Jul 10 2007

Falsification Of The Atmospheric CO2 Greenhouse Effects

Published under Climate Change

Falsification Of The Atmospheric CO2 Greenhouse Effects Within The Frame Of Physics by Gerhard Gerlich and Ralf D. Tscheuschner, arXiv:0707.1161v1 [physics.ao-ph]. [PDF]

I don’t actually recommend reading this. But one gem that they propose is that there isn’t a thing called an average temperature. Of course there is. When attempting to derive such a temperature, Gerlich and Tscheuschner arrive at a value of 87.6 C. This is clearly wrong. If the Earth were that warm, humans wouldn’t exist. They then “explain” how climatologists get their value to explain the greenhouse effect as follows:

This fictitious [greenhouse] effect is based on the assumption that one should have an average effective temperature of -18 [degrees] C. One will get this if one weights the solar constant with a factor of 0.7 and inserts a quarter of the solar constant into the “radiative balance” equation. The factor of a quarter is introduced by “distributing” the incoming solar radiation seeing a cross section σEarth over the global surface ΩEarth.

[Added August 7, 2007: This post has been linked from a comment in a Scienceblogs.com post. For those who don't believe in the greenhouse effect, please explain how the average temperature on the moon is lower, in spite of the fact that it has a lower albedo.]

Actually, the value of -18 C falls right out of the equations, not the other way around. Assume that the sun radiates at a certain temperature such that it can reasonably be modeled by the blackbody curve for some effective temperature. This shouldn’t be that hard to do, even Gerlich and Tscheuschner do so in their paper. By the time this radiation reaches the Earth, it’s intensity has decreased according to the 1/R2 law. Again, this is exactly what Gerlich and Tscheuschner do. At the Earth, this value is roughly constant - 1367 W/m2. Gerlich and Tscheuschner do not use this number, they keep their equation in terms of the temperature of the sun, radius of the sun, and distance from the Earth to the sun. It doesn’t matter, the finals answers will end up the same.

Therefore, the total energy absorbed by the Earth is related to its albedo and its radius.

EA = (1-A)S0πR2

The term 1-A is the percentage of incoming solar radiation absorbed by the Earth; the albedo (A) is the percentage reflected. S0 is the solar constant. And πR2 is the cross-sectional area of the Earth that absorbs radiation. The dark side of the Earth cannot absorb radiation from the sun.

The total energy emitted by the Earth is related to its temperature and its radius.

EE = σT44πR2

Because the Earth emits radiation from its entire surface and not just the side facing the sun, the surface area of the Earth is used (4πR2) is used instead of the cross-sectional area. The σT4 term is the blackbody emission for an object at a given temperature.

Setting the two equations equal - assuming the energy absorbed equals the energy emitted - and simplifying, we see that

(1-A)S0 = 4σT4

So, for a given A and S0, we can find the effective temperature. In the case of the Earth, the albedo (A) is about 0.3, so 1-A is 0.7, which magically explains where that factor comes from that Gerlich and Tscheuschner couldn’t explain. The factor of 4 is just a consequence of the fact that the Earth can only absorb radiation on the side facing the sun, but emits in all directions. When the values are plugged in, we (and Gerlich and Tscheuschner) get a value of -18 C.

So, Gerlich and Tscheuschner couldn’t figure out where the magical values of 0.7 and 0.25 [1/4] came from, but they are just misleading their readers. They [should] know how to compute an effective temperature. And they [should] know that such a value exists, and is physically meaningful.

(20 votes, average: 3 out of 5)

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• 60 Responses to “Falsification Of The Atmospheric CO2 Greenhouse Effects”

1. Ralf D. Tscheuschneron 12 Jul 2007 at 10:22 am

Dear Anonymous -

Just have a look at foonote 22 (Version 1.0 of the paper) and you will be happy.

Best wishes,
Gerhard Gerlich und
Ralf D. Tscheuschner

2. N. Johnsonon 12 Jul 2007 at 10:36 am

Thanks for the comment. I did see that footnote, but the actual text of the paper makes it seem that the value of 0.7 is a consequence of “fixing” the temperature at a certain value, and not that it’s related to the albedo. Hiding this fact in a footnote just offers the reader a chance to miss this important information.

[I'm not really anonymous. I just choose not to prominently display my name on this site.]

3. Eli Rabetton 19 Jul 2007 at 2:43 pm

As I recall you have to neglect convection to get this result. In simple terms radiation alone is not enough to distribute the heating near the surface vertically. I’ve seen this several times so it must be fairly well known.

4. Alan Siddonson 25 Jul 2007 at 12:36 pm

For my own two cents…

Every radiation budget you find has the earth’s surface receiving about 168 solar watts, which is only sufficient to heat it to -40ºC. This glaring deficiency in the “heat-retaining” idea of greenhouse theory led to people accepting the main outlines of the Kiehl-Trenberth version, which has the atmosphere magically producing 324 watts that get “back-radiated” down to the surface, bringing it up to a desired 390 watts, or +15ºC. That back-radiation violates the first and second laws of thermodynamics is of no concern to climatologists. It matters only that such a miracle meets the demands they place on greenhouse theory. What really causes the earth to warm is a matter of indifference to them.

I may disagee with certain conclusions that Gerlich and Tscheuschner reach, but I can only applaud their courage to challenge the palpable nonsense that greenhouse theory has become, and has been for a long time. There ought to be more people like them. But sadly, I predict that their paper will go ignored. Despite apparent differences, both climatology skeptics and alarmists are largely members of the same club, a self-insulated debating society. They have more in common with each other than with genuine skeptics like Gerlich and Tscheuschner. A possible exception is professor Marcel Leroux, who I’d expect to comment if he learns about this.

5. Andre Bijkerkon 26 Jul 2007 at 6:53 am

The Stephan-Bolzmann derative:

(1-A)S0 = 4σT4

is valid only for perfect black (gray) bodies, with also the property of being a perfect conductor, hence the heat at the equator is the same as at the poles. But planets are more near perfect insolators, not exchanging a lot of heat. Therefore the planetary “black insulator body” temperature should be calculated as integral over the radiation equations over the indivual lattitudes. The outcome is that the practical blackbody temp is a few degrees less for a perfect insolator than for a perfect conductor.

