Dodgy NASA Pages Undermines Scientific Education
The educational troubles at NOAA are well known. Now I have stumbled by accident into one set of misleading NASA-hosted pages allegedly set up in order to help teachers and students understand the greenhouse effect.
The starting page of the set “Measuring the Temperature of the Sky and Clouds” by a Forrest M. Mims III and part of teacher-focused “My NASA Data” website, claims to describe a project where “you will learn about the greenhouse effect by measuring the temperature of the sky and clouds far overhead with an infrared thermometer“.
The project is described across four pages and it might be easily misinterpreted as showing that clouds are warmer than the cloudless atmosphere because of the greenhouse properties of water vapor (one needs to read the text very carefully). But that’s not the real problem.
The real problem is that it is claimed that:
The temperature in outer space approaches absolute zero, which is -273 degrees Celsius. But you will measure a much warmer temperature if you point an infrared thermometer at the sky directly overhead (the zenith). Depending on the season and your location, the temperature will likely be near or below zero degrees Celsius. While this is very chilly, it’s far from being as cold as absolute zero. The difference is caused mainly by water vapor in the sky that has become warm by absorbing infrared radiation emitted by the Earth below. The warmed water vapor returns some of the infrared back to the Earth, and this helps keep the Earth warmer than space.
The statement above is wrong. Says who? Says mainstream scientific consensus on the behavior of atmospheres. Here’s an excerpt from a University of Texas page explaining it all:
Of course, we know that the atmosphere is not isothermal. In fact, air temperature falls quite noticeably with increasing altitude. In ski resorts, you are told to expect the temperature to drop by about 1 degree per 100 meters you go upwards. Many people cannot understand why the atmosphere gets colder the higher up you go. They reason that as higher altitudes are closer to the Sun they ought to be hotter. In fact, the explanation is quite simple. It depends on three important properties of air. The first important property is that air is transparent to most, but by no means all, of the electromagnetic spectrum. In particular, most infrared radiation, which carries heat energy, passes straight through the lower atmosphere and heats the ground. In other words, the lower atmosphere is heated from below, not from above. The second important property of air is that it is constantly in motion. In fact, the lower 20 kilometers of the atmosphere (the so called troposphere) are fairly thoroughly mixed. You might think that this would imply that the atmosphere is isothermal. However, this is not the case because of the final important properly of air: i.e., it is a very poor conductor of heat.
Note that there is not a single mention of any greenhouse property of anything. Later on the UTexas text contains a reference to water vapor but for different reasons than the greenhouse effect:
As air rises, expands, and cools, water vapour condenses out releasing latent heat which prevents the temperature from falling as rapidly with height as the adiabatic lapse rate would indicate
So if the ground is at whatsoever temperature and you point a thermometer to the sky, you’ll read “the temperature through a cone-shaped column of the troposphere“, as determined by the properties of air and water vapor. The value you will read will be far above absolute zero independently from the greenhouse effect.
It is rather worrying to see such a poorly-designed experiment getting NASA approval (well, that might explain a few things…) and who knows how many pupils have now got it all wrong. Hopefully, there’s two or twenty science teachers out there capable to use critical reasoning.
The Unbearable Nakedness of CLIMATE CHANGE 
Here’s my explanation that I promised of why the experiment with the infrared thermometer pointed at different angles shows higher temperatures at lower angles:
First off, I have to explain how infrared thermometers work. The basic idea is to compare the amount of infrared light at two different wavelengths; the ratio of those two wavelengths depends on the temperature. (IR thermometers can also sample three, four, or more wavelengths to obtain better readings, but the basic principle is established with the two-wavelength sample.) So let’s call those two wavelength bands the hot band and the cold band. Obviously, the greater the ratio of hot IR to cold IR, the higher the temperature.
Note that the IR thermometer does NOT measure the absolute amount of IR — it looks only at the ratio of the two wavelength bands. If it *did* rely on the absolute amount of IR, then you could point it at a warm object and get a higher reading as you moved closer to the warm object, because you’d be getting more IR when you’re closer.
So, the measured temperature is a function of the ratio of the amount of hot IR to the amount of cold IR.
Now get a piece of paper and draw a horizontal line in the middle of the page; this represents the earth’s surface. Put a little dot in the center of the line to mark the observer. Now draw two more horizontal lines above the first one, spaced apart somewhat. (If you want to be snazzy, mark the lower line in red to indicate that it consists of warmer air and the upper line in blue to indicate that it consists of cooler air.)
Draw a vertical line from the observer upwards; this represents his line of sight and the portion of air sampled by the IR thermometer.
Draw a line at an angular altitude of 30º from the ground; this represents his line of sight at an angle and the portion of air sampled by the IR thermometer at that angular altitude.
The IR thermometer collects all the IR coming in along its line of sight and compares the brightnesses in the two bands. So, if you point it straight up, along the vertical line, it will see all the IR coming from all the atmosphere directly above it. However, to simplify things, we’re going to concentrate on just two chunks of air: the bit of warmer (lower) air and the bit of cooler (higher) air.
Each of these parcels of air is emitting IR, but its emissions come from two different sources. One is straight blackbody radiation — the air emits IR because it is warm. The other is reflected IR from the surface of the earth. This is a smaller amount of IR but it’s still significant.
