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.
We like tha moon
http://tinyurl.com/djag
No… the moon is cold because its solid surface has a relatively high emissivity. The Earth is warmer because its gaseous atmosphere has a low emissivity (very low if it was only nitrogen and oxygen). According to Trenberth et al. the atmosphere emits195 W/m2 wheras the Earths surface emits 40 W/m2 by radiation. (102 W/m2 is transported by convection to the atmosphere where it is then radiated to space). Greenhouse gases increase the emissivity of the atmosphere. Therefore the GH gasses effectively cool the Earth!…
Well, it took me seconds to come up with a very different albedo figure for the moon at various space/physics sites (0.12 not 0.07); and it seems the same may be true about the stated temperatures…
If a non GHG atmosphere did not radiate EM energy, it would be a perfect insulator. In space N2/O2 would never cool.
In contradiction to the notion that N2/ O2 do not absorb / radiate EM radiation, here is their absorption spectra. As can be seen, both N2 and N2 absorb EM radiation. N2 has quite a broad spectrum. So if neither radiate, then the atmosphere is going to get very hot from absorption of EM radiation by N2 and O2, and this heat will conduct to the surface..
http://www.coe.ou.edu/sserg/web/Results/Spectrum/n2.pdf
http://www.coe.ou.edu/sserg/web/Results/results.htm
The nuance of the first line of the Haiku is apparently a bit different…….
I may translate it into “Oh beautiful full moon!” from the original “Mei-getsu ya”.
Compare the spectra of H2O with CO2 and N2. Which one is most unlike the other two? If climate science is to be believed, you would think N2 would be most unlike the other two. Imagine the surprise I got when I took the time to check.
http://www.coe.ou.edu/sserg/web/Results/Spectrum/h2o.pdf
http://www.coe.ou.edu/sserg/web/Results/Spectrum/n2.pdf
http://www.coe.ou.edu/sserg/web/Results/Spectrum/co2.pdf
I must diagree with Willis’ theorem as so well written above. I have tried to give an alternative view on the Greenhouse effect here…
http://wstannard.wordpress.com/the-greenhouse-effect-2/the-earths-energy-balance/
I hope I have written this as well as Willis writes. My conclusion is that there is no greenhouse effect that warms the planet. There appears some misunderstanding of the S-B equation (the Earth is not a blackbody) and the fact that the most of the radiation from the Earth is emitted by the atmosphere, not the Earth’s surface!
I would be interested in further comments.
Isn’t there also the issue of radioactive decay warming the Earth but not the moon? I imagine that would account for some of it.
Whence the warming at night to maintain a smaller range? From what energy source? I suspect the real reason Venus varies so little over its long day/night cycle (1K, or thereabouts) is that the radiative “short-circuiting” by such a dense and high-CO2 atmosphere is almost perfect. On Earth, not so much, but some. I.e., energy from the dayside is being transferred radiatively by CO2 to the nightside.
You heard it here first!
>:)
Can you equate rates of heating and cooling on earth and moon, when earth’s surface is mostly water? Surely you have to leave the oceans in place when the atmosphere is removed for your thought experiment?
Also, or moreso, probably, by H2O ↔ H2O radiative transfer. Doesn’t work so well in the low-humidity deserts, whether Saharan or Antarctican, of course. There, CO2 is on its own, and is too sparse to achieve much.
Wow… as an aside, imagine if we could increase the Moon’s rotation to about the same as Earth’s – perhaps by a deliberate glancing meteor blow. The faster rotation would reduce the temperature swings to about the same as Earth’s, thank to the calculations above…
Then we could terraform the Moon. One speculative proposal is here: http://www.lunar-union.org/planetary-engineering/terraforming_moon.html. If this is combined with a faster rotation then it would be much better.
I am amazed at the information on that other website that points out that Titan, with a surface gravity similar to the Moon’s, has an atmospheric pressure about the same as Earth…
An excellent post showing the importance of T^4 in regulating the temperature of the Earth (and other heavenly bodies).
will do some mooning tonite too, willis
we were promised our first hot day of summer today in my part of australia, but it didn’t live up to the predictions:
9 Jan: Sydney Morning Herald: Dan Cancarrow: Brisbane has hottest day of the summer
The city hit a high of 33.8 degrees at 1.21pm, exceeding last month’s average of 27.6 degrees and the summer’s previous maximum of 32.9 degrees…
Last month’s average maximum was almost two degrees (1.9) lower than the city’s long term December average of 29.5 degrees.
http://www.smh.com.au/queensland/brisbane-has-hottest-day-of-the-summer-20120109-1pqqi.html
***how precise is 249km? is there a miles equivalent that rounds off?
9 Jan: Ninemsn: AFP: Species lag in climate change shift
Fast-track warming in Europe is making butterflies and birds fall behind in the move to cooler habitats and prompting a worrying turnover in alpine plant species, studies published on Sunday say…
The papers, both published by the journal, Nature Climate Change…
A team led by Vincent Devictor of France’s National Centre for Scientific Research (CNRS) found that from 1990 to 2008, average temperatures in Europe rose by 1C.
This is extremely high, being around 25 per cent greater than the global average for all of the last century.
***To live at the same temperature, species would have to shift northward by 249km, they calculated…
The data derives from observations made by a network of thousands of amateur naturalists, amounting to a remarkable 1.5 million hours of fieldwork.
???The study was not designed to say whether these species are suffering as a result of warming, which is one of the big questions in the climate-change saga.
However, the risk of population decline is clear, the authors say.
Species that lag behind a move to a more suitable habitat accumulate a “climatic debt”.,,
The second study looked at 867 samples of vegetation from 60 mountaintop sites across Europe in an assessment of the hottest decade on record.
Seen at local level, there was little apparent change during the 2001-2008 study period.
But when the picture zoomed out to continental level, it was clear that a major turnover was under way…
(Michael Gottfried, a University of Vienna biologist) “Many cold-loving species are literally running out of mountain. In some of the lower mountains in Europe, we could see alpine meadows disappearing and dwarf shrubs taking over within the next few decades.”
The research was the biggest plant-count of its kind in Europe, gathering 32 researchers from 13 countries.
http://news.ninemsn.com.au/technology/8400117/species-lag-in-climate-change-shift
obviously a “we have the numbers” ploy…even if it ain’t CAGW’s fault!
Clear presentation, very poor physics.
The surface of the Moon is colder than the surface (and lower layer of the atmosphere) of the Earth for the same reason a man without a blanket, during a cold night, is colder than a man under a blanket.
Both Moon and Earth radiate into the space as much heat as they receive from the Sun. The only difference is that the heat exchanging surface of Earth is the upper atmosphere (which is cold), while on the Moon it is… well, the surface of the Moon.
Build a dome on the Moon, fill it with air, and it will be soon warmer inside than outside. The heat exchanging surface in this case will be the shell of the dome, not the ground surface inside the dome.
I think Mr. Eschenbach should write less and think a bit more before writing.
… the greenhouse effect has done the heavy lifting to get the planet up to its current temperature,…
What a lovely way of putting it! Thank you Willis.
This means that the greenhouse effect warms the earth in two ways,
does he mean slows down the cooling.
Moon gets hot without greenhouse gases and cools rapidly at night time,
Earth does not get so hot and cools down slowly.
Matt says:
January 8, 2012 at 11:49 pm
You’re right, Matt, I used a number from my memory, and it is slightly larger. NASA states that the lunar Bond albedo is 0.11. It makes only a trivial difference. Instead of 318 W/m2, it makes the lunar calculated radiation 304 W/m2. Doesn’t change my discussion or my points in the slightest. I’ve updated the head post with the correct figures.
w.
tokyoboy says:
January 8, 2012 at 11:53 pm
Reading Basho always makes me wish I could read Japanese.
Many thanks,
w.
[I’ve updated the head post to reflect your translation. —w.]
ferd berple says:
January 9, 2012 at 12:01 am
ferd, the N2 is the most unlike the others because the line strength is many, many orders of magnitude weaker than that of the others.
In addition, at least from what you show, N2 only absorbs in a narrow window, and in that window (according to your citations) N2 absorbs 10 orders of magnitude less than CO2.
Five-hundred-thousand years of reconstructed history of the earth’s surface temperature demonstrates that atmospheric temperature is controlled by a chaotic system which is bounded by a relatively narrow band of temperatures we define as “Warm Periods” and “Ice Ages”. It has remained in that defined band of temperatures despite horrific asteroid bombardment, cataclysmic volcanic disruptions, enormous variations in cosmic radiation and significant variations in solar energy input as the earth perpetuates from circular to elliptical orbits about the sun every 100,000 years or so. So, what exactly is the mystery here? We’re on the long, slow, and slippery slope to another Ice Age by fits and starts with the Roman Warming Period probably marking the apex of our current cycle. To quote Willis’ previous great story, “Everything else is turtles on top of turtles.”
Alexander Feht says:
January 9, 2012 at 12:30 am
Clear presentation, very poor reading skills. I said nothing comparing the moon without an atmosphere to the earth with its current atmosphere. So you are railing against a position I did not take.
Alexander and others, listen up. If you disagree with something I said, QUOTE MY WORDS THAT YOU DISAGREE WITH. Otherwise, you may end up like this example, where you are attacking a straw man and abusing me for something I NEVER SAID.
I’m perfectly serious here, folks. If you do not quote what you object to, you may get no answer from me at all. I’m tired of people who jump on my case without bothering to reveal what I said that has them in such a snit. And yes, Alexander, I’m looking at you.
If you think I’m wrong, well, I might be wrong, I’ve been wrong before. But my friends, how about before you conclude that I’m wrong, you ask about whatever it is I said that makes you think I’m wrong? Then I can clarify my words, or I can admit I’m wrong, or I can show why I don’t think I’m wrong. QUOTE MY EXACT WORDS so we can all understand what you are objecting to, and we can discuss it.
w.
Re-run the moon with the thermal inertia equivalent of the earth’s and rotation like the earth’s.
Thanks
JK
The SB equation is misused by IPCC style Climate Science.
The Moon is a perfect example of this.
1. why after 14 Earth days does the dark side never reach absolute zero but stays some 90K above?
2.why after 14 Earth days does the Sunlit side never reach its predicted radiative max but stays well below it.
The guilty little secret of IPCC Moon Science is that to get anywhere near realistic temperature figures they have to include a substantial contribution from a GROUND HEAT FLUX in addition to the radiative fluxes.
The guilty little secret of IPCC Earth Science is that they refuse to include a ANY contribution from a GROUND HEAT FLUX in addition to the radiative fluxes.
If they did so they would find little use for the so called greenhouse effect.
Willis Eschenbach says:
January 9, 2012 at 12:55 am
ferd, the N2 is the most unlike the others because the line strength is many, many orders of magnitude weaker than that of the others.
Perhaps you misread the reference? From what I see, N2 line strength is 10-28, CO2 is 10-23, which is 5 orders of magnitude. However, N2 has 10 orders of magnitude wider spectrum (600 cm-1 versus 50 cm-1). In addition, there are 4 orders of magnitude more N2 in the atmosphere than CO2. So, on this basis it is hard to see that N2 absorbs/radiates significantly less than CO2.
In contrast to CO2, H2O line strength is 10-19 which if 4 orders of magnitude stronger than CO2. As well it has a much, much wider spectrum than CO2. The absorption strength and spectra of water so overwhelms CO2 as to make it CO2 a joke when you consider the amount of H2O in the atmosphere as compared to CO2.
correction n2 spectrum is 10 times, not 10 orders of magnitude wider.
jim karlock says:
January 9, 2012 at 1:08 am
Thanks, Jim. I’m happy to be finished writing this, I need to discuss and answer questions and objections, and I have a day job. However, I do invite you to re-run the moon as you suggested and report your results.
w.
Some goose says;
January 9, 2012 at 12:30
“The surface of the Moon is colder than the surface (and lower layer of the atmosphere) of the Earth for the same reason a man without a blanket, during a cold night, is colder than a man under a blanket”
Bull, say both men don’t radiate heat by themselves, like the earth and moon, then that bloke under the blanket will get no warmth from the atmosphere and will be silly, sorry chilli.
Willis, did your calculations take into account the temperature difference between the equator and the poles? From looking at Figure 1, if you used that for your data it will not give an accurate result because it reflects an average of temperature between the equator and poles. The temperature difference between the lunar equator and poles is greater than between night and day averages.
“Most notable are the measurements of extremely cold temperatures within the permanently shadowed regions of large polar impact craters in the south polar region,” said David Paige, Diviner’s principal investigator and a UCLA professor of planetary science. “Diviner has recorded minimum daytime brightness temperatures in portions of these craters of less than -397 degrees Fahrenheit. These super-cold brightness temperatures are, to our knowledge, among the lowest that have been measured anywhere in the solar system, including the surface of Pluto.”
http://www.sciencedaily.com/releases/2009/09/090917191609.htm
All sounds very convincing. If the moon had an IR transparent atmosphere, ie. no GHG’s, then there would still be a temperature rise due to adiabatic compression as happens on Jupiter with its atmosphere of hydrogen and helium. You cannot ignore this temperature increasing physical phenomenon. It has been understood for over 100 years and is the reason for the Fohne Effect, by which the Chinook supplies warm winds to the Canadian and American prairies and the cause of warm katabatic winds. (not to mention diesel engines and refrigerators).
It is also true that to take the temperature of a system that system must be at equilibrium. The earth, with its turbulent atmosphere never is. To use the S-B equations on a system then, again, that system MUST be at equilibrium which the Earth never is.
I have three points.
– IR radiation is the only way how surface temperature is transferred from solid surface to atmosphere. If a moon had atmosphere that is perfectly transparent to IR, that atmosphere would do nothing with its surface temperature.
– Earth albedo of 0.3 is partially given by clouds and other atmospheric effects. You can’t simply imagine Earth without atmosphere but having the same albedo.
– Albedo of a solid body affects not only its absorption of incoming radiation but also its release of outgoing radiation. It’s not correct to assume Earth absorbs energy according to its albedo and then cools down as fast as Moon does.
The Moon also effects the climate on Earth in some subtle ways !
Firstly the Moon and the Earth both orbit their common centre of mass, which is about 4000 km outside the centre of the earth. This causes the Earth to shift by up to 8000 km nearer the Sun every lunar month. I found very recently that that there is a clear signal for this effect in the SORCE TIM solar radiation data – see http://clivebest.com/blog/?p=2996 I think this may be the first time anyone has noticed this !
Secondly – there is a very slight effect from extra reflected solar radiation from the full moon on Earth. Thirdly there are tidal effects on the Earth’s atmosphere as well as the Oceans. At the North and South poles these “tides” have been measured by changes in surface ozone.
Finally the moon stabilises the Earth’s rotation axis. It seems likely that without the moon’s stabilising gyroscopic effect the earth’s axis would be more chaotic. The seasons rely on the axis being tilted to the orbital plane of the earth-sun by about 23 degrees. Computer simulations show that the moon’s tidal effect has probably stabilised this tilt over billions of years. By comparison, the axis of Mars seems to be affected by a chaotic effect caused by the influence of other planets in the solar system.
Yet another quality thought provoking post by Willis. Thanx mate.
A couple of things I’m trying to get my head around..
* The moon hides behind the earth for a few days each cycle (is it 7 days?) where it receives no insolation at all. All emission no absorption. Is this reflected in the first graph, if so where?
Yes true and you made this point at an earlier thread, but here is my thought.
If the only thermal interaction between the atmosphere and the surface is via conduction, it is possible to have an average SURFACE temperature of a half a degree (no more than the S-B T) whilst at the same time having an average ATMOSPHERE temperature somewhat higher than the S-B T. No laws of conservation are broken.
The outgoing radiation is still from the surface and it is still the same as the incoming radiation at equilibrium but this doesn’t stop the atmosphere (as a whole) from being warmer.
*
The question is how much energy is transported. The moon may absorb an AVERAGE insolation of 304Wm2, but at its equator at noon it would receive the full gamut, well over 1000Wm2.
And since warming by conduction is ALWAYS faster than cooling by conduction when a gas is involved, we’d need to work out how much energy is transported.
I contend that the warmth from most of that 1000Wm2 will be transported with strong temperature inversions at night, even more so at the poles.
So if I was standing on the moon with a GHG-less atmosphere, it may be quite cold under my moon boots, but where my body is (up to 2 metres above the surface) I’d say the temperature would be quite warm.
And so it is with our present Earth. All the SB calculations may say an average -18DegC at the surface under my shoes, but this isn’t the same at the 2 metre height when an atmosphere is present, GHGs or not.
I’d appreciate your thoughts.
Bryan says:
January 9, 2012 at 1:15 am
Because the radiation becomes so small at low temperatures. Note that even after 14 days the moon is still cooling, but quite slowly. At that point it is radiating at only 2.5 W/m2, so very slow cooling would be expected.
At mid-day (lunar day) the sun provides about 1368 W/m2. Given the moon’s albedo, that corresponds with a blackbody temperature of about 109° at the surface. The sensors used for the temperature record above are slightly buried in the regolith, so I find the fact that they haven’t quite reached the maximum predicted temperature to be unsurprising.
I find that the temperature figures that I show in Fig. 1, from NASA data, is quite realistic, and agrees with theoretical calculations. It requires no ground heat flux to make sense.
If you are talking about the earth, as far as I know the ground heat flux is on the order of a tenth of a watt per square metre. I’ve run the numbers myself from a couple of directions, and it’s just not all that large. Even assuming that there are many more deep sea vents than are generally thought, there still isn’t enough heat coming from the inside of the planet to make much difference. If there were, we could sleep on the ground to stay warm …
Ground heat flux is a couple orders of magnitude too small to be responsible for the Earth’s surface temperature.
Regards,
w.
Neither Moon nor Earth are ideal black bodies; therefore, Boltzmann’s formula cannot be applied directly in both cases. If this is what Mr. Eschenbach is trying to explain, anybody who was paying attention to his physics teacher in the 8th grade of the high school knows that.
As to his answer to my post (senile emotional outbursts notwithstanding) Mr. Eschenbach does directly compare Earth and Moon. There is no other way to understand his words, however one reads them: “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 then again, in the next paragraph: “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.”.
While these statements are no more than a textbook explanation of the obvious, the general position Mr. Eschenbach takes in his article, from the very beginning, is to compare apples (a non-black body without an atmosphere) and oranges (a non-black body with an atmosphere — and a biosphere to boot). Such a comparison is misleading, since Earth’s and Moon’s mechanisms of heat exchange with the space are completely different.
I need to clarify the last paragraph of my last comment where I said..
Should read ” ….at the 2 metre height when a GHG-less atmosphere is present”
This seems (to me) to be a outstandingly thoughtful and clearly written post; appreciation and thanks are extended to Willis for his work in writing it.
Obviously these calculations aren’t easy, and very often it happens that simple lines of reasoning, that give a correct result, are evident only after long and intricate calculations have been carried to completion.
This does not mean that the long intricate calculations can be skipped, if only we are clever enough to think of the simple lines of reasoning at the beginning. Rather, both styles of calculation are essential: simple at the beginning (necessarily) and simple too at the end (hopefully), and complicated in the middle stages, where all the difficulties are worked through.
An extended discussion of those complicated middle stages can be found on the American Institute of Physics (AIP) web site The Discovery of Global Warming: Basic Radiation Calculations.
Broadly speaking, the next step in making Willis’ model of the moon’s temperature look more like models of the earth’s temperature would be to “churn” the top layers of moon dust, analogous to the vertical convective transport processes of the earth’s atmosphere; the AIP web page gives a description of how these effects are modeled in-detail.
Recommended.
Thank you, Willis. Very interesting.
“If you are talking about the earth, as far as I know the ground heat flux is on the order of a tenth of a watt per square metre. I’ve run the numbers myself from a couple of directions, and it’s just not all that large. Even assuming that there are many more deep sea vents than are generally thought, there still isn’t enough heat coming from the inside of the planet to make much difference. If there were, we could sleep on the ground to stay warm.”
Am I OK to assume there is no heat transfer, other than the .1 Wm/2, between the oceans waters and the ocean floor, either way?
Alexander Feht says:
January 9, 2012 at 2:26 am
Thanks, Alexander. You agree (I think) that both my statements are true, since you call them a “textbook explanation of the obvious.”
Nor does either statement depend on the presence or absence of an atmosphere. They are statements that are true because of differences in albedo. They are just statements of how much energy is entering either system.
As to your claims about my “general position”, even I couldn’t say what that might be, so how could you? Let me suggest that you stick to my specific positions, as expressed in my specific words, and leave speculation about my “general position” alone.
I am looking at the moon without an atmosphere to try to get an estimate of the temperature fluctuations of the Earth if it had no atmosphere. It’s called a “thought experiment”. Apples and apples.
I am also trying to understand the different ways that an atmosphere can affect a planet, and how planetary rotational speed affects the average temperature.
So I fail to see the problem. I specified several times I was interested in the effect of, how did I put it … OK, I called it a “perfectly transparent GHG-free atmosphere” on a planet. I’m working towards understanding the vagaries of that as it applies to temperature variations.
All the best, thanks for quoting what I said. As it turned out, both things you quoted me as saying were 100% true. That’s why quoting the words of someone you disagree with is so important—it allows us all to see and agree (or agree to disagree) on the exact issues in question.
w.
Don’t forget the ocean. Without it there would not be much energy to be backradiated during night.
I dont think its the swing of temperatures that affect the average, but the speed of the rotation. i.e. the amount of time the dark side has to cool.
If the moon was fixed, [the] hot side would get to 90, the dark side to -270, with an average of round about where that green line is above.
If the moon rotated at 1000 rpm, my guess would be that the average temperature would be closer to 90 than to -77
caveat – I am not a scientist (IANAS)
Why ignore gravitational compression of the atmosphere ?
Isn’t that what sets the adiabatic lapse rate independently of the effect of greenhouse gases ?
Nice summary of basic science.
Science of doom covered the same topic around a year and a half ago with the useful point that as the thermal capacity and inertia of the surface/atmosphere increases, the maximum temperature falls, but the minimum and average rise while the AMOUNT of energy emitted stays the same.
http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/
It shows up the warming from atmospheric pressure nonsense for what it is, it is the equalising effect on the temperature range from a energy transporting atmsophere that raises the average temperature.
But as the point is made above, that atmospheric effect of energy distribution can never raise the temperature ABOVE the S-B limit, it needs a GHG effect to do that…
Willis Eschenbach says of the Moons temperature record
“it requires no ground heat flux to make sense”
“after 14 days the moon is still cooling, but quite slowly. At that point it is radiating at only 2.5 W/m2, so very slow cooling would be expected.”
I thought you might be interested in a new peer reviewed paper by Gerhard Kramm and Ralph Dlugi.
They calculate the Moons Ground Heat Flux to be 16.2W/m2 – hardly negligible (page 990)
They go on to confirm their previous work that
” the greenhouse theory is a set of merit-less conjectures with no physical support.”
Of particular interest is ;
The energy reservoir diagram Fig 11
The irradiance overlap area Fig 5
On a more humorous note they find further errors in the Halpern et al G&T comment paper
See page 1316
Wrong formula
Wrong units
Which of course leads to silly numbers.
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233
What is the average value of the earth’s surface (ground) temperature? It is different from the temperature measured by weatherstations which measure air temperature just above the ground. But it is the earth’s surface that emits and receives the Boltzmann radiation, solar and back radiation. Surely a different T. In direct sunlight the earth surface is warmer than the air (eg tarmac gets hot, damp areas less so because of latent heat) but on a clear night the air will cool quicker than the surface. Is there a tendency for the average surface temperature to be higher than the average surface air temperature? I’ve never seen this discussed [anywhere] in the AGW debate.
that was my favorite heinlein.
but: ” the bigger the temperature swings, the lower the average temperature.”
an average is an average. extremities don’t change the average by any mathematical process.
antwaher = anywhere in my previous comment
“While these statements are no more than a textbook explanation of the obvious, the general position Mr. Eschenbach takes in his article, from the very beginning, is to compare apples (a non-black body without an atmosphere) and oranges (a non-black body with an atmosphere — and a biosphere to boot). Such a comparison is misleading, since Earth’s and Moon’s mechanisms of heat exchange with the space are completely different.”
Thanks for staying with it Alexander, however, I did not see Willis position as you did. It’s a big universe out there. Haven’t we always quantified our knowledge of it in within physical laws known here on earth. I only saw a comparison between earth and the moon using known physics. Of course you would rather compare apples and oranges by using the analogy of humans and extraterrestrial bodies.
You said;”The surface of the Moon is colder than the surface (and lower layer of the atmosphere) of the Earth for the same reason a man without a blanket, during a cold night, is colder than a man under a blanket.” That didn’t work for me at all.
Kasuha claims that IR radiation is the only heat transfer available between atmosphere and surface. It is not, how about conduction and the one that transfers most heat to the upper atmosphere, convection. Radiation is the smaller part.
“I am looking at the moon without an atmosphere to try to get an estimate of the temperature fluctuations of the Earth if it had no atmosphere. It’s called a “thought experiment”. Apples and apples.” — so says Mr. Eschenbach.
In the article above, I see “Earth without an atmosphere” mentioned once (“In that case the moon and the Earth without atmosphere would be roughly equivalent, both simply radiating to outer space”) — but after that, having stated the obvious again, Mr. Eschenbach continues to talk mostly about the effects of the greenhouse gases and atmosphere (just look over several following paragraphs in the above article). I don’t see here any “thought experiment” comparing our Moon with “the Earth without an atmosphere” (however meaningless such a comparison would be).
One of the main conclusions in the above article is as follows: “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.” Really? My reading skills are developed enough to see that Mr. Eschenbach is talking about atmospheric effects again.
Not to mention that it’s news to me how this can be news to anybody. Let it be, though. I have better things to do.
Alexander Feht says: January 9, 2012 at 12:30 am
I assume you are referring to a dead man – since any fool would realize that a living man has an internal heat source that makes your comparison completely irrelevant.
So do you have details of the experiment performed on two dead bodies left out overnight, one with a blanket and one without? If so, please quote a reference.
If not, please apologize for calling Willis “senile”.
Bryan says:
January 9, 2012 at 1:15
1. why after 14 Earth days does the dark side never reach absolute zero but stays some 90K above?
—————————–
The Cosmic Background Radiation is about 2.76 K, so you can rule out reaching absolute zero, or 0 K.
From David on January 9, 2012 at 12:26 am:
That could be problematic. The Moon is phase locked (aka tidally locked) to the Earth. As it is too solid, hard, and rocky to deform, even if its rotation around its axis could be sped up then it would want to return to the preferred state, with the same face always facing the Earth.
You could try speeding up the revolution of the Moon around the Earth to shorten the day/night transition period on the Moon. With simple Newtonian physics and a two-body problem, for a stable orbit the force of gravitation must equal the centripetal force.
Gravitational force: F=GMm / r^2, F=Force, G=gravitational constant, M=mass of Earth, m=mass of Moon, r=radius of orbit (distance between centers of mass)
Centripetal force: F=mv^2 / r, v=speed (technically scalar component of the velocity).
GMm / r^2 = mv^2 / r – divide both sides by m
GM / r^2 = v^2 / r – multiply both sides by r^2
GM = rv^2
Thus radius times velocity squared equals a constant (G times mass of Earth). Quick and dirty, since it takes about 4 weeks (28 days) for the Moon to revolve around the Earth, a lunar day is around 28 days, we want to shorten that to around 1/28 of current amount, the Moon would have to revolve around the Earth about 28 times faster. To keep it in orbit the radius would shorten by the square of that factor, 28^2=784, current radius divided by 784. The average distance between the centers of mass of the Earth and the Moon (called a lunar distance) is 384,400 kilometers. 384,400/784=490 kilometers, the approximate radius needed for a lunar day of around an Earth day.
However as the mean radius of the Moon is 1,737 km, and that of the Earth is 6,371 km, this method of shortening the lunar day should not be recommended.
So basically we’re stuck with the current length of a lunar day. When we get around to colonizing the Moon, if we’re going to grow plants by daylight then we’ll need something that’ll work with 2 week long on/off cycles. But since algae will get boring quickly, we’ll likely be using artificial lighting as well.
Willis, thanks for pointing out what impact temperature swings have on the average temperature through the T^4 relationship from S-B. This nicely extends previous discussion here and elsewhere, e.g. at Roy Spencer’s blog (http://www.drroyspencer.com/2011/12/why-atmospheric-pressure-cannot-explain-the-elevated-surface-temperature-of-the-earth/) .
So, you cannot only show the impact of rotational speed of a celestial body to its average surface temperature, but the presence of an atmosphere, be it a “normal” one with GHGs, or a theoretical one completely void of GHGs, will raise average surface temp through smoothing out the temp swings over the surface through redistribution of heat vertically via convection and horizontally via wind from pressure gradients.
Wouldn’t that imply that even a perfect non-GHGs atmosphere would exhibit an adiabatic temp profile? I think it would, however, Roy expects an isothermal atmosphere. This question remains unanswered. Any reasoning from you towards either direction?
Nice analysis but you’ve ignored at least 99% of the moon and 99% of the Earth both of which little to nothing is known. Sure vulcanologists etc. have some information about the centre of the Earth but they infer far too much from far too little information.
Markus says of the Ground Heat Flux
” I’ve run the numbers myself from a couple of directions, and it’s just not all that large. Even assuming that there are many more deep sea vents than are generally thought, there still isn’t enough heat coming from the inside of the planet to make much difference.”
I think you have a different definition of ground heat flux to that used by Gerhard Kramm and Ralph Dlugi.
See Fig 12
http://www.scirp.org/journal/PaperInformation.aspx?paperID=9233
Willis wrote:
Small point, shouldn’t that be average temperature?
Alexander Feht says:
January 9, 2012 at 3:57 am
“One of the main conclusions in the above article is as follows: “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.” Really? My reading skills are developed enough to see that Mr. Eschenbach is talking about atmospheric effects again.”
No, Alexander, try again. Willis talks about the effect that the T^4 term has when the temperature varies drastically, as in the case of the moon, compared to the effect it has when the temperature varies less, as in the case of the Earth. In the case of the drastic variation, a lower average temperature is necessary to allow the planetary body to radiate enough. It’s a mathematical thing.
steveta_uk says:
January 9, 2012 at 3:58 am
I assume you are referring to a dead man – since any fool would realize that a living man has an internal heat source that makes your comparison completely irrelevant.
Earth has an internal heat source. Which makes your comment completely irrelevant.
“It shows up the warming from atmospheric pressure nonsense for what it is, it is the equalising effect on the temperature range from a energy transporting atmsophere that raises the average temperature.”
As I understand it the gravitational pull on the mass of the atmosphere whether containing GHGs or not sets up the baseline lapse rate via compression of the atmosphere. Simply put, the solar energy passing through is slowed down by the compression due to the greater opportunity for collisions between more densely packed molecules.
In principle that is no different from the Radiative GHE because both methods achieve their heating effect by slowing down the flow of solar energy through the atmosphere. Neither scenario is a breach of the Laws of Thermodynamics.
GHGs can try to alter that gravitationally induced lapse rate but in fact changes in the non radiative processes will work as a negative system response.
Additionally, non radiative energy transfer processes can temporarily cause a different lapse rate (either steeper or shallower) from the gravitationally induced one by redistributing energy across the surface but not for long.
The strange thing is that I’m sure that in my schooldays it was the gravitational effect that was termed the Greenhouse Effect and it applied to every planet with an atmosphere whether with or without GHGs.
The term ‘Greenhouse Effect’ has only more recently become identified solely with radiative processes.
As regards the S-B equation that isn’t really relevant because it is only applied after the surface temperature has been set by the combination of the Gravitational GHE (which is fixed) and any net Radiative GHE after the negative system responses to the latter have been played through.
The gravitational effect involves the total mass of the entire planet including atmosphere whether GHGs or not whereas the radiative effect on Earth only involves water vapour plus a miniscule amount of non condensing GHGs. Thus one would expect the radiative component to be far smaller than the gravitational component.
The AGW viewpoint needs to recognise the gravitational component and properly quantify it as compared to the net (after negative system responses) effect of the radiative component and furthermore limit the figure to that attributable to the non condensing GHGs alone.
The condensing GHGs (water vapour) seem to neutralise their own effects via the negative system response of the water cycle and might well deal with the non condensing portion too.
I do not accept the proposition that there can be a positive system response to non condensing GHGs from the water cycle. No evidence has come to light supporting that assumption and with evaporation having a net cooling effect it is implausible to my mind.
Furthermore, adding non condensing GHGs to a transparent relatively non radiative atmosphere such as one containing mostly Oxygen and Nitrogen like Earth’s actually increases the ability of that atmosphere to radiate out to space. Oxygen and Nitrogen cannot do that to any significant degree so in theory they should produce an even hotter surface temperaure. Oxygen and Nitrogen conduct 100% of their energy downward because they cannot radiate to space. In contrast, non condensing GHGs radiate 50% of their energy out to space.
Willis says…”As a result, while the greenhouse effect has done the heavy lifting to get the planet up to its current temperature,…”
How much of the GHE on earth is actually due to the oceans where the residence time of energy is far far longer then any GHG? Also, although the average albedo of earth is higher then the moon’s, is it higher at laditudes where TSI is stongest?
Here a nice article showing influence of heat capacity, applied to moon.
You can get average temperatures from 169K to 291K.
http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/
The real problem is that averaging temperatures over time and/or space has no meaning. As you say, averaging T^4 and later take the fourth root gives you something more coherent.
If you consider a cold earth with a short temperature range (min and max close from each other) then average temperature is MORE than for a hot earth with a wide temperature range!
In a similar way, if you reduce the number of weather stations and keep only stations close from towns, you reduce the measured temperature range-> you increase the average temperature!
Willis, I’ve added this to my archive of informative articles. The cooling rate of the moon is useful; I’ve never seen it used to calculate what the earth temperature might be without an atmosphere. As Izen says (January 9, 2012 at 3:36 am), Science of Doom also has an article on Lunar temperatures but from a different perspective. He(she) takes into account various hypothetical heat capacities of the Moon.
Don’t you always find that when you post a straightforward article like this that you get a lot of flack from the anti-science brigade who insists on misreading or misinterpreting what has been said. Well maybe not misreading – some of them don’t appear to have read it.
But take heart from the poem:
“If you can bear to hear the words you’ve spoken, twisted by knaves to set a trap for fools….”
.
It always amazes me how rude ignorant people are! About 2 years ago I entered into a discussion on the temperature of the moon on this blog. Armed with just the albedo of the moon and the peak daytime temperature I estimated cooling rates etc to come to a temperature profile that would be consistent with the SB laws. I was met with a barrage of abuse until someone put up the actual figures which were within a couple of degrees of mine (so I was a bit lucky!).
Willis has come at it from the opposite direction and come to the same picture, as one might expect. I think he did a really good job. How anyone can argue with the Physics is a bit beyond me.
A couple of things that Willis did not emphasise.
The profile would be different it there was a liquid surface with a high thermal capacity. This would also flatten the profile just as it does on earth.
I think it is also worth clarifying the reason why the SB equation appears to favour the less dramatic profile. To make it simple: if the surface were at an average of 150K a drop of 50K would take it to 100K and an increase of 50 would take it to 200K. Because of the fourth power law the relative rates of radiation would be 1:16 at the extremes giving you an average of just over 8 over the cycle (assuming a square wave). A constant temperature of 150K would only give you relative figure of about 5 so the temperature would have to increase in order to radiate at 8.5 average. Willis makes this point but uses a formula which may not make it absolutely clear what is going on.
Thanks again Willis. I enjoyed your post as I normally do. Congratulations on not losing your temper (too much!). I would not be so patient.
If daylight on earth lasted 14.5 days, instead of 12 hours, there would be huge temperature swings on earth that I would like someone to calculate.
Similarly, if daylight on the moon lasted 12 hours instead of 14.5 days, temperature swings on the moon would be much smaller.
The different lengths of a day on the moon and on the earth probably account for more of the temperature swings than the presence or not of an atmosphere. How much more? Could someone tell me?
