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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Bart
January 11, 2012 3:08 pm

“There is no preference for heavier or lighter side to end up facing the Earth.”
Of course, when the pivot is at the CG, there is no heavier or lighter side. So, you need to be very careful about what you mean by “side”, and where the dividing line is between them.

January 11, 2012 3:08 pm

Thanks, the PDF is absolutely awesome. I know about half of the physics in it well enough to teach it myself (not in context), recognize the other half (a lot of it) from Things I Should Have Learned In Grad School (but never worked with and hence don’t really “know”), and am learning some of it (and a lot of terminology) for the first time. And then, there is putting it altogether in a new milieu. A lifesaver. Do you mind if I ask what your background is?
Anthony, I have no idea if you are still tracking this (I don’t see how you could be, given the number of hours in the day — hell, I can’t even keep up with WRITING on this blog much longer as teaching is about to overwhelm me and I’m stealing time from other stuff already) but if you are: Could you create a permanent link to this document somewhere toplevel on your site? This one free online book could improve the quality of discussion immeasurably — it should be “required reading” in a sense before people post physics-y stuff (much of which is nonsense).
I plan to go through it as I have time and see if I can’t improve the quality of my own contributions. I might buy some of the other actual textbooks eventually, but this one is plenty to get started.
rgb
[Moderator’s Comment: Your comment has been brought to Anthony’s attention. -REP]

DeWitt Payne
Reply to  Robert Brown
January 11, 2012 5:08 pm

Robert Brown,
I’m a retired industrial Analytical Chemist, BS Caltech 1965, Ph.D. University of Texas, Austin 1970. My graduate work was in electrochemistry but I spent most of my career in industry doing atomic spectroscopy (flame and graphite furnace atomic absorption/emission, plasma emission and mass spec. and wavelength dispersive x-ray emission spectrometery plus a lot of other minor stuff including for a time thermometer calibration against an NIST calibrated standard platinum resistance thermometer and a Wheatstone Bridge). I actually went to a lecture by Hansen on global warming at a Pittsburgh Conference some time in the 1980’s, possibly 1989 after he had testified before Congress in 1988. But I was extremely skeptical. After I retired I started going into the subject in more detail, which was quite difficult at the time (i.e. no Science of Doom). At this point, I’m what would be called a lukewarmer/fatalist. I think climate sensitivity is near the low end of the IPCC range and I don’t think we’ll (the planet as a whole) spend any serious money doing anything other than talking about it even if it weren’t.

January 11, 2012 3:35 pm

It wasn’t that bad an analogy, and the conclusion and reasoning process are still “almost” correct, but you are right. A symmetric dumbbell will line up along the radius from tide because of the tidal torque. An asymmetric dumbbell will have the heavier side preferentially point down — although there may be a barrier in between — because of the asymmetry in the nearside and farside tidal torque. I hadn’t thought about the barrier — that would indeed make near or far side heavier a coin flip. That makes my conclusion even stronger, of course — either the moon won a (biased) coin flip to end up heavy side towards the earth or it simply stabilized symmetrically and pulled a bit more mass towards the near side while still plastic due to tidal asymmetry.

gbaikie
January 11, 2012 3:47 pm

Willis Eschenbach says:
“Tim, I don’t understand Nikolov’s theory. And by your own admission, you don’t understand Nikolov’s theory. In fact, nobody I can find understands Nikolov’s theory. I asked Nikolov to explain it in clear simple terms. He disappeared. I asked if anyone can explain the core of it in a few clear sentences. No one has been able to do so.”
Over in page 3:
Ned Nikolov says:
January 11, 2012 at 9:08 am
To All:
Just want to let you know that Karl Zeller and I are working on our official reply to the blog comments. Due to unexpected work load last week, we could not finish it as planned. The article is now coming along pretty well, and we’ll be able to share it with you soon.
Thank you for your patience!
-Ned
Link here [don’t know if works- or near [95%] bottom of comments:
http://wattsupwiththat.com/2011/12/29/unified-climate-theory-may-confuse-cause-and-effect/#more-53875
My comments. I spend way too much time here:) – the traffic is so high, there isn’t chance I can read half of them.