So let’s make up some numbers for what kinds of IR the air layers are emitting. The actual values are not important; it’s the ratios between them that matter.
The lower layer first: let’s say that it’s emitting 100 units of hot IR and 50 units of cool IR as blackbody radiation. It is ALSO emitting reflected IR from the ground; let’s say that amounts to 30 units of hot IR and 0 units of cold IR — the greenhouse effect is based on the selective absorption of certain bands of IR. Thus, the lower layer is emitting a total of 130 units of hot IR and 50 units of cold IR.
Now we do the same thing for the upper layer. let’s say that it’s emitting 20 units of hot IR and 15 units of cold IR as blackbody radiation. It is ALSO emitting reflected IR from the ground; let’s say that amounts to 12 units of hot IR and 0 units of cold IR. Thus, the upper layer is emitting a total of 32 units of hot IR and 15 units of cold IR.
But wait! There’s another problem! The lower layer of air absorbs IR not only from the ground but also from the upper layer. In other words, it intercepts a fraction of the downward IR and re-emits it at the emission temperature. Only half of this re-emitted IR goes downward (simple geometry), so a percentage of the hot IR coming down gets intercepted and half of that is re-emitted downward. Let’s say that 20% of the downward IR is intercepted, so that 10% is re-emitted downward and 10% goes up. Therefore, the 32 units of hot IR is cut down to 29 units of hot IR.
The thermometer therefore collects a total of 159 units of hot IR and 79 units of cold IR — the sum of those two layers (yes, I’m ignoring the other layers, but that doesn’t affect this explanation.) That ratio — 159 to 79 — allows it to calculate a temperature.
Now let’s do the same calculation for the line of sight at a 30º angular altitude. At first glance, you’d think that we’d get exactly the same results, but there’s a big difference: the line of sight is twice as long (sin 30º = 0.5) So there will be twice as much IR coming through all that, and all our numbers are simply doubled, right? So let’s walk through them again:
The lower layer is now emitting (by blackbody radiation) 200 units of hot IR in our direction and 100 units of cold IR. It is reflecting 60 units of hot IR and no units of cold IR, so its net contribution is 260 units of hot IR and 100 units of cold IR — exactly the same ratio as we found with the vertical line, so it should produce exactly the same temperature.
The upper layer yields exactly the same doubled results: 64 units of hot IR and 30 units of cold IR.
But now comes the kicker: some portion of the hot IR from the upper layer is absorbed and re-emitted in the lower layer. This time, however, it’s NOT 10% that is absorbed and re-emitted, because that IR must travel through twice as much lower air. Twice as much air to pass through means that the absorption ratio is doubled from 20% to 40%. Thus, 20% of the hot IR is lost to scattering. That amounts to 13 units of hot IR lost. Hence, the grand total of IR that the thermometer reads adds up to 311 units of hot IR and 130 units of cold IR.
Now look at the two ratios we get for the vertical line of sight and the angled line of sight:
vertical is 159:79 = 2.01
30º LOS is 311:130 = 2.39
Thus, the thermometer gets a higher ratio at the lower angular altitude, and so computes a higher temperature. And the ONLY reason why it computes a higher temperature is the greenhouse effect.
Erasmussimo – thanks for the effort but…even taking as good your explanation of the way an IR thermometer workks… (a) naive question, where’s that explanation in the NASA page? (b) Less naive question: you say that “the actual values are not important”…but they are: if the “blackbody radiation” is overwhelmingly larger than all others, for example, and not simply a 100/30 ratio, then the result changes more than a bit.
(c) Even less naive question: your explanation is incompatible with the statement “The cloud is warmer because it is much lower in the sky than the portion of the clear sky being measured at left”. Note specifically: “the cloud IS warmer”: no reference to intervening IR exchanges between layers of air there. Also “The temperature below 4,000 meters is warmer than the temperature above 4,000 meters. This explains why the infrared thermometer measures an “average” temperature that demonstrates the decline of temperature with elevation.” – no reference to layers of air either.
(d) You have also failed to discuss the NASA’s page claim about water vapor being the reason why the thermometer doesn’t read “absolute zero” when pointed straight upwards.
But a real world example should clinch it. Are you aware that even in the driest continent still “the atmosphere is warm and radiates strongly in the infrared, often swamping the astronomical object being observed. For the infrared astronomer, the sky itself glows brightly day and night”? And so (e) do you have any evidence that an IR thermometer such as the one used by the NASA author would simply measure the same temperature in every direction in Antarctica, due to the lack of water vapor (hence, of GH effect) in the atmosphere?
There’s more. The NASA page points to a site about IR thermometers where it is explained there’s two kind of them, only one of which uses the two-color setup you describe. (f) How can we tell which one was used by the NASA author?