Fourth Root of the Mean Fourth Powers
The important thing to keep in mind here is that 304 W/m² is a measure of average energy flow. Thus the ‘characteristic temperature’ calculated by the Stefan-Boltzmann equation is not an average temperature, it is a special average based on the fact that energy flow is proportional to the fourth power (T⁴) of the absolute temperature. Thus the characteristic temperature is equivalent to a fourth root of the mean fourth powers average. This might be considered analogous to an RMS average except fourth powers are involved instead of squares. An average of this type tends to emphasize higher values and thus will yield a higher result than a simple average of surface temperatures. Of course, this only makes sense when using absolute temperatures.
markus says:
January 9, 2012 at 2:58 am
“If you are talking about the earth, as far as I know the ground heat flux is on the order of a tenth of a watt per square metre. I’ve run the numbers myself from a couple of directions, and it’s just not all that large. Even assuming that there are many more deep sea vents than are generally thought, there still isn’t enough heat coming from the inside of the planet to make much difference. If there were, we could sleep on the ground to stay warm.”
Am I OK to assume there is no heat transfer, other than the .1 Wm/2, between the oceans waters and the ocean floor, either way?
————————————————-
Markus, I think this is a good question. Of course the crustal thickness in the oceans is reduced relative to the land, therefore one would assume a higher heat flux. When discussing the affect of a small incidence of energy one must consider that energy is never lost, therfore one must consider the residence time of the material recieving that influx. The law is very simple. At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system. The residence time of the earths energy flux into the ocean depths is likely thosands of years, so I suspect it is not properly appreciated.
The darkside of the Moon does receive reflected sunlight and thermal radiation from the Earth. It is estimated to be about 0.095 W/m2 so a tiny amount, but enough to raise the Moon’s darkside temperature about 32C above the cosmic background radiation level. That still leaves lots of unaccounted for energy in the Moon’s darkside temperature.
Converting some of this data from W/m2 to Joules/second/m2 helps some in the understanding. There are accumulation rates and rates of energy loss. The numbers for the Moon are different than the Earth but not that much different.
Willis,
I love your work and the uncomplicated way you present it so even people like myself can understand what you are talking about. Anthony Watts has the best climate site on the internet because of contributors like yourself. Thanks for sharing your thoughts and ideas with the world.
I no longer bother studying long-winded theoretical articles that miss the most basic reality. The lack of real understanding, in this article, of the thermodynamics of the atmosphere is well brought out in the admission at the end:
“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.”
Those who respect the Standard Atmosphere (which my Venus/Earth temperature comparison confirmed as the equilibrium state of the atmosphere), are not so easily flummoxed by the supposedly “incredibly complex system”. The stability is clearly, even obviously, due to the weight of the atmosphere itself, in hydrostatic condition and thus exhibiting a stable, negative vertical temperature lapse rate with altitude throughout the troposphere. That stable thermal condition predominates over all other atmospheric processes, conditions or mechanisms, including the difference between night and day. Clouds and water vapor don’t rule, the hydrostatic condition does (transient and local deviations like temperature inversions notwithstanding).
I thought the lunar discussion was very good, and credit to Willis for the article. Didn’t feel it followed through though when the greenhouse subject came into the piece. Why is the average higher with greenhouse warming? Explaining that in terms of the still constant energy emission would have polished it off nicely.
gnomish says:
January 9, 2012 at 3:43 am
“but: ” the bigger the temperature swings, the lower the average temperature.”
an average is an average. extremities don’t change the average by any mathematical process.”
There are different averages here. Bigger temp swings do not change average radiation, but radiation is not proportional to temperature, it’s proportional to the 4th power of temperature.
.
distribute radiation
16 16 16 16 and you get temp 16^0.25 16^0.25 16^0.25 16^0.25 = 2 2 2 2 for an average of 2.
Take the same radiation and distribute it
31 31 1 1 and you get temps of 31^0.25 31^0.25 1^0.25 and 1^0.25 = 2.36 2.36 1 1 with and
arithmetical average of 1.68, a 16% drop in average absolute temperature .
Another point that could be addressed- in the real universe nothing acts like a black body when radiation is constantly changing, as on all rotating planets. A blackbody at earth distance would absorb and radiate 1368 watts at the equivalent of the equator at noon, for an equivalent temperature of 394 K, and would radiate at 2.7 watts- the temp of the “big bang” at night. For real bodies you need to also apply Newton’s law of heating/cooling.
Alexander Feht says:
Alexander I cannot see what you are referring to even I (uneducated save for a high school diploma) can see throughout that he is comparing the moon and the earth without an atmosphere the only atmosphere he really goes into detail about is the imaginary one he introduces in the thought experiment I think you are letting a strong dislike for Willis get in the way of reading his posts objectively.
I do not think that Lunar gravity is sufficient to hold an atmosphere. Increasing rotational speed only makes it worse. Apart from that the above thoughts will work (?).
So the moon should be absorbing more energy than the Earth..
You are absolutely right!, but it is not only about “temperature”:
Prof.Piers Corbyn uses the Sun-Moon-earth relation in his forecasting method:
http://www.weatheraction.com/
Richard Holle too:
http://research.aerology.com/aerology-analog-weather-forecasting-method/
@Willis
You state that actual measured average temperature of the moon is -77C.
That’s the predicted S-B temperature. The actual average temperature measured by Apollo missions is -23C. The raw data for the thermal conductivity experiments deployed by Apollo 15 & 17 is available somewhere on NASA website. I tracked it down once. It’s public. The gist is that at any depth in the regolith of more than about 50cm the temperature is a constant -23C. This of course is the average of the surface temperature. Both experiments were at mid-latitude locations.
@Willis
Also, broadband average albedo of the moon as measured by CERES satellite is 0.137 not the 0.11 you cite. Not a huge difference but it does make the S-B temperature several degrees higher.
@- Stephen Wilde says: January 9, 2012 at 3:23 am
“Why ignore gravitational compression of the atmosphere ?
Isn’t that what sets the adiabatic lapse rate independently of the effect of greenhouse gases ?”
The adiabatic lapse rate is set by gravitational compression and atmospheric mass independently of the surface temperature.
Which is why the temperature can vary between day and night with little variation in surface pressure or lapse rate…
It simplifies these ‘thought experiments’ if you envisage a superconducting surface that is isothermal. The oceans on Earth go some way towards this with the transfer of energy from the equator to the poles via solid/liquid/vapor phase changes and currents.
From David on January 9, 2012 at 12:26 am:
“Wow… as an aside, imagine if we could increase the Moon’s rotation to about the same as Earth’s – perhaps by a deliberate glancing meteor blow. The faster rotation would reduce the temperature swings to about the same as Earth’s, thank to the calculations above…”
Not even close. The moon reaches 75% of its maximum daytime temperature just a few hours after sunrise. The ocean is what makes the huge difference. Until one completely understands the difference between how water and rocks heat and cool nothing about the earth’s climate will be clear. Everything becomes clear after that. One of the most crucial facts to understand is that the ocean cools primarily through evaporation not radiation. If the ocean doesn’t cool by giving off longwave thermal radiation then it wont’ be warmed that way either. Therefore greenhouse gases that produce downwelling longwave radiation have little effect on the ocean. It’s an entirely different story over dry land where longwave emission is the primary means of cooling. Once you accept that all the observations start making perfect sense. Anthropogenic CO2 is largely a land-based phenomenon. The earth is largely a water world. Thus CO2 plays a more limited GHG role than it would over a world not covered by an ocean.
Willis, since comments on your “Estimating Cloud Feedback From Observations” post have long been closed, I’ll draw your attention here to this very pertinent paper, that says, amongst other things:
The violent positive feedback generation of tropical thunderstorming, derived ab initio from thermodynamics! With, as you posit, major negative feedback consequences for surface temperature increases.
Ah! The beautiful and very different male brain.
I so often notice that WWUT posts that dive headlong into abstract theories replete with abstract equations and (yikes!) math . .
rarely entice the (ahem) gentler sex to weigh in.
Vive la difference. (Are we allowed to say that anymore?)
Love you guys.
Advice to Willis: Leave discussion on Greenhouse Theory to scientists who are qualified to discuss the problem!
Willis, I am retired scientist who worked in frontline sciences for 40 years and has been privileged to be part of the team which made a major scientific discovery. During my first degree, I had to study thermodynamics and behaviour of the gasses when heated or cooled, and do the same in even more rigorous way during my PhD. However, I would not dare to write a discussion article and express my views on such a difficult topic, since I would be insulting the intelligence of the specialists in the field who really know what they are talking about. Let me ask you a very simple question: “What is it that your lengthy article is adding or correcting to what the scientists like Alan Siddons, Martin Hertzberg, Claes Johanson and Hans Schrender have very eloquently discussed in several chapters of the brilliant book called Slaying the Sky Dragon?” If the answer is ‘nothing’, then you are wasting people’s time and indulging in ‘am I cleaver amateur scientists’ – your own words in one of your presentation slides, less word ‘clever’ that I have added. If you do feel qualified, then publish critique of the book and then make the paper available for others to read.
Darko Butina, UK
Just noticed, my link is to the same paper mentioned by Bryan, above. Lots going on in it, and Tallbloke, your mod, is hosting a conversation between the authors and Gilbert, author of the “Pot-Lid” hypothesis, which integrates virial theory (Potential gravitational energy and kinetic energy trade off 1:1) with hydrostatics, which introduces a “leak” in the KE side of all the energy required to vaporize the water in a given air parcel.
Miskolczi has been strongly invited to weigh in.
David says:
January 9, 2012 at 12:26 am
An excellent post showing the importance of T^4 in regulating the temperature of the Earth (and other heavenly bodies).
I have a heavenly body and I’m hot stuff!
Willis,
As always a very thought provoking thread.
One of the main points of interest that I’ve previously not thought about (until this thread) is the idea that the swings in the Moon’s surface temperature as it rotates would be less extreme if the Moon rotated faster? If that is so, does anyone, NASA for example, have a computer model that has been validated via a scale model of an instrumented rotating sphere receiving a net surface radiative flux of ~300 W/m2? Even if the experimanetal flux had to be somewhat less than ~300 W/m2 flux of the earth/Moon surely we could conduct this type of rotating sphere in a vacuum subjected to a constant radiative flux experiment? It wouldn’t have to be that expensive an experiment surely (compared with the LHC)? Anthony are you reading this?
If this rotating sphere experiment showed that the difference in lit/unlit surface temperature extremes reduced when the speed of rotation of the sphere was increased then wouldn’t we then be in a position to have proven that when modelling the Earth we must allow for the fact that the earth rotates (which the GCMs don’t appear to do but rather instead rely on scaling the incident heat flux by the factor of 4 ratio of flat disk to sphere surface area?). When the Earth rotates presumably the ‘dark side’ is also subject to less ‘solar wind’ than the ‘lit side’?
Another (frivilous) thought. If it’s so cold on the dark side of the Moon, does that mean if we colonised the Moon that we could could have self sustaining super-conductors when on the dark side and that we could replenish our energy supplies using those high efficency PV cells whenever our habitat rotates into the sunlit side of the Moon again?
KevinUK
A fellow senile CAGW skeptic.
How terribly sexist! i’m not sure which one should be most offended, though.
By all means, step in and join Pamela Grey, Aussie, and Kim. And others. But please, no tortured obscure haiku?!? Satori is rare and not to be routinely relied on in science. 😉
Paraphrasing Hawking: Unified Field Theory is easy. The real mystery of the universe is women!
8-0 !
Willis
I always enjoy your posts.
I am not sure if this is in anyway on topic as it is something that I do not understand and hence the reason I am raising it here. A guy called Bill Illis posted this link on another thread
http://img40.imageshack.us/img40/4605/greenhousebylatitudec.png
now if I understand this it shows there is no so-called greenhouse warming in the tropics. In fact, given the amount of heat energy received in the tropics it may well be that the atmosphere acts to cool the surface in the tropics. However, given that this relatively large and thermodynamically important part of the earth does not seem to exhibit greenhouse warming, what, if anything does it say about the concept of average temperatures?
I apprecate that I may well be mixing apples and pears but this is something that has been bugging me ever since I saw it
also given your day job and other calls on your time if this is just too daft to deserve a response so be it
kind regards
Dolphinhead
>> son of mulder says:
January 9, 2012 at 3:42 am
but on a clear night the air will cool quicker than the surface. <<
The surface will cool quicker than the air, which is the reason we can have frost form at air temperatures above freezing.
Your point about a difference between air temps and surface temps is still valid.
Isn’t it great how the GHGs prevent such wide swings in temperature as experienced by the moon, except of course in deserts, where it doesn’t do a very good job.
I wonder why that is?
Cool Willis! It is nice to get a description of the effects of rotation on absorption/radiation.
This line fascinates me: In fact, anything that reduces the variations in temperature would raise the average temperature of the moon.
If I apply this to the earth, bodies of water, high RH%, low altitude, all stabilize the local temps by minimizing day/night cycling. I then have to wonder about the effects of long-term LOD Δs, even a few seconds could accumulate a gain or loss in the mean temperature over centuries.
The Moon’s orbit is inclined, so it never lacks insolation even when the Earth is between it and the sun. As a little reflection will show, since that is when the moon is full.
Even during the full lunar eclipse, the moon receives some insolation, because the Earths atmosphere refracts some light onto it.
Darko,
http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-858507
Instead of appealing to authority and claiming that Willis has nothing to add that hasn’t already been said already in ‘Slaying the Sky Dragon’ (not all of us have forked out our hard earned dosh to read it) would you actually like to point out what is wrong in Willis thread here instead? I know that will take more work, but hey that’s sort of the whole point of this blog. You appear to have the scientific training/qualificatins to do so, so why not ‘fill your boots’?
Just out of interest what was the ‘major scientific discovery’ your team discovered? Did it involve applying the scientic method? If so, how did the scientific method assist you in your ‘major discovery’? Who subsequently confirmed your discovery. Did you make all your data and analytical methods i.e. code available to third parties in order that they could confirm or rufute your discovery/findings? What difference did your ‘major discovery’ make to the way we lead our day to day lives?
I’m asking these questions after Googling ‘Darko Butina’ which returned the following link
http://www.chemomine.co.uk/DB-exec-summary.htm
If thats you I’m not sure I’d call discovering an anti-migraine drug a ‘major scientic discovery’. Did your team get a Nobel Prize for that?
KevinUK
Nice general analysis Willis, and for the most part I think you’ve pinned down the importance of the greenhouse atmosphere of Earth in terms of keeping it far warmer than it would be otherwise. Thank god you didn’t say it was gravity and the ideal gas law!
In your calculations, as other have pointed out, you’ve forgot to mention that the Earth emits more energy in LW from the surface than it receives in SW solar at the surface, which of course is not true in the case of the moon, which pretty much emits exactly back what it gets from the sun. Of course, you of all people should not forget the incredible heat sink that the ocean is for our planet and the effects the ocean has on maintaining temperature at night. I also think you might be a bit off in your calculation as to how fast the Earth would cool at night without an atmosphere. Precise Measurements have been made of the moon’s rate of cooling of the lunar surface during a lunar eclipse, and it is somewhere around 30C an hour at peak. (see http://www.diviner.ucla.edu/blog/?p=610). So, if you stripped away the Earth’s atmosphere (and took away the ocean) I think it might not cool quite this fast, but certainly faster than the 4 to 6C or so an hour that you calculate. The backradition from the greenhouse atmosphere really slows down the rate of surface cooling at night far more than you seem to calculate, as without it, there is nothing at all to slow the LW from going right back into space.
I’m not sure I agree with the Q.E.D. conclusion. To get there you have to assume an atmosphere made up of an imaginary material. I’m not convinced that a contradiction built on a imaginary basis is all that helpful.
In other words, your conclusion that this material that has mass and conductivity but no insulating property results in a perpetual motion machine does not invalidate any real world scenario as the properties of this fictitious material are what created the contradiction in the first place.
“The adiabatic lapse rate is set by gravitational compression and atmospheric mass independently of the surface temperature”.
Yes.
“Which is why the temperature can vary between day and night with little variation in surface pressure or lapse rate…”
No. Solar insolation (or lack of it) varies the temperature at the surface and upsets the adiabatic lapse rate. Convection then starts (or stops) in order to restore the adiabatic lapse rate.
But convection cannot go further than restoration of the adiabatic lapse rate because it is set independently by gravitational compression and atmospheric mass independently of temperature.
Willis, you gotta stop raking that stick along the picket fence around the yard where all those watch dogs lie. All that yapping is keeping me awake. Sorta like economists: ask ten of them a question and you’ll get twenty answers, every one of which you can take to the bank.
On the Climateect blog, back in Mid August 2011 and Oct. 16-17, I realized for the first time how the “Toy” model, ( i.e. Constant temperature, Solar insolation divided by 4, and no day-night considerations ) was so simple a model, that it was distorting a lot of thinking. In the Toy, the Greenhouse effect can only be supported by an atmosphere with GreenHouse Gases (GHGs). But in the real world, with a day and night, temperature changes, and terratons of H20 in three phases, there are so many ways to trap heat in non-thermal mechanisms.
For instance, to the question why the lunar night side is not at abolute zero, it is largely because the lunar rock is warmed in the daylight and that warmth is conducted at a decay rate several meters down. At night, that subsurface heat reservoir is conducted back to the surface at a rate probably not too different from the theoretical SB emmision rate.
Frankly, I think any serious discussion of Greenhouse Effect and Global Warming that uses an average temperature, without day or night, to be fatally flawed. The Toy model is good for one thing: to give a MAXIMUM GHG contribution to the GHE. Once you introduce all the other heat trapping mechanisms (invisible to SB physics) the contribution of GHGs to GHE must be smaller, perhaps much smaller.
The UTC presentation covers this nicely, but I will try and rephrase it : (
Willis uses the average S-B temperature of -18˚ C ( 255 K ) which is wrong because it is incorrectly calculated using a cross section instead of properly integrating over a hemisphere. The correct number is over 100˚ C colder. A black body is considered massless and rotation of a black body will in no way change the average absorption or emission.
Adding a transparent atmosphere increases the black bodies square meter emission area while at the same time the absorption sphere area stays the same.
What I am trying to say (poorly) is that with an atmosphere there will be a point above the surface radius where the absorption and emission are exactly equal to the S-B calculations. Above that altitude the temperature will be colder than the S-B number and below that point the atmosphere will be warmer than the S-B number with the surface warmer than the atmosphere.
Increasing the volume and density of the atmosphere will increase the temperature at the surface and decreasing the volume and density of the atmosphere will lower the temperature at the surface (keeping everything else constant of course). The Ideal Gas Law will calculate the temperatures accurately.
If balancing the energy budget, does the Moon “suck” or “emit” energy, relative to earth, and which would be the predominant wave lengths of such a transfer?
Replay to Kevin-UK
The only reason that I have commented on the article by Willis is that I have great respect for the blog site and have found some great articles by the specialists in their own fields. However, the field of radiation and heat balance is so complex that it really should not be open to ‘let us all join in for our thoughts’. The topics of that nature can only be properly addressed in the scientific journals, and believe me, there are lot of journals there that have not been ‘contaminated’ by the proponents of imaginary global warming. I would highly recommend to type in Google scholar names “Gerhard Kramm and Ralph Dlugi” published in Natural Sciences in 2011, which will give you a free PDF file to read. Also, I believe that Alan Siddons has contributed in the past to this blog site – that is the guy who knows this area – check the archives or Google his name. Since it seems that Willis did some reading on the topic, I think that he would contribute much more by summarising what the separate chapters in the book were saying about greenhouse gasses, rather than to venture into this field with his own ideas.
As to the major discovery that I was part of, it did drastically improve the quality of life of millions of people in the last 20 years, since this was the first effective drug for treatment of migraine. Just ask any migraine sufferer how miserable life can be. Since drug discovery is one of the most regulated sectors, each country has its own independent regulatory bodies, one in US is called FDA, and they use their own experts in field of biology, chemistry, toxicology and statistics and check all the original data that the company has produced in ten years of the research. That is the normal way that all experimentally based sciences, outside the so called ‘climate sciences’ operate.
Darko Butina, UK
ferd berple says:
January 9, 2012 at 1:49 am
Thanks, ferd. This why I said that the very close correspondence between theoretical and observed was likely in part coincidental, because there are other temperature differences. In addition to the day/night temperature differences you have the difference between the equator and the poles. However, on the moon at least, I think that they are small enough to be ignored in a first cut. Let me give you my reasoning, you tell me what you think.
• The lunar surface warms quite rapidly as soon as the sun rises, up to about 0°C. Above that temperature the warming proceeds much more slowly, and in line with the changing strength of the sun. This implies that unlike the earth, the surface roughness of the moon means that low solar angles above the horizon are no impediment to rapid warming.
• The results of the equatorial-polar difference need to be area-averaged. This greatly reduces the effect of the lowest (polar) temperatures on the average, since they represent such a small percentage of the surface area. The amount of sunlight falling on an area decreases with the cosine of the latitude. But the surface area is also falling by cos(lat), so the effect is greatly reduced from what might be expected at first glance.
Cold indeed … but bear in mind that these represent a very, very tiny fraction of the lunar surface.
My conclusion? More funds are needed to send me to the moon to take some more temperatures … “Someday, Alice … to the moon …”
My regards,
w.
Bill Illis says:
January 9, 2012 at 5:45 am
“The darkside of the Moon does receive reflected sunlight and thermal radiation from the Earth.”
I believe you mean the nearside. When the darkside is also farside, it of course does not receive radiation from the Earth. That said, a 32K increase in nearside night temperatures is highly significant, since it is close to the 43K temperature measured in the lunar polar and other areas that exist in permanent shadow. Next step: to see whether the farside has even cooler minimum temperatures in permanent shadow.
Kasuha says:
January 9, 2012 at 2:07 am
Not true. As happens on the earth, since the atmosphere actually touches the planet’s surface, heat is transferred by conduction.
w.
clivehbest says:
January 9, 2012 at 2:15 am
Very interesting, Clive, thanks. There’s another subtle effect that I always called the “moon breeze” but is more properly termed the “terminator wind”. This is a wind that accompanies the leading edge of the moonlight (the “terminator” between light and dark). Because the lighted side is warmer, a gentle breeze blows from cold to warm across the terminator. One of my favorite winds.
w.
Willis, the flat world model that is used to calculate the average does not make sense. As you correctly say the radiation changes with the 4th power of the temperature, so temperature cannot be averaged in a linear dependency, it does not make sense.
If we would have a planet with 50°K on the sunny side and 10°K on the dark side the averaged temperature would be 30°K which is not what the planet radiates.
The total radiation would be:
sigma*50**4+sigma*10**4=2*sigma*Taverage**4
If we compute the temperature of equivalent radiation will result 42°K. So 42 is the average temperature between 10 and 50 in terms of equivalent radiation (btw. “Answer to the Ultimate Question of Life, the Universe, and Everything is 42”) .
The same for the moon, if the sunny side is 60°C (333°K sic) and the cold side -175°C (98°K) the equivalent radiation temperature is 280°K which is +7°C
So the model where the sun’s radiation is divided through 4 and spread over the whole earth is leading to wrong results.
Averaging temperature as an arithmetic mean is leading to nowhere – it is a fictive value of no physical meaning.
Thanks for the posting, let me know if you see things differently or I got my calculations wrong.
Baa Humbug says:
January 9, 2012 at 2:19 am
Thanks, Baa. The moon is only eclipsed by the earth for a few hours, not days. Bear in mind that a solar eclipse from the moon is the same as a lunar eclipse from the earth … If you look at the underlying data in the cite given under Fig. 1, it shows the thermal record of such an eclipse.
I don’t understand how this idea is supposed to work. How will the atmosphere get warmer on average than the surface? Where will the energy come from to maintain it at a warmer temperature than the surface? Why will it not warm the surface, if it is warmer than the surface?
All the best,
w.
markus says:
January 9, 2012 at 2:58 am
For the purposes of first-order calculations such as we are doing here, AFAIK the answer is yes.
w.
Stephen Wilde says:
January 9, 2012 at 3:23 am
Because gravity doesn’t add energy to the atmosphere on an ongoing basis.
w.
Honorable says:
If daylight on earth lasted 14.5 days, instead of 12 hours, there would be huge temperature swings on earth that I would like someone to calculate.
Similarly, if daylight on the moon lasted 12 hours instead of 14.5 days, temperature swings on the moon would be much smaller.
The different lengths of a day on the moon and on the earth probably account for more of the temperature swings than the presence or not of an atmosphere. How much more? Could someone tell me?
—————-
Both poles are without sun for much more than 14 days and yet the temperature does not drop to anywhere near the level on the dark side of the moon. This is mainly due to the movement of warm air from the tropics which maintains the tropopause at about 200K. The existence of this relatively stable layer of the atmosphere quite close to the ice surface (it is only about 6500 metres at the poles and the ice surface is close to this height in many places) means that further cooling is not possible. So if you are asking for a calculation for the earth as it is with an atmosphere then this is an indication of what you might get. Contrary to your guess the atmosphere is the main determinant of temperature swings but the actual calculation which would have to model air flows in response to a 14 day cycle is beyond my ability and (I would suggest) a bit pointless.
Very interesting post. A few points though, what impact does the ocean have on night time cooling rates, surely it would change the earths behaviour from the moons. Secondly, how would the albedo change without cloud cover in a clear atmosphere, surely the earth would obsorb more energy and thus have a higher average temperature. Also, the earth constantly emits heat as its core cools – I a negligiable affect day to day for us, but how signficant does it become in a clear atmosphere i.e. does it determine a minimum surface temperature at night (for example higher than the moons as its cold through) which increases the average? Just some thoughts, all things that people should already have assessed before they got all hung up on co2!
Diurnal change in temperature, convection of heat and the earths thermal mass has been something that should be considered for a long time but often neglected when considering the earths climate. In climate models I understand convection is a secondary calculation after the heat trapping of GHG’s and therefore all of the warming affect is applied to GHG’s and their affect is overstated as a result.
“The surface of the Moon is colder than the surface (and lower layer of the atmosphere) of the Earth for the same reason a man without a blanket, during a cold night, is colder than a man under a blanket. ”
A man is engine creating 100 watts- in one hour: 360,000 watts.
A dead man isn’t going to be kept warm with a blanket.
gnomish says:
January 9, 2012 at 3:43 am
Thanks, gnomish, glad someone noticed.
You’re not following the reasoning. The reason the temperature average changes is that the radiation is proportional to T^4, and radiation (energy) is conserved, while temperature is not conserved. As a result, we can have different temperature averages which all have the same average radiation.
w.
Willis,
As ever an excellent and interesting post, taking a different look at things.
Your approach here may solve a different question, the Faint Early Sun Paradox. You discussed this in your previous post on the thunderstorm thermostat http://wattsupwiththat.com/2009/06/14/the-thermostat-hypothesis/ in which you summarise the question as:
“In contrast to Earth’s temperature stability, solar physics has long indicated (Gough, 1981; Bahcall et al., 2001) that 4 billion years ago the total solar irradiance was about three quarters of the current value. In early geological times, however, the earth was not correspondingly cooler. Temperature proxies such as deuterium/hydrogen ratios and 16O/18O ratios show no sign of a 30% warming of the earth over this time. Why didn’t the earth warm as the sun warmed?”
Following the approach of this post, it is not the temperature that needed to warm 30% over 4 billion years, but the average radiation, which would need to rise from about 180 W/m2 to 240 W/m2. Or via Stefan Boltzmann from about -36C to -18C.
The length of day 4 billion years ago was around 7 hours ( http://www.ptep-online.com/index_files/2009/PP-16-02.PDF) rather than 24, so with only 3.5 hours warming and 3.5 hours cooling, the diurnal range would be perhaps only around quarter that at present. So cutting 3/4 of the cooling effects of temperature swings form the ancient Earth means that it would have around 12C less ‘swing cooling’ than we’ve got today. So it would be around -24C rather than -36C.
So given a 30% dimmer sun, the ancient earth is about -24C rather than -18C today before atmospheric effects. Nothing too paradoxical about that.
Yes – This must be one key factor in resolving the Faint Sun paradox. The negative “climate feedback” from BB radiation is highly temperature asymmetric to any change in radiation DS. So DT =DS/(4*sigma*T^3). Changes to surface radiation are mostly due to 3 things – 1) Solar radiation 2) Albedo 3) Greenhouse gases. The unique feature of the Earth is that it is 70% covered by liquid water which acts to regulate the greenhouse effect AND changes Albedo through cloud formation. There is geological evidence of liquid oceans on Earth 4 billion years ago when the sun was 30% less bright. This can only be possible if somehow the Earth self-regulates its temperature through water. One simple modle of how this could occur for “water worlds” like Earth is the following.
Let’s assume that there are simple relationships both for low clouds and net greenhouse effects upon incident solar energy. Defining x = S0/342 as the normalized solar flux on a Water World relative to that incident on Earth today and taking albedo of water as 0.1 we make the following (arbitrary) assumptions.
1. Low Cloud Cover is assumed to be driven through evaporation only by solar heating: Total cloud cover (CC) is assumed to be CC = 0.4*x. The albedo for low convection clouds is taken simply as 0.5 so. This results in a planet albedo which varies as 0.1+0.2*x. This value is chosen so that the planet albedo today is 0.3 (about the same as that on Earth).
2. The net total normalised greenhouse effect g is assumed to depend inversely on x. Water evaporation and high clouds at low x yield a high g value which the decreases as higher forcing drives evaporation leading to a lower lapse rate and more direct latent heat loss to the upper atmosphere. Today g is 0.3 and the (arbitrary) proposal is that g depends inversely on increasing x so g= 0.3/x. Therefore would therefore imply that 4 billion years ago g was 0.45.
Then the global Energy balance is simply:
(0.9-0.2x)S0 = SU(1.0-0.3/x), where S0(now) is 342 watts/m2
–> SU = ((0.9-0.2x)x*342)/(1-0.3/X)
–> Tsurf(x) = T(now)*4th root(SU(x)/SU(now))
The result of this simulation can be seen here
The objective of all this is just to show HOW a water covered planet could self regulate its temperature, and it also has something to say about climate feedbacks. IPCC GCM models assume net positive feedbacks with radiative forcing. They all average around 2 watts/m2/degC If such feedbacks are assumed to be true then the Earth would have boiled off its oceans long ago as solar radiation increased by 30%. Even more contradictory is working backwards from current average global temperature of 288K
We fix the current temperature to be the observed 288K and work backwards by subtracting DT from solar forcing every million years. It is here that we see the basic problem of assuming linear positive feedback. If the temperature falls enough so that 4sigmaT^3 = F then we get a singularity. (a href=”http://clivebest.com/blog/wp-content/uploads/2011/09/hindcast.png”>You can see this calculation here.
In the context of this simple model positive feedbacks would appear to be ruled out.
Willis says: 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.
I know what you mean, but have to take issue with the terminology. GHG never provides “extra energy”, it might act as a reservoir of energy and *return* some of that energy.
On a planetary scale, even this effect is liable to be small. The thermal (energy) capacity of gas is very low compared to rock, so even if you were to take the energy accumulated in the entire atmosphere and apply it to the surface of a planet the effect would be small (and one-time … at least until it had heated up again.
As for energy transfer, from warm areas to cold, if we think about a non GH gas, the only way it can be warmed is through conduction — contact with the planetary surface (since by definition radiation passes straight through it). This is going to limit how fast the gas can acquire energy, and since the same applies to it delivering energy (physical contact and conduction) the effect is going to be small.
Then consider the volume of gas required (GH or not) to make any significant difference, and the fact that you would need a constant supply of warm gas making contact with the ground, and the velocity of the winds required to distribute sufficient energy to make a significant difference by this mechanism.
David says:
January 9, 2012 at 5:10 am
None, as I understand the greenhouse effect.
The earth albedo varies by location, by time of day, and by time of year, so it’s hard to answer your question. Where the TSI is the strongest (tropics) the albedo is part of the dynamic system keeping the earth from overheating, This means albedo also varies by temperature. It is higher where the temperature is highest, which in turn is where the TSI is greatest, so the answer to your question is generally yes, at least in the tropics.
w.
“ferd berple says:
January 9, 2012 at 1:49 am
Willis, did your calculations take into account the temperature difference between the equator and the poles? From looking at Figure 1, if you used that for your data it will not give an accurate result because it reflects an average of temperature between the equator and poles. The temperature difference between the lunar equator and poles is greater than between night and day averages.
“Most notable are the measurements of extremely cold temperatures within the permanently shadowed regions of large polar impact craters in the south polar region,” said David Paige, Diviner’s principal investigator and a UCLA professor of planetary science. “Diviner has recorded minimum daytime brightness temperatures in portions of these craters of less than -397 degrees Fahrenheit. These super-cold brightness temperatures are, to our knowledge, among the lowest that have been measured anywhere in the solar system, including the surface of Pluto.” ”
This referring to permanent shadowed craters, which can get to 20 K.
But you correct there is average surface temperature difference as you go poleward,
a difference is not as extreme. But very significant.
Harry Dale Huffman says:
January 9, 2012 at 6:22 am
Your claim is that the temperature stability of the earth is due to the weight of the atmosphere itself … riiiight … yet another universe heard from.
w.
gbaikie says:
January 9, 2012 at 10:22 am
A dead man isn’t going to be kept warm with a blanket.
______
Actually, that depends on how long he’s been dead. The decomposition of a body can generate a lot of heat, and wrapping that dead body in a blanket certainly alters the rate at which the decomposing body loses heat. This doesn’t even take into account the added activity of the maggots etc. that will want to have a nice meal and generate even more heat under that blanket.
But in comparing the Moon as a “dead” body and the Earth as a “living” body, it certainly is true that the Moon generates far less (but not zero) LW radiation than does the Earth, and furthermore, most don’t realize that the Earth’s surface generates more LW energy than it absorbs in SW energy.
Cathy says:
January 9, 2012 at 7:24 am
Thanks, Cathy. Whenever I can discuss a question without math I do so, because I figure each equation in my post loses me about 10% of the readers. So I’m working on it … but some topics, like this one, you can’t even begin to approach without math.
All the best,
w.
darkobutina says:
January 9, 2012 at 7:58 am
Advice to Darko:
1. Your credentials are of no interest to me, only your words, and your words are both personally nasty and scientifically meaningless.
2. You have not quoted a single claim that I made that you think is wrong. In fact, you have not even contradicted a thing I said … yet you claim that I am wrong. This is a very cowardly approach, but more to the point, it is unscientific. If you have a scientific objection, break it out, because up ’til now, you’re just breaking wind.
Darko, you seem to want me to write a critique of some book that I have never read … why is it that educated fools like yourself think you are qualified to tell other people what they should write about?
I write about what interests me. If you think I’m wrong, then at least have the balls to quote my words that you think are in error and tell us what you think is wrong.
And if you are just going to whine, as in your post above, Darko, then go away and whine someplace else. Don’t go away mad. Just go away. I neither need nor like whiners.
w.
“Kasuha says:
January 9, 2012 at 2:07 am
I have three points.
– IR radiation is the only way how surface temperature is transferred from solid surface to atmosphere. If a moon had atmosphere that is perfectly transparent to IR, that atmosphere would do nothing with its surface temperature.”
The air would warmed from hot surface. Any simple gas wall heater [no fans] shows that.
[Hot poisonous gas heats an heat exchange before being vented and warmed
metal {heat exchanger} heats the air in the room]
“- Earth albedo of 0.3 is partially given by clouds and other atmospheric effects. You can’t simply imagine Earth without atmosphere but having the same albedo.”
Whatever.
“- Albedo of a solid body affects not only its absorption of incoming radiation but also its release of outgoing radiation. It’s not correct to assume Earth absorbs energy according to its albedo and then cools down as fast as Moon does.”
Any heated surface regardless of color will radiate same amount of heat [roughly].
A. C. Osborn says:
January 9, 2012 at 8:40 am
Isn’t it great how the GHGs prevent such wide swings in temperature as experienced by the moon, except of course in deserts, where it doesn’t do a very good job.
I wonder why that is?
Because the most powerful green house gas is water vapor, and the relative absence of water and water vapor is exactly what makes a desert.
“* The moon hides behind the earth for a few days each cycle (is it 7 days?) where it receives no insolation at all. All emission no absorption. Is this reflected in the first graph, if so where?”
The Moon’s day and orbit around earth about 28 days. [It’s tidal locked- always has one side facing the Earth].
The Moon orbit is such that not all orbits cross earth’s shadow. Nor does the Moon’s shadow
blocks the sun on earth with every orbit [fairly rare].
Related: during lunar eclipse
http://www.diviner.ucla.edu/blog/?p=610
Measures cooling of surface when earth is blocking the sunlight.
