The Moon is a Cold Mistress

Guest Post by Willis Eschenbach

I’ve been considering the effect that temperature swings have on the average temperature of a planet. It comes up regarding the question of why the moon is so much colder than you’d expect. The albedo (reflectivity) of the moon is less than that of the Earth. You can see the difference in albedo in Figure 1. There are lots of parts of the Earth that are white from clouds, snow, and ice. But the moon is mostly gray. As a result, the Earth’s albedo is about 0.30, while the Moon’s albedo is only about 0.11. So the moon should be absorbing more energy than the Earth. And as a result, the surface of the moon should be just below the freezing temperature of water. But it’s not, it’s much colder.

Figure 1. Lunar surface temperature observations from the Apollo 15 mission. Red and yellow-green short horizontal bars on the left show the theoretical (red) and actual (yellow-green) lunar average temperatures. The violet and blue horizontal bars on the right show the theoretical Stefan-Boltzmann temperature of the Earth with no atmosphere (violet), and an approximation of how much such an Earth’s temperature would be lowered by a ± 50°C swing caused by the rotation of the Earth (light blue). Sunset temperature fluctuations omitted for clarity. DATA SOURCE

Like the Earth, averaged over its whole surface the moon receives about 342 watts per square metre (W/m2) of solar energy. We’re the same average distance from the sun, after all. The Earth reflects 30% of that back into space (albedo of 0.30), leaving about 240 W/m2. The moon, with a lower albedo, reflects less and absorbs more energy, about 304 W/m2.

And since the moon is in thermal equilibrium, it must radiate the same amount it receives from the sun, ~ 304 W/m2.

There is something called the “Stefan Boltzmann equation” (which I’ll call the “S-B equation” or simply “S-B”) that relates temperature (in kelvins) to thermal radiation (in watts per square metre). It says that radiation is proportional to the fourth power of the temperature.

Given that the moon must be radiating about 304 W/m2 of energy to space to balance the incoming energy, the corresponding blackbody lunar temperature given by the S-B equation is about half a degree Celsius. It is shown in Figure 1 by the short horizontal red line. This shows that theoretically the moon should be just below freezing.

But the measured actual average temperature of the lunar surface shown in Figure 1 is minus 77°C, way below freezing, as shown by the short horizontal yellow-green line …

So what’s going on? Does this mean that the S-B equation is incorrect, or that it doesn’t apply to the moon?

The key to the puzzle is that the average temperature doesn’t matter. It only matters that the average radiation is 304 W/m2. That is the absolute requirement set by thermodynamics—the average radiation emitted by the moon must equal the radiation the moon receives from the sun, 304 W/m2.

But the radiation is proportional to the fourth power of temperature. This means when the temperature is high, there is a whole lot more radiation, but when it is low, the reduction in radiation is not as great. As a result, if there are temperature swings, they always make the surface radiate more energy. As a result of radiating more energy, the surface temperature cools. So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.

For confirmation, in Figure 1 above, if we first convert the moment-by-moment lunar surface temperatures to the corresponding amounts of radiation and then average them, the average is 313 W/m2. This is only trivially different from the 304 W/m2 we got from the first-principles calculation involving the incoming sunlight and the lunar albedo. And while this precise an agreement is somewhat coincidental (given that our data is from one single lunar location), it certainly explains the large difference between simplistic theory and actual observations.

So there is no contradiction at all between the lunar temperature and the S-B calculation. The average temperature is lowered by the swings, while the average radiation stays the same. The actual lunar temperature pattern is one of the many possible temperature variations that could give the same average radiation, 304 W/m2.

Now, here’s an oddity. The low average lunar temperature is a consequence of the size of the temperature swings. The bigger the temperature swings, the lower the average temperature. If the moon rotated faster, the swings would be smaller, and the average temperature would be warmer. If there were no swings in temperature at all and the lunar surface were somehow evenly warmed all over, the moon would be just barely below freezing. In fact, anything that reduces the variations in temperature would raise the average temperature of the moon.

One thing that could reduce the swings would be if the moon had an atmosphere, even if that atmosphere had no greenhouse gases (“GHGs”) and was perfectly transparent to infrared. In general, one effect of even a perfectly transparent atmosphere is that it transports energy from where it is warm to where it is cold. Of course, this reduces the temperature swings and differences. And that in turn would slightly warm the moon.