Earths insolator black body temperature A=0.3 and S0=1365 (today) would be: 251.95K not 255K

[Ed: I changed 251,95K to 251.95K to be consistent with the United States convention. And of coarse you are right that assuming a black/grey body isn't the best, but it is a fairly good assuption. Just as a perfect insolator is a fairly good assumption, even though they are "opposite" assumptions. The derived temperature is only 3 K different.]

6. Alan Siddonson 26 Jul 2007 at 11:18 am

To Andre:

Check the blackbody temperature of an earth with no albedo, receiving the full impact of the sun’s radiation. It’s 279K. A blackbody is an absolutely perfect absorber/emitter, yet the real planet earth is 288K.

Such impossibilities are overlooked, however, because greenhouse theory has hypnotized almost everyone.

[Ed: The earth isn't a blackbody, it's a greybody. It has an emissivity of about 0.65.]

7. Alan Siddonson 26 Jul 2007 at 12:40 pm

To “Ed”:

You must be kidding! An imperfectly-absorbing earth is hotter than a perfect absorber/emitter with no reflective losses - but you’re pretending not to get it? Your “correction” imposes only a greater disparity between a theoretically maximum temperature and empirical reality. Perhaps that is your point, then. If so, I agree: contemporary greenhouse theory is ludicrously unsound.

8. N. Johnsonon 26 Jul 2007 at 1:15 pm

The effective temperature of an imperfectly absorbing earth (ie the real earth) is about 256K. The effective temperature of a perfectly absorbing earth is about 280 K. So the effective temperature is much higher for the perfectly absorbing earth.

However, the surface temperature is something totally different than effective temperature. The surface temperature of the earth is higher than the effective temperature of a perfect absorber because the surface is being heated by both the sun and the atmosphere.

9. Alan Siddonson 26 Jul 2007 at 2:04 pm

To N. Johnson:

It’s closer to 255K and 279K but I won’t quibble. You’re right, though: greenhouse theory counts energy twice, thus breaking the barrier that an ideal maximum of 279K (or 280K) imposes. By cheating, greenhouse theory sees no problem with the surface receiving and emitting an equivalent solar constant of 1560 watts, although the sun radiates 1368 in our neck of the woods and earth accepts only 957 to start with.

Now it merely remains for you to patent a device that produces more output than input and make yourself the richest person in history.

Future generations will laugh at us, as we laugh today at Ptolemy’s model of the cosmos.

10. Jan Pompeon 27 Jul 2007 at 5:11 am

N Johnson

I’m wondering how the atmosphere which is normally cooler than the surface warm it. Have you found a way to turn the 2nd Law of thermodynamics on its head?

I could do with such a passive heat pump should save power bills.

11. N. Johnsonon 27 Jul 2007 at 8:10 am

The second law is only valid for closed systems; the earth is not a closed system.

12. Alan Siddonson 27 Jul 2007 at 9:49 am

Wrong and backwards, N. Inclusive of open systems and closed, energy moves from hot to cold, from relative plus to relative minus. Always. Only by completely trapping energy in an isolated system, a theoretical entity which you might be confusing with a closed system, can you prevent energy from moving to an area of less energy. A closed system does not exchange matter but does exchange energy with its surroundings. From hot to cold, not from cold to hot. It’s like water running downhill. Where are you getting your physics? The real world is not controlled by mere words.

13. N. Johnsonon 27 Jul 2007 at 11:28 am

I meant an isolated system. The 2nd Law is only valid for isolated systems.

Consider this thought experiment. If the Earth were to suddenly become an isolated system, what would happen? The surface of the Earth is warmer than the atmosphere, so the atmosphere would warm and the surface would cool until they were the same temperature, in agreement with the 2nd Law.

But the Earth isn’t an isolated system; it receives energy from the Sun. To apply the 2nd Law, you need to consider the Sun too.

Second thought experiment: how does a refrigerator work?

14. Alan Siddonson 27 Jul 2007 at 12:57 pm

“The 2nd Law is only valid for isolated systems.”

To the contrary, the second law is INVALID in isolated systems. The second applies to both open and closed systems.

Open: Energy and matter in full exchange.
Closed: Only energy is exchanged.
Isolated: No matter or energy is exchanged.

But the original question was, how could a cooler atmosphere warm the surface? That IS a reversal of the second law, as Jan Pompe rightly pointed out, like water running uphill.

It’s a key question because the prevalent belief is that greenhouse gases hinder the earth from cooling down (which I don’t dispute). But I covered the inadequacy of this view in my first post: 168 watts only bring the surface to -40 C. A positive 15 C is our aim, however. Thus we now have the Kiehl-Trenberth model that unaccountably radiates 324 watts of energy to the ground.

To sum it up, the popular view can’t explain how the earth gets warm, while the amended view can’t explain how the atmosphere gets so much energy! Neither view stands up to scrutiny, so let me repeat my praise for people who challenge it.

Look at it this way. Say you’re correct that the atmosphere radiates its energy (undiminished) in two directions. Okay, so 67 atmospheric watts get added to 168 surface watts. That’s 235 watts now absorbed by the surface. But your surface energy actually reads 390 watts. You’ve still got 155 watts to make up.

See the problem?

15. N. Johnsonon 27 Jul 2007 at 2:30 pm

“To the contrary, the second law is INVALID in isolated systems. The second applies to both open and closed systems.”

I’ve never seen this before, and don’t believe it. See for example this page. “For an isolated system, the natural course of events takes the system to a more disordered (higher entropy) state.” Or dS/dt >= 0.

I believe before we can continue discussing anything, you’ll need to provide proof that the second law is valid in open systems.

16. Alan Siddonson 27 Jul 2007 at 4:36 pm

I might well have referred you to that link myself.

“The second law of thermodynamics is a general principle which places constraints upon the DIRECTION of heat transfer and the attainable efficiencies of heat engines.”

Or more explicitly,

“Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”

This rule does not apply exclusively to isolated systems. For in point of fact, isolated systems do not flow at all. To say “the 2nd Law is only valid for isolated systems” is simply nonsense, then.

