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 …
shawnhet says:
January 18, 2012 at 2:07 pm
“…regardless of whether the system is in equilibrium if the energy from a surface of temperature 15C is conducting away from that surface without radiating, then the emissivity will appear to be less than if it is radiating as per SB.”
Stefan-Boltzmann is fundamentally a steady state theory. Emissivity is fundamentally a steady state quantity. It does not even make any sense to have the words “conducting” and “emissivity” in the same sentence. You absolutely must be in an equilibrium condition, or at the very least in something that can be plausibly called a quasi-equilibrium condition, or all bets are off.
Once you actually reach equilibrium, then and only then can you invoke SB. Let’s suppose the atmosphere absorbs at wavelengths higher than IR, but low enough that the air molecules do not achieve escape velocity. When the planet reaches a temperature such that ample amounts of that radiation are being intercepted, then an equilibrium is possible.
When that equilibrium is established, you would as easily be able to interpret it the same way you do the lower energy IR case: the above-IR-absorbing gas is intercepting the outward radiation of the surface, and the incoming flux matches the outgoing at TOA. But, in fact, what has happened is that the above-IR-absorbing gas has provided a heat sink to prevent the atmosphere from warming more.
That is, the incoming flux matches the outgoing at TOA, and you could claim that the above-IR-absorbing gas has heated the surface above what it would be without an atmosphere. But, in fact, what has happened is that the above-IR-absorbing gas has provided a heat sink to prevent the atmosphere and the surface from warming even more.
You see, the two interpretations are entirely equivalent. The steady state environment does not give you enough information to choose between the two.
But, because we know that a temperature gradient will always exist when the system is not in equilibrium, we know that the latter interpretation, that the above-IR-absorbing gas has provided a heat sink to prevent the atmosphere and the surface from warming even more, is the correct one.
Bart,
You’re neglecting the obvious trivial solution of:
dT/dt = alpha*del^2(T)
which is:
dT/dt = 0 , i.e. thermal equilibrium.
Then del^2(T) = 0 so T(r) = constant for all r. That constant would be the surface temperature. Your general solution only applies if dT/dt ≠ 0. Also, if alpha*del^2(T) ≠ 0, then T is varying with time and there isn’t thermal equilibrium.
Bart:”Stefan-Boltzmann is fundamentally a steady state theory. Emissivity is fundamentally a steady state quantity. It does not even make any sense to have the words “conducting” and “emissivity” in the same sentence. You absolutely must be in an equilibrium condition, or at the very least in something that can be plausibly called a quasi-equilibrium condition, or all bets are off”.
I disagree. A substance can both radiate and conduct heat at the same time( it makes sense to talk about both concepts together because objects in the real world can do them together). If you perform the Niclos test on 15C water, it will appear to have an emissivity of some amount, regardless of whether or not it is in equilibrium or not. A system that has a temperature, radiates and conducts heat will always have an apparent emissivity. If it is not in equilibrium, then it may not have a *stable* (ie constant) emissivity.
Cheers, 🙂
I want to bring up another point of evidence another commenter made in another thread which I found curious, but saw no resolution at the time. I do not remember the commenter’s name or where to find the post, so my apologies to him.
His question was, if the greenhouse effect on Earth raises the temperature level above -18C, but at -18C, the main greenhouse gas, water vapor, is in frozen state, how did we ever warm up?
My hypothesis here, as you can see, resolves this dilemma. It was always heating up, and the water vaporized, and stopped the rise.
Bart,
But this is heat flow in a medium, not the electric field from a point charge or the gravitational field from a point mass. The boundary condition that temperature has to decay smoothly to zero as r increases doesn’t apply. In fact, it leads to an obvious violation of the Second Law, temperature increasing without limit. I’ve also restricted the solution to atmospheric pressure high enough that local thermal equilibrium applies, i.e. to ~30km for a planet with a surface g = 9.81 m/sec², a surface pressure of 1013 hPa and a composition of a symmetric diatomic molecule with an effective molecular weight of 29 amu. The temperature above that point doesn’t matter as 99% of the atmosphere is below 30 km.