A second way that even a perfectly transparent GHG-free atmosphere would warm the moon is that the atmosphere adds thermal mass to the system. Because the atmosphere needs to be heated and cooled as well as the surface, this will also reduce the temperature swings, and again will slightly warm the surface in consequence. It’s not a lot of thermal mass, however, and only the lowest part has a significant diurnal temperature fluctuation. Finally, the specific heat of the atmosphere is only about a quarter that of the water. As a result of this combination of factors, this is a fairly minor effect.

Now, I want to stop here and make a very important point. These last two phenomena mean that the moon with a perfectly transparent GHG-free atmosphere would be warmer than the moon without such an atmosphere. But a transparent atmosphere could never raise the moon’s temperature above the S-B blackbody temperature of half a degree Celsius.

The proof of this is trivially simple, and is done by contradiction. Suppose a perfectly transparent atmosphere could raise the average temperature of the moon above the blackbody temperature, which is the temperature at which it emits 304 W/m2.

But the lunar surface is the only thing that can emit energy in the system, because the atmosphere is transparent and has no GHGs. So if the surface were warmer than the S-B theoretical temperature, the surface would be emitting more than 304 W/m2 to space, while only absorbing 304 W/m2, and that would make it into a perpetual motion machine. Q.E.D.

So while a perfectly transparent atmosphere with no GHGs can reduce the amount of cooling that results from temperature swings, it cannot do more than reduce the cooling. There is a physical limit to how much it can warm the planet. At a maximum, if all the temperature swings were perfectly evened out, we can only get back to S-B temperature, not above it. This means that for example, a transparent atmosphere could not be responsible for the Earth’s current temperature, because the Earth’s temperature is well above the S-B theoretical temperature of ~ -18°C.

Having gotten that far, I wanted to consider what the temperature swings of the Earth might be like without an atmosphere. Basic calculations show that with the current albedo, the Earth with no atmosphere would be at a blackbody temperature of 240 W/m2 ≈ -18°C. But how much would the rotation cool the planet?

Unfortunately, the moon rotates so slowly that it is not a good analogue to the Earth. There is one bit of lunar information we can use, however. This is how fast the moon cools after dark. In that case the moon and the Earth without atmosphere would be roughly equivalent, both simply radiating to outer space. At lunar sunset, the moon’s surface temperature shown in Figure 1 is about -60°C. Over the next 30 hours, it drops steadily at a rate of about 4°C per hour. At that point the temperature is about -180°C. From there it only cools slightly for the next two weeks, because the radiation is so low. For example, at its coolest the lunar surface is at about -191°C, and at that point it is radiating a whopping two and a half watts per square metre … and as a result the radiative cooling is very, very slow.

So … for a back of the envelope calculation, we might estimate that the Earth would cool at about the lunar rate of 4°C per hour for 12 hours. During that time, it would drop by about 50°C (90°F). During the day, it might warm about the same above the average. So, we might figure that the temperature swings on the Earth without an atmosphere might be on the order of ± 50°C. (As we would expect, actual temperature swings on Earth are much smaller, with a maximum of about ± 20-25 °C, usually in the desert regions.)

How much would this ±50° swing with no atmosphere cool the planet?

Thanks to a bit of nice math from Dr. Robert Brown (here), we know that if dT is the size of the swing in temperature above and below the average, and T is the temperature of the center of the swing, the radiation varies by 1 + 6 * (dT/T)^2. With some more math (see the appendix), this would indicate that if the amount of solar energy hitting the planet is 240 W/m2 (≈ -18°C) and the swings were ± 50°C, the average temperature would be – 33°C. Some of the warming from that chilly temperature is from the atmosphere itself, and some is from the greenhouse effect.

This in turn indicates another curiosity. I’ve always assumed that the warming from the GHGs was due solely to the direct warming effects of the radiation. But a characteristic of the greenhouse radiation (downwelling longwave radiation, also called DLR) is that it is there both day and night, and from equator to poles. Oh, there are certainly differences in radiation from different locations and times. But overall, one of the big effects of the greenhouse radiation is that it greatly reduces the temperature swings because it provides extra energy in the times and places where the solar energy is not present or is greatly reduced.

