely different. Then the boundary emits energy as a function of temp, leading to a definite lapse rate.

Bart
January 19, 2012 7:02 pm

Spector says:
January 19, 2012 at 6:01 pm
“…then the atmosphere *must* extract and return an average power of 156 W/m² from that *radiated* surface power…”
In the steady state. There’s a lot of water vapor in the air, so it can do this, and mkelly has neglected that and other IR emitters in the atmosphere. But, I wanted to take the opportunity to reiterate and reinforce the fact that SB and its variants are purely applicable only in steady state, and loosely applicable only in situations close to a steady state.

Bart
January 19, 2012 8:14 pm

Tim Folkerts says:
January 19, 2012 at 6:57 pm
“If you impose boundary conditions that T = 0 infinity, then I agree with your conclusions.”
Hey, it’s a start.
“There is no requirement that the edge of the atmosphere be 0 K.”
If it is anything less than the surface temperature, then there is a downhill temperature gradient into the atmosphere, and energy will accumulate.
“The condition is that the TOA cannot lose…”
What is the TOA? How do you define it?
In any case, the heat equation is assuming a continuum. That the continuum assumption breaks down at some point does not invalidate the whole equation. The equation holds for certain in the region where the continuum assumption is reasonable.
“…any energy outward and must approach the temperature of the material closer in.”
You are begging the question. You are assuming there is an equilibrium without establishing that there actually is one.
I agree it will approach the temperature of the material closer in, but not at the same time. By the time it reaches that temperature, the material closer in will have grown hotter. The exact proportions, from one radial distance to the next, are specified by the spatial part of the solution to the PDE.
“This will eventually lead to equilibration at the surface temperature.”
You present no evidence but your say-so, and are implicitly claiming the heat equation does not hold. That is an extraordinary claim, and requires extraordinary evidence.
“After a few years or decades, the atmosphere would be uniform for all practical purposes.”
You’re not getting it. The spatial part of the solution to the PDE does not change with time. The temperature distribution at any given time is precisely equal to a time modulated amplitude function times the spatial distribution function. It remains conformally identical for all time. Only the scale changes.
Perhaps you have taken classes where you looked at string vibrations? These phenomena are also described by separable partial differential equations. At a particular mode of vibration, the displacement of the string is a constant function, a so-called “mode shape“, which is modulated by a sinusoid in time. The precise shape, the nodes, the endpoints… these do not change, but the shape collapses and reemerges in its mirror image back and forth as the string vibrates sinusoidally in time.
This is the common behavior of systems described by separable PDEs. In this case, the monotonically declining temperature profile is analogous to the mode shape of a string. It can only be magnified or diminished (actually, only magnified forward in time in this case) by the amplitude function, which is a function of time only.
So, if the “mode shape” of this thing is a constant independent of radius, that means the entire spherical shell has to heat uniformly. It starts at zero everywhere. It is at 1K everywhere at the same time. It is at 100K everywhere at the same time. This is physically impossible based solely on the fact that the rate of heat conductance is finite.
“Then the boundary emits energy as a function of temp, leading to a definite lapse rate.”
A definite adiabatic lapse rate, which is exponentially related to pressure, hence linear in altitude. But, that is not the only type of lapse rate which can exist.

January 22, 2012 3:13 pm

Jelbring makes this crucial and erroneous assumption: “Equilibrium atmospheric conditions have been reached meaning that the average total energy of atmospheric molecules is constant”. This is how he mistakenly gets his permanent lapse rate in the absence of a continuing flow of energy. However, that’s not what thermal equilibrium means; it means that the temperature is uniform throughout. An isothermal atmosphere has zero lapse rate.

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