Joules Verne
January 11, 2012 3:53 pm

Quickly, various things that don’t look right:
“So if the atmosphere must have an adiabatic lapse rate, or indeed any fixed lapse rate, then the surface temperature determines both pressure and density of the atmosphere at any altitude.”
Doesn’t agree with observations. My ears should pop when a cold front blows through. 10C change in temp is like a change in altitude of 1000 meters (dry adiabatic lapse rate ~ 1C / 100 meters). That would actually be painful if it happened quickly.
@rgb
The manner sea ice forms is irrelevant. It floats. It excludes salt. Everyone knows that. Might have some effect on currents with salt variation near surface but doesn’t explain how ocean temp below thermocline is 3.0C which is the essential question. Internal heat of planetary formation is in neighborhood of 100mW/m2. It could certainly raise the temperature but it couldn’t lower it. My point is that for the ocean at depth to be 3.0C the surface must average close to that low. Agreed that low is lower than any published numbers I’ve seen but nobody had any Argo buoys measuring global ocean temp back then either so there is plenty of room for error. I really saw nothing you offered that explains how the ocean temperature is what it is only that 3.9C is lower than what is generally believed to be average global temperture during an ice age. Generally 3-6C lower than today is given for ice ages but it’s an educated guess and it’s hard to fathom how the ocean could be so much colder than the average temperature during an ice age. You said nothing to explain the discrepancy only pointed out that it exists.

DeWitt Payne
Reply to  Joules Verne
January 11, 2012 5:19 pm

Joules Verne,

Doesn’t agree with observations. My ears should pop when a cold front blows through. 10C change in temp is like a change in altitude of 1000 meters (dry adiabatic lapse rate ~ 1C / 100 meters). That would actually be painful if it happened quickly.

The pressure at the surface depends on the mass of the atmosphere above the surface and the local value of the acceleration of gravity. Period. A change in temperature won’t change the mass. It’s nothing like a sudden increase in altitude. If you go up 1 km, you’ve put about 10% of the atmosphere below you and the pressure drops from about 1,000 mbar to about 900 mbar. If you decrease the temperature by 10 K from, say, 300 K to 290 K, you’ve changed the air density by about 3%, but the pressure not at all.

gbaikie
January 11, 2012 4:15 pm

“J. Radefahrt (Ger) says:
January 11, 2012 at 3:29 am
Dear Mr. Eschenbach,
At the risk of making myself rediculous, I want to ask you something.
After a closer view to this discussion I came to the following consideration:
The main difference between Moon and Earth is that the the Earth is covered with water (7/10). So I ask myself, how is the behavior of water regarding radiation.
If my understanding of physics is right, water is able to transmit absorbed energy upwards (mostly radiation if a transparent atmosphere is assumed) as well as downwards (by convection, other thermodynamical processes and radiation) by nearly the same amounts. If so, you need the energy twice to get a radiative equilibrium at TOA, because only half of the stored energy can be radiated upwards. Calculating the required S-B-temperature of water considereing an albedo of 0.1 for surface only (no clouds!) we get approx. 320K for a fully transparent atmosphere. Assuming that Ramanathan (1997) was right and we have a net cooling of approx. 50W/m² by clouds, we get about 305K.
Summarize this with the equilibrium for solids of 255K in their parts (about 7/10 of water and 3/10 of land) we get an equilibrium that is very close to the current temperatures, approx. 286K.
Now my question: is this too easy thinking or is it really that simple?”
I think it’s generally that simple.
I don’t see evaporation being counted.
But ocean surface is cool- therefore doesn’t radiate as much energy.
Land is both hotter and cooler than oceans.
Ocean is storing storing energy and land is the spouse spending the excess
energy.
If there was more land compared to water- the planet would be cooler- and wider
swings in temperature.
Or ocean warms land, land doesn’t warm ocean.
So that is big element of missing from -33 C.
Another element is the atmosphere.
And land evaporates water also.
From there we have details.
There many details, one detail may include greenhouse gases in terms radiative energy.