This means that the greenhouse effect warms the earth in two ways—directly, and also indirectly by reducing the temperature swings. That’s news to me, and it reminds me that the best thing about studying the climate is that there is always more for me to learn.

Finally, as the planetary system warms, each additional degree of warming comes at a greater and greater cost in terms of the energy needed to warm the planet that one degree.

Part of this effect is because the cooling radiation is rising as the fourth power of the temperature. Part of the effect is because Murphy never sleeps, so that just like with your car engine, parasitic losses (losses of sensible and latent heat from the surface) go up faster than the increase in driving energy. And lastly, there are a number of homeostatic mechanisms in the natural climate system that work together to keep the earth from overheating.

These thermostatic mechanisms include, among others,

• the daily timing and number of tropical thunderstorms.

• the fact that clouds warm the Earth in the winter and cool it in the summer.

• the El Niño/La Niña ocean energy release mechanism.

These work together with other such mechanisms to maintain the whole system stable to within about half a degree per century. This is a variation in temperature of less than 0.2%. Note that doesn’t mean less than two percent. The global average temperature has changed less than two tenths of a percent in a century, an amazing stability for such an incredibly complex system ruled by something as ethereal as clouds and water vapor … I can only ascribe that temperature stability to the existence of such multiple, overlapping, redundant thermostatic mechanisms.

As a result, while the greenhouse effect has done the heavy lifting to get the planet up to its current temperature, at the present equilibrium condition the effect of variations in forcing is counterbalanced by changes in albedo and cloud composition and energy throughput, with very little resulting change in temperature.

Best to all, full moon tonight, crisp and crystalline, I’m going outside for some moon-viewing.

O beautiful full moon!

Circling the pond all night

even to the end

Matsuo Basho, 1644-1694

w.

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DeWitt Payne
Reply to  Willis Eschenbach
January 14, 2012 12:25 pm

Willis,

You presume incorrectly. Molecular motion is all that is necessary to produce the lapse rate, not bulk motion of the atmosphere. w/

I disagree. Molecular motion would effectively be diffusion. Diffusion alone would produce an isothermal atmosphere. But that’s moot because a non-isothermal irradiated sphere with an atmosphere will always have bulk motion.

George E. Smith;
January 13, 2012 4:33 pm

Can people get it through their heads, that the Stefan-Boltzmann “law” is the simple result of an ordinary integration over all wavelengths or frequencies, of the Planck formula for BLACK BODY RADIATION.
So inherent in the Stefan-Boltzmann relation, is the Planck radiation formula; and that formula relates ONLY to a completey fictitious and unrealizeable, non existing ideal theoretical object called a BLACK BODY, whose only required property is that it totally absorb any and ALL electromagnetic radiation that falls on it, arriving from any direction, from any source or multiplicity of sources of quite arbitrary properties and Temperatures.
It is not any mystical cavity or any other specific geometry; it simply absorbs any EM radiation from zero to infinity frequency or wavelength.
So the Stefan- Boltzmann law does not apply to ANY REAL OBJECT, including the moon.
It is a very good starting point for many radiation problems, since we can actually make some quite good practical approximations to a black body, and many natural objects actually are not too bad themselves at approximating a black body, at least over the range of wavelengths or frequencies, which are of importance to that Temperature range.
Remember that 98% of ALL of the thermal radiation energy emitted by ANY body, as a consequence of its Temperature (and not atomic or molecular structure) is emitted at wavelengths between half of the spectral peak wavelength and 8 times the spectral peak wavelength. That is of course for the spectrum plotted on a wavelength scale. If you like your BB graphs on a frequency or wave number scale, the numbers are something else. I don’t have any text books with solar spectra plotted on frequency scales, but I don’t mind if some like chemists seem to prefer that. And the spectral peak wavelength can be obtained from Wien’s displacement law.

DeWitt Payne
Reply to  George E. Smith;
January 14, 2012 12:22 pm

George E. Smith,

Can people get it through their heads, that the Stefan-Boltzmann “law” is the simple result of an ordinary integration over all wavelengths or frequencies, of the Planck formula for BLACK BODY RADIATION.

