User talk:Blackcloak

Welcome!

Hello, Blackcloak, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are a few good links for newcomers:

I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Where to ask a question, ask me on my talk page, or place {{helpme}} on your talk page and someone will show up shortly to answer your questions. Again, welcome! --best, kevin [kzollman][talk] 07:15, 21 February 2006 (UTC)[reply]

You have no Idea (re: Ohm's Law)

That title regards your next edit- re: intimately comment where you cut a phrase. I've seen far to many masters degree holders that were totally baffled by what would be instantly obvious to a decent tech. Too many junior engineer hires, that couldn't figure out simple common base biasing on a transistor-transistor coupled multistage amplifier. They aren't digital or something.
Disagree this was a good deletion. The topic is germane to the discussion. This thing is an educational work, introducing a related topic so intertwined with the main one on the table is hardly inappropriate.
Keep in mind you don't know whether the reader is a Mensus with seven PHDs or a grade school kid trying to figure out what something said. If it's not confusing, I suggest including it. Don't let your own knowledge cloud your judgement. There is no reason in Wikipedia to be overly terse, or narrowly focused on topic like in a printed encl. where every column-inch counts.
I have a bright neighbor, an eye doctor who gets totally lost when discussing the return path on a simple series lighting circuit--his broken headlamp gizmo that he uses for examining retinas... Some peoples brains just can't get around certain concepts. FrankB 06:57, 31 March 2006 (UTC)[reply]

Actually I do have an idea or two: Ohm's Law)

The topic is Ohm's Law, not how to correct problems with teaching masters degree students how to use Ohm's law. When you know something 'intimately,' you can apply that knowledge without having to think about it. When you come across an individual who has managed to complete a masters degree program in EE and still does not have a working knowledge of Ohm's Law, you know something about the quality of the master's program that individual claims to have completed.
"Disagree this was a good deletion." The material is not germane to the discussion at the point where it was inserted. Besides, the text is poorly written. Resistivity is its own topic, and the strain effects section adequately presents the underlying ideas of the deleted text, although in the context of making small changes to cross-sectional area and length. The link to resistivity was added. Ohm's law (the formula) does not actually contain the idea of resistivity directly. So, maybe, for the sake of being more complete, someone should write a separate section contrasting the meanings of resistance and resistivity.
I wrote the "elementary explanation ..." section, as well as others of an introductory nature. Relative to other Wikipedia entries, this entry on Ohm's Law does a far better job of helping young readers understand what is going on. That's not to say it could not be done better. My sense for clear communication is the only thing that drives my decisions as to what should be deleted or improved. (Take a peek at the Ohm's Law entries under other languages if you want to see terse.)
As for your bright neighbor (I've known my fair share of the type.), there are some simple concepts that he never learned. He probably does not understand conservation of energy, conservation of charge, and a host of other physical ideas. I would start by getting him to understand what a manometer is and how it works. Then you can tackle a battery. I wrote the 'hydrologic analogy' section in an attempt to help individuals just like your neighbor. Blackcloak, 1 April 2006

Re errors in Carbon Dioxide article

I reverted because you did not edit the current page, but an older page, probably the one you last edited on 17 January, 9.28 revision, thus restoring multiple errors that you appear to have introduced yourself at that revision (please correct me if I am wrong), and which had been corrected by user:William M. Connelley. I suggest you look back at the section "Isolation" and the Categories to see what I am talking about. Your revision at 8.35 on 17 Jan is correct, but your 9.28 revision on 17 Jan introduced multiple errors. If you want to restore those edits, I suggest you explain to the community what positive contribution they make to the article. From my perspective they look like vandalism and a waste of the time and energy of other editors. In addition, I tend to agree with user:Raymond arrit that your additions lacked clarity. On the blocking - I don't see any reason for you to have been blocked, but then I don't have administrator privileges, so you will should look elsewhere for an answer to that.Plantsurfer (talk) 08:48, 18 January 2008 (UTC)[reply]

Response to Plantsurfer

To say that I introduced multiple errors without describing the nature of the errors is distinctly unhelpful and disrepsectful. To suggest that in any way my contributions are vandalism is cleary wrong and you should review Wiki policy on the subject. When an addition lacks clarity, you should try to improve, not remove. Your reverts were not made in good faith. Regarding blocking, I don't see any indication that I have been blocked (in fact, never having been blocked, I don't know what to look for), but several attempts to edit were not accepted by Wikipedia.

Regarding your reference to the Isolation section, the part that appears to be relevant to my attempted contributions is:

Carbon dioxide is soluble in water, in which it spontaneously interconverts between CO2 and H2CO3 (carbonic acid). The relative concentrations of CO2, H2CO3, and the deprotonated forms HCO3- (bicarbonate) and CO32-(carbonate) depend on the pH. In neutral or slightly alkaline water (pH > 6.5), the bicarbonate form predominates (>50%) becoming the most prevalent (>95%) at the pH of seawater, while in very alkaline water (pH > 10.4) the predominant (>50%) form is carbonate. The bicarbonate and carbonate forms are very soluble, such that air-equilibrated ocean water (mildly alkaline with typical pH = 8.2 – 8.5) contains about 120 mg of bicarbonate per liter.

First, there is no mention of interrelationship of temperature, solubility and equilibrium. Perhaps, at the ocean surface, equilibirum (mass transport of co2)is always assumed.

While I can't say I fully understand all that is being said, I infer from the text that on a weight ratio basis, approx. .0001 of the mass of surface ocean water is the dominate species of co2. The wording suggests that the mass of this bicarbonate (OH- attached to CO2, presumably) is taken to include both the OH and the CO2, which means the CO2 (only) equilibrim mass ratio is closer to .00009. That implies that surface seawater (in equilibrium) accepts one molecule of CO2 for every 3000 molecules of water. Whereas, in the atmosphere (at sealevel, say), the corresponding ratio is about one molecule of CO2 for every 4000 molecules of 'air' (N2,O2). If you care to take the time, let me know if I'm close.

I could not figure out what you mean by 'Categories.' blackcloak (talk) 19:53, 18 January 2008 (UTC)[reply]

The errors that I am referring to were many tens of question marks replacing earlier arrows in the Isolation section and other content elsewhere. Why did you do that? I have no agenda against your substantive edits (or the arguments you raise above), and suggest that if you feel they were justified you should simply replace them. But please do it by editing the current revision of the page, not an old one, otherwise you revert other people's legitimate edits. Best wishesPlantsurfer (talk) 21:00, 18 January 2008 (UTC)[reply]

What are you talking about? Question marks? Arrows? I have no idea how any such thing happened. I certainly did not do it, at least not intentionally. So I guess I have to say I'm sorry if somehow my attempt at contributing created some spurious, albeit highly annoying, artifacts. It may have something to do with keeping the editing page open for long periods of time- like an hour. Anyway, please understand that such a thing was not done intentionally. Had I been in your position, I would have simply replaced the section with a correct older version.

Further, I did not start my edits with an old version, I used the current version, and replaced, on a paragraph by paragraph basis, usually also making further wording/phrasing improvements. If you look carefully you'll probably see the differences.

I think we both have to acknowledge that these editing/transmitting systems are not fool proof and that systematic errors occur without our knowledge. It would not surprise me at all to learn that some of these problems have to do with the way Wikipedia's servers handle the updating of text that originates through many different browsers (I use IE7(XP) and IE6(98SE).) 67.87.73.86 (talk) 00:12, 19 January 2008 (UTC) I did not notice I wasn't signed in. blackcloak (talk) 00:14, 19 January 2008 (UTC)[reply]

Talk page etiquette

I seem to be a latecomer to the greenhouse effect discussion. Thank you for your work in explaining radiation. However, please be careful to not talk down to other users. Your initial comment to me was somewhat uncivil, especially at the end; I ignored that because I was jumping in on another discussion, and was partially trying to throw a wrench in the gears becasue the discussion seemed to be endless, so I may have even deserved it. Asking Damorbel, "Let me just ask now if you've ever solved heat flow problems using the appropriate partial differential equation" is both pedantic and ineffective: in putting him on the spot over questions of his ability, the most likely result will be his agitation, and he seems to be politely agitated here as well. I am not sure if your most recent comment is to me or him, as it isn't properly indented, but it seems to be to me in what it addresses. In sounding teacherly, it makes me feel as if you are talking down to me, "If you could get over this bit of confusion, you might be able to take the next step," for example. I do in fact understand that we are talking about radiation and not conduction per se, and I've solved the toy problems in which you place a "atmosphere" layer or two in a blackbody radiative equilibrium, so your assumption of my unfamiliarity was wrong, hence the problem of assuming that others lack knowledge; I find it better to assume others to be completely knowledgeable, and move down from there if they don't understand something.

My guess is that the discussion will be endless, unless someone gives up. Unfortunately I know these behavior patterns, and I know what to expect. I am not bothered by it. I am a very patient person, until ... My question may have been pedantic and ineffective, but even when I have been less confrontational (esp. with Damorbel), I get gibberish. So what's the diff? Let him get agitated, that's his perrogative. I'll read what he has to say, but probably won't respond if he gets abusive, which I consider very unlikely. My comment below addresses the indentation issue. But to repeat, the entire paragraph was directed at Damorbel. I've learned that most of the time most others are less knowledgeable, less careful, less detail oriented, and less thoughtful than I am. I can see that you may be an exception. blackcloak (talk) 20:52, 8 March 2009 (UTC)[reply]

As for the conduction and the "blanket", I feel that my analog is an effective way to show that Damorbel's views are wrong, albeit not 100% kosher, in that radiating "layers" in the atmosphere are similar to insulators. More realistically, the CO2, H20, CH4, and pals vibrate when the proper wavelength radiation excites them, and they store that heat and can re-radiate, in a somewhat analogous way blanket stores some of my heat and re-radiates it. I was mainly trying to find an example to away from anything that sounded black body, because it seems that he puts his fingers in his ears and sings as soon as that word/phrase appears, and I was going for a new approach to communication.

