Talk:Greenhouse effect/Archive 4
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Let's start fixing the intro
The first sentence of the intro should explain what the greenhouse effect is, not what "causes" it. It is all too common for authors who are very familiar with their material to jump into the middle of a topic, not seeing any need to start at the beginning (indeed, they may not even recognize they've made a fundamental mistake). Let's start at the beginning, please. Once we have a decent first sentence, we can see where it leads. That is, should we go on to "causes" or fill in other general aspects before getting into the details. Here's a first stab at what I think should be considered. "The greenhouse effect is a term used in atmospheric science to refer to a cyclical exchange of thermal energy that occurs primarily in the lower atmosphere." This contains the heart of the idea, without adding a bunch of extraneous and marginally relevant stuff that seems to crowd in as various editors seek to include material that is best left to the detail sections. Now, this first sentence leaves undefined a couple things that need to be introduced early, and before any discussion of "causes." Those are the source of the thermal energy, what components are involved in the "exchange," and maybe even the idea that "cyclical" exchange can mean both diurnal (temporally varying component) and a physical (surface to atmosphere, even atmosphere to atmosphere) exchange. Next, the idea of a balance in the long term average flow of energy and the idea that the Earth's surface and its atmosphere radiate electromagnetic energy into space to maintain the balance. As this covers the big ideas underlying the greenhouse effect, the first paragraph should end. The second paragraph should probably discuss the borrowing of the term greenhouse and introduce the idea that the use of the term is only meant to be suggestive of similarities between what happens in a greenhouse versus what happens in the atmopshere, leaving all further discussion of actual greenhouses (vs. the atmosphere) to a detail section. The current sentence on history seems okay to me, but the paragraph on temperatures needs work as too much detail is included. Only general points about temperature should be made in the intro. blackcloak (talk) —Preceding undated comment added 09:22, 7 May 2010 (UTC).
- I think the intro could be improved, too. I would write something like
The greenhouse effect is a process by which energy leaving a planetary surface is absorbed by some atmospheric gases, called greenhouse gases, which heat up as a result. They transfer heat to other components of the atmosphere, and also re-radiate the energy in all directions, including back down towards the surface. This causes more heating of the surface and lower atmosphere than there would be if direct heating by solar radiation was the only warming mechanism.
- instead of the current first two sentences. The sentence explaining the difference with the mechanism of greenhouses I would cut altogether, since there is a section dealing with it, and the history of its discovery could go in a new section after 'basic mechanism'.
- I don't particularly like the 'cyclical exchange' wording proposed above, because it doesn't give an idea of the direction of energy flows and the word 'cyclical' confuses me. It suggests things like seasons and tides.
- Thoughts? Squiddy | (squirt ink?) 15:09, 8 May 2010 (UTC)
- Experience suggests this will lead to long arguments. For example, which heat up as a result is technically wrong - this is an equilibrium process after all (ignoring the diurnal cycle etc). I like the very simple version: The earth is warmer than it would be without the atmosphere because it receives incoming radiation from two sources: the sun and the atmosphere. But I'd be happy to move the history and the difference out of the intro to make space for more mechanism, if required 18:17, 8 May 2010 (UTC)
- I hacked around the "basic mechanism" section somewhat, because I thought it was confusing. It started off explaining stuff then dived off into complications. I think it is better to present the simplified picture first, which is fairly easy to understand, and then explain later why it isn't quite true William M. Connolley (talk) 18:26, 8 May 2010 (UTC)
- OK, I haven't been involved with this article before, and if the present wording is the result of long debates, then I don't want to stir it up unnecessarily. But I do agree with Blackcloak that 'The first sentence of the intro should explain what the greenhouse effect is, not what "causes" it.' Would it be acceptable to insert a short sentence right at the start - along the lines of "The greenhouse effect is a (process/mechanism) by which energy is (trapped/retained) at the surface and in the lower atmosphere." After that, the existing sentence about the cause follows naturally. Replacing 'Greenhouse gases trap heat within the surface-troposphere system, causing heating at the surface of the planet or moon' with 'That infrared radiation absorbed and re-emitted downwards causes a net retention of energy which would otherwise have escaped into space' would give the actual mechanism explicitly, which the current lead only implies.
- I repeat, though, that if this is going to cause endless debate I'll drop the matter, and if no-one agrees with these proposed changes I don't mind - the existing lead is OK. Squiddy | (squirt ink?) 12:17, 9 May 2010 (UTC)
- Oh sorry, I didn't intend to be that off-putting :-( William M. Connolley (talk) 12:30, 9 May 2010 (UTC)
- No apology necessary. :)
- I'm just trying to be extra-cautious, because I've come back from a long wiki-break, and things have changed quite a bit - such as the article probation stuff. I'm trying to rein in my natural gung-ho-ness and the temptation to be bold (especially in mature articles). Squiddy | (squirt ink?) 13:58, 9 May 2010 (UTC)
- Oh sorry, I didn't intend to be that off-putting :-( William M. Connolley (talk) 12:30, 9 May 2010 (UTC)
- I hacked around the "basic mechanism" section somewhat, because I thought it was confusing. It started off explaining stuff then dived off into complications. I think it is better to present the simplified picture first, which is fairly easy to understand, and then explain later why it isn't quite true William M. Connolley (talk) 18:26, 8 May 2010 (UTC)
- Experience suggests this will lead to long arguments. For example, which heat up as a result is technically wrong - this is an equilibrium process after all (ignoring the diurnal cycle etc). I like the very simple version: The earth is warmer than it would be without the atmosphere because it receives incoming radiation from two sources: the sun and the atmosphere. But I'd be happy to move the history and the difference out of the intro to make space for more mechanism, if required 18:17, 8 May 2010 (UTC)
OK, I've hacked the intro a bit, roughly per your suggestion. Re the probation, that only really applies to politically contentious stuff, which this isn't (at the moment). I've broken the paras up so the history and the not-like-greenhouses is clear. I didn't move the history into its own section, because there wasn't enough of it William M. Connolley (talk) 20:50, 9 May 2010 (UTC)
- Looks good to me. Thanks. Squiddy | (squirt ink?) 08:24, 10 May 2010 (UTC)
4th paragraph
The 4th paragraph is wrong.
- The black body temperature of the Earth is 5.5 °C.[4] Since the Earth reflects about 28% of incoming sunlight[5], in the absence of the greenhouse effect the planet's mean temperature would be far lower - about -18 or -19 °C [6][7] instead of the much higher current mean temperature, about 14 °C.
The black body temperature of the Earth is -18 or -19 °C, not 5.5 °C (which is the temperature of an ideal blackbody the same distance from the Sun as the Earth). It was correct before it was changed on Dec 9,2009. Also, reference [4] does not support that statement. I suggest the following as a possible replacement paragraph and even as a possible first paragraph since it actually defines what Greenhouse Effect really refers to.
- If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth, it would have an expected blackbody temperature of 5.3 °C. However, since the Earth reflects about 30%[1] (or 28%[2]) of the incoming sunlight, the planet's actual blackbody temperature is about -18 or -19 °C [3][4], about 33°C below the actual surface temperature of about 14 °C or 15 °C.[5] The mechanism that produces this difference between the actual temperature and the blackbody temperature is due to the atmosphere and is known as the Greenhouse Effect.
- ^ NASA Earth Fact Sheet
- ^ Introduction to Atmospheric Chemistry, by Daniel J. Jacob, Princeton University Press, 1999. Chapter 7, "The Greenhouse Effect".
- ^ Solar Radiation and the Earth's Energy Balance
- ^ Intergovernmental Panel on Climate Change Fourth Assessment Report. Chapter 1: Historical overview of climate change science page 97
- ^ The elusive "absolute surface air temperature," see GISS discussion
Note that I also modified the expected temperature and albedo based on NASA data. If the values in the current paragraph are used in Stephan's equation, 5.5°C with no reflection is associated with -16.5°C with 28% reflection and -18°C with 30%. Q Science (talk) 18:06, 10 May 2010 (UTC)
- I glanced at that when I hacked the intro :-(. I think you're right. Also, the source [4] is wikibooks, which we shouldn't use. Looking at wikibooks, they seem to have forgotten the albedo William M. Connolley (talk) 21:23, 10 May 2010 (UTC)
- Since you are clearly right, I pasted your version in. We now have 2 defns of the GHE effect. I don't much care, but someone who cares about wordings might William M. Connolley (talk) 21:28, 10 May 2010 (UTC)
- Thanks. Because of the probation, I thought it would be better here first. What do you think of having a page that presents the various "constants" - TOA insolation, albedo, surface temperature - and discusses the fact that different values come from different references? I have seen these parameters included in many of the climate change pages and it seems appropriate to have a single place to discuss the uncertainties. Q Science (talk) 22:22, 10 May 2010 (UTC)
- "The mechanism that produces this difference between the actual temperature and the blackbody temperature is due to the atmosphere and is known as the Greenhouse Effect." This is a statement about one consequence of the greenhouse effect. It doesn't say anything about what the greenhouse effect is. You're suggested rewording does not meet (what I consider) basic criteria for an introductory statement in an encyclopedia. So here is my advice for how to evaluate the suitability of a proposed sentence. Figure out precisely (i.e. you have to think, maybe even for 10 minutes or so) what question your sentence answers. Then decide whether or not that particular question is the one that should be answered in the place where you have placed it. Be hard-nosed about that evaluation, because others will think poorly of you if you make a mistake. Another way to decide whether or not a particular statement of fact (like a temperature) is appropriate, where you have placed it, is to estimate the number of other factoids (about the subject, in this case the greenhouse effect) of similar weight (in the spectrum from general to specific, bearing in mind the article's subject) that should be mentioned. If it is one among say three to six ideas of similar weight, then the introduction is the proper place. If it is one among 10 or so, perhaps it is a candidate for a detail section of its own. If it is one among 100 or so, it fits among other facts within a detail section. blackcloak (talk) 02:58, 11 May 2010 (UTC)
- All the temperature numbers are inappropriate for an introduction. General trends, ideas, explanatory information is. Besides, the conditions under which some statement is made about temperatures are insufficiently clearly defined. To make them explicitly clear would require even more discussion, which is completely inappropriate for an introduction. Move the detail elsewhere. blackcloak (talk) 02:58, 11 May 2010 (UTC)
1st paragraph of Intro still has many PROBLEMS
1) Here's how the first five sentences begin: The ..., They ..., This ..., This ..., The ... . Does anyone pay attention to style anymore? Am I really the only one to pick up on this stuff? blackcloak (talk) 05:25, 14 May 2010 (UTC)
- Does anyone pay attention to style anymore? - I don't care a great deal about style. If you do, then please make a stab at fixing it William M. Connolley (talk) 08:51, 14 May 2010 (UTC)
- Content before style. In all due time. Practicing terse-ness. blackcloak (talk) 04:38, 15 May 2010 (UTC)
2) 1st sentence: Only absorption processes are mentioned. Equally important is the emission process (yes, I see the word re-radiate). And radiation into outer space is not mentioned in the first paragraph. There is no mention of the source of the incoming radiation (yes, I know, some editors think this is obvious). The asymmetric mention of absorption is reinforced by only mentioning the transfer of energy from greenhouse gases to the "other components" (only 99% of the atmosphere). Indeed the atmosphere spends more time and uses more area (volume too) at any one time to (net) shed thermal energy into outer space. The present description doesn't do justice to night-time processes, or the idea that new source energy reaches the surface in roughly 12 hour sinusoidal bursts each day (closer to sine squared for one 12 hr cycle, and zero for the next 12 hr period, that being too much detail for an intro), while heat leaves the Earth at a relatively constant rate (much lower peak to valley variation). blackcloak (talk) 05:25, 14 May 2010 (UTC)
3) Where is the idea of balance, or long time average net zero energy out vs. energy in? The greenhouse effect operates within a severe set of constraints. And yes it is a cycle (note use of recycle in the caption of the figure), whether or not that particular word is the most appropriate one to describe the exchange of energy, with loss to outer space, between lower atmosphere and the surface. blackcloak (talk) 05:25, 14 May 2010 (UTC)
4) Last sentence: "was" should be "were" blackcloak (talk) 05:25, 14 May 2010 (UTC)
- You're right about point 4, I've changed it. Squiddy | (squirt ink?) 12:22, 14 May 2010 (UTC)
- One down, six to go. Ready to take on something a little more challenging? blackcloak (talk) 04:38, 15 May 2010 (UTC)
5) Last sentence: This transfers ... appears to refer only to the preceding "including back down to the surface" but it takes too long to deduce this. Greater clarity would make it less difficult to decipher. blackcloak (talk) 05:25, 14 May 2010 (UTC)
6) The context of the term "the greenhouse effect" is not stated anywhere. blackcloak (talk) 05:25, 14 May 2010 (UTC)
7) In the last sentence, "than it would be if direct heating by solar radiation" were "the only warming mechanism" can be taken to mean that sources other than "solar" of "warming" are playing a part. blackcloak (talk) 05:25, 14 May 2010 (UTC)
Solve these seven problems and we'll have a decent first paragraph. blackcloak (talk) 05:25, 14 May 2010 (UTC)
- Two quick comments: I also like "The earth is warmer than it would be without the atmosphere because it receives incoming radiation from two sources: the sun and the atmosphere."
- And: is it necessary to use two values for albedo in the introduction ("about" implies some uncertainty after all)?
—Apis (talk) 20:06, 14 May 2010 (UTC)
- The problem with the sentence you quote is that it is a nonsequitor. Outer space is "warmed" by both the atmosphere and the surface, i.e. by thermal energy leaving the Earth. Who's to say whether or not the net effect is warming or cooling at the surface. Another way to see there is a chicken and egg problem is to realize that the heat leaving the atmosphere and striking the surface originated at the Sun. Finally, "the sun and the atmosphere" is a daytime process. You've ignored the night-time process. blackcloak (talk) 04:38, 15 May 2010 (UTC)
- I don't understand why it's non sequitur. If you add another source of radiation the net effect must be warming, right? And the sun shines on earth during night as well, doesn't it? :)
—Apis (talk) 10:34, 15 May 2010 (UTC)- The night-time process happens on that part of the Earth's surface not illuminated directly by the Sun at any given moment. I thought that was kinda obvious. blackcloak (talk) 04:24, 17 May 2010 (UTC)
- I don't understand why it's non sequitur. If you add another source of radiation the net effect must be warming, right? And the sun shines on earth during night as well, doesn't it? :)
- The problem with the sentence you quote is that it is a nonsequitor. Outer space is "warmed" by both the atmosphere and the surface, i.e. by thermal energy leaving the Earth. Who's to say whether or not the net effect is warming or cooling at the surface. Another way to see there is a chicken and egg problem is to realize that the heat leaving the atmosphere and striking the surface originated at the Sun. Finally, "the sun and the atmosphere" is a daytime process. You've ignored the night-time process. blackcloak (talk) 04:38, 15 May 2010 (UTC)
- My explanation of why one did not follow from the other is reasonably clear. If you have a question about the explanation itself, let me hear it. I don't see the point in repeating what has already been said. If you want to understand the extent to which peak temperatures on an Earth's surface without an atmosphere are tempered (i.e. kept cooler) by the atmosphere, then do the calculation. Show us your result if you bother to do it. blackcloak (talk) 04:24, 17 May 2010 (UTC)
8) I now have identified another problem. The second sentence states "They transfer heat to other components of the atmosphere, and also re-radiate the energy in all directions, including back down towards the surface." The "they" refers to greenhouse gases. And the term "the energy" refers to "radiative energy leaving a planetary surface" in the first sentence. The problem is the implication of the second sentence is that the same energy is transfered "to other components of the atmosphere" is "also" re-radiated "in all directions ...". This would imply a violation of the law of conservation of energy. The sentence should be modified so that it may not be misinterpreted. blackcloak (talk) 03:07, 18 May 2010 (UTC)
- I've taken out the 'the' from 'the energy' - removing (any) implication that all the energy is re-radiated. Squiddy | (squirt ink?) 09:19, 18 May 2010 (UTC)
- That's still not quite right. It is the same energy that is re-radiated. You have removed the possible mis-reading that the same heat is passed to other components of the atmosphere as well as being re-radiated at the same time, but even after it is passed into these other components, it is still re-radiated in all directions in the end. The point is that the energy needed to raise the temperature of the atmosphere by a small amount is entirely negligible compared to the energy that arrives from the sun (and is re-radiated) day after day, year after year by the same atmosphere. I suggest, "They [the greenhouse gasses] transfer heat to other components of the atmosphere, and it is re-radiated in all directions, including back down towards the surface." --Nigelj (talk) 15:35, 18 May 2010 (UTC)
Does the lede need to draw the reader into this level of technical detail in the first paragraph? Not to rock the boat too much but how about the following as a way of getting across the main ideas without committing to any details of the underlying mechanisms that could be considered either technically subtle, controversial, or hard to fine tune?
- On any planet with an atmosphere, the greenhouse effect is the retention of heat by greenhouse gases in the atmosphere. A greenhouse gas is one that is transparent to incoming shortwave radiation from a nearby hot sun but relatively opaque to outgoing longwave radiation generated by the planet's warmth. In our own situation the Sun is about twenty times hotter than Earth, the incoming radiation is correspondingly twenty times shorter than the outgoing, and the main greenhouse gases impeding the latter are water vapor, carbon dioxide, methane, nitrous oxide, and ozone, which together raise Earth's temperature more than 30°C. --Vaughan Pratt (talk) 17:55, 18 May 2010 (UTC)
- There is a major problem in the first sentence. There are two problems with the third sentence. Try taking a crack at identifying them. In the last sentence, the delta 30°C is not specified relative to a defined base. Generally, in the Greenhouse Effect article we've been avoiding as much as possible discussion of Greenhouse gases, especially in the intro, as that topic has its own article. blackcloak (talk) 09:27, 19 May 2010 (UTC)
- Good points. Regarding the problem you mention with my first sentence, I'm guessing your concern is that I executed on my suggestion to omit details that "could be considered either technically subtle, controversial, or hard to fine tune." The heated discussion above over the past several days suggests that reradiation is such a detail, one reason I omitted it. Another reason is that reradiation is only one of a number of mechanisms governing the rate at which heat gradually leaks through the atmosphere to space, and focusing on it to the exclusion of equally relevant phenomena such as conversion of electron energy to molecular vibrational energy, transfer of vibration, rotation, and translation energy between different species of molecules in the atmosphere, etc. gives a misleading impression of the relative importance of reradiation, especially on Earth with its thermally massive non-greenhouse (FIR-transparent) gases, unlike Mars and Venus. Third, isotropic repropagation of both photons and phonons (both obeying Bose-Einstein rather than Fermi-Dirac statistics and both relevant here) is not confined to atmospheric physics but is integral to any insulating scenario involving radiation and (thermal) conduction, whether in solids, liquids, or gases. Fourth, this level of detail adds little of value to the basic idea that greenhouse gases trap heat; arguing that that's not what they do is like arguing that one doesn't suck cider through a straw because there's no such thing as suction---true but so far no one's complained about the phrase "When liquid is sucked through the straw" in the article Drinking straw.
- Regarding the base, I'd considered explicitly separating out Earth's effective temperature (255°K or -18°C), which is roughly where we'd be without greenhouse gases, and the conventionally accepted mean temperature (15°C). In the end I decided against it on the ground that it burdened the reader with too much information in the first paragraph when all that's needed here is some idea of the impact, leaving the details to the main article. With no mention at all of the impact readers have no idea whether it's 30°C or 1°C.
- Regarding avoiding "greenhouse gases," according to WP:LEAD "The lead should be able to stand alone as a concise overview of the article." Since the article devotes a section to greenhouse gases it seemed appropriate for the lead to say something about them. Certainly greenhouse gases have their own article, but with that argument you could eliminate almost all of the present lead since incoming shortwave radiation and outgoing longwave radiation also have their own articles and the latter even treats the greenhouse effect. Inevitably there is overlap between Wikipedia articles and one should aim for smooth reading rather than strict avoidance of duplication, otherwise you force readers to jump back and forth awkwardly between articles in order to get the picture. --Vaughan Pratt (talk) 19:09, 19 May 2010 (UTC)
- That certainly is an effort that warrants a response. I can see you have a handle on physical processes as they pertain to atmospheric science. But... you're letting your knowledge get in the way of the much more obvious. Always look for the simple stuff first. Occam's razor is a useful idea to bear in mind when writing for the reader of WP articles. Regarding the first sentence, you started to get close at one point. In the atmosphere, the O2 and N2 retain about 99 times the heat that the greenhouse gases retain (temps above freezing). If you'd read all the material in the Greenhouse gases and Greenhouse effect discussion pages, you would have seen this point raised. Greenhouse gases are the agent for the exchange of radiative energy and thermal energy (okay, a bit over simplified). Regarding the third sentence, radiation is not shorter or longer than anything; the wavelength of radiation has length, which can be longer or shorter. As for the second problem in that sentence, blackbody radiation has a power vs wavelength spectrum, with a maximum power at a specific (peak) wavelength (for a given temperature). So it is not strictly correct to say that the incoming radiation is 20x shorter in wavelength that the outgoing wavelength. It would be correct to say that the wavelength at which the peak in the power vs wavelength curve (shorthand) of the Sun is one twentieth of the wavelength of the peak in the power vs wavelength curve governing bb emission from the Earth. Regarding your comments about the importance of mentioning greenhouse gases, please explain why all three sentences must include some discussion of the term. Also note, the IPCC description of the greenhouse effect (if I recall correctly) does not mention greenhouse gases. There's a good reason for this. Greenhouse gases are the means through which energy of one form is exchanged for energy of another form. It's a detail that is not required when describing WHAT the greenhouse effect is. To describe HOW the cyclical exchange occurs you have to introduce the idea of Greenhouse gases. An intro is for "what," and not for "how." As for the base upon which 30 degree C is added, a simple addition like "relative to the Earth's effective temperature" with link should do just fine. To be consistent with my comments elsewhere, I'd rather not see any delta degrees. A mention of higher temperatures (surface and lower atmosphere) associated with the Greenhouse effect should suffice. Finally, you mentioned that Greenhouse gases trap heat. This is wrong. I ask you not to use the word trap. Greenhouse gases serve as the "conduit" or "means" by which energy flows into and out of the atmosphere (ALL the gases of the atmosphere). Heat is not trapped; it is temporarily retained or temporarily stored. In fact, I would argue that the (somewhat restrained) cooling processes by which the atmosphere loses its temporarily stored thermal energy are more (at least equally) important than the warming processes. blackcloak (talk) 03:18, 20 May 2010 (UTC)
- Again, good points. You're quite right that "radiation is not shorter or longer than anything," easily fixed by changing "shorter" to "shorter in wavelength." And your suggestion to add "relative to the Earth's effective temperature" is excellent (should have thought of that myself).
- However I don't understand why you raise the question of peak power. The first sentence of Wien's displacement law states that "the wavelength distribution of radiated heat energy from a black body at any temperature has essentially the same shape as the distribution at any other temperature" which means that the comparison is not with the peak but with the whole curve, which is simply displaced without change of shape 20x along the electromagnetic spectrum. Every point on the incoming curve has a precisely corresponding point on the outgoing one that is very close to 20x the wavelength. But why is this level of detail that's worrying you appropriate for the first paragraph, shouldn't one be looking for the simple stuff first?