The Stefan-Boltzmann expression applies to any body that has constant emissivity over the wavelength range where it emits 99% of the total energy emitted, not just unit emissivity, i.e. gray bodies as well as black. That’s a reasonable first order approximation for a solid or liquid surface. That’s why the S-B equation is usually written:
j* = εσT^4

Brian H
January 13, 2012 8:45 pm

Tom in recently clold Florida. says:
January 10, 2012 at 6:10 am
Alexander Feht says:
January 9, 2012 at 11:43 am
“P.S. Couple of people here propose a weird argument about “dead man under a blanket.” Be it known to them that plants are protected from freezing by blankets, though, last time I checked, plants had no internal sources of heat (unless you burn them)”
However, plants and the surrounding ground retain heat from the day and preventing that from escaping is what will keep the plants under the blankets warm enough to prevent damage. .

False. At night, with no incident light, plants metabolize sugars and produce CO2, like animals. They warm themselves (to varying degrees) when they do so. Larger plants, like trees, maintain a constant internal temperature of about 21°C.

Tim Folkerts
January 13, 2012 9:00 pm

I haven’t been following all this carefully, but there seems to be a misconception about rising hot air. When hot air rises, it does not stay that temperature! As a “packet” of air rises, it will necessarily be gaining gravitational potential energy, so it must be losing some other energy. The way for air to lose energy is by expanding and cooling. This cooling is the ultimate source of the lapse rate. (see http://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate)
Convecting air WILL cool as it rises. Convection cannot raise the bulk temperature of the atmosphere to the surface temperature.

Bob Fernley-Jones
January 13, 2012 10:17 pm

Concerning the sarcastic allegations of tardiness in N&Z providing an improved presentation:
Considering that on the original WUWT Ira thread here ALONE, there are over 1,000 diverse comments, I’m hardly surprised that they run a tad late on their originally anticipated improved presentation. It may well be more qualitative and pre-emptive to people condemning stuff that they simultaneously admit that they do not understand; who knows? Meanwhile I wait in anticipation, whilst feeling that some of the derivations from their basic premise, (which in itself I think is valuable), might be a bit stretched.
Hopefully we will see some clarifications within the next few weeks, and let’s be patient.
BTW, I think this here debate is the best form of “peer review”

jae
January 14, 2012 8:05 pm

Willis:
“Willis Eschenbach says:
January 12, 2012 at 10:18 pm
jae says:
January 12, 2012 at 6:48 pm
Willis, et. al.: whataboutthat zero emissivity issue? Would you have a hot planet or a cold one, considering that hot gases rise and can’t sink again if they can’t emit?
DeWitt? Anyone?
Sorry, jae, but I’ve lost what the “zero emissivity issue” was. Could you recap the bidding?”
Up in the unending comments somewhere, DeWitt explained the importance of emissivity in determining how the SB equation works. You appear to be ignoring this part of the SB equations, evidently assuming that EVERYTHING has an emissivity of unity. Not true, as DeWitt said.
SO, if the atmosphere has no emissivity, it will get infinitely hot, as I understand the issue. This could be a problem with your model.

January 14, 2012 9:13 pm

George E. Smith,
You asked:
Can people get it through their heads, that the Stefan-Boltzmann “law” is the simple result of an ordinary integration over all wavelengths or frequencies, of the Planck formula for BLACK BODY RADIATION.
To derive the power law of Stefan and Boltzmann two integrations are necessary. The first is the integration of Planck’s blackbody radiation law over all frequencies. This is what you mentioned. The second is is the integration over the adjacent half space.
Best regards
Gerhard

DeWitt Payne
Reply to  Willis Eschenbach
January 15, 2012 9:02 am

Robert Clemenzi,

However, on a longer time scale, once convection from the surface is no longer possible (because the air is too hot), conduction will heat the colder atmosphere.

Convection will always be not merely possible but always happen because the poles will always be colder than the equator. That means that the pressure at a given altitude will always be higher near the equator than at high latitudes. A pressure difference will always cause air flow. Air flow horizontally forces air flow vertically. That air circulation forms a heat engine that creates an adiabatic lapse rate. Once an adiabatic lapse rate forms, convection is easier. Conduction is so slow in air that any convection at all will overwhelm it.