And now for some interesting quotes from A Mallama and J.J. Degnan, “A Thermal Infrared Cloud-mapping Instrument for Observatories”, Publications of the Astronomical Society of the Pacific, 114:913–917, 2002 August:
“When pointed toward a perfectly clear sky with no thermal emission, an IR thermometer would register near absolute zero. In the presence of the thermally emissive air that actually surrounds the Earth, the thermometer will read much above absolute zero but still some tens of degrees Celsius cooler than the ambient air temperature on the ground”
(“thermally emissive”, in this case, in a band unaffected by water vapor, see Fig. 2)
“If clouds are present along the thermograph line of sight, the IR temperature will be warmer than for clear sky as a result of the added thermal emission. The measured apparent temperature depends on how strongly the clouds are emitting, which in turn depends on their actual temperature and optical depth. For thick, relatively warm clouds, such as low-altitude cumulostratus, the IR temperature approaches the surface air temperature. At higher altitudes, the air and the clouds are cooler because of the lapse rate of the atmosphere”
(note that they do measure warmer temperatures along the horizon than from the zenith, see the right image in Fig. 5)
Finally, and amazingly, figure 4 in this document appears to suggest that under different conditions, different considerations apply, so that a cloudy sky shows same temperature independent from elevation angle, but a cloud-free sky doesn’t.
To answer your questions:
(a) where’s that explanation in the NASA page?
Not there. They don’t try to explain it. I can understand why — the real ones are likely chock full of little corrective complexities. My explanation is a physics explanation, not an engineering explanation.
(b) you say that “the actual values are not important”…but they are: if the “blackbody radiation” is overwhelmingly larger than all others, for example, and not simply a 100/30 ratio, then the result changes more than a bit.
Indeed, you can screw around with the numbers to get any result you want, but the number set I used is consistent with the basic physics. For example, the cleanest way to get the bolometric temperature of a blackbody is to sample near the peak and a wavelength about twice as long as the peak — but if you wanted to, you could sample at very different wavelengths and get quite different results. In particular, the overlap between the water vapor absorption bands and the thermometer sampling bands will have a big impact on the actual numbers — but the principle remains the same — it’s just that the calculation gets a lot more complicated in cases of low overlap.
And yes, if the blackbody radiation swamps the re-emitted radiation, then the latter will be lost in the noise. However, we already know that, in its absorption band, water vapor gobbles up a goodly percentage of the overall radiation — up to 80% at its (narrow) peak wavelength, as I recall.
(c) your explanation is incompatible with the statement “The cloud is warmer because it is much lower in the sky than the portion of the clear sky being measured at left”. Note specifically: “the cloud IS warmer”: no reference to intervening IR exchanges between layers of air there.
Irrelevant. In the first place, I’m not explaining the cloud experiment, I’m explaining the difference in results at different angles. In the second place, the fact that he doesn’t reference intervening IR exchanges doesn’t mean that they play no role. He also doesn’t mention latent heat of water, which plays a huge role in cloud formation, the difference between dry adiabatic lapse rate and wet adiabatic lapse rate, or in fact ANY of the primary causal factors at work.
(d) You have also failed to discuss the NASA’s page claim about water vapor being the reason why the thermometer doesn’t read “absolute zero” when pointed straight upwards.
Irrelevant. The experiment I explained clearly demonstrates that the greenhouse effect produces higher temperature readings.
“Are you aware that even in the driest continent still “the atmosphere is warm and radiates strongly in the infrared,”
Yes indeed; the atmosphere is warmed by the ground. Of course, ANY object at a temperature higher than absolute zero emits blackbody radiation, so the observation that the atmosphere radiates IR is a no-brainer. Let me point out, however, that astronomers prefer to carry out infrared observations in the lowest-humidity atmospheres they can find. Water vapor is the primary thing screwing them up, because it absorbs so much IR.
“(e) do you have any evidence that an IR thermometer such as the one used by the NASA author would simply measure the same temperature in every direction in Antarctica, due to the lack of water vapor (hence, of GH effect) in the atmosphere?”
Of course: the water vapor greenhouse effect is weaker at the poles and in deserts, where there’s less water vapor. Since the thermometer is observing the greenhouse effect, and it’s much weaker in these locations, the thermometer would show a much flatter curve.
(f) How can we tell which one was used by the NASA author?
There are actually three kinds of IR thermometers, although two of them are very closely related. The first kind is the multiple wavelength band measuring device, which is rather new because it requires lots of internal computational power to calculate the temperature. The second is called a bolometer, which measures the heating effect of the IR. The third, a pyrometer, does much the same thing as a bolometer, only it uses a slightly different technique.
We know that the author is using the first type because the photograph shows a hand-held device with a temperature display. Most bolometers and pyrometers are rather clumsy devices that would be rather difficult to miniaturize. Moreover, the new type is so much more useful that I can’t imagine anybody preferring the other types (Note: pyrometers and bolometers still have utility in certain types of measurements where you’re dealing with something that deviates significantly from blackbody emissions.)
The paper from ASP is interesting but I don’t think that it offers anything significant to this discussion; it pretty much backs up what I’ve been saying here.
“Finally, and amazingly, figure 4 in this document appears to suggest that under different conditions, different considerations apply, so that a cloudy sky shows same temperature independent from elevation angle, but a cloud-free sky doesn’t.”