Regarding your analog, I think you've made it far too complicated. It opens up the opportunity for him to pick and choose what part of the analogy he wants to challenge. I think his problem in understanding has to do with the differences in the way radiative energy transfer occurs vs. the way conductive thermal energy transfer occurs. You're assuming he understands this more basic set of ideas. I see no other way to work with him other than to ask him directly to confirm that he understands the basic ideas- and preferably to repeat them in his own words so we can see that he understands them. blackcloak (talk) 20:52, 8 March 2009 (UTC)[reply]

This is not to say I mind being corrected; I love being corrected and learning. And I'm not upset that it looks like you've talked down to me. In the future, and with other people (particularly if you come across thin-skinned ones), you may just want to think about your phrasing a little more. Awickert (talk) 09:17, 8 March 2009 (UTC)[reply]

Damorbel and I have exchanged a lot of comments recently. Review Greenhouse Gas discussion. I know him to be thick skinned. Usually I am very gentle, only becoming more confrontational as the circumstances dictate. Let me assure you I do think about my phrasing carefully, and I do not apologize for the tone I have adopted. I do find that many editors are making contributions (to the articles, not the discussion) far beyond their ability to understand fully the material they are presenting. But I accept the spirit of your comment and I ask you to repeat it if you feel compelled to do so. I've given up on editing actual articles, having wasted far too much time creating material that is reverted, usually by those who have not taken the time to understand what the changes are being made. There is probably a hundred to one ratio of the time I spend creating stuff to the time some other editor spends deciding to reverse my contribution. I think this is why the real experts don't appear to be contributing to "politically charged" (there's probably a better term) articles in wikipedia. (I don't mind thoughtful editing of my contributions- I'll usually accept it even when I don't agree that an improvement has been made; I'm offended by summary dismissal, especially when the offered reasoning is flawed, cursory, dismissive, or unjustified opinion.) blackcloak (talk) 20:52, 8 March 2009 (UTC)[reply]

But if you would please indent your comment to make me know who you're talking to, I'd happily reply on the talk page of the article, if you were talking to me. Awickert (talk) 09:19, 8 March 2009 (UTC)[reply]

The rules I follow for indentation are: Indent once more that the message to which I am responding. I indented properly to respond to Damorbel. My comment followed yours because I wrote it after you wrote yours. You were responding to Damorbel as well. If I had been responding to you, I would have indented once more. Incidentally, you should not have indented your last paragraph,unless you intended to "respond" to your own comments. Now for clarity I have to indent once more than I should have to in order to respond to your earlier comment. (My additions are being made from the bottom up.) blackcloak (talk) 20:52, 8 March 2009 (UTC)[reply]

Thanks for your thoughtful response. Yes, I'm sure it gets very tedious, and you could very well be right about the (in)effectiveness of my analog. Sorry for being confused about the indent; the content was just closer to my edit, while the indent said it was a reply to Damorbel. I actually tend to try to do more on geology, geophysics, and fluid dynamics because of the very reasons you stated: politically-charged topics are a pain. Thanks for your "long-winded" response to my long-winded comment; I sympathize with your frustration, and while I'm neither an admin nor a climate scientist, feel free to leave me a message if your contributions start being hacked down by non-NPOV vandals, and I'll try to help out. Awickert (talk) 21:13, 8 March 2009 (UTC)[reply]

Greenhouse effect discussion copied here. Much of this is non-archived, deleted material

Thermal equilibrium vs steady state

The very first sentence in this article is wrong because it confuses thermal equilibrium with steady state. In thermal equilibrium there is no greenhouse effect, because that would be a violation of 2.law of thermodynamics and Kirchhoff law. This also seems to be a source of confusion for many people who discuss on this page. If Earth received certain amount of energy from the Sun and radiated this energy back to Sun, that would be thermal equilibrium; if it receives some energy from Sun and radiates this energy into [colder] space, so that its temperature stays constant, it is known as steady state.213.220.211.183 (talk) 10:45, 20 January 2009 (UTC)[reply]

Not sure what was supposed to be in thermal equilibrium but steady state makes more sense there. I've changed it.
—Apis (talk) 13:08, 20 January 2009 (UTC)[reply]

Archiving I

Persistant attempts are being made to archive discussion. In view of the active status of discussion on GHE I am sure participnts with a positive outlook will want good visibility of contributions.--Damorbel (talk) 19:49, 22 January 2009 (UTC)[reply]

They are visible in the archives. There is no reason to keep discussions here which haven't been active in more than 2 months. I suggest you stop reverting the bot. --Kim D. Petersen (talk) 19:51, 22 January 2009 (UTC)[reply]

Kim D. Petersen your view of the archives is not born out by the facts. How is it that [Archive 2] has newer material in it than [Archive 3] ? This is a shambles and shows just how damaging overenthusiastic archiving is. It is quite clear that this archiving is merely an attempt by editors to sweep all criticism of the greenhouse effect under the carpet.

I have complained about this kind of archiving before. Archiving is supposed to be agreed, I do not remotely agree with this archiving policy, it gives a strong impression that discussion is not tolerated. Archives either cannot or are not supposed to be edited, certainly contributions are a waste of time because of the archiving on your extraordinarily restrictive suggestion of a few months.

Putting discussion in an archive is for discussions that are closed, i.e. contributions are no longer expected. Given the bizarre ideas on heat transfer adopted in the article, discussion is likely to last for a long time!

I have a strong inclination to make a formal complaint about the aggressive editing in this article, anyone who remarks about the dubious quality of the GHE science is speedily ridiculed William Connolley (with his infamous "septic" observation) you and your supporters. I understand that I can modify the archive bot, in light of you completely ignoring my objections to the archiving policy, am I supposed to be so foolish that you will not change it back? Where stands the famous "open Wiki" policy on this? --Damorbel (talk) 08:23, 23 January 2009 (UTC)[reply]

The reasons that the archives are somewhat broken - is that you are interfering with the bots operation, and reverting only the bots archive of this page - which makes for general confusion to it (and duplications of archived material).

Discussions which have been inactive for 2 months are closed. And you seem to be the only person who dislikes the archives. Archiving is done so that there is room for new discussions. As a side note - please read WP:NPA. --Kim D. Petersen (talk) 12:05, 23 January 2009 (UTC)[reply]

Stuff and nonsense. Which particular deity decided that two months are sufficient? That has got nothing to do with it. The only possible reason for archiving is when the current page becomes too big to manage, date has nothing to do with that, check here -Archiving a talk page before you start creating an archive policy of your own. I'm sorry if I caused trouble because you wanted quick archiving but the bot is clearly only the agent of the problem, the real villains are those trying to make the discussion less visible; you perhaps?. Please remove the bot or replace it with one that archives material that is more than one year old, the current one is inhibiting discussion on a very controversial topic. --Damorbel (talk) 21:10, 23 January 2009 (UTC)[reply]

A discussion that has had no reply in two months is not a discussion. It's stale beyond belief. The archive bot always archives threads based on time since the last contribution. There is no sekrit conspiracy going on here. --Stephan Schulz (talk) 23:00, 23 January 2009 (UTC)[reply]

Review the dates of my entries in the most recent archive. I agree with Damorbel on this issue. blackcloak (talk) 07:18, 5 March 2009 (UTC)[reply]

Stephan Schulz, you say "A discussion that has had no reply in two months is not a discussion." I didn't see this in the guidance for talk pages Archiving a talk page which makes it very clear why the current dicussion page should not be archived in the way you want, perhaps you could give me a link, until then I see the aggressive archiving i.e. small sections of only a few pages and less than a year, as attempt to limit discussion on a controversial topic and I will seek a remedy for this anti Wiki behaviour.

Your attitude to dicussion is very negative, you clearly are very well informed about Wiki rules but your negativity shows in trying to hide discussion which is finished. Why hide it unless the file is cumbersome? Are other people, not so expert as you, not allowed to see the subjects that have been discussed without looking in the archives? It is very tedious to look in the archives; you, being an expert, should know this, that is why you arouse suspicion that your motive is to suppres discussion. --Damorbel (talk) 09:53, 25 January 2009 (UTC)[reply]

Please take a look at WP:TPG. "The purpose of a Wikipedia talk page is to provide space for editors to discuss changes to its associated article or project page. Article talk pages should not be used by editors as platforms for their personal views." Note the "discuss" bit. Talk pages are not a forum to permanently display some not-even-fringe views and their refutation. The discussion is still available to everybody in the archives, of course. That's not an unreasonable burden. I'll also point out both that there is no "discussion" that is being limited, and the greenhouse effect itself is in no way controversial. --Stephan Schulz (talk) 13:01, 25 January 2009 (UTC)[reply]

While the effect isn't so controversial, the way it is described is. The underlying problem is that the online literature also suffers from the same problem. When authors/editors try to simplify their descriptions, they tend to lose balance and necessary nuance. blackcloak (talk) 07:18, 5 March 2009 (UTC)[reply]

..and to make the archives more accessible for "not so expert" users and others, I've just added a search box to the top of the page. --Stephan Schulz (talk) 17:26, 28 January 2009 (UTC)[reply]

The WP:TPG guide , it makes it clear you should not be using your personal opinion as to what it is or isn't proper to archive. You are making it very clear that your ideas are the ones that will dominate, I intend to contest this as improper behaviour.--Damorbel (talk) 20:33, 25 January 2009 (UTC)[reply]

I beg to differ. "When a talk page has become too large or a particular subject is no longer being discussed, do not delete the content — archive it." But do whatever you see as fit, of course. --Stephan Schulz (talk) 20:48, 25 January 2009 (UTC)[reply]

Since I see no justification for archiving on the opinion of an editor of the state of a discussion or if a few months is sufficient for a file to be archived, rather than the size of the file of this talk page. I must ask you if you are willing to accept some form of arbitration on the matter.--Damorbel (talk) 20:54, 25 January 2009 (UTC)[reply]

Given that I never archived anything, that is a strange request. But WP:DR is available to you, of course. --Stephan Schulz (talk) 21:24, 25 January 2009 (UTC)[reply]

If you have "never archived anything" then you probably have never felt the need to, and I imagine you are familiar with discussion files that are quite large. However, as I suspected, you are not concerned about archiving at all but are much more concerned with your POV that "the greenhouse effect itself is in no way controversial". This is so funny, you cannot have followed the discussion, in [this section] you will find some strong views on the quality of GHE physics, how it requires energy transfer from a cold troposphere to a warm surface to warm it yet further and you maintain that this is not controversial? I think you once before asked me to provide evidence for the 2nd Law of thermodynamics, a matter about which you do not seem well informed.