DeWitt Payne
Reply to  Willis Eschenbach
January 15, 2012 9:11 am

Higher up, more potential energy, lower temperature.

Higher up there is indeed more potential energy, but not necessarily lower temperature. A one dimensional atmosphere will be (eventually) isothermal. If it weren’t, it would violate the Second Law. Heat diffusion doesn’t care about gravity. A temperature gradient causes heat diffusion. But we don’t get an isothermal atmosphere in a planet because the thermal conductivity of air is so low that any convection at all will overwhelm conduction (except, obviously, at a surface boundary). And there will always be convection in a planetary atmosphere driven by the temperature difference between the poles and the equator which creates a pressure gradient force which causes air flow. If the surface were isothermal, then eventually, so would be the atmosphere.

Robert Clemenzi
January 15, 2012 12:06 am

Tim Folkerts says:
January 13, 2012 at 9:00 pm

Convecting air WILL cool as it rises. Convection cannot raise the bulk temperature of the atmosphere to the surface temperature.

I completely agree.
DeWitt Payne says:
January 14, 2012 at 12:04 pm

Robert,
Above this mixing layer, the lapse rate should be zero and the temperature close to the peak surface temperature..

Nope. There will be a temperature inversion in the surface boundary layer at high latitudes and the potential temperature will be determined by the peak surface temperature. … That means the lapse rate in most of the atmosphere will be the adiabatic rate (that’s what a constant potential temperature means).
In the short term, you are correct, the potential temperature at every point in the atmosphere above the inversion will be very close to the peak surface temperature. This profile is generated via convection. However, on a longer time scale, once convection from the surface is no longer possible (because the air is too hot), conduction will heat the colder atmosphere. Remember, the atmosphere will get 9.8 K/km colder with increasing altitude. As heat continues to flow from hot to cold (via conduction / diffusion), additional convection will occur, but not from the surface. Instead, an isothermal layer will begin to form above the surface boundary layer and additional convection will occur from the top of this isothermal layer until, eventually, the entire upper atmosphere is isothermal and the lapse rate is zero.
I agree that there should be some air flow to the poles for the reasons you give, but if the boundary layer is not thick enough, I doubt the a jet stream could form. My rough analysis indicates that the sea level boundary level would be less than 1 km, significantly less than the height of the Rockies (and other ranges) which should limit jet stream formation.

jae
January 15, 2012 7:28 am

Willis:
Sorry to be so vague. What I’m trying to figure out is what happens to the heat absorbed by your “transparent gases” through conduction. Normally the gases rise and cool, because they can give off heat through convection/radiation. However, how do your gases get rid of the heat. They rise due to convection, but cannot lose the heat, except by conduction to neighboring molecules, because their emissivity is so low. It seems to me that your atmosphere will be backwards to the one on Earth, where it is equal to surface temp. at the surface and much hotter up higher.
Maybe this is one reason why temp. starts increasing with altitude beyond the tropopause.
As I recall DeWitt mentioned that freshly galvanized steel and some other materials that can get very, very hot because of their low emissivity.
You have to include emissivity in the SB equation! And I don’t think it works very well for gases. But I am just a chemist.

Brian H
January 15, 2012 8:31 am

Willis;

Higher up, more potential energy, lower kinetic energy.
But for a gas, temperature is a measure of the kinetic energy of the molecules. So with gas, here’s the corresponding relationship
Higher up, more potential energy, lower temperature.

Yes; taking this one step further (too far?), one might say “In a gravity well, height is (the equivalent of) heat.”

DeWitt Payne
Reply to  Willis Eschenbach
January 16, 2012 7:36 am

Willis,

Not sure what you are calling a “one-dimensional” atmosphere, but a GHG-free transparent atmosphere at equilibrium will end up isentropic (equal energy everywhere), not isothermal (equal temperature everywhere). If it were isothermal, the total energy (kinetic + potential) would increase with altitude, which is a non-equilibrium situation.