@cal: I get your point but poles might not be such good examples to the extent that when it is day, it is the equivalent of a permanent sunrise (i.e. the sun does not do so much warming). What would happen at the equator if days lasted 14.5 days followed by an equally long night. I would not be surprised if temperature would rise to 60 degrees, but that is a pure guess.
Finally, your interesting pole argument does not clarify what would happen on the moon if days lasted 12 hours instead of 29 times longer.
Joe says:
January 9, 2012 at 9:21 am
I said nothing about the gas having no insulating properties, just that it contain no GHGs. There are a number of gases that fit the bill, so I’m not talking about any “fictitious material”. So my contradiction is built on a real basis.
w.
“gravity doesn’t add energy to the atmosphere on an ongoing basis.”
It doesn’t need to.
It just slows down the flow of solar energy through the system by increasing density at the surface to produce more opportunities for molecular collisions before the energy is released back to space.
The result is an accumulation of solar energy within the system at the surface so that a higher surface temperature can be achieved.
PaulR says:
January 9, 2012 at 11:03 am
A. C. Osborn says:
January 9, 2012 at 8:40 am
Isn’t it great how the GHGs prevent such wide swings in temperature as experienced by the moon, except of course in deserts, where it doesn’t do a very good job.
I wonder why that is?
Because the most powerful green house gas is water vapor, and the relative absence of water and water vapor is exactly what makes a desert.
______
Indeed, the absence of water vapor allows for the wide swings in temperature in a desert, but try taking out all the GH gases, such as CO2, CH4, and N2O above a desert and see what happens. Despite the belief of some, you would not get much “graviational” or “ideal gas law” warming from the nitrogen and oxygen left. The temperature drop would in fact be similar (though not identical) to what the Moon experiences during a Lunar eclipse when the Earth’s shadow passes over the sunlit moon and during the course of the eclipse the temperature drops 100K. Nitrogen and Oxygen won’t stop the LW at all.
As an example of Fourth Root of the Mean Fourth Powers averaging, the maximum and minimum temperatures on the graph of Figure 1 appear to be about + 90 and -190 which have a simple average of -50 degrees C. These are equivalent to absolute temperatures of about 363 and 83 degrees K. If these values are then raised to the fourth power we have 17,363,069,361 and 47,458,321 having an average value of 8,705,263,841 and the fourth root of this is about 305 degrees K or about 32 degrees C.
This is the type of value that an average energy flow S-B characteristic temperature would return. We do not have to use the S-B constant to convert these values to energy because that constant would be removed by division when the average is converted back to temperature.
“Here a nice article showing influence of heat capacity, applied to moon.
You can get average temperatures from 169K to 291K.
http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/
The real problem is that averaging temperatures over time and/or space has no meaning. As you say, averaging T^4 and later take the fourth root gives you something more coherent. ”
The mistake is that cooler something is, the more work can be done.
Simply if surface can only be heat to say 123 C from solar energy then
if surface is 120 C then less energy is absorb compare to radiated.
So a 10 C surface can absorb far more energy than a 120 C surface.
Something with large heat capacity could take days of sunlight to reach 10 C.
Something with low heat capacity could takes hours to get to 120 C.
Are there any forbidden to mention wavelengths?…just thinking about, as IR and visible light are not the whole spectrum.
izen says:
January 9, 2012 at 3:36 am
“But as the point is made above, that atmospheric effect of energy distribution can never raise the temperature ABOVE the S-B limit.”
This misconeption seems to be repeated time and again. It is bass ackwards. The temperature can go above the S-B “limit” (of which, there isn’t one), the radiated energy cannot in equilibrium. Kirchoff’s Law:
Alan D McIntire says:
January 9, 2012 at 6:30 am
“There are different averages here. Bigger temp swings do not change average radiation, but radiation is not proportional to temperature, it’s proportional to the 4th power of temperature.”
hi alan. i know you’re trying to be some kind of helpful, but radiation is not temperature; averaging radiation does not average temperature. measuring radiation is not the same as measuring temperature (it’s a proxy, mmk?) furthermore, the word average MEANS what it does and not something whimsical that changes with the consensual breezes.
all of which really substantiates my main point that if one can’t use the language = words that by definition have definitions – you won’t be doing logic- and reason will exceed the grasp of gobbledegook.
if you want to have temperature and radiation mean the same thing- lose one of the words.
but the distinguishing characteristic of my rant is that logic can not be done without the cognitive tools- and temperature is not heat, it’s not radiation, it’s not average radiation, it’s not a function of gravity, either.
i mean- this is the first thread i’ve even seen where somebody noticed that submerging a heat source in a conductive fluid *maybe kinda sorta might* work to refrigerate it instead of heat it or insulate it.
.
“honorable says:
January 9, 2012 at 5:31 am
If daylight on earth lasted 14.5 days, instead of 12 hours, there would be huge temperature swings on earth that I would like someone to calculate.
Similarly, if daylight on the moon lasted 12 hours instead of 14.5 days, temperature swings on the moon would be much smaller.
The different lengths of a day on the moon and on the earth probably account for more of the temperature swings than the presence or not of an atmosphere. How much more? Could someone tell me?”
It’s difficult to give simple answer, but I like challenges.
The earth does have a long nite- longer than the Moon’s.
Every year the region of earth within the arctic circle enters
a 6 month long nite.
DirkH says:
January 9, 2012 at 5:00 am
Willis talks about the effect that the T^4 term has when the temperature varies drastically, as in the case of the moon, compared to the effect it has when the temperature varies less, as in the case of the Earth. In the case of the drastic variation, a lower average temperature is necessary to allow the planetary body to radiate enough. It’s a mathematical thing.
Oh, it’s “a mathematical thing”… Really. So, what’s new or interesting about this? Anyone who ever saw Boltzmann’s formula or knows what the fourth-power means, understands that the loss of heat by radiation quickly increases with the increase of temperature. So what? Why are we being treated as preschoolers here?
I don’t agree with Darko — popularizing science, even doubting the established science, is good and useful. Amateurs made more breakthroughs in science than professors and doctors ever dreamed of. But this particular amateur doesn’t explain anything new, complex or interesting.
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). Thermal insulation is just that, an insulation — it keeps cold out even if there is no heater inside. Being a Siberian, I remember how much longer after sunset the warmer air is contained in a log cabin covered by deep snow (without any fire or people heating it yet), compared to any structure open to elements.
I think you should just have gone with: The Moon is a Harsh Mistress
Your statement willis:
“But a transparent atmosphere could never raise the moon’s temperature above the S-B blackbody temperature of half a degree Celsius.” need to be clarified.
This is why:
A transparent atmosphere would be heated by convection with moon surface. The warm gas will raise due to density decrease. And heat the entire atmosphere by convection.
And what will a transparent moon atmosphere be cooled by?
Only by convection with the same moon surface. The problem is that a cooling atmosphere from below get stable. That is an inversion. A inversion will prevent warm air to decend to the surface and be cooled by surface radiation. Next day will the same surface heat a little bit warmer atmosphere than the day before.
It will not take long until this ideal moon atmosphere would reach a average temperature nearly as warm as the moon highest temperature. How close depend on the atmospheres heat capacity and length of day and night.
This only because the surface heat and cool an atmosphere, which have to obey the gas laws.
When You mention the moon temperature do You mean the moon surface temperature or the temperature of a fictive moon atmosphere?
Do I have to remind You that we do measure air=atmosphere temperature on earth. Not surface temperatures. They are very often close to each other but that is mainly due to convection=wind.
So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.
You betcha, as I showed slightly more quantitatively as the reposted topic of another thread. One that is actually almost directly applicable to the moon — much more so than the Earth.
The greater the temperature inhomogeneity over the surface, the greater the cooling, because T^4 goes up with T faster than it goes down with T. It’s really that simple.
rgb
“The result is an accumulation of solar energy within the system at the surface so that a higher surface temperature can be achieved.”
stephen – which is heavier, a pound of feathers or a pound of lead?
stephen – which is hotter?
stephen – btw- how thick is this surface? how much depth can you pull out of this superficiality? how profound can shallow be?
sure would be nice if people used words – which have definitions – so everything isn’t confabulated into one long loud munch painting.
a surface has no thickness, gravity has no temperature, pressure has no volume and sounds without definitions are grunts, not words.
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.
Sorry, I’m reading and commenting as I go. This is consistent with observations in the very dry, windless desert. To quote the Wikipedia article on Deserts, in hot deserts the daytime peak temperature can be 45C and the minimum temperature right before dawn can be 0C or even cooler. That is dead on your estimate, allowing for the fact that even in the desert there is stratospheric water vapor acting as a GHG to block a little of the heat — along with CO_2 of course.
Your estimate is of enormous consequence, in other words. It suggests that there is almost no greenhouse effect active over hot, dry deserts. Not CO_2 GH. Not H_2O GH. Not CH4 GH. They radiatively cool almost as fast as the dark side of the moon!
Almost no is not the same as no, but this points out the enormous importance of water vapor as a GHG. I’m guessing that it forms a critical aspect of glacial epoch stability — when all that water is bound up in glaciers and it is really cold all the time near the poles, I’m betting that it is always dry all of the time in all of the upper latitudes. Dry air just doesn’t trap a lot of heat at night, no matter what the hell the CO_2 levels are.
I’ve suggested — repeatedly — that direct measurements of the rates of radiative heat loss in the middle of geographically large deserts would provide us with more or less a direct measurement of the actual heat trapping of the greenhouse effect in the absence of water vapor. Your back of the envelope calculation shows why it is so important. We’re talking about a tiny effect compared to raw vacuum, not a big effect, and that includes the cooling of the thermal ballast of the lower atmosphere as well.
rgb
“I do not think that Lunar gravity is sufficient to hold an atmosphere. Increasing rotational speed only makes it worse. Apart from that the above thoughts will work (?).”
I have strong bias towards not changing the Moon [or Mars]. Because i think the Moon as it
is better in many ways than earth. But with that said.
If want “milder climate” on the Moon, you could use water.
First water is very abundant in our solar system, even compared to our water world Earth.
There easily 1000 times more water than we have in earth oceans, which is “accessible”.
The Moon itself is very water poor- compared to Earth. As is Mars. But the Moon does
have concentrated quantities of water in it’s permanently shadowed polar craters. This
amount is on order of billions of tonnes. Which is tiny compare to our amount of water.
A billion tonnes is 1 cubic kilometer- so that amount is fair size lake. Or each square kilometer
of deep ocean has about 3 cubic km of water and earth has 510 million square miles and 70%
of area is deep ocean.
So, anyhow to answer question 10 meter depth of water is roughly equal our atmosphere. So
one could pools of waters 10 meter deep and these wouldn’t varying much in day and nite cycles.
Let’s consider two superconducting spheres, one with a radius of one and the second with a radius of two. They are the exact same distance from the sun and they have the identical average solar insolation per square meter of 100 W per square meter and they radiate at 100 W per square meter. Their surface temperatures are identical.
Now let’s place the smaller sphere inside the larger sphere. The outer sphere 4π(2R)^2 has 4 times the area of the inner sphere. As a result the outer sphere has no change in insolation or radiation. The inner sphere now has 400 w per square meter insolation and correspondingly higher temperature.
A planets atmosphere acts in effect like like an outer conducting sphere, The bigger the atmosphere the higher the internal spheres temperature. It is just that simple.
Overall, good article, but you are off by a factor of two in twelve hour cooling of the middle of a hot desert. It often is 45C, around 4C/hour. Not a lot of greenhouse effect when there isn’t water vapor to help.
The rest of the article is dead on the money. The earth almost certainly self-organizes to increase the efficiency of heat loss as it warms by increasing the (surface integrated) delta T. This adds to its overall stability (provides negative feedback) — within limits.
rgb
honorable says:
January 9, 2012 at 11:14 am
@cal: I get your point but poles might not be such good examples to the extent that when it is day, it is the equivalent of a permanent sunrise (i.e. the sun does not do so much warming). What would happen at the equator if days lasted 14.5 days followed by an equally long night. I would not be surprised if temperature would rise to 60 degrees, but that is a pure guess.
Finally, your interesting pole argument does not clarify what would happen on the moon if days lasted 12 hours instead of 29 times longer.
Your second point is easier to answer than the first where, as I tried to explain, the enormous winds that would inevitably sweep round the globe would be difficult to model. However it would be pretty safe to say that the presence of the atmosphere would dramatically reduce the temperature excursion compared to the moon.
In the second case one can more easily estimate the effect since the moon is a much simpler system. One can see from the graphs that Willis has provided that the temperature drops precipitously once the day ends. Nothing changes in this respect if the day is only 12 earth hours long. However, after 12 hours, just before it can reach its current absolute mimimum it would start to warm again. It is difficult to judge exactly how low the temperature would reach before it started to rise again but my guess is that it would only be about 20 degrees more than it is with a 14 day night. The warming cycle is more difficult to guess since the shape of the upward curve on the temperature plot is due to the fact that the sun is oblique for several days whereas in the 12 hours scenario it would only be a couple of hours so one cannot just read the rate of rise off the graph. My guess would be that it would reach a temperature at least as high as the current moon surface after 3 or 4 days and then decline. But I could be wrong. One could argue that at peak sun the surface has to radiate as much as it radiates at peak sun on 14 earth day cycle therefore the temperature has to be the same. The reason I have gone for the slightly lower temperature is that the surface would start warmer so would be radiating more in the early hours. There is also a thermal capacity issue which I can only guess at. If I am right this would make the total excursion between -160 to +60. Still much larger than the earth.
Still not quite sure why you are asking the question!
If we give the moon an atmosphere that does not have greenhouse gas then the atmosphere will have zero radiation or emissions and the moons surface will radiate all the energy received from the sun at the top of atmosphere into space.The atmosphere contains some of the received energy from the sun but it cannot radiate it to space, my point is that we can’t ignore the non-radiation of the atmosphere in calculating the amount of radiation sent out into space.The energy radiated at the planet surface would be some average of the emissions at the surface and that at the atmosphere which would be lower than the amount received at the top of the atmosphere until an equilibrium at a higher temperature was reached.
Willis, I too appreciate the choice of Title. The Moon is a Harsh Mistress is a favorite SF and Political novel.
In that it is a story about:
Ecology, Politics, Economic Freedom, Manners,
And a growing collection of people, organized by computer, attempting to overthrow the “Authority”, it is a story that well suits the “Battle in support of CAGW Skepticism.”
It has a place on my “Read many times” bookshelf. TANSTAAFL!
KevinUK says:
…
“One of the main points of interest that I’ve previously not thought about (until this thread) is the idea that the swings in the Moon’s surface temperature as it rotates would be less extreme if the Moon rotated faster?
…
Another (frivilous) thought. If it’s so cold on the dark side of the Moon, does that mean if we colonised the Moon that we could could have self sustaining super-conductors when on the dark side and that we could replenish our energy supplies.”
Any rotation of planet in terms of heat can thought of as means of transporting heat to the night side of planet. So superconductor transporting electricity or heat to the nite side is similar.
As for the nite side of the Moon. It’s very easy to warm a building in a vacuum. It is also very easy to keep a building cool. So lunatics aren’t going wasting a huge amount energy, that earthlings do on heating or cooling their homes. With LED lighting there would not be much energy use to keep the lights on.
So, we left with industrial processes that need a lot energy. One could go to location on the Moon and get constant solar energy [high elevation] at lunar poles.
One also can use nuclear energy on the Moon. The Moon is perfectly safe place to store radioactive waste. And there is little reason to have much in terms of containment, a complete nuclear meltdown is not important other than wrecking the facility.
ferd berple says:
January 9, 2012 at 1:23 am
Willis Eschenbach says:
January 9, 2012 at 12:55 am
ferd, the N2 is the most unlike the others because the line strength is many, many orders of magnitude weaker than that of the others.
Perhaps you misread the reference? From what I see, N2 line strength is 10-28, CO2 is 10-23, which is 5 orders of magnitude. However, N2 has 10 orders of magnitude wider spectrum (600 cm-1 versus 50 cm-1). In addition, there are 4 orders of magnitude more N2 in the atmosphere than CO2. So, on this basis it is hard to see that N2 absorbs/radiates significantly less than CO2.
No Fred, Willis is right CO2 is 10 orders of magnitude greater than N2 as I have pointed out here several times. Your mistake is to compare one rather weak CO2 band with the solitary band for N2. In fact there are many CO2 bands, between 4 and 5 microns there are ~21,000 lines with the strongest band at ~10-18, by contrast there are ~100 lines in the same range due to N2, peaking around 10-28. There are ~1000 lines due to H2O in the same range up to 10-21. That wavelength range isn’t particularly relevant to GHE however because it’s not in the Earth’s emission band, between 12 and 18 microns CO2 has ~20,000 lines peaking around 10-19, H2O has ~500 lines up to 10-21, N2 has no lines at all!
In contrast to CO2, H2O line strength is 10-19 which if 4 orders of magnitude stronger than CO2. As well it has a much, much wider spectrum than CO2. The absorption strength and spectra of water so overwhelms CO2 as to make it CO2 a joke when you consider the amount of H2O in the atmosphere as compared to CO2.
The joke appears to be on you.
From gnomish on January 9, 2012 at 11:55 am:
In a federally funded study of the effects of climate change on the quality of feathers and lead emissions (per the funding request), from a fixed height of 6 feet, 100% of survey participants agreed that a pound of lead dropped on their bare heads was heavier than a pound of feathers. From 2 meters, 100% found a kilogram of lead to be heavier than a kilogram of feathers.
This study remains unpublished due to incompleteness of the American/metric equivalence portion, as all participants refused to be hit with slugs of lead.
Interesting post. Incidentally I know the moon orbits and keeps one face to the earth. But interestingly the idea that it has an axial rotation as opposed to just an orbit is widespread, even to a report on the BBC a few days ago. Maybe I should not be surprised by the BBC, but otherwise it’s a common mistake even in scientific discussion.
“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).”
Plants do generate heat, as does all life [including reptiles or microbes].
Though mammals generally run much hotter.
Usually “cold mistresses” are deceitful as they really are engaged in a urgent search for heat 🙂
Willis said:
“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.”
Yes, but the atmosphere is not transparent, and I don’t know anyone who says it is (looks like a straw man to me). Remember that the “S-B theoretical temperature of approx. -18 C” represents that average equilibrium radiation coming from about 5,000 KM! NOT FROM THE SURFACE AND NOT THROUGH TRANSPARENT GASES. Adjusting for lapse rate, that makes the surface about 15 C. That can be called a greenhouse effect, if one prefers to do so, but I don’t think that it is mostly due to radiation.
I don’t have near enough time to read all the comments you get and give, but to my knowledge you (or nobody else) still have not countered the points made over and over again by Harry Dale Huffman and the authors of a previous post, points that clearly demonstrate that atmospheres on planets all have positive effects on temperature and that those effects are not affected much, if at all, by the types and amounts of gases in those atmospheres.
Willis-
Thanks for another thought provoking post.
Unlike Brian H @ January 9, 2012 at 8:31am, I loved the Haiku. Perhaps if Brian H were more widely read, he would have known that the Haiku is to the Japanese as the sonnet is to the English, and that far from being obscure, Basho is one the most famous of the Haiku poets.
Brian H says:
January 9, 2012 at 8:31 am
old engineer says:
January 9, 2012 at 1:44 pm
Brian, there were no tortured obscure haiku in my post, in fact no haiku were harmed in the writing of the post. Instead, there was a famous haiku by a world renowned poet.
Don’t like it?
Sorry, de gustibus et coloribus et cetera …
w.
Joules Verne: One of the most crucial facts to understand is that the ocean cools primarily through evaporation not radiation. If the ocean doesn’t cool by giving off longwave thermal radiation then it wont’ be warmed that way either. Therefore greenhouse gases that produce downwelling longwave radiation have little effect on the ocean.
What exactly does the downwelling longwave radiation do to the surface of the water?
jae says:
January 9, 2012 at 1:37 pm
I sincerely hope that you are kidding. Near as I can tell Huffman has added nothing but pseudo-science to the mix, unless I missed something, always possible. He did say that the amazing thermal stability of the earth is due to the mass of the atmosphere, a claim so preposterous it made my ears ring. The mass of the atmosphere allows the planet to change temperature daily by tens of degrees. How on earth could just plain mass keep the Earths temperature stable to a half degree per century? jae, you’ve got to get more skeptical here. Huffman’s claims don’t pass the laugh test.
So give me Huffman’s claims in three sentences, just reading his stuff makes my head hurt. What has he said and why do you think it is important?
w.
“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.”
The -18 C surface is at roughly 40km above the Earth. Your assumption that the temperature calculated using the Stefan-Boltzmann and the Sun’s flux at the TOA is the temperature of the surface you happen to standing on is false.
Willis Eschenbach
“So give me Huffman’s claims in three sentences, just reading his stuff makes my head hurt. What has he said and why is it important?”
I believe Huffman is overstating his point.
But I think gravity does have an affect.
Gravity from point of view can have all the effect in this universe.
Energy is all about gravity, as in, no gravity basically no energy in this universe.
But in terms affects on greenhouse effect, I think has some effect.
But this effect isn’t quantified enough, as far as I have seen.
An obvious problem is what meant.
But keeping it simple, gravity affect how fast gas molecules would travel.
And gravity is buoyancy. No gravity- no mixing of gas/fluids.
More gravity more mixing.
So you can’t simply ignore gravity- not if you want
a complete theory.
So what is preventing the Earth from reaching 90C each day? Albedo, the shorter day, what exactly is cooling our planet?
Septic Matthew says:
January 9, 2012 at 1:54 pm
Matthew, please don’t encourage the Joules Verne type of lunacy. There is a school of thought out there that claims that longwave radiation can’t heat the ocean. They go so far as Joules Verne, who makes the further idiotic claim that the ocean
This, in spite of the fact that satellites can measure the ocean’s surface temperature from space by measuring the very longwave radiation from the ocean that this credulous gentleman says does not exist.
Not one of those folks can answer a simple question: where does the energy come from to keep the ocean liquid, if not from the downwelling infrared energy from the GHGs?
There’s nowhere near enough solar energy hitting the ocean to keep the ocean from freezing. Global average solar radiation hitting the surface is only about 170 W/m2. We know the global average upwelling longwave radiation from the surface is about 390 W/m2, and solar only contributes about 170 W/m2 … so what is providing the rest of the energy needed to keep the ocean from freezing? Not one advocate of that crazy idea (that the ocean can’t absorb infrared) can answer that question.
In the face of such willful blindness, any response is a bad idea. Or as they say, DFFT … don’t feed the trolls. This thread is not about oceanic IR absorption, and I don’t want to get lost in that swamp again.
w.
Physics demands, requires and wants equilibrium. That means that the heated side of the moon, once in the shade, will expend its heat until it is the same temperature as the body around it ie. space, ie. about 3 degrees Kelvin. These laws are universal across the whole… universe. The moon has no internal source of heat to disrupt this.
The Earth is incomparable for a variety of reasons. It has internal heat, it has an atmosphere that acts a a restraint on both ends of the scale, and it has oceans that are a ‘dampener’ on heat – oceans take far longer to warm up and cool down than land does.
So I really don’t see where this well-meaning article is trying to take us. The moon and earth are fundamentally different. They cannot be compared. The moon is a cold rock that will forever be trying to radiate its excess sunny side heat into space according to the laws of physics so that everything is the same, with no impediment provided by atmosphere, oceans etc. End of story.
Oh, and Harry Dale Huffman isn’t the one claiming heating by atmospheric mass, that sounds like Nikolov Zeller. Harry emphatically states that is wrong (perpetual motion like) in the link below, and also sums up his argument fairly succinctly. You asked for three sentences though, so here’s three from that comment.
“Keeping it simple, the atmospheres must be like sponges, or empty bowls, with the same structure (hydrostatic lapse rate), filled with energy by the incident solar radiation to their capacity to hold that energy. In short, compressing the lower atmosphere doesn’t heat it, it merely allows it to retain more heat energy per volume than the lower-pressure levels above. All of the energy is provided by the Sun. The pressure distribution simply dictates vertical temperature distribution, which constitutes the structure, or energy-retaining form, of the “bowl” I likened the atmosphere to.”
http://theendofthemystery.blogspot.com.au/2010/11/venus-no-greenhouse-effect.html?showComment=1325125898177#c5729196180038388134
gbaikie says:
January 9, 2012 at 2:14 pm (Edit)
Could someone please translate this for me? That’s worse than Huffman himself. There is no substance in there, nothing to measure, nothing to grab ahold of. Yes, we can’t ignore gravity … so? Where’s the meat in this sandwich?
There are a lot of people out there who handwave about how by some mysterious mechanism, a perfectly transparent, GHG free atmosphere can warm the planet above it’s S-B temperature. Can’t happen. Violates the Laws of Thermodynamics. Their claim is that by their magic method you could end up with a surface that radiates more than it absorbs. No way for that to occur, that’s perpetual motion.
Notice that none of them can explain the exact mechanism by which it’s supposed to occur. Oh, they’ll repeat the word “gravity” lots of times, and talk a lot about lapse rates and the like. But without GHGs, the only thing radiating is the surface, and the atmosphere is transparent. How can the surface possibly radiate more to space than it absorbs from the sun? No way! Sheesh, this isn’t rocket science.
There is nothing mysterious about the greenhouse effect. I don’t understand why people think it’s not real. Read my posts “The Steel Greenhouse” and its sequel, “People Living In Glass Planets“. A planetary greenhouse doesn’t require GHGs or an atmosphere, you can build the whole thing out of steel.
w.
The main difference between Earth and Moon is the core heat. The amount of thermal energy under the surface of the Earth (ie down to the core) is several orders of magnitude greater than that in the oceans, let alone the atmosphere. This has a stabilising effect on Earth’s temperatures, especially between day and night and between summer and winter. I have explained this in far more detail on this page of my site: http://climate-change-theory.com/explanation.html
Willis,
Not that you need my help, but I agree with pretty much 100% of what you have said here. I haven’t had time to read every word carefully in the thread, but everything I have read from you seems spot on.
Willis: This, in spite of the fact that satellites can measure the ocean’s surface temperature from space by measuring the very longwave radiation from the ocean that this credulous gentleman says does not exist.
Thank you for the response.
“nano pope says:
January 9, 2012 at 2:19 pm
So what is preventing the Earth from reaching 90C each day? Albedo, the shorter day, what exactly is cooling our planet?”
Might start with what is warming our planet.
The sun has a max heat at your distance from it.
The moon at same distance can reach more than 90 C.
One way say it is moon has 1360 watts per square meter of sunlight.
And earth below our atmosphere has about 1000 watts per square meter.
Moon reaches a surface temperature of about 123 C
The earth reach a surface temperature of about 180 F [ 82 C]
Convert 123 and 82 into Kelvin:
396 K and 355 K
cube 396 times it by .0000000567
and get 1394 watts per square meter of sunlight.
cube 355 K times it by .0000000567
And get 900 watts per sq meter.
Earth get more than 900 watts, but air convection takes heat away from surface.
And if you wonder why I say earth can get to 82 C, I am talking about surface
temperature not air temperature in the shade- which how we measure temperature
on earth.
90 C or 363 K
is 987.4 watts per square. If stop most of the air convection losses, you should
be able to get 90 C on the earth surface at noon and a clear sky. higher elevation
should make this easier.
The other element is earth radiate heat gained into space.
So what stops us from getting 90 C air temperature?
Well that would generally mean the surface temperature was as hot or
hotter, and therefore the whole planet would radiating roughly 4 times as much energy
as it got from the sun. It can radiate the same energy as it received from the
sun.
So on small scale if limit conviction and conduction of heat during noon on clear day
you get 90 C. But it’s not noon everywhere nor is it daylight.
In other words the heat is spread out. And though it might possible to get 90 C
not possible on earth to get 120 C from sunlight regardless of how stop heat loss-
need more solar power per square meter [not have 300 or so watts blocked by
our atmosphere].
You can’t have a GHG-free atmosphere, not on Earth anyway. CO2 is a GHG and is a natural component of all that goes on, given off by plants in the dark. Likewise methane, naturally emitted by decomposition.
Sorry if I am repeating other’s points, I don’t have time to read them all.
1) thank you for demonstrating why it is STUPID to compute an average temperature of the earth by dividing the incoming energy over the whole surface. Can we throw out the ridiculous comparison between a bare ball and the earth as we know it now??
2) emissivity is also important. The earth has a much lower emissivity due to the atmosphere, about .7-.8 including the earth underneath it!!!
3) you totally ignored conductivity and thermal mass of the totally transparent atmosphere. Effect will be low admittedly, but will still be there. you also completely ignored the the thermal flows in the surface.
4) I believe vacuum is a great insulator don’t you?? On the moon we have large areas convered by dust and VACUUM!!! That’s right, there is no air to fill the spaces in the dust particles so we end up with a very nice INSULATOR compared to the ground down here!!
5) Sloppy.
Willis Eschenbach says:
January 9, 2012 at 10:30 am
David says:
January 9, 2012 at 5:10 am
Willis says…
”As a result, while the greenhouse effect has done the heavy lifting to get the planet up to its current temperature,…”
How much of the GHE on earth is actually due to the oceans where the residence time of energy is far far longer then any GHG?
None, as I understand the greenhouse effect.
——————————————————————————————————–
Willis, please understand that of course I do not literally mean GHE, when I refer to the oceans; except in the context of thermal capacity. At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system. (You may henceforth refer to this as David’s law. (-;) There is a fairly exact correlation between residence time of energy and thermal capacity. As the ocean thermal capacity is thosands of times that of the atmosphere, it appears logical that it is a more effective GHLiquid, then any GHG; your thoughts in this regard are welcome aand requested.
Also, although the average albedo of earth is higher then the moon’s, is it higher at laditudes where TSI is stongest?
Willis responds…”The earth albedo varies by location, by time of day, and by time of year, so it’s hard to answer your question. Where the TSI is the strongest (tropics) the albedo is part of the dynamic system keeping the earth from overheating, This means albedo also varies by temperature. It is higher where the temperature is highest, which in turn is where the TSI is greatest, so the answer to your question is generally yes, at least in the tropics.” W. ———————————————————-
At first glance this does not appear logical to me. In general the oceans are a blackbody, absorbing whatever radiation reaches the surface with little reflectivity. The NH has a great deal of landmass north of the tropics, as well as year round snow and ice in the artic, antarctic as well as tremendous winter albedo beyond year round ice. Additionally the incident angle of sunligh creates ever greater reflectance as one moves further from the tropics. A further factor is the poles appear to have a great deal of consistent.cloud cover as I look at the global map on the right side of WUWT home page. For these reasons I would have to see actual meauserment to accept your assertion here, as I suspect that the tropics. especially the southern tropics have the lowest albedo as well as the greatest TSI, especially in January when the earth is thee million miles closer to the sun and TSI is close to 100 W/m2 greater then in July. Your thought here are appeciated as well.
clivebest had the basic idea for feedbacks, but got them backwards. Current albedo is 0.3
so we get (1- 0.3)* 342 watts =239.4 watts. With a sun 70% as luminous as at present, and with a surface covered with oceans and no clouds, we would have received
0.7 (sun)* 1 (no clouds)* 324 watts, same as now. albedo wasn’t 0.45 back then, it was closer to zero.
Alan – that’s right. Albedo reduces for lower incident radiation . As the Earth’s ocean surfaces get heated, the Earth starts to “swet” – producing clouds which reduces the Albedo. The second assumed mechanism at work is that the H2O greenhouse effect increases at lower solar radiation levels ( by higher humidity in upper atmosphere). As more convection clouds form – this leads to more rain out from the atmosphere and a drop in upper atmosphere humidity.
The overall idea is that there is a balance point which is a play-off between the two effects Albedo and GHG. Water covered planets have an infinite sink of water available to maintain this balance over a wide range of solar radiative forcing levels. This has nothing to do with CO2 which is assumed to be absent in the atmosphere.
Is the reference value of 0.11 a visible spatial average or full spectrum, spatial average?
“GHG free atmosphere can warm the planet above it’s S-B temperature. Can’t happen. Violates the Laws of Thermodynamics. ”
Well, I am not convince any atmosphere, roughly with 1 atm with any quantity of greenhouse gases can warm above it’s S-B temperature. You are apparently assuming this could occur.
I am not.
And don’t think Huffman idea can do this either.
Or anything can do this.
Unless warming include huge asteroid strikes or huge, huge super volcano. Supernovas, and huge solar output, also might work.
Great post Willis. I really enjoy reading all your articles.
I’d like to take the no GHG atmosphere thought experiment a bit further. Since the atmosphere cannot radiate, it can only exchange heat with the surface. Eventually the upper atmosphere would warm through conduction. The lapse rate would be near zero which would shut down any vertical convective heat transport since the warming/cooling due to adiabatic expansion would almost instantly kill any vertical motion. It would be the ultimate in super stable atmospheres. There would be no Hadley cells to transport heat from the equator to the poles. There could be some horizontal transport due to surface temperature differences (day – night and equator – pole) but I would guess that they would be very weak without any vertical motion to drive it. I imagine that the atmosphere, particularly up high would have a very uniform temperature over the entire globe.
Is it the green house gases that give us our weather? With GHG back into our thought experiment atmosphere, there is cooling of the upper atmosphere due to outgoing long wave radiation. This give rise to a vertical temperature gradient (lapse rate) which can drive convective transport. I believe convective transport is the source of almost all of our weather.
As a paraglider pilot who likes good thermals, I certainly appreciate green house gases!
Regards
Peter Spear
Very interesting article with many good points.
Only problem i see is that the Earth albedo is for a great part clouds and with no GHG gasses you would not have clouds. So the albedo would be much smaller hence much more energy absorbed, and the SB average temperature would be more like -5 to 0C.
There is a small difference that have big impact Willis. You write:
“But without GHGs, the only thing radiating is the surface, and the atmosphere is transparent. How can the surface possibly radiate more to space than it absorbs from the sun?”
You are right in that statement regarding the surface. But the ATMOSPHERE will be warmer than average temperature since it is heated more efficient than cooled by convection. Surface will radiate as much as before but with less temperature swing due to the heat capacity of the atmosphere. The difference is where to measure the temperature, surface or atmosphere.
Doug Cotton says:
January 9, 2012 at 3:02 pm
Thanks, Doug. Core heat provides only tenths of a watt per square metre at the surface. That’s far too small to stabilize anything. If core heat were a major player we wouldn’t have ice and snow staying on the ground, they’d melt from the bottom.
w.
John Billings says:
January 9, 2012 at 3:16 pm
John, that’s why this is what is called a “thought experiment”. I’m trying to see the various reasons why the earth stays so much warmer than its S-B temperature. Part of that is to disentangle the warming due to the atmosphere itself, from warming due to atmospheric GHGs.
w.
Genghis says: January 9, 2012 at 12:04 pm: “Let’s consider two superconducting spheres … ”
Genghis, you have some good thoughts, but you also make a couple mistakes. With the two concentric shells (separated by a vacuum), you will find that the temperature will be the same for both shells. If the outer shell could transfer energy via radiation to the inner shell and warm it up, that would violate the 2nd law of thermodynamics.
Most of the radiation from the inside of the outer shell will not hit the inner sphere, but will indeed return to some other part of the outer shell itself. If you look more carefully, I am sure you will find both surfaces do indeed radiate 100 W/m^w and are at the same temperature.
(If you add an atmosphere, then the situation could change a bit, and the inner surface could indeed be warmer than the outer shell by an amount related to the lapse rate.)
gbaikie says:
January 9, 2012 at 3:28 pm
The earth is well above the S-B temperature. You get your own opinions, but not your own facts.
w.
[I]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). Thermal insulation is just that, an insulation — it keeps cold out even if there is no heater inside. Being a Siberian, I remember how much longer after sunset the warmer air is contained in a log cabin covered by deep snow (without any fire or people heating it yet), compared to any structure open to elements.[/I]
Some plants can insulate themselves with snow or “blankets”, but that is rare on the surface. They can come back from roots that are below the freeze line in the soil using it as a blanket. The mechanism really is about anti-freeze in the sap. Conifers have the best version of anti-freeze running in the plant family, which explains their hardiness. It is all about keeping ice crystals from damaging the cells.