So Stephan Schulz why are you, not having archived anything, writing so much about it? Is this not another form of Wiki abuse, defending [Kim Petersen] perhaps, someone who reverts my edits without good reason. Both of you appear to me as [vested contributors] determine to keep any questions about matters that you see as "not controversial" out of Wikipedia by any means to hand.

Blackbody

Just to keep your "not controversial" POV on the boil; all predictions of the equilibrium temperature of the Earth give about 254K, but then they freely admit that this is assuming that Earth emits "like blackbody". This is quite impossible, most of the heat radiation leaving the Earth comes from H2O & CO2, neither of which emit with anything approaching blackbody efficiency, the spectrum alone should tell you that; these gases also emit high in the troposphere at a temperature far below the surface temperature which in real physics means - much less efficient than at the warmer surface! This is what I mean when I say the GHE physics, as portrayed in the this and the GHE article itself, is seriously oversimplified and grossly mistaken.--Damorbel (talk) 22:17, 25 January 2009 (UTC)[reply]

Sorry, but if you are so scientifically illiterate that you cannot distinguish between net energy transfer and transfer of radiation, then you probably should not edit this article. I will not restart this discussion unless you find a reliable source (i.e. a peer-reviewed paper in a recognized scientific journal or a decent quality textbook) that denies the existence of the greenhouse effect, or even one that claims that the second law of thermodynamics makes it impossible. There are people that claim that the earth is flat, and who make passionate arguments about it. That does not mean that the spherical earth is controversial. And given that I have a grand total of 4 (four) edits to this article, two of which fix simple vandalism, calling me a "vested contributor" seems to be a bit far-fetched. --Stephan Schulz (talk) 23:13, 25 January 2009 (UTC)[reply]

Stephan Schulz, what do you mean by "net energy transfer"? Is there any other kind? Radiation is an electromagnetic field, the 2nd para. of the link is "EM radiation carries energy and momentum, which may be imparted when it interacts with matter" the key phrase being may be imparted, the may refers to a number of conditions, 1) the matter must not reflect the radiation and 2)the matter must be cooler than the radiation source if it is to have its temperature raised. The first condition is covered by [Fresnel equations] and [Reflection (physics)] and the 2nd http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html Stefan-Boltzmann Law which describes the heat tranfer by radiation as being proportional to the difference of the 4th power of the temperatures of source and sink. William Connolley claimed his toy model explained this (I have tried to copy this but since it is archived I can only get the text) here "Its easiest to see this with a toy model. Assume we have solar radiation at S1=(1-a)S/4 per sq m (S/4 cos of the area of a sphere vs cross section, "a" is the albedo). Assume the planet radiates LW at rT^4. Assume a layer of atmos of temp U transparent to SW and of emissivity 0<=e<=1. Then the radiaton downwards at the sfc is S1 + erU^4. The radiation up is rT^4; hence rT^4=S1+erU^4. The radiative balance of the layer gives you 2erU^4=erT^4 hence 2U^4=T^4. Hence T^4=S1/(r(1-e/2)) if I haven't got my numbers wrong. Draw a picture, it will make more sense William M. Connolley (talk) 19:26, 1 May 2008 (UTC)" Here he makes the mistake made by all those who say GHE is "not controversial", his eqation rT^4=S1+erU^4 sums the radiation from the Sun and the troposphere, this ignores the fact that the temperature of the troposphere is lower than the surface so it cannot contribute to the warming of the surface, the surface is only cooled by radiation to the CO2 in the troposphere, thus the GHE doesn't exist and, to answer William Connolley's question "if I haven't got my numbers wrong", - he has!

It is of course quite possible that you didn't follow this discussion because it has been hidden in the [Greenhouse effect/Archive 2] (2 archives ago) made on 29 Oct 2008. At the time of the discussion (12 Oct.) William Connolley demanded I stick to his Radiation budget idea, which ignores temperature difference, so it is clear that he is unaware of the "temperature difference" aspect of heat transfer, but I don't feel obliged to convince you of the imperative of a heat source being at a higher temperature than the sink for heat transfer to take place; if you are convinced that it is just the idea of a "scientifically illiterate" person then there isn't a lot I can do for you. If this is the quality of thought common among Wiki editors, then there is not much to be expected from Wikipedia. --Damorbel (talk) 09:58, 26 January 2009 (UTC)[reply]

Sorry, but that is plain wrong. By your argument, I would only be able to ever see things when it's above 100 degrees Fahrenheit. Whenever a photon flies from a source a and hits a target b, it imparts some energy coming from a to b. The second law of thermodynamics is a statistical emergent property. If b is warmer, much more photons (and those of higher energy) fly from b to a than vice versa. That's why, while there is a transfer of energy from a to be, the net transfer of energy is from b to a. Will is right and you are wrong. --Stephan Schulz (talk) 11:14, 26 January 2009 (UTC)[reply]

Regarding Damorbel's two conditions in his comment about whether or not energy and momentum can be transferred by radiation, his #1 is wrong and his #2 is correct. Radiation does interact with matter, even when the the dominant part is reflection. Energy and momentum are transferred, just think of the Doppler effect. There are simple experiments that show this. As for #2, Damorbel is fairly careful to say a temperature rise can only occur if energy is added to the receiving body, which statistically only occurs when the source body is warmer than the receiving body. If we're talking about the atmosphere and the Earth's surface, we're basically considering the case when it is nighttime (to keep things simple). Damorbel does err when he says "... but I don't feel obliged to convince you of the imperative of a heat source being at a higher temperature than the sink for heat transfer to take place." Energy (implied by the vernacular heat) does take place (in both directions), but had he said "net heat transfer" he would have been correct. blackcloak (talk) 19:19, 11 March 2009 (UTC)[reply]

What you are talking about is Thermography which detects thermal radiation, this is done for objects down to >0K but it is sometimes necessary to cool the sensor and perhaps the whole device to avoid too much interfering radiation from hotter objects near by that are "glowing" much more brightly at, shall we say, 10K? Lots of devices detect IR, see some types here; the [microbolometer] is particularly interesting even when not cooled, it is easy to offset for observing cold objects. None of this is any reason to imagine that the tropospheric CO2 radiating at 254K will warm the surface that is already at a temperature above 254K. Like most greenhouse afficionados (William Connolley also, as I pointed out to him) you are confusing radiation and heat transfer. This is what the Stefan-Boltzmann Law saves you from doing, it separates out the radiation, which is always detectable, and heat energy transfer, which is another matter altogether since heat is transferred only from hot to cold. Here of course the second law of thermodynamics applies, if you think it doesn't, then what you are proposing, like many before you, is neither more nor less than Perpetual motion. I suggest you chew that one over. You may have something but I doubt it. --Damorbel (talk) 14:51, 26 January 2009 (UTC)[reply]

You keep deviating. Neither Thermography not bolometers are relevant here. There is an energy transfer associated with any form of radiation. A photon emitted by a hotter body is indistinguishable from one emitted by a colder body at the same frequency. Both carry the energy E=hf, and, if absorbed by another body, will raise its temperature by this amount. Again, the second law is a statistical law. Energy is transferred via radiation from a hotter to a colder body because the hot body emits, on average, more energy in the form of photons, not because the colder one emits no photons. In the case of the greenhouse effect, there is no net energy transfer from the cold troposphere to the warmer earth. But there is a radiative transfer that reduces the net transfer in the other direction. Why or how would a photon emitted by a greenhouse gas molecule in the troposphere and hitting the ground not warm it? Where is the energy going? --Stephan Schulz (talk) 15:39, 26 January 2009 (UTC)[reply]

You said "I would only be able to ever see things when it's above 100 degrees Fahrenheit" Thermography and bolometers are ways of detecting radiation, just as your eyes are. The greenhouse effect is about increasing the temperature of the earth's surface, differeent physics are needed to explain it.

Photons emitted by a black body have a Maxwell–Boltzmann energy distribution, just like the particles in a gas; some have energy above the average; the temperature is determined by the average translational energy, the "hotter" photons from the cool body are fewer in number than those from the warm body thus the hot body cools and the cold one is warmed up; heat flows from hot to cold!

I think that one of the confusing claims for the Greenhouse effect is that it says the radiation balance temperature would be 254K "for the surface". But the radiation from the Earth is about 60% from the troposphere at, shall we say 250K? It is entirely necessary if the energy budget is to mean anything that it is made clear which surfaces are the radiating ones.

Another confusing thing is the explanation from the IPCC. The IPCC has this diagram in its explanation of the physics of global warming, if you take the difference between the (390-40)=350W/m^2 of surface radiation absorbed in the troposphere and the 324 W/m^2 "back" radiation you will see thst the IPCC thinks radiation is responsible for only a "net" heat transfer of 26W/m^2 only 2W/m^2 more than the 24W/m^2 of the thermals and much less than the 78W/m^2 of water evaporation and 40W/m^2 through the spectrum "window". So even the IPCC, who believe the "back radiation" warms the Earth 33°C, do not make a very good case for it since this 26W/m^2 includes the water vapour component also.