Open: Energy and matter in exchange.
Closed: Only energy is exchanged.
Isolated: No matter or energy is exchanged.

http://www.engineersedge.com/thermodynamics.htm

What you seem not to grasp is that the terms open, closed and isolated are merely mental conveniences employed to deal with a full context on one side or to simplify the picture on the other. I’ve never seen anyone but you claim that the real earth is exempt from thermodynamic laws due to not being an isolated system. Really, you’ve got the whole thing backwards.

You also seem vague about the first law, however. So I ask again: Where do the extra 155 watts come from?

17. Jan Pompeon 27 Jul 2007 at 4:50 pm

Mr Johnston,

On the site you linked you also have “Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow.”

You might also look up “perpetuum mobile of the second kind”

There are various expressions of the second law depending on context as is clearly shown on the hyperphysics site. The entropy version applies to isolated systems the heat flow to open and closed systems and the planet is of the latter sort.

18. Al Tekhassion 29 Jul 2007 at 9:41 pm

Dear ATMOZ, you said:
“When attempting to derive such a temperature, Gerlich and Tscheuschner arrive at a value of 87.6 C.”

I think you clearly misunderstood the paper. The temperature of 87.6C is a temperature that a back-insulated flat grey disk (A=0.7) will reach under typical direct insolation of 1367 [W/m2]. You can verify this result experimentally by yourself in a good sunny day.

Unfortunately, the Earth is not a flat disk, as it also appears in your example. If you could read the paper attentively, you would find out that G&T also reproduced your -18C result, but they explained (rather poorly IMO) that the way you construct averages is mathematically incorrect. On pages 63-64 they presented a _correct_ way to obtain spatial averages on a globe for the T^^4 function, which leads to an average temperature of -129C. Obviously, incorrect way of calculating averages (as in your presentation) cannot lead to correct result, and the number -129C is also not any close to reality. As result, the intermediate conclusion of the paper was that “evidently, something must be fundamentally wrong here”. And it obviously is.

Cheers,
- Al Tekhasski

19. N. Johnsonon 29 Jul 2007 at 10:57 pm

If their correct way of averaging is really correct, why does it lead to an obviously wrong answer?

20. Adrian Brockon 30 Jul 2007 at 5:38 am

The argument about the 2nd law of thermodynamics
looks very confused to me.

There is no immediate problem for the
“Kiehl-Trenberth” model
http://www.cgd.ucar.edu/cas/trenberth.papers/KiehlTrenbBAMS97.pdf
since contrary to what is being argued, the net
heat flow is from the surface to atmosphere.

Thermals (24) + Evapo-transpiration (78)
+ surface radiation absorbed by atomsphere (350)
- “back radiation” (324) = 128 Wm-2

i.e. the surface warms the atmosphere, not vice-versa

Two problems I do see are (there are others):

to the surface? Surely these numbers should be
the same? The radiation is isotropic.

2) If you add those two numbers together you get
519 Wm-2. The atmosphere would have to be
at 36 degrees celsius to produce that radiation.

So it does break the second law in the sense
that heat is flowing from the surface
to the atmosphere even though the atmosphere
is supposedly hotter?

Clearly all of the 324 is NOT radiation
and must be some other mechanism.

If you make the radiation isotropic, i.e.
195 is radiated downwards then the
“some other mechanism” is tranferring 129 Wm-2

But this is also clearly wrong since the
total radiation of the atmosphere would be
390 Wm-2 (15 degrees celsius)
which is the same temperature as the
surface. Contrary to observation.

You can take these calculations with
a “pinch of salt”, all they really show
is that the model is incomplete.

On the -18 degrees celsius, that is the temperature
of an equivalent blackbody radiating the
same 235 Wm-2 as the Earth+atmosphere does.

Reading anything else into that number is just weird.

If the Earth had no atmosphere you might
want to claim it as a global average
surface temperature,
but the real temperatures would be more
like those of the Moon.

21. Mac Lorryon 30 Jul 2007 at 8:28 am

I believe the point the authors are making is not that there is no measurable average temperature, but that calculating it has been grossly over simplified, and thus, produces wrong results.

The idea introduced in the paper is that every square meter must be considered individually to arrive at the correct value for that square meter. When that is properly done the gain from the so-called greenhouse effect disappears. The validity of that argument is supported by the argument that the atmospheric greenhouse effect doesn’t exist.

Maybe some would be more willing to accept the average temperature claim if the authors first demonstrated that the atmospheric greenhouse effect doesn’t exist. The question then follows as to the error in calculating the average temperature.

22. N. Johnsonon 30 Jul 2007 at 9:21 am

to the surface? Surely these numbers should be
the same? The radiation is isotropic.

I don’t know why this is, and I haven’t read the paper thoroughly to know why.

2) If you add those two numbers together you get
519 Wm-2. The atmosphere would have to be
at 36 degrees celsius to produce that radiation.

So it does break the second law in the sense
that heat is flowing from the surface
to the atmosphere even though the atmosphere
is supposedly hotter?

To get the 519 W/m2, you add 350 W/m2 from the radiation emitted by the earth not transmitted through the atmosphere, 67 W/m2 absorbed from the downwelling radiation, 24 W/m2 from convection (thermals), and 78 W/m2 from evapotranspiration/latent heat.

23. Adrian Brockon 30 Jul 2007 at 1:20 pm

“To get the 519 W/m2, you add 350 W/m2 from the radiation emitted by the earth not transmitted through the atmosphere, 67 W/m2 absorbed from the downwelling radiation, 24 W/m2 from convection (thermals), and 78 W/m2 from evapotranspiration/latent heat.”

Yes, that is the conservation of energy (1st law of thermodynamics) using the numbers in the model.

I’m arguing it is wrong, since the model says the energy you quote is balanced by the thermal radiation of the atmosphere (either to space or the Earth).

But to do that (Stefano-Boltzmann) would need the atmosphere to be at 36 degress celsius.

A more reasonable figure for the atmospheric temperature -40C would mean a thermal radiation intensity of 168 Wm-2

Of course, it is very debatable whether Stefano-Boltzmann really applies to the gases in the atmosphere, but it is the model that claims all energy is radiated away.

24. N. Johnsonon 30 Jul 2007 at 2:20 pm

Remember that the vertical temperature profile is not constant. The temperature corresponding to the outgoing infrared flux will be lower than the temperature for the surface downwelling flux.