- That certainly is an effort that warrants a response. I can see you have a handle on physical processes as they pertain to atmospheric science. But... you're letting your knowledge get in the way of the much more obvious. Always look for the simple stuff first. Occam's razor is a useful idea to bear in mind when writing for the reader of WP articles. Regarding the first sentence, you started to get close at one point. In the atmosphere, the O2 and N2 retain about 99 times the heat that the greenhouse gases retain (temps above freezing). If you'd read all the material in the Greenhouse gases and Greenhouse effect discussion pages, you would have seen this point raised. Greenhouse gases are the agent for the exchange of radiative energy and thermal energy (okay, a bit over simplified). Regarding the third sentence, radiation is not shorter or longer than anything; the wavelength of radiation has length, which can be longer or shorter. As for the second problem in that sentence, blackbody radiation has a power vs wavelength spectrum, with a maximum power at a specific (peak) wavelength (for a given temperature). So it is not strictly correct to say that the incoming radiation is 20x shorter in wavelength that the outgoing wavelength. It would be correct to say that the wavelength at which the peak in the power vs wavelength curve (shorthand) of the Sun is one twentieth of the wavelength of the peak in the power vs wavelength curve governing bb emission from the Earth. Regarding your comments about the importance of mentioning greenhouse gases, please explain why all three sentences must include some discussion of the term. Also note, the IPCC description of the greenhouse effect (if I recall correctly) does not mention greenhouse gases. There's a good reason for this. Greenhouse gases are the means through which energy of one form is exchanged for energy of another form. It's a detail that is not required when describing WHAT the greenhouse effect is. To describe HOW the cyclical exchange occurs you have to introduce the idea of Greenhouse gases. An intro is for "what," and not for "how." As for the base upon which 30 degree C is added, a simple addition like "relative to the Earth's effective temperature" with link should do just fine. To be consistent with my comments elsewhere, I'd rather not see any delta degrees. A mention of higher temperatures (surface and lower atmosphere) associated with the Greenhouse effect should suffice. Finally, you mentioned that Greenhouse gases trap heat. This is wrong. I ask you not to use the word trap. Greenhouse gases serve as the "conduit" or "means" by which energy flows into and out of the atmosphere (ALL the gases of the atmosphere). Heat is not trapped; it is temporarily retained or temporarily stored. In fact, I would argue that the (somewhat restrained) cooling processes by which the atmosphere loses its temporarily stored thermal energy are more (at least equally) important than the warming processes. blackcloak (talk) 03:18, 20 May 2010 (UTC)
- There is a major problem in the first sentence. There are two problems with the third sentence. Try taking a crack at identifying them. In the last sentence, the delta 30°C is not specified relative to a defined base. Generally, in the Greenhouse Effect article we've been avoiding as much as possible discussion of Greenhouse gases, especially in the intro, as that topic has its own article. blackcloak (talk) 09:27, 19 May 2010 (UTC)
- The amplitudes are also different. The image used here in wikipedia is misleading because the curves are normalized. The -18C curve should have the same area as the solar curve. At 15C, the Earth's surface is radiating more energy than it gets from the Sun (when averaged over a full day). At noon, the solar integral should be about 4 times the -18C value. Q Science (talk) 15:48, 20 May 2010 (UTC)
- While QS is correct when pointing out the scaling also occurs in radiated power, the difficulty avoided by the more complex language is for another reason. There is indeed a one-to-one correspondence between points of the distribution curve at different temperatures, but there is no one-to-one correspondence between the wavelength of an absorbed quanta and the wavelength of any (corresponding) re-emitted quanta. In passing through a (atmospheric) thermal reservoir, all information about the wavelength of the incident radiation is lost. Any correspondence is purely statistical. That is why the additional language is important. If you had read the contributions from Damorbel on this subject, you'd understand why including the slightly more complicated language has the potential for reducing the likelihood of conveying a misunderstanding of the underlying physics. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Regarding "the O2 and N2 retain about 99 times the heat that the greenhouse gases retain (temps above freezing)," I now see what was bothering you: you were interpreting "by" in "retention of heat by greenhouse gases" to mean that the heat was retained in the greenhouse gases when I intended it to mean that GHGs caused retention of heat (which they obviously do since Earth would be 30°C cooler without them). One way to disambiguate this would be to change "by" to "due to" or similar.
- Again, I object the use of the word "cause"d. One incorrect implication of the (mis)use of this term in your statement is that doubling the GHG component would double the retained heat. There is no causal relationship between (the amount of) GHGs and (the amount of) heat retained in the atmosphere. It is correct, on the other hand and without being strictly accurate, to say that the rate at which energy enters the atmosphere from radiation passing through the atmosphere is proportional the concentration of GHGs in the atmosphere. And that, the rate at which (thermal) energy leaves the atmosphere and is lost to outerspace is proportional to the concentration of GHGs. We know this is true because if there were no GHGs in the atmosphere, the rate of energy transfer into and out of the atmosphere by radiative processes would go to zero (ignoring other radiative energy transfer processes for the moment) and there would be no atmospheric temperature change due to EM radiation in the atmosphere (but there would be conduction and convection processes still at work). The original source of all energy that heats the surface and the atmosphere is the Sun. The concentration of GHGs governs the rates of atmospheric heating and cooling (a bit oversimplified). blackcloak (talk) 07:54, 30 May 2010 (UTC)
- It is closer to 2,000 times the heat, not 99. Also, what does "temps above freezing" mean? Q Science (talk) 15:48, 20 May 2010 (UTC)
- CO2 at 0.04% isn't the only GHG, water vapor at 1% is a much larger component. Blackcloak's point about freezing presumably refers to the latent heat of fusion of water which is equivalent to an 80°C rise in temperature, making ice a very effective refrigerant from a thermal capacity standpoint. Even though only 2% of Earth's water is frozen (a process that began some 50 million years ago when the Azolla event is conjectured to have lowered the CO2 level sixfold), melting it all would take the same amount of heat needed to warm the oceans about 1.5°C, and even more considering that most of the ice is well below freezing. --Vaughan Pratt (talk) 16:47, 20 May 2010 (UTC)
- It is closer to 2,000 times the heat, not 99. Also, what does "temps above freezing" mean? Q Science (talk) 15:48, 20 May 2010 (UTC)
- I don't think I understand everything VP is saying here, but my mention of the freezing temperature of water was just to point out the highly non-linear relationship of maximum (dew point) water vapor concentrations vs. temperature. Below freezing atmospheric maximum concentrations of water vapor are about .5% or less, while above freezing the concentrations can be something like 4%. I realize there is a pressure component to all this, which I am brushing over when pointing out major trends. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Were AR4 not to have mentioned greenhouse gases as you claim then presumably section 2 of this article could have been deleted; all I was doing was following WP:LEAD's guideline that the lead provide an overview of the article. However Frequently Asked Question 1.3 of AR4 is "What is the Greenhouse Effect?" and although "greenhouse gas" does not appear in the first paragraph of the answer it appears five times in the next three paragraphs. FAQ 7.1, 9.1, and 10.3 all talk about greenhouse gases. Furthermore sections 2.3.4 and 2.3.7 of AR4 talk explicitly about greenhouse gases while section 2.3.8 is titled "Observations of Long-Lived Greenhouse Gas Radiative Effects" while other sections refer to the World Data Centre for Greenhouse Gases. Without greenhouse gases there would be no greenhouse effect, so I don't understand your reluctance to mention GHGs.
- I think it is accurate to say there would be no greenhouse effect without the Sun. (The inverse is also true.) I think it would be correct to say there would be a greenhouse effect with no greenhouse gases, where the greenhouse effect (cycling energy into and out of the atmosphere, exchanging energy with the surface) would be less prominent (i.e. lower heat transfer rates) as heat transfer processes would be restricted to conduction and convection. More generally, because other authors might be confused by the true nature of GHGs, it doesn't mean we have to be. And, again more generally, while the Earth's case certainly involves GHGs, if you agree that a high level description of The Greenhouse Effect involves a number of high level ideas (the cyclical exchange of energy between atmosphere and surface; the source of the energy is the Sun; the sink of the energy is outer space; conservation of energy governs the long term average energy exchanges processes; radiation, convection and conduction all participate in the cyclical exchange of energy) then what possible high level purpose does introducing a detail, one associated with rates of temperature change, like GHGs serve? It's just a matter of carefully placing GHGs where they belong in the hierarchy of pertinent ideas. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Regarding your "Heat is not trapped; it is temporarily retained or temporarily stored," one could say the same of stoats hunted for their fur. On the one hand you argue for simplicity, on the other you argue for fine semantic distinctions that are going to be completely lost on the first-time reader, not to mention the experts. You will find the terms "trapped," "prevented from escaping," "insulation," and so on used in many texts treating the greenhouse effect.
- There are a lot of misguided authors/conveyors of "information" out there. Just look at the Texas Board of Education. You have to decide if accuracy (and honesty and truth) is more important than mimicry. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Regarding your "An intro is for 'what,' and not for 'how'," can you back this up with a source? It's news to me and I don't see anything to that effect in WP:LEAD.
- That is a challenge. I just work off of first principles. And sometimes I have to figure out the principles from scratch (read common sense). But if I run across something published on this subject, I'll come back here with a reference. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Regarding your "I would argue that the (somewhat restrained) cooling processes by which the atmosphere loses its temporarily stored thermal energy are more (at least equally) important than the warming processes," you have my undivided attention. If it bears on the lead I would be even more interested. --Vaughan Pratt (talk) 07:37, 20 May 2010 (UTC)
- Why don't we work on this one the other way around. Suppose you took a course on atmospheric science at some famous university during which you learned all you now know about global warming, greenhouse gases, thermal storage and transfer processes etc. Then suppose your final exam had one essay topic: You've learned all about the processes that lead to warming the atmosphere. Now defend the proposition that surface/atmospheric cooling processes are more prominent, pervasive, important to the maintenance/existence of life, and stabilizing than warming process. blackcloak (talk) 07:54, 30 May 2010 (UTC)
- Were AR4 not to have mentioned greenhouse gases as you claim then presumably section 2 of this article could have been deleted; all I was doing was following WP:LEAD's guideline that the lead provide an overview of the article. However Frequently Asked Question 1.3 of AR4 is "What is the Greenhouse Effect?" and although "greenhouse gas" does not appear in the first paragraph of the answer it appears five times in the next three paragraphs. FAQ 7.1, 9.1, and 10.3 all talk about greenhouse gases. Furthermore sections 2.3.4 and 2.3.7 of AR4 talk explicitly about greenhouse gases while section 2.3.8 is titled "Observations of Long-Lived Greenhouse Gas Radiative Effects" while other sections refer to the World Data Centre for Greenhouse Gases. Without greenhouse gases there would be no greenhouse effect, so I don't understand your reluctance to mention GHGs.
Suggested fix for the 8 concerns above
With Blackcloak's good suggestions my suggested first paragraph now reads as follows.
- On any planet with an atmosphere, the greenhouse effect is the retention of heat due to greenhouse gases in the atmosphere. A greenhouse gas is one that is transparent to incoming shortwave radiation from a nearby hot sun but relatively opaque to outgoing longwave radiation generated by the planet's warmth, thereby raising the planet's surface temperature in order to maintain the balance between incoming and outgoing radiation. In our own situation the Sun is about twenty times hotter than Earth, the wavelength of the incoming radiation is correspondingly twenty times less than the outgoing, and the main greenhouse gases impeding the latter are water vapor, carbon dioxide, methane, nitrous oxide, and ozone, which together raise Earth's temperature more than 30°C.
In the end I didn't follow the suggestion to add "relative to Earth's effective temperature" because effective temperature is a single value whereas the Earth's temperature without GHGs varies greatly with latitude. The delta on the other hand is relatively (though of course not completely) independent of latitude and is therefore a less problematic indicator of the magnitude of the greenhouse effect in that regard.
I believe the paragraph also addresses all eight of the concerns Blackcloak raised in the previous section. Instead of "The ..., They ..., This ..., This ..., The ..." the sentences now begin "On ..., A ..., In ..." (assuming that was a serious concern to begin with---I'm with WMC on that detail of style). Emission to outer space is covered by the reference to Outgoing longwave radiation while the source of the incoming radiation is identified as the Sun (I agree with Blackcloak that it is appropriate to mention this, however obvious it may seem). Balancing incoming and outgoing radiation is now mentioned. A context is supplied, namely "On any planet with an atmosphere." The rewrite should also take care of the other concerns he listed. --Vaughan Pratt (talk) 17:59, 20 May 2010 (UTC)
- This has lost what (for me) was the 'penny-drops' concept - radiation comes in 'unhindered', but on the way out, at a different frequency, is absorbed and part of it is re-emitted back down. I know this is implicit in the word 'opaque', but you need a decent level of physics education to know that.
- I'd rather that the first couple of sentences keep to a very short, simple, and (if necessary) intentionally vague description. We are after all taking three bites of the cherry anyway - the first and third paras of the lead, and then a whole 'basic mechanism' section. Ideally (IMHO) the first attempt ought to be comprehensible to the widest possible audience. Squiddy | (squirt ink?) 21:19, 20 May 2010 (UTC)
- I agree that I don't think this is better. It seems to turn the whole concept onto those two strangely specific, but to my eye uninformative articles, shortwave radiation and outgoing longwave radiation. Secondly it seems in a hurry to summarise not just the article, but some of the rest of the lede - the 30°C figure appears a few paras later as 33°C. I don't see that the existing opening is nearly as awful as you guys seem to think it is. --Nigelj (talk) 21:49, 20 May 2010 (UTC)
- The first sentence is awkward to read. Furthermore, it is incorrect. The second sentence would be a good building block for a leadoff. The third sentence also is awkward, to the point that I can't figure out what it's trying to say (probably you're trying to make a single sentence do too much work). Squiddy's remarks should be taken on board. Short Brigade Harvester Boris (talk) 00:25, 21 May 2010 (UTC)
- If people feel the present version is fine as is then by all means leave it be. I was just responding to what I had taken to be reasonable complaints.
- Regarding my first sentence, I'd certainly be interested to know what's awkward about it, what's wrong about it, and how you'd fix it. My third sentence can easily be broken into two; is there anything else awkward about it besides its length?
- Regarding my omission of "radiation re-emitted back down," every point in the troposphere radiates FIR in all directions, a not atypical situation in radiation physics. The only departure from perfect isotropy in this case is a preference for vertically upwards radiation over downwards. Referring to this bias as "radiation re-emitted back down" comes across to physicists if not climatologists as a violation of the second law of thermodynamics, which Gerlich and Tscheuschner exploited mercilessly in their anti-global-warming article. In the FIR part of the EM spectrum the action of the troposphere is simply as a thermal insulator, and it propagates heat the same way any other thermal insulator does, scattering phonons and photons in all directions as the heat flows through. Unless you clarify things by pointing out that all thermal insulators propagate heat "backwards" in the sense used in climatology you create the impression that the troposphere somehow works differently from an ordinary thermal insulator when in fact it obeys Fourier's law just like any other homogeneous thermal conductor when suitably adjusted for its dependence of density on altitude. --Vaughan Pratt (talk) 04:44, 21 May 2010 (UTC)
- The first sentence is awkward to read. Furthermore, it is incorrect. The second sentence would be a good building block for a leadoff. The third sentence also is awkward, to the point that I can't figure out what it's trying to say (probably you're trying to make a single sentence do too much work). Squiddy's remarks should be taken on board. Short Brigade Harvester Boris (talk) 00:25, 21 May 2010 (UTC)
- I agree that I don't think this is better. It seems to turn the whole concept onto those two strangely specific, but to my eye uninformative articles, shortwave radiation and outgoing longwave radiation. Secondly it seems in a hurry to summarise not just the article, but some of the rest of the lede - the 30°C figure appears a few paras later as 33°C. I don't see that the existing opening is nearly as awful as you guys seem to think it is. --Nigelj (talk) 21:49, 20 May 2010 (UTC)
Molecular Interaction
Could a more detailed description of the infrared absorption / release phenomenon for greenhouse gases like carbon dioxide and water vapor (or a link to it) be added? A series of molecular level drawings (or short animation) depicting the energy storage / release mechanism (change in bond angle, electron shell level change) would be real helpful.
From the fourth bullet point:
To maintain its own equilibrium, it re-radiates the absorbed heat in all directions, both upwards and downwards. This results in more warmth below, while still radiating enough heat back out into deep space from the upper layers to maintain overall thermal equilibrium.
It is not clear to me why thermal infrared energy absorbed and re-emitted by a greenhouse gas molecule is treated differently than thermal infrared energy striking either a greenhouse or non-greenhouse gas molecule and changing its kinetic energy. Either process is going to slow the transmission of energy back into outer space.
- The difference between a GHG molecule like CO2 and a non-GHG molecule like O2 is that the electron energy levels of the latter have a much simpler structure. There are therefore many fewer infrared wavelengths that match possible energy transitions in the latter (via the relation E = hν), with the result that most IR radiation fails to interact with non-GHG molecules and just goes on by without changing their energy. --Vaughan Pratt (talk) 05:45, 21 May 2010 (UTC)
And I still believe that a higher concentration of greenhouse gases in the atmosphere raises the by volume specific heat of the atmosphere. Gas mixtures with higher specific heats require more energy to achieve the same change in temperature which is not addressed in the article. —Preceding unsigned comment added by 24.3.11.218 (talk) 03:08, 21 May 2010 (UTC)
- It's not addressed because it's generally considered negligible. Doubling the CO2 level in the atmosphere doubles the rate at which photons are captured by CO2 in any small neighborhood, yet it raises the
specific heatthermal capacity of the atmosphere less than 0.1% because the CO2 level increases only from 0.04% to 0.08% by volume. --Vaughan Pratt (talk) 05:45, 21 May 2010 (UTC)- "Doubling the CO2 level in the atmosphere doubles the rate at which photons are captured by CO2" Yes Vaughan, but the rate of emission is doubled also; the energy (of the CO2) remains the same, it is defined by the temperature which is the same as the other atmospheric gases. --Damorbel (talk) 06:17, 21 May 2010 (UTC)
- Of course, and I would have said so had it been relevant to 24.3.11.218's concern about specific heat. Rate of photon exchange has negligible bearing on specific heat, for example O2 and CO2 have very different such rates but very similar specific heats. The only reason I even brought up photon exchange was to contrast it with specific heat, which when measured by weight as usual actually decreases very slightly with increasing CO2. (The by-volume specific heat increases, but again only very slightly. I struck out "raises the specific heat of the atmosphere" in favor of (total) thermal capacity because that's a more useful measure of the (very slight) thermal impact of adding CO2 since the addition isn't at the expense of the other gases whose total mass remains essentially the same, and it also eliminates the weight-vs-volume consideration.) --Vaughan Pratt (talk) 15:04, 21 May 2010 (UTC)
- "Doubling the CO2 level in the atmosphere doubles the rate at which photons are captured by CO2" Yes Vaughan, but the rate of emission is doubled also; the energy (of the CO2) remains the same, it is defined by the temperature which is the same as the other atmospheric gases. --Damorbel (talk) 06:17, 21 May 2010 (UTC)
Here is the key sticking point. Even if the IR radiation is not absorbed and re-emitted by non-greenhouse gasses by changing electron energy levels, it seems to me that the infra-red radiation still would undergo an electro-magnetic to kinetic energy conversion whenever it strikes a non-greenhouse gas. This is the part I don't understand. When someone says that "most IR radiation fails to interact with non-GHG molecules and just goes on by without changing their energy" does that imply that taking a cannister of non-greenhouse gas and shooting a bunch of long wave infra-red energy at it will not raise the temperature of the gas inside - ie there is no electro-magnetic to kinetic energy conversion?
- Yes, that is exactly what it means. BTW, long wave IR radiation does not have enough energy to affect electron levels. Instead, it affects rotation and vibration. Diatomic molecules (O2 and N2) absorb rotational and vibrational energy outside the long wave IR band. Q Science (talk) 20:27, 21 May 2010 (UTC)
- What do you mean by "instead"? It does both. While electrons in orbit about a nucleus have energy levels spaced on the order of an electron volt, in the vicinity of visible light, those in a bond have lower energy and I'd expect 0.1 eV to be possible (but I'd appreciate a pointer to exact numbers for typical GHG bonds if anyone has them). The picture you have would entail photons directly shifting the energy of protons (the only charged particles in the nucleus), but every proton is shielded by a cloud of electrons whereas the electrons in a bond are out there as sitting ducks for any FIR ~ 0.1 eV photon to pick off. --Vaughan Pratt (talk) 17:38, 22 May 2010 (UTC)
- You can use spectralcalc.com to plot an approximate spectrum. (It does not apply pressure broadening, adjust for differences in concentration, or adjust for temperature.) Just select the molecules you want. Set the range from 200 cm-1 (0.02 eV) to 2500 cm-1 (0.31 eV) to approximate a 15C blackbody. Q Science (talk) 18:54, 22 May 2010 (UTC)
- That only tells you (a bit indirectly) the probability of capture of a given frequency of photon by a given gas, which doesn't address electronic vs. vibrational levels. To complete the picture for photon absorption in GHG's that I was giving above, photons don't just pick any old low-energy electron in a bond, they wait for a bond which besides having a suitable electron configuration is also at a positional extreme in its vibration (of whatever kind), this being where the kinetic energy is at a minimum, called the Franck-Condon principle (not the playwright but respectively German and American physicists). This permits vibronic coupling in which the photon shifts the electronic and vibrational energy levels simultaneously. Photons avoid bonds with high kinetic energy because changing the kinetic energy of nuclei requires vastly more momentum than available from the photon. Instead the shifted electron level induces an instantaneous shift in the potential energy in the bond, like adjusting the spring constant in a spring, which is subsequently reflected quarter of a cycle (~.01 picoseconds) later in the bond's shifted kinetic energy. (Disclaimer: this is at best a semiclassical approximation.)
- Incidentally photon emission is not the exact mirror image of photon absorption, due to Kasha's rule that emitted photons come mainly from the lowest excited electron state, which should be taken into account when theorizing about re-radiation. On the other hand phonons as the bosons mediating direct exchange of vibrational energy between colliding molecules of air are like photons in lacking momentum and there is therefore also a Franck-Condon principle for phonons, with electrons participating via the van der Waals force. --Vaughan Pratt (talk) 20:13, 22 May 2010 (UTC)
- You can use spectralcalc.com to plot an approximate spectrum. (It does not apply pressure broadening, adjust for differences in concentration, or adjust for temperature.) Just select the molecules you want. Set the range from 200 cm-1 (0.02 eV) to 2500 cm-1 (0.31 eV) to approximate a 15C blackbody. Q Science (talk) 18:54, 22 May 2010 (UTC)
- Okay, now we are getting somewhere. Why doesn't the article just come out and say that greenhouse gases slow the transmission of long wave thermal infra-red energy from the surface of the earth back to space by transforming that energy (which travels at the speed of light) into molecular rotational and vibrational energy which propagates back up through the atmosphere at a much slower speed. This difference in the speed of propagation of energy causes solar energy to be retained beyond the diurnal and seasonal cycles resulting in an increased atmospheric temperature over time.
- The article instead seems to imply that the warming effect from greenhouse gases is a result of the directional trajectory of the energy transfer rather than a change in form and speed. —Preceding unsigned comment added by 24.3.11.218 (talk) 21:47, 21 May 2010 (UTC)
- Agreed (in principle if not with your exact details), but try telling that to the editors here with climatology backgrounds. My early background by training is physics, not climatology, but when I tried making simplifications like that the climatologists said I was wrong. Physicist Richard A. Muller gets similar reactions when writing about paleoclimatology. I think if you locked twenty good climatologists and twenty good physicists in a conference center like MSRI for a week the public would find climatology more believable thereafter, or at least more consistently explained. Either that or Wikipedia is not getting its fair share of climatologists---more likely the latter when one considers that the 2009 (fourth) edition of John T. Houghton's book Global Warming: the complete briefing contains more science and less politics than the report of the panel he (co)chaired from 1988 to 2002 and should be more than sufficiently clear yet accurate for those with a basic scientific interest in global warming. For greater technical depth there is C. Donald Ahrens' Meteorology Today now in its 9th edition, and if that's still not enough then Roland B. Stull's Meteorology for Scientists and Engineers (2nd ed.) will give you a real workout while coordinating tightly with Ahrens' book. --Vaughan Pratt (talk) 22:01, 22 May 2010 (UTC)
- The article instead seems to imply that the warming effect from greenhouse gases is a result of the directional trajectory of the energy transfer rather than a change in form and speed. —Preceding unsigned comment added by 24.3.11.218 (talk) 21:47, 21 May 2010 (UTC)
Also, --I struck out "raises the specific heat of the atmosphere" in favor of (total) thermal capacity because that's a more useful measure of the (very slight) thermal impact of adding CO2 since the addition isn't at the expense of the other gases whose total mass remains essentially the same, and it also eliminates the weight-vs-volume consideration.)--
The addition of CO2 is at the expense of oxygen. That's what happens when oil, coal, and other hydrocarbons are burned - oxidation. Consider the burning of methane (CH4). The combustion equation is CH4 + 2O2 = CO2 + 2H2O. So for every mole of CH4 that is burned, 2 moles of oxygen are removed from the atmosphere, two moles of water vapor and a mole of carbon dioxide are added to the atmosphere. And so if we were doubling CO2 levels from 380 ppmv to 760 ppmv by burning hydrocarbons it is reasonable to expect that O2 levels would decrease by 760 ppmv and water vapor would increase by some fraction of 760 ppmv.
- Water vapor remains approximately constant. Therefore, the extra water vapor simply condenses into the oceans. Q Science (talk) 20:27, 21 May 2010 (UTC)
- Excellent points. So what's the net effect on the thermal capacity of the atmosphere (defined say as the number of exajoules to heat it one degree) attributable to burning a typical year's mix of fossil fuels: does it go up or down? --Vaughan Pratt (talk) 03:32, 22 May 2010 (UTC)
- Running a back of napkin way simplified calculation using the known mixing ratios of the various gases in the atmosphere and the known mass of the atmosphere and then burning enough methane gas to double the CO2 content to 760 ppm gives a rise in the specific heat of the atmosphere from about .029165 kJ / mole * Deg K to .029171 kJ / mole * Deg K. There are about 1.8x10^20 moles of atmospheric gas, so the difference in energy to heat the two atmospheres by one degree Kelvin is about 3.08 exajoules more for the atmosphere with twice as much CO2.
- Did you remember to take out the oxygen Q Science mentioned? --Vaughan Pratt (talk) 18:04, 23 May 2010 (UTC)
- Actually Q Science implied that the extra water vapor should be removed from the "new" atmosphere. I am not sure if I agree that the water vapor content of the atmosphere is relatively constant (positive feedback loop from increased atmospheric temperature causing more water evaporation). Anyway, I ran the calculations removing the Oxygen from the atmosphere that is consumed by combustion, and added all of the water vapor and CO2 to the atmosphere that is generated by combustion. Like I said, this was a quick and dirty calculation to get a rough idea of the effect of specific heats from different atmospheric mixing ratios.