You have the numbers wrong. The diagram should be read as, instead of loosing 350W/m^2, the surface is only loosing a net of 26W/m^2. For the blackbody temperature, consider a body receiving 168W/m^2 (-40°C) vs one receiving (168+324)= 492W/m^2 (32°C, 90°F). (168+67) = 235W/m^2 provides -19°C, 390W/m^2 provides 15°C. Q Science (talk) 20:34, 28 January 2009 (UTC)[reply]

By the way the diagram in the article doesn't include any estimate of the evaporation, convection and thermal components of atmospheric heating. I think the IPCC diagram is much more clear on the deficiencies of the GHE and should replace the current one.--Damorbel (talk) 17:53, 28 January 2009 (UTC)[reply]

Damorbel, why do you keep insisting that the atmosphere is 254K? Anyone with a thermometer will tell you that it is warmer than that *near the surface*, and about 190K near the tropopause. As I pointed out above (2 sections up), in the mornings, the surface of the planet is typically cooler than the atmosphere a few hundred feet above it. In fact, if it wasn't for the heat coming from the atmosphere, the surface would be even colder. Q Science (talk) 17:48, 26 January 2009 (UTC)[reply]

254K (-19°C) is the temperature identified by some promoters of the Greenhouse effect as the average temperature the Earth would have in the absence of the Greenhouse effect.--Damorbel (talk) 15:21, 28 January 2009 (UTC)[reply]

The key word is "average". If you want to understand how things work, then actual temperatures are more important. Everyone knows that temperature distribution is far more important than the average temperature. For instance, if the tropics cool a lot (because of more clouds) and the poles warm (because of changes in circulation), then it is possible for the ice caps to melt even though the global average temperature decreases. That is why I cringe every time I see that year A is hotter than year B. It is nothing but a lot of political hype (propaganda). Valid data always contains a distribution (such as Global Warming Map.jpg) and error bars. Q Science (talk) 18:56, 28 January 2009 (UTC)[reply]

Q Science, I tend to agree with you. I suggest the distribution of temperature is of limited value when trying to determine if anything man made is happening. No doubt you made the point about the ice caps melting for effect; the real point is, they receive so little solar radiation [1], why aren't they a great deal colder than they are? The answer is fairly simple, the sea (4000m deep under the North pole) and the atmosphere bring heat from the tropics. The effect is quite general, I think it is about lattitude 45 that the incoming radiation from the sun equals the outgoing Earth radiation. If the Earth had no tilt the poles would get almost zero sunlight and, without sea and atmospheric heat, only conduction from the interior would keep them above 3K (CMB). I don't (yet) understand your map link.

I cite 254K for the "temperature without greenhouse gases" because it is used as a political tool to justify the whole AGW industry. Although used (or a close figure) by every AGW aficionado, it is completely spurious because it is derived on the asumption that Earth radiates "as a black body", this is patently absurd, even in the infrared, just look at the spectrum of the evil GHGs. [2] --Damorbel (talk) 09:00, 29 January 2009 (UTC)[reply]

It's not patently absurd. The Earth, minus its atmosphere, would radiate like a black body, where the equation describing the phenomena has the emittance factor (with suitable wavelength and angle dependencies). See the Military Handbook of Infrared Technology for a good starting point. Observing the spectral content of the power leaving the Earth and passing through the atmosphere results in a wavelength dependent modification to the surface-only flux. If you want me to react to your comment in a more direct fashion, you're going to have to explain what aspect of emitted radiation doesn't follow well understood behavior, in your opinion. blackcloak (talk) 07:35, 5 March 2009 (UTC)[reply]

Blackcloak, if you gave me a link to "the Military Handbook of Infrared Technology" I would like to see it, but let us not resort to authority, military or otherwise. A blackbody is a concept devised by Kirchhoff to describe a surface that absorbs all radiation falling on it (heating it); it also emits according to its temperature. The Earth doesn't absorb all the radiation landing on it so it isn't a black body. What do you mean "well understood behavior"? Do you mean that the radiation of the Earth does not come from the atmosphere with a spectum rater like this [[3]] ? This IR spectrum is far from black. Kirchhoff's blackbody model is the most efficient radiator possible, the only restriction to its efficiency is that of a theorertical emitter/absorber of Electromagnetic radiation, it doesn't take into account the limitations that the Fresnel equations impose. Check this link and you will see that, because of total internal reflection, any material with a Refractive index greater than one does not allow 100% of radiation to pass to a material with a lower RI. No material has an RI of one and so a fraction radiation is stopped at the interfaces of liquids and solids with gas or vacuum. I suggest your "well understood behaviour" is just a popular myth (Myth - "A widespread but untrue or erroneous story or belief".)

Here is the ref, you can buy it through NTIS: /Handbook of Military Infrared Technology. Michigan Univ., Ann Arbor. ProductType: Technical report NTIS Order Number: AD-646 082 Please call 1-800-553-6847 or 703-605-6000 to order (NTIS Order Number AD-646 082 ). Please select a media type! Media Count: N/A Date: 1965 Author: W. L. Wolfe/ It's easy to google. Regarding your comments, the bb wiki article is one of those over simplifed versions that does not properly reflect the true complexity of the underlying material, which you'll begin to understand after you read some of the more indepth sources. One of your basic points is correct, although nothing more than a red herring: that the (simple, wiki-like) BB equation does not carry the required complexity to properly describe real world objects, and that the simple version only describes an idealized radiator. That's why we have fudge factors. blackcloak (talk) 23:05, 5 March 2009 (UTC)[reply]

Continuing on a bit. Let's first see if we agree on some of the basics. Do you agree that the diagram [[4]] you reference in some respects is a gross misrepresentation of the actual bb curves having different peak emission wavelengths? Specifically, do you agree that a bb curve of a perfect emitter at some temperature T1 everywhere in wavelength emits more radiation than a perfect emitter at some temperature T2<T1? Said more simply, should the blue bb curve be scaled so that everywhere in wavelength it rises above the black bb curve? Should the violet bb curve be scaled so that everywhere in wavelength it rises above the properly scaled blue bb curve? Should the red bb curve be scaled so that everywhere in wavelength it rises above the properly scaled violet bb curve? When this is done, will not the red curve (representative of the sun) so dominate the graph that the IR (scaled blue, black and violet bb curves) be lost within the line width of the red bb curve? The correct answer to each of the above questions is yes, so if you answer otherwise, there is no point in going on to the next step. As before, no need for windy answers. Simple yeas and nays are all I'm looking for. blackcloak (talk) 02:45, 7 March 2009 (UTC)[reply]

With respect to

Should the red bb curve be scaled so that everywhere in wavelength it rises above the properly scaled violet bb curve?

The correct answer is NO. The red curve is the energy at the top of the atmosphere. Therefore, it is the energy from the Sun divided by R^2. Actually, it is a bit more complicated because the Sun is not a point source. There is also the problem that the energy in a photon depends on its wave length. Q Science (talk) 07:58, 22 May 2009 (UTC)[reply]

There are two possible explanations for your comment: 1) you misunderstood my comment or you failed to understand the referenced (in the article) diagram or, 2) you do not understand the underlying physics. Since I can not determine the proper direction to head, I will ask you to 1) review carefully the diagram and my comment, and 2) study the curves (in the two figures) on page 18 of the Handbook of Military Infrared Technology. I (attempted to) put into words what those figures show graphically. If someone like you is having difficulty sorting through and/or identifying improperly presented material, I think it should be obvious to everyone that the complexity of this body of knowledge is beyond the casual/occasional wiki editor. (And to me that means that there needs to be a place within (at least) certain wiki articles for the exchange of knowledge that is NOT directly related to the improvement of the wording of the article, but is geared to the improvement of the knowledge of potential editors. And BTW, no editor on the wiki GW topics fully meets my expectations of accuracy and understanding. I don't consider myself an expert on this topic, and I have a whole bunch of questions for which I have not found/seen an answer.) blackcloak (talk) 19:02, 23 May 2009 (UTC)[reply]

For 2 metal plates 10 feet away, the radiation curve for a 5,500K plate is always greater than the curve for a 288K (15C) plate. However, if the 5,500K plate is moved 93 million miles away, and it is increased in size to match the area of the solar disk, then the energy recorded with a bolometer is approximately 4 times what is recorded for the 288K plate only 10 feet away. (The factor of 4 is because the Earth rotates.) The graph you are referring to is an attempt to compare the energy from the sun to the energy emitted from the earth as measured from an instrument in orbit around the earth. Therefore, the area under the solar curve should be equal to the area under the 288K curve. I agree that the 3 longwave curves are totally misleading as drawn. (They should never overlap.) In addition, the shape and scale of the red curve is probably very wrong. (This is where the equations get weird.) However, the red (solar) curve should NOT be drawn higher than the other (colder) curves because the distance to the 2 sources is different. You should attempt to use the blackbody equation to plot the curves yourself (I have) instead of trusting a book. There are several interesting issues that none of the sources I've read really cover. Q Science (talk) 20:39, 23 May 2009 (UTC)[reply]