The temperature seen for both escaping and backradiation will correspond with the temperature of the atmosphere at one optical depth. It appears the numbers in the cartoon are for cloudy situations (higher optical depth). So the number I come up with (from the cartoon) are surface temperature of 15C, a temperature of 2C for the back radiation, and a temperature of -41C for the escaping radiation; 390W/m2, 324W/m2 and 165W/m2 respectfully.

It also appears that the values of 235W/m2, 324W/m2 and 390W/m2 are all measured (or derived) values of fluxes.

25. Adrian Brockon 31 Jul 2007 at 3:08 am

“The temperature seen for both escaping and backradiation will correspond with the temperature of the atmosphere at one optical depth.”

Ok, but naively the radiation will correspond to half that amount (half the radiation goes upwards/downwards).

In practice you’d have to integrate over the whole atmosphere and take into account re-absorbtion (and the different temperatures in different layers).

Some of the radiation emitted near the surface will be absorbed before it reaches space and vice-versa. It will be re-radiated at the temperature of that layer of the atmosphere.

You’d also have to do the same for the incoming IR in the solar radiation.

I haven’t tried to do the calculation.

“It also appears that the values of 235W/m2, 324W/m2 and 390W/m2 are all measured (or derived) values of fluxes.”

My understanding from reading the paper is that 235 (outgoing to space) and 390 (temperature of Earth) are “observed averages”.

You seem to claim the 324 is based on the temperature at one optical depth. The paper claims it depends upon the cloud cover as well (~46WM-2 - page 201 - again all downwards?)

And also says at the top of page 200: “We must rely on model calculations to determine the surface radiative fluxes” to which I’d prefix “Unfortunately!”

26. N. Johnsonon 31 Jul 2007 at 8:56 am

“You seem to claim the 324 is based on the temperature at one optical depth. The paper claims it depends upon the cloud cover as well.”

Yes, of course. Cloud cover will dramatically increase the optical depth in the infrared.

I haven’t done the whole radiative transfer calculation for Earth either. I have done it for other planetary bodies (Titan and Uranus).

A simple calculation assuming an atmosphere of N slabs can be seen in section 1.6 of Fundamentals of Atmospheric Radiation by Bohren and Clothiaux.

Chapter 12 of Radiative Transfer in the Atmosphere and Ocean by Thomas and Stamnes is titled The Role of Radiation in Climate, and it has a good section (12.2) on the greenhouse effect.

27. Mac Lorryon 01 Aug 2007 at 7:05 am

Doing the black (gray) body calculations for a disk or sphere always produces the wrong results when applied to Earth. Those calculations produce an average temperature below the observable value. The missing energy has been attributed to the greenhouse effect, but if it doesn’t exist then there must be some other mechanism at play.

From reading the paper I conclude the atmosphere is acting as a non-radiating heat sink. For example, during the day the ground’s temperature rises above that of the air, which heats the air by conduction and convection. Being transparent to infrared, the nitrogen and oxygen in the air can neither absorb nor emit infrared. That trapped energy cannot be lost to space by radiation, but is transferred back to the ground when the temperature of the ground is below that of air. By this means the observable temperature of the Earth may be explained in a way that corresponds to the laws of thermodynamics.

If the atmosphere is acting as a non-radiating heat sink, then it seems logical that introduction of so-called greenhouse gases would cause the heat sink to leak energy into space and cool the Earth rather than warm it. That would seem counter to the observed warming trend. One possible explanation is the Sun / cosmic ray / cloud mechanism now being studied at CERN.

28. N. Johnsonon 01 Aug 2007 at 8:03 am

“From reading the paper I conclude the atmosphere is acting as a non-radiating heat sink.”

If the particles in the atmosphere have a temperature above absolute zero, you’re theory is wrong. Everything with a temperature radiates.

29. Mac Lorryon 01 Aug 2007 at 8:58 am

“If the particles in the atmosphere have a temperature above absolute zero, you’re theory is wrong. Everything with a temperature radiates.”

But not everything has the emissivity of a black body.

Kirchhoff’s law of thermal radiation states that “at thermal equilibrium, the emissivity of a body (or surface) equals its absorptivity.” Thus, any material that has a very low absorptivity (is transparent) at a given wavelength also has a very low emissivity a that wavelength.

The absorptivity and emissivity of nitrogen and oxygen at infrared and visible wavelengths is a small fraction of 1 (black body). Thus, these gasses acquire heat from the ground during the day by conduction and convection at a far higher rate then they can radiate that heat away at night. In practical terms then, the atmosphere acts as a non-radiating (low emissivity at the important wavelengths) heat sink.

That’s why your black body calculations never produce the observable average temperature of the Earth. The Earth is not a simple black body, but has a means of storing heat in a low-emissivity sink.

30. N. Johnsonon 01 Aug 2007 at 9:46 am

O2 and N2 may be ‘transparent’ to infrared radiation, but H2O, CO2, N2O, and other are not.

Greenhouse theory correctly predicts the thermal structure of the atmosphere. Period.

31. saurabhon 08 Aug 2007 at 10:23 am

Forgive me - thermodynamics is many years behind me, and this is far from my domain. But what, if not the atmosphere, makes the surface warmer than it should be if it were behaving as a perfect blackbody, or a graybody? Why is the temperature on Venus 700 K?

I like Mac Lorry’s comment - “these gases acquire heat from the ground during the day by conduction and convection at a far higher rate then they can radiate that heat away at night. In practical terms then, the atmosphere acts as a non-radiating (low emissivity at the important wavelengths) heat sink.” This seems consistent, to my idiot comprehension, with thermodynamic constraints.

32. N. Johnsonon 08 Aug 2007 at 11:47 am

It is the atmosphere that warms the surface. That’s the point. See Bad Greenhouse and the first two steps in The CO2 problem in 6 easy steps.

33. Lazaron 08 Aug 2007 at 4:11 pm

Gerlich (section 3.9.1):

Second law of thermodynamics: Heat cannot move itself from a cooler body into
a warmer one. A heat transfer from a cooler body into a warmer one cannot happen
without compensation.