There’s nothing amazing in that at all, if you think in terms of the greenhouse effect. Remember, clouds have lots of water vapor in them. Water vapor reflects IR. Imagine yourself in a megapolis like Los Angeles on a cloudy night. Everywhere you look upwards, you see reflected city lights; the clouds are bright. The same thing is happening with IR, even when there’s no city light. The clouds are bright with IR, so wherever you look, you see lots of IR. So long as the cloud layer has a lot of optical depth in IR, it doesn’t matter how high it is, or how cold the clouds themselves are: you’ll see the same IR spectrum at all angles. That’s another manifestation of the greenhouse effect.
Erasmussimo – “Sagan was *assuming* the greenhouse effect in his calculations”
You missed the important bit…”runaway” was the word…
Exercise for tomorrow: (a) Compute the surface temperature of Venus using Sagan’s values for Γ (and the cloud top’s temperature), but in an atmosphere with a zc as for Earth’s. (b) Compute then the surface temperature of Earth using terrestrial values for Γ (and the cloud top’s temperature), but in an atmosphere with a zc as for Venus’s. (c) Finally, explain what has happened to the “runaway greenhouse effect”.
Informed guesses:
The value of (a) is not far off from Earth’s surface temperature
The value of (b) is not far off from Venus’s surface temperature
The whereabouts of (c) are sadly unknown.
In other words, the surface temperature of Venus is due to a combination of the infrared and optical properties of its atmospheric constituents WITH the great height of the atmosphere’s “top”. There is no “runaway greenhouse effect” on Venus, rather a combination of GH, atmospheric mass and planetary gravity.
“You missed the important bit…”runaway” was the word…”
Ah, so you agree that the greenhouse effect is real, and that it increases temperatures. Your only quibble is that there is no runaway greenhouse effect on Venus. Great. We agree.
The exercise you suggest is nonsensical for a variety of reasons. First, applying Sagan’s calculations to the earth’s atmosphere is meaningless because the earth’s atmosphere contains water vapor, which has latent heat. Sagan himself notes that Venus’s atmosphere does not contain significant amounts of water, and hence the “latent heat of condensible materials will not be important in determining the overall atmospheric structure.” Moreover, Sagan notes that the Venusian atmosphere is in convective equilibrium — that’s certainly NOT true of the earth’s atmosphere!
Next, the runaway greenhouse effect arises from a number of factors that do not exist on Venus, so that comparison is useless. By the way, I’m not claiming any certainty with regard to runaway greenhouse effect; we know that there are positive feedbacks and negative feedbacks, and our evidence currently suggests that the positive feedbacks outweigh the negative feedbacks, so some degree of amplification is highly likely, but the magnitude of that amplification is still very much in doubt.
As for your “informed guesses”, I’ll point out that, since the calculations are meaningless, so are the guesses.
” the surface temperature of Venus is due to a combination of the infrared and optical properties of its atmospheric constituents WITH the great height of the atmosphere’s “top”.”
The TOA for Venus is lower than that of earth. The surface temperature of Venus is readily calculated using basic blackbody calculations modified to take into account the greenhouse effect and Venus’ rotation.
” There is no “runaway greenhouse effect” on Venus, rather a combination of GH, atmospheric mass and planetary gravity.”
Of course there’s no runaway greenhouse effect on Venus. The high surface temperature of Venus is due almost exclusively to the greenhouse effect. Perhaps you interpret “runaway” to mean “large”; if so, you are incorrect. Most uses of the term “runaway greenhouse effect” refer to positive feedbacks that are so large that the temperature climbs very quickly to very high values.
no, really, why don’t you give it a try, values exactly as on Venus but with O2 and N2 instead of CO2, what would the surface temperature be? Or build an Earth with CO2 in the atmosphere in place of N2 and O2, what would the surface temperature be? Leave H2O aside, at least for now.
You can define “denialist” in any way you chose, but I use it to refer to people who deny scientific results without any scientific basis. Your denial of the basic physics of blackbody radiation above would fall under that definition.
” you must know full well by now the UTexas page explaining why atmospheres have adiabatic lapse rates, and how large they are”
Yes, I learned about adiabatic lapse rates in 1976 in a planetary atmospheres course. Were you alive at that time?
“an intriguing Carl Sagan scientific communication in the pages of the Astrophysical Journal (1967) estimating the surface temperature of Venus without a single mention of the “runaway greenhouse effect”.”
The reason Mr. Sagan never mentions the greenhouse effect is that he is *reporting* the temperature, not explaining it. He doesn’t mention ANY causal factors — he just reports on the best current knowledge of Venus’ radius and surface temperature. However, the greenhouse effect is implicit in Table 1. Read the values of the adiabatic lapse rate in the middle column. Note that the values are high for low values of CO2 and low for high values of CO2. What does that tell you? I won’t bother explaining it to you — it’s complicated and if you can’t understand simple physics of blackbody radiation, the physics required to understand that table is way too complicated.
Erasmussimo – let’s each of us use our own vocabularies, shall we. And so I use “denialist” to refer to people called Erasmussimo. Now let’s hope the teacher doesn’t find out, oh wait, we’re not children any longer, shouldn’t we behave as adults? Perhaps we should 😎
And so one day, when you will decide as much, you will come around to understand that there’s no meaning in calling me a “denialist [because I] deny scientific results”, since I am not denying any of them, at all.