It's now clear to me that you understand the basic physics, and we seem to agree that the graph doesn't properly display some things. We can move to the next level. I disagree with your comment about the bb curve for the incident radiation, a curve obviously copied from the other three as it has the same amplitude and shape. Overlaying a bb curve, even if properly scaled to represent power levels at a particular point, isn't really accurate, while it may have some instructive value. The basic problem as I see it is bb curves are emission curves (usually from a surface, but if not, from/within a volume. In my understanding, it is not proper to describe a radiance pattern in space (even after integrating over 2pi steradians of solid angle) as a bb emission pattern. This is particularly apparent in the Sun/Earth case because the Sun's rays as seen to leave a plane (perpendicular to a line betwen the two centers) above the Earth's atmosphere is quite collimated. Since radiance/brightness properly describes what is happening to the nature of the changes in the light field as it reaches the Earth, those terms should be used, and graphs that show the power distribution (probably best to integrate over 2pi steradians) as the rays enter the atmosphere, pass through the atmosphere and reach the Earth's surface. There should be an entirely separate graph/s that show/s the progression of brightness/radiance for the radiation field that leaves the surface and heading out into space. Indeed there should be two graphs (at least) showing what happens to the power leaving the Earth/atmosphere, namely one corresponding to high noon, and another for the dead of night. These will be dramatically different because the nighttime one will represent the case of cooling both the atmosphere and the Earth's surface, exclusively. Also because of the distribution of power from the Sun over the spherical Earth's surface, there should be some acknowledgment that at each point on the Earth, a greater portion of the day is spent cooling the Earth (surface and atmophere) than warming the Earth/atmosphere. (Indeed, in any one day, there is a large area that only cools.) So, to narrow the discussion to the first change I would make, separate the bb curves from the radiance/irradiance curves, only describing the emission processes and put in the large difference in scale from Sun and Earth. Then in other graphs (bb curves left out, but the shape being bb-like, with absorption detail, scale factors chosen properly, with actual solid-colored areas shown equal, requiring plots vs. freq or equiv.) show how EM power moves through the atmophere. I could have said more, but this leaves you with plenty to respond to, should you choose to do so. blackcloak (talk) 04:52, 27 May 2009 (UTC)[reply]

I agree that the curves in [[5]] are not perfect but what they make very clear is that radiation from gases does not even approximate BB (the curves are for transmission, but the general shapes are close to the H20/CO2 radiation curves). It is clear that gases radiate much less efficiently than a BB thus they will need to be much hotter to radiate equal power.

I referenced the Wiki article on BB because it was handy, however the science of radiation is available elswhere, you can find a translation of Kirchhoff's original papers here [6] you can choose the format, PDF etc., that suits you. Planck's works are still in publication, you can browse them in Google Book here [7]

In his 1916 paper "Strahlungs-Emission und - Absorption nach der Quantentheorie." (Emission and absorption of Radiation according to Quantum Theory"), Einstein acknowledges his debt to Planck (and thus Kirchhoff) for giving him the fundamental ideas that lead to his quantum theory. Among those ideas a very important one is that, in the absence of any other energy source (or sink), i.e. energy equilibrium, the absorption and emission coefficients of any material object are the same. Now since the Earth is quite clearly not black (because it reflects - has an albedo) it cannot be "black" in the radiation sense either. In Kirchhoff's work you will find that he explicitly states that wavelength has no effect. It is not that the Earth isn't black, it is the ratio of emissivity to absorptivity that must be unity, should by some miracle the Earth radiate like a BB then it would absorb like a BB also - it would also be rather difficult to see!--Damorbel (talk) 16:21, 7 March 2009 (UTC)[reply]

Since you did not answer yes to any question, I must presume that you do not understand the point I was making by asking the questions. Saying the curves are not perfect is meaningless. I must therefore conclude that you do not understand one key underlying characteristic of bb radiation. Until this situation is corrected, there is no point in going on to more difficult material. Quite the contrary, we have to fall back to simpler material. You'll have to let me know if you want me to do that. Let me just ask now if you've ever solved heat flow problems using the appropriate partial differential equation. blackcloak (talk) 07:17, 8 March 2009 (UTC)[reply]

Archiving II

I see the phantom archiver is back! Poor WIKI! Does nobody respect its Help:Archiving a talk page these days? You Greenhouse Effect editors seem terrified that the inconsistencies in the "science" of the global warming hypothesis shall ever see the light of day!--Damorbel (talk) 13:57, 5 March 2009 (UTC)[reply]

Blackbody II

I don't see what the big scientific issue is, though I see what the ideological one is. Yes, Earth is not a perfect black body, though it is a good 1st order approximation, as pointed out by blackcloak above. No, an academic contest of who can be more pedantic won't help. I'm going to try to address some issues introduced above, and suggest a re-set to the scientific discussion, unless the ideological gap is determining the arguments, in which case this will likely be my last post here.

In my opinion, 254K isn't a hoax to make the GHG's seem bad. It's an approximation of an atmosphere-free planet which, to me, it makes GHG's seem really good, because we'd be freezing without an atmosphere.
The poles would not be only heated from the geotherm if there were no tilt. The axial tilt actually changes over time, but doesn't mean that every relevant Milankovich cycle, the poles get down to the -100's degrees C without solar contribution. Atmospheric circulation carries heat from the equators to the poles, and would regardless of the tilt.

I just don't get why we're arguing about a 1st order approximation versus a more precise measurement... and what we're trying to get at by doing so. Awickert (talk) 03:58, 7 March 2009 (UTC)[reply]

I'm taking the time to do this because Damorbel is expressing certain ideas that are clearly wrong. But because he is being civil and persistent, and because he may be teachable, he deserves to understand where the underlying differences are, if not to reach a better understanding of the science. Now, two points to consider for you: 1) Without GHG there would be no life (as we know it), so there'd be no we to be freezing. 2) Rather than the average temperature, which would be decidedly chilly by our standards, the daily excursion of temperature, at least on the equator, would be pretty extreme. You might like to calculate it. About 50% more power would reach the surface directly, and then the peak power would be four times the average (as used in all the long term average calculation, as above). Figuring out what the actual daily excursion of temperature would be would require a good understanding of average thermal conductivity. And then there's the question of wind and convection. Finally, if you don't want to argue or be challenged by these kinds of things, go find something else to read. I landed here; you may well end up elsewhere. blackcloak (talk) 05:33, 7 March 2009 (UTC)[reply]

Whoa, okay, sorry. So what I was trying to do is to mix up a conversation that didn't look like it was going much of anywhere. I overall agree with you and disagree with Damorbel, and wanted to try try to point at some things listed above to reach some common ground and improve the specificness of the discussion in order to finally improve the article in some way. I agree with you on the above blackbody / exact spectrum comment. We say more or less the exact same thing in our #1, so we agree again. I also agree with your #2, if I interpret correctly the large equator-pole temperature gradient to be related to no atmosphere, as opposed to the non-GHG ocean/atmosphere system you mention later, which would be extremely efficient in transporting heat because of the larger pole-equator temperature gradients which would set up some pretty big convective instabilities. I bet in the no atmosphere case, it wouldn't be too hard to assume a 1E-6 m^2/s thermal diffusivity of rock and calculate the effects of lateral variations with a balance of incoming and outgoing solar radiation as a function of latitude. Don't feel like doing it just for a talk page, though. : )

The main idea is that I feel that you two are arguing a 1st-order approximation vs. a more precise approximation, but somehow Damorbel thinks that the more precise approximation reverses the result, which totally makes no sense at all.

Awickert (talk) 06:43, 7 March 2009 (UTC)[reply]

Awickert, whether it's a big scientific issue or not depends on your attitude. The current science of the Greenhouse Effect usees this BB assumption to build all the horrible predictions about the fate of the planet. Central to this is that GHGs cause the surface temperature to be 33°C higher than the 254K predicted by the BB assumption. If you abandon the BB assumption you get a much more acceptable 282K without the need for GHGs; sad to say, the whole Greenhouse Effect is just a mistake based on a false assumption, like so much other "old hat" science!--Damorbel (talk) 16:37, 7 March 2009 (UTC)[reply]

This is beyond ridiculous. We have direct measurements of upwelling and downwelling longwave radiation that show the absurdity of your argument. Short Brigade Harvester Boris (talk) 16:46, 7 March 2009 (UTC)[reply]

Harvester, I'm not sure what your measurements are about. The absorption and emission of EMR takes place because some molecules have a dipole moment that converts EMR to mechanical force whereby absorbed EMR heats up the molecule. If the EMR is too weak or the wrong frequency, the dipole in the molecule causes radiation to be produced (emitted) the loss of energy (to radiation) cools the molecule. This is why absorption and emission are symmetrical. Even when no tranfer of energy is taking place the dipole produces a radiation field, the molecule is in the middle of the radiation field produced by many other molecules; all these molecules have energy according to the Maxwell-Boltzmann distribution; in equlibrium there is, on average, no exchange of energy between the molecules--Damorbel (talk) 18:39, 7 March 2009 (UTC)[reply]

So what I think that you're saying is that the greenhouse effect is completely unrelated to surface temperature. So therefore, if I ignore atmospheric absorption (by definition, greenhouse gas effects), you are saying that preferential absorption and emission from the solid Earth would result in a 33 degree C warming. So why do you think that the emission from rock and water would be greater than the absorption by rock and water to actually increase the temperature this much? The Greenhouse has a great proven mechanism of molecules preferentialy absorbing outgoing long-wave radiation. Your model would have to compensate for this loss by increased input and reduced output such that steady-state would have a 33 degree C higher temperature than the blackbody assumption. Awickert (talk) 17:04, 7 March 2009 (UTC)[reply]

What do I say about the Greenhouse Effect? That putting the Earth's equilibrium temperature at 254K is not correct because that assumes the Earth radiates as a black body which has the maximum efficiency theoretically possible. In real life not much happens with 100% efficiency, so the thinking individual should be a bit suspicious of a claim so high. Kirchhoff's law of thermal radiation states that emissivity and absorptivity are equal, the explanation I gave to Harvester (above) is rather more plausible than the black body assumption needed to predict an equilibrium of 254K, I recommend it to you. If you accept my explanation then there is no longer a need for the Greenhouse effect with it's bizarre requirement for heat to flow from a cold troposphere to a warm surface, a requirement that breeches the second law of thermodynamics. --Damorbel (talk) 19:02, 7 March 2009 (UTC)[reply]

You don't answer my question. By repeating your earlier statements, I am therefore lead to believe that you think that rock and water are better absorbers than emitters of radiation such that the steady-state temperature would be higher. From this, it seems that you believe that therefore, Earth would be hot with no heat-trapping atmosphere, which seems unlikely given a Venus/Mars comparison, in spite of their comparative distances from the Sun.