Joules Verne
January 11, 2012 4:52 pm

Phil. says:
January 11, 2012 at 11:40 am
“As far as I’m aware it’s always been explained by the high density, cold salty water produced in the North Atlantic and Southern Ocean descending to the ocean floor and circulating, it takes about 1500 years to return to the surface if I recall.”
Certainly. But we aren’t talking about the ocean floor. We are talking about 90% of the ocean being at 3.0C. Little of it either colder or warmer. Given there’s nothing special about density at that temperature the only explanation for why it’s 3.0C at 300 meters deep at both the equator and under the north pole is that’s the average surface temperature. There is no refrigeration coming from below so it must be coming from above. So how cold must the average surface temperature be to drive almost the whole enchilada down to 3.0C?
I think what happens is that during glaciations temperature plunges over ice to far below 0C and this cold dense air flows out over the warmer ocean sucking the heat out of it so that when you average that surface air temperature it is indeed in the neighborhood of 4C. I think the guesstimates of glacial era surface temperature only 3-6C below today is wrong by long shot and that it gets bitter cold year-round over a sizeable fraction of the earth’s surface. Not only won’t the children know what above-freezing temperatures are in Alaska they won’t know what they are in Pennsylvania either. The average temperature in Pennslyvania is about 15C or so. During a glacial epic that would plunge into negative numbers. There wouldn’t be a summer at all because there wouldn’t anything but snow and ice turning all the warmth from sun straight away year round.
This is how you get an average surface temperature of 3.9C. The air over the ocean doesn’t need to drop more than the commonly given number of 3-6C for the average surface air temperature of the earth drop by a lot more simply because over land it can plummet by tens of degrees no problem over pretty much 30% of the earth’s surface plus however much more is exposed when sea level goes down by 100 meters and how much more is covered by sea ice than today. Ten thousand years of bitter cold air rolling off the larger continents and sea ice sheets should be quite sufficient to cool the ocean down to 3.0C.

gbaikie
January 11, 2012 5:07 pm

“The manner sea ice forms is irrelevant. It floats. It excludes salt. Everyone knows that. Might have some effect on currents with salt variation near surface but doesn’t explain how ocean temp below thermocline is 3.0C which is the essential question. ”
Well it explain why arctic ocean lacks ice on it’s bottom.
But to your question, the ocean below the thermocline hasn’t always been 3 C- it’s been cooler and it’s been warmer. Of course it has a limit to how cold it can get.
I would say that if earth temperature remain the same for next 1000 years, the temperature would rise, and of course maybe rise full degree if given 10,000 year of current warmth.
But the reason it’s 3 C is we have been living in Ice age period for last few tens of millions years- that’s why is isn’t warmer. And if we had no interglacial periods as we in currently, it would be colder.

sky
January 11, 2012 5:18 pm

Bomber_the_Cat says:
January 10, 2012 at 5:39 pm
“If you had said that solid bodies radiate energy related to their temperature and emissivity then this would have been correct. Mass and specific heat have nothing to do with it.”
I had in mind a fixed volume of material, not necessarily solid, with a constant stream of energy keeping it at a constant temperature. That temperature is certainly related to its heat capacity C in Joules per K! The specific heat is simply C per unit mass and mass density must enter into consideration of total energy flux through any fixed volume. Emissivity is simply a comparitive measure of radiation relative to a blackbody at the same temperature. Since the atmosphere is not a blackody, and I wanted to point to the to the far-infrared radiation of “inert” constituents, I avoided reference to tless fundamental concept.

gnomish
January 11, 2012 5:25 pm

Joules Verne says:
“My ears should pop when a cold front blows through. 10C change in temp is like a change in altitude of 1000 meters (dry adiabatic lapse rate ~ 1C / 100 meters). That would actually be painful if it happened quickly.”:
but if it did work, you could take everyone outside and it might pop out the blockage that prevents them from hearing what you said.

gnomish
January 11, 2012 5:27 pm

but i suppose if the gravity.heat meme got flushed as it should be there would only arise in its place something worse – possibly ‘dark.phlogiston’.