I guess “bringingwildpredictionscometoearthist” would be a far more appropriate use of the English language.
Fine — use the word any way you want to. I’ll use it to mean what is most commonly meant by the word. You’re welcome to assign your own meaning to it, but when I use it, you’ll have a clear idea of what it means.
“shouldn’t we behave as adults? Perhaps we should”
I would be pleased if you would. I therefore propose a truce: I will, for the next 24 hours, refrain from making any snide, denigrating, insulting, or otherwise derogatory comments regarding you, either explicitly or implicitly. After that, it’s straight tit for tat: if you maintain high standards of civility, so will I. If you start getting snide or rude, I’ll respond in kind.
Deal?
Deal it is. But “denialist” remains a person that “denies” the reality of something that is real, i.e. has happened or is happening. And we still haven’t figured out what, if anything, I am denying 😎
Good! I’d much rather discuss this in a civilized manner. I look forward to a productive discussion with you.
Erasmussimo – “being lower is not the ONLY contributing factor to the measured temperature of the cloud. You are denying a phenomenon — the greenhouse effect”
I am not denying anything. It’s you the one showing a complete misunderstanding of what the GH effect is about.
Nobody that knows a thing about the GH effect will ever claim that the higher temperature measured pointing the thermometer at the cloud is due to anything else than “because it is much lower in the sky than the portion of the clear sky being measured at left”. That’s a direct quote from the NASA author, and I have provided you with a link with all the formulae and reasoning needed to understand what is going on.
ps I am sure you know plenty of people with a scientific background capable of addressing this point. Please ask any of them to explain you the GH effect. Hint: it’s a property of the whole atmosphere, not a local increase in temperatures.
“Please ask any of them to explain you the GH effect. Hint: it’s a property of the whole atmosphere, not a local increase in temperatures.”
Actually, I don’t need to ask anybody — I hold a master’s degree in physics and have a very clear idea of exactly what’s going on here. Let me explain it to you:
During the day, the earth’s surface warms due to sunlight. The surface therefore emits blackbody radiation in the infrared. That blackbody radiation travels upward into the sky. Some of it escapes the atmosphere. Some of it collides with dust in the atmosphere, some with water droplets, and some is collides with greenhouse gases such as water vapor, CO2, or methane. These latter photons are absorbed and re-emitted in random directions; for our purposes we need merely note that half of the newly re-emitted photons go up and half go down. The ones that go down hit the surface of the earth, warming it. This phenomenon of IR photons hitting greenhouse gases and being partially reflected back downward is called “the greenhouse effect”.
Clouds form when a parcel of air is pushed upward so that its temperature, pressure, and water vapor content exceed the dewpoint, forcing some of the water vapor to condense into tiny droplets of water. Those droplets of water, in turn, intercept photons — perhaps you may have noticed that thick enough clouds are not transparent. Because they’re not transparent, they intercept ALL of the IR traveling upward from the surface of the earth. That in itself warms the cloud by some small amount. However, for our purposes, the important point is that the cloud re-radiates IR in all directions — including downward.
Thus, on a partially cloudy day, only SOME of the IR emitted by the surface is intercepted by the cone of blue sky air, but ALL of the IR emitted by the surface is intercepted by the cloud. Perhaps you are aware of the fact that “all” is greater than “some”. Because the cloud is intercepting MORE IR than the blue sky air, it ends up radiating MORE IR than the blue sky air. Which means that, when you point an IR thermometer at a cloud, you’ll get a higher reading than you get when you point that same thermometer at blue sky.
If you wish to learn more about this, I’ll be happy to explain it to you — I’ve taught physics and this stuff is really quite elementary. I’ll even overlook your snide comments.
We are not talking about the concept of the GH effect. We are talking about a specific NASA-supported “experiment”: and there is no way to measure the GH effect by pointing a thermometer at the free sky and at a cloud because the adiabatic lapse rate in the troposphere will mask it all: i.e., even if the GH effect did not exist, still the cloud would appear warmer. That’s what you should be arguing for or against.
I present a detailed explanation of how the physics behind the experiment works and your response is to simply declare that there is no way to measure the effect. Your argument is that the adiabatic lapse rate will mask it all. I suspect that you haven’t the faintest idea of what “adiabatic lapse rate” means. We’re talking about measuring infrared radiation with an infrared thermometer. Got that? We’re talking about INFRARED RADIATION. That’s radiation of the infrared sort. Light. Electromagnetic radiation. Photons. Got that?
So, follow the photons. Start with the IR photons emitted by the surface of the earth. Follow them up to the cloud, where all of them are intercepted, and most are re-emitted. Follow them back down to the thermometer. Golly-gee, when they hit the thermometer, we get a reading!
Now follow IR photons traveling from the surface of the earth through atmosphere with no clouds. Some of the IR is intercepted, but much of it shoots straight out into space. So when we point our thermometer at the blue sky, there aren’t as many photons bouncing back to our thermometer, and we measure a lower temperature.
High school kids can understand this science. Why can’t you?
Erasmussimo – it would have been easier had you been a high school kid, then, as you cannot understand the science presented in the NASA page under review. I see little point in discussing this further, as you are going to refuse agreeing on anything at all (haven’t you noticed, you are making a point that is not made at all by the NASA author, that explicitly and in no uncertain words links the cloud’s temperature to its height, not to the GH effect).