Dead men don’t care and the embalming fluid might work as an anti-freeze though…
Hardly any scientists believe that the earth’s core and mantle is producing any energy, other than a modest amount of fission. Unlike a sun, no nuclear fusion can occur. However, I am aware of a theory that does claim creation of energy in planetary bodies, above and beyond any nuclear fission contribution. This theory – subquantum kinetics – has made a number of verified predictions (according to it’s author) See http://etheric.com/LaViolette/Predict2.html. The extension of a mass-luminosity curve to planets, was discovered by Laviolette, and is consistent with his theory of “genic energy”. Not mainstream science, but consider the fact that a Greek physicists says that LaViolette deserves 2 Nobel prizes for his mass-luminosity discovery, and it makes you wonder.
Lance Endersbee has shown that a 21 year moving average of Sea Surface Temperature vs. CO2 at Mauna Loa has a correlation of .9959, while a 1 year moving average is poorly correlated. Meanwhile, Nir Shaviv has shown that solar activity considered as modulated by cloud cover (a function of solar magnetic effects) correlates very well with the short term change in SST (see http://www.sciencebits.com/calorimeter).
This huge gap in timescale between Shaviv’s results and Endersbee’s 21 year moving average results is a mystery, I believe. That’s where LaViolette’s theory may come in. Although I’m not a domain expert, I will say that I don’t expect the heating that Shaviv has studied to take decades to ‘average out’ via various thermal transport processes. However, in LaViolette’s theory, a new source of energy (which he calls ‘genic energy’) is being created WITHIN planetary and solar bodies. Such heat could conceivably take years to work it’s way to the surface of the planet.
Meanwhile, subquantum kinetics has, as it’s governing equations, reaction diffusion equations such that one or more of the ‘substrates’ is electric potential. I will speculatively wonder whether or not the sun’s magnetic effects not only modulate incident radiation on the Earth, but also modulate internal genic energy production within the earth, but via the electric potential substrate which is fundamental in subquantum mechanics.
Some more info on LaViolette’s theories here: http://blog.hasslberger.com/docs/PioneerEffect.pdf . He also has a book out on the subject, now in its 3rd edition. LaViolette has a bachelor’s degree in physics, but a Ph.D. in systems theory. His father was a Ph.D. physicist, who reviewed his work.
See also “Comparison of Subquantum Kinetics to Conventional Physics and Astronomy” at http://www.etheric.com/LaVioletteBooks/SQK-c.html.
G’morning W.
Being a shift worker my responses can be sporadic.
I’ll do my best to express my view. The last time I told my late dad “but you don’t understand” I got a very solid clip across the ear with the response “then explain yourself clearly” lol
There is no “extra” energy. But there is energy “locked away” by the atmosphere unable to radiate it away.
At our hypo planet with non-GHg atmo, the highest insolation is at the equator, tapering down towards the poles.
The air nearest the surface at this point will warm, rise and spread only to be replaced by cooler air from aloft. This process continues for 14 days of daylight.
Although the surface cools at night as your post details, the atmosphere cannot cool as quickly due to the temperature inversion phenom. The air nearest the surface will cool to be the same T as the surface but this will cause it to act like a barrier against the still warm air aloft because it is more dense and cannot rise away from the surface as it did when warming.
At the next dawn the process begins again, but this time we have air aloft that is already warmer than it was 28 days ago (due to not being able to radiate) and is suplemented by further warming for the next 14 days.
The key is temperature inversion. This slows down the conductive cooling rate of the atmosphere which (I think) will accumulate heat until it is at a temperature somewhat very close to that of the surface at noon, the warmest part of the day.
The AVERAGE temperature of the SURFACE will still be much the same as that of the planet with no atmosphere as S-B dictates, but the AVERAGE temperature of the ATMOSPHERE will be much higher.
Hence my statement…
I think commentor coldlynx is saying much the same thing at 11:46am and 3:38pm jan 9th
regards
coldlynx says:
January 9, 2012 at 3:38 pm
The transparent GHG free atmosphere is warmed by conduction from the surface. Where your claim hits the reef is that since heat only flows from warmer to colder, the surface can never warm the atmosphere to be warmer than the surface is. No matter what the difference in efficiency is, the surface can only warm the atmosphere up to the temperature of the surface. Beyond that it can’t go, you can’t make heat flow uphill.
w.
One issue is the differing values for how warm or cold the Moon gets.
I’ve seen numbers saying the surface (really rocks and soil) reaches temperatures of 95C in the sunshine period and I’ve seen values of 120C.
These two different values make quite a difference in how I view the issue. If temperatures reach 120C then its temperature is hotter than it should be and its energy accumulation rate is not much different than the Earth. If it is 95C, then it is cooler than it should be and it cools off and warms up much faster than the Earth.
(Note the Earth would be very different if the rotation rate was 27 days times 24 hours. If the surface temperature accumulated energy at the same rate is does now for 13.5 days, the Earth’s surface temperature would approach 150C near the end of the 13.5 times 24 hours day and of course all the water would boil off and the Land surface would be baked so much that much gas would be liberated from it and we would be on our way to a mini-Venus affect).
Robert Brown says:
January 9, 2012 at 12:00 pm
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.
Sorry, I’m reading and commenting as I go. This is consistent with observations in the very dry, windless desert. To quote the Wikipedia article on Deserts, in hot deserts the daytime peak temperature can be 45C and the minimum temperature right before dawn can be 0C or even cooler. That is dead on your estimate, allowing for the fact that even in the desert there is stratospheric water vapor acting as a GHG to block a little of the heat — along with CO_2 of course.
Your estimate is of enormous consequence, in other words. It suggests that there is almost no greenhouse effect active over hot, dry deserts. Not CO_2 GH. Not H_2O GH. Not CH4 GH. They radiatively cool almost as fast as the dark side of the moon!
______
The estimate of 4C an hour for cooling at the surface over a desert on earth is a pretty good one, but probably not realistic for the cooling seen at the lunar surface when the sun sets. Measurements during lunar eclipses indicate peak rates of about 30C an hour or around 100C during the full length of the eclipse (around 4.5 hours). Of course desert areas are not completely devoid of all water vapor in the atmosphere as the relative humidity is just so low, and you do have the other greenhouse gases which also help to keep the desert from cooling anywhere near the rate seen on the moon. If you left the nitrogen and oxygen, but took every greenhouse gas molecule out of the atmosphere above a desert on earth, you’d certainly get cooling at greater than 4C an hour when the sunlight went away. It might not reach the 30C an hour peak rate as seen on the moon, but, it would be higher than 4C an hour, such that the high to low temperature swing would also be greater.
Willis, if you could please respond to the cogent thoughts expressed here. David says:
January 9, 2012 at 3:24 pm
Willis, for some reason you ignored most of my comment. Is it that bad??
But your statement:
“So give me Huffman’s claims in three sentences, just reading his stuff makes my head hurt. What has he said and why do you think it is important?”
Was handled wekk by nano Pope above.
To repeat the rest of my comment in slightly different, maybe more understandable?, words:
On this planet, the radiative equilibrium occurs at about 5 km above the surface, NOT AT THE SURFACE. Average T is -18 at 5 km, but it is higher at the surface. The atmosphere is not transparent on this planet or any other one, everyone agrees (that’s why your arguments about clear atmospheres are just “out there”). It can be AND IS hotter at the surface of Earth than the equilibrium radiation at 5 km above it dictates, by an amount determined by the lapse rate, which is determined ONLY by heat capacity and gravity.
We agree that that extra heat constitutes the “greenhouse effect.” What we don’t agree on is WHY it is warmer at the surface than at 5 km. I say Huffman and all the others with EMPIRICAL DATA from other planets with atmospheres (as well as some really simple observations here on Earth) trump the radiation cartoons (which, BTW, don’t accomodate your excellent explanation about why “average radiation” doesn’t mean much).
Bill Illis says:
January 9, 2012 at 4:41 pm
One issue is the differing values for how warm or cold the Moon gets.
I’ve seen numbers saying the surface (really rocks and soil) reaches temperatures of 95C in the sunshine period and I’ve seen values of 120C.
These two different values make quite a difference in how I view the issue. If temperatures reach 120C then its temperature is hotter than it should be and its energy accumulation rate is not much different than the Earth. If it is 95C, then it is cooler than it should be and it cools off and warms up much faster than the Earth.
_______
Beside the distance from the sun, which is of course the same as the earth’s distance from the sun, it is the composition of the lunar rocks and soil and their location on the surface of the moon that are the prime factors determining how warm the surface gets in any region. Heat is conducted down through the lunar soil up to several meters, mainly through conduction. Temperatures on the Moon can go as high as 130C during the lunar day and down to around -170C at night. As experiments have shown that the surface of the moon cools off at up to 30C an hour during a lunar eclipse when the sunlit sides goes into earth’s shadow, and around a 100C drop over the course of a 4 to 5 hour eclipse, we see that the lunar soils are giving up LW rather rapidly in the absence of sunlight. If we know the range of the moon’s temperature is 130C down to -170C, and it can cool at 100C in 5 hours, we see that the moon’s first few meters of rocks and soil (the depth to which heat is conducted during the lunar day) cool pretty rapidly, and the surface devoid of sunlight on the moon reaches its coldest in less than 24 hours.
R. Gates says:
January 9, 2012 at 5:26 pm
—————————
This is why I think we need to move the discussion down to the Quantum level and in time. The Moon’s rocks are accumulating energy and then giving up energy when the Sun sets at specific rates. These are far, far, far lower than what we expect compared to the radiation coming in and in the absence of solar radiation. The same is true for the temperature of the atmosphere on Earth at 2 metres high, and especially at the tropopause. The numbers are in the range of 0.01 to 0.00001 joules per second versus the Sun’s energy at 1362 joules per second. It is hard to square.
“This, in spite of the fact that satellites can measure the ocean’s surface temperature from space by measuring the very longwave radiation from the ocean that this credulous gentleman says does not exist.”
argo does that, right? i don’t think it is possible for a satellite to do that.
get out your ir radiation colormometer and see if you can possibly read the surface of some water. i tried it. you can’t because the vapor layer blocks the transmission. so what you read is a layer of vapor – not any surface, ok?
dunno why you continue to conflate a layer of ATMOSPHERE with a PLANETARY SURFACE. atmosphere is not a surface. infrared radiation is not measuring temperature of any ocean. draw the distinction.
the ocean of water is not frozen. there is no sposedta about it. the layer of atmosphere is warmed by it as well. so the model must be wrong.
kadaka (KD Knoebel): Apropos of what orbit the Moon would need to be in in order to have a 24-hour day: It would be a geostationary orbit, which is to say about 22,200 miles above the equator (26,200 miles from the Earth’s center, some 42,000 km). You can see a derivation of this in the wikipedia article for “geostationary orbit.” I suppose the fact that the Moon is so much more massive than the man-made moons currently in geostationary orbit would alter this some–presumably making the geostationary orbit slightly higher.
But the tides we’d get with the Moon in that orbit would make any sea level rise from global warming look insignificant. The tidal “force” is proportional to the cube of the distance between the two bodies. Moving the Moon from its current 239,000 miles to 22,200 miles would increase the tidal force more than 1200 times. So I think I’ll go with increasing the Moon’s speed of rotation; it may be tidally locked now, but how long would it take to become tidally locked again after having been spun up? A long time, I’m thinking, by which time maybe we can find another asteroid to spin it up. (Guess we’d need to evacuate the Moon that time.)
BTW, the idea of altering planetary orbits to solve climate change problems (or was it terraforming?) comes up in one of Jules Verne’s books, if I’m not mistaken.
Willis makes the mistake…”If you are talking about the earth, as far as I know the ground heat flux is on the order of a tenth of a watt per square metre. I’ve run the numbers myself from a couple of directions, and it’s just not all that large.”
If the difference in temperatures between the two objects is large then the conductive effect will also be large. And according to your graph the difference is around 270C. Enormous. Your assumption of the effect of a hypothetical non-GHG atmosphere is flawed on the basis you’re ignoring conduction.
The Moon receives in excess of 1200 W/sq m during the day and nothing at night. This radiation heats the surface to over 107 degrees and up to 123 degrees if you believe NASA
Given the Lunar day is some 27 Earth days the surface heats, probably quickly given how the Earth can heat over the course of a summer day, to the maximum temperature associated with the radiative flux of ~1200 W/sq m. During the long lunar night it is no wonder the temperature continues to plunge to low levels.
The tendency to reduce solar input based on geometry of spheres to an average value is nonsense !!
There is no demonstrated mechanism which validates reducing energy input to an “average” value – especially on a planetry scale – the planetry object is either illuminated and heating or not illuminated and cooling.
This rubbish of quartering the solar insolation and using this to calculate temperatures is simply wrong. All the observational evidence says so.
The Earth is fortunate – our atmosphere and oceans distribute thermal energy from equatorial and near latitudes to polar regions and the upper atmosphere.
Before the Earth has a chance to become “overcooked” because the insolation during the day is much more than 170 W/sq m on average – I estimate that at the equator it reaches a maximum of 4 times that and at 75 degrees latitude the maximum summer figure is about 170 W/sq m.
The thing that make Earth habitable are the period of our day (24 hours), the oceans and the fact that at the equator and near latitudes the surface is mostly water and our convecting atmosphere.
Surely all that matters is the relationship between the radiation from the Sun and the temperature the SB equation sayis associated with that radiation – averaging over the sphere means nothing except to tell you the average rate of energy loss to balance input.
The quarter of the insolation thingie is incorrect – I cannot understand why it persists when common sense and logic dictate there is no demonstrated mechanism how a sphere illuminated on one side and approximated by a disk can reduce the radiative flux from a distant star – this is like saying that I can reduce the radiation from my heater by keeping my back turned – which is obvious nonsense as I have a overheated front and cold kidneys.
Will says:
“The true surface of the Earth is exactly the same as it’s S-B temperature, -18º C. Also known as the effective emission height. ”
And it is BELOW that ‘true’ surface that the GHG effect creates a thermal gradient that results in the solid surface of the planet that most of us reside on being warmer than the effectived emission height.
On average… -grin-
“The earth is well above the S-B temperature. You get your own opinions, but not your own facts. ”
If Earth had a atmosphere as thin or 1/10 the atmosphere as Mars, would you still chose to measure the Earth temperature as air temperature and in the shade?
What is the planetary standard for measuring S-B temperature?
Willis : Just a heads-up, but it would appear that NASA’s page on the Lunar Thermal Environment has been updated (and by up-dated I mean completely re-written) since last Friday.
It now gives the average temperature at the Lunar Equator of 206K (-67C) which strikes me as being somewhat at odds with your quoted average planetary temperature of -77C.
http://www.diviner.ucla.edu/science.shtml
A read through of a the Unified Climate Theory thread on Tallbloke’s blog would indicate that this change was made partially at the prompting of a certain scientist with the initials “NN”
We live in interesting times…….. (:-
Will says:
January 9, 2012 at 4:23 pm
What I and others have been consistently calling the “S-B temperature” in this thread is the temperature the planet would have if it were an airless blackbody. It is the temperature calculated from the S-B equation for the amount of solar energy hitting the planet.
Do try to keep up, there’s a good fellow. And don’t get all snarky when you’re not following what’s going on. It just makes people point at you and laugh, like they’re doing now.
w.
Baa Humbug says:
January 9, 2012 at 4:30 pm
It is the same thing coldlynx said, with the same problem. You are claiming that a surface can warm an atmosphere to a temperature warmer than the surface. To do that, you’d have to have heat flowing from the cooler object (the surface) to the warmer object (the atmosphere).
If you can make heat flow from cooler to warmer, you have a working plan. If not …
w.
Will says:
January 9, 2012 at 4:36 pm
We are discussing a “perfectly transparent GHG-free atmosphere”. It’s a GHG FREE GAS, GET IT? Like I said above, do try to keep up.
Will, let me go over this once again. If I said something you disagree with, QUOTE IT. You have an entire straw-man factory going of things that we’re not discussing and claims I never made. QUOTE WHAT YOU DISAGREE WITH and you won’t look so clueless.
w.
Willis:
OMG: YOU SAID THIS:
What I and others have been consistently calling the “S-B temperature” in this thread is the temperature the planet would have if it were an airless blackbody. It is the temperature calculated from the S-B equation for the amount of solar energy hitting the planet.””
WOW. Please explain how this relates to the real Planet Earth, sir! I am very confused about this statement. We do NOT have an “airless body!” WTF ARE YOU SAYING, MAN?
David, you say:
January 9, 2012 at 4:49 pm
Since you gave me a nice link, I went to take a look. I found this:
David says:
January 9, 2012 at 5:10 am
David, I’m not here to try to figure out what you mean. You asked me how much of the greenhouse effect was due to the ocean.
I said none.
You said “I do not literally mean greenhouse effect when I refer to the oceans” …
David, there’s dozens and dozens of comments in this thread. I have to do constant, on-the-spot triage—no answer, quick answer, or detailed answer.
When you ask me a question about how much greenhouse effect is due to the oceans, I’ll answer, quick answer. When you then say you didn’t really mean “greenhouse effect”, you meant mumble mumble, you immediately go to the “no answer” pile. I can’t be waiting for you to decide what it was you really meant, or read your explanation of how when you clearly and plainly said “greenhouse effect”, you didn’t mean greenhouse effect. I didn’t even finish reading the rest of your post. If you are using words that sloppily, I’m not interested. No time, my friend.
The good news is I have a very short memory. So if you want to encapsulate whatever you are interested in, and put it into some substantial, meaty sentences, and post it, hey, I may give it a shot. My intention is to judge each post on that post’s content alone, and I’m always looking for interesting posts. That’s why I play this game, after all.
w.
Surely the “day” side of the Moon is radiating way more than 304 W/sq m – up to almost equal to the solar constant whilst the “dark” side is radiating way less – say even as low as a few watts/sq m if NASA is correct for the minimum temperatures.
These would be the absolute maximum and minimum and there would be variation over the whole sphere.
jae says:
January 9, 2012 at 7:56 pm
Jeez, jae, cut back on the coffee or something. The whole post was about the moon, and about the effect of transparent GHG-free atmospheres. It was not about “the real Planet Earth” other than being an attempt to isolate some of the mechanisms involved.
In this discussion and in this context, what do you think “S-B temperature” means? Get a clue, man.
I have been clear about the usage. Up your reading level, and stop being so damn picky. It doesn’t look good on you.
w.
I’m sorry I didn’t make myself clear.
I am claiming the equator can warm the atmosphere to a level higher than the higher lattitudes hence the atmosphere will have a higher average temperature than the average temperature of the surface.
The warmest “object” is the equator. Lets forget about averages for a moment.
It is the equator which warms the atmosphere (the most) via conduction. And because the atmosphere is a gas, it rises and spreads when warmed via conduction. However, the reverse, i.e. cooling via conduction, cannot happen as quickly, or as efficiently if you will, due to temperature inversion.
I do not claim the atmosphere will ever get warmer than the surface at the equator at noon, but the average temperature of the (whole) atmosphere will be warmer than the average temperature of the (whole) surface.
Why would the atmosphere…say 100mtrs altitude at the poles… be warmer than the surface at the poles? Because temperature inversion will not allow the atmosphere to cool down to the level of the surface.
The energy comes from the very warm equator via conduction. Is distributed globally via thermals and convection and conduction but it cannot cool as efficiently without radiation, hence the build up.
regards
I simply must disagree with this “But a transparent atmosphere could never raise the moon’s temperature above the S-B blackbody temperature of half a degree Celsius.”
Clearly the Moon gets much hotter than half a degree celsius so I find the reference to blackbody temperatures misleading. The radiation from the Sun at ~1367 W/sq m hits half of the sphere of the moon thus the blackbody temperature of this part is ~ 381 K whilst illuminated. As the Moon spins the radiation levels drop very low on the dark side – I’ll say again I see no reason to even consider an average – it is basically meaningless and a tool of the AGW to use to justify their theories.
This clearly demonstrates that probably the most significant feature of the Earth with respect to habitability is the period of rotation followed by the amazing properties of water followed by a freely convecting atmosphere.
jae says:
January 9, 2012 at 4:54 pm
No. I sometimes only answer someone’s first point, particularly if it contains some misunderstanding.
I wasn’t impressed, but I’ll answer it.
I’m talking about the Moon. I’m talking about what the Earth (or the Moon) would be like with a perfectly transparent GHG free atmosphere. Why should I discuss the altitude of Earth’s average effective radiation level? Can you stick to the topic of the thread?
That’s why I don’t go further than the first misunderstanding. They multiply.
w.
nano pope says:
January 9, 2012 at 2:35 pm
The “atmospheres” must be like sponges, or empty bowls? Really?
Go away, and come back with some science. That’s bad poetry.
Near as I can tell, for unknown reasons Huffman has renamed what scientists call “heat content” as “retained energy”. And then he waxes rhapsodic about how at the lower levels of the atmosphere are denser so they can have greater heat content per unit volume.
But so what? Where’s the explanation of how that heats things? That’s entry level understanding of heat content, but how does that help?
I still don’t understand his claim. Maybe if we tried “Complete the sentence”.
For example complete this sentence:
“According to the Huffman hypothesis, the perfectly transparent GHG-free atmosphere ends up warmer than the surface because ______________________________.
Or maybe that’s not what Hoffman’s claiming, I don’t know, I can’t make heads or tails of it. And you quoting him doesn’t help. If you can’t explain it in your own words, you don’t understand it yourself, so what use is your quoting Huffman?
w.
Willis,
Here is an interesting thought experiment. Set a large rock out on a sunny day on the moon (or lifeless earth-like object). Once it reaches ~ 100 C, place the rock into a very well insulated container. By reducing convection and conduction and radiation, you could in principle keep that rock very close to 100 C all thru the night. By repeating this process, you could in principle heat an arbitrarily large amount to rock up toward the MAXIMUM daily temperature. As long as the RADIATING SURFACE is at the S-B effective temperature, other parts could be warmed up to a level approaching the MAX temperature with clever insulation.
In a similar way, in principle it could be possible to heat the IR-transparent atmosphere up to a temperature approaching the MAXIMUM surface temperature.
* heat some air by putting it into thermal contact with the hottest part of the ground
* move the hottest air so it is out of thermal contact with the ground
* repeat.
So I can sort of see the argument some are making. It could be possible to get the whole bottom layer of the atmosphere to ~ the maximum surface temperature. (I do think the lapse rate would keep the whole atmosphere from getting this warm
**********************************************
In practice I don’t think this would work without some sort of “compartments” in the atmosphere. The actual atmosphere would warm and rise in the “hot afternoon” areas (but cool as it rose). This would tend to set up a circulation with hot air rising, spreading out at high altitudes, and heading toward the night side, where the air would fall, warming the cool side. The cool air from the night side would “get sucked back” toward the hot side to replace the rising warm air. Only if you could “trap” the hot air in some compartment above the ground (out of thermal contact with the ground as well) (akin to putting the rocks in the well-insulated containers) would this even start to work effectively (at least in my opinion).
” So in an equilibrium situation like the moon, where the amount of emitted radiation is fixed, temperature swings always lower the average surface temperature.”
Hi Willis,
Seems like a mistake here before you’ve even got started. The moon has thermal mass, so incoming radiation and outgoing radiation only balance in the longer term. Temperature variation is a short-term effect, You can only assume the balance when working in the long-term, so to do so whilst working in the short-term invalidates the analysis.
Regards
Dave
TimTheToolMan says:
January 9, 2012 at 6:42 pm
Tim, thanks for quoting my words so the topic of discussion is clear.
What’s not clear is, what are the “two objects” you mention whose difference in temperature is large? Where is that shown in my graph? And what does that have to do with ground heat flux?
Finally, your statement that “If the difference in temperatures between the two objects is large then the conductive effect will also be large” is incomplete. The amount of heat conducted per unit area through some material (the “conductive effect”) is given by
q = ∆T / (L / k)
So you are right that q is a function of ∆T, the temperature difference. But it is also a function of L, the thickness of the material the heat is flowing through, as well as k, the “thermal conductivity” of the material the heat is flowing through.
As a result of those factors, particularly the thickness of the mantle and its thermal conductivity, the amount of ground heat flux is quite small.
w.
I hate to bring up argument four from Radiating the Ocean but I cannot agree with the argument :-
“In addition there are losses of sensible heat (~ 30 w/m2) and evaporative losses (~ 70 w/m2). That’s a total loss of 390 + 30 + 70 = 490 w/m2.
But the average solar input to the surface is only about 170 watts/square metre.
So if the DLR isn’t heating the ocean, with heat gains of only the solar 170 w/m2 and losses of 390 w/m2 … then why isn’t the ocean an ice-cube?”
As I see it the answer is simple – during the day with the Sun vertically overhead the insolation is probably 4 times 170 W/sq m – 1367 W/sq m solar constant (Trenberth) X 51 % heats the Earth’s surface (IPCC). The physics I was taught said the insolation could be calculated by multplying half the solar constant by the latitude of the point of interest at the equinox. So for where I live, 27 S, the maximum insolation in those circumstances would be in excess of 600 W/sq m.
Clearly this would fry us all – although we can get fairly warm here – if it weren’t for the the energy soaked up by evaporating water, a convecting atmosphere and the relatively short day length – ie 12 hours max baking followed by 12 hours cooling.
There seems to be supported by the solar panels I once foolishly bought. They are rated at 175 W per 1.3 sq m. This reduces to ~135 W/sq m which is a reasonable 25 odd % efficiency.
Accordingly I believe the role of the Sun has been downplayed to support AGW theory. Also I refuse to believe 99% of the atmosphere comprised of Nitrogen and Oxygen does not become heated and as such radiate IR and therefore rendering the amount of IR from CO2 negligible – negligible to my way of thinking is less than 1 % and CO2 is way less than that.
Anything is possible says:
January 9, 2012 at 7:43 pm
It may be that they are giving a (max + min)/2 kind of “average” temperature, rather than the average of the instantaneous temperatures that I have given.
Thanks for that link, wish I’d had it when I started this discussion.
Dang … way cool … thanks for the heads-up.
w.
I think that the oceans are responsible for most of the radiation at night, they warm the air above the surface which begins to convect as the land has lost the “stored heat”. Thus a “land breeze” sets in drawing the cooler air to the ocean to be warmed and humidified and this warmed air helps slow heat loss and provides the radiative effect observed at night – explains why the seaside is always warmer than inland at night and why Palm Trees can grow in Scotland – I’ve seen ’em.
I still think trace gases with low specific heat properties cannot create a large radiative effect and the amount of energy is small compared to latent heat of water – just my opinion but AGW theory seems to be the CO2 tail wagging the water dog.
It’s all a cycle as someone once said.
Baa Humbug says:
January 9, 2012 at 8:25 pm
That I don’t understand. Bear in mind that all the atmosphere can do is reduce the cooling from temperature swings and differences. These include the temperature difference from equator to pole.
But it does this, as you say, by warming the poles with heat from the equator. Once the poles have warmed enough the equilibrium is restored.
But that doesn’t mean that the atmosphere is warmer than the surface. Because if the atmosphere is warmer than the surface, it will simply warm the surface until it is no longer warmer than the surface. Heat flows from warm to cold until they are equal.
w.
Hi Willis,
Good article.
The surface temperature of some part of the moon can exceed 120 C (1368 W/m2 for a black body) if the emissivity of the surface for long wavelength IR is both less than 1 and less than the absorptivity for solar radiation. There are a number of substances known to have this property (see the table here: http://www.redrok.com/concept.htm ). Freshly galvanized metal plate, for example, has an absorptivity for solar radiation of 0.65 and an emissivity for thermal IR of 0.13. If we put a sheet of galvanized metal at the lunar equator, it would absorb 0.65 * 1368 = 889.2 W/m2. But to radiate that much energy, it would have to have a temperature of (889.2/(5.67E-7 * 0.13))^0.25 = 1805 K or 1532 C. Of course, that’s more than 1000 C above the melting point of zinc, but you see the point. There’s even a company, Almeco-TiNOX Solar ( http://www.almeco-tinox.com/en/products/solar_absorber/coating_technology ) that makes a coating that they claim has a solar absorptivity of 0.95 and a thermal emissivity of less than 0.04. Now that would get really hot!
@ Rosco
The “Palm Trees” you see in Scotland are Cordyline australis, they grow in a wide range of climates in New Zealand, from snow level to sea level, this has been discussed previously in WUWT
I have photographs of them from my time in Scotland and Ireland, along with various Hebe spec. that have been grown from stock from New Zealand.
They are not a true “Palm tree”
“But a characteristic of the greenhouse radiation (downwelling longwave radiation, also called DLR)……”
The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. This is an iron-clad law. There is no getting around it. Any so-called radiative ‘down-welling’ is totally irrelevant to warming. The Greenhouse Theory is based on this nonsensical idea.
You must distinguish between Greenhouse Theory and atmospheric insulation. Two very different concepts.
They have had 100 years to prove the Greehouse Theory – and have utterly failed. Forget it guys. It is bogus. Move on.
Willis
I disagree and here is why.
1. Even in that perfectly transparent atmosphere there is an atmosphere.
2. The temperature of the surface is about 60-90C during the day.
3. Atmospheric molecules would still physically impact the surface of the Earth.
4. The molecules would pick up that thermal energy and pass it on to other molecules.
5. Convection would heat the atmosphere
6. Due to the frequency of collision of molecules it would take time for the atmosphere to mechanically radiate its warmth back to the ground with the ground colder at night.
And thus without an IR at all you have been abel to average the temperature. To what?
The molecules are absorbing energy from a body radiating at 70-90 c and the molecules are incapable of emission in the IR so how do the get rid of their now excess mechanical energy?
@- CRISP says: January 9, 2012 at 10:12 pm
“The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. This is an iron-clad law.”
So does The 2nd Law of Thermodynamics: mean that you cannot tranfer heat from a cold body (a coat) to a hotter body (me) without doing work. ?
No point in wearing a coat on a cold morning then…..
Willis You write
“… the surface can never warm the atmosphere to be warmer than the surface is. No matter what the difference in efficiency is, the surface can only warm the atmosphere up to the temperature of the surface. Beyond that it can’t go, you can’t make heat flow uphill. ”
Right. But that is not what I say. (Or try to…, sorry for my english). I say the average atmosphere temperature will be warmer than average surface temperature, and below maximum surface temperature. The “daily” temperature swing will pump up the temperature in the atmosphere to above average surface temperature and below maximum surface temperature.
Just because of how heat is transported. From warm to cold. And warm gases have less density than cold gases.
For those who’d like to watch an actual NASA lecture given to potential future Astronauts headed to the moon, with lots of great detail about the thermal, geological, radiation, environment, I highly recommend:
http://www.lpi.usra.edu/lunar/moon101/mendell/
Isn’t dividing solar output by 4 a form of averaging that will tend to distort the black-body calculations and result in too high a calculated temperature? Isn’t the radiation perpendicular to the sun 1368 W/m2, but as you approach the poles and terminator this drops off COS. Don’t you need to raise this to the 4th power and then average? Similarly, isn’t using a constant albedo also a form of averaging. Surely the moon’s surface is not homogenized. Both these factors would appear to increase the calculated BB temperature above what could be expected without averaging.
I enjoy your posts, Willis.
Two points. First, Luna’s surface warms rapidly after dawn partially because the thermal conductivity of the regolith is very low. The effective surface depth is thin because it’s not rock or compacted; it has good vacuum insulation between the particles. I haven’t a clue how to quantify this, or whether it would significantly change the thought experiment results.
Second, I was around when the exterior coatings were chosen for the Space Station. That’s another example of a body at the same distance from the sun. The coatings were chosen to have absorption and emission characteristics to keep the structure at about 20C without active thermal control. Turns out that’s not terribly difficult.
Rosco says:
January 9, 2012 at 9:16 pm
Also I refuse to believe 99% of the atmosphere comprised of Nitrogen and Oxygen does not become heated and as such radiate IR
It does seem remarkable that the emissivity of N2/O2 would be 0.0000. That would seem imply that N2/O2 would never cool in space. Hardly seems possible.
You may call me stupid but I wonder, if one wants to know the equilibrium of the atmosphereless Earth, why does this one not recognize that there were no clouds, therefore no cloud albedo effect and so only 10% albedo – the pure albedo of the surface? Clouds are, as I understand it, an effect bound on an atmosphere.
The next thing I wonder is, why does one not consider that a comparison with a grey/black body is only feasible for 3/10 of the Earth surface because of the Oceans? The behavior of absorption and emission by water is seriously different to a solid surface, because short wave radiation is mostly absorbed below the area where the emission of longwave radiation (upwards) is possible, what means that not all heat can be radiated upwards what finally reduces the emission coefficient significantly (radiation is emitted in both directions up and down within the water).
In result this would give a much higher equilibrium as the stated 255K, possibly higher as the current average (what would make sense regarding to Ramanathan who stated a net cooling by clouds by about 50 W/m²).
“Hardly any scientists believe that the earth’s core and mantle is producing any energy, other than a modest amount of fission. ”
Sorry, wikipedia says the opposite. http://en.wikipedia.org/wiki/Geothermal_gradient
“The Earth’s internal heat comes from a combination of residual heat from planetary accretion (about 20%) and heat produced through radioactive decay (80%).[2] ”
Lord Kelvin had estimated the age of the earth at ~40 million years, by calculating how long it would have taken to cool off to present temperature (and unaware of nuclear fission).
The same wikipedia article says, “An estimated 45 to 90 percent of the heat escaping from the Earth originates from radioactive decay of elements within the mantle.[8]”. Given such a degree of imprecision, and given that I doubt the composition of the earth, including nuclear species, is anything like certain, I still hope LaViolette’s ideas are pursued.
How much of the GHE on earth is actually due to the oceans where the residence time of energy is far far longer then any GHG?
Willis answers,,,”None, as I understand the greenhouse effect.”
——————————————————————————————————–
Willis, please understand that of course I do not literally mean GHE, when I refer to the oceans; except in the context of THERMAL CAPACITY. At its most basic only two things can effect the energy content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system. (You may henceforth refer to this as David’s law. (-;) There is a fairly exact correlation between residence time of energy and thermal capacity. As the ocean thermal capacity is thosands of times that of the atmosphere, it appears logical that it is a more effective GHLiquid, then any GHG; your thoughts in this regard are welcome aand requested.
Also, although the average albedo of earth is higher then the moon’s, is it higher at laditudes where TSI is stongest?
Willis responds…”The earth albedo varies by location, by time of day, and by time of year, so it’s hard to answer your question. Where the TSI is the strongest (tropics) the albedo is part of the dynamic system keeping the earth from overheating, This means albedo also varies by temperature. It is higher where the temperature is highest, which in turn is where the TSI is greatest, so the answer to your question is generally yes, at least in the tropics.” W. ———————————————————-
At first glance this does not appear logical to me. In general the oceans are a blackbody, absorbing whatever radiation reaches the surface with little reflectivity,(especially in the tropics) The NH has a great deal of landmass north of the tropics, as well as year round snow and ice in the arctic, antarctic as well as tremendous winter albedo (land mass snow cover) beyond year round ice. Additionally the incident angle of sunligh creates ever greater reflectance as one moves further from the tropics. A further factor is the poles appear to have a great deal of consistent.cloud cover as I look at the global map on the right side of WUWT home page. For these reasons I would have to see actual meauserment to accept your assertion here, as I suspect that the tropics. especially the southern tropics have the lowest albedo as well as the greatest TSI, especially in January when the earth is thee million miles closer to the sun and TSI is close to 100 W/m2 greater then in July. Your thought here are appeciated as well.
At its most basic, “only two things can effect the heat content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.”
It therefore follows that any effect which increases the residence time of LW energy in the atmosphere, (water vapor for instance) but reduces the input of SW energy entering the oceans, causes a net reduction in the earth’s energy balance, proportioned to the energy change involved
relative to the residence time of the radiations involved.
So a further question for anyone wishing to do more then a thought experiment. Sunlight radiating on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. (TOA of plus 100 W/m2 ) Despite the increased insolation the atmospheres average temperature is about 4 degrees cooler in January. Because the Northern Hemisphere has more land, which heats easier then water, most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the ATMOSPHERE is 4 degrees higher in July. In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There is not only 7% more intense radiation, there is more ocean to receive this radiation. So, the atmosphere in January loses energy two ways. One to the oceans until it is re radiated as LWIR, and two, to space due to greater NH albedo resulting in greater total earth albedo during the NH winter months. Now the question. Is the earth gaing or losing energy in January, and please quantify your answer?