- Did you remember to take out the oxygen Q Science mentioned? --Vaughan Pratt (talk) 18:04, 23 May 2010 (UTC)
- Running a back of napkin way simplified calculation using the known mixing ratios of the various gases in the atmosphere and the known mass of the atmosphere and then burning enough methane gas to double the CO2 content to 760 ppm gives a rise in the specific heat of the atmosphere from about .029165 kJ / mole * Deg K to .029171 kJ / mole * Deg K. There are about 1.8x10^20 moles of atmospheric gas, so the difference in energy to heat the two atmospheres by one degree Kelvin is about 3.08 exajoules more for the atmosphere with twice as much CO2.
- 24.3.11.218, please sign your edits by adding 4 tilde's (~) to the end. Please. Q Science (talk) 20:44, 23 May 2010 (UTC)
- If instead I add to the atmosphere is about 30,000 metric tons of CO2 (yearly production) and calculate (assuming all the CO2 is generated by burning methane) the energy difference for heating the atmosphere by 1 degree K is only about 30 gigajoules (3 x 10^10 joules). Compared with .9 yottajoules (.9 x 10^24 joules) of yearly solar energy received by the Earth this isn't even a rounding error. —Preceding unsigned comment added by 24.3.11.218 (talk) 06:33, 23 May 2010 (UTC)
- You might want to redo this after changing "metric" to "mega." We add close to 30 gigatonnes of CO2 to the atmosphere in each year just from consuming and flaring fossil fuels, not counting biofuels, forest fires including slash-and-burn, etc. On the other hand nature has been trying to keep up with this increase and based on the Keeling curve seems to be taking down ~60% of it, so that the net annual increase is only 16.7 gigatonnes of CO2 = 4.55 Gt of carbon (GtC). By 2020 however, extrapolating the (remarkably regular) Keeling curve under the assumption of business as usual, the annual increase in atmospheric CO2 should be about 5.4 GtC. --Vaughan Pratt (talk) 18:04, 23 May 2010 (UTC)
- I used metric tons (a metric ton is 1000 kg) to spare myself a bunch of unit conversions.
- You might want to redo this after changing "metric" to "mega." We add close to 30 gigatonnes of CO2 to the atmosphere in each year just from consuming and flaring fossil fuels, not counting biofuels, forest fires including slash-and-burn, etc. On the other hand nature has been trying to keep up with this increase and based on the Keeling curve seems to be taking down ~60% of it, so that the net annual increase is only 16.7 gigatonnes of CO2 = 4.55 Gt of carbon (GtC). By 2020 however, extrapolating the (remarkably regular) Keeling curve under the assumption of business as usual, the annual increase in atmospheric CO2 should be about 5.4 GtC. --Vaughan Pratt (talk) 18:04, 23 May 2010 (UTC)
- If instead I add to the atmosphere is about 30,000 metric tons of CO2 (yearly production) and calculate (assuming all the CO2 is generated by burning methane) the energy difference for heating the atmosphere by 1 degree K is only about 30 gigajoules (3 x 10^10 joules). Compared with .9 yottajoules (.9 x 10^24 joules) of yearly solar energy received by the Earth this isn't even a rounding error. —Preceding unsigned comment added by 24.3.11.218 (talk) 06:33, 23 May 2010 (UTC)
- True in general, though oxidation of CH4 is a significant source of water vapor to the mid to upper stratosphere. See e.g., [1] Short Brigade Harvester Boris (talk) 03:58, 22 May 2010 (UTC)
- BTW, atmospheric gases should NOT be referenced as having a "concentration" of xx ppmv. Instead, refer to a "mixing ratio" of xx ppm. This is because the pressure is variable. Q Science (talk) 20:27, 21 May 2010 (UTC)
- Not only is "level" shorter than either but people are less likely to pick on you. ;) --Vaughan Pratt (talk) 03:32, 22 May 2010 (UTC)
I can only assume that the volume of the atmosphere does not change (which is why I use specific heat by volume), otherwise the Power emmission for the Earths atmosphere (Power = epsilon * sigma * Area * Temperature ^ 4) increases not only for increased temperature but also for increased surface area (surface area of sphere = 6 * volume / radius).
Of course, my assumption on fixed atmospheric volume may be incorrect. Molar densities (moles per cubic meter) of oxygen, water vapor, and carbon dioxide are pretty close (44.656 for O2, 44.627 for Water Vapor, and 44.922 for CO2). And so burning methane adds more moles of gasses to the atmosphere (1 mole CO2 and 2 moles water vapor) than it takes in (2 moles oxygen). Those extra moles gotta go somewhere, either higher atmospheric pressure and density in the same volume or larger volume. —Preceding unsigned comment added by 24.3.11.218 (talk) 18:28, 21 May 2010 (UTC)
- The main take-away point here is that changes in either the specific heat (however defined) or the thermal capacity of the atmosphere attributable to burning fossil fuels are essentially irrelevant to global warming. First the changes are tiny compared to the thermal impact of increasing GHGs. Second such changes, to the extent that they're felt at all, don't impact annually averaged temperature but only the amplitude of the diurnal and other oscillations of the surface temperature, and that negligibly. --Vaughan Pratt (talk) 03:32, 22 May 2010 (UTC)
Heat and infrared radiation
From the Article: Each layer of atmosphere with greenhouses gases absorbs some of the heat being radiated upwards from lower layers. To maintain its own equilibrium, it re-radiates the absorbed heat in all directions, both upwards and downwards. This results in more warmth below, while still radiating enough heat back out into deep space from the upper layers to maintain overall thermal equilibrium.
This is an inaccurate description in a couple senses. First "heat" is not a form of energy. From the definition of heat in Wikipedia: Heat is a flow of energy, rather than a form of energy. The term the article is reaching for is thermal energy or atmospheric internal energy.
Second, greenhouse gasses do not absorb thermal energy radiating from the Earth. Greenhouse gasses convert one energy form - long wave infrared electromagnetic radiation to another form translational and rotational kinetic energy.
Third, after this radiation is converted to kinetic energy, the reverse process does not take place. There is no re-radiation. If that were the case then greenhouse gasses would have little effect on how long energy is retained by the Earth's atmosphere since all electromagnetic radiation travels at the speed of light.
And so the paragraph should read: Greenhouses gases in the atmosphere receive some of the long wave infra-red radiation from the Earth's surface. The remaining radiation escapes at the speed of light into space. This energy is converted from electromagnetic radiation to rotational and translational kinetic energy of the greenhouse gas molecules. This increase in kinetic energy of the green house gas molecules is shared by the remaining non-greenhouse gasses. The increased kinetic energy of the molecules in the atmosphere raises the temperature of the atmosphere. 24.3.11.218 (talk) 03:23, 7 June 2010 (UTC)
- Read the heat transfer article and notice the inconsistencies with the heat article. Also, the heat article does mention that the term is often used to mean thermal energy. As for your proposed rewrite, 1) "receive" should be "absorb", "This energy" should be "Absorbed energy", the sentence begining "This increase in kinetic energy..." should be replaced with "Some of the energy of greenhouse gases molecules may be transfered to other molecules of the atmosphere, and some may be re-emitted as infrared radiation."
- This is a point that is unclear from anything that I have read on greenhouse gasses. When visible light strikes a solid object some wavelengths of radiation energy are converted to kinetic energy of the whole molecule (heating) and others are converted to kinetic energy of individual electrons. Certain wavelengths cause electrons in the outer shells of the molecules to jump back and forth between electron shells. This "jumping back and forth" of electrons produces an electromagnetic wave that radiates outward from the molecule. This is otherwise known as a reflection.
- What you seem to be saying is that when long wave infra-red radiation strikes greenhouse gas molecules, a mix of both actions happens. My first question is what is the energy storage mechanism in a greenhouse gas molecule - what else changes about the molecule besides its translational and rotational kinetic energy. I always picture rotational kinetic energy of molecules as they are spinning freely in one direction in space. But I have heard mention of dipole moment and vibration. A dipole moment exists in a molecule whenever shared electrons spend more time around one atom of the molecule than others. The molecule therefore has a positive bias on one end and a negative bias on the other end. In the presence of an electric field molecules with dipole moments will align themselves in the field (positive bias towards negative field source and negative bias toward positive field source). Is the electric field from the infra-red radiation causing these molecules to try to spin back and forth (vibrate) along an axis? And does this vibration of a dipole molecule create its own electromagnetic radiation - in essence reflecting the incoming radiation? Again a short series of illustrations or a short "cartoon" would be real helpful in understanding what is happening on the molecular level.24.3.11.218 (talk) 13:49, 8 June 2010 (UTC)
The last sentence should read "In places where there is a net increase in the average energy of the molecules in the atmosphere, there will be an corresponding increase in the temperature." Now, this handles the warming up side, but says nothing about the cooling side. Basically each sentence you provided has a correpsonding one that describes the cooling process. You're only giving us half the story. Your statement that "Third, ... does not take place" is simply and totally wrong.
- The only way for re-radiation (photon emission) to take place is for charged particles to generate a time varying electromagnetic field. It was pointed out to me that the long wave infrared radiation does not cause individual electrons in greenhouse gas molecules to jump back and forth between electron shells (energy levels), and so I didn't know how else the reflected radiation could be created until I came across a mention of dipole moment.24.3.11.218 (talk) 13:49, 8 June 2010 (UTC)
You seem to be saying that absorbed energy, by the atmosphere, must cycle back to the Earth, heat the surface by conduction, and then radiate this energy back towards outer space to complete the cycle. The concentration levels of greenhouse gases govern energy transfer rates, in both directions. And now I'm going to raise a subject not covered anywhere in WP that I know of. This is the asymmetry in energy transfer rates between greenhouse gases and the two energy sources: radiation and non-greenhouse gases. When the atmosphere is bathed by EM radiation originating from the surface, the absorption process is a "volume" process with Beer's law 1/e absorption distances of the order of kilometers. This means that an IR photon heading for outer space will pass nearby, potentially aborbing, greenhouse gases billions of times before reaching outer space, it if does. A key point here is that all those billions of opportunites to be absorbed occur within tens of microseconds. Energy transfer from "energized" greenhouse gases to neighboring molecules occurs when there are random encounters under just the right conditions. Most importantly, there are 99 opportunities to transfer energy to a non-greenhouse gas molecule for every one opportunity to transfer energy to a (another) greenhouse gas molecule. Now, when it comes to cooling, after a greenhouse gas molecule has released its excess energy in the form of IR radiation, half of which heads towards outer space and half of which heads for the surface, it may accept its next packet of energy. This time, on random encounters with neighboring molecules, an energized non-greenhouse gas molecules is 99 times more likely to come in contact with another non-greenhouse gas molecules than a greenhouse gas molecule. While the random encounters that may result in the transfer of thermal energy between (both into and out of) greenhouse gas molecules and non-greenhouse molecules are roughly symmetrical, the average time for a packet of energy entering a greenhouse molecule via IR radiation and via a nearby non-greenhouse gas are markedly different. There is for instance a strong pressure dependence. We can estimate, roughly, the ratio of the energy capture rate to the energy release rate at 1 to 3 (really just a guess) by estimating, at any one instant, the ratio of the Earth's surface that is above the average global temperature to the surface area that is below the average global temperature. blackcloak (talk) 05:59, 8 June 2010 (UTC)
- Actually, most of the surface radiation is absorbed in less than one meter from the surface. It is only at the edges of the absorption bands that the 1/e absorption distance is on the order of kilometers. As a result, infrared cameras are designed to operate in one of the "windows" where there is little atmospheric absorption. Q Science (talk) 21:32, 8 June 2010 (UTC)
- I'll assume you know more than I do on this matter, but will comment on your points anyway. The only way a 1/e in band absorption distance of 1 meter at the surface makes sense to me is if the vast majority of greenhouse gas molecules that have just absorbed a photon re-radiate that energy rather than transfer it to a neighboring non-greenhouse gas molecule. So let's see if you can answer the following questions. 1) Near the surface, under nominally equilibrium conditions (high noon), what percentage of greenhouse-gas-molecules excited by in-absorption-band (say H20) photons transfer (at least of portion of) the just absorbed energy to a non-greenhouse gas molecule (as opposed to re-radiating the energy)? 2) What is a typical average time that a greenhouse gas molecule (surface, equilibrium) stays in an excited state, after absorbing a packet of energy (either from a photon or by interacting with a nearby molecule), before radiating the energy (or a portion) or transfering the energy (or a portion) to a nearby molecule? 3) Is there a net energy flow consideration present here that leads to your understanding of 1/e absorption distance of a meter and mine of a much longer distance? Specifically, under equilibrium conditions, a volume r^3 of atmosphere near the surface that receives x energy per unit time and discharges x energy per unit time would force 1/e "effective" distances to be far greater than r. The net energy change in the volume is zero even though absorption and emission processes may occur frequently (your picture) within the volume. I could go on, but let's see how you respond. (BTW I didn't introduce the idea of "walking out of the absorption bands" on successive absorption/emission cycles to avoid (what may be necessary) complexity.) blackcloak (talk) 06:49, 9 June 2010 (UTC)
- At equilibrium, the energy transferred to non-Greenhouse gas molecules is identically equal to the energy transferred the other way. By definition. At high noon, since the atmosphere is getting warmer there is a net heat flow into the atmosphere. Because this also changes the atmospheric density, this warmer air will convect so that the measured temperature two feet above the ground will not change. As a result, the ratio of absorbed to emitted radiation is based only on the temperature difference between the surface and the air.
- As for the amount of time between absorbing a photon and simply re-emitting it, it must be longer than the amount of time between collisions. If it was not, then the atmospheric pressure would not broaden the emission bands. In fact, the atmosphere would cool very quickly at night if every collision caused a photon to be emitted. Instead, the night time atmospheric temperature tends to be much warmer than the ground.
- Time between collisions - Using the RMS velocity from v=sqrt(3p*N/V*m) and the molecule sizes I was able to pull from the internet, a rough calculation for collision times between molecules is about 2.6E-12 sec for Oxygen, 1.9E-12 sec for Water vapor, 2.5E-12 sec for Nitrogen, and 3.0E-12 sec for Carbon Dioxide at standard pressure (101 kPA). This would be in a container with only a single gas, but since the numbers are all of the same magnitude figure about 2 to 2.5 picoseconds. Those numbers reflect the speed at which two molecules traveling head on will reach each other. Because the average space between molecules is about 11 molecules wide (10.3 for Carbon Dioxide, 12.0 for Water Vapor, 11.4 for Oxygen, 11.3 for Nitrogen), the odds of collision between two molecules is about (1/11^3)= 1 in 1331. So multiplying those times by the odds of collision yields a time between collisions of 3.8E-9 sec for Oxygen, 3.3E-9 sec for Water Vapor, 3.5E-9 sec for Nitrogen, 3.3E-9 sec for Carbon Dioxide. At lower pressures higher up in the atmosphere the collision times get longer - for instance at thirty thousand meters the pressure is about 1 kPA and the collision times are 30,000 times longer.24.3.11.218 (talk) 20:20, 9 June 2010 (UTC)
- You're going to have to be a lot more verbose if you want us to follow your logic. I can, for instance, envision many more photon/GHG molecule absorption/emission cycles between collisions, such that, when the collisions do occur and the GHG molecule is in an excited state, there is a probability that energy will be transferred to a non-GHG molecule. Radiation remitted at the absorbed wavelength simply moves on where, on its next interaction with a GHG molecule, it may happen to be absorbed just before a non-GHG molecule strikes the GHG molecule (or emitted just after a collision with a non-GHG molecule that involved kinetic or vibrational energy transfer and subsequent "broadening"). On the cooling very quickly at night comment, only one in 99 collisions have the possibility of transfering energy that can then be radiated. The low percentage of GHG molecules in the atmosphere becomes the bottleneck through which energy must flow to cool all the molecules of the atmosphere. Also slowing down cooling is omnidirectional emission and re-absorption by other GHG molecules. blackcloak (talk) 03:46, 10 June 2010 (UTC)
- I agree - "As for the amount of time between absorbing a photon and simply re-emitting it, it must be longer than the amount of time between collisions." This is false. The photon / absorption emission times should be 1 divided by the frequency of the radiation. So for Carbon Dioxide those times for 2.6, 4, and 13 micron radiation are 8.67E-15, 13.33E-15, and 43.37E-15 seconds. These are significantly less than the collision time for CO2. And so there are thousands of absorption / emission cycles between collisions. And as you get higher up in the atmosphere (lower pressure) those thousands become millions. Also, I don't think CO2 represents a bottle neck in the cooling process. When two molecules (of any material) collide it is not a perfectly elastic collision, ie some energy is lost in the form of thermal radiation. This thermal radiation is absorbed by neighboring molecules and heat is transferred through the atmosphere. At high altitudes (lower pressure and less density) the thermal radiation has a greater chance to avoid striking neighboring gas molecules and make it back into space.24.3.11.218 (talk) 05:47, 10 June 2010 (UTC)
- If you were correct, then there would be no absorption and, therefore, no absorption spectra. Please note that one over the frequency is simply how long it takes a photon to travel one wavelength, not how long the energy is stored before it is emitted. Also, note that dividing the speed of light by the frequency gives the wavelength. Q Science (talk) 13:36, 10 June 2010 (UTC)
- For CO2 how big are the photons, meaning how many wavelengths long are the energy "packets" that are absorbed at the various frequencies of infrared radiation? I have always assumed (bad idea) that a single photon is represented by a single wavelength.24.3.11.218 (talk) 15:58, 10 June 2010 (UTC)
- I don't think anyone knows. According to the CO2 page, the largest dimension of a CO2 molecule is about 230pm, while the wavelength of the absorbed IR radiation is close to 4um and 15um, about 4 orders of magnitude larger than the molecule. Q Science (talk) 17:59, 10 June 2010 (UTC)
- By the Heisenberg Uncertainty Principle the (lateral) diameter of a photon is inversely proportional to the angular diameter of the source. This is because the smaller the source the closer the known lateral momentum of the photon is to zero, and hence the less certain is its lateral position. Photons from distant stars can be many meters wide in that sense, the basis for the Hanbury Brown and Twiss effect and the Narrabri Stellar Intensity Interferometer (at least that was how Hanbury Brown explained it in a course I took from him in 1964). --Vaughan Pratt (talk) 22:12, 21 June 2010 (UTC)
- You're going to have to fill in the details of your argument before I will believe you. Earlier comments refute your simple statement about absorption, and you have said nothing to refute the counter arguments. Perhaps the best way to resolve this is for you to provide a reference. blackcloak (talk) 06:05, 11 June 2010 (UTC)
- If what you're saying is correct ("When two molecules (of any material) collide it is not a perfectly elastic collision, ie some energy is lost in the form of thermal radiation."), this is a new idea to me. You do not qualify your statement, so I can assume I can choose an extreme set of circumstances. Let's consider two O2 molecules, both molecules in their lowest electronic energy state, both molecules having kinetic energy (finite temperature), that collide, their relative speed being very slow. You are saying that there will be some, however small, energy released in the form of electromagetic radiation. This means that two "particles" become three "particles" (do I hear four) with energy and momentum conserved, and since the O2 molecules are in their lowest electronic state before the collision they can not be in anything but their lowest electronic state after the collision. This I am finding hard to believe, but I can learn. (Are you new to Wikipedia, or just not signed in?) blackcloak (talk) 08:29, 10 June 2010 (UTC)
- If you sum the momentums of the two O2 molecules before and after collision the sums will not be exactly the same. See the Wikipedia article Elastic_collision. Molecular collisions are sometimes inelastic and sometimes superelastic but rarely ever perfectly elastic. Here is the important sentence from the fourth paragraph - "Averaged across the entire sample, molecular collisions can be regarded as essentially elastic AS LONG AS black body photons are not permitted to carry away energy from the system." In the case of the atmosphere black body radiation IS permitted to carry energy away from the system.24.3.11.218 (talk) 13:46, 10 June 2010 (UTC)
- Thanks. I read the reference and noticed that very little was said about inelastic collisions. I also read the article on inelastic collisions which did not have much more to say about the collisions of gas molecules. The superelastic case interests me because it appears to violate conservation of energy. Could it be that very long wavelength IR radiation bathes the atmosphere and is therefore available as an energy source in molecular collisions? Specifically, during the collision of two gas molecules the colliding molecules can accept energy from the sea of very long wavelength photons, leading to a higher energy content of the molecules just after departure from the collision compared with their energy just before collision. This would lead to the conclusion that the general case of two gas molecules colliding should be seen as a four particle problem. Two molecules collide and during the collision a photon is absorbed, the two molecules possibly exchanging energy. Then as the two molecules depart a photon is emitted. blackcloak (talk) 19:32, 10 June 2010 (UTC)
- I don't know how long a collision between molecules lasts (10^-18, 10^-20 sec?) compared to the time required for absorption so I can't venture a guess on how often superelastic collisions occur. I would say not that often at all for small molecules like CO2, N2, and O2.24.3.11.218 (talk) 21:44, 10 June 2010 (UTC)
- Just for grins, use the Lennard-Jones potential distance supplied for an Argon dimer. The (diagram in the) article estimates the interaction distance as something like 4.5E-10m, double for in then out (maybe double again for two identical colliding molecules). Use your RMS velocity for CO2 to compute the interaction time (maybe double again to account for accelerations). Compare with average collision times. Then as the icing, compute the number of wavelengths over which the energy would spread if all the emerging photon energy were contained within the computed interaction time. My guess is that something like 300 wavelengths (100 too few, 10,000 too many) would be reasonable. blackcloak (talk) 06:39, 11 June 2010 (UTC)
- RMS Velocity of CO2 at STP is about 560 m/sec. Double that for two CO2 molecules headed straight at each other is 1120 m/sec. Interaction distance is 2 x 4.5E-10 = 9E-10 (In then out). And so time of interaction would be 9E-10 / 1120 = 8.03E-13 sec. Hmmmm - longer than I would have thought. Number of wavelengths over that period of time for 13 micron radiation is 19 wavelengths, 60 wavelengths for 4 micron radiation, and 92 wavelengths for 2.6 micron radiation. Recognize that this is not a calculation of photon packet size for infrared radiation released when two CO2 molecules collide. The packet sizes would have to be smaller than these numbers since we can always increase the pressure of the gas while maintaining constant temperature. Therefor we can increase the RMS velocity without shifting the frequency spectrum of the emissions. It would be logical to conclude that there is a pressure for CO2 at which there are fewer emission wavelengths meaning the RMS velocity is too high and the interaction time is to short. Another way to look at it is that at high enough pressure, CO2 molecules lose degrees of freedom in their molecular vibrations. For instance, at 20 Degrees C and thirty atmospheres of pressure (about 3000 kPA) phase change to liquid begins to occur for CO2. At that point for CO2 I come up with about 3 wavelengths per packet of emission for CO2 for 13 micron radiation. Recognize that RMS velocity is an average and so some molecules are going slower and some are going faster. As the pressure is increased beyond that point the average number of wavelengths per packet becomes smaller. At about 35000 kPA the average number of wavelengths per packet becomes less than 1 for 13 micron radiation. I am not sure what happens on Venus. On Venus CO2 is in supercritical fluid mode (temp 480 Deg C, pressure 9000 kPA). 24.3.11.218 (talk) 09:19, 11 June 2010 (UTC)
- By the way. I was using the Lennard-Jones potential for Argon to do the calculations above for a very, very rough idea. The proper way to do it would be to solve the partial differential equations E = 1/2*m*(dr/dt)^2 + C12/r^12 - C6/r^6 for all six atoms in the collision of two CO2 molecules. I found a listing of C12/C6 parameters for carbon, nitrogen, oxygen, and hydrogen here - http://www.csb.yale.edu/userguides/datamanip/autodock/html/Using_AutoDock_305.a.html What you would find is that after collision as the molecules are separating, the translational kinetic energies of the whole molecules can be smaller (inelastic collision) or bigger (super-elastic collision) than before. Energy is added to or subtracted from bond vibrations. These bond vibrations accept and bleed off electromagnetic radiation.24.3.11.218 (talk) 00:33, 12 June 2010 (UTC)
- By the way, I got the absorption distance by "simply" plotting the HITRAN spectra and assuming 1% water vapor. Q Science (talk) 17:57, 9 June 2010 (UTC)
- Take a look at the paragraphs above starting with "To complete the picture" that I wrote back on May 22 concerning the mechanisms involved in GHGs absorbing and reemitting photons. Basically absorption is via vibronic coupling in which the photon interacts with electrons in bonds (not in shells, that would take too much energy) which in turn interact with the nuclei at each end of the bond via the shells, on a time scale of 0.01 picoseconds and an energy scale of a tenth of an electron volt. Probability of absorption is highest when the wave functions of the bond electron and the mechanical bond vibration are in phase (the vibronic principle). Also note Kasha's rule for emission. More details if interested. --Vaughan Pratt (talk) 21:14, 8 June 2010 (UTC)
- Irrg. When two atoms are bonded together the electrons that are shared between the atoms do not suddenly go stationary. They still orbit the atoms. An electron that is shared will spend part of its time circling one atom in its outer shell and part of its time circling the other atom in its outer shell. And so when you say photons interact only with electrons in bonds do you mean that an electron will only be affected by infrared radiation when the electron is at some crossover point between circling one atom and circling another? Also, does the picture shown here Rotational–vibrational_coupling accurately depict what is happening to the bond - stretching and then collapsing?24.3.11.218 (talk) 00:29, 9 June 2010 (UTC)
- Thanks for redirecting me to your earlier comments. I went back and read them more carefully. You appear to be responding to my mention of tens of microseconds as the time for interaction with greenhouse gases. Your .01 picosecond figure is the time scale for the energy exchange of a photon and a single atom. My figure was intended to refer to the length of time a photon spends in transit through the entire atmosphere when the photon originates at the surface. In other words, a photon leaving the surface and headed directly outward into space has an exposure-to-being-absorbed-by-a-greenhouse-gas-molecule time of the order of 10 microseconds. If it lives longer, it enters outer space and may take (provided it also misses the Moon and the Sun) thousands to millions of years to be absorbed. blackcloak (talk) 06:49, 9 June 2010 (UTC)
- (Just back from a conference whence the delay, sorry.) No, I was fine with your microseconds---unimpeded radiation takes some 50 microseconds to traverse the troposphere and more than twice that to pass through the stratosphere. I only mentioned the subpicosecond time scale for absorption for perspective. --Vaughan Pratt (talk) 01:38, 20 June 2010 (UTC)
- Thanks for redirecting me to your earlier comments. I went back and read them more carefully. You appear to be responding to my mention of tens of microseconds as the time for interaction with greenhouse gases. Your .01 picosecond figure is the time scale for the energy exchange of a photon and a single atom. My figure was intended to refer to the length of time a photon spends in transit through the entire atmosphere when the photon originates at the surface. In other words, a photon leaving the surface and headed directly outward into space has an exposure-to-being-absorbed-by-a-greenhouse-gas-molecule time of the order of 10 microseconds. If it lives longer, it enters outer space and may take (provided it also misses the Moon and the Sun) thousands to millions of years to be absorbed. blackcloak (talk) 06:49, 9 June 2010 (UTC)
Convection and radiation from Earth's surface then and now
Apropos of the discussion in the previous section of the papers of Wood and Abbot in Phil. Mag. a century ago, it's interesting to compare what we know today with what they knew then about heat transport from the Earth's surface. Although Wood had little to say on the subject other than that GHGs don't play a big role, Abbot makes the point that without GHGs (which Abbot rightly takes to be mainly water vapor) convection "would be only a small factor," but with them it becomes "the main agent in removing heat from the earth's surface."
