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.
Discover more from Watts Up With That?
Subscribe to get the latest posts sent to your email.
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 @ur momisugly 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
@ur momisugly 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 @ur momisugly 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.