What I mean by a one-dimensional atmosphere is that the atmosphere is isotropic for the two axes parallel to the surface and non-isotropic along the axis perpendicular to the surface.
As far as isentropic vs. isothermal. Consider a one-dimensional transparent atmosphere with a fixed constant surface temperature. Let’s make the atmosphere isothermal. I can think of no way for an isothermal one-dimensional transparent atmosphere to become non-isothermal. Without air movement, and there certainly wouldn’t be any driving force for convection, there is no way for the upper atmosphere to cool. You have to do work to move energy from high to low. If you take a packet of air at high altitude and move it to the surface, it will be warmer than the air around it, which means it takes force to make it happen. Force times distance is is energy. With no temperature difference, there is no free energy to do the work. You can only move energy without doing work if there’s a temperature gradient. If you start with an atmosphere where the temperature increases with altitude, there is a gradient and energy will flow downward toward the surface. But it will stop when the gradient becomes zero. The same must also be true for an atmosphere where the temperature decreases with altitude. There is no requirement that the atmosphere be isentropic. In fact, entropy tends to increase and an isothermal atmosphere has higher entropy than an isentropic atmosphere.
As far as your bullet analogy: It would be true if the mean free path of an atmospheric molecule were measured in kilometers. But then you’d also get stratification by molecular/atomic weight. In fact, that does happen at altitudes greater than about 100 km where the mean free path is on the order of 1 km.. But in the troposphere, a molecule bounces off another molecule every nanosecond or so. The molecules at high altitude already have high gravitational potential energy and it would take years, possibly a lot of years (I haven’t run the numbers) for a molecule at high altitude to diffuse to the surface. In the process, it would have equilibrated with the molecules in its vicinity along the way. That’s required by local thermal equilibrium.
Here’s another example: Take a long tube full of air at 1 atmosphere parallel to the surface and perfectly insulated. Allow the contents to equilibrate and become isothermal. Now rotate the tube until it’s perpendicular to the surface. The pressure will decrease at the top and increase at the bottom. That will cause a temperature gradient as well. But you’ve done work to make that happen. Even if you try to pivot the tube around its center of gravity there will still be a force involved because the center of gravity will shift as the tube rotates.
In the absence of turbulence, energy movement is driven by temperature gradients, not potential temperature gradients. There’s a lot of discussion of this at Science of Doom on the Venus threads as well as at Nick Stoke’s web site. When you add dimensions to the atmosphere and allow temperature to vary on the horizontal axes, you get circulation driven by the pressure gradient force and that circulation can do work. Then you get an isentropic vertical temperature profile.

Robert Clemenzi
January 15, 2012 11:56 am

Willis Eschenbach says:
January 15, 2012 at 10:13 am

If it were isothermal, the total energy (kinetic + potential) would increase with altitude, which is a non-equilibrium situation.

Yet the tropopause is isothermal, stable, and about 10km thick. Above that is the stable stratosphere, about 30km thick, where the temperature increases with increasing height. In general, in the current atmosphere, the total energy increases with increasing height. Even in the current troposphere, the total energy increases with increasing height.
I am not sure what you mean by non-equilibrium situation.

jae
January 15, 2012 6:18 pm

But, Willis, you say:
“Thanks, Jae. A transparent GHG-free atmosphere will gain energy until the lowest layer is at the temperature of the surface and the temperature profile above that is the dry adiabat, decreasing upwards at the lapse rate.”
Yes, but is this limited to the AVERAGE surface temperature. I don’t think so. This averaging habit gets us into all kinds of questions. At high noon on your planet with transparent gases, the surface is MUCH higher than average. Why do all the gases have to be average?

jae
January 15, 2012 6:21 pm

Also, Willis, I’m not sure the lapse rate paradigm applies to a “transparent atmosphere,” since it is “weird.” WHY does temperature increase beyond the troposphere?