We can evaluate Abbot's statement in light of Figure 7 on p.206 (PDF page 10) of http://www.geo.utexas.edu/courses/387h/PAPERS/kiehl.pdf , "Earth's Annual Global Mean Energy Budget", by Jeff Kiehl and Kevin Trenberth. They give the net transport of heat by convection from the Earth's surface as 24 W (per square meter) for "thermals" and 78 W for evaporation, totaling 102 W. Net radiation upwards (surface radiation less back radiation) is 390 - 324 = 66 W, a mere two thirds of the convective transport. Without GHGs the full 390 W would be radiated to space, almost four times the convective transport.
On p.34 of his paper Abbot says "It is very difficult to estimate how fast the heat of the earth's surface escapes by convection, because neither the difference of temperature between the surface and the air nor the rate of motion of the air is well known." What did not occur to Abbot is that if he'd approximated the global annual precipitation as one meter (pretty close to what we know now, and not far off what Abbot would have taken it to be), the convective transport attributable to evaporation alone would be 540*4.2/31.55 = 72 W (540 calories latent heat of vaporization for water, 4.2 joules per calorie, 31.55 million seconds per year), which alone exceeds the radiative loss when GHGs are taken into account, though not when they're neglected as Abbot points out. Abbot attempted to estimate convection a different way, neglecting latent heat altogether, but did not take potential temperature into account; had he done so his estimate may well have been be considerably lower, though hitting Kiehl and Trenberth's figure of 24 W on the nose would have been impressive given what they knew in 1909. --Vaughan Pratt (talk) 08:30, 26 June 2010 (UTC)
CO2 bonds
(I started a new section partly because the preceding one was getting inconveniently long to edit and partly because it was drifting towards the title of this one.--Vaughan Pratt (talk) 01:38, 20 June 2010 (UTC))
The two bonds in CO2 can be depicted reasonably accurately in ASCII (with the caveat below) as the Lewis structure O=C=O because the bond angle is exactly 180 degrees. The bonds are oppositely polarized and therefore cancel exactly so that CO2 has no net dipole moment. The bonds are covalent rather than ionic, meaning that each bond shares valence electrons contributed by both adjacent atoms, as illustrated in the third electron sharing diagram or Lewis diagram about 1/3 of the way down the page at http://www.cem.msu.edu/~reusch/VirtualText/intro2.htm (the = in O=C=O is a synonym for four dots :: representing four shared electrons---some authors use the term Kekule structure when writing = instead of ::). This sharing brings the shells of both the oxygen and carbon atoms up to their complement of 8 by adding 2 to each oxygen atom and 4 to the carbon atom.
The exact geometry of the bonds is a bit more delicate than the above, which unfortunately the carbon dioxide article does not go into. See instead http://www.chem1.com/acad/webtext/chembond/cb07.html , in particular the picture just below the heading "Multiple bonds between unlike atoms" about quarter of the way down the page. Whereas two of the four shared carbon electrons are 2s (spherical with no angular momentum, the other two being 2p, dumbbell-shaped with angular momentum 1), the two shared oxygen electrons are 2p. Each oxygen atom first bonds to the carbon atom by pairing one of its 2p electrons with a 2s carbon electron to form an sp-hybrid pair, or σ-bond, aligned along the main O=C=O axis (about which these two hybrids spin). This leaves four 2p electrons to pair up to complete the bonds. The remaining 2p electron in one of the oxygen atoms aligns parallel to one of the two 2p electrons in the carbon atom, with both perpendicular to the main O=C=O axis, forming a π-bond. The same happens with the other oxygen atom, but oriented perpendicular to all the other bonds. The upshot is that the two σ-bonds (two electrons each) are in one axis (end-to-end and hence parallel to each other) and the two π-bonds (two electrons each) are in the two remaining axes orthogonal to each other. --Vaughan Pratt (talk) 01:38, 20 June 2010 (UTC)
- See Degrees_of_freedom_(physics_and_chemistry)#Degrees_of_freedom_in_physics From the model it appears that the 3N degrees of freedom for a CO2 molecule are thus -
- 1. Translation of molecule in any x, y, or z direction (3 Degrees)
- 2. Rotation of molecule about the center of mass along z axis, rotation about center of mass along y axis (2 Degrees) Note - Rotation about center of mass along x axis is ignored except for high energy physics.
- 3. Compression / expansion (like a spring that is compressed and then returns to shape) of pi and sigma bonds along x axis (1 Degree)
- 4. Pi and sigma bond synchronized bending (like a piece of spring steel that is bent and returns to shape) in y or z direction (2 Degrees)
- 5. Pi and sigma bond asynchronous bending - one pi and sigma bond group bends in y direction and the other bends in z direction (1 Degree)
- For the three wavelengths of infrared radiation that are in the absorption spectrum of CO2 (13 micron, 4 micron, and 2.6 micron) is it known which degree of freedom is affected by each wavelength? —Preceding unsigned comment added by 24.3.11.218 (talk) 19:02, 20 June 2010 (UTC)
- Those are Far-IR, not UV, bands. Q Science (talk) 22:46, 20 June 2010 (UTC)
- Dangit. Fixed the question.24.3.11.218 (talk) 16:43, 21 June 2010 (UTC)
- See the last sentence of Carbon dioxide in Earth's atmosphere#Current_concentration which attributes the 14.99 μ peak to bending and the 10x stronger 4.257 μ peak to asymmetric stretching. If there were a resonance for symmetric stretching it would be at 7.205 μ but if you look at http://www.spectralcalc.com/spectral_browser/db_intensity.php (the resource User:Q Science mentioned earlier) you'll see that the lines there are a million times weaker than the other two resonances because symmetrically stretching a molecule like CO2 with no dipole moment creates no net change to the electric field and hence that vibration mode cannot interact significantly with radiation. The closest things to the 2.6 μ you asked about are a couple of intermediate-strength resonances centered on 2.692 μ and 2.767 μ but I don't know what vibration modes they correspond to. Translation is unbound and hence not resonant; the other modes you asked about refine the standard explanations and it would interesting to calculate them to see whether they can be matched up to the observed lines. --Vaughan Pratt (talk) 20:22, 21 June 2010 (UTC)
- Is it correct to assume that for the 4.257 micron wavelength the asymmetric stretching occurs because the pi bonds between the carbon atom and the oxygen atoms are in different planes? Meaning that depending on the orientation of a CO2 molecule with respect to an incoming infrared beam, one pi bond may stretch and contract more than the other? Also this would imply that the sigma bonds are not stretched and compressed by the incoming radiation, but are rather "pulled along" by the stretching and compressing of the pi bonds. By the way, thanks for the link to the chem1 site. It really helps generate a mental picture of what goes on.24.3.11.218 (talk) 16:13, 22 June 2010 (UTC)
- No and yes. Asymmetric stretching can be accounted for naively simply in terms of the higher electronegativity of oxygen relative to carbon. In order to model it with any precision however I'd expect you'd need the detailed bond geometry. It's a great question, I'll run it by my AMO and physical chemistry friends when I get a chance. Meanwhile browse MO diagram#Carbon_Dioxide_MO_Diagram for additional detail not in chem1---the first image there lists the atomic orbitals that can form individually within each of the atoms, the second lists the molecular orbitals for the CO2 molecule, and the third diagrams these, locating the higher energy orbitals higher in the diagram. Whereas chem1 gives only three bonds, this gives four, chem1 has omitted the MO3/4 (anti)bonding case of two 2p orbitals parallel to the main (z) axis. --Vaughan Pratt (talk) 04:55, 23 June 2010 (UTC)
- I slept on this last detail and in the morning it occurred to me that the MO4 bonding shown in the third diagram at MO diagram#Carbon_Dioxide_MO_Diagram could only happen for an excited state of the CO2 molecule, in which one or both of the carbon atom's 2s electrons are excited to 2pz (z the principal O=C=O axis). The picture at chem1 is presumably the correct one for the ground state of CO2, taking the MO4 orbital to be zero. --Vaughan Pratt (talk) 16:37, 23 June 2010 (UTC)
- Is it correct to assume that for the 4.257 micron wavelength the asymmetric stretching occurs because the pi bonds between the carbon atom and the oxygen atoms are in different planes? Meaning that depending on the orientation of a CO2 molecule with respect to an incoming infrared beam, one pi bond may stretch and contract more than the other? Also this would imply that the sigma bonds are not stretched and compressed by the incoming radiation, but are rather "pulled along" by the stretching and compressing of the pi bonds. By the way, thanks for the link to the chem1 site. It really helps generate a mental picture of what goes on.24.3.11.218 (talk) 16:13, 22 June 2010 (UTC)
- See the last sentence of Carbon dioxide in Earth's atmosphere#Current_concentration which attributes the 14.99 μ peak to bending and the 10x stronger 4.257 μ peak to asymmetric stretching. If there were a resonance for symmetric stretching it would be at 7.205 μ but if you look at http://www.spectralcalc.com/spectral_browser/db_intensity.php (the resource User:Q Science mentioned earlier) you'll see that the lines there are a million times weaker than the other two resonances because symmetrically stretching a molecule like CO2 with no dipole moment creates no net change to the electric field and hence that vibration mode cannot interact significantly with radiation. The closest things to the 2.6 μ you asked about are a couple of intermediate-strength resonances centered on 2.692 μ and 2.767 μ but I don't know what vibration modes they correspond to. Translation is unbound and hence not resonant; the other modes you asked about refine the standard explanations and it would interesting to calculate them to see whether they can be matched up to the observed lines. --Vaughan Pratt (talk) 20:22, 21 June 2010 (UTC)
- Dangit. Fixed the question.24.3.11.218 (talk) 16:43, 21 June 2010 (UTC)
- Those are Far-IR, not UV, bands. Q Science (talk) 22:46, 20 June 2010 (UTC)
- I suppose that you both realize that the molecular orbitals are around 10 to 20 eV which corresponds to electromagnetic radiation of 124 nm to 62 nm. (green = 550 nm, uv = 200 to 400 nm) The CO2 absorption band at 15 um corresponds to 0.0827 eV. This is the reason that bending, stretching, and rotation are thought to be associated with IR absorption. Q Science (talk) 17:31, 23 June 2010 (UTC)
- The energy needed to destroy a CO2 molecule by breaking its bonds is indeed far greater than needed to make it vibrate. The former is irrelevant here. --Vaughan Pratt (talk) 19:49, 23 June 2010 (UTC)
- I suppose that you both realize that the molecular orbitals are around 10 to 20 eV which corresponds to electromagnetic radiation of 124 nm to 62 nm. (green = 550 nm, uv = 200 to 400 nm) The CO2 absorption band at 15 um corresponds to 0.0827 eV. This is the reason that bending, stretching, and rotation are thought to be associated with IR absorption. Q Science (talk) 17:31, 23 June 2010 (UTC)
- I think you are having as much trouble decyphering the diagrams as I am. There are four bonds shown on the chem1 site and four bonds shown in the MO-Diagram. On the chem1 site there are 2 sigma bonds and 2 pi bonds. The two sigma bonds are MO1 = A1 (Carbon 2s) + A8 (Oxygen 2pz) and MO4 = A2 (Carbon 2pz) + A7 (Oxygen 2pz). The two pi bonds are MO5 = A3 (Carbons 2px) + A9 (Oxygen 2px) and MO7 = A4 (Carbon 2py) + A11 (Oxygen 2py). A couple of things don't make sense about those diagrams. Why is the Carbon 2s electron being used in the sigma bond instead of the fourth Carbon 2p electron? Second, the chem1 site shows both sigma bonds being sp hybrids, but the MO-Diagram shows only one sigma bond being an sp hybrid - which is right? Also it appears that the energy diagram is showing what energy level the molecular orbitals (bonds) exist at and how much energy is required to break each of these bonds. For instance the MO1 bonding orbital has a low energy state and the MO2 anti-bonding orbital has a much higher energy state. With all of this my original question becomes when infrared radiation strikes a CO2 molecule at a given angle is it conceivable that the MO5 bond will be resonated (be stretched and compressed) by the radiation while the MO7 bond is left unaffected - hence asymmetric stretching? 24.3.11.218 (talk) 18:51, 23 June 2010 (UTC)
- When you say that the two sigma bonds are MO1 and MO4, are you referring to chem1? If so then you're overlooking that carbon contributes two 2s electrons, one to each oxygen atom. What I'm seeing at chem1 are two MO1's and no MO4's, not an MO1 and an MO4. --Vaughan Pratt (talk) 19:49, 23 June 2010 (UTC)
- No I am referring to diagram #2 on the MO page which lists the atomic orbitals that are combined to form the molecular orbitals. The type of bond (pi or sigma) is shown on diagram #1 on the MO page. If you count the bonds on diagram #1 there are two pi bonds and two sigma bonds which is the same number as depicted on the chem1 site. 24.3.11.218 (talk) 21:14, 23 June 2010 (UTC)
- Yes but the two sigma bonds at chem1 are both spz (if we take z to be the principal axis as per the MO page---the chem1 page takes it to be x), whereas the two sigma bonds at the MO page are spz for MO1 and pzpz for MO4. A sigma bond is one that is cylindrically symmetric about the principal (here z) axis, which describes both MO1 and MO4, as discussed earlier on the chem1 page under the heading "More about sigma and pi bonds." The MO page is listing the four ways in which two electrons can form a bond, whereas the chem1 page is listing the four pairings of two electrons each in an actual configuration. Two of the latter pairings are MO1, I don't see any MO4 pairing in the CO2 molecule on the chem1 page. --Vaughan Pratt (talk) 22:05, 23 June 2010 (UTC)
- Also note that MO7 on diagram 2 of the MO page is just MO5 viewed from a different angle and extremely close to the blue side of that bond, which only looks bigger than the red side due to perspective, and likewise for diagram 1. --Vaughan Pratt (talk) 22:24, 23 June 2010 (UTC)
- Going back to your earlier questions:
- 1. "Why is the Carbon 2s electron being used in the sigma bond instead of the fourth Carbon 2p electron?" What 4th carbon 2p electron? In its ground state the six electrons of a carbon atom consist of two each of 1s, 2s, and 2p electrons. The carbon atom shares all of these with the two oxygen atoms except the two 1s electrons.
- My mistake. I asked about the wrong electron. The question should have been why does the second 2s electron in Carbon have to move to the 2pz level to form the sigma bond MO-4? On the chem1 site, both sigma bonds are sp hybrids.24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- (Sorry, only noticed this just now.) I don't think MO4 can exist in the ground state of CO2, which seems to be what chem1 is describing. Radiation heats CO2 by exciting it to higher states, and presumably MO4 describes the pairing of a 2pz oxygen electron with a carbon electron that's been excited from 2s to 2pz (just guessing). --Vaughan Pratt (talk) 08:10, 7 July 2010 (UTC)
- My mistake. I asked about the wrong electron. The question should have been why does the second 2s electron in Carbon have to move to the 2pz level to form the sigma bond MO-4? On the chem1 site, both sigma bonds are sp hybrids.24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- 2. "The chem1 site shows both sigma bonds being sp hybrids, but the MO-Diagram shows only one sigma bond being an sp hybrid - which is right?" Both are right as they're describing different things.
- I don't think so. If you go to diagram 1 on the Molecular Orbital page, the bond types (sigma or pi) are listed alongside the atomic orbitals that form the bonds. On diagram 1 there are listed two sigma bonds and two pi bonds, the same as the chem1 site. The chem1 site shows both sigma bonds as sp hybrids. Diagram 2 on the MO page shows one sigma bond (MO1) being an sp hybrid and the other (MO4) being a combination of the two p orbitals A2 (Oxygen 2pz) and A7 (Carbon 2pz).24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- (Again sorry for the delay.) By "different things" I mean that chem1 is referring to two instances of the same type of sigma bond whereas the MO page is referring to two types of sigma bond. The former type is sp (so two sp's), the latter two types are sp and pp (so one sp and one pp). --Vaughan Pratt (talk) 08:06, 7 July 2010 (UTC)
- I don't think so. If you go to diagram 1 on the Molecular Orbital page, the bond types (sigma or pi) are listed alongside the atomic orbitals that form the bonds. On diagram 1 there are listed two sigma bonds and two pi bonds, the same as the chem1 site. The chem1 site shows both sigma bonds as sp hybrids. Diagram 2 on the MO page shows one sigma bond (MO1) being an sp hybrid and the other (MO4) being a combination of the two p orbitals A2 (Oxygen 2pz) and A7 (Carbon 2pz).24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- 3. "Is it conceivable that the MO5 bond will be resonated (be stretched and compressed) by the radiation while the MO7 bond is left unaffected - hence asymmetric stretching?" That one's above my pay grade. At this small-molecule scale ground truth resides in the wave function. Whether the relevant solutions to the relevant Schroedinger equation(s) admit a classical or even semiclassical interpretation like the one you're asking for is best answered by an expert (which I'm certainly not), and then only in the context of the solution itself. For the time being settle for what I said in the beginning: asymmetric stretching is one of the vibration modes resulting from the interaction of the radiation with the differing electronegativities of carbon and oxygen. --Vaughan Pratt (talk) 23:03, 23 June 2010 (UTC)
- Can we at least agree to what is meant by asymmetric? One pi bond stretches and compresses by a certain amount while the other stretches and compresses by a lesser amount if at all.24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- The first thing to realize is that for each of the main vibration modes of the CO2 molecule, the energy alternates between kinetic and potential. The former reaches its maximum when the three nuclei each reach their maximum momentum relative to the center of mass of the molecule, while the latter reaches its maximum when those nuclei reach their maximum displacement from their respective maximum-momentum or "rest" positions, 90 degrees in phase or the order of 0.01 picoseconds later. Whereas the kinetic energy resides in the nuclei, which are nearly 4000 times more massive than the electrons, the potential energy resides in the stretched electronic bonds, which can be pictured as single rubber rods, even the double bonds.
- For both kinds of stretching, symmetric and asymmetric, all three nuclei move along the principal O=C=O axis. For asymmetric stretching, when the carbon nucleus moves distance d to the right, each of the two oxygen nuclei move 3d/8 to the left to conserve momentum since the carbon atom has 3/4 the mass of one oxygen atom. This stretches the double bond (sigma and pi together) on the left while compressing the double bond on the right, the meaning of "asymmetric" in this context.
- The difference between the sigma and pi bonds is itself an asymmetry, but it is not what makes the main stretching modes asymmetric. Since the sigma bonds (MO1 in the MO diagram) are lower they would be stronger, so presumably more of the potential energy is stored in the sigma component of the stretched double bond than in the pi component. However I would also expect the potential energy to be split fairly equally (exactly?) between the stretched double bond and the compressed one.
- I have no idea what additional vibration modes are created by the difference between the sigma and pi bonds. Modeling CO2 in more detail than its main vibration modes is a complex business, which is why I can't tell you what impact the difference between pi and sigma bonds has on CO2 resonances. In any event it would be theoretical and speculative, and experimental data on CO2 line strengths takes precedence over theoretical predictions when computing the impact of greenhouse gases on global warming. --Vaughan Pratt (talk) 22:08, 25 June 2010 (UTC)
- Can we at least agree to what is meant by asymmetric? One pi bond stretches and compresses by a certain amount while the other stretches and compresses by a lesser amount if at all.24.3.11.218 (talk) 18:56, 25 June 2010 (UTC)
- No I am referring to diagram #2 on the MO page which lists the atomic orbitals that are combined to form the molecular orbitals. The type of bond (pi or sigma) is shown on diagram #1 on the MO page. If you count the bonds on diagram #1 there are two pi bonds and two sigma bonds which is the same number as depicted on the chem1 site. 24.3.11.218 (talk) 21:14, 23 June 2010 (UTC)
Greenhouses must be cooler inside
Sorry to be a nag about this, but it just occurred to me that, if this article is correct that the greenhouse effect has nothing to do with how greenhouses work, then greenhouses must be colder inside than out. This is because a single-paned greenhouse reflects 8% of the incident sunlight while the more modern double-paned variety reflects 16%. In contrast the ground outside receives 100% of the incident sunlight and therefore will get hotter.
All else being equal, if you were to drop a greenhouse over a portion of a field full of vegetation, the greenhouse would partially shade that portion, which would then cool down relative to the surrounding temperature were it not for the greenhouse effect whereby the windows impede outgoing radiation.
The article argues that "the air continues to heat because it is confined within the greenhouse, unlike the environment outside the greenhouse where warm air near the surface rises and mixes with cooler air aloft." The editor who wrote this on 12 January 2008 would appear to be unfamiliar with the concept of potential temperature, which predicts that if you mix warmer air below with cooler air above, the air at the ground afterwards typically gets warmer rather than colder, due to the temperature increasing when the pressure is increased adiabatically, often by an amount greater than the lapse rate. A related phenomenon is the perceived temperature of wind, which seems cold not because it brings down cold air from high up but because it evaporates moisture; see the lead and picture at Chinook wind for dramatic examples of just how much hotter cold air aloft can get when it comes down to the ground. (My thanks to User:Q Science for bringing this concept to my attention a couple of months ago, though its relevance to greenhouses hit me only today.)
Contrary to what many (but by no means all) climate scientists believe, including textbook authors such as Abraham Oort, José Peixoto, and Daniel Schroeder cited in the present article, trapping of infrared radiation by the plastic or glass windows of a greenhouse is the only reliable mechanism by which a greenhouse can get warmer than outside (but once it does get warmer then of course it becomes important to avoid losing that heat by convection to what has now become the cooler outside). Furthermore the trapping effect must be quite substantial in order to overcome the 8-16% of insolation lost to reflection and heat the greenhouse (but we knew this as long ago as 1767 when Horace-Bénédict de Saussure built a triple-pane hotbox, which loses close to 24% of the insolation to reflection yet gets extremely hot).
The article's statement "greenhouse gases act to warm the Earth by re-radiating some of the energy back towards the surface" equally (but poorly) describes the mechanism by which any thermal insulator works, whether greenhouse gases, greenhouse windows, or R-30 insulation in building walls. A better description is in terms of a thermal gradient with individual photons and phonons creating an overall energy flux by flying in all directions with a bias that creates an overall trend from warmer to cooler, consistent with the thermal gradient in all three of these kinds of insulation and with the second law of thermodynamics. "Re-radiation back down" is a confused and confusing way of describing the process of thermal conduction and its associated gradients.