So why don’t you find one of your eminent experts and submit your considerations to them, then come back here with a link or two where some discerning body will have explained what exactly is that thermometer reading. There’s plenty of blogs where you can ask such a question safely and in the company of fellow AGW believers….tamino, deepclimate, greenfyre, perhaps even realclimate.
You write:
” you cannot understand the science presented in the NASA page under review.”
Right. I have a master’s degree in physics and you don’t and you’re saying that you understand the page and I don’t? Right.
You claim that the experiment in question has nothing to do with the greenhouse effect. Perhaps you should read the first sentence on that web page:
“Purpose: In this project you will learn about the greenhouse effect by measuring the temperature of the sky and clouds far overhead with an infrared thermometer. ”
You claim that the web page attributes the higher temperature of the cloud to its lower altitude. That’s true — but it doesn’t mean that the lower altitude of the cloud is the ONLY factor at work. Indeed, if you read the other pages about that experiment — pages 2, 3, and 4 — you’ll find an extended discussion with a number of links that make it very clear that the experiment is about the greenhouse effect — not the adiabatic lapse rate.
If you refuse to learn basic physics, that’s your problem, not mine.
Erasmussimo – You haven’t realized it yet, but that has been my point from the beginning. The page’s purpose is to “learn about the greenhouse effect”, then it proceeds to do nothing of the sort. The fact that the measured temperature is nowhere near absolute zero relates to a property of every atmosphere, with or without the greenhouse effect.
I repeat: in the author’s words, a cloud will make the reading higher “because it is much lower in the sky than the portion of the clear sky being measured (far from the cloud)”. I agree with that, the UTexas page agrees with that, NASA agrees with that, I suspect Carl Sagan would agree with that too, so you are the only human being I know of that is arguing otherwise. That’s why it would make more sense for you to find something more solid backing up your stance. Who knows, perhaps you will find somebody else? Good luck with that…
” The page’s purpose is to “learn about the greenhouse effect”, then it proceeds to do nothing of the sort.”
If you had bothered to read the ENTIRE experiment and its accompanying materials, the author clarifies the point in several ways. The most telling is an additional experiment he provides in which two students measure the IR temperature at different angular altitudes. The experiment produces higher IR temperatures for lower angular altitudes — an effect that the author ascribes to the greater amount of water vapor at lower angles. Here’s the key quote:
“The apparent temperature of the sky straight overhead (the zenith) is cooler than when an IR thermometer is pointed at lower angles in the sky. This is because the IR sensor in the thermometer is looking through more water vapor when the instrument is at an angle away from the vertical.”
So you are quite wrong claiming that the experiment is not about the greenhouse effect. I agree that the wording of that single statement, taken out of context, is misleading. But if you read all four pages and carry out some of the exercises, the meaning of the experiment is quite clear. You are making a big case out of a single poorly worded statement. I have already demonstrated that many of your own statements are much less correct than this one, so your criticism is very much a matter of the pot calling the off-white coffee mug black.
“The fact that the measured temperature is nowhere near absolute zero relates to a property of every atmosphere, with or without the greenhouse effect.”
False. A completely transparent atmosphere — one that has no interactions with light — would have zero effect on the radiative balance of the planet. The surface temperature would be solely a function of the amount of EM incident upon its surface. And if you had an atmosphere that interacts with light, then you get a greenhouse effect.
“you are the only human being I know of that is arguing otherwise. ”
That is not what I’ve been saying. You’re playing a straw man game.
This sentence is wrong: “The apparent temperature of the sky straight overhead (the zenith) is cooler than when an IR thermometer is pointed at lower angles in the sky. This is because the IR sensor in the thermometer is looking through more water vapor when the instrument is at an angle away from the vertical.”
It is wrong because, as explained clearly in the UTexas page, when an IR thermometer is pointed at lower angles in the sky it is measuring temperatures nearer to the surface and thereby warmer because of the adiabatic lapse rate. It is quite simple and straightforward and there is no way to tell, in that setting, if there is any GH effect at all. It is true that measured values would be different if the air were very humid or very dry, but that is related to latent heat and not to the GH effect.
Please refrain from accusing me of making straw-man arguments, or at least don’t do that a few sentence after talking about a “completely transparent atmosphere”, whose “adiabatic lapse rate” would be a completely different exercise to compute.
“when an IR thermometer is pointed at lower angles in the sky it is measuring temperatures nearer to the surface and thereby warmer because of the adiabatic lapse rate. ”
I just spent half an hour writing up a simple, clear explanation why this statement is false. Unfortunately, it was so clear and simple that it didn’t amount to anything! It’s like the classic case of doing a long derivation and reaching the conclusion that x = x. So I’m going to have to do this the hard way and explain the difference between the adiabatic lapse rate and the measured lapse rate in humid air. That’s a bit of a mess, and I’ll need some time to organize my thoughts, but I’ll have the results here within 16 hours.
This post presents a misunderstanding of what the linked pages actually say. First, let me observe that the NASA page presents a simplified but nevertheless clear and reasonably accurate representation of the truth as we know it.