January 11, 2012 5:44 pm

I’ll second Robert Brown’s praise for Rodrigo Caballero’s Lecture Notes on Physical Meteorology ( http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf ).
If only such a text could be required reading before people get to comment on physical meteorology issues! They number of posts would go way down, and the quality of the posts would go way up.

Bart
January 11, 2012 6:10 pm

Robert Brown says:
January 11, 2012 at 3:35 pm
“That makes my conclusion even stronger, of course — either the moon won a (biased) coin flip to end up heavy side towards the earth or it simply stabilized symmetrically and pulled a bit more mass towards the near side while still plastic due to tidal asymmetry.”
I’m not taking issue with your conclusion, whatever it is. I just wanted to help out with the concept.

jae
January 11, 2012 6:25 pm

Willis, you say:
“Tim, one further thought. You say that Nikolov made the point that “atmosphere density determines its temperature”. In that regard, NASA refers to the “equation of state“. In one of its variants, that says
p = R ρ T
where p is pressure, R is the gas constant, ρ (rho) is density, and T is temperature. Solving for T gives us
T = p / (R ρ)
This means that density does not uniquely determine temperature as you say. You need to know pressure as well.
So if you still claim that Nikolov showed that only density is needed to determine temperature, you’ll have to explain how that works.”
But, oh guru, you have not REFUTED ANYTHING that Nikolov has said. Maybe you just play the game of thought experiments about impossible transparent atmospheres. Which, BTW, would probably get VERY, VERY HOT, since the emissivity you define is zero! Hell, I agree with DeWitt Pain for once!

Bart
January 11, 2012 6:38 pm

Tim Folkerts says:
January 11, 2012 at 5:44 pm
“If only such a text could be required reading before people get to comment on physical meteorology issues!”
With that said, I would like to point out something to which I referred here. I keep seeing this misconception, and indeed, we argued it extensively on that other thread. As Dr. Caballero states on page 117:

Kirchhoff’s law means that black body radiance
is the maximum radiance a body can emit at a given temperature.

So, we have people saying that the temperature of the Earth’s surface works out to a corresponding blackbody radiance of greater than 300 W/m^2, while emissions from the Earth are 240 W/m^2. But, according to Kirchoff’s law, the greater than 300 W/m^2 figure is merely the maximum radiance you can get, not the minimum. So, it is entirely possible to have the Earth’s surface emitting a little more than 240 W/m^2 (scaling by the ratio of radii squared at the surface and TOA to reach 240 W/m^2) even though the temperature there would induce greater radiance from a blackbody.
Of course, there is the issue of thermodynamic equilibrium, too, but the point is that blackbody relationships do not rule out the possibility of non-GHG heating of the atmosphere.

jae
January 11, 2012 7:03 pm

Timbo Folkerts:
““If only such a text could be required reading before people get to comment on physical meteorology issues!”
Aha, VERY typical obnoxious progressive approach, you have there, sir: YOU, like the Big Gov. that is about to swallow all our freedoms and dictate just what we can and cannot THINK AND DO, even, get to pick what I read, BEFORE I get to comment!??? Have you absolutely NO understanding of freedom and the American Way? I really doubt it, considering such a biased, facist, self-righteous, elitist, simplistic, overbearing, egotistic, etc…..statement!
Just who the hell do you think you are, anyway?