- I agree completely. If you want to get down to the nitty gritty infrared radiation emitted from the earth's surface travels at the speed of light back up through the atmosphere. Upon striking greenhouse gases this electromagnetic energy is converted to vibrational kinetic energy of the molecule. This kinetic energy progresses much more slowly through the atmosphere as the molecules collide into each other. If there were no greenhouse gases the infrared radiation would leave the earth at the speed of light and have little (it could hit a plane) if any effect on the earth's temperature. The reason the earth's atmosphere is warmed by the infrared radiation and stays warm has nothing to do with re-radiation. It is because of the energy conversion. 24.3.11.218 (talk) 21:13, 23 June 2010 (UTC)
- The effect on temperature of having no GHGs is pretty much what's described at Effective temperature, which assumes a reflecting planet at solar wavelengths (the albedo) and unit emissivity at terrestrial wavelengths (so in particular no GHGs to impede terrestrial radiation). I wouldn't neglect re-radiation, rather it should be taken into account by viewing all air molecules as being essentially in equilibrium with the ambient radiation (taking the overall gradients into account), with the GHGs absorbing and emitting more photons than the non-GHGs (the root cause of the induced thermal impedance), and with all gases exchanging energy at a steady rate via collisions (in equilibrium with the phonon flux as well as the photons if you will). --Vaughan Pratt (talk) 21:49, 23 June 2010 (UTC)
The one remaining shred of support for the theory that greenhouses don't work by trapping infrared is R.W. Wood's February 1909 article in Phil. Mag. As I pointed out earlier Wood's claim was thoroughly refuted in a longer and more carefully researched paper in the July issue by Charles Abbot, then director of the Smithsonian Astronomical Observatory and later Secretary of the Smithsonian from the Depression to the end of WWII. In April I wrote to Taylor and Francis asking if they'd mind my posting Abbot's article. They didn't reply, so on the assumption they have no objection to this Fair use of the article I've posted it at http://boole.stanford.edu/Wood/AbbotReplyToWood.pdf . (The bottom line of p.33 is the relevant citation for Abbot's article in Vol 18, cf. the bottom line of p.32 citing Wood's article in Vol 17. Each Phil. Mag. volume consists of 6 monthly issues but the issue numbers increase steadily with No. 1 of the 6th Series being Jan 1901, No. 121 being Jan 1911, etc. which makes No. 103 July 1909, the date of Abbot's reply to Wood). --Vaughan Pratt (talk) 19:37, 23 June 2010 (UTC)
- Please fix the link to AbbotReplyToWood.pdf and then delete this request. Q Science (talk) 04:09, 24 June 2010 (UTC)
- Oops, sorry. (It was missing ".stanford.edu" which I didn't notice because my machine was adding it automatically.) Incidentally deleting anything from talk pages is generally frowned on, and I would think not appropriate in this case. --Vaughan Pratt (talk) 17:35, 24 June 2010 (UTC)
- Thanks. I have now read Abbot's paper and I think that he actually agreed with Wood's description of how a greenhouse works. He clearly states that
- convection is the main agent in removing heat from the earth's surface
- Thankts for Abbott. On a quick skim it looks to me like A and W both agree that GH's are warm due to preventing convection. This is the point that Wood's paper is valuable on. H and A differ about how atmospheric physics works, but this is not of great interest. So I don't see which bit of A you think thoroughly refutes W's claims. Could you quote the section you mean? William M. Connolley (talk) 06:29, 26 June 2010 (UTC)
- No one is disputing that if you open up either a solar oven or a greenhouse it cools down---even de Saussure knew that in 1767. Abbot writes at the start of his third paragraph (after pointing out that more panes of glass can produce much higher temperatures) "Agreeing with Professor Wood that the main function of the cover of a 'hot-box' or 'hot-house' is to prevent loss of heat by convection..." (with which we all agree) and then warms up gradually to his main point at the bottom of p.32: "as shown below there is reason to think that 'trapping' is more important perhaps than Professor Wood thinks."
- What Abbot argues on pp. 33-34 is that Wood's picture of rock salt as being transparent to infrared is flawed. Abbot calculates that, far from blocking the 0% that Wood implies, rock salt blocks 65% of what glass blocks when convection from the outside of the cover is neglected, and "maybe much more" when one takes into account that the glass heats up considerably (observed in practice) therefore accelerating convective losses from the glass, which Abbot points out can be "a considerable factor" (I've noticed this myself in doing similar experiments). On that basis Abbot expects to see only a small difference between the glass and rock salt cover "although it would seem strange that [Wood] observed no difference at all." (What I find strange myself is that Wood reduced the temperature from 65 C to 55 C by introducing an additional sheet of glass without bothering to analyze the factors contributing to this considerable reduction, instead stopping because this had made the boxes the same temperature which was the effect he wanted. This alone should disqualify Wood's experiment in the mind of any serious experimenter, quite apart from the lack of enough relevant parameters to duplicate Wood's experiment.)
- Had Abbot agreed with Wood that 'trapping' plays no significant role in greenhouse warming, he would not have spent more than a page arguing that Wood's experiment cannot show the big difference Wood was expecting. Abbot's point is that Wood should have been looking for a small difference, which he clearly wasn't doing. --Vaughan Pratt (talk) 07:05, 26 June 2010 (UTC)
- then warms up gradually to his main point at the bottom of p.32: "as shown below there is reason to think that 'trapping' is more important perhaps than Professor Wood thinks." yes indeed, but you are omitting the start of that same sentence: "For the dependence of the earths temperature on the atmosphere". Thus it is clear that Abott is (correctly) objecting to Wood's asserting that the GHE is of little importance in the atmosphere. Over the mechanism *in a greenhouse* they agree William M. Connolley (talk) 19:22, 26 June 2010 (UTC)
- You're omitting the middle. Here's the sentence in full. "For the dependence of the temperature of the earth's surface on the atmosphere, some numerical data can be assigned also, and as shown below there is reason to think that 'trapping' is more important perhaps than Professor Wood thinks."
- Now if the analysis in the next five paragraphs had been about the atmosphere I'd agree with you that this is what the second half of that sentence is referring to. But those paragraphs say nothing about the atmosphere; instead Abbot does the analysis that Wood should have done before doing his experiment. What Abbot seems to be wanting to show is that Wood had theorized incorrectly that his experiment would show a big difference if IR opacity played an important role. When Wood had convinced himself that the difference was below his expectation of what it would have to be if IR opacity played a role, he considered his experiment complete. The point of Abbot's analysis is to show that Wood's experiment can be predicted not to show a big difference even when IR opacity is significant (this is foreshadowed in paragraph 3, "it is interesting to see if this [the small difference] could be predicted"). The inference I drew here was that Wood should have performed his experiment more carefully.
- I would agree that Abbot's argument isn't clearly stated. It's too bad he isn't around to disambiguate it for us. --Vaughan Pratt (talk) 01:01, 27 June 2010 (UTC)
- Incidentally one can get more insight into what Abbot was thinking from the five paragraphs (¶4-6 on p.33, ¶7-8 on p.34) calculating what Wood should have expected to see.
- ¶4 Cites Compan's experiments on ratio of convective to radiative heat loss in still and moving air. (Interesting that Compan's 4/3 ratio for convection/radiation in still air is so close to Kiehl and Trenberth's ratio of 102/66 for the same thing in the atmosphere.)
- ¶5 Points out that a coverless box at Wood's location would be exchanging radiation principally with water-vapor layers, whose temperature Abbot estimated at 0° C.
- ¶6 With energy units chosen to make the radiation in ¶5 100 units, and using Compan's 4/3 ratio from ¶4, Abbot computes convection+radiation for coverless, glass, and salt boxes as respectively 97+100, 0+0, and 0+70. (Abbot figured "about" 70% transmission for the rock salt window by using a figure of 19% absorption, citing Kayser's Handbuch d. Spectroscopie, Vol iv, p.485 with the assumption of 1 cm thickness, and adding another 10% for reflection. I digitized the NaCl transmission curve at http://www.internationalcrystal.net/optics_16.htm on the assumption of 1 cm thickness, customary for such curves, and convolved it with Planck's law at 55 °C to get 68.5% transmission, in fair agreement with Wood's estimate.)
- ¶7 Infers that the glass and salt boxes impede 197 and 127 units respectively. Points out that, unlike the salt cover, the glass cover will get hot (I can confirm this myself, having done a number of experiments aimed at understanding Wood's experiment better) whence more heat will be both convected and radiated away from the outer surface of the glass cover than the salt cover, whence the ratio of 127/197 is likely to considerably underestimate the actual ratio.
- ¶8 On the basis of these calculations Abbot agrees with Wood that the salt cover would warm the box almost as efficiently as the glass one. However he finds it strange that Wood observed no difference at all, and speculates that perhaps the glass cover that Wood placed over the salt cover might have been a factor. At this point he turns to consideration of the atmosphere.
- Like Abbot I found it strange that Wood saw no difference, which is why I designed some experiments aimed at getting more insight into what was going on, starting with http://boole.stanford.edu/WoodExpt/ . I now find it very strange. --Vaughan Pratt (talk) 00:38, 3 July 2010 (UTC)
- I looked at your experiment page (pretty neat) and I realized something - Saran Wrap is not a very good substitute for rock salt because of its thermal conductivity. The thermal conductivity for .015mm Saran Wrap is .00247 J / cm^2 * Deg K, for 3/32" thick rock salt it is .459 J / cm^2 * Deg K, for 3/32" glass it is .479 J / cm^2 * Deg K, and for 6" of air space it is .0199 J / cm^2 * Deg K. Thermal conductivity of a material in a single direction is simply density of material * specific heat of material * thickness of material. You are forgetting about the third method of heat transfer, that being conduction. With the saran wrap most of the temperature drop from 355 degree K black body temperature to 300 degree K outdoor temperature occurs across the saran wrap because of its high resistance to heat flow (low thermal conductivity). With rock salt and glass the opposite is true - most of the temperature drop occurs in the air space because glass and rock salt have a lower resistance to heat flow (high thermal conductivity). This is not strange at all. If it helps, think of heat flow, temperature drop, and thermal resistance like current flow, voltage drop, and electrical resistance in an electrical circuit. And so the statement from your web page - "Very preliminary tests conducted on November 28, 2009 indicate that boxes 2 and 3 run respectively 15 and 20 degrees hotter than box 1, with a significantly larger drop across box 3's window than box 2's. While this is consistent with the expected differences obtained by theoretical considerations, it is so far removed from what Wood found as to raise the very interesting question of how he was unable to observe any significant difference between his two boxes." seems odd. If Wood was using rock salt and glass of equal thickness, the difference in thermal conductivity between the two is insignificant. FYI, The formula for the intermediate temperature inside the box at the air / glass boundary is T = (T-Inside + T-Outside * k-Glass / k-Air) / (k-Glass / k-Air + 1). The variables k-Glass and k-Air represent the thermal conductivities for the materials and the variables T are temperatures. ((24.3.11.218 (talk) 06:08, 5 July 2010 (UTC)
- Ah, excellent, someone who talks my language. Yes, you're absolutely right, other than for your claim that I was forgetting about thermal conductivity which I was indeed following closely. If you go back to my posts in this talk page of last year (now archived) you'll see that I took that into account; the only reason I stopped doing so is that it started to dawn on me that no one was following at that level of detail so why bother.
- Nevertheless I did address it in the second of my three experiments, the one with double glazing, which attempts to simulate the combination of FIR-transparency of rock salt and its thermal conductivity with an air-gap between two layers of saran wrap aimed at introducing a thermal resistance equivalent to that of a slab of rock salt of some thickness. Since a second sheet of saran wrap adds another 8% reflectivity in each direction, for parity the FIR-opaque glass/perspex window had to be doubled too.
- But one can't simply extrapolate a 1/4 " air gap to 6" because at around 1" convection sets in and tends to short-circuit what should have been a high thermal resistivity. This is why all the air gaps between two thicknesses of anything in my experiments are on the order of 1/4" to at most 1/2", so that the published data for thermal conductivity of air are meaningful. In my most serious attempt at getting this right I tediously fabricated a celluloid lattice aimed at inhibiting even lateral convection without significantly blocking normally incident insolation---mere transparency is not enough when every extra interface reflects another 4%. --Vaughan Pratt (talk) 07:22, 7 July 2010 (UTC)
- I looked at your experiment page (pretty neat) and I realized something - Saran Wrap is not a very good substitute for rock salt because of its thermal conductivity. The thermal conductivity for .015mm Saran Wrap is .00247 J / cm^2 * Deg K, for 3/32" thick rock salt it is .459 J / cm^2 * Deg K, for 3/32" glass it is .479 J / cm^2 * Deg K, and for 6" of air space it is .0199 J / cm^2 * Deg K. Thermal conductivity of a material in a single direction is simply density of material * specific heat of material * thickness of material. You are forgetting about the third method of heat transfer, that being conduction. With the saran wrap most of the temperature drop from 355 degree K black body temperature to 300 degree K outdoor temperature occurs across the saran wrap because of its high resistance to heat flow (low thermal conductivity). With rock salt and glass the opposite is true - most of the temperature drop occurs in the air space because glass and rock salt have a lower resistance to heat flow (high thermal conductivity). This is not strange at all. If it helps, think of heat flow, temperature drop, and thermal resistance like current flow, voltage drop, and electrical resistance in an electrical circuit. And so the statement from your web page - "Very preliminary tests conducted on November 28, 2009 indicate that boxes 2 and 3 run respectively 15 and 20 degrees hotter than box 1, with a significantly larger drop across box 3's window than box 2's. While this is consistent with the expected differences obtained by theoretical considerations, it is so far removed from what Wood found as to raise the very interesting question of how he was unable to observe any significant difference between his two boxes." seems odd. If Wood was using rock salt and glass of equal thickness, the difference in thermal conductivity between the two is insignificant. FYI, The formula for the intermediate temperature inside the box at the air / glass boundary is T = (T-Inside + T-Outside * k-Glass / k-Air) / (k-Glass / k-Air + 1). The variables k-Glass and k-Air represent the thermal conductivities for the materials and the variables T are temperatures. ((24.3.11.218 (talk) 06:08, 5 July 2010 (UTC)
- then warms up gradually to his main point at the bottom of p.32: "as shown below there is reason to think that 'trapping' is more important perhaps than Professor Wood thinks." yes indeed, but you are omitting the start of that same sentence: "For the dependence of the earths temperature on the atmosphere". Thus it is clear that Abott is (correctly) objecting to Wood's asserting that the GHE is of little importance in the atmosphere. Over the mechanism *in a greenhouse* they agree William M. Connolley (talk) 19:22, 26 June 2010 (UTC)
- Thankts for Abbott. On a quick skim it looks to me like A and W both agree that GH's are warm due to preventing convection. This is the point that Wood's paper is valuable on. H and A differ about how atmospheric physics works, but this is not of great interest. So I don't see which bit of A you think thoroughly refutes W's claims. Could you quote the section you mean? William M. Connolley (talk) 06:29, 26 June 2010 (UTC)
Evacuated-tube collectors and low emissivity techniques
I like your page, well done. You seem to think that there is no way to stop convection. Might I suggest trying evacuated-tube collectors which can get to over 100C. (Some references say 125C to 175C.) Since the tubes remain cool to the touch, I assume that that means that the borosilicate glass tubes are IR transparent.[2] Q Science (talk) 03:14, 3 July 2010 (UTC)
- Thanks. Regarding evacuated-tube collectors, these have an internal coating that inhibits radiative heat loss (low emissivity at those wavelengths). The outer pyrex tube remains cold (so cold it can't melt snow that falls on it in winter) because there is relatively little outbound radiation to heat it. Even 1/16" Pyrex (borosilicate glass) is almost completely (> 95%) opaque between 5 and 15 microns, so it would heat up a lot were it not for the low-E coating.
- (Now that I think of it, I suppose the low-E coating could be as simple as just an inner glass tube with a water-cooled outer surface. With that approach the evacuation might not even be needed, in which case the tubes could be replaced by much cheaper flat glass plates, with the whole thing being less delicate. Hmm, very interesting, have to think about that some more.)
- If you don't mind I'll rename your "break" heading to something more reflective (no pun intended) of the contents of this subsection, depending on where it goes. --Vaughan Pratt (talk) 00:04, 4 July 2010 (UTC)
- You can always rename my breaks, or even move them to a more logical place. I just don't like having a lot of text in the edit window.
- Hmm, I wasn't aware of the low-E coating. However, that would explain the temperatures over 150C (the maximum I feel safe computing). Q Science (talk) 01:42, 6 July 2010 (UTC)
- I should clarify that I didn't mean "low-E" in its proprietary sense (as used in the window trade) but only in the sense that the coating had been designed to reduce emissivity. There are plenty of ways of reducing emissivity without either infringing the extant low-E patents or driving up costs. --Vaughan Pratt (talk) 06:57, 7 July 2010 (UTC)
- Hmm, I wasn't aware of the low-E coating. However, that would explain the temperatures over 150C (the maximum I feel safe computing). Q Science (talk) 01:42, 6 July 2010 (UTC)
Suggested rewording of effective-temperature paragraph in lead
I had trouble with the wording below, not the point or content. The article seems generally clear, correct, and well written to me. I did, however, have trouble with the wording in the 4th paragraph of the intro, and since the page is (probably wisely) locked for editing, suggest something like the following:
If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth, it would have an expected blackbody temperature of 5.3 °C. However, since the Earth reflects about 30%[4] (or 28%[5]) of the incoming sunlight, it is not an ideal blackbody, and it's surface temperature would be approximately [delete the planet's actual blackbody temperature is about)] -18 or -19 °C [6][7] were it not for the greenhouse effect, or about 33°C below the actual surface temperature of about 14 °C or 15 °C.[8] [delete The mechanism that produces this difference between the actual temperature and the blackbody temperature is due to the atmosphere and is kown as the greenhouse effect.] —Preceding unsigned comment added by 97.89.22.141 (talk) 23:45, 4 July 2010 (UTC)
- Wouldn't it be simpler just to refer to the Effective_temperature#Planet article, which takes albedo for granted but not greenhouse gases? While albedo is relevant to Earth's temperature it is irrelevant to the effect of greenhouse gases. Dragging the reader through the math of albedo is therefore an unnecessary distraction, especially when that math has been absorbed into the concept of effective temperature of a planet. --Vaughan Pratt (talk) 03:56, 5 July 2010 (UTC)
- This confuses incident radiation and blackbody radiation. Kirchhoff's Law states: emissivity = absorptivity. Ergo, a greybody in incident radiation will acquire the same temperature as a blackbody in the same incident radiation. The expected mean temperature for the absorbing mass component of planet earth, is on the order of 5.5 °C because the reduced emissivity of a greybody makes up for precisely the same amount of heat lost to reflection (See - http://greenhouse.geologist-1011.net for the math). —Preceding unsigned comment added by 58.111.253.225 (talk) 04:37, 5 July 2010 (UTC)
- That's false. The heat lost by reflection is lost by reflecting the 5700 °C heat of the Sun, whereas the relevant emissivity is that for 1/20 of that temperature, which could be quite different. If you disagree then your complaint is not with me but with the article on effective temperature, which I'm simply proposing to refer to.
- But while there's at least a grain of truth in what you say yourself (as it happens the emissivity of ice for example at 10-15 microns is not hugely different from that at half a micron), the greenhouse article you cite is, to put it charitably, highly creative as it claims on the basis of a complex chain of arguments that (if I've understood it correctly) there is no greenhouse effect. The same reasoning employed there, in particular his "The problem with this defense is that no amount of heat congestion can result in an average power output exceeding the average power input" shows that resistors cannot warm their neighboring components. If Mr. Casey can boil his reasoning down to one paragraph then it becomes worth analyzing, if not then it's a waste of time wading through a huge number of pages in which there's room to hide a dozen or more fallacies. --Vaughan Pratt (talk) 05:57, 5 July 2010 (UTC)
The greenhouse effect is appropriately named
The third sentence of the lede reads "This mechanism is fundamentally different from that of an actual greenhouse, which works by isolating warm air inside the structure so that heat is not lost by convection." The sole scientific basis for this claim would appear to be Robert W. Wood's paper Note on the Theory of the Greenhouse in the February 1909 issue of the Philosophical Magazine. The current acceptance of Wood's claim by climatologists appears to be quite recent, dating from a 1990 paper by Jones and Henderson-Sellers and enthusiastically endorsed in a more recent paper by Gerlich and Tscheuschner.
Having argued unsuccessfully against this viewpoint in the past, I'm now resorting to John Cook's very effective format, now installed as an app on innumerable iPhones, for refuting material for which there is no scientific support.
Wood's position vs The literature and an experiment you can easily do yourself.
1. "Wood proved that greenhouses are not warmed by trapping infrared."
Wood's very brief note offers only two measurements: 65°C for both boxes without extra glass, 55°C with, along with the disclaimer "I do not pretend to have gone very deeply into the matter." In view of the many possible parameters influencing the observed temperature in Wood's brief experiment, these observations are eminently repeatable and thus you can test their reliability yourself. Moreover, the fact that they address directly the impact of greeenhouse materials on greenhouse temperature cannot be denied. Wood's results conform to the findings of de Saussure and Fourier that heat leaving a greenhouse does so by conduction. Only Tyndall and Arrhenius managed to misunderstand Fourier because they were trying to reshape the laws of heat transfer to fit with aethereal wave propagation - something Fourier tried to distance his work from. It's really quite obvious if you read Tyndall's very confused account of gaseous opacity (which he confused with absorption) and then concluded with the assertion that all heat radiating within a material was lost to aethereal wave propagation (See - http://tyndall1861.geologist-1011.mobi - p. 285), saying: "When a molecule of alum, on the contrary, approaches a neighbour molecule, it produces a swell on the intervening ether of space, and thus lost as regards conduction. This lateral waste prevents the motion from penetrating the alum to any great extent, and the substance is what we all a bad conductor"
2. "Wood's note went unnoticed until Jones and Henderson-Sellers brought it to the attention of the climate scientists in their 1990 article 'History of the Greenhouse Effect'."
Wood's note was certainly noticed in 1909. It found full support and was confirmed in the July 1909 issue of Phil. Mag. by Charles Abbot (then director of the Smithsonian Observatory, later secretary of the Smithsonian Institute from 1928 to 1944) in a paper that went into far more technical detail than Wood's. For more details see http://boole.stanford.edu/Wood/. For Abbot's reply see Abbot's paper.
Abbot (1909) actually agreed with Wood: "Agreeing with Professor Wood that the main function of the cover of a hot-box or hot-house is to prevent loss of heat by convection, it is interesting to see if this can be predicted." We can see from the text of Abbot (1909) that this reply is not a refutation, but a confirmation extended by calculations which further confirm Wood's results, acknowledging that "salt hinders...65% as much as the glass", and going on to state: "In view of these figures we may agree with Professor Wood that a salt cover is nearly as efficient as a glass one for a hot-box, although it would seem strange that he observed no difference at all." The expected difference due to absorption by glass over and above salt is on the order of fifteen degrees Celsius (See - http://greenhouse.geologist-1011.net - for calculations that go into far more technical detail than Abbot). This led Abbot to conclude by replacing Arrhenius' "Greenhouse Effect" with the "blanket-effect", whereby only the variation of temperature is affected by atmospheric absorption.
3. "Greenhouses are warmed by preventing exchange of air with the cooler outside."
This does not undermine the analogy with greenhouses at all because Earth's atmosphere is warmed in exactly the same way, there being no "cooler outside" with which to exchange air.
4. "The contribution of infrared trapping by glass is neglible."
The contribution can be calculated and shown to be non-negligible, as done by Abbot in his confirmation of Wood. It can also be assessed empirically with an experiment you can easily perform yourself without any expensive equipment. Simply interpose a sheet of glass between the Sun and a sheet of paper on a sunny day, with a one-inch gap to prevent any conduction from the paper to the glass, and wait a few minutes for the glass to warm up. With white paper there is relatively little appreciable warming, whereas with black paper the side of the glass facing the paper warms appreciably. This effect can be enhanced by
- (i) doing the experiment in parallel with two sheets of glass in front of respectively the white and black paper and switching the two sheets of warmed glass back and forth between your hands to better feel the difference;
- (ii) allowing more time (on the order of 10-20 minutes) for the glass to heat up;
- (iii) separately wrapping each of the glass and the paper in saran wrap (as loosely as possible without too much leakage, and no more than once around to minimize reflection losses) to reduce cooling, incidentally further reducing the already minimal opportunity for significant convection between the paper and the glass;
- (iv) using thicker glass (and waiting proportionately longer for the increased thermal mass to respond); and
- (v) using perspex in place of glass (4x better insulator resulting in a greater thermal drop across the window).