You keep saying that SB and by extension the Planck function, only applies when there is thermal equilibrium. That also isn’t true. The necessary condition is that local thermal equilibrium applies. That is always true for solids and liquids at normal temperatures at all times, i.e. not things like Bose-Einstein condensates, and is true for gases (Planck function times the emissivity at the emitting wavelength) as long as energy transfer by collision is much more probable than energy transfer by radiation. That also means the kinetic energy of the gas molecules must obey Maxwell-Boltzmann statistics. If SB only rarely applied, IR thermometers wouldn’t work as they rely on surface emitting according to the grey body version of SB (emissivity < 1).
Oh, and wind velocity roughening the sea surface can increase emissivity as well as lower it. It depends on the velocity.
Bart says: January 18, 2012 at 1:28 pm
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
============
Solids above 0K radiate energy dependent on their temperature and their emissivity.
The sb curve just shows where the radiation peaks in the spectra.
It is not dependent on equilibrium – the body just radiates little at low temperatures and much at high temperature – instantly.
So if the planet warms then it radiates more – it cannot not radiate more. if 200w/sqm are received then the body cannot radiate more than this.
In your case you suggest that it gets hotter but does not radiate more – can you explain this please?
If a an instant change in radiation of gigawatts occurs the gas will still be heated only by surface. The surface will slowly heat from its current temp at a rate dependent on the thermal mass and the energy input. The non-ghg will only follow this temp it can never exceed the sb temperature. if this pulse disappears the surface temperature will start to cool.
If you chande its emmitance to 0 then it could reach the temperature of the suns. unfortunately if its emmitance is zero its reflectance is 100% so in fact it will neither heat nor cool
DeWitt Payne says:
January 18, 2012 at 2:49 pm
“You’re neglecting the obvious trivial solution of:
dT/dt = alpha*del^2(T)
which is:
dT/dt = 0 , i.e. thermal equilibrium.”
Doesn’t work. T must go to zero at infinite radius.
DeWitt Payne says:
January 18, 2012 at 3:24 pm
“…it leads to an obvious violation of the Second Law, temperature increasing without limit.”
No. Temperature will increase until it either A) excites the emission of higher frequency photons or B) the particles achieve escape velocity, and the atmosphere wafts away into space.
“The necessary condition is that local thermal equilibrium applies.”
There is no thermal equilibrium of any kind.
jjthoms says:
January 18, 2012 at 5:23 pm
“It is not dependent on equilibrium – the body just radiates little at low temperatures and much at high temperature – instantly.”
You guys really need to read up on SB. A surface will only radiate heat to the degree it cannot dissipate it otherwise.
An N2 atmosphere will grab heat energy off the surface faster than the relaxation IR emission time of the surface molecules. The time between collisions of surface atmosphere with solid surface material is just 0.00000000015 seconds which is many times shorter than the relaxation IR emission rate of the surface at SB temperature.
In other words, the atmosphere near the ground will heat up faster than the surface can emit in the IR. The energy accumulating in the atmosphere will be conducted up through the atmosphere until an equilibrium is reached between atmospheric collisional energy exchange at the surface and the amount that is being conducted up through the atmosphere.
One would to crunch the numbers to see if there is literally enough time difference between the two so that more energy ends up being stored in the atmosphere than is being stored at the surface accounting for both the SB temperature and the collision energy exchange rate.
But if the N2 atmosphere always grabs energy off the surface faster than the relaxation time, then there is NO IR emission. Now it can’t always be faster, there is always some really fast IR emission, but these amounts are small compared to the overall numbers.
In other words, the surface HAS to be warmer because it is has BOTH solar radiation and N2 atmosphere energy exchange. It will heat up until IR emission becomes very fast. In other words, many times hotter than the SB temperature.
Time and distance are important considerations in this debate but they are almost always ignored.
DeWitt Payne says:
January 18, 2012 at 3:24 pm
“The boundary condition that temperature has to decay smoothly to zero as r increases doesn’t apply.”