January 11, 2012 7:34 pm

Bart,
Thanks for making some clear, specific statements — it makes it much easier to see where we seem agree and where we seem to differ. (NOTE: all numbers below are approximate, ie within a few percent. I don;t want to get bogged down in whether a number should be 288 or 289 or 290)
> while emissions from the Earth are 240 W/m^2
This is easy to figure, based on the observed average solar irradiance (341 W/m^2) and the albdeo (~0.3). The absorbed power is then 341 W/m^2 * 0.7 = 240 W/m^2. Since the earth is pretty good long-term stability, it must (over the long term) emit this much radiation or it will have significant warming or cooling. I think we both agree with this number.
>But, according to Kirchoff’s law, the greater than 300 W/m^2 figure is merely
>the maximum radiance you can get, not the minimum.
There are (at least) two different issues closely related to this subject.
1) A blackbody produces the maximum thermal radiation — more than any real object at the same temperature, (ie the emissivity of any real object is strictly less than 1 which is one of the key results of Kirchhoff’s Law). So a blackbody at 288 K will emit 390 W/m^2, but any actual object will emit less energy than this maximum.
As I understand you, this is the point you are making here. And I agree with this 100%.
The experimentally determined emissivity of the earth for thermal IR is indeed less than 1, as required. But it is not much less than 1. One reference on one of these recent threads listed water’s emissivity as ~ 0.97. Most soils seem to be 0.9 or above (http://www.icess.ucsb.edu/modis/EMIS/html/soil.html). Limestone and marble and gypsum are also 0.9+. So typical materials at 288 K would emit at least 0.9 * 390 W/m^2 = 350 W/m^2.
You claim that the earth might only emit ~ 240 W/m^2, but if this is based on emissivity arguments, then the emissivity would have to be 240/390 = 0.62, which is way below any realistic estimates. Emissivity will not provide a solution.
2) The other big issue is that the non-uniformity of the earth’s temperatures will affect the power emitted. This is a big issue itself, and we could go into it more, but I think it has been discussed quite a bit already.

gbaikie
January 11, 2012 8:25 pm

“Humidity decreases sharply with height, dropping to near zero above 2–3 km.”
Yeah never snows above 2–3 km. And skiing is like exposing
your skin to the vacuum space.
And living in Denver- that’s Edge City, dude. Edge City.
Anyone saying they live above 3 kms, are LYING.

DeWitt Payne
Reply to  gbaikie
January 11, 2012 8:59 pm

gbaikie,
Are you sure you’re replying on the correct thread? I mean you’re right and all, but I can’t find to post to which you’re referring. The scale height of water vapor is ~2 km so at most it will drop to 1/e of the surface value at 2 km and probably less because it probably wasn’t 100% RH at the surface. OTOH, on the Antarctic plateau, which is about 3 km high, during the depths of a glacial period it takes thousands of years to accumulate enough snow to make 1 m of ice. But there you’re a long way from open ocean and the temperature is really cold.

January 11, 2012 8:34 pm

At this point, I’m what would be called a lukewarmer/fatalist. I think climate sensitivity is near the low end of the IPCC range and I don’t think we’ll (the planet as a whole) spend any serious money doing anything other than talking about it even if it weren’t.
I agree with the first — there’s a rather lot of evidence that this is the case, if one looks at it. The second — we’ve already been spending moderately serious money on it, alas. Not yet the tens to hundreds of billions of dollars some people would like to spend, over years, but the bill is almost certainly up in the billions. California alone has spent, or caused to be spent, rather huge amounts of utterly wasted money with its carbon-linked taxes and requirements imposed on electrical generating companies to invest in non-cost-optimal generation methods. And then there is Europe, and to a lesser extent other countries and locales. We just haven’t given the UN what it wants, and I agree, that is at least currently quite unlikely and the probability is diminishing in time at this time.
Climategate 1 and 2 really have had an effect. Not on the religious — nothing can impact the religious beliefs of a dedicated warmist — but on anyone that has actually read the exchanges. Science at its worst. Or I should say “science” at its worst.
The online book you sent me really is brilliant. I already am starting to understand a lot of the terms used in discussions on WUWT, such as “lapse rate”, which I found frankly puzzling before. I’ve skimmed the whole thing through the chapter on radiation — should be required reading before claiming “there is no GHE” on WUWT — and am just getting to its discussion of turbulence. Some of the figures — e.g. his presentation of albedo — are a bit maddening because he changes scale from microns to nanometers to inverse centimeter wavenumbers with abandon, and I’m used to thinking in nanometers almost exclusively because I “focus” on visible light when teaching optics (heh, heh). But the idea is straightforward, and the numbers he presents are invaluable. I do wonder that his coupled ODEs here omit actual absorption by the clouds — I’m assuming that it is so low that it is negligible, but given multiple scattering and many, many opportunities for absorption, that doesn’t really make sense. Doesn’t sunlight warm clouds as well as reflect off of them? It certainly warms (and evaporates) water on the ground.
But still, more than enough to not only get started, but make some good progress towards understanding the physics of climatology.
rgb