Utter bollocks.
Heat is the net transfer of energy between two macroscopic systems due to a difference in temperatures. Heat is not a substance (heat is not caloric), heat is not moved or transferred… heat is the net transfer of energy.
Serway & Beichner, “Physics for Scientists and Engineers” fifth edition: “Heat is defined as the transfer of energy across the boundary of a system due to a temperature difference between the system and its surroundings.”
The second law correctly stated wrt the transfer of heat from a colder system to a warmer one in the form of the Clausius statement: “It is impossible to construct a cyclical machine whose sole effect is the continuous transfer of energy from one object to another object at a higher temperature without the input of energy by work.”
Heat refers to the net flow of energy between two systems and the second law refers to macroscopic systems. So we are talking about the net flow of energy between macroscopic systems. Consider two molecules at different temperatures seperated by some distance. Both radiate energy. It is physically possible for a quanta of energy to be radiated from the colder molecule and absorbed by the warmer molecule, but that the reverse does not occur. Since the second law relies on the statistical properties of macroscopic systems this does not violate the second law. Consider two macroscopic systems, adjacent and at different temperatures. It is physically possible (and highly probable) that energy radiated by the colder system will be absorbed by the warmer system. But it is extremely unlikely that this energy will exceed the energy flowing in the other direction. In other words, the net energy flow will be from the warmer system to the colder one. This is what is meant by the seccond law of thermodynamics.

It is a fact that greenhouse gasses absorb IR emitted from the surface.
It is a fact that these gasses radiate energy in all directions, not caring which direction is ‘down’ or what the temperature of ‘down’ is.
It is a fact that these gasses radiate energy toward the surface.
It is a fact that some of this energy will be absorbed by the surface.
It is a fact though, that energy transferred to the atmosphere from the surface exceeds the energy transferred from the atmosphere to the surface, this is how the second law applies.
Without an atmosphere, the energy from the surface would be radiated directly to space.
That the atmosphere radiates some energy to the surface results in warming relative to the above scenario.
Do not be confused by Gerlich’s conflation of net energy flow with energy flow in one direction.

34. Lazaron 09 Aug 2007 at 10:50 am

Stefan’s law of the net loss of energy by radiation for a system at temperature T to surroundings at T0:
P = o*A*e*(T^4 - T0^4).
Energy flows in both directions.
The system looses a lot of energy per unit time to surroundings, and gains a lesser amound of energy per unit time from its surroundings.
The net difference between energy gained and lost determines the rate of cooling.
The larger the T0, the slower the system cools.
Because the larger the T0, the more energy the surroundings radiate, and therefore more energy the hotter system absorbs from its surroundings.
Therefore the net energy loss of the system is less.
So it cools slower.
When T=T0, the energy gained per unit time is equivalent to the energy lost per unit time, so the temperatures remain equal at equilibrium.
“When an object is hotter than its surroundings, it radiates more energy than it absorbs, and its temperature decreases.” [Serway & Beichner]
Which is why objects cool faster radiatively the lower the temperature of their surroundings.
The perpetuum mobile violation of the second law the professor cites should in fact refer to the net energy flow, not energy flow in one direction.
The trick the professor does is to confuse energy with heat.
Which might confuse those illiterate in thermodynamics.
And satisfy many denialists.

35. Lazaron 09 Aug 2007 at 12:16 pm

I’m sorry I can’t resist.
Gerlich and Tscheuschner quote;

The carbon dioxide in the atmosphere lets the radiation of the Sun, whose maximum lies in the visible light, go through completely, while on the other hand it absorbs a part of the heat radiation emitted by the Earth into space because of its larger wavelength. This leads to higher near-surface air temperatures.”

and rejoind,

The second statement is falsified by referring to a counterexample known to every housewife: The water pot on the stove. Without water filled in, the bottom of the pot will soon become glowing red. Water is an excellent absorber of infrared radiation. However, with water filled in, the bottom of the pot will be substantially colder.

A fine theory with a few slight problems.
The specific heat of co2 gas at 1 atm and 25 C is ~ 0.85 kJ / kg.K
… whereas the specific heat of liquid water is more than four times greater, 4.2 kJ / kg.K
The thermal conductivity of co2 is 0.09 W / m.K
… of water is more than six times greater, 0.58 W / m.K
The latent heat of vaporization for water is 2.26E6 J / Kg
… for atmospheric co2 is irrelevent.
The temperature differential between the earth’s surface and atmospheric co2 is at most 100 C.
… between a glowing red hot pot and room temperature water is about 500 C.
The composition of the atmosphere is approximately
0.036% co2.
… the ‘atmosphere’ of the pot is about 100% water.

In other words, liquid water in a pot is excellent at conducting and convecting thermal energy away from the material of the pot, has a huge capacity to store such energy before it’s temperature even approaches boiling point, at which point takes even more energy away in the form of latent heat of vaporization.
0.036% atmospheric co2 does not share those qualities, unforutnately, the non-negligible effect is radiative.

[Response: Excellent point.]

36. Al Tekhassion 09 Aug 2007 at 5:38 pm

N.Johnson asks: “If their correct way of averaging is really correct, why does it lead to an obviously wrong answer?”

They are using the same assumption that there is no atmosphere, just as your “average balance” does. If you need a 1370W/m2 insolation and a planet without atmosphere, the Moon would be a good candidate to compare notes. What is the “yearly-averaged global temperature” of Moon’s surface?

[Response: A page at UCAR says, "On the sunlit side, the average temperature is around 100 deg C, while on the dark side the average temperature is a very cold -173 deg C." You also need to take into account that the Earth and the Moon have different albedos. About 0.3 for the Earth and about 0.12 for the Moon.]

37. Al Tekhasskion 10 Aug 2007 at 7:05 am

This is not the question I asked, and this is not the answer that the bold Stefan-Boltzmann approach gives you. Sakhara desert also has pretty high daytime temperatures, and surface in Antarctica is also cold. You also need to realize that 0.7^^0.25 is only 5% different from 0.88^^0.25. Even then, it means that Moon must be warmer than Earth with no atmosphere. [Ed: Exactly!] Now, the question again: which way of radiative balance calculation gives you more realistic number for Moon: the Gerlich&Tscheuschner, or the “classic climatological” ?
Cheers,
- Al Tekhasski

[Response: Neither one will give you the "right" answer because the Moon rotates so slowly that the radiation absorbed on the light side does not get effectively radiated throughout the whole surface area of the sphere. But just blindly pluging in values gives:

Gerlich&Tscheuschner: -123C
"classic climatological": -4C

The G&T equation gives a fairly close value to the temperature of the dark side of the moon, but not for the average temperature. If we assume a global average of the temperature of the light side plus the temperature of the dark side divided by two, that gives us a global temperature of the moon of about -36.5C. Clearly, the "classic climatological" value is much closer than G&T.]