To what you say: heat does not flow from low temperatures to high temperatures, true. The outer atmosphere is cooler than the inner atmosphere, true. However, the heat within the atmosphere reduces the temperature gradient between the surface and outer space. This reduces both diffusion of heat and the magnitude of convective instabilities. As an analog to the diffusion, it is like lying under a blanket. Although the temperature outside the blanket is colder than the blanket, which is colder than you, because the blanket has a low thermal conductivity and (for the analog) reduces temperature gradients, you (the heat producer, analog to the Earth surface heated by solar radiation and then losing that heat) are able to maintain a higher steady-state temperature.

The Second law of thermodynamics wears green breeches, in fact. I picked them out myself.

Awickert (talk) 19:35, 7 March 2009 (UTC)[reply]

The tranferring of thermal energy via an IR process is different from what happens in the transfer of thermal energy via conduction. You're using the terms for conduction and you should not be. If you could get over this bit of confusion, you might be able to take the next step. One key point: IR arriving at the Earth's surface may have originated from one of two "sources," the sun or the warm atmosphere. The molecules of the surface can not distinguish between the quanta arriving from the sun vs. from the atmosphere. Those molecules can not therefore decide which ones to absorb and which to "reject" (reflect?). Therefore they will absorb the same percentage of the quanta arriving from the atmosphere as the quanta arriving from the sun, at any given wavelength. Each absorbed quanta of a given wavelength has a corresponding energy (to repeat, independent of the source of that quanta), and results in a corresponding increase in the amount of thermal energy within the aborbing material (notice I did NOT say that there is a corresponding increase in temperature because the outflow-emission- may exceed the inflow, in which case the material is actually cooling). blackcloak (talk) 07:45, 8 March 2009 (UTC)[reply]

I didn't answer your question? The explanation I gave is sufficient for an understanding of absorption and emission which is what the Greenhouse Effect is mostly about and it is where the mistaken assumptions are. Try rephrasing your question and I'll try to answer it. I have read what you said about rocks etc. and I don't see how its relevant. What you said about the Second law of thermodynamics is not very serious, we all like a bit of humour. Perhaps I'm dull but "wears green breeches etc." is just stupid; stuff like that is generally written by vandals.--Damorbel (talk) 20:50, 7 March 2009 (UTC)[reply]

Damorbel, it is not sufficient. The reason it may vary from a blackbody would be because of the non-ideal properties of the materials, as you say. I ask why this would cause a higher steady-state temperature than a blackbody: in particular, for this to happen, the absorption would have to be more efficient than the emission. You do not acknowledge my counterargument about heat transport, which is central to the importance of greenhouse gases. If you respond, please also do so. The "green breeches" was an play on words based on what you said, "a requirement that breeches the second law of thermodynamics". Lighten up! Awickert (talk) 21:02, 7 March 2009 (UTC)[reply]

Awickert, "The reason it may vary from a blackbody" is not because it is a less than perfect blackbody (which is quite sufficient reason to blow the assumption out of the water and the GHE with it) but because it is real "stuff", gas, liquid or solid; none of these can ever match the emissive properties of the hypothetical blackbody, assuming they do just gives the wrong answer that is wildly incorrect(254K) when calculating the Earth's equilibrium temperature. Those who blab on about global catastrophies etc. using this as an argument are really quite dangerous in their ignorance.

Thanks a million for spotting my breach of the spelling rules, perhaps I was deaming of playing a breeches role!--Damorbel (talk) 10:43, 8 March 2009 (UTC)[reply]

What are you arguing about here? There is just so much text. FWIW, the Earth's surface is pretty well a block body in the IR (though not in the visible, obviously). And very little solar energy comes in IR William M. Connolley (talk) 09:13, 8 March 2009 (UTC)[reply]

The proper way to say this is: A very small portion of the total solar energy entering the Earth's atmosphere is in the infrared. But this too may not be sufficiently easy to understand unambiguously since the wavelength above which the radiation is not considered infrared is not specified. So, using figure 2-6 pg 19 of the Handbook of Military Infrared Technology, I (think I)learn that setting the trip point wavelength at 1 micron, using 5700 deg K, about 65% to 70% of the power exiting the sun is of a shorter wavelength. I take this to mean that approx 30% of the solar energy arriving at Earth's atmosphere is in the infrared (wavelength longer than one micron). In this area of the spectrum, the percentage moves rapidly as a function of trip point wavelength. I quickly estimate the following two additional data points (using the figure). At the 2 micron trip point, the infrared content is approx. 6-7%. At the .78 micron trip point (roughly the furthest into the red that the human eye can see) the infrared energy content is about 45%. blackcloak (talk) 21:51, 8 March 2009 (UTC)[reply]

William, short and simple: the assumption that the Earth radiates like a blackbody is needed to get its equilibrium temperature to be 254K, all agree on that. The blackbody assumption is untenable because the mechanism that causes the albedo (reflection) is the same mechanism that stops the Earth radiating in any part of the spectrum with the efficiency of Kirchhoff's blackbody. This is not to say the spectrum doesn't resemble a blackbody spectrum, let us say in the IR (do you think the peaks and valleys of CO2 and H2O in the IR resemble a blackbody?).--Damorbel (talk) 10:43, 8 March 2009 (UTC)[reply]

I'm just trying to throw a wrench in the gears by getting Damorbel to say why he thinks that a GHG or (for that matter) atmosphere-free Earth would be ~~warmer than~~ as warm as the Earth with the atmosphere, and hopefully stimulate some thinking to end the discussion and possibly even (maybe) improve the article. If it doesn't look like any improvement will come, I'd say close the discussion and move to user talk pages if anyone wishes to continue, as per WP:NOTFORUM. Awickert (talk) 09:31, 8 March 2009 (UTC)[reply]

Awickert, you are going too fast, i.e. changing the subject. Lets just stick to blackbodies and reality, two different matters, just as different is heat balance in the atmosphere. As you note this is not a forum. The second paragraph of Greenhouse Effect states "In the absence of the greenhouse effect and an atmosphere, the Earth's average surface temperature[7] of 14 °C (57 °F) could be as low as −18 °C (−0.4 °F), the black body temperature of the Earth". This is incorrect. --Damorbel (talk) 10:57, 8 March 2009 (UTC)[reply]

OK - excellent. So it sounds like we should try to make the statement more correct than the simple "black body" approximation. To do so, we'd need to couple the [8] incoming solar spectrum outside the atmosphere with an approximation of the outgoing spectrum from the Earth if it had no GHG's in the atmosphere. As far as I know, there aren't measurements of this, because we do have an atmosphere which hugely perturbs the signal; do you know of any studies or approximations that have been made as to the radiation spectrum given off by the surface of Earth? Awickert (talk) 16:40, 8 March 2009 (UTC)[reply]

Awickert, it's a lot simpler than that. Kirchhoff's law of thermal radiation states that, for an isolated body, the absorptivity and emissivity are equal, wavelength independent etc. (check with my answer to Harvester below). This means that the equilibrium temperature of a symetrical body like a rotating planet, is dependent only on the distance from the heat source, in our case the Sun. This would still be true if the planet was made out of glass.--Damorbel (talk) 19:57, 8 March 2009 (UTC)[reply]

Sorry - you miss my point, possibly my fault, so I'll try to be more clear. What I'm trying to say is a little more specific. We know the spectrum of incoming solar radiation and can therefore calculate the incoming power flux from the sun. We know that the Earth is more correctly a gray body than a black one, but we also know that even without the idealized blackbody radiation curve, the radiation as a function of wavelength should be given by some empirical function (like in the measured solar example above) if we knew what it was. So imagine we can replace the ideal with the real. Then we should be able to calculate the temperature in a gray body balance.

I'll let you and Boris chew through your bottom comments and instead use a simplified thought experiment on albedo to try to explain why I think surface composition is important. We observe high-albedo objects to be colder than low-albedo objects. My favorite example is that the black T-shirt has a higher equilibrium temperature, where heat in equals heat out, than the white T-shirt. That means that if the incoming and outgoing radiation is in a balance, with equal absorptivity and emissivity (let's assume no variations with wavelength), the higher albedo object is cooler. It is absorbing and emitting less, and therefore has a lower equilibrium temperature, showing that therefore the composition of the surface is important for a radiation balance.