At some point, the atmosphere effectively ends. So, you could argue that the concept of temperature simply breaks down at the ragged edge as the atmosphere reaches a particular density. But…
The edge of the atmosphere is going to be defined by the point at which the greater part of molecular velocities exceeds the escape velocity, which is proportional to the square root of the gravitational potential. If the temperature has to be uniform to, and is limited by, that distance, then you are imposing a uniform temperature limit which depends on gravity.
Out of the frying pan and into the fire. In disagreeing with me, you find yourself having to agree with N&Z!
Bart,
Wrong. The energy density must go to zero at infinite radius. And it does. For a gravitationally bound exponential atmosphere the total energy density in J/m3 goes to zero exponentially with altitude. Here’s a plot of altitude vs the the logarithm of total of gravitational plus kinetic energy/m3 calculated using 1005 J/kg K for the heat capacity of air at a constant temperature of 255 K at all altitudes. Note that the pressure at 100 km is 0.00188mbar so 99.9998% of the atmosphere is below 100 km. The US 1976 standard atmosphere has the temperature at 100 km equal to 195.1 K, not all that far from 255 K.
At 100 km altitude, the escape velocity has decreased by all of 87 m/s to 11,093 m/s. For nitrogen, the RMS velocity at 255 K is 476.6 m/s. The altitude at which the escape velocity is 476.6 m/s is 3.5E6 km.
Wrong again. You’re simply asserting that with no proof whatsoever. You need to read up on the concept of Local Thermal Equilibrium and Kirchhoff’s Law. If LTE didn’t apply in the atmosphere, then it wouldn’t be possible to accurately calculate IR absorption and emission spectra. But it does and you can. If LTE didn’t apply to solids and liquids, it would be possible to construct perpetual motion machines of the second kind where the free energy is extracted from a single reservoir at one temperature.
Bill Illis,
Here we go again. The emission rate is determined solely by the number of excited atoms or molecules and the decay rate. In the case of an excited molecule, that’s the Einstein A21 coefficient. The half life is approximately the inverse of the decay rate. But a molecule in an excited state can emit at any time. There are lots of molecules even if only a few percent are excited at any time. A solid or liquid surface has many more surface sites capable of emitting and absorbing radiation. At local thermal equilibrium the number of molecules or surface states in the excited state is constant. Local thermal equilibrium isn’t something rare. Most things are in LTE. Also, most of those collisions are elastic and don’t involve energy exchange so the actual lifetime of an excited state is longer than a nanosecond. But LTE requires that the collisional lifetime is orders of magnitude shorter than the radiative lifetime. See http://noconsensus.wordpress.com/2010/08/17/molecular-radiation-and-collisional-lifetime/ . As I remember, I’ve pointed this out to you before.
The only way the surface can heat to a temperature hotter than a black body is if the emissivity for long wavelength radiation is less than the absorptivity for short wavelength radiation. This isn’t true for most things that cover the surface of the Earth like water and dirt.
DeWitt Payne says:
January 18, 2012 at 7:47 pm
“The energy density must go to zero at infinite radius.”
Of course the energy density goes to zero at infinite radius. The particle density goes to zero at infinite radius. But, the temperature also has to decline to the background radiation temperature of the cosmos, which is as close to 0K as makes little difference (actually, our hypothetical planet doesn’t have to inhabit this universe, so I’ll just proclaim it 0K).
“At 100 km altitude, the escape velocity has decreased by all of 87 m/s to 11,093 m/s. For nitrogen, the RMS velocity at 255 K is 476.6 m/s.”
That is the RMS deviation from orbital velocity. If that is the highest point you reach on an Earth orbit, you could be going as fast as 7844 m/s (circular orbit) +/- 477 m/s 1-sigma.
But, we’re not talking about Earth here, you know. Go ahead and specify that the hypothetical non-GHG planet has an Earth-like mass, if you like. But, then, it is still not required to have anything like the same temperatures.
That’s my whole point. It’s not going to have the same temperatures. If it has no other mechanism to radiate energy away, then eventually you could reach 2550K, or 25500K. You’re sure to start losing your atmosphere, then.
Doesn’t sound plausible? Of course it doesn’t! But, that is because we have an implausible atmosphere on a thought planet that doesn’t even radiate in the ultraviolet or beyond!