As an aside it may be worth mentioning that a mixing ratio of 1‰ (1000 ppm) by volume of atmospheric CO2, if deposited on the Earth's surface as dry ice, will be almost exactly 1 cm thick. Hence the current 0.39‰ level of CO2 would freeze out as a 0.39 cm thick layer of dry ice, roughly the thickness of a typical sheet of glass. The absorption spectrum of dry ice (not line-by-line of course) bears a sufficient resemblance to that of glass, both consisting of triatomic molecules, as to further justify comparing the glass of a greenhouse with the current level of atmospheric CO2. The main differences are the far greater thermal mass of Earth's atmosphere (mainly nitrogen and oxygen) relative to the components of a greenhouse, which greatly impacts the rise time of the response without however any comparable impact on its equilibrium value, and the impact of atmospheric water vapor on absorption, which is considerably greater than that of current atmospheric CO2, though it is not changing as dramatically on a decadal time scale: the CO2 mixing ratio in units of ppmv as measured at Mauna Loa is remarkably well modeled by the formula 260 + exp(t) ppmv where t is time in units of 60.0 years with t=0 defined as 1718 AD. --Vaughan Pratt (talk) 09:20, 29 April 2010 (UTC)
- You have missed the point of de Saussure's and Wood's experiments, they prove that the atmosphere cools the surface of the Earth when the Sun is shining. In particular, Wood showed that convection, not radiation, is the mechanism of that cooling. Q Science (talk) 18:30, 29 April 2010 (UTC)
- Let me state my understanding of the respective points of their experiments and you can tell me where I've misunderstood them.
- In the case of (Horace-Bénédict, not Ferdinand) de Saussure, my understanding is that he was a keen mountaineer who was also very interested in the botany, geology, and physics of the Alps. He found it particularly puzzling that the peaks were colder than the valleys despite the former having less air between them and the Sun. Without knowing about either wavelengths or greenhouse gases (this was a third of a century before Young's wave theory of light supplanted Newton's corpuscular theory) he somehow hit on the idea that both the atmosphere and glass allowed solar energy to penetrate but blocked terrestrial heat from escaping, a principle he could not explain but which he was nevertheless able to demonstrate with great success with a series of ever more efficient hotboxes, the precursor of one design of modern solar ovens, which developed essentially the same temperature inside the box whether at an icy mountain peak or thousands of feet below in a warm valley, a key detail of his experiment.
- In the case of Wood, my understanding is that his main concern in his note lay not with the Earth's atmosphere but with the mechanism by which greenhouses attain their temperature. He maintained that capture of outgoing far infrared radiation by the glass played a negligible role and that the considerable warmth was attributable essentially solely to retention of the warmed air. The only connection with global warming was via a throwaway remark at the end of his note, which I had not understood as his primary goal.
- If you see their respective experiments as having the common goal of proving that the atmosphere cools the surface of the Earth by day (obviously true when compared with the far higher temperature of the Moon's airless surface by day) then unless I'm overlooking something it seems to me that you and I have very different understandings of the objectives of these two experimenters. In any event I take it that you reject Abbot's carefully conducted and documented experiments in favor of Wood's one carelessly conducted and minimally documented experiment. --Vaughan Pratt (talk) 22:25, 29 April 2010 (UTC)
- Unfortunately, the Abbot article is behind a paywall. As a result, I have no idea if I would reject Abbot's experiments or not. I would not want to speculate about the goals of either experiment, but we both appear to agree that their experiments prove (demonstrate) that convection (by the atmosphere) cools the daytime surface. At night, radiation cools the surface. As a result, the atmosphere is warmer than the surface and radiation from the atmosphere helps to keep the nights warm. However, the temperature will continue to drop all night unless dew, fog, or frost form to stabilize the temperature.
- Hopefully fixed, see Abbot's paper. --Vaughan Pratt (talk) 17:55, 24 June 2010 (UTC)
- Unfortunately, the Abbot article is behind a paywall. As a result, I have no idea if I would reject Abbot's experiments or not. I would not want to speculate about the goals of either experiment, but we both appear to agree that their experiments prove (demonstrate) that convection (by the atmosphere) cools the daytime surface. At night, radiation cools the surface. As a result, the atmosphere is warmer than the surface and radiation from the atmosphere helps to keep the nights warm. However, the temperature will continue to drop all night unless dew, fog, or frost form to stabilize the temperature.
- As for "the peaks being colder than the valleys", are they? Really? Yes, the measured temperatures are colder, but the potential temperatures are much higher. That means that if a volume of air from the top of a mountain was compressed to the same pressure found in the valley, its temperature would be higher. Most references miss this, the heat per unit volume decreases with height, but the heat per unit mass increases. Q Science (talk) 06:37, 30 April 2010 (UTC)
- In view of the public's interest in the greenhouse effect I agree that it's unfortunate that access to Abbot's refutation of Wood is limited to those affiliated with a subscribing institution. I've written to Taylor and Francis to make a case for releasing Abbot's paper to the public. In the meantime you could visit the nearest institution with a subscription and read the article in their library, either online or from their stacks if they have the two 1909 volumes. As a stopgap that should be well within the bounds of fair use, here's a paragraph from the paper in which Abbot points out, based on experiments carried out at the Observatory 12 years earlier, that Wood is attempting to measure temperatures that can be hugely dependent on the exact circumstances (a point I've noticed myself in my own experiments along these lines), implying that the measurement of temperatures that might differ by only 2-3°C (the climate sensitivity for CO2 doubling---CO2 can be assumed to double from its pre-industrial level of 280 ppmv to 560 ppmv around 2060 AD assuming business as usual) requires far greater care than Wood appears to have taken.
- It may interest some to know that much higher temperatures can be reached within a "hot-box" than that observed by Professor Wood, if precautions are taken to diminish the loss of heat by convection from the warmed outer surface of the cover. On November 4, 1897, the thermometer recorded 118°C within a circular wooden box 50 centimetres in diameter, 10 centimetres deep, insulated in feathers, covered with three superposed and separated sheets of plate glass and exposed normally to the sun rays in the yard of the Astrophysical Observatory at Washington. The temperature outside was 16°C.
- Abbot goes on to agree with Wood (as I do) that it is essential to retain the warmed air in the hotbox (equally essential for the Earth---imagine if it lost its air!), but questions Wood's claim that there is no significant contribution attributable to trapping far infrared.
- Agreeing with Professor Wood that the main function of the cover of a "hot-box " or "hot-house" is to prevent loss of heat by convection, it is interesting to see if this [the 118°C] could be predicted. Published experiments on the cooling of solids in dry air and in vacuum give the relative rates of loss by convection and radiation under known circumstances. Planck's radiation formula for the "black body" enables computations to be made of the losses by radiation for different temperatures of source and sink. The transmission of glass, salt, and the water vapour of the atmosphere, and the effective temperature of the latter are approximately known. I have attempted to compute from such data the relative hindrance which salt and glass covers would interpose to the loss of heat by convection and radiation combined from a "black" surface at 55°C. For the dependence of the temperature of the earth's surface on the atmosphere, some numerical data can be assigned also, and as shown below there is reason to think that "trapping" is more important perhaps than Professor Wood thinks.
- Note that Abbot is willing to do the math here, unlike Wood whose forte was experimental physics, at which he was an acknowledged expert.
- Regarding your point that Earth's surface is cooler at noon than it would be in the absence of an atmosphere, I agree fully there. However your explanation of how it happens suggests that the Earth radiates away its heat at night faster than during the day when in fact the opposite is the case. Here's how I would explain the situation, you can tell me which parts of it you disagree with.
- My understanding is that the atmosphere serves primarily as a thermal reservoir that dampens out what would otherwise be the wild gyrations of temperature that the moon experiences during one moon day (29.5 earth days). That the damping effect is considerable can be seen from the fact that a gram of air absorbing a watt of heating power warms at a rate close to 1°C/sec (about a quarter the specific heat of water, which would require 4.2 seconds, a calorie being 4.2 watt-seconds or joules). The atmosphere weighs 5148 exagrams (1018 grams), and Earth as a whole steadily intercepts .175 exawatts from the Sun, so if all that power went into uniformly heating the whole atmosphere with no loss to radiation it would warm at an hourly rate of .175/5148 * 3600 = 0.122°C/hr. However Earth steadily radiates .175 exawatts uniformly back out to space in all directions so overall the warmings and coolings cancel (if they didn't Earth would adjust its temperature until they did). But because the Sun is heating only one side of the Earth at any instant there is a daily temperature fluctuation. This is felt very strongly down at the surface where the relevant conversion of solar to terrestrial radiation takes place, but the 0.122°C/hr sensitivity of the total atmosphere to heating implies that, averaged over its total height, any given vertical column of air cannot vary by much more than 2-5°C between day and night. The surface varies by considerably more than that from day to night, but without that huge damper above it would vary far more.
- I don't see any connection between atmospheric thermal damping, which is independent of wavelength, and either of the mechanisms addressed by de Saussure and Wood, both of which are heavily wavelength-dependent (even though de Saussure didn't know this). I also don't understand your claim that "the atmosphere is warmer than the surface," which is not what my account above would predict but seems somehow to be a corollary of your understanding. Can you cite radiosonde or other data supporting your claim? If not then your understanding would seem to be producing strange predictions.
- In view of the public's interest in the greenhouse effect I agree that it's unfortunate that access to Abbot's refutation of Wood is limited to those affiliated with a subscribing institution. I've written to Taylor and Francis to make a case for releasing Abbot's paper to the public. In the meantime you could visit the nearest institution with a subscription and read the article in their library, either online or from their stacks if they have the two 1909 volumes. As a stopgap that should be well within the bounds of fair use, here's a paragraph from the paper in which Abbot points out, based on experiments carried out at the Observatory 12 years earlier, that Wood is attempting to measure temperatures that can be hugely dependent on the exact circumstances (a point I've noticed myself in my own experiments along these lines), implying that the measurement of temperatures that might differ by only 2-3°C (the climate sensitivity for CO2 doubling---CO2 can be assumed to double from its pre-industrial level of 280 ppmv to 560 ppmv around 2060 AD assuming business as usual) requires far greater care than Wood appears to have taken.
- As for "the peaks being colder than the valleys", are they? Really? Yes, the measured temperatures are colder, but the potential temperatures are much higher. That means that if a volume of air from the top of a mountain was compressed to the same pressure found in the valley, its temperature would be higher. Most references miss this, the heat per unit volume decreases with height, but the heat per unit mass increases. Q Science (talk) 06:37, 30 April 2010 (UTC)
- Almost ALL the radiosonde data supports that (really). The only exceptions I know of are the polar summers, the oceans, and small islands. Basically, the land cools at night, but most of the atmosphere does not. Only the atmosphere near the surface cools at night. NOAA provides the data, but their plots are almost useless. You will need to download the data and plot it yourself. This page plots some data for Tucson, 2000. Q Science (talk) 06:33, 1 May 2010 (UTC)
- Thanks for those pointers. You're absolutely right about the radiosonde data, the night time lapse rate would indeed appear to be negated for the first km or so, where the temperature goes up a few degrees, sorry to have doubted you there. What I was claiming only kicks in after the first km or so where it plunges far more than it went up.
- Doubt (skepticism) is healthy. You should doubt what everyone says until your personal research convinces you otherwise. What most people miss is that the "Greenhouse effect" is what occurs in that bottom kilometer or so. It also provides visual confirmation of how far infrared radiation travels before being totally absorbed. Q Science (talk) 07:25, 2 May 2010 (UTC)
- Whoa, now you've really lost me. My impression was that only about 10% of the greenhouse gases are in the bottom kilometer. Are you saying the other 90% has no effect? I would have thought 90% of the greenhouse effect was well above the bottom kilometer. I also don't understand the concept of "totally absorbed." I thought that the heat absorbed by GHGs was reradiated (isotropically); what does "total absorption" mean when reradiation is occurring? --Vaughan Pratt (talk) 07:26, 3 May 2010 (UTC)
- Doubt (skepticism) is healthy. You should doubt what everyone says until your personal research convinces you otherwise. What most people miss is that the "Greenhouse effect" is what occurs in that bottom kilometer or so. It also provides visual confirmation of how far infrared radiation travels before being totally absorbed. Q Science (talk) 07:25, 2 May 2010 (UTC)
- I can also see how the surface could lose heat by radiation to the upper atmosphere or space itself without violating the second law of thermodynamics, since the bottom km or so of atmosphere isn't going to interact strongly with the radiation yet has a high thermal capacity and so can stay warm while the surface cools. My mistake when I predicted the opposite was to overestimate the interaction of the surface's escaping night time radiation with the bottom kilometer or two of the atmosphere relative to the latter's thermal mass. (You may have another way of explaining the same thing, or you may even find my explanation wrong.)
- Assuming we agree up to this point, what I'm missing in this picture is any exchange between the surface and what's below. During the day the subterranean heat flow will be downwards, but one would expect this to reverse at night, i.e. the ground below should contribute conductively to the cooling of the surface along with the radiative cooling to the sky. Do you have a calculation (even approximate) giving the ratio of heat lost by the surface downwards to the ground below to that lost upwards to the sky? What you've said so far seems to assume that it's zero. --Vaughan Pratt (talk) 05:39, 2 May 2010 (UTC)
- Zero is the normal assumption. Diurnal penetration is measured in millimeters. Seasonal changes are what determine how deep foundations have to be (how deep the freeze line is). The oceans are handled differently because of the deep ocean currents. Q Science (talk) 07:25, 2 May 2010 (UTC)
- The diurnal damping depth is about 10 cm, actually. Short Brigade Harvester Boris (talk) 11:03, 2 May 2010 (UTC)
- Thanks, Boris. So assuming a specific gravity of (say moist) soil of 3, that would seem to suggest that around 300 kg/m^2 of soil is being cooled overnight. Assuming 10°C cooling over a period of 40000 seconds (i.e. overnight), and 1 kj/kg specific heat, that should come to around 300*10/40000 kW/m^2 or 75 watts/m^2 of downwards flux. (If the cooling is only 4°C then only 30 W/m^2, etc.) Does that sound plausible? If it's anywhere near right then Q's assumption of zero seems a bit low. --Vaughan Pratt (talk) 07:26, 3 May 2010 (UTC)
- 75 W/m2 is a little high, but within the range of reasonable values. When using round number terms I typically quote a value of 50 W/m2. Short Brigade Harvester Boris (talk) 08:00, 3 May 2010 (UTC)
- Well, 50 W/m^2 is exactly what my calculation gives for dry soil at a specific gravity of 2. It's also what it gives if instead I reduce the 10°C cooling to 6.7°C. And 10 cm is a pretty round number too. Anyway it's plenty close enough to ask the next question: what is the net upward flux from the surface at night? One can't just use the 340 W/m^2 given by Stefan's law (which presumably is what Q's reference later on to 120°C in 10 minutes is doing) because much of that is offset by the net downward radiation from the atmosphere (one source of the atmosphere's damping effect on diurnal surface fluctuation). Moreover the air only cools the surface convectively as long as the nearby lapse rate remains positive (meaning higher air is cooler)---once it goes negative the air must warm the surface convectively, another source of temperature damping. Given all that, how could the ratio of downward to upward cooling of the surface be anything like zero as Q claims? --Vaughan Pratt (talk) 18:10, 3 May 2010 (UTC)
- 75 W/m2 is a little high, but within the range of reasonable values. When using round number terms I typically quote a value of 50 W/m2. Short Brigade Harvester Boris (talk) 08:00, 3 May 2010 (UTC)
- Thanks, Boris. So assuming a specific gravity of (say moist) soil of 3, that would seem to suggest that around 300 kg/m^2 of soil is being cooled overnight. Assuming 10°C cooling over a period of 40000 seconds (i.e. overnight), and 1 kj/kg specific heat, that should come to around 300*10/40000 kW/m^2 or 75 watts/m^2 of downwards flux. (If the cooling is only 4°C then only 30 W/m^2, etc.) Does that sound plausible? If it's anywhere near right then Q's assumption of zero seems a bit low. --Vaughan Pratt (talk) 07:26, 3 May 2010 (UTC)
- The diurnal damping depth is about 10 cm, actually. Short Brigade Harvester Boris (talk) 11:03, 2 May 2010 (UTC)
- Zero is the normal assumption. Diurnal penetration is measured in millimeters. Seasonal changes are what determine how deep foundations have to be (how deep the freeze line is). The oceans are handled differently because of the deep ocean currents. Q Science (talk) 07:25, 2 May 2010 (UTC)
- Your intriguing concept of "potential temperature" is news to me and I would have to see more details before I could say anything sensible about it. What laws govern it, can the potential temperature of snow on a mountain peak be shown equal to that of water in a valley stream, is the potential temperature of the bottom of the ocean the same as at the surface, etc.? When you say "Most references miss this," what's one that doesn't? --Vaughan Pratt (talk) 23:51, 30 April 2010 (UTC)
By the way I should clarify that by "The greenhouse effect is appropriately named" I did not mean that retention of warm air is never useful. I'm not taking sides on the question, the problem is with the article, which does take sides by insisting dogmatically that "This mechanism is fundamentally different from that of an actual greenhouse, which works by isolating warm air inside the structure so that heat is not lost by convection," with sources cherry-picked for that side---there are plenty of reputable sources both for the other side and for "it is both." The statement is obviously true when a cool front is coming in and chilling one's plants. And it is just as obviously false when there is little air movement, as on a cold but windless morning when there are no updrafts (thermals are largely an afternoon phenomenon). In the latter case the 16% of reflection of incoming sunlight by a double-glazed glass or plastic roof and walls would make a greenhouse a thermal liability were it not for the way the opacity of glass to IR retains the 84% of heat that does get inside. Anyone who's been inside the Louvre pyramid entrance on a sunny day will know what IR trapping feels like: the huge area of glass way above feels like a radiator, you feel the radiation decreasing as the escalator down takes you away from it. This is a known issue for the Louvre, see page 19 of [3], "In summertime and by day: hot roof •greenhouse effect and air layering under the roof •near the floor: air mixing. Hot air penetrates by the corridor from the Pyramid." --Vaughan Pratt (talk) 18:57, 15 July 2010 (UTC)
break
Potential temperature applies to compressible fluids (like the atmosphere). Many references on weather prediction stress the importance of this. It also helps explain Chinook winds. Global Warming references ignore it.
The known top of atmosphere (TOA) solar energy is 1361 W/m2 which should produce a maximum blackbody temperature of 120°C, very close to the 118°C measured by Abbot. However, because the absorption and emission frequencies are different, any amount of incident energy can produce any temperature because it is the ratio of absorptivity to emissivity that is important. In the special case of a black body, the ratio is 1 (by definition).
Yes, the atmosphere is a thermal reservoir. However, it is also a radiator. Of the 340 W/m2 (average) TOA, 30% is reflected back to space, 40 W/m2 (12%) is radiated from the surface to space, 20% is directly absorbed by the atmosphere, 8% of the surface radiation is absorbed by the atmosphere, and the rest (30%) is moved to the atmosphere by convection and evaporation. Eventually, the atmosphere radiates about 58% of the TOA energy back to space. Another 324 W/m2 of radiation is exchanged between the surface and the atmosphere in addition to the 340 W/m2 from the Sun.
The numbers you use are a bit misleading. Because the Earth is 75% covered with water, a significant amount of energy is used to evaporate water. As a result, about 46% of the solar energy absorbed by the surface is added to the atmosphere without changing the atmosphere's temperature. However, the basic analysis is good.
As to the mechanisms addressed by de Saussure, Wood, and Abbot, they all showed that convection is the main method of cooling the surface while the Sun is shining. It is well known that night time cooling is by radiation, that is why cloudy nights are warmer than clear nights. Q Science (talk) 06:16, 1 May 2010 (UTC)
- That's a very nice concept, I appreciate the introduction. (The natural-sciences portion of my background is physics, not atmospherics and oceanography, and I hadn't run into it before.)
- However I don't understand the sense in which it makes peaks the same temperature as valleys. Most of the thermal capacity of a mountain resides in solid material, whereas potential temperature seems only defined for and relevant to fluids. Since the actual temperature of a mountain peak is comparable to that of the air just above it, the peak itself is cold and I don't see any notion of potential temperature applicable to it that would justify saying that the peak is the same temperature as the valley. Furthermore even if the potential temperature of the air at the peak is greater than in the valley, this is cold comfort (so to speak) to anyone who forgot to bring their parka on the climb. I don't see the physical relevance of potential temperature to any phenomenon save that of fluid stability, for example helping to explain why there are only three cells (Hadley, Ferrel, and polar) instead of hundreds (though it wouldn't seem to explain why there are three rather than zero).
- Potential temperature applies to the air, not the solid surface. Q Science (talk) 20:56, 1 May 2010 (UTC)
- Exactly so, which would suggest that the peak itself really is colder than the valley, regardless of what one might think of the air immediately above it. --Vaughan Pratt (talk) 05:30, 2 May 2010 (UTC)
- Potential temperature applies to the air, not the solid surface. Q Science (talk) 20:56, 1 May 2010 (UTC)
- The temperature of the surface is not used in climate calculations. Instead, the temperature of the air some distance from the surface is used. (I have seen the value 1.5m suggested as typical.) On a recent day, the air temperature was about 80°F, some wood boards were 133°F, and a dark wall was 161°F. Guess which was closest to the reported value. Q Science (talk) 07:25, 2 May 2010 (UTC)
- Were these all in the shade? If in the sun it's obviously an unfair comparison. If in the shade, how much variation is possible between air and solids?
- In any event I don't see the relevance of potential temperature to global warming. Are you claiming that NOAA is reporting temperature incorrectly? --Vaughan Pratt (talk) 07:39, 3 May 2010 (UTC)
- The temperature of the surface is not used in climate calculations. Instead, the temperature of the air some distance from the surface is used. (I have seen the value 1.5m suggested as typical.) On a recent day, the air temperature was about 80°F, some wood boards were 133°F, and a dark wall was 161°F. Guess which was closest to the reported value. Q Science (talk) 07:25, 2 May 2010 (UTC)
- Regarding "can produce any temperature," I presume you don't mean that literally. (How would you reach 7000°C for example?)
- Yes, I really mean ANY. For instance, the thermosphere is 2,500 °C during the day. Q Science (talk) 20:56, 1 May 2010 (UTC)
- That's still not 7000°C, which I picked because the Sun itself is only 5700°C. Moreover I suspect the 2,500°C might result not from any wavelength-dependent emissivity/absorption ratio but from the solar wind, which is 30-150 times hotter than the Sun, though I'm not sure how much of the solar wind reaches the thermosphere. --Vaughan Pratt (talk) 05:30, 2 May 2010 (UTC)
- Yes, I really mean ANY. For instance, the thermosphere is 2,500 °C during the day. Q Science (talk) 20:56, 1 May 2010 (UTC)
- Also in January (namely at perihelion) the TOA insolation is more like 1410 w/m^2 which should increase the achievable temperature by 2-3 degrees however achieved. But this is a bit of a digression.
- The atmosphere indeed participates in the .175 exawatts that Earth radiates away, as you say (and I won't quibble about the percentages), but it does so steadily day and night (with a slight increase by day). You were claiming it does so only at night. The daily temperature fluctuation at the surface results from switching off the insolation at night, not from any variation in radiation from the Earth's atmosphere, which actually decreases at night.
- Sorry, I was not clear. We agree on this. Q Science (talk) 20:56, 1 May 2010 (UTC)
- Regarding "misleading numbers", you may have overlooked the "if" in "if all that power went into uniformly heating the whole atmosphere." I was deriving only an upper bound on how much the average temperature of the earth's atmosphere can vary daily. Your point about some of the energy going elsewhere than the atmosphere while good is only relevant to improving that bound, which while nice would be additional work that isn't needed to make the basic point that the atmosphere strongly damps out temperature fluctuations.
- Regarding "convection is the main method of cooling the surface while the Sun is shining," how is daylight relevant? How can the Earth be cooled by radiation at night (which it is as you point out) if not also by day? Are you claiming that the ratio of convective cooling (however you define that) to radiative cooling of the surface somehow decreases at night? Also I don't see how "cloudy nights are warmer than clear nights" supports this: the only difference between night and day that I'm aware of is that clouds cool by increasing the albedo locally (i.e. by reflecting sunlight), while continuing to warm by trapping radiation just as at night.
- Convection plus radiation cools the surface more than 120°C during the day. (The difference between the hot box experiments and the local environment is typically more than 120°C.) Radiation alone cools the surface about 15°C during the night, but only 1 or 2°C when there are clouds. Also, notice that the 120°C cooling is over about 10 minutes, but the 15°C is over the entire night. Q Science (talk) 20:56, 1 May 2010 (UTC)
- In what naturally arising outdoors situation could 120°C cooling over 10 minutes occur? That would be very interesting to see. --Vaughan Pratt (talk) 18:18, 3 May 2010 (UTC)
- Convection plus radiation cools the surface more than 120°C during the day. (The difference between the hot box experiments and the local environment is typically more than 120°C.) Radiation alone cools the surface about 15°C during the night, but only 1 or 2°C when there are clouds. Also, notice that the 120°C cooling is over about 10 minutes, but the 15°C is over the entire night. Q Science (talk) 20:56, 1 May 2010 (UTC)
- That is what the temperature would be if convection and radiation were not occurring, even in winter. Q Science (talk) 19:06, 3 May 2010 (UTC)
- I think someone else is going to have to step in here to convince me that de Saussure, Wood, and Abbot all showed the same thing, I'm afraid I'm not seeing it.