Firstly, when anyone quotes the above mentioned assumption that backradiation is similar day or night they might do well to read about Prof. Nahle’s experiment* and his night time measurements of downwelling radiation which ranged from 61.93 W/m^2 down to 46.45 W/m^2 seven hours later. These are far less than the values shown on any energy diagram.
Nahle explains how his instrument views a verticle cone which has a 5km diameter at the top, where it seems most of the measured radiation is coming from. Now his instrument cannot tell what height the radiation is actually coming from, so I suspect it assumes an average within its range which goes up to 30,000 metres. But if most radiation is coming from that 5km diameter disc, that’s a lot of square metres to divide by. How on Earth (or in the atmosphere) can we expect any high degree of accuracy in such measurements?
Secondly, on the subject of Moon temperatures, the use of the S-B equation should be averaged over the full day/night cycle, which is 26.3 Earth days on the Moon. This gives far lower values than the proverbial -18 deg.C figure.
Remembering that the incident solar radiation is of course zero at night, it might appear that the temperature on the Moon would also be zero (0 deg.K) at night.
But neither the Earth’s outer surface nor the Moon’s act like strict blackbodies. The reason is that a perfect blackbody has no conductive interface with its surrounds. The Earth and even the Moon do. These surrounds are not just the atmosphere in the case of the Earth, but also the deep ocean waters and the underground mass of crust, mantle and core. The Sun’s radiation causes thermal energy to flow into these regions during the day, and that energy then takes a similar time to flow back out at night, these rates depending upon conductivity or convection in the oceans. Of course some of this energy (especially in the oceans) may take months or years to fully exit again.
The common error in all the blackbody calculations relating to surface temperatures is that this conducted energy, and the rates of conduction and rates of revolution of the planet or moon, all seem to be ignored when they should not be. Blackbody theory is also incorrectly applied to gases and even individual molecules in the atmosphere. Whenever it is applied, there should always be a compensating calculation which not only deducts energy transferred by means other than radiation, but which also deducts a compensating calculation based on the temperature of the surrounding region into which the radiation is directed. How then can we calculate positive radiation from a cool layer of the atmosphere to a warmer layer below, let alone to a warmer surface? Nahle suggests that the only radiation he is measuring is actually coming from warm globules of air radiating to cooler regions below them, as is possible in the stratosphere where there is temperature inversion.
* http://climate-change-theory.com/RadiationAbsorption.html
Doug Cotton:
“Firstly, when anyone quotes the above mentioned assumption that backradiation is similar day or night they might do well to read about Prof. Nahle’s experiment* and his night time measurements of downwelling radiation which ranged from 61.93 W/m^2 down to 46.45 W/m^2 seven hours later”
Nahle’s experiment is bogus, as usual. For one thing, his instrumentation is inadequate for the task. His IR thermometer only goes down to -60C or 117 W/m2, so I don’t see how gets values below the lower limit of the device. See my reply to Jack Frost at Science of Doom: http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/#comment-15176 for more detail. You might have to wait a while, though. It was long with a lot of links so it’s stuck in moderation.
DeWitt:
Your sheet of metal does not have to radiate the amount you calculate for the simple reason that thermal energy will be conducted from its underneath side into the surface, as it is hotter than the surface.
You, like others, ignore the conduction of thermal energy from the thin outer layer of the surface deep into the underground regions where it builds up during the day, and then exits out of the surface at night. On Earth much of that exiting of energy from the surface to the atmosphere is by means other than radiation anyway. This is why the Earth’s surface cools at a slower rate than the atmosphere at night, the atmosphere being colder all the time of course. The crust itself is doing most of the insulating at night, not the atmosphere, and there is certainly no warming effect by any backradiation at night, nor even any resultant slowing of the rate of cooling of the surface.
There is no atmospheric greenhouse effect.
It does if I insulate the other side. Actually, the thermal conductivity of the Lunar surface is quite low so even without extra insulation, there would be little heat lost from the back side.
CRISP expounds: “The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. ”
I hear this unsupported assertion quite often. I tell you what, Crisp, the more fundamental version of the 2nd law is “the entropy of a thermally isolated system can only increase”. When you can calculate that the entropy change for the atmosphere/earth system & show that it increases when the cool atmosphere radiates back a fraction of the energy it receives, then I will put some confidence in your conclusions.
RE: CRISP: (January 9, 2012 at 10:12 pm)
(i)”The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work.”
Not quite. You cannot move against an opposing force without doing work. No force opposes the motion of a photon emitted in the cold upper atmosphere before it is absorbed, either in the atmosphere or on the ground. Back radiation is real. The only thing that thermodynamics says is that there must be *more* energy in the up-welling radiation than in the down-welling radiation if the upper air is cooler than the surface.
The primary problem with the carbon-dioxide greenhouse alarm is an assumed broad-spectrum effect for a narrow-band process. When the amount of CO2 is doubled, the effect of almost all of the added CO2 is hidden behind the mask created by the CO2 that was already in the atmosphere. It would be like adding to the length but not to the blocking width of a tree in the middle of a stream.
ferd berple says: “It does seem remarkable that the emissivity of N2/O2 would be 0.0000. That would seem imply that N2/O2 would never cool in space. Hardly seems possible.”
Then it is a good thing that science is not limited by your imagination. Or my imagination. Or Willis’s imagination.
The simple fact is that N2 has a MUCH lower emissivity than CO2, and consequently N2 would cool MUCH slower than CO2 out in space. Just like a polished piece of aluminum would cool MUCH slower than the same piece of Al painted black. (Actually any color paint would do, since the IR properties are mostly independent of the visible properties).
Hi Willis, do you perhaps know if anyone ever measured the change in humidity over the years?
After looking at the daily average readings from about 20 weather stations all over the world I am finding a change of about -0.02%RH per annum.
http://www.letterdash.com/HenryP/henrys-pool-table-on-global-warming
So, if this estimate is not far from correct, then the average global humidity is now about o.75% RH lower than it was 37 years ago.
If I am not mistaken (at 15 degrees C) that translates again to a loss of about 0.1% in absolute humidity.
You see how that compares with the increase in of CO2? (0.01% increase over the last 50 years)
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. .
Interesting post. Incidentally I know the moon orbits and keeps one face to the earth. But interestingly the idea that it has an axial rotation as opposed to just an orbit is widespread, even to a report on the BBC a few days ago. Maybe I should not be surprised by the BBC, but otherwise it’s a common mistake even in scientific discussion.
The term is “gravitationally locked”. 4 billion (or so) years ago the moon orbited much closer to the earth than it is today — its orbit is receding by a few centimeters a year — and the tidal force exerted on the Earth (proportional to 1/r^3 where r is the radius of the orbit) was many times stronger. Tides in the Earth’s early oceans would have been spectacular things, rising five or ten meters nearly everywhere. Tidal heating was much larger as well, and the gravitational coupling turned the moon’s initial axial rotation, whatever it was, into heat.
Eventually the moon “froze” into an orbit such that its period of revolution around the Earth and period of rotation around its axis are the same, deformed by the tide so that it is very slightly football shaped, with the long axis pointing towards the Earth (the Earth is similarly a flattened sphere due to its rotation with a running tidal bulge that faces/opposes the moon).
I don’t think this is a common mistake in scientific discussion, though, at least not discussion among physicists or astronomers. However, I occasionally teach astronomy, and in the process learned a disturbing thing. Some bright (but cynical) person did a survey of Harvard students out on the quad one fine day. One of the survey questions asked the students to draw the relative position of the Earth, Moon and Sun when the Moon is new, first quarter (half full), full, last quarter (half full).
Only one student in four, if I recall correctly, could do so. A sad commentary, really. Readers are invited to try it for themselves, now, without looking. Bear in mind that you’ve been looking at the Moon as it proceeded through its phases for your entire life, and you have in addition the powerful guide of your reason (your answer must make sense and “work”).
Astronomy is really an amazing thing to study — it’s simple enough that anybody can understand it (even without a lot of calculus or physics in their background), it includes things like the Stephen-Boltzmann law and Wien’s Law (both used, along with I = P/(4\pi R^2), as key steps in determining how far away everything is from the Earth), and amazingly good backyard telescopes are cheap and readily available as never before. Amateurs can photograph things like the Orion nebula or the planets as easily as professionals did 100 years ago, with telescopes that are vastly surpass the ones they used to make awesome discoveries way back then.
rgb
@- HenryP says:
January 10, 2012 at 5:57 am
“Hi Willis, do you perhaps know if anyone ever measured the change in humidity over the years?”
The Radiative Signature of Upper Tropospheric Moistening
http://www.sciencemag.org/content/310/5749/841
Identification of human-induced changes in atmospheric moisture content
http://www.pnas.org/content/104/39/15248.abstract
Data from the satellite-based Special Sensor Microwave Imager (SSM/I) show that the total atmospheric moisture content over oceans has increased by 0.41 kg/m2 per decade since 1988.
“The molecules are absorbing energy from a body radiating at 70-90 c and the molecules are incapable of emission in the IR so how do the get rid of their now excess mechanical energy?”
Has anyone explains what actually happens to an atmosphere that can’t absorb/emit radiation (IR). Is this even realistic. What happens if you replace the zero absorption/emission atmosphere with a slightly more realistic pure N2 atmosphere??
I have strong bias towards not changing the Moon [or Mars].
Good, because it wouldn’t work. The problem is this. The molecules of a gas at some temperature at equilibrium have what is called a Maxwell-Boltzmann distribution of velocities. That is, some molecules are moving fast, some slow, most in between, at any given temperature. The MB distribution is a peaked curve with a long tail at higher speeds, and it is different (scaled) for particles of different mass because of equipartition, where 1/2 m v-average-squared = 1/2 kT as a constraint (if you like) on the distribution so that more massive molecules move more slowly, on average at the same temperature are lighter ones
A planet has something called an “escape speed” — a speed for which something thrown away from the planet’s surface will never fall back, assuming no drag forces. It is easily computed, for the Earth it is 11.2 km/second. For the moon it is 2.4 km/second, as it is smaller and has weaker gravity.
The Moon’s escape speed is sufficiently low that any of the lighter molecular mass gases have a relatively short lifetime on the hot side. Too large a fraction of the MB distribution consists of molecules up above escape speed, and as those molecules fly off into space the remaining ones re-thermalize and constantly replace them. It is actually a cooling mechanism, as the lost molecules carry away energy from the high tail of the MB distribution (a similar mechanism is used as an ultracooling refrigerator in certain physics experiments).
Perhaps you could give the Moon an atmosphere of Krypton, or Xenon (no, not the warrior princess:-), if you had enough.
Incidentally, this is why the atmosphere of Mars is mostly CO_2, and why the Earth’s atmosphere has very little hydrogen or helium or methane in it. The molecular weight of CO_2 is greater than that of O_2 (by the weight of one carbon atom) or N_2 (by the weight of one oxygen atom). It takes longer for CO_2 to outgas from a planet via the MB tails because the whole distribution is shifted down so a lot smaller percentage of the molecules are above the escape velocity cutoff.
The Earth is constantly losing atmosphere in exactly this way, lightest molecules first. To some extent, the atmosphere lost is replaced by outgassing from the Earth’s crust and e.g. biological activity, but once it is gone it is gone. I do not know precisely how atmospheric density is thought to have varied over the last 4 billion years (aside from a few things about the very early atmosphere in the pre-oxygen era) but I’m guessing that on average it is decreasing and will continue to decrease gradually over geological time. This could even be an explanation for why the Earth has been gradually slipping into long term cooling and glaciation, although I’m not certain it is a necessary explanation given periods of geological glaciation hundreds of millions or billions of years ago.
I am most skeptical about “heating” of the atmosphere due to pressure and PV=NkT — that is nonsense, frankly. However, Willi’s observation above that an atmosphere acts like “thermal ballast” and helps reduce the hot side and cold side temperature differential, which de facto increases the mean temperature of the planet closer to that predicted by SB for a uniform ball seems to me to be dead on the money. The more uniform the temperature of a planet, the warmer the planet on average. The denser the atmosphere, the better the greenhouse gas trapping, the more uniform the temperature, where the latter may well be less important than the former — I’d have to do some estimates of the heat gain/loss of a volume of atmosphere (plus the top whatever centimeters of the surface) to get a feel for it.
Note well that while PV=NkT may not be the mechanism at all, it is closely related to the heat capacity of a gas, and the heat capacity of a gas may be an important component of the mechanism. It would be very interesting to see if this assumption can actually quantitatively reproduce the NZ curve for the planetary temperatures as a function of atmospheric density. I intuitively think that it might.
rgb
Robert Brown says:
January 10, 2012 at 6:20 am
Standing on north pole facing the sun.
New moon is at your 12 o’clock.
Quarter moon is 1:30 and 10:30
Half moon is 3:00 and 9:00
Three quarter moon at 4:30 and 7:30
Full moon at 6:00
That a bunch of Harvard students would get that wrong is no surprise. Formal education is a racket that enriches the employees of the institution at the expense of others including students, parents, and taxpayers. The system became increasingly worse as independent low-cost access to learning materials became more widespread. The internet greatly accelerated the decline into a real sham.
That question you posed, by the way, is typical of what appears on the math portion of the Scholastic Aptitude Test which was dumbed down considerably in 1995. I aced the math portion of the SAT back in 1978 before it was dumbed down. So maybe I’m just a bigoted autodidact with an IQ exceeded only by my cynicism but it’s rare that I’m wrong so that’s not a very likely scenario.
Tidal locking (the more commonly used term than gravitational locking) is a little misleading as it has nothing to do with tides on the earth. It begins with a mass distribution asymmetry in a smaller rigid body. The earth’s moon is large enough to form a gravitationally shaped sphere but the mass distribution won’t be perfect even if the spheroid shape is perfect. The heavier side of the smaller body will eventually come to constantly face the larger body.
CRISP expounds: “The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. ”
That is true but it should be qualified by saying there is no net transfer of energy from the colder to the warmer.
What actually happens is that the rate of net energy transfer from the warmer object (the earth) to the colder object (the cosmos) is slowed down. So it is quite correct to say that the cooler cannot heat the warmer but the rate at which energy moves from warmer to cooler can be throttled by intermediaries like London Fog coats and London fog banks. The principle is pretty much the same for both. It’s passive insulation in both cases. If you believe that insulation such as clothing works to limit the loss rate of body heat then you should also believe that insulation such as water vapor works to limit the loss rate of the earth’s heat. Word.
(If you add an atmosphere, then the situation could change a bit, and the inner surface could indeed be warmer than the outer shell by an amount related to the lapse rate.)
How’s that again? Read your own argument. If the system is at equilibrium it will all be at the same temperature, including the atmosphere, including gravity, under any and all circumstances. The only circumstance whereby the inner (idealized) sphere can be warmer than the outer one is if it starts out warmer and one hasn’t waited long enough for the system to equilibrate.
To put it microscopically, the average energy per molecule per degree of freedom will eventually be 1/2 kT, period. As you said, heat will always flow from warmer areas to cooler ones until equilibrium is reached. So very good reply, except for the parentheses…;-)
rgb
http://www.tech-know.eu/uploads/Greenhouse_Effect_on_the_Moon.pdf
To quote NASA’s analysis,
During lunar day, the lunar regolith absorbs the radiation from the sun and transports it inward
and is stored in a layer approximately 50cm thick. As the moon passes into night, the radiation
from the sun quickly approaches zero (there is still a bit of radiation from the earth) and, in
contrast with a precipitous drop in temperature if it was a simple black body, the regolith then
proceeds to transport the stored heat back onto the surface, thus warming it up significantly over
the black body approximation.
The above is a link to a article several years old regarding this subject. The quote is from the article.
Willis, thanks for a very interesting, thought-provoking article!
Also, very good and polite answers to comments, many not so thoughtful or polite.
clivehbest says:
January 10, 2012 at 1:40 am
“Alan – that’s right. Albedo reduces for lower incident radiation . As the Earth’s ocean surfaces get heated, the Earth starts to “swet” – producing clouds which reduces the Albedo.”
The earth sweats more as it gets warmer.
That’s the simplest and most insightful analogy I’ve seen in quite some time. A wonderful way to convey the concept of negative feedback in the climate system. Thank you.
Tom_R says:
January 9, 2012 at 8:38 am
>> son of mulder says:
January 9, 2012 at 3:42 am
but on a clear night the air will cool quicker than the surface. <<
"The surface will cool quicker than the air, which is the reason we can have frost form at air temperatures above freezing. Your point about a difference between air temps and surface temps is still valid."
Good point but it depends on the surface. I have a high emissivity surface (white mineral roofing) on a flat (1:14 slope) roof with an unobstructed view of the night sky. Beneath it is R-30 insulation. The shapes that appear (rafters, underlying heat sources, etc) in the dew and frost patterns are fascinating and I see it every morning from above. I've observed frost on it when the air temperature as recorded by sheltered a min/max thermometer 10 feet above it never fell below 40F during the night. I've never really seen, that I can recall, frost forming on natural surfaces when air temperature is above freezing.
The estimate of 4C an hour for cooling at the surface over a desert on earth is a pretty good one, but probably not realistic for the cooling seen at the lunar surface when the sun sets. Measurements during lunar eclipses indicate peak rates of about 30C an hour or around 100C during the full length of the eclipse (around 4.5 hours).
Noted. I looked to see if I could find an online computation of the cooling rate T(t) of an ideal spherical surface layer subject to some reasonable assumptions — largely because I am too lazy to WILLINGLY compute it myself, it’s a fair amount of work — just to see if one can get an order of magnitude feel for what it “should” be. Basically integrating Stefan-Boltzmann, taking the heat capacity of the surface up to some depth as the reservoir. Since linear (in T) loss would produce an exponential, the T^4 suggests T(t) is much faster than exponential, so that it loses a lot of heat very quickly, but the rate of heat loss drops dramatically with the temperature. So it isn’t completely crazy for the cooling rate to be down to 4C/hour at the terminator, if the temperature there is already down close to 0C.
I have so very much to do, and this is so very entertaining and distracting, but if I actually finish some good fraction of my chore list I’m thinking seriously of building a model of my own. I downloaded rttm so that I could take a look at it and might use it as a base, but sometimes there is virtue in trying something on your own first and then looking at what others have done — it’s easier to find something they might have missed or a mistake they might have made if you aren’t “trapped” into following their reasoning.
Hopefully rttm does the full computation “right” — spatially and temporally inhomogeneous, with emissivity a function of temperature and wavelength, and so on. I’m still very curious about how they manage the interaction between convection and GH — bulk convective cooling leading to vertical heat transport is roughly the same order as radiative cooling (more, if there is a wind) and carries hot air up through the densest layer of atmosphere near the ground to where it can radiatively cool more efficiently, an effect that should be strongly enhanced by water vapor carrying heat of vaporization up above that dense layer. In other words, I’m curious as to whether they correctly integrate in the vertical direction and hence end up with too much GHE trapping.
I suppose the best analogy is that the atmosphere isn’t like a “greenhouse”, a ground surface and an IR absorber over head. It is more like a ground surface and a large collection of very thin (much weaker) IR aborbers placed so that air and moisture can easily convect up through them, while their integrated thickness is the same as that of the original IR absorber in the glass house model. Convection carries warm air from the ground up through most of the plates to create a warm layer at the top that cools much faster via radiation than the air or ground at the bottom, then falls back both cooling the plates themselves in an irregular pattern (reducing surface trapping of the heat) to pick up still more heat from the surface and lift it, warming the plates where it lifts. This would very quickly create inhomogeneous surface temperatures fed by semistable convective rolls that cools the surface with a wind much faster even than the initial convection, and much faster than radiation. To make matters even more complicated, the “plates themselves” are just CO_2 molecules and get lifted up to cool faster than pure radiative transport in a static atmosphere would would allow, so there is a constant turnover of the plates as they absorb down low and emit up high.
The point — consistent with the entire discussion so far and Willis’ analysis above — is that any system with an atmosphere self-organizes(creating hot and cold differential) to lose heat faster than any static model (or model with the wrong transport dimensionality) predicts. The entire “greenhouse” is turning over all the time it is functioning, cooling the ground by direct wind-driven convection and evaporation and carrying heat to the top to be lost through a much thinner greenhouse than one expects just looking at lower troposphere. The correlation of the current not-so-warming trend with a surprising drop in stratospheric H_2O seems to add some weight to the possibility that this isn’t being done right. As the stratosphere dries, it could significantly enhance cloud-top cooling — even “resonantly” enhance, as it leaves a growing hole right where water vapor is radiating.
rgb
@Robert brown –
even more outrageous than the survey on the quad about the moon-
ask around to see what color people think the sun is – and it’s not like they had no acquaintance with it.
but most of em will grab the yellow crayon.
Tidal locking (the more commonly used term than gravitational locking) is a little misleading as it has nothing to do with tides on the earth. It begins with a mass distribution asymmetry in a smaller rigid body. The earth’s moon is large enough to form a gravitationally shaped sphere but the mass distribution won’t be perfect even if the spheroid shape is perfect. The heavier side of the smaller body will eventually come to constantly face the larger body.
Joules, you are dead right about your cynicism exceeding your IQ. I tend to think that it isn’t a matter of formal education being “a racket” as much as it is due to a mix of a distribution in abilities plus the fact that most humans — and I’ll include myself in this — are not Holmesian in our powers of observation or inference. Also, for most people using their brain seems to hurt, or tire them out, or something. It isn’t pleasant. Not entirely unreasonable — the brain burns between 1/4 and 1/3 of one’s daily calorie intake, and we are bio-programmed to conserve energy (the reptile within is a lazy bastard that wants only to bask on a rock, get laid, and eat, in complete safety).
As for your remark, it has nothing to do with tides on the Earth, it has everything to do with tidal “force” on the moon where I’m pretty sure you understand why I put the word force in quotes, as the “tidal force” is just the (negative of the) difference between the actual gravitational force and the total force acting on different mass points in an extended object accelerating in some way relative to the attracting body. Points on the tidally locked moon (good catch, I was just trying to convey the idea but you are right) that are closest to the Earth experience too much gravitational attraction to be in “orbit” and hence require less force to oppose the moon’s gravitational attraction in the opposite direction. Points on the opposite side from the earth experience too little Earth gravitation to be in orbit, the difference is again made up by Moon gravity at expense of the e.g. normal force holding them up. Hence a bulge in both directions, that isn’t quite symmetric (in higher order than third), so indeed the heavier side ended up facing the Earth.
However, tidal forces also likely contributed to the bulge and mass asymmetry in the first place. The moon isn’t (or wasn’t, as it became locked) a completely rigid body.
Tidal force from the moon (and sun, much larger but much farther away) acting on the earth is very, very slowly slowing it down so that at some point in the very distant future the Earth and the Moon, given time, they would be mutually tidally locked. I can’t remember if the Sun will have done the red giant thing or not by then, so this may not happen before there is no more Earth or Moon to lock. In any event, human-derived life forms will either be extinct or have moved on by then to other worlds, and the molecules that currently make “me” up will probably not even be locally contiguous any more, so I suppose it isn’t worth worrying about:-)
rgb
Willis, I don’t know if you’re still reading the comments. But this is simple, straightforward, thank you. It may be stating the obvious, but often the obvious is overlooked. The bigger the temperature difference (slower the rotational speed), the lower the AVERAGE temperature.
Would make for an interesting hypothesis for partially explanating the “small sun, warm earth” paradox of the early solar system.
Septic Matthew says:
January 9, 2012 at 1:54 pm
Joules Verne: One of the most crucial facts to understand is that the ocean cools primarily through evaporation not radiation. If the ocean doesn’t cool by giving off longwave thermal radiation then it wont’ be warmed that way either. Therefore greenhouse gases that produce downwelling longwave radiation have little effect on the ocean.
“What exactly does the downwelling longwave radiation do to the surface of the water?”
It is completely absorbed in the first few microns of depth. It raises the evaporation rate and in most cases is not propagated downward. The result is called the cool skin layer which is approximately one millimeter deep and is 1-2C cooler than the water below it. The cool skin layer is the predominant condition of the ocean surface. Breaking waves can obliterate it but it reforms in about 10 seconds.
Someone in this thread mentioned that the greenhouse effect is caused by increased residence time of insolation due to the variable properties of the earth’s surface. The global ocean is actually the big kahuna in this regard. The ocean is transparent to incoming shortwave which penetrates at the speed of light to a depth of about 100 meters depending on water clarity. Water is opaque to longwave so the solar energy absorbed at depth must be mechanically transported to the surface for emission. This increases the residence time of the insolation and causes the ocean to warm up higher than its S-B temperature in order to attain equilibrium. The majority of greenhouse warming is accomplished by a liquid ocean that covers 70% of the earth’s surface not the tenuous and far less effective greenhouse gases in the atmosphere. Take away the ocean and leave everything else equal and the earth would be a giant deep freezer.
“I’m trying to see the various reasons why the earth stays so much warmer than its S-B temperature.”
Basically, you assume the “blue dot” is a black body with no atmosphere and no gravity, i.e., the only contribution to the thermodynamics is the radiative piece – while ignoring the “elephant in the room”, namely, the atmosphere and gravity.
Yet it’s trivial to estimate the effect of gravitation on the atmosphere and it’s influence on the Earth’s surface temperature by using the equations of state starting at the -18C degree surface (roughly 40km above the Earth) and dropping down to the surface.
Then there’s the single point temperature measurement on the Moon which clearly shows a constant “nightly” decay in temperature which is only interrupted by “dawn”. Do you have any idea what the data is telling you?
@rgb
You have the cart before the horse. The fixed tidal bulge only emerges after tidal lock has occured. Before that point the tidal bulge travels along the surface. And yes the moon isn’t completely rigid but it’s pretty darn close and in the context of my missive it is completely rigid as higher density features within it don’t migrate. A small tidal bulge occurs as it flexes but the ellipsoid shape today happened after it became tidally locked not before. In its early history it was presumably a perfect sphere and was not tidally locked with the earth so the tidal bulge that, if fixed, could serve as a torque point to keep it tidally locked, was free to migrate across the surface. It is density asymmetry that produced the initial torque and established exactly which portion of the moon came to permanently face the earth. After that it became ellipsoid to the degree that it isn’t quite a completely rigid body.
From Mike Maxwell on January 9, 2012 at 6:42 pm:
Thanks for catching that. I should have gone with the rotational speed, not the actual speed. Geostationary orbit does seem boring though, with the Moon always directly above about the same spot always, thus depriving about half of the planet from ever seeing it at all, with nights that are exceptionally dark. An additional complication, there are also biological cycles that appear tied to the Moon’s transit around the Earth, such as the menstrual cycles of human females, and many predators and other creatures are nocturnal and need that limited amount of moonlight to see. With a fixed moon, who knows how such would be disrupted.
The mass of the satellite factors out, so that wouldn’t be an issue. What could be important is the exosphere, Earth’s highest atmospheric layer. The upper boundary in theory is about 190,000 km from Earth, about half the distance to the Moon. With the considerably closer position of a geostationary Moon, this could lead to atmospheric drag and the Moon eventually spiraling into the Earth if we don’t speed it up again. That would also be close enough that the Moon could start trapping particles from the Earth’s atmosphere.
Except in a geostationary orbit there wouldn’t be any lunar-influenced tides, as the moon would be in a fixed location relative to the surface of the Earth. No moving Moon, no such tides. The constant fixed pull would draw water to “under” the Moon leading to a fixed amount of rise, eventually the Earth itself would distort into a bulge. But we wouldn’t have those tides, and we likely would have less earthquakes and volcanic activity without the regular lunar-induced flexing of the Earth’s thin crust, although geological forces may become more pent-up leading to more powerful events.
…reduces the Albedo
I believe you meant to say that clouds increase the albedo, reflecting more heat away during the day and trapping more heat at night.
Clouds are a source of great controversy — and uncertainty — in climate models. Catastrophic models all assume that warmer temperatures leads to more water in the atmosphere leads to more net heat trapping, positive feedback. However, this is a vastly oversimplified statement. A number of studies have shown that a wetter environment cools faster because evaporation is a very efficient heat loss mechanism. One of the big factors in UHI effects is the lack of green plants, that cool the ground as they absorb energy, store some of it, use some of it to pull water out of the ground, and then respire, releasing water vapor. Warm water vapor gets carried up in convective rolls to the cooler upper atmosphere where it condenses, releasing the heat of vaporization into the air far above most of the greenhouse effect, and eventually falls as rain that is usually cooler than the ground or ocean it falls on, cooling it by conduction. Clouds over the polar circle regions at night trap heat so that they cool more slowly (net warming). Clouds over the tropics tend to reflect more heat and have a net cooling effect.
This all isn’t simple and cannot be reduced to a sound bite that should convince us that it is “settled science” that more water vapor in response to enhanced CO_2 GHE produces net positive feedback and a higher climate sensitivity, independent of the state of the sun. Nor should it blind skeptics to the possibility that it could lead to positive feedback and a higher climate senstivity. The fact is that at the moment we do not know for certain what it does, in part because the major oscillations have a major effect on the distribution of clouds at different times of the year. Spencer argues, not unconvincingly, that negative-to-neutral is most consistent with the data, but we really don’t have very good data for even a single whole cycle of many of the major oscillations, and one would really want to observe them over several cycles (or come up with really good models that match and predict them) before calling anything settled. Or both. Say, 100 years or so more observation and computational work and we’ll “settle” this, maybe, depending on what we eventually conclude about the role of the Sun and how chaotic everything turns out to be.
But I do think that by far the most plausible suggestion is that the Earth nearly always self-organizes in response to additional forcing to increase, not decrease, its differential rate of cooling. That is, there are good empirical reasons to think that overall feedback is consistently negative — as Willis has noted, the total temperature variation we’re seeing over the entire Holocene (excluding the Younger Dryas) is order of a couple of degrees C, order of 1% of the mean temperature of the Earth, with variations per century that are only rarely half as large and are usually smaller. Hard to imagine this without negative feedback — one would expect much larger and more violent excursions of temperature in the past. To put it another way, if the system were near a “critical point” one symptom that is expected is a gradual divergence in the response to fluctuations (fluctuation-dissipation theorem). The “susceptibility” of the system should increase. This is the warmist prediction of “more violent storms” and the like — a strong and rapid increase if we were near a “catastrophe” (or “phase” transition to a much warmer phase).
The data, however, shows nothing of the sort. There has been basically no statistically significant change in the frequency or violence of storms, or in the total energy being dissipated in storms. What variation there is is reasonably well understood as response to the major climate oscillations, notably ENSO and factors such as rainfall in West Africa and tropical SSTs and the prevalence of upper atmospheric shear. In a way, it is almost surprising — storms arise from temperature differentials and cool the planet faster than it would cool without the storm (they consume free energy, if you like). The flatness in the total energy dissipated in tropical storms actually argue for the flatness of tropical temperatures across a suprisingly long period. Even given a very simple linear response model one would expect at least a linear growth in strength.
This argues for strong negative feedback, or perhaps more likely confounding factors that have nothing to do with global temperature. Complex system, not well understood, not settled science.
rgb
Agile Aspect says:
January 10, 2012 at 9:10 am
If it’s all about gravity then why does the lapse reverse and the atmosphere become increasingly warmer with increasing altitude past the thermopause?
Gravity does not cause heating once the compressive and expansive forces become equal. Since the acceleration of gravity here is constant at 32 feet/sec^2 and the composition of the atmosphere, and indeed the surface pressure, is relatively constant, there is nothing that gravity is doing to cause a temperature gradient in the atmosphere. The sun heats the surface, the surface heats the atmosphere, and the farther away from the source of heat the colder it gets. It’s the same “lapse rate” principle you experience as you move further away from a campfire.
The lapse rate reverses at altitude even though pressure keeps on decreasing because very energetic portions of solar spectrum which don’t penetrate to lower altitudes are absorbed by the very thin air and heat it up to thousands of degrees at the farthest reaches of the atmosphere.
This increases the residence time of the insolation and causes the ocean to warm up higher than its S-B temperature in order to attain equilibrium. The majority of greenhouse warming is accomplished by a liquid ocean that covers 70% of the earth’s surface not the tenuous and far less effective greenhouse gases in the atmosphere. Take away the ocean and leave everything else equal and the earth would be a giant deep freezer.
Well said (and quite correct), sir!
Add to that the fact that water achieves its greatest density around 4C, not coincidentally the mean temperature of the ocean and the actual temperature of nearly all of the oceans, starting a few hundred meters below the surface. If water achieved the greatest density like “normal” fluids, at its freezing point of 0C, the oceans would freeze from the bottom up, not the top down, and the earth would be a giant deep freezer because once the surface froze (at the poles) it would never, ever melt to the depths, and the high albedo of ice would slowly, surely, cover the entire planet with ice.
Much of the surface cooling away from the oceans involves water as well, especially where there are green plants, lots of trees, that sort of thing. Changes in land use (cutting down vast stretches of forest, introducing goats to e.g. the Sahara) very likely did in the past and continues to have today a profound anthropogenic effect on the local warming or cooling of comparatively large areas of the planet’s surface, the UHI effect being just the tip of the iceberg, so to speak, in this regard. There’s also the “cultivated field vs forest full of trees” effect, the “didn’t this desert used to have plants on it?” effect, the “let’s irrigate the hell out of things” effect, the “let’s dam this river and make a big lake” effect, the “Oh hell, it’s too much of a hassle to site our weather station out in the middle of a forest-surrounded field, let’s just put it here next to the building on the southwest side” effect, and many more effects that all contribute to measured warming and cooling.
That’s why UAH rocks. Satellites rock. We can’t trust anything global about temperature data before roughly 1979, and have to take local data with several well-considered grains of error-bar salt. Even the satellite data is worrisome (for well-enough understood reasons) but at least there is a chance that we can build up a century’s worth of consistently measured data with the same general systematic and random errors over time.
In the meantime, the ocean is this enormous reservoir of slowly turning over heat, with timescales ranging from hourly to a thousand years all contributing. It is thermal ballast that both dampens major temperature fluctuations and (via evaporation) self-organizes in response to increased heating and coupling with atmospheric circulation into structures with increased net cooling — like hurricanes, thunderstorms, plain old rain, large patches of cumulus clouds, updrafts, downdrafts, high and low pressure center circulations that take heat and move it around, generally to cool the hot spots, warm the cool spots, and increase the rate of over all cooling as it does so.
rgb
Willis,
Very nice exposition on T^4. I’m not at all disturbed that you got some details wrong. That will be (and is being) corrected. The essence of your argument is the important thing.
@rgb
There are two great attractors in the earth’s climate system. One is a completely frozen world and the other is a completely unfrozen world. Antarctica was covered by temperate forest more than several million years ago. The warm attractor historically dominates over the cold attractor by time factor of about 10:1. It’s all about albedo. Ice and snow breed more ice and snow by reflecting solar energy which would be absorbed if it was ocean or rocks instead of snow and ice. Non-condensing greenhouse gases probably take on their main role in the climate when the cold attractor is in force. On a frozen world the sinks that normally scrub volcanic CO2 from the atmosphere are missing in action thus the gas is free to accumulate. Dark volcanic ash lighter than water is also free to accumulate on the snow/ice surface during any partial melt turning it darker and darker as time goes on. Then all in a rush the worm turns and snow/ice being retreating in positive feedback situation. CO2 sinks don’t reactivate as quickly so the system overcompensates and we get back to the “normal” state of affairs where CO2 level is several times greater than today, global average temperature several degrees C higher, and green plants love those conditions which is why there are temperate forest remains under the Antarctic ice sheet.
At this point in history we are at the precipice between one or the other attractor taking over. If it weren’t for the sun being hotter today than it was during the last big ice age hundreds of millions of years ago there wouldn’t be any interglacials today – it would be all glacial all the time until volcanic activity described above could work long enough to throw the system into the other state.
Robert Brown says:
January 10, 2012 at 9:40 am
“Add to that the fact that water achieves its greatest density around 4C
That’s not true for saltwater. A very common mistake that I constantly have to correct.
Seawater at modern salinity level keeps on increasing in density until it reaches its freezing point of about -2C.
The ocean below the thermocline is a relatively constant 3C. This constant temperature water comprises about 90% of the mass of the ocean. The average temperature of the global ocean is 3.9C.