January 11, 2012 8:54 pm

“Humidity decreases sharply with height, dropping to near zero above 2–3 km.”
Yeah never snows above 2–3 km. And skiing is like exposing
your skin to the vacuum space.
And living in Denver- that’s Edge City, dude. Edge City.
Anyone saying they live above 3 kms, are LYING.

Sarcasm aside, this seems like a good time to trot out the book Dewitt linked above. To quote 3.12:
Since temperature decreases with height in the troposphere, e_s will also decrease with height, and we expect the atmosphere to become drier with height. We can estimate the rate of decrease by assuming an atmosphere with a dry-adiabatic lapse rate…
e_s is the saturation vapour pressure. Basically, the higher you go, the colder it gets. On average the drop in humidity is shown (in the book) to be roughly exponential in height, with a length scale of ~2 km. So no, it doesn’t drop near zero above 2-3 km — it drops to 1/e its value on the ground. Around 4 km it is 1/e^2. Around 6 km, 1/e^3. So somewhere around 3 or 4 miles up, the humidity on average is 1/10 what it is on the ground, maybe 7 miles up it is 1/100th of what it is on the ground, until you hit the tropopause and everything changes.
Note well that I didn’t say up “from sea level” and I don’t think this is intended to refer to actual clouds or what happens when you reach the temperature for e_s and the air is saturated. So you’re both right, but “dropping to near zero” is a strong exaggeration and neglects both clouds and precipitation and the fact that one starts to compute the lapse rate from ground level and temperature — really you’re using the variation of temperature with height above the ground to determine e_s to express how the humidity drops off with T (and hence height). It also really doesn’t handle how warm wet air masses lift up to rise over mountain ranges, dropping moisture as they go, and blow out on the other side much drier — often dry enough to leave deserts underneath. And so on. A rule like this has to be sanely applied.
rgb

DeWitt Payne
January 11, 2012 9:05 pm

Robert Brown,
This moderation thing can be maddening at times. You end up replying to someone who has already been answered, but you haven’t been able to see it.

I’m used to thinking in nanometers almost exclusively because I “focus” on visible light when teaching optics (heh, heh).

I could tell you weren’t an organic chemist. Since all they know, mostly, is IR spectrometry, it’s all reciprocal centimeters to them.