January 15, 2012 9:03 pm

Dear Willis,
it might be that some confusion arose because you stated:
“The key to the puzzle is that the average temperature doesn’t matter. It only matters that the average radiation is 304 W/m2. That is the absolute requirement set by thermodynamics—the average radiation emitted by the moon must equal the radiation the moon receives from the sun, 304 W/m2.
But the radiation is proportional to the fourth power of temperature. This means when the temperature is high, there is a whole lot more radiation, but when it is low, the reduction in radiation is not as great. As a result, if there are temperature swings, they always make the surface radiate more energy. As a result of radiating more energy, the surface temperature cools. So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.”
What you called the average temperature of the Moon (close to 273 K) is the temperature of the planetary radiative equilibrium (see my Eq. (6)). The true average surface temperature of the Moon, however, is given by the Eq. (5). Thus, the actual surface temperature is swinging around the true average surface temperature. This is illustrated in your Figure 1. Consequently, the temperature of the planetary radiative equilibrium for the Earth in the absence of its atmosphere of T_e = 255 K is a senseless house number ,as already argued by Gerlich & Tscheuschner (2009) and Kramm & Dlugi (2011).
In front of these findings it is interesting to mention the paper of Kondratyev and Moskalenko (The role of carbon dioxide and other minor gaseous compounts and aerosols in the radiation budget. In: Houghton, J.T., ed., The Global Climate. Cambridge University Press, Cambridge/ New York, 1984, pp. 225-235). The authors, for instance, argued that their calculations for a standard model atmosphere showed that the total greenhouse effect amounts to 33.2 K, with the following contributions from optically active gaseous components: H2O – 20.6K; CO2 – 7.2 K; N2O – 1.4 K; CH4 – 0.8 K; O3 – 2.4K; NH3 + freons + NO2 + CCl4 + O2 + N2 – 0.8 K. If the true average surface temperature for the Earth in the absence of its atmosphere is around 200 K or somewhat lesser as we determined than the argument of Kondratyev and Moskalenko is rather worthless.
Nevertheless, one can find this argument even in the 3rd report of the Enquete Commission of the German Parliament “Vorsorge zum Schutz der Erdatmosphaere” (Precaution for protecting the Earth’s atmosphere) from May 24, 1990 (11/8030). Scientific members of this Enquete Commission were:
Wilfried Bach,
Paul Crutzen,
Rudolf Dolzer,
Hartmut Grassl,
Klaus Heinloth,
Peter Hennicke,
Klaus Michael Meyer-Abich,
Hans Michaelis,
Wolfgang Schikarski,
Wolfgang Seiler, and
Reinhard Zellner.
Some of these members had strong connections to the German nuclear power lobby. Nuclear power was considered in that report as one of the favorite possibilities to fight against global warming caused by so-called greenhouse gases. Thus, it is not surprising to me to recently find the following arguments on the website of the German Atomic Forum:
“Nuclear energy has an excellent carbon footprint.
Along with water and wind energy, nuclear power has far and away the lowest CO2 emissions of all energy sources. If one includes the entire life cycle of nuclear energy utilization (including the extraction and conversion of uranium, reprocessing of fuel assemblies, construction and demolition of power plants, waste management), then at 5 to 33 g CO2-equivalent per kilowatt hour produced, the greenhouse gas emissions are negligible. For the sake of comparison: electricity production based on fossil energy sources such as natural gas, crude oil, hard coal or lignite is responsible for 399 to 1,231 g CO2 per kWh.”
Best regards
Gerhard

Robert Clemenzi
January 16, 2012 12:16 pm

DeWitt Payne says:
January 16, 2012 at 7:36 am

When you add dimensions to the atmosphere and allow temperature to vary on the horizontal axes, you get circulation driven by the pressure gradient force and that circulation can do work. Then you get an isentropic vertical temperature profile.

But only in the boundary / mixing layer, not above it.
Understanding this reveals the primary flaw in the AGW theory – if it wasn’t for the greenhouse gases, the atmosphere would be much hotter than it is today. Therefore, since greenhouse gases cool the atmosphere, adding more will not make the atmosphere warmer.

DeWitt Payne
Reply to  Robert Clemenzi
January 16, 2012 2:39 pm

Robert Clemenzi

But only in the boundary / mixing layer, not above it.

Wrong.
The pressure gradient force increases with altitude. That causes air at high altitude, not at the surface, to flow from the equator toward the poles. This flow is then diverted by coriolis force until it becomes parallel to the direction of rotation. It’s called a geostrophic wind. But that flow reduces surface pressure at the equator and increases it at higher latitudes. That causes surface winds to blow towards the equator. But now you have a convergence zone near the equator with winds flowing from both hemispheres towards the equator. That will cause a vertical flow which replaces the air mass moving away from the equator. Pretty soon you probably get Hadley cells in both hemispheres. Actually it’s more complex than that, but basically you get air flow at most altitudes, not just the surface layer.