So the deep ocean is 3C not because that’s the low density point. Think again. I can’t think of any way for the ocean to be that temperature except for that being the average of its surface temperature over timespans long enough to cover a complete glacial/interglacial cycle.
Feel free to offer a different explanation. I’m all ears.
After that it became ellipsoid to the degree that it isn’t quite a completely rigid body.
Don’t mind me, I just teach this stuff. Or if you prefer, can you provide evidence to back your description?
The correct description for the cause of tidal locking is breathing mode deformations that transforms angular momentum into heat, because the lobes lag the point of greatest tide and create a net torque. An asymmetric perfectly rigid rotator in orbit, OTOH, experiences no net torque IIRC as it rotates. Similarly, the thing slowing the earth today is again the asymmetric lagged tidal deformation, which cannot transform (axial) rotational angular momentum to the locked moon any more and is therefore transferred to the (orbital) angular momentum of the moon, increasing its orbital radius by a few cm a year as the earth’s rotation slows.
See, in particular, the description in:
http://en.wikipedia.org/wiki/Tidal_locking
or
http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/tides.html
which has simple, easy to understand pictures that permit you to directly see the vector lines of effective force responsible for the net torque.
So I would have to assert that your description does not correspond to the accepted physics of tidal locking, nor does it make sense. Obviously, since this elastic bulge is directly responsible for the torque leading the tidal locking in the first place, the moon was ellipsoidal right up to where the locking occurred, and sufficiently elastic for some time afterwards for the moderate transfer of additional mass to the fourth order “stronger” tide, on the side facing the earth.
Note well that this all probably happened a long time ago when the moon was much closer and tidal heating was correspondingly much greater (and where the moon may have still had substantial heat left over from its formation). The Earth, OTOH, will not become tidally locked to the moon before the sun goes Red Giant. I couldn’t remember the timescales, but they are apparently order of 10^10 years for tidal locking of the Earth, and the sun won’t last that much longer.
rgb
Robert Brown says:
January 10, 2012 at 6:58 am
“I am most skeptical about “heating” of the atmosphere due to pressure and PV=NkT — that is nonsense, frankly.”
Then you must be skeptical that the atmospheric pressure and density decreases exponentially with height too.
If you ignore gravity, then the pressure of the atmosphere would be constant (from the surface to x equals infinity.)
Try doing the math.
@RGB
“Much of the surface cooling away from the oceans involves water as well, especially where there are green plants, lots of trees, that sort of thing.”
Not as much as you might think. Half the surface (higher latitudes) half the time (winter months) is too cold for much transpiration to occur. Another interesting tidbit there is that plants transpire less as CO2 level rises. That’s because evaporation occurs during gas exchange. Stomata iris open and closed in response to need to take in CO2. Water evaporates from the plant when the stomata are open. Higher level of CO2 in the atmosphere makes the gas exchange happen faster and stomata spend less time open and thus water use is conserved.
This is important because the way I see it we’re going to run out of fresh water for agriculture before we run out of diesel for the farm equipment. What a happy coincidence it is, and the height of irony for global warming boffins, that burning fossil fuel to grow more plants lowers the fresh water requirements for growing those plants. A positive feedback in beneficial consequences! The benefits of higher atmospheric CO2 are legion and the downside practically nil as far as I can determine. If we weren’t raising atmospheric CO2 through fossil fuel consumption we’d need invent some other way to do it. Just sayin’.
Joules Verne says:
January 10, 2012 at 9:31 am
“…altitudes are absorbed by the very thin air and heat it up to thousands of degrees at the farthest reaches of the atmosphere.”
Why is the air “very thin”? And my bet is you would freeze before you melted at the altitudes you are talking about. A molecule may be “hot” but there are so few of them you would neverfeel it.
CRISP says:
January 9, 2012 at 10:12 pm
You are correct about heat, because heat is the net flow of energy.
However, what you seem to have missed is that energy (not net energy, but energy) can certainly flow from a cooler object to a warmer object by radiation. Surely you don’t think that when radiation hits an object, it only transfers its energy to the object if it is warmer than whatever emitted the radiation, do you? Whatever the radiation is absorbed by, it increases the energy of that object, regardless of the relative temperatures of the emitter and the absorber. And no, this doesn’t violate the 2nd Law. The second law is discussing heat flow, not energy flow.
w.
“ferd berple says: “It does seem remarkable that the emissivity of N2/O2 would be 0.0000. That would seem imply that N2/O2 would never cool in space. Hardly seems possible.”
Then it is a good thing that science is not limited by your imagination. Or my imagination. Or Willis’s imagination.
The simple fact is that N2 has a MUCH lower emissivity than CO2, and consequently N2 would cool MUCH slower than CO2 out in space. Just like a polished piece of aluminum would cool MUCH slower than the same piece of Al painted black. (Actually any color paint would do, since the IR properties are mostly independent of the visible properties).”
Most of space in this galaxy has little light from stars, and average value of the universe is dark.
The other aspect is there is a lot time. Blackness, lots of time, so anything which is going radiate any energy it could have radiated it.
BUT the energy of any gas in the vast darkness of space or the gas in our bright world we living in
is mostly related to it’s velocity. how fast is it going relative to something else.
So in vast area of darkness, gases which have no energy to radiate, can form a galaxy of light and energy- and not only can this occur, this is what has already occurred- hundreds of billions of times.
Dennis Ray Wingo says:
January 9, 2012 at 10:42 pm
They get rid of the excess energy the same way that they picked it up, by conduction. Do you think they can’t get rid of it and thus the atmosphere will end up warmer than the surface?
That’s the part where heat flows from cold to hot … you’ll have to think about that part a bit more. The mere fact that something being warmed may not lose energy easily does not mean that it can be heated to 90°C by an object at 80°C, doesn’t work that way …
w.
Doug Cotton says:
January 10, 2012 at 4:26 am
And if anyone reads Prof. Nahle’s work they might do well to remember that Nahle’s claims are nonsense. People measure DLR around the planet all the time. There’s hundreds of papers on the subject. Nobody gets results like Nahles. I don’t know whether he can’t read the instructions on his instruments or what, but his results are way, way, way out of line with the rest of the global researchers.
He (and Doug Cotton, I guess) also subscribe to the foolish theory that DLR can’t warm the ocean … but neither he nor Doug have explained how, if their claim is true, the ocean doesn’t freeze. The ocean emits about 390 W/m2 of thermal (infrared) radiation. It gains about 170 W/m2 from the sun, and about 320 W/m2 from DLR. Total gained equals total lost, so the temperature doesn’t change and the ocean is in general equilibrium?
If you don’t believe in DLR … then where is the 320 W/m2 coming from to keep the ocean in equilibrium?
w.
Can anyone tell me why there is a lapse rate with the greenhouse effect, it is not something that I dispute but I don’t know why it should be.How can a few molecules absorbing and emitting I R radiation lead to this?
Bruce says:
January 10, 2012 at 8:52 am
Thanks, Bruce. I know that it is somewhat simplistic, but I wanted to clarify that an atmosphere can warm a planet independent of GHGs by reducing the temperature swings. I’d never thought much about that.
w.
Joules Verne says:
January 10, 2012 at 9:31 am
“If it’s all about gravity then why does the lapse reverse and the atmosphere become increasingly warmer with increasing altitude past the thermopause?”
But it’s not all about gravity.
The Stefan-Boltzmann Law gets you -18 C or roughly 40km below the TOA. To get to the surface from 40km, you need the equation of state of the atmosphere.
The radiative piece and the gravity piece are only 2 terms in the full Navier-Stokes equation.
And the goal is explain the surface temperature.
If you actually calculated the gravity piece in the thermopause starting at the top of the thermosphere then you would notice it’s contribution is insignificant. 80% of the atmosphere is below the tropopause.
In addition, the molecules are so far apart and there is virtually no mixing so it’s hard to define a temperature. It’s argued by some that this is the cause of the temperature inversion.
M. Simon says:
January 10, 2012 at 9:44 am
Thanks for that. I knew I’d take heat because my calculations were merely a first cut, and as such will only give a ballpark figure. I just wanted to discuss the T^4 and what the implications of that are, and the Moon was a perfect example.
w.
Joules Verne says:
January 10, 2012 at 9:56 am
True, Joules, and there’s a good discussion of the subject here. At the salinity of most sea water (around 35 psu), the maximum density is at slightly below freezing.
w.
Agile Aspect says:
January 10, 2012 at 10:08 am
Sorry, you don’t get to make up what he is skeptical about, because he told you what he’s skeptical about, which is the “heating” of the atmosphere by gravy. That has nothing to do with pressure and density decreasing with height.
w.
“The sun heats the surface, the surface heats the atmosphere, and the farther away from the source of heat the colder it gets. It’s the same “lapse rate” principle you experience as you move further away from a campfire.”
The heat near the campfire gets higher with more molecules (increased air density) around the flames because it is moved away more slowly due to the higher number of molecular collisions.. So the density of the air around the campfire increases the temperature gradient from fire to you. In effect the denser atmosphere obstructs radiation leaving conduction relatively more important. Conduction is a slower process than radiation so the temperature rises.
In exactly the same way the denser the Earth’s atmosphere the hotter the surface becomes and the steeper the temperature gradient upward because space remains at the same temperature but the surface gets hotter. Just as with the campfire the solar energy hitting the surface is moved away more slowly because the role of slow conduction is enhanced relative to that of fast radiation.
Density of the air at the surface is a result of the strength of the gravitational pull and the mass (not composition) of the atmosphere.
So, indirectly, through pressure and then density, gravity does determine the lapse rate and it is mass dependent and not composition dependent so Oxygen and Nitrogen are involved despite being relatively non radiative.
That is what relegates radiative processes to a secondary role and explains why it is gravity rather than radiation that sets the lapse rate.
Gravity and density alter the balance between fast radiation and slow conduction. If one reduces radiation and increases conduction the heat content and temperature will rise given the same energy input.
Opps – should have read:
The Stefan-Boltzmann Law gets you -18 C or roughly 40km above the surface of the Earth. To get to the surface from 40km, you need the equation of state of the atmosphere.
Willis Eschenbach says:
January 10, 2012 at 11:35 am
“Sorry, you don’t get to make up what he is skeptical about, because he told you what he’s skeptical about, which is the “heating” of the atmosphere by gravy.”
The question was directed to him so I’ll wait for him respond with an answer. He’s a big boy – he can speak for himself.
“That has nothing to do with pressure and density decreasing with height.”
Those are your words – but your understanding of physics is atrocious.
Willis says:”…Whatever the radiation is absorbed by, it increases the energy of that object, regardless of the relative temperatures of the emitter and the absorber…”
The “absorbed by” is assumed, not a fact. It could just as well be reflected. Granted it was a few years ago but I was taught that a higher energy object does not absorb lower energy radiation. It is already there. You can’t take a down elevator to go up to the next floor so to speak.
That is why q/a = 0 when T1=T2 even to the 4th power.
mkelly says:
The “absorbed by” is assumed, not a fact.
This is a question that I have thought about without a neat agreed answer as yet.
The radiation(or photons) from the colder object certainly reaches the hotter surface.
The first law says it can be exchanged for an equal quantity of radiation of radiation from the hotter surface.
The second law says however that the radiation cannot increase its quality.
This means that it cannot occupy higher levels or floors in your analogy.
The interaction of photons and surface is much more complicated than the simple picture usually presented.
What the SB equations tell us is compatible with the ‘colder’ radiation being absorbed then contributing to the flow of the ‘hotter’ radiation, which will be more intense at every wavelength.
Alternatively and equally valid way of looking at it is to consider the colder objects radiative effect as being radiative insulation slowing down heat loss from the warmer object.
Stephen Wilde says:
January 10, 2012 at 11:39 am
Density of the air at the surface is a result of the strength of the gravitational pull and the mass (not composition) of the atmosphere.
So, indirectly, through pressure and then density, gravity does determine the lapse rate and it is mass dependent and not composition dependent so Oxygen and Nitrogen are involved despite being relatively non radiative.
No, it depends on g and the composition, adiabatic lapse rate = -g/Cp and Cp depends on composition.
The temperature inversion in the stratosphere is due to OTHER heating sources. For example, this discussion has been mostly assuming that the atmosphere is transparent — in fact the ozone in the stratosphere absorbs a lot of UV light, leading to heating. Farther up, the incoming solar wind warms the thermosphere.
The ocean emits about 390 W/m2 of thermal (infrared) radiation.
It gains about 170 W/m2 from the sun, and about 320 W/m2 from DLR. total = 490 W/m2
There has to be a typo here somewhere. Totals don’t match.
Stephen Rasey,
“There has to be a typo here somewhere. Totals don’t match.”
Not a typo. Convective energy transfer from the ocean surface to the atmosphere of about 100 W/m2 was left out. Most of this transfer will be latent heat in the form of evaporation of water from the surface and condensation in the atmosphere, but some is sensible heat in the form of hot air rising upward from the ocean surface (when the sun is shining).
What I never see mentioned is that when a body emits a quanta of energy the energy level of that body decreases. So the very fact that the Earth’s surface “heats” the atmosphere means the energy level decreases. Clearly during the day the incoming energy can overwhelm the rate of heat loss and the surface warms but once the influx reduces the rate of heat loss overwhelms heat gains.
Atoms in the atmosphere also loose energy as they radiate but AGW never seems to explain this to lay people as they continually refer to “greenhouse gases” trapping heat.
As I see it – in the absence of an external or internal source of energy everything in the universe is coolind down.
I find it hard to buy the radiative imbalance by “trapping” heat – I really doubt we have the technology or the finances to say we have evidence of this outside of theory – and the theory I have seen do not model the world – especially the “radiation balance” based on the geometry of a disk to a sphere – I consider this a fraud.
There is little chance a set of equations based on insolation varying from pole to pole, varying by inclination of the Sun on a sphere that is continually rotating and not understandin so many things about energy circulation, changes in atmospheric and oceanic conditions, changes in Biomass etc etc could ever be truly scientifically constructed – the simple variation of the insolation and rate of heat loss over the globe soon becomes mind boggingly complex..
So a further question for anyone wishing to do more then a thought experiment.
Sunlight radiating on the Earth when it’s about 3,000,000 miles closer to the sun in January, is about 7% more intense than in July. (TOA of plus 100 W/m2 ) Despite the increased insolation the atmospheres average temperature is about 4 degrees cooler in January. Because the Northern Hemisphere has more land, which heats easier then water, most people state that the Earth’s average temperature is about 4 degrees F higher in July than January, when in fact they should be stating that the ATMOSPHERE is 4 degrees higher in July.”
Or there no attempt to measure surface temperature. And the air temperature in the shade
is fairly good measurement of average ATMOSPHERIC temperature. And very sensible if
you concerned about knowing the weather.
” In January this extra SW energy is being pumped into the oceans where the “residence time” within the Earth’s ocean land and atmosphere is the longest. There is not only 7% more intense radiation, there is more ocean to receive this radiation. So, the atmosphere in January loses energy two ways. One to the oceans until it is re radiated as LWIR, and two, to space due to greater NH albedo resulting in greater total earth albedo during the NH winter months. Now the question. Is the earth gaing or losing energy in January, and please quantify your answer?”
Well most important aspect is most heating in done in tropics. So in terms distance in terms thousands of miles- most of heating isn’t widely separated.
So it seems to answer your question one should focus on the bouncing ball- ignore the rest of the world. The area it bounces in is easily more than 1/2 of the world and easily receives more than 1/2 the energy.
At the equinox the sun is centered over the equator and the line of Capricorn and Cancer is the furthest it travels in winter and summer.
The lines Capricorn and Cancer defines what is called the tropics, within these boundaries lies 40% of the earth surface. Meaning obviously, that 30% of the remainder of earth surface area is north or south of these lines.
So Tropics is 40% of earth, and the area the ball bounces covering even larger area of the Earth.
Now, how large in diameter is the ball?.
The sun shines on half the planet 180 degree of sphere. It’s shines most energy per square during 9 am and 3 pm, 90 degree of that 180. Or a 1/4 of earth circumference. Earth is about 40,000 km in circumference, so make the ball about 10,000 km in diameter.
So at fall/spring the ball extends 5000 km north and south of equator. Summer and direct above Tropic Cancer the ball extend 5000 km further north and 5000 km south. Same in winter when sun is over Tropic of Capricorn.
Tropic of Cancer: 23° 26′ 22″ N And Tropic of Capricorn: 23° 26′ 22″ S and they north and south of the equator by about 2600 km.
So ball is about twice diameter of region of tropic, and tropics is always within it.
From Havana, Cuba [near tropic Cancer] to Chicago, US is about 2100 km, so at equinox the edge of ball reaches US-Canadian border. London is just above it. Southward covers most Argentina and easily all of Australia. Part of Australia is in tropic, when sun furthest away some it will be within ball diameter.
So during summer in northern Hemisphere every continent [other than Antarctic] has large percentage land area within the ball. But most area covered by ball at highest it gets is still mostly ocean area- so when highest amount land area is in this more intense sunlight- it is still mostly ocean area.
In in terms total land and intensity of sunlight summer in northern Hemisphere gets the most warming on land. During spring/fall there huge loss of land area getting this kind intensity, and ball goes further south less land gets this intense sunlight [Antarctic never gets it nor northern areas in arctic circle]
As guess, in terms of ocean area it is something like northern hemisphere summer: 60% equinox: 75% winter 80%.
With land it’s as about 90%, 50% and 40%
So, 7% more intense in January. Means land area is not getting as hot- and not radiating more energy into space. So since less energy goes into space, earth warms. But land area fairly insignificant. Unless one is concerned glacier advance or retreat- glaciers at moment are insignificant.
In terms of ocean one going from around 60 to 80% of surface [big amount of area added- not a big difference in area]. I would say bigger cooler ocean [than tropics] which can warmed higher amount solar energy. Such heating will not significantly affect how much is radiated- result being warmer average ocean temperature. And some more evaporation.
So adding significant amount of long term warming ocean warming, but doubtful you could measure it in a year or decade.
Then you must be skeptical that the atmospheric pressure and density decreases exponentially with height too.
If you ignore gravity, then the pressure of the atmosphere would be constant (from the surface to x equals infinity.)
Try doing the math.
Then you must be skeptical that the atmospheric pressure and density decreases exponentially with height too.
If you ignore gravity, then the pressure of the atmosphere would be constant (from the surface to x equals infinity.)
Try doing the math.
Sorry, keyboard fart.
You mean, try doing the math like I did here:
http://www.phy.duke.edu/~rgb/Class/intro_physics_1.php
See 8.1.9, Variation of Pressure in Compressible Fluids. Oh snap. I did not just do that.
Now, please explain what that has to do with heating of the atmosphere. Would that be nothing? I think it would, wouldn’t it.
Like I said somewhere, this thread or somewhere else, if you want to do fluids “right”, look at Navier-Stokes. Otherwise, get out of the game. Everything one does is heuristic or bullshit. PV=NkT per se has so cosmically nothing whatsoever to do with global warming that it is rather a pity that it has been brought into the discussion. Heat capacity might. Buoyancy might. Turbulence might. Convection definitely does. Radiation definitely does.
Look, it is perfect possible for a column of gas to be at a uniform temperature even though there is a pressure gradient. You can also compress a gas to (say) ten atmospheres at room temperature. You just have to remove the heat as you do so (and the volume will contract. Yeah, I have chapters for thermo as well but I took them out of the textbook because we stopped covering them in our standard series for premeds.
Just FYI — I’ve taught general intro physics for 30 years this fall, interspersed with teaching things like graduate E&M and quantum mechanics. I don’t use lecture notes to teach — I just walk in and do it. I have (as you can see) written my own textbook on the subject, and another at the graduate E&M level. I am not an idiot when it comes to basic physics and I don’t just “do the math”, I derive the relations from first principles wherever possible as I teach or write (in a level appropriate way). So seriously — if anyone wants to try to show me how PV=NkT can warm anything at all without the input of actual energy from somewhere else I’m all ears. Impress me.
rgb
Willis writes “As a result of those factors, particularly the thickness of the mantle and its thermal conductivity, the amount of ground heat flux is quite small.”
I wasn’t actually talking about the earth, I was talking about the moon. My quote put the argument into context. The two objecst are fairly obviously, then, the moon’s surface and its hypothetical non-GHG atmosphere.
I would suggest that a non-GHG atmosphere could make the moon much warmer than it is now on average.
Furthermore the amount of warming is proportional to the density of the atmosphere. I’m not sure you understand the Nickolov argument. I dont think he’s saying that the pressure increases the temperature in some sort of “it does work” way, rather he’s suggesting that the pressure defines the temperature (which is also dependent on the level of insolation) and not so much the levels of GHGs that help put the energy there.
I once read a quote from someone in a thread long ago who suggested that it was only a requirement for “enough” GHGs to exist in the atmosphere to get the GHG effect and that more didn’t necessarily keep making it hotter. Yes, I’ve seen the arguments made to show why it might make it hotter and I’m far from convinced. I think you are far from convinced on this too…
“because he told you what he’s skeptical about, which is the “heating” of the atmosphere by gravy.”
Aah! Bisto
Re the opening of Willis’s article:
Translation:
I had a bit of a rant over on Ira’s article on the N&Z “Unified Climate Theory” article, and Richard Courtney responded with an explanation why addition of any kind of atmosphere to a moon/planet results in warming and transfer of energy from tropics to poles etc. Tim Folkerts and others also touched on the difference between average surface temperature and effective radiative temperature and/or that both are of dubious concept on an airless moon, especially if its day lasts around four earth weeks. With these new understandings, I was able to write a long and brilliant article.
Here is a précis:
On the moon, there is a very slow moving hotspot under the sun, which, because of surface curvature, reduces in energy level by ~41% at 45 degrees latitude/longitude, and 100% at the terminators. (where it goes dark). It is nonsense to apply the Stefan-Boltzman emission law based on average surface temperature, (or the so-called effective radiative temperature spread over the entire surface of the moon), under these conditions. The reason is that the emission rate is proportional to the fourth power of temperature (K), thus the heat loss is relatively extremely high over a small area, and it makes no sense to try to average it over the entire surface.
[Bob, you screwed up the blockquoting on your rant, I fixed it for you. Thanks for referring to my article as “long and brilliant”. —w]
“I am most skeptical about “heating” of the atmosphere due to pressure and PV=NkT — that is nonsense, frankly.”
Well as you mention with lower gravity worlds there problem with keeping a gas atmosphere.
And If we are talking about formal paper- I agree it was severely lacking. But we aren’t talking about a formal paper.
I will interested in looking the more formal version- though no doubt that could also have problems.
“However, Willi’s observation above that an atmosphere acts like “thermal ballast” and helps reduce the hot side and cold side temperature differential, which de facto increases the mean temperature of the planet closer to that predicted by SB for a uniform ball seems to me to be dead on the money.”
Yes.
But I think willis is slightly underestimating the effect.
And as I recall that was one aspect of informal paper aforementioned.
Or increased gravity would increase the efficiency of cooling- and it is
the cooling of sun lit surface- which results in lowers radiative losses. And thereby maintains SB global balance.
The rate cooling may considered unimportant, and some extent this could be the case, but also no gravity means no cooling.
So if you talking 1/2 gravity compared 1 gravity or 1/2 density vs 1 density, well it
might not have much effect.
But what about 1/100th gravity as compare 10 times gravity?
Or we 1/100th gravity is much closer to none:)
I posted a couple days ago that I thought 1/2 earth atmosphere and same gravity, would be generally warmer- because the surface would get more solar energy. And I didn’t mention it at time- but a hotter surface also would transfer more energy in less time to the atmospheric gases.
mkelly says:
January 10, 2012 at 12:15 pm
That would have been believed a few hundred years ago, not a few years ago. Radiation doesn’t know about the temperature of what it strikes. It just strikes the object and some percentage of it is absorbed depending on the absorptivity of the object. There is no magical reflective force to prevent it from doing that.
w.
TimTheToolMan says:
January 10, 2012 at 2:48 pm
Obviously, Tim, that wasn’t obvious, or I wouldn’t have asked.
I’m quite positive that I don’t understand Nikolov’s argument. I’ve been asking people over and over to explain it. I asked him, and he disappeared, hasn’t been heard from since. Why do you defend someone who is unwilling to defend himself?
In any case, so far everyone has done like you do, make vague handwaving statements about his work with nothing in them to measure, nothing in them to falsify, no scientific meat at all. So I have to conclude that you don’t understand Nikolov’s claims either, or you could explain them.
Hey, me too! I once read a quote from someone in a thread long ago who said that precise citation of your sources is very important in scientific discussions …
w.
“From Mike Maxwell on January 9, 2012 at 6:42 pm:
kadaka (KD Knoebel): Apropos of what orbit the Moon would need to be in in order to have a 24-hour day: It would be a geostationary orbit, which is to say about 22,200 miles above the equator (26,200 miles from the Earth’s center, some 42,000 km). You can see a derivation of this in the wikipedia article for “geostationary orbit.” …
Thanks for catching that. I should have gone with the rotational speed, not the actual speed. Geostationary orbit does seem boring though, with the Moon always directly above about the same spot always, thus depriving about half of the planet from ever seeing it at all, with nights that are exceptionally dark.”
Good point Mike, never though of it like that.
As for problem moon stuck in sky, I would be more worried by tidal effect.
So put moon at say 50,000 to 100,000 miles.
Now, you get lunar day which better than 28 day and longer than 24 hours.
Though would wreck the Geostationary satellite launch business.
Give us lots of solar ellipses, and might get solar power from form moonlight,
but if not that, it certainly good enough light to read book.
Willis writes “So I have to conclude that you don’t understand Nikolov’s claims either, or you could explain them.”
You’d be right, I’ve not looked into them in great detail but I can be fairly sure of one thing.
It doesn’t help the discussion when you (and Ira and others) go down a clearly irrlevent path suggesting that the pressure does work on the gas and pointing out this is wrong. At some point you need to actually think about what people are saying and importantly meaning, or you may as well not bother commenting at all.
Nickolov at the very least has made the point that atmosphere density determines its temperature and you have effectively agreed yourself in your hypothetical atmosphere example. Perhaps that is a starting point for you to think about his meaning.
TTTM,
Then he’s wrong. The temperature determines the density. Conversion of gravitational potential energy to kinetic energy can provide a lot of kinetic energy. That’s how we think suns ignite. But it’s a one time thing. If fusion doesn’t start at the core, all that energy eventually radiates away and the ball of gas cools down. In the fullness of time if nothing else happens, it will reach the temperature of the CMB, 2.72 K and its density will reach a maximum. You can even see it on the Earth. The height at 100 mbar near the poles in winter, ~15.5 km is about 1km lower than at the equator( ~16.5 km). The difference is the temperature, not the force of gravity.
“So the deep ocean is 3C not because that’s the low density point. Think again. I can’t think of any way for the ocean to be that temperature except for that being the average of its surface temperature over timespans long enough to cover a complete glacial/interglacial cycle.
Feel free to offer a different explanation. I’m all ears.”
Well, ice floats.
Though given enough pressure ice might not float.
But don’t oceans have that much pressure.
As long as the atmospheric “greenhouse effect” is discussed absent any comprehension of the physical fact that ALL substances with a temperature above Kelvin zero–not just solid bodies and GHGs–radiate energy whose flux density is related to its mass density and specific heat, all explanations of that effect will remain misguided. If the conventional explanation had solid physical basis, it would not require any tortured reasoning or oversimplifications to command the assent of the scientifically literate. Despite a plethora of postures, such command remains patently absent.
G’morn from sunny Brisbane Willis
You replied
I don’t think we’re quite on the same wavelength, which is my fault for not articulating my thoughts well enough. With your permission I’ll try an analogy, I like analogies.
I’ve designed a new cook top element. It consists of an inner ring R1 and a larger outer ring R2. Total diametre is 6 inches. Each ring has a seperate on/off temperature dial.
I switch on R1 and turn the dial to a point where R1 heats up to 60c. I leave R2 dead.
What is the AVERAGE temperature of the cook top element? (R1+R2)/2=30c
Now I take a pot with a base diametre of 6 inches, fill it with tap water and place it on the element.
Will I be able to heat that water to a temperature above 30c? I believe I can.
Even though heat is coming from R1 only (at 60c), conduction and convection within the pot of water will ensure the temperature of the water will be well above THE AVERAGE element temperature of 30c.
If I understand your point, you are saying this (what is now) warmer water will also conduct to the cold R2 heating it up until equilibrium, hence the average temperature of the WHOLE element goes above 30c thereby the average temperature of the water can never be above that of the element.
This maybe true…HOWEVER
In the case of a gaseous atmosphere, once the parcel of atmosphere near the cold surface (at the pole) cools down, it can no longer convect/advect like the water in the pot does. Warm parcel of gas above, cool parcel of gas below, equals temperature inversion.
Therefore, so long as some portion of the atmosphere aloft is blocked from contacting the ground, it can not conduct away its heat nor will it radiate away its heat due to it being a non-GHG gas.
So, by the time we get to daybreak on day 2, we have an atmosphere that is now warmer (as an average) than it was the day before. The whole process begins again with the very warm equator warming more of the atmosphere etc etc.
When will this process stall? that’s a discussion all of its own, but it won’t stall before the AVERAGE temperature of the WHOLE atmosphere is higher than the AVERAGE temperature of the WHOLE surface.
A few of dot points that keep sticking in my mind when I think about the above
* The surface is 2 dimensional but the atmosphere is 3 dimensional
* During the warming phase, convection and overturning warms the whole atmosphere (effectively overcoming the 3rd dimesion)
* During the cooling phase, denser cooler parcel of atmosphere near the surface prevents the parcel above from contacting the surface. The cooling phase cannot overcome the 3rd dimension.
* temperature inversions on Earth are most common at the poles and at coastal upwelling zones (from wiki “In the polar regions during winter, inversions are nearly always present over land.”)
http://en.wikipedia.org/wiki/Inversion_%28meteorology%29
* Conductive warming of the atmosphere is ALWAYS quicker/more efficient, than conductive cooling. This leads to residual warmth left in the atmosphere at the end of a day/night cycle.
p.s. At the wiki link I provided above it also says this..
I though that was interesting.
best regards
Dewitt writes “Then he’s wrong. The temperature determines the density. ”
…and density defines temperature (in a planetary/gravity situation) is what they’re saying in my brief readings of this. I certainly dont proclaim to be an expert and I’m not defending this argument but I will defend against those who put up strawman arguments.
sky says:
January 10, 2012 at 4:20 pm
” ALL substances with a temperature above Kelvin zero–not just solid bodies and GHGs–radiate energy whose flux density is related to its mass density and specific heat”
Sky, this is simply not correct. 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.
The pressure-depth relates to the actual depth via the integral of densities above the point of measurement. That requires knowing the density profile at the time of measurement. But if the dive takes several days (and the probe drifts), the previous densities no longer apply.
I’m surprised that inertial sensors weren’t used as an adjunct measurement system; to provide a second point of reference and to measure lateral translation, sideways drift due to current as different depths. The inertial sensors detects acceleration and orientation (6 degrees of freedom). They could be re-zeroed when surfaced.
Previous comment should have made it clear that I am writing about Argo probes that are ostensibly used to determine the amount and of energy and its convective flow within the major thermal mass of the climate system; the oceans.
But I got ahead of myself in commenting…
Willis, further my article @ January 10, 3:03 pm, and your footnoted comment thereon; Thanks for your support; it is invaluable. I repeat my précis on your article, and seek your help:
If we consider Earth without an atmosphere or vegetation etc, it is rather speculative as to what the surface conditions might be compared with the moon. It would be affected by how long in this condition, increased gravitational attraction to uninvited planetoids and stuff, what volcanism, continental masses versus what we currently know as ocean crust, tectonics, lack of sedimentation and precipitative erosion, thermal/albedo qualities of the regolith and, and…
Putting all that aside, if we consider the hotspot under the sun to be plus/minus 20 degrees in longitude, that would give a rotation period of ~ 2.7 hours based on the current day length. Intuitively, I feel that it would take less than 2 hours for the hotspot to reach equilibrium T. Does this seem to be reasonable conjecture to you?
TimTheToolMan says:
January 10, 2012 at 3:59 pm
Tim, you can’t explain Nikolov’s hypothesis. Neither can anyone else. Nikolov might be able to, but he has bowed out.
So you can’t say that “Nickolov at the very least has made the point that …”, because Nikolov hasn’t shown anything at all. He has not shown that density determines temperature, that’s fifty steps beyond what he has shown.
And yet you and others still claim that somehow he is right, despite the fact that no one can explain to me how his hypothesis is supposed to work!
Man. Talk about the Emperor and no clothes. You sit there and freely admit that you don’t know what Nikolov is saying, but you still believe him. Not only that, you advise me to “think about his meaning”, even though you can’t explain his meaning. Why the hell should I think about his meaning, when no one (including Nikolov) will say what he means?
I swear, sometimes I think I’ve fallen down the rabbit hole, where I’m supposed to believe six impossible things before breakfast …
w.
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.
w.
Willis writes “Man. Talk about the Emperor and no clothes. You sit there and freely admit that you don’t know what Nikolov is saying, but you still believe him. ”
Whereas I specifically said “I certainly dont proclaim to be an expert and I’m not defending this argument but I will defend against those who put up strawman arguments.”
So no I dont believe him because I’m not defending his argument. But Frankly to make a strawman argument about pressure (not) doing work and therefore Nikolov having nothing useful to say diverts any useful discussion towards something you know you didn’t understand in the first place.
You appear to agree that hypothetically adding non-GHG atmosphere to the moon will raise the average temperature of the moon. Ira doesn’t believe this. Whether you follow through with that thought or not is up to you. I expect less than 1% of WUWT readers read my posts but far more will read your thoughts…
Willis writes “This means that density does not uniquely determine temperature as you say. You need to know pressure as well.”
Where did I say uniquely?
ferd berple says:
January 9, 2012 at 12:01 am
First, ignore all wave numbers larger than 2,500 cm-1 and smaller than 50 cm-1, then look at the line strengths.
H2O and CO2 have line strengths greater than 1e-19
N2 has a line strength around 1e-28
O2 has two groups of lines, one around 1e-28 and another at 1e-24.
These are very big differences. In addition, those plots assume zero pressure. When typical pressures are considered, the individual lines merge into bands, which explains how CO2 and H2O bands overlap even though individual lines do not.
BTW how do you get CO2 is 1e-23? I don’t see it in the plots.
Willis Eschenbach says:
January 9, 2012 at 10:13 am
Baa Humbug says:
January 9, 2012 at 2:19 am
Easy, the atmosphere will be the same temperature as the maximum daytime temperature. Only the lower few meters would conduct heat back to the surface. Above that, the atmosphere would be at a constant (isothermal) temperature. The only reason the Earth’s atmosphere cools with increasing height is because Greenhouse gases radiate energy. The troposphere is cooled by water vapor. The stratosphere and mesosphere are cooled by CO2.
Robert Brown says:
January 10, 2012 at 9:19 am
—————————-
Thank you for your educational comments. I have a question if you are willing. My questions at this point are primarily directed at the effects of water vapor and clouds on SWR entering the ocean and on the earths energy budget relative to the “residence time” of changes in SWR versed changes in LWR. You spoke of water vapor (clear sky) in this comment and its well known GHE. My question is related and will contain an assertion or two which are also questions.
As I understand it, about 98% of TSI energy lies between about 250 nm in the UV and 4.0 microns; with the remaining as 1% left over at each end. Spectral graphs often have superimposed on them the actual ground level (air Mass once) spectrum; that shows the amounts of that energy taken out by primarily O2, O3, and H2O, in the case of H2O which absorbs in the visible and near IR perhaps 20% of the total solar energy is capture by water VAPOR (clear sky) clouds are an additional loss over and above that. So water vapor alone prevents 20% of TSI from reaching the surface. It is understandable how this is thought to heat the atmosphere, however, is it preventing that energy from impacting a part of the earth/ocean/atmosphere system which has an even longer residence time then the GHG atmosphere, thus lowerering earths energy budget, despite increasing energy to the atmosphere? This is stated within two assertions as follows…
1. At its most basic only two things can effect the heat content of any system in a radiative balance. Either a change in the input, or a change in the “residence time” of some aspect of those energies within the system.