January 11, 2012 9:10 pm

Just who the hell do you think you are, anyway?
You’re right. I’m so ashamed. What was I thinking?
It is a god-given right of all Americans (especially) to say whatever they like, quite independent of whether or not they can tie their own shoes or count to twenty without removing them. Especially on something perfectly obvious. Like climate. Hell, everybody understands climate. Just look out your window! Who needs math?
Not ,I>just climate, of course. People should (and generally do) feel free to comment on anything at all. I actually got a Christmas card — I am not making this up — from somebody who has figured out the unified field and solved all of physics. Complete with a convenient link, so that you can go “how nice, the theory of everything and all I have to do is click this link and download, and I can too understand how the mass ratio of the proton and the electron can be computed — exactly — just from a bizarre series of trig functions, taken to just the right powers. Turns out that “the universe is an integrated whole” based on a “special inversive geometry” such that the “physical is not infinitely divisble”. Who knew?
And here I wasted all of that time reading Leonard Susskind and learning that it was all a strange kind of holography tied together (so to speak) with string theory.
Aw, shucks. I don’t feel right about keeping this to myself. And I’m sure it is relevant to the climate debate, somehow. You too can visit naturefromscratch.com and learn all about it, if you are ever really, really bored.
In the meantime, I fully retract my suggestion. Please, do not ever, anyone, actually try to learn the physics of climate before commenting on it. That way you can let your imaginations roam free, unconstrained by any unseemly requirement of actual correspondence with reality…;-)
rgb

January 11, 2012 9:12 pm

JAE,
The comment about “required reading” was a bit tongue-in-cheek. The wide-open discussions can be wild and fun, but it also pretty much ensures that no real progress will be made and that discussions will revert to the lowest common denominator.
In every field, experts are required. Many activities have “hurdles” that must be cleared before you are allowed to take part: teaching school, driving a car, prescribing medicine, being a police officer, voting, even cutting hair. All of these require special training/education/certification.
This is no legal requirement to engage in scientific discussions. An informed, engaged scientifically interested populace is a great thing. But just like training and knowledge and experience are valuable for someone to be an effective police officer or pilot or barber, training and knowledge and experience are valuable for someone to discuss science.
The decision has been made in WUWT to allow rather free-ranging discussions of science, and that is a perfectly legitimate decision. Other blogs have a more restrictive policy for commenting on scientific, which can have its place. The more focused you want the discussion to remain on science, the more you need to moderate repeated claims that are simply repeating ideas that have already been discussed and rejected.
PS It is not a “Progressive” approach at all, since Progressivism typically favors the masses over the elite. It is a very Conservative in fact, supporting traditional idea and approaches, trying to keep people from drifting to radical ideas.
PPS. This will be my only post replying to this topic. I REALLY don’t want to let the discussion become hijacked by side issues like this.
PPPS Thanks for the support, RGB. I just saw your comment: “I’ve skimmed the whole thing through the chapter on radiation — should be required reading before claiming “there is no GHE” on WUWT”

January 11, 2012 9:35 pm

The pressure at the surface depends on the mass of the atmosphere above the surface and the local value of the acceleration of gravity. Period. A change in temperature won’t change the mass. It’s nothing like a sudden increase in altitude. If you go up 1 km, you’ve put about 10% of the atmosphere below you and the pressure drops from about 1,000 mbar to about 900 mbar. If you decrease the temperature by 10 K from, say, 300 K to 290 K, you’ve changed the air density by about 3%, but the pressure not at all.
Or, as I pointed out elsewhere, one can have an air column where the pressure varies exponentially that is all at a constant temperature. Or, the dry adiabatic lapse rate is predicated on having a constant “potential temperature”, which in turn means that a parcel of dry air has constant entropy as it rises or falls, expanding or contracting to match the (exponential) pressure. This implies that the temperature itself falls in just the right way to keep the entropy or potential temperature constant.
The connection between the height, the pressure, and the temperature is not fixed in stone, in other words. Rising and falling air isn’t precisely isoentropic/adiabatic. The atmosphere inverts over Antarctica (for example) so that the lapse rate is negative and the air gets warmer as one goes up. Moisture screws everything up. So do other things.
Caballero’s book rocks!
rgb

Bart
January 11, 2012 11:15 pm

Tim Folkerts says:
January 11, 2012 at 7:34 pm
“The experimentally determined emissivity of the earth for thermal IR is indeed less than 1, as required. But it is not much less than 1.
Apples and tennis balls. The experiments do no duplicate the same environmental conditions.

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