Robert Clemenzi
January 16, 2012 9:03 pm

DeWitt Payne says:
January 16, 2012 at 2:39 pm

Pretty soon you probably get Hadley cells in both hemispheres.

I agree. The questions are – How thick those cells will be? What determines the cell thickness?
Currently, there are 3 circulation cells in each hemisphere. At the surface, two flow toward the equator and the one between them (the Ferrel cell) flows toward the pole. How many would there be without greenhouse gases?
I assume that the number of cells is related to their thickness and that the thickness is related to the lapse rate. Unfortunately, I have no data or theory to support that. However, I think it should be an odd number – 3, 5, 7, etc.

Stephen Wilde
January 18, 2012 2:04 pm

All this talk about pressure patterns and lapse rates and circulations is all well and good but does anyone ever stop and think what it is all FOR, what it all DOES ?
The surface pressure distribution and the relative sizes and intensities of ALL the climate zones are as they are for a reason.
They shift over time beyond normal seasonal variation but who is thinking about WHY they do so ?
The truth is that they adjust the rate of energy flow from surface to space with the consequence noted by Nikolov and Zeller that the effects of GHGs in the air are reduced to ZERO.
GHGs don’t warm the Earth nor cool the Earth. Their ability to radiate upward negates the effect of their ability to radiate downward or if there is a slight discrepancy then changes in convection and the speed or size of the water cycle wipe it out.
The only climate effect resulting from any necessary adjustment is a miniscule shift in all the phenomena that you chaps are just beginning to get a grip on. But you are all blinded by detail. All that matters is the big picture as far as changes in climate are concerned.
The detail is just weather.
Pressure and solar input rule the entire system and anything that seeks to disturb the basic equilibrium is negated by climate zone shifting.

DeWitt Payne
Reply to  Stephen Wilde
January 18, 2012 2:33 pm

Stephen Wilde,

Their ability to radiate upward negates the effect of their ability to radiate downward or if there is a slight discrepancy then changes in convection and the speed or size of the water cycle wipe it out.

In a word, no. The upward radiation in the wavelength regions where greenhouse gases absorb strongly comes from higher in the atmosphere than the downward radiation. Since the maximum emission at any wavelength is constrained to be less than for a blackbody with emissivity equal to 1 and blackbody emission depends exponentially on temperature, that means emission upward will be less than emission downward as long as temperature decreases with altitude. That’s why the value of the lapse rate is important. An increase in total annual precipitation with temperature will act to reduce the increase in temperature, i.e. a negative feedback, but it can’t eliminate it. It’s also difficult to imagine an increase in annual precipitation without an increase in total water vapor in the atmosphere. That’s a positive feedback.

Stephen Wilde
January 18, 2012 11:34 pm

“An increase in total annual precipitation with temperature will act to reduce the increase in temperature, i.e. a negative feedback, but it can’t eliminate it. It’s also difficult to imagine an increase in annual precipitation without an increase in total water vapor in the atmosphere. That’s a positive feedback”
The atmospheric changes that result in a faster or larger water cycle also change the atmospheric heights so I think it can be eliminated.
An increase in annual precipitation need not involve an increase in total water vapour if condensaton increases in parallel with more evaporation i. e. a faster water cycle.

DeWitt Payne
Reply to  Stephen Wilde
January 19, 2012 8:01 am

Stephen Wilde,

The atmospheric changes that result in a faster or larger water cycle also change the atmospheric heights so I think it can be eliminated. An increase in annual precipitation need not involve an increase in total water vapour if condensaton increases in parallel with more evaporation i. e. a faster water cycle.

You think? And why should this convince me? Hand waving is insufficient. What you’re saying is that the evaporation rate will increase enough to restore the energy balance without a change in surface temperature. The rate of evaporation is a function of temperature and wind velocity. So the average wind velocity is going to magically increase without the temperature changing? I don’t think so. If that could happen, it could and would happen now. In fact, it probably does to some extent. Which could be one reason why there’s ‘weather noise’ in the global average temperature anomaly. But that’s noise around a mean value. Increasing greenhouse forcing will shift that mean.

January 19, 2012 9:38 am

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