- On an unrelated note, is there a reason to insert a dummy section heading ("===break===") when reindenting to the left margin? It has the downside of creating a strange sectioning structure in the TOC. Isn't the reindenting itself sufficient indication? --Vaughan Pratt (talk) 14:01, 1 May 2010 (UTC)
- When editing, I find shorter sections easier to manage. I use the "break" just to add the "edit" link. The same thing was done in section 8 of the TOC. Q Science (talk) 20:56, 1 May 2010 (UTC)
- Perhaps Wikipedia could benefit from greater independence of its logical sections and its editable sections. The alternative might be for discussants (editors of talk pages) like us to sectionize their contributions better, with the added benefit of a more easily navigated talk page. --Vaughan Pratt (talk) 17:39, 2 May 2010 (UTC)
- When editing, I find shorter sections easier to manage. I use the "break" just to add the "edit" link. The same thing was done in section 8 of the TOC. Q Science (talk) 20:56, 1 May 2010 (UTC)
- Concerning the .175 exawatt number, I should point out that I lumped the reflected and reradiated components together when I said Earth radiated .175 exawatts. From the standpoint that reflected radiation is still radiation this is technically correct (and is why I said "intercepted" rather than "absorbed"), but if one follows the convention of distinguishing the visible radiation coming from Earth as reflection and using "radiation" to refer only to that attributable to the Earth's temperature (i.e. making the natural distinction between a reflector and a radiator), then assuming an albedo of 0.30[4], approximately 0.12 exawatts of insolation is absorbed by the Earth and reradiated at 20x the wavelength. The 0.112°C/hr upper bound on how fast the total atmosphere could possibly warm then drops to .08°C/hr because the reflected component doesn't warm the Earth. If we consider that only half the atmosphere is being heated at one time and that it gets 12 hours of sunlight then an average 2°C rise on the hot side is possible, and as much as 4-5°C at the hottest part, which is still consistent with my range of 2-5°C for the daily temperature fluctuation of a full column of atmosphere averaged over all its molecules. (Averaging over altitude instead of molecules or mass unduly weights the thermosphere and gives a larger but meaningless temperature rise.) And looking at the radiosonde data it would appear that most of this fluctuation is concentrated in the lowest kilometer or so, with the air higher up varying hardly at all between day and night. --Vaughan Pratt (talk) 18:41, 1 May 2010 (UTC)
- Yes, the response to diurnal heating/cooling is concentrated near the surface. We have an OK-ish article on the planetary boundary layer. Short Brigade Harvester Boris (talk) 08:07, 3 May 2010 (UTC)
Likely size of Wood's experiment
Sorry to go on about this, but another thought occurred to me. The article's statement that the greenhouse effect has nothing to do with greenhouses rests entirely on one experiment by Robert W. Wood in the Feb. 1909 issue of Phil. Mag., plus some untested and questionable speculation about convection. (This presumably means it has nothing to do with the interaction of glass or plastic with IR either.) Hence it's important to know whether Wood's experiment could ever be repeated, since if it can't be then it may well have been in error, particularly in light of Abbot's surprise at the outcome as noted above.
One reason I've been unable to repeat exactly Wood's experiment myself, http://boole.stanford.edu/WoodExpt notwithstanding, is difficulty in procuring a sheet of sufficiently transparent rock salt of a size equal to the glass sheet I was using in the other box, namely a foot square. So I approached a famous Stanford chemist (who shall rename anonymous by his request) who worked extensively with infrared-transparent windows during his long career, and asked him what he would recommend. He said that the IR-transparent windows ordinarily used in physics and chemistry labs of the kind Wood would have had access to were typically a couple of centimeters wide, and that IR-transparent materials such as rock salt lacked the mechanical strength needed to support much larger windows. I asked whether a one-foot rock-salt window would be feasible and he laughed and said he'd never heard of such a thing.
The most likely size for the boxes used in Wood's experiment therefore would have to be on the order of one inch. On that assumption it becomes more feasible to duplicate the experiment.
At the same time however it raises the interesting question as to the implications of a heat experiment on a one-inch box for the behavior of a greenhouse. One obvious difference is that convection is much more restricted in such a small space, and perhaps there are others. In any event I think we have to stop thinking in terms of a one-foot box, which is what I'd been assuming for my attempts at duplicating Wood's experiment at http://boole.stanford.edu/WoodExpt . --Vaughan Pratt (talk) 21:24, 9 July 2010 (UTC)
Suggested simplification of lead
I wonder whether it would be possible to convey the essential concepts underlying the greenhouse effect with less technical overhead. If so this would make the Wikipedia article on the subject accessible to a much broader audience. As one possible direction for this I've written User_talk:Vaughan_Pratt/Sandbox#Suggested_greenhouse_effect_lead. Let me know of any problems you see generally with simplifying the article along these lines, and specifically with the proposed simplification itself. --Vaughan Pratt (talk) 06:13, 5 July 2010 (UTC)
- I am not sure such a simplification can be made. Consider an Earth with an atmosphere containing no greenhouse gases. Some heat transfer by conduction would still occur between the surface of the Earth and the atmosphere itself. In essence the phonon flux is shared by the Earth's surface and the air molecules that collide with it. And so the statement - "This heating of the atmosphere by the outgoing radiation further warms the planet's surface above its effective temperature, the temperature the planet would theoretically achieve by direct radiation in the absence of greenhouse gases but taking reflection or albedo into account." In reality the effective temperature of the Earth's surface is the temperature it would be at if there were no atmosphere at all - like the moon. And I am not even sure I like the "further warms the planet's surface" reference because it implies an energy disequilibrium as if the presence of CO2 and water vapor magically add energy back into the Earth surface from nowhere. I still think the best way to describe the effect is an energy conversion - long wave infra-red electromagnetic radiation leaving the earth's surface at the speed of light strikes greenhouse gas molecules. The electromagnetic energy is converted into kinetic energy (vibration) of the bonds in the greenhouse gas molecules. This kinetic energy progresses much more slowly through the Earth's atmosphere than electromagnetic energy. 24.3.11.218 (talk) 08:09, 5 July 2010 (UTC)
- It's easy to get the reasoning wrong here (and the error might be mine, but you need to persuade me of that). I understand your point about conduction, and you can throw in convection too. But how can an atmosphere that is completely transparent to radiation have a temperature different from what the temperature of the surface would be without any atmosphere? By what mechanism would the lapse rate be other than zero? How could the top of the atmosphere be even one degree colder than the bottom?
- The point is that an atmosphere without greenhouse gases is not completely transparent to all radiation. You need to remember that the incoming radiation from the sun is broad spectrum (short wavelength ultra-violet to long wavelength infra-red). Without an atmosphere, the surface of the moon receives the entire spectrum (including a fair amount of x-ray and some gamma rays because of its very small electromagnetic field). Short wave radiation is scattered (Rayleigh scattering by the Earth's atmosphere - this is why the sky appears to be blue. By most estimates 25% of the direct radiation from the sun never reaches the Earth's surface because of scattering. And so the effective temperature, as I understand it, is the temperature if all of the incoming energy reaches the surface. —Preceding unsigned comment added by 24.3.100.36 (talk) 00:56, 25 September 2010 (UTC)
- And what would the temperature be? Q Science (talk) 02:06, 6 July 2010 (UTC)
- The effective temperature (under its usual assumption of unit emissivity). The point is that an atmosphere that cannot absorb or emit radiation cannot interact with space, whence the only temperature it is capable of "measuring" is the surface of the planet, which therefore determines the thermal destiny of the atmosphere. Under those conditions the blanketing effect of the atmosphere is purely reactive (in EE jargon), namely on account of its thermal mass interacting with the surface via conduction and convection, with no resistive component. It will therefore exert a damping effect on the diurnally varying surface temperature analogous to an inductor in series or a capacitor in parallel, but like them it cannot itself heat up (above the temperature of the surface) the way a resistor does. --Vaughan Pratt (talk) 04:43, 6 July 2010 (UTC)
- "The point is that an atmosphere that cannot absorb or emit radiation cannot interact with space." An atmosphere that has no greenhouse gases can still absorb and emit radiation (though at shorter wavelengths).24.3.11.218 (talk) 17:31, 24 July 2010 (UTC)
- Both sentences are true. --Vaughan Pratt (talk) 17:53, 24 July 2010 (UTC)
- I see it more like charging an ideal capacitor thru a diode. It can get as hot as the warmest spot on the planet, but has no way to return that heat to the surface. As a result, there is no thermal dampening (once the capacitor is fully charged). Q Science (talk) 14:43, 6 July 2010 (UTC)
- It's true that in the absence of diurnal variation the atmosphere can get no warmer than the surface (and you don't need a diode in the model to account for that). But if there's variation then when the surface cools down at night it gets below the temperature of the air, which then belatedly follows it down like a transmission line (capacitor and inductor together), via all of convection, conduction, and back radiation. What's the diode in that situation? --Vaughan Pratt (talk) 06:49, 7 July 2010 (UTC)
- "The point is that an atmosphere that cannot absorb or emit radiation cannot interact with space." An atmosphere that has no greenhouse gases can still absorb and emit radiation (though at shorter wavelengths).24.3.11.218 (talk) 17:31, 24 July 2010 (UTC)
- The effective temperature (under its usual assumption of unit emissivity). The point is that an atmosphere that cannot absorb or emit radiation cannot interact with space, whence the only temperature it is capable of "measuring" is the surface of the planet, which therefore determines the thermal destiny of the atmosphere. Under those conditions the blanketing effect of the atmosphere is purely reactive (in EE jargon), namely on account of its thermal mass interacting with the surface via conduction and convection, with no resistive component. It will therefore exert a damping effect on the diurnally varying surface temperature analogous to an inductor in series or a capacitor in parallel, but like them it cannot itself heat up (above the temperature of the surface) the way a resistor does. --Vaughan Pratt (talk) 04:43, 6 July 2010 (UTC)
- It's easy to get the reasoning wrong here (and the error might be mine, but you need to persuade me of that). I understand your point about conduction, and you can throw in convection too. But how can an atmosphere that is completely transparent to radiation have a temperature different from what the temperature of the surface would be without any atmosphere? By what mechanism would the lapse rate be other than zero? How could the top of the atmosphere be even one degree colder than the bottom?
- We are discussing "an atmosphere that is completely transparent to radiation". In this case, heat can return to the surface via conduction ONLY. If the atmosphere is transparent, then there can be no back radiation. Once the surface cools the lower atmosphere, that atmosphere will be denser than the warmer air above, thus, no night time convection. The diode covers this (almost) "one way" transmission of heat. Since convection produces a rapid flow of heat and conduction is much slower, the atmosphere will eventually obtain the peak surface temperature with a diurnal inversion near the surface, much like the ripple on a power supply filter capacitor. Q Science (talk) 08:01, 7 July 2010 (UTC)
- The diode exists only in your reasoning, not in the counterfactual situation created by postulating no GHGs. In that situation the oscillating temperature of the Earth's surface will induce oscillations in the temperature of the atmosphere that, by your conduction-only argument and the very low conductivity of air, will decay rapidly with altitude while remaining symmetric about the effective temperature. What diode did you have in mind that would break that symmetry? --Vaughan Pratt (talk) 19:36, 8 July 2010 (UTC)
- We are discussing "an atmosphere that is completely transparent to radiation". In this case, heat can return to the surface via conduction ONLY. If the atmosphere is transparent, then there can be no back radiation. Once the surface cools the lower atmosphere, that atmosphere will be denser than the warmer air above, thus, no night time convection. The diode covers this (almost) "one way" transmission of heat. Since convection produces a rapid flow of heat and conduction is much slower, the atmosphere will eventually obtain the peak surface temperature with a diurnal inversion near the surface, much like the ripple on a power supply filter capacitor. Q Science (talk) 08:01, 7 July 2010 (UTC)
- Starting with a cold atmosphere, convection will rapidly warm the atmosphere to the point that the temperature decreases at the adiabatic lapse rate from the surface to "the edge of space". The point being that there will be no stratosphere, no region where the atmosphere warms with increasing altitude. From this starting point of a "cold" atmosphere to one with an adiabatic controlled thermal gradient, heat flow is mostly one way, into the atmosphere.
- At that point, some parts of the atmosphere will be warmer than others. Via conduction and diffusion, the colder parts will get warmer and the warmer parts will get colder. This will be much slower than convection. However, as warm parts get colder, they will be reheated again via convection from the surface. At some point in time, the temperature of the entire atmosphere should obtain the same temperature. Once that occurs, then I agree with you that oscillations at the surface will affect the lower atmosphere. However, consider that when the surface warms the next morning heat will be added back to the atmosphere (via convection) faster than it was taken out by conduction. Think of two series diode/resister pairs, one pointing up and the other pointing down. One pair has a large resister and the other a small resister. This is (almost) equivalent to modeling the difference as a diode leakage current. Q Science (talk) 20:09, 8 July 2010 (UTC)
- Ok, now I see your diode, thanks for clarifying this (I was a bit slow). But in that case I was also missing the point about adiabatic lapse rate even without diurnal variation. Wouldn't the equilibrium situation in the absence of diurnal variation be the dry adiabatic lapse rate, DALR, since no GHGs implies no water vapor? That's a chilling 9.8 °C/km, so the TOA would have to be below 200 K. Ignoring the adiabatic lapse rate made my theory of the TOA temperature way wrong.
- At 9.8 °C/km, the atmosphere would be 0 K at 38 km (assuming a surface temperature of 100 °C). The current stratopause is at 47 km, the mesopause is at 84.8 km.
- Well, since it's implausible that the DALR will drop abruptly from 9.8 °C/km to zero at some point, it must do so more gradually. The linear approximation would have to be valid only at low altitudes. At what rate does the DALR decrease with altitude? A better question might be, what is the right formula for DALR taking pressure into account? --Vaughan Pratt (talk) 18:45, 9 July 2010 (UTC)
- At 9.8 °C/km, the atmosphere would be 0 K at 38 km (assuming a surface temperature of 100 °C). The current stratopause is at 47 km, the mesopause is at 84.8 km.
- Actually, it would cause the atmosphere to condense. Think liquid nitrogen clouds, and above those, oxygen. The actual DALR formula takes gravity and pressure into account. Q Science (talk) 22:19, 9 July 2010 (UTC)
- Got it. (But won't the nitrogen clouds (77.3 K) be higher than oxygen (90.2 K)?) So as soon as the oxygen condenses (assuming something it can nucleate on) it's only nitrogen from then on up until that too condenses and then the atmosphere comes to an abrupt halt (ignoring the solar wind and all that). But then there won't be any atmosphere at all at 38 km for it to be 0 K. The atmosphere will never get colder than 77.3 K, at wherever nitrogen condenses (easily computed if it weren't my bedtime). --Vaughan Pratt (talk) 06:48, 10 July 2010 (UTC)
- Actually, it would cause the atmosphere to condense. Think liquid nitrogen clouds, and above those, oxygen. The actual DALR formula takes gravity and pressure into account. Q Science (talk) 22:19, 9 July 2010 (UTC)
- Note to self, check references :( Sorry. Actually, water condenses in the real atmosphere at temperatures above 0°C, but there is still some water vapor even at -70°C (at the tropopause). Q Science (talk) 07:21, 10 July 2010 (UTC)
- Your cloud analogy breaks down in light of the 1-4% density of water vapor in air, vs. 100% density of nitrogen when that's the only remaining gas. Nitrogen molecules on their own don't have to fight their way through some alien viscous gas to find each other or a hospitable nucleus the way water molecules do in air, they can nucleate on each other. Supersaturation is largely (entirely?) confined to the situation where the partial pressure of the vapor in question is well below the total pressure. --Vaughan Pratt (talk) 17:17, 10 July 2010 (UTC)
- Note to self, check references :( Sorry. Actually, water condenses in the real atmosphere at temperatures above 0°C, but there is still some water vapor even at -70°C (at the tropopause). Q Science (talk) 07:21, 10 July 2010 (UTC)
- So I take it your point about the diode is that diurnal variation takes the edge off the DALR by reducing it. By how much? What's the right way to look at this, I'm having trouble calculating it. --Vaughan Pratt (talk) 22:36, 8 July 2010 (UTC)
- Consider a metal rod with a heater under one end and ice at the other. At steady state, there will be a thermal gradient, similar to the DALR. If the rod is removed from the heat source and sink and placed in an insulated box, the temperature gradient will eventually approach zero and the entire rod will have the same temperature. However, if just the ice was removed, the entire rod will eventually heat up to the maximum temperature, the temperature of the heater. In the absence of a heat sink at altitude, the atmosphere does the same thing. It will heat up to the maximum temperature of the surface. The diurnal variation complicates this, and actually reduces the peak temperature a small amount, but the temperature is much closer to the peak surface temperature than it is to the effective balckbody temperature (because of the "diode" effect).
- Remember, the DALR defines temperature/density stability. If a parcel of air is colder than predicted by the DALR, then it is denser than the air below it, and it will sink. If it is warmer, then it rises. Once the atmosphere has warmed to the DALR gradient, the peak surface temperature can no longer produce convection from the surface. However, conduction will still cause heat to flow from hot to cold. As long as any part of the atmosphere is colder than the peak surface temperature, heat will flow. Initially, heat will flow by conduction to a few meters above the surface. Once the temperature there is high enough, the parcel will begin to rise, cooling at the DALR. When it stops rising, its "potential temperature" will be higher than the surface temperature. This process will continue until the entire atmosphere is at the same temperature and the environmental lapse rate (ELR) is zero. Q Science (talk) 05:22, 9 July 2010 (UTC)
- Ok, so is the bottom line that I was right (that the TOA temperature will equal the surface temperature) but for the wrong (too simplistic) reason? --Vaughan Pratt (talk) 18:45, 9 July 2010 (UTC)
- Remember, the DALR defines temperature/density stability. If a parcel of air is colder than predicted by the DALR, then it is denser than the air below it, and it will sink. If it is warmer, then it rises. Once the atmosphere has warmed to the DALR gradient, the peak surface temperature can no longer produce convection from the surface. However, conduction will still cause heat to flow from hot to cold. As long as any part of the atmosphere is colder than the peak surface temperature, heat will flow. Initially, heat will flow by conduction to a few meters above the surface. Once the temperature there is high enough, the parcel will begin to rise, cooling at the DALR. When it stops rising, its "potential temperature" will be higher than the surface temperature. This process will continue until the entire atmosphere is at the same temperature and the environmental lapse rate (ELR) is zero. Q Science (talk) 05:22, 9 July 2010 (UTC)
- However, the temperature would be close to the peak surface temperature (112°C), not the blackbody average (5°C). The bottom line is that the atmosphere would be hot enough to boil water if something did not allow it to lose heat. Q Science (talk) 22:19, 9 July 2010 (UTC)
- Ah, excellent point. (You always were better at this adiabatic lapse rate/potential temperature stuff than me.) But where did those temperatures come from? Assuming the usual figure of 0.3 for albedo, I calculate 87 °C for the peak (but that's still above b.p. at such a low pressure) and −18 °C for the average.
- However, the temperature would be close to the peak surface temperature (112°C), not the blackbody average (5°C). The bottom line is that the atmosphere would be hot enough to boil water if something did not allow it to lose heat. Q Science (talk) 22:19, 9 July 2010 (UTC)
- 0.3 is for the Earth as a whole. Some places are very different. Q Science (talk) 05:24, 10 July 2010 (UTC)
- That doesn't explain 112°C and 5°C. Where did those come from? --Vaughan Pratt (talk) 05:45, 10 July 2010 (UTC)
- 0.3 is for the Earth as a whole. Some places are very different. Q Science (talk) 05:24, 10 July 2010 (UTC)
- 112°C is for 100% absorption at noon because it represents the peak temperature. 98% would be more accurate, but this is a theoretical model and I used the average insolation and did not account for the annual variation due to an elliptical orbit. In other words, it is close enough for an example. The 5°C is the expected temperature of a blackbody in the same orbit as the Earth. I should have used 1°C, the blackbody temperature of the moon. Sort of. The actual value varies from equator to pole for geometric reasons. Q Science (talk) 07:21, 10 July 2010 (UTC)
- Ok, so you're saying for each of these numbers there exists a particular albedo and latitude it could arise at. Got it. Presumably these would be different albedos and/or latitudes for different numbers. My inclination would have been to standardize on one for both numbers. With no albedo, 100% absorption, and the sun right overhead, the peak should be 121 °C, so 112 °C is feasible somewhere. --Vaughan Pratt (talk) 16:45, 10 July 2010 (UTC)
- 112°C is for 100% absorption at noon because it represents the peak temperature. 98% would be more accurate, but this is a theoretical model and I used the average insolation and did not account for the annual variation due to an elliptical orbit. In other words, it is close enough for an example. The 5°C is the expected temperature of a blackbody in the same orbit as the Earth. I should have used 1°C, the blackbody temperature of the moon. Sort of. The actual value varies from equator to pole for geometric reasons. Q Science (talk) 07:21, 10 July 2010 (UTC)
- There's a really nice paradox in there: instead of warming the atmosphere, GHGs actually cool it (by allowing the atmosphere to radiate its heat away to space). I've never seen this paradox pointed out before, but it's a beautiful one! Can you source this, or did you just figure it out yourself? If the latter, that's brilliant, even if it has a flaw somewhere.
- By the Conventional Wisdom there is clearly a flaw in your reasoning. We "know" that adding GHGs warms the atmosphere (I put quotes around "know" there, whereas I would not put quotes around it in the statement "we know that CO2 is rising," which the Keeling curve clearly demonstrates). Yet you've just shown that adding GHGs cools the atmosphere, at least initially.
- I see three possibilities.
- 1. There's some flaw in your reasoning: your "diode" doesn't work quite the way you think it does.
- 2. At some point with increasing GHGs the cooling reverses and becomes warming.
- 3. There's some flaw in the greenhouse effect theory: GHGs don't warm the atmosphere.
- I'm inclined to go for 2. Just my intuition, there's obviously a differential equation hidden in there waiting to be solved, but I'm guessing that it shouldn't take too much GHG to leak heat to space faster than your diode effect can replace it. As the GHGs increase, at some point the cooling will turn round and go into warming mode. The nature of the turn-around point or points (which might slide up or down rather than happen at a single GHG level) and the associated temperature(s) there would be extremely interesting. --Vaughan Pratt (talk) 00:01, 10 July 2010 (UTC)
- By cooling the atmosphere, GHGs heat the surface. Remember, 50% up and 50% down. Q Science (talk) 05:24, 10 July 2010 (UTC)
- I didn't say they didn't heat the surface, I'm just pointing out that the atmosphere is cooled by GHGs. (They better cool it, or the atmosphere would be even hotter than your 112°C after adding GHGs.) Greenhouse-effect theory says GHGs warm the atmosphere. --Vaughan Pratt (talk) 05:45, 10 July 2010 (UTC)
- By cooling the atmosphere, GHGs heat the surface. Remember, 50% up and 50% down. Q Science (talk) 05:24, 10 July 2010 (UTC)
- On further reflection I'm dubious that the atmosphere can reach the full peak temperature. Even if we accept your point that most (say 90%) of the atmosphere by mass is thermally isolated from the ground, the bottom 10% is going remain relatively well coupled to the surface fluctuations via the convection that will begin the moment the surface starts to rise above −18 °C. This convection will circulate air that is not terribly far from the surface temperature at all times including say 9 am while the surface is not yet close to 87 °C, which will tend to cool the bottom 10% or 800 meters. The 87 °C air rising off the surface at noon then has to punch its way through that cool bottom kilometer, and it seems implausible it will still be 87 °C by the time it's risen to the more thermally isolated part of the atmosphere. In other words I'm claiming a cooler bottom kilometer than you're expecting by virtue of better coupling to cold temperatures than you're assuming. --Vaughan Pratt (talk) 00:18, 10 July 2010 (UTC)
- Convection occurs when a parcel of air is less dense than the air above it. Since cold air is more dense than warm air, cold air sinks. At night, the surface radiates heat, the surface gets colder than the air above it, and convection stops. The next morning, convection starts as soon as the surface is about 5°C above the night time minimum. It just doesn't go very high. Even with today's real atmosphere, the surface temperature peaks at 10°C to 20°C higher or lower than the atmosphere only two meters (2m) above. Thermal conduction in a gas is extremely slow. Q Science (talk) 05:24, 10 July 2010 (UTC)
- It just doesn't go very high. Well, at this point I don't trust anyone's intuition, I'd need to see the differential equation and a likely solution thereof. --Vaughan Pratt (talk) 05:45, 10 July 2010 (UTC)
- Even with today's real atmosphere, the surface temperature peaks at 10°C to 20°C higher or lower than the atmosphere only two meters (2m) above. This just went above my pay grade. With even a very light breeze there must be considerable mixing in the bottom few meters. So is it that the rate of transfer of heat from the surface to 20°C cooler air contacting it doesn't heat the air terribly quickly? --Vaughan Pratt (talk) 06:18, 10 July 2010 (UTC)
- Convection occurs when a parcel of air is less dense than the air above it. Since cold air is more dense than warm air, cold air sinks. At night, the surface radiates heat, the surface gets colder than the air above it, and convection stops. The next morning, convection starts as soon as the surface is about 5°C above the night time minimum. It just doesn't go very high. Even with today's real atmosphere, the surface temperature peaks at 10°C to 20°C higher or lower than the atmosphere only two meters (2m) above. Thermal conduction in a gas is extremely slow. Q Science (talk) 05:24, 10 July 2010 (UTC)
break and outdent
Higher: touch a hot car on a warm day, or walk bare foot on a black road.