So my thought from above is that if we could find quantified somewhere the full absorption/emission spectrum for the surface of the Earth, then we could do the gray body balance to figure out what the true equilibrium temperature would be in absence of a greenhouse (and also just using today's values, and ignoring Earth's lovely dynamic tendancy to sometimes go hothouse or snowball and have glacial cycles and other changes in albedo). Awickert (talk) 20:33, 8 March 2009 (UTC)[reply]

Let's examine Damorbel's contention that a glass earth would reach the same temperature as our own earth, which has an average absorption coefficient somewhere in the neighborhod of .9 (pure guess). I'll further simplify things by removing the atmosphere and making the surface perfectly smooth. Now glass has an index of refraction of about 1.5 which would be responsible for some loss of energy to space by Fresnel losses. Damorbel has pointed out this loss mechanism before, so he should accept this. But as the index is increased to some higher value we should expect greater losses to space (effectively the albedo increases), and further, as we increase the index evermore, we would expect a monotonic increase in the reflected/albedo power. So taking this to the limit, the index going to infinity, no power is actually absorbed by the glass planet, and the planet can not heat up. Since everything went monotonically to this condition, all intermediate values for the index must produce an intermediate equilibrium temperature. So by this logic I don't think Damorbel's glass planet contention can be correct. blackcloak (talk) 06:06, 10 March 2009 (UTC)[reply]

From what little sense I can make of this looooooooooooooooong and tortured thread, Damorbel misunderstands Kirchoff's law. K's law sez absorptivity equals emissivity at a given wavelength and Damorbel is instead equating shortwave absorptivity to longwave emissivity (different wavelengths). But there's so much circuitous argumentation that it's hard to divine what the point is. Short Brigade Harvester Boris (talk) 18:44, 8 March 2009 (UTC)[reply]

Harvester, you may complain about the length of this discussion but had you followed it you would know that Kirchhoff's law of thermal radiation is not as you describe, I gave a link [9] to where you can find a translation of his paper. If you open this link you can find it in 4 or 5 different formats, the PDF is a image of a book containing English translations of a number of important works in thermal radiation. If you take the PDF (20MB!) you will find on p89 ff that Kirchhoff explains why wavelength is irrelevant. You would realise this for yourself if you thought a little; absorption and emission of IR take place on a molecular/atomic scale, much shorter than light waves. The optical effects of colour sensitivity observed are due to molecular resonance or optical interference, not the processes of absorption and emission. And don't forget, both Planck and Einstein chewed over Kirchhoff's work before it was codified into Quantum Mechanics. Well perhaps they got it wrong, but lets save that for another thread. --Damorbel (talk) 19:42, 8 March 2009 (UTC)[reply]

Rubbish. Once again, how do you get around the fact that your supposition is contradicted by observations? The shortwave albedo of melting snow (which is what's outside my window at the moment) is around 0.7 implying absorptivity around 0.3, while its longwave emissivity is around 0.95. Short Brigade Harvester Boris (talk) 19:57, 8 March 2009 (UTC)[reply]

Sir Boris, Kirchhoff's law of thermal radiation applies to an isolated body, for isolated blocks of ice think of comets. For starters your snow is lying on ground immersed in an atmosphere; next, it is a solid and because it has a refractive index >1 (Optical properties of water and ice RI=1.3) radiation cannot escape 100% (Fresnel equations). I don't know where your expertise lies but all solids and liquids have an RI >1 (gases also but not much), it is one of the limiting considerations for the efficiency of light emitting diodes [10] With these considerations alone you have to do some serious calculations to find the relevance (or lack of it) of your observations.

You have to understand exactly what is meant by Kirchhoff's law. In addition your melting ice is not in thermal equilibrium, nor is anything else that is changing state/temperature. Think of a shiny metal sphere in space, shiny metal has a very low absorptivity/emissivity (~0.1) so its temperature would change (due to radiation) much more slowly than if it had a coating of carbon black, but the equilibrium temperature would be the same.--Damorbel (talk) 13:26, 9 March 2009 (UTC)[reply]

Most people call him Boris (or Sir). But The blackbody assumption is untenable because the mechanism that causes the albedo (reflection) is the same mechanism that stops the Earth radiating in any part of the spectrum with the efficiency of Kirchhoff's blackbody sounds very much like you think that the SW albedo has something to do with the LW (IR) albedo, and it doesn't. Snow is white in the visible and black in the infra red. The Earth's *surface* is a black body in the IR, or close enough. What happens in the atmosphere is complicated by all the spectral windows and stuff William M. Connolley (talk) 20:52, 8 March 2009 (UTC)[reply]

Yes William, water and ice absorb and radiate in IR, with various peaks in the near IR, the strongest is about 3 microns. However the spectrum is very bumpy so that alone means it can't be a BB; the BB spectrum is smooth. The other limitation is total internal reflection, the Fresnel matter I have referred to elsewhere; this traps radiation, meaning that it cannot match theoretical BB efficiency, i.e. its emissivety (absorptivity) is less than 1 for this second reason also. The second reason is independent of the first so the 2nd multiplies the 1st.--Damorbel (talk) 21:05, 9 March 2009 (UTC)[reply]

I've now read parts of the referenced 20MB doc, Kirchhoff's part starting on page 76. I'll try to bring out some of the statements in the paper relevant to the discussion thus far. Starting with the first paragraph, K says in relevant part, "A luminous body in space sends out light rays that are independent of the bodies on which they fall; similarly all heat rays which a body sends out are independent of the bodies which form its environment." (K distinguishes carefully between heat rays and light rays while acknowledging that the only difference between them is their wavelength (range). I won't for brevity make this distinction.) This means that no aspect (esp. the temperature of some surface receiving radiation) of the radiation coming from the source is governed by the surface that eventually receives (absorbed or reflected) the radiation, or by any propery of any object within the source's environment. In particular, this must mean that a warm (T1) atmosphere radiates into 4pi, half the power directed to the surface, no matter what the temperature (e.g. T2<T1) of the receiving surface is. K then goes on to carefully show that incoming and outgoing power must balance (law of equivalence of heat and work) if the surface is to be maintained at a constant temperature, and that temperature dependent non-linearities (my term) are removed from consideration if the constant temperature condition is maintained. K mentions Carnot's law to describe how we know the amount of heat that must be removed or added in order to maintain the constant temperature condition. When an object is in an enclosed box, the energy per unit time (power) emitted must equal the power absorbed. K then goes on to describe "perfectly black" bodies. He points out the necessity of matching indices, medium to body (surface), to avoid the reflection issue. K then talks about perfect reflectors, pointing out that no radiation can leave such an object. By doing the prep work, he is preparing the reader for various thought experiments, in particular providing a framework for describing the limiting cases. Next K sets up various geometries comprising various elements, including a "source" body (initially a perfectly black body, and later a body of absorptance less than 1) and an enclosure. He conjectures that "The ratio between the emissive and the absorptive power is the same for all bodies at the same temperature." By this he says, in my interpretation, that if you have any number of objects within an enclosure, all of them at a constant temperature, even when each may have a different absorption coefficient, for all cases the ratio of absorptive power to emissive power is the same. The proof for the perfectly back body is first presented using some fairly complicated, and very general, mathematics. K appears to be using a technique known as proof by contradiction, whereby he assumes two cases of differing emissive power, and shows the difference between them must be zero. K then discusses polarization and plates and different colors. This gets us to page 89 where section 12 (and section 13) starts and presumably where Damorbel is directing our attention. With this section the case for a non-perfect (emissivity<1) body begins. The proof is complicated, but follows the same general approach of the perfect black body case. The difference appears to be that one case (given wavelength and temperature) is described by an emittance and the other by an absorptance (time the ratio), the difference proved to be zero. The conclusion being, written in a modern fashion, E(λ)/A(λ) is a constant at a given temperature. Having gone through the process suggested by Damorbel, I'm wondering why he says wavelength is irrelevant. Clearly wavelength is important; K keeps the lamdas inside the integrals. Maybe someone else has more insight into this than I have. But enough for now. blackcloak (talk) 06:50, 9 March 2009 (UTC)[reply]

Blackcloak, on page 90 (top) Kirchhoff writes "only let the body C no longer be a black body, but arbitrary", this doesn't mean "E(λ)/A(λ) is a constant" as you suggest, it means E(λ)/A(λ) is not a constant, it is arbitrary. He goes on to explain on page 94 how all bodies that emit (thermally) begin to do so at the same temperatures but with very different intensities according to their absorptivities (emissivities). His explanation for the apparent exceptions is that the emission is due to another process, phosphorescence. Kirchhoff's work is quite remarkable since it is the point of departure for Quantum mechanics, if there was something odd about it you would think it would have been noticed by now. --Damorbel (talk) 15:02, 9 March 2009 (UTC)[reply]

Damorbel: Now that you're directing our attention to page 94, let me first comment on the earlier issue to which you failed to respond appropriately. K's section 15 begins on that page by saying "When any given body-a platinum wire for instance-is gradually heated, it emits, up to a certain temperature, only rays whose wave lengths are greater than that of the visible rays. At a certain temperature, rays of the wave length of the extreme red begin to be visible; as the temperature rises higher and higher, rays of a shorter and shorter wave length are added, so that for every temperature, rays of a corresponding wave length come into existence, while the intensity of the rays of greater wave lengths increase." Had you read and understood this passage you would have been able to answer in the affirmative all my questions above about the graph you referenced except for the last one. For the last question you would have to understood one other basic fact- the T^4 dependence. Now, two respond to two of your points just above. My summary stated correctly what the K paper indicates: arbitrary means choosing a value for the emissivity that is less than one. For example, the case being addressed would include an emissivity described by the following equation .5+.4*sin(λ/λο). This is a unitless quantity. And now that you've stimulated me to go back and understand what K is saying a little better, I find that his A is the modern absorptivity. K states that it is a ratio, and that it is wavelength dependent, even though he uses a confusing term to name A, namely "absorptive power." A is a function of wavelength. So it would be correct to write something like A(λ)=.5+.4*sin(λ/λο). Agreed? As for E, what K calls "emissive power" and refers to it as an intensity, the text (notation) and the diagram do not serve to clearly communicate unambiguously what the units of E are (radiance vs brightness) but whatever they are, they are the same as K's e. I think K's w is an area, which suggests that K's I has units of power per unit area per unit solid angle (brightness), and E has units of power (watts). K is clear that E is a function of wavelength. So, Damorbel, I think you are right, E(λ)/A(λ) can not properly be thought of as a constant because e(λ)=E(λ)/A(λ) is function of wavelength. But it should be understood that K's statement that you quote regarding an arbitrary black body refers to the emissivity not being exactly 1, or equivalently, the absorptance not being exactly 1, and rather that it is somewhere between 0 and 1 as in the example A(λ)=.5+.4*sin(λ/λο). Now before I drop this subject, I want to point out how K creates much of this confusion. Look at sections 2 and 3. K calls E an "emissive power," A an "absorptive power" that is the ratio of two intensities, and then in section 3 K states that the ratio of these two "powers" "is the same for all bodies at the same temperature." A ratio of powers suggests the result is unitless, but that does not appear to be the way K is thinking about it. For me to sort this all out would take time I'm not willing to spend at the moment. Finally, as for your introduction of phosphorescence, this was in the nature of an aside where K briefly mentions work by Draper, and apparently K concedes that there are some (probably very minor, in my opinion) caveats to the general rule. I would be surprised if the ratio of the phosphorescence power to the total emitted power from a body is above .0001, and therefore irrelevant, especially since not many materials exhibit this property. Besides, energy is energy, no matter the wavelength. Power emitted by a phosphorescence process is removed from the body, that part of the power emitted by phosphorescence not being available to increase the level of black (or gray) body emission. blackcloak (talk) 05:06, 10 March 2009 (UTC)[reply]