But you aren’t even clear about your complaint. You declare that “This statement is wrong.” But you refer to a paragraph containing six sentences. Let me therefore ask you precisely what you mean to criticize.
Is this statement incorrect: “The temperature in outer space approaches absolute zero, which is -273 degrees Celsius.” ?
Is this statement incorrect: “But you will measure a much warmer temperature if you point an infrared thermometer at the sky directly overhead (the zenith). ”
Is this statement incorrect: “Depending on the season and your location, the temperature will likely be near or below zero degrees Celsius.”
Is this statement incorrect: “While this is very chilly, it’s far from being as cold as absolute zero.”
Is this statement incorrect: “The difference is caused mainly by water vapor in the sky that has become warm by absorbing infrared radiation emitted by the Earth below.”
Is this statement incorrect: “The warmed water vapor returns some of the infrared back to the Earth, and this helps keep the Earth warmer than space.”
If you would but state your objection more precisely I will be happy to help clarify your confusion.
Erasmussimo – the thermometer will measure a temperature above -273 whatever the composition of the atmosphere above it. The main bulk of the difference will not be caused by water vapor, as any visit to the Atacama desert will confirm 😎
Therefore nobody can “learn about the greenhouse effect by measuring the temperature of the sky and clouds far overhead with an infrared thermometer”.
I am sure at UTexas they’ll be happy to hear from you about how amazingly wrong their computations about the adiabatic lapse rate are.
You failed to answer my request that you specify what you thought was incorrect about the paragraph in question.
This statement:
“The main bulk of the difference will not be caused by water vapor, as any visit to the Atacama desert will confirm”
is true but irrelevant to the point made by the NASA page. Yes, the earth’s temperature exceeds absolute zero — that’s because the earth is warmed by the sun. However, water vapor and carbon dioxide increase the temperature of the earth’s surface due to the greenhouse effect. Every desert on this planet undergoes rapid cooling at night because the absence of water vapor in the atmosphere overhead.
You write:
“nobody can “learn about the greenhouse effect by measuring the temperature of the sky and clouds far overhead with an infrared thermometer”.”
This statement is false. The measurement of the IR temperature of a cloud versus that of blue sky clearly shows a higher value for the cloud. That demonstrates that the cloud is radiating more IR than the blue sky. Clouds are composed of a mixture of small water droplets immersed in a parcel of air at 100% humidity. It is this high-humidity air that is reflecting all the infrared light. And reflecting IR back to the earth’s surface is what’s called “the greenhouse effect”. In other words, the measurement directly demonstrates the greenhouse effect at work. Your statement denying this is false.
You write:
“I am sure at UTexas they’ll be happy to hear from you about how amazingly wrong their computations about the adiabatic lapse rate are.”
This statement is balderdash: adiabatic lapse rate has nothing to do with this experiment.
> higher value for the cloud
Even the NASA author steers very clear from inferring the greenhouse effect from that measurement: “The cloud is warmer because it is much lower in the sky than the portion of the clear sky being measured at left. ”
Please try again.
“Even the NASA author steers very clear from inferring the greenhouse effect from that measurement: “The cloud is warmer because it is much lower in the sky than the portion of the clear sky being measured at left. ”
This pair of sentences belies befuddled logic. Yes, the cloud is warmer than the blue sky. Yes, being lower contributes to a warmer measured IR temperature. However, being lower is not the ONLY contributing factor to the measured temperature of the cloud. You are denying a phenomenon — the greenhouse effect — that has been known for over a century, that has been measured in the laboratory countless times, and for which there exists a complete theoretical explanation. Sheesh…
Alan,
over 50% of the suns output is in infrared.
http://hyperphysics.phy-astr.gsu.edu/Hbase/wien.html
Just a hint, it’s the area under the curve that counts.
That University of Texas page also screwed up:
“The first important property is that air is transparent to most, but by no means all, of the electromagnetic spectrum. In particular, most infrared radiation, which carries heat energy, passes straight through the lower atmosphere and heats the ground. ”
Nope, most of the infrared does NOT pass through the atmosphere to heat the ground. Only a fraction of the sun’s energy is infrared, and part of this is absorbed directly by the atmosphere. The larger fraction that is NOT infrared passes through the atmosphere and is absorbed by the ground. The ground in turn reradiates the energy in the infrared, which AGAIN the atmosphere partially absorbs, else there’d be no greenhouse effect.
I suspect that poor quality statements like that Texask paper are due to the large increase in climatologists over the last decade thanks to government grant. With the inflated number of climatologists, we also get a deflaction in the average ability and intelligence of climatologists.
Mr. McIntire, you are not correct in stating that “most of the infrared does NOT pass through the atmosphere to heat the ground”. If you will examine this graph of atmospheric transparency:
you will note that indeed most of the solar infrared does pass through the atmosphere. The yellow curve shows what reaches the ground; you can readily see that the yellow curve does not drop to low values across the entire IR spectrum; the regions of large absorption are at 940 nm, 1375 nm, and 1950 nm. These are pretty long wavelengths and don’t represent a lot of energy. It’s in the near IR wavelengths, carrying most of the IR energy, that absorption is only about 20%.