But, it still has a mechanism to prevent “temperature increasing without limit”, which was your objection, and which has been satisfied.
“If LTE didn’t apply in the atmosphere, then it wouldn’t be possible to accurately calculate IR absorption and emission spectra.”
In Earth’s atmosphere. Fine. So what?
“You need to read up on the concept of Local Thermal Equilibrium and Kirchhoff’s Law.”
OK.
Since that is precisely what I am asserting, that the surface is not radiating at its local kinetic temperature, it means I am asserting it is not at LTE. It cannot be, because a significant portion of the absorbed heat is flowing into the atmosphere.
DeWitt Payne says:
January 18, 2012 at 8:07 pm
“The only way the surface can heat to a temperature hotter than a black body is if the emissivity for long wavelength radiation is less than the absorptivity for short wavelength radiation.”
Kirchoff’s law states that blackbody irradiance is the most you can get for a given temperature.
J = (J/sigma)^(1/4)
It is a lower bound for temperature given radiance.
Duh… I used an inequality sign, and it turned into an html tag. Should have been
J .LTE. sigma*T^4
implies
T .GTE. (J/sigma)^(1/4)
DeWitt Payne says:
January 18, 2012 at 8:07 pm
“The only way the surface can heat to a temperature hotter than a black body is if the emissivity for long wavelength radiation is less than the absorptivity for short wavelength radiation.”
—-
The surface temperature of every body that has an atmosphere that we know about in the universe, is hotter than its blackbody temperature.
****
Joel Shore says:
January 18, 2012 at 11:47 am
You are right that when there is convection, the lapse rate is going to be close to the adiabatic lapse rate but you have the cause-and-effect partly backwards: Lapse rates less steep than the adiabatic lapse rate (such as in the stratosphere) are stable and hence convection is suppressed. Lapse rates steeper than the adiabatic lapse rate are unstable to convection and hence convection occurs, transporting heat upward until the lapse rate is equal to the adiabatic lapse rate.
*****
Thanks Joel (I do appreciate your comments). I have to think more about this. It’s not just me — after reading this blog & others, seems quite a few obviously smart people aren’t sure about this either. I agree any real, surface-heated world should develop some lapse-rate in their atmospheres. As you say, sorting out cause & effect seems to be the problem.
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
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?
If CO2 is radiating all this energy and by definition this emission has to be spectral, then where is the huge & incredibly bright emission line introducing an additional 150 W/m2 into the surface, and that which should also be exiting the TOA? The ENTIRE output spectrum of the Earth only comes out to 240 W/m2, so this additional 150 W/m2 from spectral emission must be huge! Where is it?
It doesn’t exist.
Instead, where it could exist, is a huge hole in the spectrum, a LACK of energy power. It HAS TO EMIT to be said to cause heating by radiative emission in the first place. Yet where it could emit, it doesn’t exist. And at the bottom of the notch where CO2 should be emitting to add all this extra power, is a smooth blackbody curve corresponding to something like -80C. There is a very small emission peak right at 15um, but it’s barely worth noting.
So fine, let’s pretend to go with the standard theory of GHG back-emission. And so I’ll ask: where is the emission?
Pointing an IR sensor at the sky and it telling you the temperature, converted to some power units, is simply taking a temperature reading! It has NO relevance to the huge emission line we should see from CO2 & GHG’s causing all this heating. That’s a fraudulent interpretation of the measurement! The IR sensor is measuring a rough black-body that has LACK of emission flux at GHG wavelengths.
It might just be that simple. Fine: radiation causes heating, we all know that. So show me the radiation from CO2 then. Oh it doesn’t exist? Well then what the heck…non-existent, non-observable spectral emission from GHG’s causes heating. Wonderful.
Now someone might try to back-track and say that the downward IR gets all absorbed by the time it reaches the ground, and because it is all absorbed this is why it is causing heating. But wait, it can’t very well be said to be heating the surface then, can it, if it never GETS to the surface. And additionally, you can’t hide at the bottom of the atmosphere anyway – the TOA is free to space and there’s NO reason we shouldn’t see the 150 W/m2 of GHG spectral emission there. But it’s not there either, is it; except for a tiny little pip at 15um with maybe a couple of Watts in it.