38. Al Tekhasskion 12 Aug 2007 at 7:49 pm

Comment deleted. My patience has run out. I see no need to attempt to continue this conversation. A better use of my time would be to argue with someone who questions the law of gravity.

39. Mac Lorryon 16 Aug 2007 at 6:06 pm

Apart from the validity of the greenhouse effect itself, Gerlich and Tscheuschner’s criticism is valid that there’s no well supported, formal and correct explanation of the greenhouse effect in the scientific literature. If such does exist, it’s apparent from the variety of explanations quoted by Gerlich and Tscheuschner that it’s not well known. Maybe somebody should fix that.

[Response: It is well known. It is in the literature. The "greenhouse effect" is simply a consequence of the physics of radiative transfer. I suggest looking up Chandrasekhar.]

40. Mac Lorryon 17 Aug 2007 at 3:57 am

In section 3.5.5 Gerlich and Tscheuschner address “Two approaches of Radiative Transfer” including those of Chandrasekhar. The most relevant part of their conclusion to this section seems to be “It cannot be overemphasized, that differential equations only allow the calculation of changes on the basis of known parameters.”

That seems to be the crux of the matter. There are many confounding factors when modeling the Earth’s Climate. If you have the patience I would like to present some that confound me and see if you or other commentaries can clear them up.

[Response: In that section, G&T attempt to say that the source function can either be scattering or absorption. That is simply not true. It can include both. The fact that in some cases one of the terms is negligible and can be ignored is no different than simplifying other problems in physics.

There are many difficult problems to understand about the climate system. Of the ones that are well known, some can be explained easily to non-experts, and others cannot. I thought the greenhouse effect would be one of the easy ones; it appears it isn't. But I would suggest you do this experiment that should show that the addition of greenhouse gases, in particular CO2, will raise the temperature of the atmosphere.]

41. Mac Lorryon 17 Aug 2007 at 8:27 am

Well I have seen the Coke bottle experiment before and I believe the C02 would get hotter than the N2/O2, at least initially. The question is, what’s the temperature of the Coke as that represents the surface of the Earth. Woods’ 1909 experiment suggests the Coke would get warmer in the bottle with the N2/O2 atmosphere. Also, as the Coke warms up the amount of water vapor in the atmosphere of each bottle will increase. Being water vapor is a better absorber of infrared than C02 it’s important to know how much water vapor can be held by the C02 atmosphere compared to the N2/O2 atmosphere. If CO2 holds more water vapor, than any measured effect is likely the result of the water vapor rather than the CO2. If N2/O2 holds more water vapor then there’s no easy explanation as to why the CO2 atmosphere would be warmer. It’s these kinds of confounding factors that make the atmospheric greenhouse effect difficult to prove experimentally, but it also makes for an interesting discussion, well at least at my end.

I do agree with Lazar’s comment on August 8, 2007 4:11 pm. The second law is not violated if there is no net transfer of heat (energy) between a cold region and a warm region. That point does undermine Gerlich and Tscheuschner’s paper considerably. However, I still have some confounding factors I would like to post and see what other’s make of them.

[Response:"If CO2 holds more water vapor... If N2/O2 holds more water vapor"
Neither a CO2 nor N2/O2 atmosphere hold any water vapor! [1, 2, 3] The saturation vapor pressure is almost totally dependent upon the temperature. [4, 5, 6] In effect, the atmosphere in the CO2 bottle will contain more H2O than the N2/O2 bottle because it has a larger temperature.]

42. Mac Lorryon 17 Aug 2007 at 9:37 am

I have numbered my paragraphs to make discussion easier. I first want to establish my basic understanding of the problem before I get into the factors that confound me.

1. The heat radiating ability of CO2 is proportional to it’s amount (mass), rather than it’s concentration in a non-radiating gas (N2/O2). Obviously, there is a diminishing effect as more CO2 is added, but the amount in Earth’s atmosphere has not reached that level. If that’s not the case, then we don’t need to be concerned about adding more.

2. I agree with above comments that the second law is not violated when there is no net flow of heat (energy) from a colder material to a warmer material. For there to be no net flow, however, the temperature difference between the two materials cannot increase. It’s been calculated that the greenhouse effect contributes about 33 deg. C. to the temperature of the Earth’s surface, thus, the greenhouse gas would also have to rise in temperature by a minimum of 33 deg. C (on average of course).

3. Being the re-radiating material is a gas thinly dispersed in a non-radiating gas (N2/O2) only the boundary nearest the Earth’s surface would experience the full rise in temperature of 33 deg. C. A temperature gradient would exist from the boundary nearest Earth’s surface to the boundary furthest from Earth’s surface. The effect would be a temperature inversion. That is, the atmosphere is warmer near the Earth’s surface and gets progressively colder with altitude.

4. Only the troposphere and the mesosphere exhibit such a temperature inversion. The mesosphere is located from about 50 km to 90 km altitude above Earth’s surface and is said to be heated by gravity waves and the burning up of millions of meteors daily. Gases in the mesosphere are rarefied with an average pressure of about 0.05 kPa.

5. The CO2 in the mesosphere can have only a minimal effect on the temperature of the surface for two reasons. First, the amount (not concentration) of CO2 is very small, only about 1/2000 as much as is in the troposphere. Second, most infrared emitted from the surface has been absorbed by greenhouse gases in the troposphere. That’s because of their far greater amount and their proximity to the radiating surface.

6. The greenhouse warming of the surface is due to the greenhouse gases in the troposphere with the warmest re-radiating boundary being the interface with the surface. Not that it has any effect, but every time you exhale, the CO2 in your breath begins to absorb and re-radiate infrared, as does the water vapor.