2. It therefore follows that any effect which increases the residence time of LW energy in the atmosphere, but reduces the input of SW and LWIR energy entering the oceans, (such as clear sky water vapor and clouds) causes a net reduction in the earth’s energy balance, proportioned to the energy change involved, relative to the residence time of the radiations involved. (To paraphrase an old maxium, one unit of energy absorbed into the oceans is worth two in the atmosphere)
Thanks for your thoughts.
Bob Fernley-Jones says:
January 10, 2012 at 7:05 pm
Willis, further my article @ January 10, 3:03 pm, and your footnoted comment thereon; Thanks for your support; it is invaluable. I repeat my précis on your article, and seek your help:
Thanks, Bob. The application of S-B is not “nonsense”. It gives us a theoretical upper limit to the temperature of the planet. We know that if the planet’s average temperature exceeds that value, it is emitting more than it is radiating. This is useful information, because it helps us identify impossible claims.
Indeed it would, which is why I described my calculations as “back of the envelope”. I wanted to at least get some sense of the size of the swings. The size I arrived at (±50°C) seems like a reasonable first cut, given the modern values of about half of that.
You are talking about a main difference between the earth and the moon, which is that the moon rotates slowly enough that there is little thermal lag in the system. I just took a years worth of data for the nearest weather station to me (Santa Rosa, CA). Peak insolation, of course, is at noon. Peak temperature is at about 2:45 PM. So that’s the lag between the peaks here. Not sure if that helps.
All the best,
w.
TimTheToolMan says:
January 10, 2012 at 8:30 pm
You said that “atmosphere density determines its temperature”. But density doesn’t determine temperature, so your statement is untrue. The word “determines” contains the claim of exclusivity, because if other things were required, then density couldn’t determine the temperature. You’d need other information.
w.
Robert Clemenzi says:
January 10, 2012 at 11:36 pm
Ah, there’s the problem. You misunderstand the reason the atmosphere cools with increasing height. The reason that the atmosphere has a dry adiabatic lapse rate (cools with increasing altitude) has nothing at all to do with greenhouse gases. It is a consequence of the distribution of energy within the air parcel, a distribution created by gravity. In fact, the size of the lapse rate is given by
Dry Adiabatic Lapse Rate = g / Cp
where g is the force of gravity, and Cp is the specific heat of air at constant pressure. For the Earth’s atmosphere, it is a decrease of 9.8°C per kilometre of elevation. Note that the equation says nothing about GHGs, as they are immaterial in calculating the dry adiabatic lapse rate.
w.
Tim, you started out the Nikolov discussion by saying the following:
Now, for you to tell me that I don’t understand Nikolov, that must mean you understand him, no? Otherwise, how could you claim I don’t understand him?
You go on to say what you think Nikolov is saying, something about pressure and temperature.
Now, you say the following:
Indeed, you are defending his argument, you are sounding very much like you believe him about how “pressure defines the temperature” whatever that means, and you are telling me I don’t understand Nikolov’s argument in the bargain.
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.
I don’t know whether or not Nikolov has anything useful to say because I can’t understand his theory, and neither can you, and neither can anyone else I can find. I find it absolutely ludicrous that so many people are defending something that they don’t understand … humans are indeed strange.
w.
David says:
January 10, 2012 at 11:40 pm
I’m not Robert Brown, David, and I hope he answers you. My problem with what you say is that I have no idea what you are calling the “residence time” of energy. It would be extremely useful if you could answer the following two questions:
How do you measure the residence time of energy, and in what units is it measured?
All the best,
w.
Willis Eschenbach says:
January 11, 2012 at 1:02 am
The cooling of 9.8°C per kilometre of elevation is simply the dry adiabatic lapse rate and I agree that this has nothing to do with greenhouse gases. However, the actual troposphere lapse rate is 6.5 °C per kilometre of elevation. It is water vapor that cools the atmosphere and creates this difference.
To be perfectly clear, without greenhouse gases to cool the atmosphere, the atmosphere would be a single, very hot, temperature and there would be no convection to cool the surface.
Sorry, I was not clear. Without greenhouse gases, the actual troposphere lapse rate would be zero because the atmosphere would be isothermal. The greenhouse gases cool the atmosphere producing a lapse rate of 6.5 °C per kilometre of elevation.
Robert Clemenzi,
Absent greenhouse gases, the atmosphere would not be isothermal. The surface temperature at the poles would still be colder than the surface temperature at the equator. That means the pressure will decrease less rapidly with altitude at the equator than at the poles. That causes what’s called a pressure gradient force which in turn causes air circulation. Any air circulation will move heat around and force the lapse rate towards the adiabatic rate. The free energy for the work needed to do this comes from the temperature difference between the poles and the equator. The circulation will increase the temperature at the poles and decrease the temperature at the equator by turbulent convection from the moving air. So even without greenhouse gases in the atmosphere, you would still have something like the Hadley cells and probably jet streams in the Northern and Southern hemispheres as well.
@Willis
Do you have no shame? That was SO out of context. It was preceded by saying the ocean primarily cools by evaporation. I have given the facts many times which anyone may verify in the literature. Ocean cools approximately 70% via evaporation, 25% via radiation, and 5% via conduction. Given that’s true, and it IS true, then IF the ocean doesn’t cool by longwave radiation then neither is it heated by it.
You have continually failed to dispute this, Willis. Calling it lunacy doesn’t make it that. It just makes you look weak in that you lack a substantive rebuttal.
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?
Willis writes “I find it absolutely ludicrous that so many people are defending something that they don’t understand … humans are indeed strange.”
Well thats your perception I suppose. I’m not defending his argument, I’m merely stating its not the strawman argument you’ve put forward (as did Ira) and given my own brief understanding of what his argument is very roughly about. I dont pretend to understand the detail and yet you put words into my mouth about exclusivity.
Anyway, it seem to me that its inescapable that the amount of atmosphere has a bearing on the temperature of a planet but its clearly not a simplistic argument that you’ve been putting forward. There is time of rotation, heat capacity as well as inversion layers vs convection and all sorts of possibilities beyond PV=nRT.
So whilst Nickolov may not even be correct about his version of the detail (whatever that is) the idea may still have merit.
Robert Brown says:
January 10, 2012 at 10:01 am
After that it became ellipsoid to the degree that it isn’t quite a completely rigid body.
“Don’t mind me, I just teach this stuff. Or if you prefer, can you provide evidence to back your description?”
Don’t mind me either. I’m just an engineer who designs mechanical systems around these principles. Teaching is a hobby.
The tides slow down the rotation they don’t lock it into any specific configuration. This loss in angular momentum is of course translated into heat. If it weren’t for mass distribution asymmetry what would determine which half of the moon would come to rest facing toward us? And if there is no preference for any particular face how can a particular face be the chosen one? In other words why isn’t it the other side of the moon that’s facing us?
One might also want to think about Mercury which is tidally locked into 1.5 revolutions per orbit and how that’s possible.
Per your request (and an answer to my question about Mercury’s wonky tidal lock):
Planetary Siences
Imke De Pater, Jack Jonathan Lissauer
http://books.google.com/books?id=RaJdy3_VINQC&pg=PA34&lpg=PA34&ots=oozdNGPUJ0&dq=tidal+lock+mass+distribution+asymmetry
Willis Eschenbach says:
January 11, 2012 at 1:02 am
You are shedding more heat than light (pun intended) on this gravitational heating canard. In your missive above you quote the formula for dry adiabatic lapse rate. Dry adiabatic lapse rate is for a RISING parcel of air i.e. a parcel of air in vertical motion. Nicholov and Huffman are describing static parcels of air. In a static parcel the decrease in temperature is simply a matter of moving farther away from the source of heat which in dry air the heat source is the surface until you reach thermopause where highly energetic regions of solar spectrum begin to directly heat the air in a big way. At the edge of the atmosphere the temperature is in the thousands of degrees!
Stephen Rasey says:
January 10, 2012 at 1:31 pm
“There has to be a typo here somewhere. Totals don’t match.”
No typo. That’s how climatre boffin accounting works.
The totals actually match. 390W out, 320W in, for a NET radiative out of 70W. Net radiative in is 170W. The rest leaves by evaporation and conduction not radiation. The actual numbers aren’t right. Actual heat budgets obtained from ocean studies (use google=scholar for “ocean heat budget”) put evaporative (latent) heat loss at about 150W/m2, radiative loss at 50W/m2, and conductive at less than 10W/m2.
Latent heat loss is the big Kahuna. Willis needs to somehow get a mental grasp on the fact that evaporative cooling is about 3 times more efficient than radiative cooling. Thermal radiation is a bit player over the ocean because of evaporative efficiency. Greenhouse gases are therefore also a bit player over the ocean because their modus operandi is thermal radiation. I have explained the physics involved many times in many ways. It is explained by theoretical physics and it is evidenced in ocean heat budget studies. There is no possible informed dispute. One has to be ignorant of the material properties of water (i.e. the physics), in denial of the observations made in myriad ocean heat budget studies, or just bullheaded and incapable of admitting an error.
Willis Eschenbach says:
January 11, 2012 at 1:42 am
David says:
January 10, 2012 at 11:40 pm
Robert Brown says:
January 10, 2012 at 9:19 am
—————————-
Thank you for your educational comments. I have a question if you are willing. …
… 2. It therefore follows that any effect which increases the residence time of LW energy in the atmosphere, but reduces the input of SW and LWIR energy entering the oceans, (such as clear sky water vapor and clouds) causes a net reduction in the earth’s energy balance, proportioned to the energy change involved, relative to the residence time of the radiations involved. (To paraphrase an old maxium, one unit of energy absorbed into the oceans is worth two in the atmosphere)
I’m not Robert Brown, David, and I hope he answers you. My problem with what you say is that I have no idea what you are calling the “residence time” of energy. It would be extremely useful if you could answer the following two questions:
How do you measure the residence time of energy, and in what units is it measured?
All the best w.
—————————————————————
Thanks for the questions Willis. By rsidence time I am referring to how long energy stays within a defined area. For earth that area is the ocean, the land, and the atmosphere. As for energy I am referring to heat, as measured in W/m 2, which of course can be latent heat as in the evaporation – condensation process. Each wavelength of incoming TSI has a different residence time within the atmosphere, land and ocean. This residence time is of course affected by it own inherent properties as well as all of the material it encounters. This can be very short, almost instant in the case of some SWR reflecting from a cloud and again lost in space, to very long for some photons of SWR which reach deep into the ocean from 660 to 3,000 feet (200 to 900 meters), where only about 1 percent of sunlight penetrates. This layer is known as the dysphotic zone (meaning “bad light”). http://www.scienceclarified.com/
How much the particular W/L of energy which comprises this 1% changes from solar cycle to solar cycle is unknown. I think that the long-term balance of these flows across the ocean surface partially determines the oceanic (and therefore the atmospheric) temperature. As a result, small sustained imbalances can cause gradual temperature shifts of the entire system. Solar spectrum energy mostly goes deep into the oceans which reflect only2-3% of sunlight; and the part of the earth which is mostly oceans is in the tropics where most of the solar energy arrives.
Absorption by the the H2O in the atmosphere does warm the atmosphere; in fact it is one of the major warming influences; and that results in long wave Ir radiation in an isotropic pattern, so only half of that radiative LWIR energy comes down to the surface. The other half goes upwards to space; so there is a net energy loss to the surface of about 1/2 of the amount that H2O vapor absorbs from the solar input. Reducing the total amount of solar energy that reaches the earth surface results in it getting cooler.; that can be seen instantly in a partial eclipse of thes sun; or standing in the shadow zone of a cloud. There is no H2O vapor, liquid or solid phase phenomenon anywhere in the atmopshere where an increase in water results in an increase in the ground level solar energy. Additionally evaporation conduction of latent heat may vary far more then realized and would decrease the ability of LWIR to warm the oceans. (Newell & Dopplick’s (1979) calculations that tropical temperatures cannot rise any further.) I hope this answers your questions. A simple analogy given below perhaps help clarify.
1. On a highway if ten cars per hour enter the highway, and the cars are on the road for ten hours before exiting, there will be 100 cars on the road and as long as these factors remain the same the system is in balance. If you change the INPUT to eleven cars per hour, then over a ten hour period the system will increase from 100 cars to 110 cars before a balance is restored and no further increase occurs. The same effect as the increase in INPUT achieves can be realized by either slowing the cars down 10% or by lengthening the road 10%. In either case you have increased the energy in the system by ten percent by either increasing the residence time or the input.
2. Now lets us take the case of a very slow or long road with the same input. Ten cars per hour input, 1000 hours on the road, now you have ten thousand cars on the road. Now lets us increase the input to eleven cars per hour just as we did on the road with a ten hour residence time. Over a 1,000 hour period we have the same 10% increase in cars (energy) However, due to the greater capacity on that road, the cars (energy) have increased 100 times relative to the 10 hour road with a 10% increase in input. (1,000 car increase verses a 10 car increase.) Any change in the input or the residence time of this 1,000 hour road will have a 100 times greater effect then on the 10 hour road if the input change endures for 1,000 hours. The ocean of course is the 1000 hour road, the atmosphere is the 10 hour road.
mkelly says:
January 10, 2012 at 10:50 am
“Why is the air “very thin”?”
Because there isn’t enough of it to stay dense all the way to the moon.
“And my bet is you would freeze before you melted at the altitudes you are talking about. A molecule may be “hot” but there are so few of them you would neverfeel it.”
You’d lose that bet because the same sun that heats that rarefied air would boil you in your own juices. We’re talking 1500W/m2 up there IIRC. That’s why satellites and such that have to maintain a fixed orientation have highlyt reflective foil all over them and/or must rotate so that they are not frozen on one side and roasted on the other.
You appear to be confusing heat capacity with absolute temperature. The air there is rarefied so it has little heat capacity and thus little ability to transfer heat to solid objects with far higher heat capacities. That’s the saving grace which allows the International Space Station, for instance, to orbit through air with a temperature in the thousands of degrees. But it still is very hot and that fact still isn’t explained within the context of Huffman’s or Nikolov’s gravitational hypothesis.
Izen says
http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-859566
Henry@Izen & Stephen Wilde
Hi Izen. Thx. The comparison of my own terrestial data (in RH) with that of the upper troposhere is a bit of a problem. It is not very likely to be comparable. As far as your 2nd reference is concerned, is seems the report is heavily biased towards ACC (AGW) and in this respect their own findings are actually somewhat contradictory.
Namely, for example, why then blame the increase in CO2 (your carbon footprint) rather then the increase in water (vapor) in the atmosphere CC?
I have posed this very possibility of an increase in water vapor due to human activities as a cause for man made warming,
http://www.letterdash.com/HenryP/more-carbon-dioxide-is-ok-ok
as very much more likely and probable than it being caused by an increase in carbon dioxide.
Both you Izen and the report say that the water content of the atmosphere has increased by 0.41 kg per m2 per decade since 1988. The units that are used here are a bit of a problem to me. How much is that in kg/m3?
1m3 air is about 1.2 kg so I don’t think the unit used is a mistake on the part of the report.
(0.41 kg per m3 per decade would mean the entire atmosphere being added with water in 3 decades and that is unlikely).
I also could not access the measuring- and standardisation procedures.
I must say that I doubt the results and I will tell you why.
I have looked very carefully at my own datasets \ 20 weather stations
http://www.letterdash.com/HenryP/henrys-pool-table-on-global-warming
and in my opinion I think that the warming is caused by more heat being slammed into the SH (oceans) due to a lesser appearance of clouds there, rather than more intense heat from the sun….
Does Stephen Wilde agree with me on that?
That being the case, (is it correct?) then my reported esitmate of -0.02% RH/ annum globally since 1974 actually makes sense (to me).
By the way Willis it was this understanding…”Solar spectrum energy mostly goes deep into the oceans which reflect only2-3% of sunlight; and the part of the earth which is mostly oceans is in the tropics where most of the solar energy arrives.” which promted me to question your assertion that albedo is greater in the tropics. ( By greater I mean the % of radiation affected) Yes the earths albedo is higher then the moons on average, but I support the possibility that it is relatively lower in the tropics and the SH. In general the oceans are a blackbody, absorbing whatever radiation reaches the surface with little reflectivity. The NH has a great deal of landmass north of the tropics, as well as year round snow and ice in the artic, as well as tremendous winter albedo beyond year round ice. The polar SH of course has antarctia with its very high albedo. Additionally the incident angle of sunligh creates ever greater reflectance as one moves further from the tropics. A further factor is the poles appear to have a great deal of consistent.cloud cover as I look at the global map on the right side of WUWT home page. For these reasons I would have to see actual meauserment to accept your assertion here, as I suspect that the tropics. especially the southern tropics have the lowest albedo as well as the greatest TSI, especially in January when the earth is thee million miles closer to the sun and TSI is close to 100 W/m2 greater then in July. Therefore the earths albedo varies far more then the moons, and may be lowesest as a % of TSI, where that energy enters the earths largest heat capacity, the oceans. This may indeed be a great factor in the earths relative warmth compared to the moon’s, despite the moons lesser but evenly distributed albedo.
Willis Eschenbach says:
January 10, 2012 at 11:31 am
“True, Joules, and there’s a good discussion of the subject here. At the salinity of most sea water (around 35 psu), the maximum density is at slightly below freezing.”
Of course it’s true. I wouldn’t have written it down if it weren’t. But that’s just an interesting factoid. The $64,000 question it raises is how the global ocean can possibly have an average temperature of 3.9C with 90% at a constant temperature of 3C with just a comparatively thin surface layer any warmer than that. It isn’t because the rocks underneath are cold. Those get up to millions of degrees G (degrees Gore).
The only explanation I can come up with is tha 3.9C is the average temperature of the surface taken over a complete glacial/interglacial cycle. IMO what we have to worry about is the fact that the Holocene interglacial and our civilization is a thin warm temporary skin floating on a bucket of icewater. Any disturbance in the force which keeps the thin warm layer from mixing too fast with the cold layer will plunge the globe back into the freezer in no time flat. Interglacials tend to have steep shoulders on them: http://en.wikipedia.org/wiki/File:Ice_Age_Temperature.png
Stephen Wilde says:
January 10, 2012 at 11:39 am
“The sun heats the surface, the surface heats the atmosphere, and the farther away from the source of heat the colder it gets. It’s the same “lapse rate” principle you experience as you move further away from a campfire.”
The heat near the campfire gets higher with more molecules
——————————————————————————
It was an analogy. Don’t try to pick it apart into quantum mechanics please. Energy tends to dissipate with distance from the source usually because it is expanding into a larger volume and where this implies an inverse square relationship between distance and intensity. The energy source is the earth. It is expanding into a sphere of infinitely larger volume. Energy density will therefore decrease with distance. Gravity has nothing to do with it. It could be a vacuum and this will still hold true so it has nothing to do with air pressure or density either.
@Robert G. Brown
So did you have time to digest the fact that seawater increases in density all the way down to its freezing point yet?
This puts a whole different spin on ocean temperature and how it got that way. A lot of people, even physics professors at Duke University, just assume it got that way because 3C is the highest density point so it sinks and becomes isolated because nothing colder can sink below it. Au contraire. Colder seawater is quite free to below it. So this must be the equilibrium temperature instead of the highest density point. Unless I’m wrong but so far no one jhas explained any other way for the ocean to be a relatively constant 3C everywhere below a few hundred meters deep. I’d really like to hear the opinion of a newly enlightened Duke University physics professor on this matter.
Izen says
http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-859566
Henry@Izen & Stephen Wilde
Hi Izen. Thx. The comparison of my own terrestial data (in RH) with that of the upper troposhere is a bit of a problem. It is not very likely to be comparable. As far as your 2nd reference is concerned, is seems the report is heavily biased towards ACC (AGW) and in this respect their own findings are actually somewhat contradictory.
Namely, for example, why then blame the increase in CO2 (your carbon footprint) rather then the increase in water (vapor) in the atmosphere CC?
I have posed this very possibility of an increase in water vapor due to human activities as a cause for man made warming,
http://www.letterdash.com/HenryP/more-carbon-dioxide-is-ok-ok
as very much more likely and probable than it being caused by an increase in carbon dioxide.
Both you Izen and the report say that the water content of the atmosphere has increased by 0.41 kg per m2 per decade since 1988. The units that are used here are a bit of a problem to me. How much is that in kg/m3?
1m3 air is about 1.2 kg so I don’t think the unit used is a mistake on the part of the report.
(0.41 kg per m3 per decade would mean the entire atmosphere being added with water in 3 decades and that is unlikely).
I also could not access the measuring- and standardisation procedures.
I must say that I doubt the results and I will tell you why.
I have looked very carefully at my own datasets \ 20 weather stations
http://www.letterdash.com/HenryP/henrys-pool-table-on-global-warming
and in my opinion I think that the warming is caused by more heat being slammed into the SH (oceans) due to a lesser appearance of clouds there, rather than more intense heat from the sun….
Does Stephen Wilde agree with me on that?
That being the case, (is it correct?) then my reported estimate of -0.02% RH/ annum globally since 1974 actually makes sense (to me).
AAAARgh
it (the system) does not want to notify me of follow up comments
I try again
Willis,
The adiabatic lapse rate fixes the relationship of pressure to volume, PV^γ = constant. The value of γ is determined by the ratio of the gas heat capacity at constant pressure to the heat capacity at constant volume, which in turn is related to the degrees of freedom of movement of the gas molecules or atoms. 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.
Joules Verne,
The ocean behaves like the atmosphere only it’s much less compressible. If the temperature decreases with depth, you don’t get convection because the warmer water above is less dense than the colder water below. At the poles, the surface water becomes colder with higher salinity because of evaporation so it’s denser than the water below and it sinks. That forces upwelling everywhere else, more some places than others. But the surface of the ocean in the tropics and mid-latitudes is warmed by sunlight. Turbulent convection, also called eddy diffusion, carries that heat from the warm surface toward the colder depths. But since you also have upwelling cold water, you get a steady state where the warm surface water transitions to the cold deep water. That’s the thermocline. The depth of the thermocline varies seasonally, but the average depth remains constant.
DeWitt Payne says:
January 11, 2012 at 7:57 am
Good points. Thanks. I wonder how high up those effects would reach. The jet stream is currently located just below the tropopause, a feature that is created by greenhouse gases. Perhaps a similar stream would occur at the top of the nightly temperature inversion. Or perhaps the inversion would be more permanent because cold air might tend to flow from one pole (winter) to the other (summer).
DeWitt Payne says:
January 11, 2012 at 7:57 am
“Absent greenhouse gases, the atmosphere would not be isothermal. The surface temperature at the poles would still be colder than the surface temperature at the equator. That means the pressure will decrease less rapidly with altitude at the equator than at the poles. That causes what’s called a pressure gradient force which in turn causes air circulation. Any air circulation will move heat around and force the lapse rate towards the adiabatic rate. ”
Very true, DeWitt. Even on Venus with a practically opaque atmosphere (the opposite condition) the convection bands are clearly visible, and there heating would be from near the top of the atmosphere. It seems that in the real world with a fluid atmosphere convection is ubiquitous and you can’t escape the adiabatic lapse rate. Probably even true of the ocean. I’d caution Nicol that composition does affect lapse rate tho’.
Joules Verne says:
January 11, 2012 at 6:43 am
@Robert G. Brown
So did you have time to digest the fact that seawater increases in density all the way down to its freezing point yet?
This puts a whole different spin on ocean temperature and how it got that way. A lot of people, even physics professors at Duke University, just assume it got that way because 3C is the highest density point so it sinks and becomes isolated because nothing colder can sink below it. Au contraire. Colder seawater is quite free to below it. So this must be the equilibrium temperature instead of the highest density point. Unless I’m wrong but so far no one jhas explained any other way for the ocean to be a relatively constant 3C everywhere below a few hundred meters deep.
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.
I would suggest that a non-GHG atmosphere could make the moon much warmer than it is now on average.
I think that I agree, although I’m still working on just how and why. However, one has to be careful — Willis proposes a perfect non-GHG atmosphere, which is probably self-contradictory — matter is made up of charge, charged matter interacts with the electromagnetic field and has a heat capacity, and surrounding the earth with a neutral dielectric material with a heat capacity would suffice to trap some radiation and warm the surface, I think. I suspect that this is probably the reason that there is (if the NZ curve is correct — I do not have any easy way to check that it is) that there is an heuristic/empirical relationship between density of atmosphere and surface temperature. The GHE simply takes into account nonlinear gain due to the nonlinear susceptibility of real gas components, but I suspect there would be some linear gain without it. But I haven’t worked out the algebra, so I’m not certain.
I also don’t know about the word “much”. Without a physical argument in mind, it is difficult to estimate an order of magnitude, don’t you think?
The reason I think that it would increase surface temperatures (on average) is simple enough. Sunlight comes in (warming the atmosphere and the surface in some differential way, we’ll assume warming the surface a lot more than the atmosphere because the latter is “transparent”. As the surface warms, it further warms the atmosphere, setting up convection rolls that lift heat from the surface up and bring cold “air” down to be warmed, gradually mixing to distribute the heat in the atmosphere. The surface has a low albedo; the atmosphere doesn’t really have much albedo. The surface has a high emissivity, the atmosphere has an emissivity (it is made up of dielectrically polarizable charges) but we’ll assume that it is relatively low as well, see “transparent” to first order.
This daytime cooling of the surface reduces its peak temperature and hence lowers (on average) its loss of energy during the day, while at the same time storing some of the energy in the atmosphere. Quite possibly a lot of energy (a significant fraction of the total incident heat, that is). Nighttime falls, and the cycle reverses. The surface radiatively cools until it is cooler than the air, but the vast convective rolls established during the day have an inertia and air continues to flow over the surface, warming it, especially if there is differential cooling (regions with different emissivity and albedo and hence different temperatures during the day) to continue to help drive convective heat transfer. Even a non-GHG will also return some of the heat absorbed during the day to the ground (slowly, sure, with its lower emissivity) in the absence of convection, just as the non-GHG will slowly cool via radiation on its own while still absorbing some of the heat given off by the ground and slowing its transit out of the system.
Even without this, however — even if the wind stops and one straight up radiatively cools the surface, chilling a surface layer of air by (slow) conduction — one cools from a lower initial temperature and so the cooling is slower, less net energy is lost than would have been otherwise. Temperature inhomogeneity favors cooling, and by storing heat in the atmosphere it acts as “thermal ballast” to reduce the daytime peak temperature and nighttime minimium temperature by absorbing and storing some of the heat in the day and giving it back at night.
Make the air wet air, add water to the mix, even ignoring its GHG properties, and things change once again. Oceans act as major thermal reservoirs that do two things. One is drive trade winds and climate oscillations — large scale thermal imbalances, sustained over long times, cause huge chunks of atmosphere to get caught up moving in giant closed loops. These loops enormously enhance the process above, again quite independent of GHE per se. You have two big reservoirs for solar thermal energy buffering the daytime-nighttime fluctuation, reducing (mostly) thermal variation, and hence (mostly) increasing the global mean temperature. You’ve put the planet in a blanket (the atmosphere) with a hot water bottle that warms more during the day than it cools at night) the oceans compared to the land, buffering surface temperatures on a decadal timescale on Earth.
I don’t remember who it was — Willis, Crosspatch, somebody — who suggested on this or some other thread that the ocean was the elephant in the room that was being carefully ignored compared to the GHE. I’d say it is one of an entire menagerie — there are multiple reservoirs of heat, all of which contribute to net warming outside of the GHE, so that the warming we observe from the theoretical “vacuum Earth” estimate is not all CO_2 based GHE warming, it has to be split up. I’m guessing — guessing in moderate ignorance, mind you — that it should probably be split up roughly in thirds — if the anomaly in the mean is 30C, 10C of that is probably due to the ocean, 10C is probably due to the fact that we have an atmosphere, and 10C is due to GHGs. But it could be 15, 10, 5; 10, 5, 15, or some other split — without solving a complete coupled hydrodynamic radiative model including the ocean and continental structure and solar coupling including any GCR variation-linked effects and… it would be very difficult, I think, to soundly justify one split or another outside of pulling numbers out of your ass. Maybe not — its the kind of thing were I’m open to argument.
But any assertion that Earth lacking CO_2 altogether would suddenly revert to the same mean temperature as the moon seems as though it would be obviously false. It might kick us into permanent glaciation, but I can’t see the tropics ever permanently freezing. We probably got close to this in the last glacial epoch, where it is asserted (based on evidence) that atmospheric CO_2 dropped to 180 ppm, roughly 1/2 what it is today. Again, if there were a hole pointing downward toward true coldside catastrophe, atmosphere freezing out and all that, this sort of event would likely have triggered it. Negative feedback from the ocean and the rest of the atmosphere very likely blocked that kick; clearly the Earth was able to warm back to interglacial (current) average temperatures.
rgb
“I have looked very carefully at my own datasets \ 20 weather stations
http://www.letterdash.com/HenryP/henrys-pool-table-on-global-warming
and in my opinion I think that the warming is caused by more heat being slammed into the SH (oceans) due to a lesser appearance of clouds there, rather than more intense heat from the sun….
Does Stephen Wilde agree with me on that?”
Hi Henry.
I agree that the data does seem to show that cloudiness globally did decrease during the late 20th century warming spell.
I put it down to more poleward/zonal jetstreams. Svensmark thinks it is due to less cosmic rays but I think he is wrong.
Since the late 90s the jets started to become more equatorward/meridional and global cloudiness is on the rise again. Meanwhile ocean heat content has stopped rising.
Interesting that the sun is less active too !!!
“You’ve put the planet in a blanket (the atmosphere) with a hot water bottle that warms more during the day than it cools at night) the oceans compared to the land, buffering surface temperatures on a decadal timescale on Earth.
I don’t remember who it was — Willis, Crosspatch, somebody — who suggested on this or some other thread that the ocean was the elephant in the room that was being carefully ignored compared to the GHE”
Yay !!!
http://climaterealists.com/index.php?id=1487&linkbox=true&position=5
“The Hot Water Bottle Effect”
It has taken nearly four years but is now entering the vernacular.
The tides slow down the rotation they don’t lock it into any specific configuration. This loss in angular momentum is of course translated into heat. If it weren’t for mass distribution asymmetry what would determine which half of the moon would come to rest facing toward us? And if there is no preference for any particular face how can a particular face be the chosen one? In other words why isn’t it the other side of the moon that’s facing us?
Consider a pendulum with a frictionless pivot. Give it a kick strong enough to make it rotate. It will rotate forever, because gravity points down and exerts no torque, and we made the pivot frictionless so that won’t slow it down either.
Now add damping — damping that is strictly due to oblate deformation in the case of the moon as both I and all of the textbooks assert and as you can easily understand by looking at simple mechanical pictures and can even test by comparing to a derived equation, but we might as well damp from the pivot for our pendulum example, or if you want a less extreme example a rotating disk with one side heavier than the other. As long as the pivot exerts a damping torque, it slows down, until eventually the energy needed to rotate the heavy side up and over the top is less than the energy available in the system, The heavy side rises one final time and — a-l-m-o-s-t — makes it over the top, but then it falls back. It now behaves like an ordinary damped oscillator until it comes to rest, heavy side down.
In the specific case of the moon, the tidal forces on the near side and far side of the moon aren’t quite symmetric. They are only symmetric to leading order. If you want, I’ll either derive or point you to a derivation of the effective tidal force/field on the near side vs the far side, but I assure you that they aren’t identical and that the near side field is slightly stronger. Given that the moon is still (probably) internally molten at this point — remember, the damping force that slowed it down was plastic deformation that heated the interior of the moon right up to the point where the moon was fully tidally locked — slow oscillations as it came to rest “shook” a small surplus onto the side facing the Earth.
Personally I doubt very much that there was much average azimuthal asymmetry in the mass distribution of the moon throughout the damping process. The moon was far closer to the Earth, and the tidal forces acting on it were far stronger (perhaps 8x stronger or thereabouts, again, can’t remember the number and am too lazy to look it up, but it is known because it is computable from current observations and models). The moon was almost certainly still a relatively thin and uniform crust surrounding a more or less equilibrated interior of hot squishy rock, kept uniform by the fact that it was literally turning on a spit that exposed the entire moon to the same deforming forces, day after day after day as the tidal bulge chased the Earth-centered antipodes. So I think that most of the asymmetry was created during the cooling process. But I don’t much care — the simple, easy to understand, and obviously correct physical argument above shows you how the heavy side would always end up “down”, facing the earth, even though “down” is an inverse fourth-order down compared to gravity (second order) and tide (third order).
rgb
David says: (a bunch of stuff).
Yeah, sounds very reasonable, and I pretty much agree. Both the ocean and the atmosphere without considering GHGs contribute to the net warming of the surface relative to an atmosphere-free planet, both by doing nonlinear stuff that traps heat (in the case of oceans) on their own and by acting as thermal ballast in a vast circulatory model that stores some of the heat that would otherwise “just” warm the surface and radiate back to space.
Personally I have little feel for the relative magnitudes (as I just mentioned on another thread) but the total “anomaly” of the Earth above the “vacuum Earth” purely radiative temperature really needs to be split up among GHGs, atmospheric convection (for a wet atmosphere with all sorts of complexity including vast decadally-sustained patterns of atmospheric heat and moisture transport) and the ocean, or maybe between GHGs, Water, Air (GH neutral, mostly), and GHGs with a strong absorption in the IR. There is no way that all 30C (or whatever) is “just” due to CO_2. It may well be that most of it is not due to CO_2. As I said, I don’t have a good idea of the physics here — I’m still a, what was the word my Ph.D. advisor used after prelims, oh, yeah dilletante of climate science, learning as I go.
rgb
The only explanation I can come up with is tha 3.9C is the average temperature of the surface taken over a complete glacial/interglacial cycle. IMO what we have to worry about is the fact that the Holocene interglacial and our civilization is a thin warm temporary skin floating on a bucket of icewater. Any disturbance in the force which keeps the thin warm layer from mixing too fast with the cold layer will plunge the globe back into the freezer in no time flat. Interglacials tend to have steep shoulders on them: http://en.wikipedia.org/wiki/File:Ice_Age_Temperature.png
Just to show that I don’t always argue with you, I agree about the Holocene being a warm skin floating on a bucket of icewater, and the Younger Dryas is evidence (if hypothesized Oceanic Conveyor Belt explanations are correct) that disturbing the surface transport of stored heat can quickly kick the Earth back into the deep freeze. I also think that there is considerable cause for 40,000 year concern — the general trend of glaciation seems to be getting colder during the glacial eras, and it doesn’t need to get much colder at the coldest to enable various catastrophes that seriously compromise the human race’s ability to survive at current population levels with anything like our current technology.
As for the 3.9C — oceanic ice forms in a complex way, because salt water has a lower freezing point than fresh water. Hence ice nucleation is around fresh-direction fluctuations and both grows preferentially with fresher water (increasing the salinity of what it leaves behind) and actually sheds salt, which remains liquid, more dense, and melts its way through the ice. Icebergs are hence (mostly) fresh water, and I’m guessing that they precipitate out at the point where the greatest freshwater density occurs (nucleation point) as that is where there would begin to be a chemical potential favoring separation. The heat of fusion goes back into the ocean, carried by the (more saline) salt water. Ocean ice still floats, which forms a nice insulating layer with more or less fixed temperature on the lower surface of the ice. So the ocean still doesn’t freeze from the bottom up, but from the top down, and otherwise I agree about the downwelling and upwelling and OCB and all that.
I would expect that another contributor to bottom temperatures — possibly weak, but not necessarily negligible — is outgoing heat from the Earth. Yes, the energy flux is low, but the energy released at the bottom doesn’t go anywhere in a hurry, so there a long time for heat to accumulate. It might be enough to keep the bottom above freezing even if the ocean otherwise froze all the way down to near the bottom.
I don’t think that it is “just” the average temperature over the entire glacial cycle. For one thing, it doesn’t really match the average surface temperature during glaciation, let alone during the warmer glaciation. Also, IIRC the timescale for oceanic equilibration is supposed to be roughly 1000 years. That means (if true) that it is well-equilibrated as far as things like average temperature are concerned during the Holocene, and the rapid plunge is somewhat more likely to be due to flipping from one global pattern of heat transport (primarily confined to the surface) to another more than major bulk alteration of overall heat content.
But I’m happy to be educated, here, if you know better.
rgb
The adiabatic lapse rate fixes the relationship of pressure to volume, PV^γ = constant. The value of γ is determined by the ratio of the gas heat capacity at constant pressure to the heat capacity at constant volume, which in turn is related to the degrees of freedom of movement of the gas molecules or atoms. 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.