Someone might also try to back-track and say that because all the 15um radiation is absorbed and you don’t see it, that’s why it causes heating. But that’s still inconsistent with the spectral reading at the TOA – it should still be seen at the TOA – and it also implies that LACK of emitted radiation is what causes heating. But that’s what non-GHG’s implicitly do in the first place – not radiate and therefore trap heat – and the whole GHG theory is based on the idea that GHG’s spectrally radiate.
This whole theory is shot; full of holes. With this OP and the comments here in it, and the other one by Robert Brown and the comments in that one, it is clear to anyone reading that GHG Theory is dead. I still think my treatises give a good explanation of why it is dead.
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
Read only the last of those links if you want – it is a very short paper with a succinct summary of the paradigm shift.
Of course, I give thanks to the book that started it all:
http://www.amazon.com/Slaying-Sky-Dragon-Greenhouse-ebook/dp/B004DNWJN6
Joe Postma,
I suggest you get a textbook on atmospheric radiative transfer and study it so you can make an intelligent argument. Your post above clearly shows you have no idea about how the atmospheric greenhouse and molecular absorption and emission of radiation in the atmosphere works. 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. Grant Petty, A First Course in Atmospheric Radiation, 2nd edition, Sundog Publishing would be a good choice. Or, for a start, read Chapter 5 in Caballero’s Lecture Notes on Physical Meteorology: http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf
In the spectrum from the surface looking upward, there is no line because the emission cannot exceed the intensity of a black body at the same temperature. In fact, the emission is calculated by multiplying the emissivity, which cannot exceed 1, at a wavelength by the Planck function for that wavelength and temperature. The emission spectrum in the CO2 band closely matches a blackbody at the surface temperature because the emissivity/absorptivity is almost exactly one for center of the CO2 band.
Bart,
The kinetic temperature of a gas is not a field emanating from a point source. It’s a function of the local RMS velocity and mass of the molecules. 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. And only then if the molecule velocity exceeds the escape velocity. 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. Below 100 K, the temperature varies such that in the stratosphere the temperature is higher than 200K. At 50km altitude for the 1976 US standard atmosphere, the temperature is 270K, not far from the surface temperature of 288K.
If a gas molecule at high altitude were actually in an orbit that didn’t intersect the surface, it’s velocity would be at least an order of magnitude greater than the RMS kinetic velocity at 255 K. That would imply a temperature of ~30,000K. But of course, they don’t have to be in a non-intersecting orbit, and the vast majority aren’t. They stay approximately where they are because they are continually bouncing off other molecules.
I don’t have time to address ALL the issues that keep coming up, but let me make one correction to what I said earlier about lapse rates.
IN PRINCIPLE, in a perfectly transparent atmosphere that cannot radiate or absorb EM radiation (and also is not being hit by a solar wind or other energy inputs) that sits around a planet with a perfectly uniform temperature, the equilibrium temperature profile of the atmosphere will be uniform at the temperature of the surface. There will be NO lapse rate.
If the atmosphere DID get a temperature gradient, thermal conduction would slowly bring the temperature profile back to a uniform temperature throughout.
IN PRACTICE, having a lapse rate would be the expected situation. Small amounts of convection would set up a lapse rate. As long as there is convection (which would always happen on a rotating planet that goes from cooler nights to warmer days day), the atmosphere will get some sort of lapse rate.
Furthermore, the power flow through the atmosphere when the temperature gradient is as large as 10 K/km would be less than 1 mW/m^2. So all that is needed to maintain a lapse rate is an energy flow out of the top of the atmosphere of around 1 mW/m^2. Even the tiniest amounts of GHGs or dust would be enough to do that. It is possible that even non-GHGs would be able to radiate that amount. In any case, the actual radiation by GHGs from high in the stratosphere is way more than 1,000x higher than needed to maintain a lapse rate.
Joe Postma says:
In addition to what DeWitt Payne said, I should just add that I assume the 150 W/m^2 that you are referring to is a measure of the total radiative forcing due to the greenhouse effect. That being the case, it is not all due to CO2…In fact, the largest portion is due to water vapor and clouds.