Given the misconception with water vapor in my prior post this is a good place to stop and get feedback.

[Response: I don't have time to fully respond now, but I wanted to publish this so you didn't think it got lost in a spam filter. Hopefully by tomorrow I'll get something up. Thank you for your interest in getting the science right!]

43. jim Kreitzbergon 19 Aug 2007 at 11:29 pm

Thank you Lazar.
You did a lot better job falsifiying the greehouse theory than G&T. Much is easier to understand when you realize energy is one thing and heat another.

44. Lazaron 20 Aug 2007 at 3:58 pm

Jim,

You may care to expand on where and how I ‘falsify the greenhouse effect’.

Heat is a numerical value for the net flow of energy.
A change in internal energy (Joules) = heat(ing) (Joules) - work done (Joules)
Joules are units of energy.
And the above is the first law.
The idea that the absorption by an object of energy at a fixed wavelength depends, not just upon the wavelength, but on the temperature of the object from which it was emitted… is obviously incorrect. Consider a blackbody at 2000 K in proximity to a blackbody at 1900 K. The one at 2000 K would need to ‘know’ that the radiation hitting it came from a cooler blackbody, and then selectively choose to not absorb it, that is, to cease behaving as a blackbody.
Now G&T cloud the issue by moving the discussion away from energy to heat, and then claiming that energy is not heat.

45. Lazaron 20 Aug 2007 at 6:09 pm

“It’s been calculated that the greenhouse effect contributes about 33 deg. C. to the temperature of the Earth’s surface, thus, the greenhouse gas would also have to rise in temperature by a minimum of 33 deg. C (on average of course).”

… I suppose the temperature rise at any given altitude relative to ‘no greenhouse gas’ would no more and no less than 33 C due to the lapse rate?

[Response: See Stoat: Why does the stratosphere cool under GW.]

46. jimon 20 Aug 2007 at 7:54 pm

Hi Lazar,
Ok let’s see, yes i agree the temp of the source does not matter but that IR will not increase surface temp. The energy at the surface is vibrating at a faster rate than the surroundings that’s why IR is radiating away from the surface. THe IR coming back to the surface will not create more heating. It will increase the distance or slow the rate of cooling as you say. So co2 doe not trap or block IR it vibrates faster in response to the IR.
I thought this is what you were saying.
I’m a lay person so it’s easy for me to misinterpret what is being said.
Thanks, jim

47. Lazaron 20 Aug 2007 at 8:07 pm

“[Response: See Stoat: Why does the stratosphere cool under GW.]”

Thanks.

I’m a bit confused regarding increasing ghg’s… the effective radiating level (height or pressure) remains constant and the lapse rate increases, specific heat decreases?

Or does in fact the erl just move higher, the lapse rate remains constant?

In the former, I think the tropospheric temperature increases below the erl but decreases above, whereas in the latter, equilibrium temperatures at all levels of the troposphere increase.

[Response: With increasing GHGs, the effective radiating height will increase and the lapse rate will remain constant. Thus as you say, the temperature at all levels of the troposphere will increase (assuming a fully convective troposphere). However, if I recall correctly, the effective radiative height is at about the tropopause, so the temperature above the effective radiating height (the stratosphere and above) will decrease. As William Connolley said at the above link, not many people really understand and can explain this. Unfortunately, I am not one of them. I'll ask around tomorrow to see if I can further my understanding.]

48. Lazaron 21 Aug 2007 at 5:07 am

Jim

“Ok let’s see, yes i agree the temp of the source does not matter but that IR will not increase surface temp. The energy at the surface is vibrating at a faster rate than the surroundings that’s why IR is radiating away from the surface. THe IR coming back to the surface will not create more heating. It will increase the distance or slow the rate of cooling as you say. So co2 doe not trap or block IR it vibrates faster in response to the IR.”

Yes, spot on.

“I thought this is what you were saying.
I’m a lay person so it’s easy for me to misinterpret what is being said.”

I think Jim it was I who misinterpreted you. I’m a lay person too, so I just ask questions all the time! Thanks for the response.

49. Lazaron 21 Aug 2007 at 7:09 am

a third option… the lapse rate decreases but the erl moves upward to compensate, which… if the erl shifts enough… would increase temperatures across the board, and the increase would be greater at higher altitudes.

Ugh. I’ve now read all three versions at credible sources, two versions at RealClimate. I’ll have a look for some papers.

50. Lazaron 21 Aug 2007 at 7:09 am

Jim, yes, you are correct. I think it was I who misread you. I’m a lay person too, so I ask questions all the time! Thanks for the response.

51. Lazaron 21 Aug 2007 at 7:41 am

Ok, lapse rate is a feedback to the effect of ghg’s in increasing surface temperature, but the response appears unknown… it may increase, decrease or remain ~ constant, therefore amplify or counteract surface warming, it may increase or decrease the erl. There do not appear to be measurements of the change in lapse rate. What do the GCMs say?

[Response: I have no idea. I'm not a modeler. You should ask over at RealClimate.]

52. Lazaron 21 Aug 2007 at 5:19 pm

The GCMs say lapse rate feedback is negative at about -1 W/m^2/K, which is a decreasing lapse rate. So energy is transported aloft more effectively, which decreases the surface temperature whilst increasing the temperature of the erl… this is why models predict more warming in the troposphere. Yes, the erl moves upward. It appears then that Gavin’s description (increasing lapse rate, pivoting around fixed erl), as far as I understand it, is incorrect, whereas Spencer Weart, Raypierre Humbert, Tamino and yourself have it right.

53. joe90on 31 Aug 2007 at 6:16 am

Perhaps those who claim there is no such thing as a greenhouse effect at all, might try explaining why the Moon’s average temperature is around 30C colder than that of the Earth.

54. jaeon 07 Oct 2007 at 10:30 am

“Perhaps those who claim there is no such thing as a greenhouse effect at all, might try explaining why the Moon’s average temperature is around 30C colder than that of the Earth.”

That’s easy. The moon has no water vapor to store the heat for awhile. If you call that a greenhouse effect, fine, but I simply call it heat storage by a very efficient heat storing molecule. If you compare the average July temperatures in arid places with those at the same latitude and elevation in humid places, the average temperatures are HIGHER in the arid places. Where is the greenhouse effect here? See here for more information: http://www.esnips.com/web/climate

55. Fred Stapleson 10 Oct 2007 at 5:46 am

I have been reading the G and T paper, and some of their most important points seem to have been lost in these posts.