Ah, this is exactly what I was missing in this discussion. OK, assuming I know what \gamma is and all about these relations (true enough) where can I learn about it in the direct context of the atmosphere and convection? Also assume that I can do algebra and calculus and even PDEs and all that (I’m a theorist and can do serious math if my interest motivates it). Ideally a reasonably easy introduction — baby steps first…
By the way, your replies are cogent and clearly well-founded. I appreciate it. I’m guessing that this is directly related to the way the atmosphere stores dayside/tropical heat and moves it in all-scale convective rolls north and south, coriolis and continent deflected into major surface patterns, which is the way even a GHG-free atmosphere would reduce the moon-like rates of cooling by establishing more uniform overall temperatures, right?
rgb
Robert Brown,
My introduction to Physical Meteorology was by reading Rodrigo Caballero’s Lecture Notes on Physical Meteorology ( http://maths.ucd.ie/met/msc/PhysMet/PhysMetLectNotes.pdf (big, 28 MB pdf). He’s located at University College Dublin. This is something of a work in progress and has expanded over the years. I presume that it will eventually end up as a textbook. If you have access to a University library, you could probably find other textbooks like Curry and Webster’s Thermodynamics of Atmospheres and Oceans, Ray Pierrehumbert’s Principles of Planetary Climate or Grant Petty’s A First Course in Atmospheric Thermodynamics. If I were going to buy one myself, I would go with Grant Petty because I find his A First Course in Atmospheric Radiation indispensable. It may not be as rigorous as Goody and Yung, but it’s very accessible. Both of Petty’s books can be ordered direct from the publisher, Sundog Publishing, at a significant discount.
As to your question on circulation, I think that’s correct, but I’m still learning about this stuff too.
Robert Brown says:
January 11, 2012 at 12:16 pm
“Consider a pendulum with a frictionless pivot. “
Bad analogy, unless you specify where the pivot is located. In space, the “pivot” is at the CG. A better analogy would be a dumbbell supported at the midpoint. There is no preference for heavier or lighter side to end up facing the Earth. It is a flip of the coin.
The theory for this is exceedingly well-established, as hundreds of satellites relay upon gravity gradient stabilization (which you can google for voluminous references) for attitude control or attitude assist.
“…rely upon…” a
I should have mentioned, what happens is, the minimum energy configuration is one in which the object ends up rotating about its major axis of inertia, whereas the minor inertia axis ends up parallel to the Earth-to-satellite vector.
“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.
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]
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.
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.
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.
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.
Joules Verne,
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.
“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.
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.
“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.
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.
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.
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’.
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.
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.
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!
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:
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.
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?
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.
“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.
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.
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
“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
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 could tell you weren’t an organic chemist. Since all they know, mostly, is IR spectrometry, it’s all reciprocal centimeters to them.
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
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”
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
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.
Joules Verne says:
January 11, 2012 at 3:12 am
Failed to dispute it? First I’ve read those estimates of how the ocean loses energy, Joules. I fear I must confess that I pay little attention to your writings. Your current claims are a perfect example of why.
The amount of surface cooling by radiation can be estimated by the Stefan-Boltzmann equation. It is also measured directly by satellite, as well as from ships and buoys. Heck, you can buy a handheld remote thermometer and measure how much cooling radiation is being emitted by objects around you. As a global average, surface cooling by radiation is on the order of 400 W/m2.
The amount of cooling by evaporation can be estimated by looking at the amount of precipitation, which globally averages out to about a metre of water per year. The energy needed to evaporate that amount of water works out to about 80 W/m2.
Finally, we have direct cooling by conduction. This is estimated to be on the order of 20 W/m2.
So your numbers are a long, long, long ways from the generally accepted (and in some cases measurable) values for radiation, evaporation, and conduction. Here are the differences:
Radiation – Best estimate 80%, you say 25%. Since this one is actually measured, and the measurements generally agree with the S-B equation, I have great confidence in the larger value.
Conduction – Best estimate 4%, you say 5%.
Evaporation – Best estimate 16%, you say 70%. This last one is also demonstrably wrong. Here’s why it is incorrect.
At 16% of the heat loss being from evaporation (best estimate), that’s the energy needed to evaporate one metre of water. Your claim, on the other hand, is that 4.4 metres of water were evaporated. But what goes up must come down, so where did the water go? Our estimates of global precipitation may be off, but we are not underestimating global precipitation by a factor of four. We’ve been measuring rain for a long time now. We’re not that far wrong. So your estimate must be in error.
Anyhow, Joules, your numbers are so wildly different from the generally accepted measurements and estimates that I mostly just skip over your stuff. Sorry, our worldviews just don’t intersect enough to discuss things.
Details of the derivation of the numbers I gave above are available here.
All the best,
w.
Willis Eschenbach @ January 11, 12:01 am
Willis, as to your first point:
Perhaps my use of the word “nonsense” was not the best, but what I meant was that to apply S-B over a supposed average temperature of an airless sphere based on average insolation spread over that sphere does not give a sensible result. The energy emission (cooling) under the solar-noon hotspot is very much greater than elsewhere, per T^4, and on the dark side, the rate of cooling drops exponentially to a very much lower level.
Averaging temperature within an opaquish atmosphere like Earth may be “good enough” without worrying about T^4, but definitely NOT in an airless situation. Furthermore, if I understand your figure correctly, it shows for a single point on the surface, the longitudinal variation in surface T over several lunar days. However, the latitudinal T variation between those terminations MUST surely be significantly greater because of the greater time spent in darkness or close to darkness!
As to your third point, (I’ll come back to the second):
[1] I’ve had a quick read of your source paper referenced in your figure in your article, and note that everything back in 1972 was rather tentative, for example, quote: ”One possible explanation of the poor data was that these heat flow experiments were considered as trial runs for the “real” experiments that were scheduled for the Apollo 18 mission and beyond”.. What strikes me is that I think there must be something rather strange and unexpected in the thermal properties of the lunar regolith, according to that early data.
[2] Thanks muchly for your Santa Rosa data of a lag of ~2.75 Hrs to peak air T. (I guess in a generally cloud free sky). It gives me a good feeling for my intuition that the surface T on an Earth with no atmosphere would stabilize in less than 2 hours. That’s putting aside the puzzling lunar regolith affected data. (or might that puzzle be a human intuitive smoothing of the data?).
BTW, CASUALLY, when you say noon, is that real noon or what we do in Oz, where we add an hour for summer time. (non importante)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Willis, I’ve just noticed the time, and I have to go, and don’t have time to come back to your second point.
Best.
Willis says…”The amount of cooling by evaporation can be estimated by looking at the amount of precipitation, which globally averages out to about a metre of water per year. The energy needed to evaporate that amount of water works out to about 80 W/m2….”
Willis, is the 80 W/m2 averaged to the entire globe, or to the 70% of the globe which the oceans cover?
With regard to evaporation it appears reasonable to note that the evaporation rate increases dramatically as T rises. The point being that a considerably larger portion of a GHG 1.7 W/m2 increase in DLWR would be used in evaporation then the global averages discussed here. This is part of the limitation of temperature in the tropics I referenced here http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-860609 in response to your questions to me.
Kind regards
David
RE: Joules Verne: (January 10, 2012 at 7:36 am)
“CRISP expounds: ‘The 2nd Law of Thermodynamics: You cannot tranfer heat from a cold body (upper atmosphere) to a hotter body (the earth surface) without doing work. This applies to absolute transfer, not to nett transfer, despite the bullshit being spouted by the alarmists. ‘
“That is true but it should be qualified by saying there is no net transfer of energy from the colder to the warmer.
“What actually happens is that the rate of net energy transfer from the warmer object (the earth) to the colder object (the cosmos) is slowed down. So it is quite correct to say that the cooler cannot heat the warmer but the rate at which energy moves from warmer to cooler can be throttled by intermediaries like London Fog coats and London fog banks.”
Imagine two indefinitely large plates facing each other. The cooler upper plate is radiating according to the Stefan-Boltzmann law, 150 watts per square meter and the warmer, lower plate is radiating 200 watts per square meter. These are *absolute* radiant energy emission levels. The upper plate is receiving a net 50 watts per square meter and the warmer, lower plate is losing the same *net* amount. When both plates reach the same temperature, both plates will emit energy at the same rate and the net radiant energy transfer will be zero.
@rgb
Just so you don’t think I’m arguing with everything you say… I share your loathing of inverse centimeter wavenumbers. These bastids are obviously an insididious invention of some mad scientist bent on sowing confusion, uncertainty, and dread (CUD) amongst the cow-ering masses.
I spent what little of last night that I semi-slept in a learning-dream state chewing over Caballero’s book and radiative transfer, and came to two insights. First, the baseline black-body model (that leads to T_b = 255K) is physically terrible, as a baseline. It treats the planet in question as a nonrotating superconductor of heat with no heat capacity. The reason it is terrible is that it is absolutely incorrect to ascribe 33K as even an estimate for the “greenhouse warming” relative to this baseline, as it is a completely nonphysical baseline; the 33K relative to it is both meaningless and mixes both heating and cooling effects that have absolutely nothing to do with the greenhouse effect. More on that later.
I also understand the greenhouse effect itself much better. I may write this up in my own words, since I don’t like some of Caballero’s notation and think that the presentation can be simplified and made more illustrative. I’m also thinking of using it to make a “build-a-model” kit, sort of like the “build-a-bear” stores in the malls.
Start with a nonrotating superconducting sphere, zero albedo, unit emissivity, perfect blackbody radiation from each point on the sphere. What’s the mean temperature?
Now make the non-rotating sphere perfectly non-conducting, so that every part of the surface has to be in radiative balance. What’s the average temperature now? This is a better model for the moon than the former, surely, although still not good enough. Let’s improve it.
Now make the surface have some thermalized heat capacity — make it heat superconducting, but only in the vertical direction and presume a mass shell of some thickness that has some reasonable specific heat. This changes nothing from the previous result, until we make the sphere rotate. Oooo, yet another average (surface) temperature, this time the spherical average of a distribution that depends on latitude, with the highest temperatures dayside near the equator sometime after “noon” (lagged because now it takes time to raise the temperature of each block as the insolation exceeds blackbody loss, and time for it to cool as the blackbody loss exceeds radiation, and the surface is never at a constant temperature anywhere but at the poles (no axial tilt, of course). This is probably a very decent model for the moon, once one adds back in an albedo (effectively scaling down the fraction of the incoming power that has to be thermally balanced).
One can for each of these changes actually compute the exact parametric temperature distribution as a function of spherical angle and radius, and (by integrating) compute the change in e.g. the average temperature from the superconducting perfect black body assumption. Going from superconducting planet to local detailed balance but otherwise perfectly insulating planet (nonrotating) simply drops the nightside temperature for exactly 1/2 the sphere to your choice of 3K or (easier to idealize) 0K after a very long time. This is bounded from below, independent of solar irradiance or albedo (or for that matter, emissivity). The dayside temperature, on the other hand, has a polar distribution with a pole facing the sun, and varies nonlinearly with irradiance, albedo, and (if you choose to vary it) emissivity.
That pesky T^4 makes everything complicated! I hesitate to even try to assign the sign of the change in average temperature going from the first model to the second! Every time I think that I have a good heuristic argument for saying that it should be lower, a little voice tells me — T^4 — better do the damn integral because the temperature at the separator has to go smoothly to zero from the dayside and there’s a lot of low-irradiance (and hence low temperature) area out there where the sun is at five o’clock, even for zero albedo and unit emissivity! The only easy part is to obtain the spherical average we can just take the dayside average and divide by two…
I’m not even happy with the sign for the rotating sphere, as this depends on the interplay between the time required to heat the thermal ballast given the difference between insolation and outgoing radiation and the rate of rotation. Rotate at infinite speed and you are back at the superconducting sphere. Rotate at zero speed and you’re at the static nonconducting sphere. Rotate in between and — damn — now by varying only the magnitude of the thermal ballast (which determines the thermalization time) you can arrange for even a rapidly rotating sphere to behave like the static nonconducting sphere and a slowly rotating sphere to behave like a superconducting sphere (zero heat capacity and very large heat capacity, respectively). Worse, you’ve changed the geometry of the axial poles (presumed to lie untilted w.r.t. the ecliptic still). Where before the entire day-night terminator was smoothly approaching T = 0 from the day side, now this is true only at the poles! The integral of the polar area (for a given polar angle d\theta) is much smaller than the integral of the equatorial angle, and on top of that one now has a smeared out set of steady state temperatures that are all functions of azimuthal angle \phi and polar angle \theta, one that changes nonlinearly as you crank any of: Insolation, albedo, emissivity, \omega (angular velocity of rotation) and heat capacity of the surface.
And we haven’t even got an atmosphere yet. Or water. But at least up to this point, one can solve for the temperature distribution T(\theta,\phi,\alpha,S,\epsilon,c) exactly, I think.
Furthermore, one can actually model something like water pretty well in this way. In fact, if we imagine covering the planet not with air but with a layer of water with a blackbody on the bottom and a thin layer of perfectly transparent saran wrap on top to prevent pesky old evaporation, the water becomes a contribution to the thermal ballast. It takes a lot longer to raise or lower the temperature of a layer of water a meter deep (given an imbalance between incoming radiation) than it does to raise or lower the temperature of maybe the top centimeter or two of rock or dirt or sand. A lot longer.
Once one has a good feel for this, one could decorate the model with oceans and land bodies (but still prohibit lateral energy transfer and assume immediate vertical equilibration). One could let the water have the right albedo and freeze when it hits the right temperature. Then things get tough.
You have to add an atmosphere. Damn. You also have to let the ocean itself convect, and have density, and variable depth. And all of this on a rotating sphere where things (air masses) moving up deflect antispinward (relative to the surface), things moving down deflect spinward, things moving north deflect spinward (they’re going to fast) in the northern hemisphere, things moving south deflect antispinward, as a function of angle and speed and rotational velocity. Friggin’ coriolis force, deflects naval artillery and so on. And now we’re going to differentially heat the damn thing so that turbulence occurs everywhere on all available length scales, where we don’t even have some simple symmetry to the differential heating any more because we might as well have let a five year old throw paint at the sphere to mark out where the land masses are versus the oceans, and or better yet given him some Tonka trucks and let him play in the spherical sandbox until he had a nice irregular surface and then filled the surface with water until it was 70% submerged or something.
Ow, my aching head. And note well — we still haven’t turned on a Greenhouse Effect! And I now have nothing like a heuristic for radiant emission cooling even in the ideal case, because it is quite literally distilled, fractionated by temperature and height even without CO_2 per se present at all. Clouds. Air with a nontrivial short wavelength scattering cross-section. Energy transfer galore.
And then, before we mess with CO_2, we have to take quantum mechanics and the incident spectrum into account, and start to look at the hitherto ignored details of the ground, air, and water. The air needs a lapse rate, which will vary with humidity and albedo and ground temperature and… The molecules in the air recoil when the scatter incoming photons, and if a collision with another air molecule occurs in the right time interval they will mutually absorb some or all of the energy instead of elastically scattering it, heating the air. It can also absorb one wavelength and emit a cascade of photons at a different wavelength (depending on its spectrum).
Finally, one has to add in the GHGs, notably CO_2 (water is already there). They have the effect increasing the outgoing radiance from the (higher temperature) surface in some bands, and transferring some of it to CO_2 where it is trapped until it diffuses to the top of the CO_2 column, where it is emitted at a cooler temperature. The total power going out is thus split up, with that pesky blackbody spectrum modulated so that different frequencies have different effective temperatures, in a way that is locally modulated by — nearly everything. The lapse rate. Moisture content. Clouds. Bulk transport of heat up or down via convection. Bulk transport of heat up or down via caged radiation in parts of the spectrum. And don’t forget sideways! Everything is now circulating, wind and surface evaporation are coupled, the equilibration time for the ocean has stretched from “commensurate with the rotational period” for shallow seas to a thousand years or more so that the ocean is never at equilibrium, it is always tugging surface temperatures one way or the other with substantial thermal ballast, heat deposited not today but over the last week, month, year, decade, century, millennium.
Yessir, a damn hard problem. Anybody who calls this settled science is out of their ever-loving mind. Note well that I still haven’t included solar magnetism or any serious modulation of solar irradiance, or even the axial tilt of the earth, which once again completely changes everything, because now the timescales at the poles become annual, and the north pole and south pole are not at all alike! Consider the enormous difference in their thermal ballast and oceanic heat transport and atmospheric heat transport!
A hard problem. But perhaps I’ll try to tackle it, if I have time, at least through the first few steps outlined above. At the very least I’d like to have a better idea of the direction of some of the first few build-a-bear steps on the average temperature (while the term “average temperature” has some meaning, that is before making the system chaotic).
rgb
@ Robert Brown
You asked me if I was still following this thread and mentioned earlier about some PDF you thought should be posted as a primer – can you clarify?
Your post above is being elevated to a story.
Anthony
Robert,
You can do a lot of what you want to do with a spreadsheet. I’ve done several of those examples already. I’ve never tried axial tilt, though. I think that goes way beyond a spreadsheet. For the lunar model, what you really need is a diffusive heat transfer surface. If you use a well-mixed (effectively superconducting for the purposes of the model) surface layer with finite heat capacity, the surface temperature doesn’t fall fast enough when the sun goes down and heats up too slowly as the sun rises. I haven’t done that either. It works moderately well for a water covered sphere, though. The moon rotates slowly enough that the thermal lag time is small compared to the rotation rate. That’s not true for a sphere rotating in 24 hours. Then you do see lag.
It would appear that the “superconducting” Earth/Moon is a fiction resulting from an attempt to make the characteristic temperature of the average energy flow equal to the average temperature. The simple thing to do is to simply accept the fact that the Stefan-Boltzmann characteristic temperature of the average energy flow is *NOT* the average temperature. It never was. The solar and terrestrial heat balance equations are based on average energy flows, not average temperatures. It even might make more sense to calculate the greenhouse effect as a function of average energy flow rather than temperature. Radiant energy flow is proportional to the fourth power of the temperature. Thus most of this flow originates from the warmer parts of the planet.
For those critics of wavenumber plots, the point to keep in mind is that the wavenumbers in cycles per centimeter or ‘kayzers’ is a measure of frequency and thus have the advantage of uniform radiant energy density. One only need multiply those values by the speed of light in cm/sec to get the frequencies in Hz.
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?
jae,
Zero emissivity would mean zero absorptivity, which would mean perfect reflectivity and no power absorbed. But of course like a perfect black body, a perfect reflector can’t exist. Its temperature would be undefined. Any measured temperature would actually be the temperature of the surroundings.
jae says:
January 12, 2012 at 6:48 pm
Sorry, jae, but I’ve lost what the “zero emissivity issue” was. Could you recap the bidding?
w.
DeWitt writes “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.”
Right. So if the atmosphere’s existence alters that average surface temperature (as Willis was getting to) and the density of that atmosphere determines just how much the surface temperature changes then the atmosphere (and its density) sets the average surface temperature.
Of course you can look at it the other way around too as is the traditional way of seeing it… but perhaps Nickolov is onto something afterall. Of course this doesn’t say anything about how changing GHG concentrations effects that, but what it does suggest is that its not necessarily valid to attribute surface temperature to GHG concentration as a primary influence.
TTTM,
Atmospheric and oceanic circulation redistributes energy. They aren’t a source of energy. This redistribution raises the linear average temperature because radiation is proportional to the fourth power of the temperature. When you redistribute energy, the cold areas warm a lot, but the hot areas only cool a little. The heat capacity of the surface and the atmosphere, particularly the ocean surface reduces the diurnal temperature difference. Again, that raises the linear average temperature. But what sets the average temperature of the planet as observed from space is the amount of incident solar energy absorbed by the surface and the atmosphere. Are you contending that if a large screen were placed at L1 between the sun and the Earth that blocked some fraction of solar radiation, the Earth wouldn’t cool because gravity would keep it warm? Because that’s how I read your comment.
Willis,
Since it seems that you have not totally withdrawn from this thread, could I draw your attention to my post above that you have neglected, please?
Bob Fernley-Jones @ January 11, 11:21 pm
OR, you could handle it at the Robert Brown thread, whatever turns-you-on.
Robert Clemenzi says:
January 11, 2012 at 1:56 am
Huh? Why would the lack of GHGs mean that there would be no dry adiabatic lapse rate?
w.
David says:
January 11, 2012 at 5:21 am
I tend to think of things in terms of flows rather than capacities. For example, on average the ocean is both absorbing and losing (to radiation and latent and sensible heat transfer) about half a kilowatt per square metre.
What difference does it make how long the energy resides in the ocean? It could be a minute or a month or no time at all. What we know is that there is a constant, unceasing flow. The ocean is constantly absorbing, and constantly emitting/losing, about half a kilowatt per square metre. Is that energy that has been in the ocean a while, or energy that just now got there? It doesn’t matter.
w.
@Willis
This thread is getting long and you are talking with a number of respondents, maybe you missed my response to you at http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-860181
regards
Willis wonders “Huh? Why would the lack of GHGs mean that there would be no dry adiabatic lapse rate?”
I’d be interested to hear how you think a non-GHG atmosphere would behave. Say the earth with a comparable N2-only atmosphere. Obviously there is no “correct” answer because its not something we’ve seen but there is still interesting application of logic in explaining an answer.
RE: jae: (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?”
Regarding *hypothetical* atmospheres that are not emissive at sub-luminal wavelengths but may, perhaps, be emissive and absorptive in the optical range, I think one can assume such an atmosphere would continue to heat and expand until it was lost to outer space or became hot enough to balance its net absorption by *optical* emission. There is also a usually negligible background emissivity and absorptivity created as a result of the transient fields that are generated during molecular collisions.
One could assume that the effect on planetary temperatures might be neutral if the back radiation reaching the surface from such an atmosphere matched the incoming energy being absorbed by that atmosphere. Contact heating might be another issue, but I would presume the hot emissive layers would be in the upper atmosphere.
Willis Eschenbach says:
January 12, 2012 at 11:33 pm
The dry adiabatic lapse rate (DALR) is produced by gravity and the properties of the gas. The actual (measured / environmental) lapse rate (ELR) is what matters. This is the lapse rate controlled by (created by) greenhouse gases. When convection occurs, the air parcel cools at the DALR until the parcel is the same temperature (actually, density) as the surrounding air. To determine how high that will be, a lapse rate plot is used .. using the actual measured values. For any given surface temperature, a line is plotted with the same slope as the DALR. Assuming that clouds will not be produced, air will rise to the height where the 2 lines intersect.
At the present, the ELR is about 6.5 K/km. Without greenhouse gases, it would be about 0 K/km. But more important, the temperature at the top of the nightly boundary layer inversion would be about the maximum surface temperature. As a result, once the system stabilized, there would be no convection above the boundary layer because the surface temperature could never be greater than the temperature at the top of the inversion. As pointed out above, cold air from the poles would make this a bit more complicated.
The DALR is 9.8 K/km and not affected by the presence of greenhouse gases.
To be clear, air cools at the DALR only if the air is hotter than what is above it.
Robert Clemenzi,
That could only happen if the surface were isothermal. It won’t be. The poles will be cold and the equator hot. You’ll still get Hadley cells. Any circulation will force the atmosphere to have a non-zero lapse rate. See for example: Axisymmetric, nearly inviscid circulations in non-condensing radiative-convective atmospheres
Rodrigo Caballero, Raymond T Pierrehumbert and Jonathan L Mitchell. QJRMS 134(634):1269-1285 (2008) (moneywalled unless you’re affiliated with an academic institution)
@Willis Eschenbach says:
January 12, 2012 at 11:33 pm
…
Huh? Why would the lack of GHGs mean that there would be no dry adiabatic lapse rate?
Indeed, why would it? I am of the same opinion, but respected scientists belief otherwise, as I referred to earlier in this thread: http://wattsupwiththat.com/2012/01/08/the-moon-is-a-cold-mistress/#comment-858399 .
Any idea how to disprove that there will be an isothermal atmosphere?
@Robert Clemenzi says:
January 13, 2012 at 2:19 am
… a line is plotted with the same slope as the DALR. Assuming that clouds will not be produced, air will rise to the height where the 2 lines intersect.
Two lines with the same slope intersect? You surely lost me here!
I believe that the adiabatic lapse rate, dry or otherwise, presumes a state of continuous convection. Warm air must be able to rise, cool, and descend to enforce this temperature regime. Note that the decreasing-temperature lapse rate stops at exactly that altitude where the atmosphere literally runs out of steam: water vapor. I believe water vapor plays two important roles. First, condensation assisted, runaway upward convection allows humid air to rise to the top of the troposphere. Next, based on MODTRAN modeling, it appears that most radiant energy escaping to outer space actually originates in the atmosphere, not from the ground.
The most likely suspects for this are water molecules which would otherwise be liquid or solid at atmospheric temperatures due to their unusually strong polar electrical attraction. This same attraction may indicate they would have a tendency to emit LWIR photons during collision events that are characteristic of the nature of these collisions rather than the structure of the water molecule. As such, they would be unlikely to be reabsorbed by the next water molecule they encounter. These collision event photons created as a result of the unbalanced shape of the water molecule may be what allows the air in the troposphere to cool and return to the surface.
The stratosphere should give us a good example of what the Earth’s atmosphere would be like without water vapor.
The stratosphere is the way it is because oxygen absorbs in the UV. The excited oxygen molecule reacts with another oxygen molecule to create ozone. Ozone absorbs even more strongly over a wider wavelength range than oxygen. That means the stratosphere warms with altitude. That’s a temperature inversion or negative lapse rate which means no convection. If there were no oxygen in the atmosphere, the structure would be very different. Of course we wouldn’t be here to observe it either. It has little or nothing to do with water vapor. Water vapor in the troposphere acts to reduce the lapse rate from the DALR towards the moist adiabatic rate.
PeterF says:
January 13, 2012 at 5:51 am
Sorry. The actual lapse rate will have a slope of about 6.5 K/km, the DALR will have a slope of 9.8 K/km. Plots of this type are used by glider pilots to determine how high thermals will go. They also indicate why parking lots produce better thermals than other surfaces.
I have several web pages providing a number of examples. But to really understand this, try running my program. You will be able to add DALR and ELR lines to animations showing actual soundings. Be sure to check out the South Pole – I was very surprised.
DeWitt Payne says:
January 13, 2012 at 8:32 am
I agree, to a point. The questions are “How thick will the mixing layer be?” and “How far will the mixing layer extend from the pole?” However, since this is an academic question to make a point, the actual answers don’t really matter. Above this mixing layer, the lapse rate should be zero and the temperature close to the peak surface temperature..
This is where we disagree. I my theory, water vapor in the troposphere acts to increase the lapse rate from an isothermal 0 K/km towards the DALR. This is what creates the ELR (the observed lapse rate). That said, when water vapor condenses to form clouds, then I agree with you that “water vapor in the troposphere acts to reduce the lapse rate from the DALR towards the moist adiabatic rate”.
Robert,
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. At mid-latitudes you’ll get winds similar to the trade winds on Earth. That means the lapse rate in most of the atmosphere will be the adiabatic rate (that’s what a constant potential temperature means). The reason is that the pressure gradient force increases with altitude. That makes winds at high altitude that blow initially towards the poles. But coriolis force makes the winds eventually blow parallel to the equator. That’s what drives the jet streams in the Earth’s atmosphere. There is some question of things like baroclinic stability, which is starting to get beyond my pay grade. But you will still get rising air near the equator, which will cool adiabatically as it rises and descending air at higher latitudes, which will heat adiabatically as it descends at higher latitudes. That air will be warmer than the surface at those latitudes, creating a temperature inversion and heat transfer from the air to the surface.
Dear Willis,
I beg your pardon, but your arguments regarding the power law of Stefan and Boltzmann are incorrect. This power law can only be applied on a small surface element, but not on a global mean temperature. Consequently, a local energy flux budget has to be considered. It reads:
(1 – a_s) R_s – eps_s sigma T_s^4 + G = 0 (1)
where R_s is the incoming solar radiation, a_s is the surface albedo in the solar range, eps_s is the surface emisssivity in the infrared range, sigma is Stefan’s constant, and G is the ground heat flux. Vasavada et al. (1999, Icarus 141, 179–193) have used such an equation to determine the surface temperature of Mercury and Moon. Their Figure 2 that illustrates the temperature at the Moon’s equator is in substantial agreement with your Figure 1.
It is clear that for surface elements on the dark side of the Moon the local energy flux budget reads
G – eps_s sigma T_s^4 = 0 (2)
Thus, for a local surface temperature of 130 K the ground heat flux amounts to G = 16 W/m^2. This is the reason why G is often ignored. However, this is a big mistake because this ground heat flux establishes a connection between a heat reservoir within the soil that is neither changed owing to the penetration of heat into the upper soil layer during the Moon’s daytime nor owing to the emission of infrared radiation during its nighttime. In other words: there is a certain depth within the soil at which the temperature does not change during the rotation. At this depth the exchange of heat is negligible.
Ignoring G in Eq. (1) would lead to
(1 – a_s) R_s – eps_s sigma T_s^4 = 0 (3)
This means that on the sub-solar point we would have a temperature of about 390 K. Whereas the surface temperature on the dark side of the Moon would be equal to zero.
Equation (1) can be rearranged to provide:
T_s = ([G + (1 – a_s) R_s]/(eps_s sigma))^0.25 (4)
Globally averaging all results providing by Eq. (4) yields the mean surface temperature, {T_s}, of the Moon. Based on on our estimates we found for the Moon
{T_s} = 1/A INT_A ([G + (1 – a_s) R_s]/(eps_s sigma))^0.25 dA = (200 +/- 10) K (5)
Here, INT_A means the integral over the entire surface A. This value of {T_s} = (200 +/- 10) K is in substantial agreement with observations. It is 70 K lower, on average, than the temperature of the planetary radiative equilibrium given by
T_e = ((1 – a_E) S/(4 eps_E sigma))^0.25 (6)
The derivation of this formula requires assumptions that are, by far, not fulfilled.
In case of the Earth in the absence of its atmosphere formula (6) provides
T_e = 255 K
if a_E = 0.3 and eps_E = 1 are chosen. However, the planetary albedo of a_E = 0.3 is only valid for the entire Earth-atmosphere system. Without an atmosphere a_E of the Earth would be much smaller (perhaps, comparable with the Moon’s albedo).
The difference between the globally averaged near-surface temperature of {T_ns} and T_e = 33 K which is often used to quantify the so-called atmospheric greenhouse effect. However, T_e and {T_ns} have completely different meaning. They have the same unit, namely Kelvin, but I can also express my blood temperature in Kelvin. Only the globally averaged temperature provided by Eq. (4) is comparable with {T_ns}.
Best regards
Gerhard
Robert Clemenzi says:
January 13, 2012 at 2:19 am
I’m not discussing whether it “matters”, robert, that’s a different question. I’m saying that the derivation of the dry adiabatic lapse rate doesn’t depend on GHGs, it depends on gravity and Cp, the specific heat of the atmosphere. So I find the claim that if there were no GHGs there would be no lapse rate to be totally incorrect. If there were no gravity there would be no dry adiabatic lapse rate, but GHGs are immaterial.
w.
Spector says:
January 13, 2012 at 6:01 am
Your belief is incorrect. Molecular motion is all that is necessary to produce the lapse rate, not bulk motion of the atmosphere.
w.
Willis,
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.
Gerhard Kramm says:
January 13, 2012 at 11:59 am
[SELF-SNIP: see note below.]
[EDITED TO ADD: Upon reflection, I think we may be having a terminology problem. You seem to be using ground heat flow to mean the temporary storage of energy in the ground, whereas I understand it as the heat flowing from the moons interior. In that case I have deleted my own comments and must reconsider your comments … but I’m working now, it might be a bit of time. I tried to strike my comments out, but I couldn’t get the “strikeout” html to work. Thanks, —w.]
Gerhard Kramm says:
January 13, 2012 at 11:59 am
You are correct about your equations, Gerhard, but you neglect to say where I am “incorrect” as you seem to think.
I have compared what I will call the “theoretical” S-B lunar surface temperature (blackbody temperature based on absorbed insolation) and the actual surface temperature. I have used the actual surface temperature and the S-B equation to show that averaged over the monthly cycle, the total radiation from the actual temperatures is close to the incoming radiation, as we would expect.
Now, you say this is wrong. But like many others, you have not quoted a single thing I said that you claim is wrong. Instead, you’ve put out your own version, which I agree with.
So where is it that I am wrong, Gerhard? I ignore the storage term G because over the long term that I am discussing (monthly averages), it has to average out to zero. So it drops out of your equations … why is that a problem?
Thanks,
w.
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.
George E. Smith,
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
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.
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.
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”
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.
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
jae says:
jae, thanks, we’re moving forward, but once again I say, QUOTE MY WORDS if you disagree with them. You say:
Now, perhaps I am ignoring something. But where and what? I don’t have any idea, and you haven’t given us a clue. You just make the random claim that I’m “assuming that EVERYTHING has …”, but you offer nary a hint as to what I have written that has you so exercised.
I can defend my own words. I can’t defend myself against your kind of content-less accusations.
Give it another shot, and put some substance in it.
Thanks,
w.
DeWitt Payne says:
January 14, 2012 at 12:25 pm
Thanks, DeWitt. Consider again that the dry adiabatic lapse rate is defined as g / Cp. There’s nothing in there about circulation. The lapse rate is a result of the difference in potential energy between the bottom and the top of the atmosphere, or any vertical column of air. Air at the top has high potential energy, it’s at the top of a gravity well. As a result, it has low kinetic energy.
It’s like a bullet fired up into the sky. As the bullet rises, it slows. It’s trading kinetic energy for potential energy. Here’s the relationship for the bullet as it rises up thousands of metres:
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.
This has nothing to do with circulation.
w.
Robert Clemenzi,
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.
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.
Tim Folkerts says:
January 13, 2012 at 9:00 pm
I completely agree.
DeWitt Payne says:
January 14, 2012 at 12:04 pm
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.
Willis;
Yes; taking this one step further (too far?), one might say “In a gravity well, height is (the equivalent of) heat.”
jae says:
January 15, 2012 at 7:28 am
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.
At that point the atmosphere is isentropic, and the lowest layer of the air is the same temperature as the surface, so no further convection or energy exchange will take place.
HTH,
w.
DeWitt Payne says:
January 15, 2012 at 9:11 am
Thanks, DeWitt. 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.
Note that in normal life, isothermal IS isentropic, because the temperature distribution is the energy distribution as well, but this is not true for gases under gravity.
w.
Willis,
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.
Willis Eschenbach says:
January 15, 2012 at 10:13 am
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.
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?
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?
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
DeWitt Payne says:
January 16, 2012 at 7:36 am
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.
Robert Clemenzi
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.
DeWitt Payne says:
January 16, 2012 at 2:39 pm
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.
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.
Stephen Wilde,
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.
“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.
Stephen Wilde,
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.
If you’re still on the fence: grab your favorite earphones, head down to a Best Buy and ask to plug them into a Zune then an iPod and see which one sounds better to you, and which interface makes you smile more. Then you’ll know which is right for you.
RE: Willis Eschenbach: (January 13, 2012 at 12:25 pm)
REF: [I believe that the adiabatic lapse rate, dry or otherwise, presumes a state of continuous convection. Warm air must be able to rise, cool, and descend to enforce this temperature regime.]
“Your belief is incorrect. Molecular motion is all that is necessary to produce the lapse rate, not bulk motion of the atmosphere.”
As best I can determine, the dry and moist adiabatic lapse rates establish maximum slopes by which the atmosphere can cool with increasing altitude before convective activity begins. I find no indication that lesser rates of temperature decrease are not stable if no combination of lower level heating and upper altitude cooling are attempting to increase the cooling gradient with altitude. In our own atmosphere we have the example of the stratosphere where temperatures actually increase with altitude.
“What you’re saying is that the evaporation rate will increase enough to restore the energy balance without a change in surface temperature.”
There will be a change in the temperature of the topmost molecules of the oceans or on land containing moisture that then evaporate earlier than they otherwise would have done.
There will be no effect on the energy content of the system as a whole which includes the bulk of the oceans.
There will be a slight shift in the surface pressure distribution including the permanent climate zones but imperceptible compared to natural variability.