Best not to disparage the intelligence of others unless you’re quite sure of your facts.
Now, let’s see. Imagine the atmosphere had no ‘greenhouse gasses’ at all: no CO2, no water vapour, no methane etc, just good old nitrogen and oxygen. Would the infrared thermometer now read near absolute zero? Not a chance. You have tonnes of atmosphere above your head, and that is going to radiate in the infrared because it is at a certain temperature, as all bodies do. Just as the hot gasses on the sun (hydrogen and helium – non-greenhouse gasses) radiate in the visible spectrum because they are hotter, so the cooler gasses in our atmosphere (oxygen and nitrogen) will radiate in the infrared. Black body radiation, not ‘greenhouse emission’.
The atmosphere gets cooler at higher altitude because of the reduction in air pressure. It is gas pressure that makes the surface of Venus so hot: Venus has such dense clouds of sulphur dioxide that very little of the sun’s energy gets through to the surface in the form of light: it’s pretty dismal at the Venusian surface. But boy is it hot, because of the density of the atmosphere! Ignorant people, including some scientists, believe it is so hot on Venus because of the ‘greenhouse effect’ due to its atmosphere being mostly CO2. That’s nonsense. Those who can calculate it can work out what the temperature would be on Venus if the CO2 was replaced by some other non-greenhouse gas of similar molecular weight. It would barely affect the oppressively hot temperature at all. The temperature in the Venusian atmosphere is pretty much determined by the atmospheric pressure. Going up through the Venusian atmosphere to where the atmospheric pressure is similar to Earth’s, the temperature there is similar to Earth’s.
Heck! You don’t even need an outside source of heat for a gas to warm: a massive enough body of gas will collapse gravitationally and reach temperatures such as those of the sun, or hotter. No nuclear fusion required. This can be demonstrated by the virial theorem.
ScientistsForTruth, your science is false. It is true that gravitational contraction — not pressure, contraction — can generate heat. But that does not mean that gravitational contraction is the only source of heat. Venus is hot because the sun is shining on it, and sunlight is hot. The high temperatures in Venus have been shown to be due to large amounts of CO2 in the atmosphere. If you wish to learn more about this, I will be happy to provide you with references on the science at work here.
Hasn’t the old idea that Venus is hot due to the ‘greenhouse effect’ been pretty much considered false? Or at least brought into question?
I remember reading an article that expressed the theory that much of Venus’ heat is likely internally generated. Likely because the planet was until recently a boiling cauldron of molten lava due to a very large meteor or comet strike.
AhmNee, the only people questioning the greenhouse effect on Venus are denialists. Among scientists, the greenhouse effect provides the only plausible explanation for the high temperatures on the planet.
There is a denialist ***** out there claiming that the high temperatures are due to the high atmospheric pressures. It has never dawned on this ********** that an aerosol can contains gas under high pressure, but is not hot.
If internal heat were the source of the high surface temperature, the next question would be, where is that heat coming from? The primary cause of the earth’s internal heat is radioactivity — but if that were the cause of Venus’ high surface temperatures, then we’d see a very different spectral signature from Venus. It’s the spectral signature that’s the dead giveaway — Venus’ high surface temperature is due to a greenhouse effect.
(this comment has been slightly edited as this blog is very different from the average believer’s site – mm)
Erasmussimo – this habit of yours of spreading the “denialist” label as if butter on toast doesn’t reinforce your argument, to the contrary it seems like the climate equivalent of Samuel Johnson’s “patriotism” quote.
As for the reason why a planet with a high surface pressure (and enough incoming solar radiation) would have a large surface temperature regardless of the GH effect, you must know full well by now the UTexas page explaining why atmospheres have adiabatic lapse rates, and how large they are (note: this doesn’t mean the GH effect does not exist).
In that respect, replying to AhmNee…no, the “old idea that Venus is hot due to the ‘greenhouse effect’” has not been shown false. And I do not think the Cytherean heat is “internally generated”, even if the collision hypothesis is very, very fascinating indeed.
For anybody interested, in this blog I have posted a few links on the topic with commentaries by me and others, including an intriguing Carl Sagan scientific communication in the pages of the Astrophysical Journal (1967) estimating the surface temperature of Venus without a single mention of the “runaway greenhouse effect”.
BTW, I went back and looked at the *entire* paper (there’s only one more page, preceding the page you link to). Here’s the basic purport of the paper:
The authors set out to determine an estimate of the surface temperature of Venus that is independent of the measurement obtained by microwave measurements. They start with the measured temperatures at the top of the atmosphere and work down from there — I mistakenly thought they were starting at the bottom and working up. Using the measured limb temperatures and basic planetary atmosphere structure, they calculate possible surface temperatures for Venus based on three variables: CO2 concentration, whether a specific temperature correction is made, and whether the observed top of atmosphere is 44 km above the surface or 65 km above the surface. The inclusion of the CO2 concentration is important ONLY because of its greenhouse gas behavior. The end result best matches the microwave measurement at a CO2 concentration of 50% and an atmospheric depth of 65 km.
Sagan was *assuming* the greenhouse effect in his calculations. It’s such an obvious factor to include that he didn’t bother to mention it.