Lower: When there is a temperature inversion just before dawn, there is almost no breeze.
This happens almost every day. You just need to plot some real data and you can see this for yourself. Q Science (talk) 06:52, 10 July 2010 (UTC)
- Certainly, but that's why I said 9 am and not 6 am. 9 am is sufficiently after dawn as to start kicking up some thermals,yet still well before the peak temperature of noon. What I'm having trouble calculating is the rate at which heat is being exchanged with the atmosphere from a surface that is being warmed halfway between dawn and noon. --Vaughan Pratt (talk) 17:32, 10 July 2010 (UTC)
- I was focusing on your claim of "at all times". You might find this useful. Use the Read File button to view the South Pole data. Q Science (talk) 20:15, 10 July 2010 (UTC)
- You showed me that data a while back, it was a real eye-opener for me, thanks for that! (But does how the South Pole data bear on whether the whole atmosphere can achieve the peak at the equator, whether your 112 °C or my 87 °C? I thought the South Pole was cold.) By "at all times" I meant that at all times there would be a layer of air (0.5-2 meters maybe?) just above the surface that was roughly tracking the surface temperature, and that within say an hour each side of the peak surface temperature, where the bottom centimeter of air is going to be constantly warming to 87 °C and rising, this meter layer would cool that rising air. By the time the meter layer has had a chance to get close to 87 °C the surface has started to cool. I just don't believe the couple of hours during which the surface attains 87 °C is long enough to heat this meter of air to 87 °C before the whole neighborhood starts to cool down again. It may be possible to model this situation electrically with an RLC network along with your diode, except it would be a number of diodes distributed over the network. --Vaughan Pratt (talk) 22:08, 10 July 2010 (UTC)
- I was focusing on your claim of "at all times". You might find this useful. Use the Read File button to view the South Pole data. Q Science (talk) 20:15, 10 July 2010 (UTC)
- It wouldn't be much per day, but over 4.5 billion years, well, it adds up. Q Science (talk) 23:36, 10 July 2010 (UTC)
- Well, it's a good question how long it would take to get into a reasonable equilibrium, but I imagine it would be pretty well settled into its final state in less than a century. But let's say a million years just to be on the safe side. My point is that the temperature oscillation within a few meters of the surface will never get above some temperature in a range something like 77-83 °C, say 80 °C for definiteness, even though the surface itself hits 87 °C for say an hour. This is because there's enough convection between 6 am and noon to cool that layer back down. Hence the atmosphere above that layer will never see more than 80 °C, or whatever that limit turns out to be. There's no way for the atmosphere say 3 meters to creep up to 87 °C even in a million years since it will never see air that hot. It will max out at 80 °C, by my argument above about there not being enough time for the 87 °C peak to overcome the 6 am to noon convection process that's cooling that layer down with air from the 6 am to noon surface temperature, which rises because it's increasing in temperature relative to what it had been overnight. Overnight should be long enough to cool the bottom 0.5-2 meters (just a guess) by conduction.
- If there's a flaw in my reasoning, the most likely one would be that I'm not giving the 87 °C surface sufficient credit for how fast it can pump 87 °C air into the 0.5-2 m layer. The layer is defined by however much the overnight temperature can cool the air by conduction. The cooling starts gently by say 4 pm (depending on how conductive the subsurface is) and continues to 6 am, so 14 hours say. That's a lot of conductive cooling, much longer than the noontime convective heating at the full peak surface temperature. My difficulty is believing that the entire 0.5-2 meter layer can be heated by convection to 87 °C in that short time. But perhaps it's possible, in which case you'd be right. --Vaughan Pratt (talk) 03:59, 11 July 2010 (UTC)
- It wouldn't be much per day, but over 4.5 billion years, well, it adds up. Q Science (talk) 23:36, 10 July 2010 (UTC)
- Of course, if the hottest latitude was part way to the pole, then there may only be 6 hours of cooling. At any rate, it is not worth more time. The point is that the atmosphere will be considerably warmer than the expected blackbody average, and we both agree on that. Q Science (talk) 08:00, 11 July 2010 (UTC)
- That would be in summer assuming a significant inclination to the ecliptic; in winter the opposite would happen.
- At the equator it might turn on closer examination that the TOA is within half a degree of the peak surface temperature, or it might be 10 degrees less, but I can see it will be very hot up there round the clock, contrary to my first guess before I understood your "diode". (Still neglecting solar wind and other effects of course.) As you say we agree.
- What's a good source for your diode analogy? Or did you come up with that? It's very good. --Vaughan Pratt (talk) 06:09, 12 July 2010 (UTC)
- Of course, if the hottest latitude was part way to the pole, then there may only be 6 hours of cooling. At any rate, it is not worth more time. The point is that the atmosphere will be considerably warmer than the expected blackbody average, and we both agree on that. Q Science (talk) 08:00, 11 July 2010 (UTC)
- It's my way of describing what the lapse rate graphs show. It is of limited value because most people don't know what a diode is. Q Science (talk) 07:34, 12 July 2010 (UTC)
- By the way, what's your take on Planetary boundary layer? My impression is that it's driven largely by the radiosonde data, does that seem fair? The article says "As Navier-Stokes equations would suggest" but doesn't say how. The same thing in terms of potential temperature and adiabatic lapse rate might be a lot more obvious than a mysterious reference to Navier-Stokes. You seem to have a deeper understanding of how that layer should operate than is apparent from the article. Do you have sources that would support a more theoretical treatment in that article? --Vaughan Pratt (talk) 06:51, 12 July 2010 (UTC)
- Radiosonde data "describes" the two, somewhat difference boundary layers, 2 to 5 inches, and then the next kilometer or so. I have lots of references. Most are either too basic to explain anything or lots of differential equations. There appears to be a lot of describing without really understanding. Q Science (talk) 07:34, 12 July 2010 (UTC)
- That's very interesting. Not sure what else to say without references. --Vaughan Pratt (talk) 08:05, 12 July 2010 (UTC)
- Radiosonde data "describes" the two, somewhat difference boundary layers, 2 to 5 inches, and then the next kilometer or so. I have lots of references. Most are either too basic to explain anything or lots of differential equations. There appears to be a lot of describing without really understanding. Q Science (talk) 07:34, 12 July 2010 (UTC)
break2
- Regarding "further warms the surface," I don't understand your complaint. First it's a true statement, second the warmth is not coming from "nowhere," it is coming from energy that would otherwise have been lost to space. You're arguing that it's impossible for people to accept that placing a warm body next to a cold one will make the latter warmer unless you first go into a detailed theory of heat transport. I would have thought they could, especially if they didn't want to be dragged through a complicated theory when a simple one suffices.
- I understand what "further warms the surface" implies - energy radiated by the sun is received by the Earth's surface and re-radiated. Because this re-radiation is "slowed down" by greenhouse gases, the energy is retained within the atmosphere beyond the di-urnal cycle resulting over time in a warmer atmospheric and surface temperature. I just think the phraseology could be misconstrued by the uninformed.24.3.11.218 (talk) 21:49, 5 July 2010 (UTC)
- A true statement should be at risk of being misconstrued only when it is not stated clearly. What's a clearer way of saying that warming the atmosphere further warms the surface, that is just as true? Since such warming of the surface occurs via all three of radiation, conduction, and convection, going into details puts you at risk of turning a true statement into a false one by implying that only one of the three is involved. --Vaughan Pratt (talk) 23:11, 5 July 2010 (UTC)
- "What's a clearer way of saying that warming the atmosphere further warms the surface, that is just as true?" A clearer way is this - Long wave infrared energy leaving the Earth's surface at the speed of light is converted to kinetic energy in greenhouse gas molecules. This kinetic energy is shared by the rest of the atmosphere and progresses much more slowly through the atmosphere by conduction and convection. There is no generalized further warming of the Earth's surface by the atmosphere. Whatever energy the surface loses, the atmosphere gains and vice versa whether it be by re-radiation, conduction, or convection. The point that is missing from your statement is the time scale over which energy from the sun traveling at the speed of light is impeded from leaving the Earth at the same speed. It really boils down to a comparison of dQ / dt for energy received by Earth and dQ / dt leaving the Earth. Because greenhouse gases lower dQ / dt for energy leaving the Earth, there is a heat build up that is shared between the atmosphere and the surface.24.3.11.218 (talk) 16:04, 7 July 2010 (UTC)
- This might be fine for some part of the body of the article, assuming it's a good way of looking at things, but I disagree that it's appropriate for the lead because it assumes a technical sophistication that most people trying to grasp the basic idea simply don't have. If you don't believe this we could do a little experiment where we first ask people whether they'd expect the surface to get warmer if the air got warmer, and then bounce your explanation off them and see what they think. --Vaughan Pratt (talk) 00:04, 9 July 2010 (UTC)
- A little off subject but is there a saturation mixing ratio for greenhouse gases - meaning is there a point at which adding more greenhouse gases has no additional warming effect? I did some specific heat calculations for various atmospheric mixing ratios a while back, and adding more CO2 and water increased the specific heat of the atmosphere. By these calculations, adding more CO2 and water should result in a cooler atmosphere. But these calculations don't take into account wavelength dependent radiative heating. Does anyone know of a chart that shows this for various compounds and elements - thermal capacity for a substance at constant temperature (J / kg * Deg K) at various wavelengths of incident radiation?24.3.11.218 (talk) 17:07, 7 July 2010 (UTC)
- Your first question is something of a trick question because it doesn't specify whether the atmospheric pressure is to remain constant. If the atmosphere was pure methane (and we disqualified anything worse than methane as an additive) then you could add methane forever, which would raise both the pressure (on account of the increasing mass) and the temperature (because the overall thermal resistance to radiation and conduction would continue to increase, I would guess). In that case there is no limit. Think of Venus, where the surface pressure is nearly 100 times that on Earth because perhaps as much as all of the planet's carbon has been boiled out of the ground by its 460 C temperature. (If it's all then we have a much more accurate idea of the total carbon on Venus than on Earth, since we can only guess at the amount of buried carbon on Earth.)
- By changing the mixing ratio I don't mean just adding more CO2 (or methane in the case you describe). Consider the methane combustion equation - CH4 + 2*O2 = 2*H2O + CO2. When 1 mole of methane is burned, 2 moles of O2 are pulled from the atmosphere, 2 moles of water vapor are added to the atmosphere, and 1 mole of CO2 is added to the atmosphere. Now what happens to atmospheric thickness, pressure, and temperature when a lot of this happens (say doubling CO2 concentration from 380 parts per million to 760 parts per million)? Well from the ideal gas law PV = nRT. Dividing both sides by mass and rearranging terms we get P = density * R * T / Molecular Weight. The atmospheric pressure equation is P = atm-thickness * density * g (acceleration due to gravity). Setting the two equal yields - T = atm-thickness * g (acceleration due to gravity) * Molecular Weight / R. Now what happens to T if we add a lot of CO2 and water and subtract a lot of oxygen from the atmosphere? The molecular weight of the atmosphere would increase and a balance of two things would happen - temperature would increase and/or atmosphere thickness would decrease. Of course all of this assumes that the thermal capacities (specific heats) of the gas components of the atmosphere are wavelength independent (which they are not). Never mind my question - I relooked and found my mistake. 24.3.11.218 (talk) 16:12, 9 July 2010 (UTC)
- Your first question is something of a trick question because it doesn't specify whether the atmospheric pressure is to remain constant. If the atmosphere was pure methane (and we disqualified anything worse than methane as an additive) then you could add methane forever, which would raise both the pressure (on account of the increasing mass) and the temperature (because the overall thermal resistance to radiation and conduction would continue to increase, I would guess). In that case there is no limit. Think of Venus, where the surface pressure is nearly 100 times that on Earth because perhaps as much as all of the planet's carbon has been boiled out of the ground by its 460 C temperature. (If it's all then we have a much more accurate idea of the total carbon on Venus than on Earth, since we can only guess at the amount of buried carbon on Earth.)
- "What's a clearer way of saying that warming the atmosphere further warms the surface, that is just as true?" A clearer way is this - Long wave infrared energy leaving the Earth's surface at the speed of light is converted to kinetic energy in greenhouse gas molecules. This kinetic energy is shared by the rest of the atmosphere and progresses much more slowly through the atmosphere by conduction and convection. There is no generalized further warming of the Earth's surface by the atmosphere. Whatever energy the surface loses, the atmosphere gains and vice versa whether it be by re-radiation, conduction, or convection. The point that is missing from your statement is the time scale over which energy from the sun traveling at the speed of light is impeded from leaving the Earth at the same speed. It really boils down to a comparison of dQ / dt for energy received by Earth and dQ / dt leaving the Earth. Because greenhouse gases lower dQ / dt for energy leaving the Earth, there is a heat build up that is shared between the atmosphere and the surface.24.3.11.218 (talk) 16:04, 7 July 2010 (UTC)
- A true statement should be at risk of being misconstrued only when it is not stated clearly. What's a clearer way of saying that warming the atmosphere further warms the surface, that is just as true? Since such warming of the surface occurs via all three of radiation, conduction, and convection, going into details puts you at risk of turning a true statement into a false one by implying that only one of the three is involved. --Vaughan Pratt (talk) 23:11, 5 July 2010 (UTC)
- I understand what "further warms the surface" implies - energy radiated by the sun is received by the Earth's surface and re-radiated. Because this re-radiation is "slowed down" by greenhouse gases, the energy is retained within the atmosphere beyond the di-urnal cycle resulting over time in a warmer atmospheric and surface temperature. I just think the phraseology could be misconstrued by the uninformed.24.3.11.218 (talk) 21:49, 5 July 2010 (UTC)
- PV=nRT assumes that no energy enters or leaves the system. (This is called "adiabatic".) In addition, the volume of the atmosphere is not defined. Therefore, that equation is not appropriate in the case. Q Science (talk) 16:44, 9 July 2010 (UTC)
- PV=nRT assumes that there is no NET flow of energy into or out of the system. Energy can still enter and exit the system but whatever enters the system has to be equal to whatever leaves it. I am disregarding the exothermic energy generated by combustion and talking exclusively about mass (CO2 and H2O) entering the system. —Preceding unsigned comment added by 24.3.11.218 (talk) 17:37, 9 July 2010 (UTC)
- PV=nRT assumes that no energy enters or leaves the system. (This is called "adiabatic".) In addition, the volume of the atmosphere is not defined. Therefore, that equation is not appropriate in the case. Q Science (talk) 16:44, 9 July 2010 (UTC)
- My point was that the energy of the system WILL change because there is no way to stop it. Q Science (talk) 21:57, 9 July 2010 (UTC)
- But if you insist that the pressure remain constant, then once the atmosphere was all methane, anything you did to the atmosphere thereafter that kept the pressure constant would only reduce the temperature.
- If that wasn't your question feel free to rephrase it. :) --Vaughan Pratt (talk) 00:04, 9 July 2010 (UTC)
- I agree that the more detailed the description of the process the deeper the insight. But I also feel that Wikipedia articles should begin with the simplest explanation that is technically correct and flesh it out with more detail later. For example Ohm's law can initially be explained without talking about electrons but simply treating current as a flow of electricity. Here we can say that longwave radiation heats the atmosphere without initially making molecules part of that explanation. And we can say that warming the atmosphere adds to the warmth of the surface without committing to one of the several possible approaches to analyzing the ambient radiation into its directional components, and without implying that convection and conduction don't contribute, which they do to a larger degree than usually acknowledged. --Vaughan Pratt (talk) 14:34, 5 July 2010 (UTC)
- We cannot say that longwave radiation emitted by the Earth's surface warms the atmosphere which in turn further warms the Earth's surface. Longwave radiation that is emitted by the Earth's surface results in a cooling of the surface. It is because this longwave radiation is converted to kinetic energy of massive particles (as opposed to mass less photons) that energy is retained by the Earth's surface and atmosphere.24.3.11.218 (talk) 16:33, 7 July 2010 (UTC)
- Your reasoning here points up the complexity of the warming action of GHGs. It might be best to analyze the situation perturbationally, by considering GHGs as a perturbation of the atmosphere's transmissivity (absorptance, transparency, opacity, whatever). Without the GHGs the Earth settles into its effective temperature, warmed by the incoming shortwave radiation and cooled by the outgoing longwave radiation. Adding GHGs does nothing to that cooling action itself, rather it captures the outgoing radiation already present before the GHGs, which then warms the atmosphere, initially warming the GHGs but almost immediately warming the nearby non-GHGs (mainly oxygen and nitrogen). This warmed atmosphere then further warms the surface of the Earth, by all three of radiation, convection, and conduction. Since convection plays a major role in exchanging heat between the surface and the atmosphere it is incorrect to represent radiation as the dominant transport mechanism in achieving any long-term thermal equilibrium between the surface and the atmosphere, as I pointed out earlier based on Figure 7 in the Kiehl-Trenberth paper.
- Apropos of the short-term impact of radiation "it is interesting to see" (as Abbot would say) how quickly some radiation can reach the atmosphere. The strongest CO2 absorption lines have a strength on the order of 10−18 cm−1/(molecule.cm−2) and a width at low altitude (induced by pressure broadening) on the order of .1 cm−1 or 3 GHz (the clock cycle of a CPU). There are 6.022*1023/22.4 = 27*1021 molecules in a liter of air at STP, hence 27*.000389*1021/1000 = 1016 molecules of CO2 in a cc of air. So a photon of frequency chosen uniformly at random in a 1 cm−1 band (to simplify the math while giving the photon a fighting chance of not being captured) centered on one of those lines has a chance of 1016*10−18 = .01 of being captured by an atmospheric CO2 molecule in a one-centimeter journey towards outer space. Its mean free path before being captured by a CO2 molecule is therefore on the order of a meter.
- After the photon is captured by a CO2 molecule, it can either be re-emitted or the energy stored in the bonds can be converted into translational / rotational kinetic energy upon colliding with other greenhouse and non-greenhouse molecules. As you progress further up through the atmosphere this energy should become equally divided amongst all of the atmospheric components by the equipartition theorem Law_of_equipartition. My question is if the mean free path is 1 meter for a photon hitting a CO2 molecule, what is the distance required from the Earth's surface to assure equipartion of this energy among all atmospheric components?24.3.11.218 (talk) 17:44, 24 July 2010 (UTC)
- CO2's strongest line, at 2361.465809 cm−1, has strength 3.524*10−18 cm−1/(molecule.cm−2) and pressure-broadened halfwidth .0734 cm−1 according to HITRAN2008. So had we picked a photon at random plus or minus a half-width from that line, the mean free path would be something like 2*.0734/3.542 meters or about 4 centimeters! (I may not have treated the line shape consistently with HITRAN's definition of line strength, which I haven't taken the trouble to parse, but even so this should still be within a factor of two.) Since I'm sure hardly anyone is following by this point, let me repeat that, indented to catch the eye.
- Photons on the strongest CO2 line have a mean free path in today's air of 4 centimeters.
- I understand. However, the photon of energy once captured by a CO2 molecule can either be re-emitted or if an intermolecular collision happens, the vibrational energy of the bond (or some portion of it) is converted to translational and / or rotational kinetic energy. Given intermolecular spacing of about 11.4 molecules for CO2 at STP (about the same for O2, N2, and H2O), the odds of collision are about 1 / (11.4 ^ 3) = 1 / 1480. And so, how far up through the atmosphere do you have to go before the photon of energy has been equally divided (full diffusion) amongst all of the atmospheric molecules?24.3.11.218 (talk) 13:20, 10 July 2010 (UTC)
- That's not how to compute molecular collision odds, which depend crucially on the diameter of the molecule (with diameter the Planck length, collisions would be so rare and mfp so long as to seriously violate the ideal gas law). Kennard 1938 at http://amsglossary.allenpress.com/glossary/search?id=mean-free-path1 gives the mean free path of air molecules at sea level (so presumably STP) as the order of 100 nm. I got 70 nm when I calculated it; as Kennard says, the concept of molecular diameter is fuzzy, though I would say the real fuzziness enters at drawing the line between close encounters and collisions.
- Okay. Mean free distance is equal to Boltman's Constant * Temperature / SQRT(2) * pi * Pressure * (Molecular Diameter)^2. This can be re-expressed as Mean free distance = Volume of container / Avogadro's # * SQRT(2) * pi * # moles in container * (Molecular Diameter)^2. So I can see where the mean free distance for CO2 is between 70 and 100nm. The intermolecular spacing distance for CO2 at STP is on the order of about 3.8nm so the odds of collision are about 1 in 25.24.3.11.218 (talk) 15:34, 11 July 2010 (UTC)
- Once the photon is absorbed, it's been fully assimilated into the atmosphere. Granted photons are re-emitted later, but in general not with the same wavelength, making it meaningless to say that an emitted photon is part of a photon absorbed earlier (so no notion of "full diffusion"). In that connection it's worth pointing out that the re-emitted photons have a vastly shorter mean free path than those radiated by earth, less than a foot, because by Kasha's rule and its corollary the Kasha-Vavilov rule most of them are sitting right on top of the strongest absorption lines. --Vaughan Pratt (talk) 16:24, 10 July 2010 (UTC)
- If an intermolecular collision occurs before the CO2 molecule has a chance to re-emit (at whatever wavelength), then the vibrational bond energy could be converted to kinetic translational energy of the molecule it collides with. Meaning a super-elastic collision would take place - Translational momentum of the colliding molecules after the collision is greater than translational momentum of the colliding molecules before the collision. If that happens often enough then the photons of energy received by greenhouse gas molecules are replaced by phonon flux in all atmospheric gas molecules and the Earth's surface. This is what I meant by diffusion - I might have chosen the wrong word.24.3.11.218 (talk) 15:34, 11 July 2010 (UTC)
- That process is fine, my concern was with tying diffusion to a photon. Photons are bosons and therefore don't have identities, so (correct me if you think I'm wrong) it shouldn't make sense to attribute part or all of a former photon to part or all of a subsequent phonon. Bosons aren't real in that sense. --Vaughan Pratt (talk) 06:19, 12 July 2010 (UTC)
- If an intermolecular collision occurs before the CO2 molecule has a chance to re-emit (at whatever wavelength), then the vibrational bond energy could be converted to kinetic translational energy of the molecule it collides with. Meaning a super-elastic collision would take place - Translational momentum of the colliding molecules after the collision is greater than translational momentum of the colliding molecules before the collision. If that happens often enough then the photons of energy received by greenhouse gas molecules are replaced by phonon flux in all atmospheric gas molecules and the Earth's surface. This is what I meant by diffusion - I might have chosen the wrong word.24.3.11.218 (talk) 15:34, 11 July 2010 (UTC)
- That's not how to compute molecular collision odds, which depend crucially on the diameter of the molecule (with diameter the Planck length, collisions would be so rare and mfp so long as to seriously violate the ideal gas law). Kennard 1938 at http://amsglossary.allenpress.com/glossary/search?id=mean-free-path1 gives the mean free path of air molecules at sea level (so presumably STP) as the order of 100 nm. I got 70 nm when I calculated it; as Kennard says, the concept of molecular diameter is fuzzy, though I would say the real fuzziness enters at drawing the line between close encounters and collisions.
- I understand. However, the photon of energy once captured by a CO2 molecule can either be re-emitted or if an intermolecular collision happens, the vibrational energy of the bond (or some portion of it) is converted to translational and / or rotational kinetic energy. Given intermolecular spacing of about 11.4 molecules for CO2 at STP (about the same for O2, N2, and H2O), the odds of collision are about 1 / (11.4 ^ 3) = 1 / 1480. And so, how far up through the atmosphere do you have to go before the photon of energy has been equally divided (full diffusion) amongst all of the atmospheric molecules?24.3.11.218 (talk) 13:20, 10 July 2010 (UTC)
- Photons on the strongest CO2 line have a mean free path in today's air of 4 centimeters.
- (If correctly taking the line shape into account turns out to change that by more than a centimeter I'll edit it accordingly.) Fortunately most of CO2's millions of lines are millions of times weaker than this or even 1 ppm of CO2 would cook us! --Vaughan Pratt (talk) 21:31, 8 July 2010 (UTC)
- We cannot say that longwave radiation emitted by the Earth's surface warms the atmosphere which in turn further warms the Earth's surface. Longwave radiation that is emitted by the Earth's surface results in a cooling of the surface. It is because this longwave radiation is converted to kinetic energy of massive particles (as opposed to mass less photons) that energy is retained by the Earth's surface and atmosphere.24.3.11.218 (talk) 16:33, 7 July 2010 (UTC)