They accept the conventional “bare rock” temperature of 255 degrees K, but they make the obvious point that if this rock is then wrapped in a transparent (to incoming solar radiation) atmosphere with low thermal conductivity, its surface temperature must increase.

They then argue that all heat transfer mechanisms to the atmosphere – conduction, convection, vaporisation, and radiation - will be from hot to cold, surface to atmosphere, never in the other direction.

They then seek to demolish the conventional “back-welling” radiation theories with a variety of arguments.

First, there is no sign of back radiation from the glass in a real greenhouse (since, in our globally warmed UK climate, I live in a glass room for most of the year, I agree).

Next, there is no sign of such an effect from thermal conductivity measurements on CO2.

Their “pot on the stove” argument does not depend on the efficiency of the heat transfer mechanism (Lazar, August 9th): the point they are making is the direction of the heat transfer. The water cools the surface of the pan – it does not heat it.

Their principal argument rests on the second law of thermodynamics. I wonder if they are themselves confusing heat with energy when they rule out any possible “back-warming” effect. The atmosphere cannot warm the surface directly, but back-radiation from CO2 energy absorption might slow down the heat loss from the surface (but not if the warmer air rises and cools, as one of their references argues).

They mention in passing that CO2 does not figure in the calculation of the temperature lapse rate, which explains the 33 degrees K temperature drop from the surface to the tropopause from adiabatic gas laws without reference to radiation or greenhouse gasses.

In my day, the fifties, that was the end of the argument. Modern textbooks start from this calculated lapse rate and propose an AGW effect by a perturbation “higher is colder” argument. Sadly G and T do not address this theory – if they read these posts perhaps they will.

56. Magnus Won 27 Oct 2007 at 4:37 am

http://www.sourcewatch.org/index.php?title=Gerhard_Gerlich

57. Dipl.-Physiker Jochen Ebelon 17 Nov 2007 at 1:31 pm

The paper by Albert Einstein (1916) to the interaction between gas and radiation is missing in the references.

Therefore following questions:

How do the authors explain
- the existence of the tropopause
- the decrease of temperature with altitude (almost adiabatic)
- the nearly isotherm temperature in the zone above the tropopause

without of the greenhouse effect?

To the translation in german and the comments to the paper:
http://www.ing-buero-ebel.de/Treib/Auszug.htm

Best regards
Dipl.-Physiker Jochen Ebel

58. froston 19 Nov 2007 at 9:33 am

@ Marcus W, Oct 27:

While an argument based on the Guilt By Association fallacy may be effective in political arenas, this is a science blog. If the president of Exxon/Mobile himself were to post here you would still be expected to address his points objectively, without reference to his place of employment.

[Response: You are absolutely correct.]

59. Bill McClenneyon 25 Nov 2007 at 8:22 am

As a geologist, I have been interested in climate change for a variety of reasons, particularly sedimentation. Over the years since the Vostok core data has been available, I have found the following quite useful. Forget the argument as to whether CO2 concentrations increase first or temperature. It actually doesn’t matter when you step back and take a hard look at the data. Notice in those 4 Pleistocene ice ages recorded (also one quarter of the Pleistocene ice ages BTW) that long slow slide into each 100k year long ice age. And then notice the abrupt terminatations of each ice age. Let us assume that GHGs are the cause of these terminations. According to Dan Ponti of the USGS this summer, his recent sediment coring in the LA Basin has produced data which he believes records 16 climate change events each with a commensurate 120 meter rise in sea levels. Or about 400 feet. This matches what was published in the LA Basin hydrogeology “Bible” in 1961 for the last four such events, later found in the Vostok data as well. So, 400 feet of sea level rise accompanies these abrupt terminations. Now all we need is a source of these GHG “eruptions” that can trigger a 400 foot change in sea level. It seems to me that the amount of GHGs needed after each 100k year long deep freeze would be ever so slightly more than that needed today to achieve either the 2-foot rise predicted by committees or the 20-foot rise predicted by Al Gore. Anyone have any idea what natural source of GHGs appears in nature every 100k years that does not also leave a layer of volcanic ash with it? And at the bottom of these 100k year long deep freezes, would there be (a) a sudden explosion in quantities and types of life forms capable of emitting vast quantities of GHGs or (b) less numbers and types of life forms capable of emitting quantities of GHGs to near instantaneoulsy cause sea levels to rise 400 feet in the most punctual series of events known from the geologic record? I don’t have the answer but the question does beg one. If GHGs are somehow being emitted in quantities that make today’s arguments pale by significance, where are they coming from, and why on a 100k year clock in the Pleistocene 16 times, and dozens of times in the Pliocene on a 41k year clock but with less extremes in terms of climate change? Until we can explain such vexing things from our past, I shall remain in doubt when reading such esoteric calculations as I see in today’s discussions that go back and forth attempting to address the present and the future while category ignoring the 800 pound gorilla in the debate. A most intriguing thing to include in your deliberations is that by many sources of data, we are presently sitting about 100 feet below the typical sea level high stand when these sixteen most recent terminations have occurred. How will we be able to distinguish an anthropogenic effect, that even if Gore is right, still leaves us 80 feet short of normal? Which of the 5 20 foot rises in sea level we still have to go in the present interglacial will Mr. Gore’s 20-footer be?

60. ScienceofDoomon 06 Apr 2010 at 4:19 pm

Much later than everyone else..

A new article - “On the Miseducation of the Uninformed by Gerlich and Tscheuschner (2009)”

http://scienceofdoom.com/2010/04/05/on-the-miseducation-of-the-uninformed-by-gerlich-and-scheuschner-2009

It follows the less comprehensive - “On Having a Laugh – by Gerlich and Tscheuschner (2009)”

http://scienceofdoom.com/2010/03/14/on-having-a-laugh-by-gerlich-and-tscheuschner-2009/

I also wrote to the editor of International Journal of Modern Physics B asking if they were interested in a paper on the many flaws and how it got published in the first place. No doubt there are many more people who have already asked this question..

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