Blackcloak, I hope you had as much fun with this as I did. It opened up my experience quite a lot. One of the nice things about these guys like Kirchhoff etc. they did not beat about the bush, printing must have been more expensive in those days! I find the works of Planck and Einstein's 1905 paper on the photo electric effect and the 1916 one on absorption and emission not too bad either.--Damorbel (talk) 15:31, 10 March 2009 (UTC)[reply]

For all practical purposes, the atmosphere IS phosphorescent. Thus, E/A = 1 only if you take the integral over all frequencies and over a long period of time. Q Science (talk) 05:51, 10 March 2009 (UTC)[reply]

Q Science, there are lots of ways of producing light, Phosphorescence is one of them, light emitting diodes another, lasers yet another. This discussion is about thermal excitation of polar materials (or not polar as the case may be) Phosphorescence is so different that it is completely irrelevant to the discussion. --Damorbel (talk) 15:31, 10 March 2009 (UTC)[reply]

One of the definitions of Phosphorescence is that energy is absorbed at one frequency, stored for more than 1 ms, and then emitted at another frequency. If the atmosphere absorbs IR photons by water molecules and then emits IR photons by CO2 molecules several hours later, then the atmosphere IS phosphorescent, by definition. Terminology aside, the main point is that, for the real atmosphere, the frequency of radiation absorbed does not equal the frequency of radiation emitted. Though, in the case of Earth's atmosphere, the amount of radiation emitted is actually greater than the amount absorbed. Q Science (talk) 02:05, 11 March 2009 (UTC)[reply]

A timely reminder from Q Science that radiation plays only a minor role in transporting heat into the atmosphere. The wiki article uses this diagram [11] which is a complete travesty of the thermal reality. This one from the IPCC [12] is a bit better in that it shows net radiation = 26W/m2 (350-324), 24W/m2 by "thermals" (convection etc.) 78W/m2 by evaporation and 40W/m2 by direct radiation to space, 50% more than is transported to the atmosphere by absorption! The IPCC diagram is far from perfect with 324W/m2 "tranported" by back radiation. --Damorbel (talk) 07:44, 11 March 2009 (UTC)[reply]

Q Science, I would have thought that phosphorescence arose from the absorption of energy by the phosphorescent material itself, is that not different from heat absorbed at the equator, transported by air and water currents over the Earth's surface, evaporated, radiated etc. into the atmosphere before being finally radiated into space? Surely this is different from the phosphorescent figures on your wristwatch?--Damorbel (talk) 07:44, 11 March 2009 (UTC)[reply]

I don't think that our knowledge is good enough to know that the molecule that absorbs a photon is the same one that emits a photon via phosphorescence. Therefore, it is not clear if they are the same or not. I read once that only substances that have phosphorescence can be used for lasers because they must be able to store energy long enough that a photon can stimulate an emission, and CO2 is definitely used for lasers. Q Science (talk) 08:43, 22 May 2009 (UTC)[reply]

I think I've got it in as few words as possible. My short and sweet argument is that absorption and emission are both functions of wavelength, and if a body absorbs on one spectrum and emits on another, and is more efficient at emitting at that arbitrary wavelength than absorbing at the other wavelength, it will require a ~~higher~~ lower temperature for an equilibrium between absorption and emission to be maintained [addition 06:01, 10 March 2009 (UTC)→] than it would if it absorbed and emitted equally, because the rate of power output is proportional to T^4. This is why, as I said before, the material properties are important. This is backed by observations. Awickert (talk) 21:03, 9 March 2009 (UTC)[reply]

And here's a nice little figure for water to go with it. Awickert (talk) 21:10, 9 March 2009 (UTC)[reply]

Archiving III

I remain very unhappy with the archiving in place on this talk page, it seems to work with some embedded software that is hidden, at least I haven't been able to find it. I understand that I can change it but efforts to find out how have just met with little help at all from someone who likes to sign with Cyrillic characters (Miza?).

I have put a lot of work into this talk but the current attitude of other editors of instant deletion of changes to the article (Peterson?) is not my style so I will not adopt this deeply unpleasant practice.

If I can find the time I will take it up on an appeal to the admin. but this is unlikely to happen.--Damorbel (talk) 08:07, 11 March 2009 (UTC)[reply]

== End of Greehouse Effect material

KimDabelstein removed this text from the discussion page on Greenhouse effect!

Blackcloak, the rigorous approach is to treat this radiation as Electromagnetic radiation (EMR). The important matter is how heat is transferred between two places by this radiation. You no doubt know that thermal radiation comes from the accelerating of electrical charge by thermal vibrations. To be rigorous, thermal radiation is not heat, it is an electric field with an amplitude related to the temperature of the source object (no radiation at 0K). Two objects exchange heat when one is hotter than the other because the hotter one has an EM wave with a higher amplitude. The amount of heat transferred depends on the difference in the amplitudes between the hotter and the colder. If both objects have the same temperature, the EM waves have the same amplitude and there is no heat transfer. It must be realised that, with the same temperature producing the same electric field, there is no electric field between the two objects, that is why there is no heat transfer.

The "back radiation" concept is rather well illustrated by this diagram used by the IPCC [[13]] you will see that Greenhouse gases are shown as producing "324"(W/m²) back radiation. Now W/m² is power/m² whereas radiation field is measured in v/m (volts per metre). As illustrated the diagram states that "energy is being transferred from the greenhouse gases to the surface at 324 joules per second", this could only be true if the surface was at 0K (i.e. colder than the GHGs). Frequently the argument is put that there is a "net tranfer" of the difference between 350W/m² and 324W/m² i.e. 26W/m². But this will not wash either, not only because it is not justified by Electromagnetic radiation theory (see above) but because the surface and tropospheric temperatures are very different at different locations, i.e. it is not an electromagnetic phenomenon.

There is a lot more wrong with this diagram [[14]]. The incoming solar radiation is reflected, scattered and absorbed by the atmosphere clouds and the surface, why doesn't this happen to the "back radiation"? The whole concept of "back radiation" is so full of holes it cannot be used to support any hypothesis about the Earth's surface temperature. --Damorbel (talk) 15:20, 26 June 2009

I don't choose to start correcting the multitude of errors you've made because I know you won't participate in the discussion using strictly scientific quantities and concepts. For now I'll simply say that the word "heat" does not have a useful definition in this context. If you are going to talk about bb radiation, solar radiation, or back radiation, you can only use the word power, or related terms like energy, irradiance and radiance. Every sentence that contains the word "heat" will simply be ignored because it can not contain a scientifically rigorous idea. Put another way, if you can not express your concepts without using the word "heat," then you are simply not able to work within the scientific lexicon. At the very least you should tell us what the units of heat are, as you are using the term, so we can start to understand you. blackcloak (talk) 03:10, 28 June 2009 (UTC)[reply]

Traps

Were you directing the following at me? Sorry, I was trying to help. Q Science (talk) 04:59, 21 July 2009 (UTC)[reply]

You need to learn not to spring traps.

Yes. I'd hoped others would recognize the trap in the provided link to entropy. And, yes, I figured you thought you were being helpful, in a robotic sort of way. On related matters, I've read some of your contributions, and I do think they make a genuine effort to work with Damorbel, admittedly a really tough nut. blackcloak (talk) 03:46, 22 July 2009 (UTC)[reply]

Baiting him is so easy, there is no sport in that. (Yes, I know you disagree.) However, I have learned a lot trying to figure out WHY he thinks that way. Without his help (and to some extent yours also), I would have never understood why both sides have it wrong. The following is where we agree.

No one bothers to explain that greenhouse gases actually spend most of their time cooling the atmosphere. Most simple minded explanations of the greenhouse effect don't bother to mention that increases in greenhouse gases lead to faster cooling [1]

Unfortunately, nothing like that is allowed in any of the articles. Q Science (talk) 06:21, 22 July 2009 (UTC)[reply]

I'm surprised to see that you chose not to include my comment regarding the use of the word "trapped." I would have thought you had reached an understanding of this point. blackcloak (talk) 15:55, 22 July 2009 (UTC)[reply]

Baiting and setting traps are just shorthand ways to describe a process of eliciting a direct response from someone who walks away from a challenge. blackcloak (talk) 15:55, 22 July 2009 (UTC)[reply]

Sorry, I wasn't thinking of it that way. (I assumed it was mostly for fun.) Q Science (talk) 18:17, 22 July 2009 (UTC)[reply]

You provocatively suggest that I "have it wrong" somewhere. Care to explain? BTW I've avoided entering your exchanges with Damorbel to avoid confusion. You allow yourself to become ensnared in time consuming discussion when you and Damorbel do not share a common understanding of basic physics. blackcloak (talk) 15:55, 22 July 2009 (UTC)[reply]

We all have it wrong somewhere, otherwise there would be more agreement. My current OR suggests that "radiative forcing" does not exist in the way presented. As a result, the entire theory is based on nonsense. Q Science (talk) 18:17, 22 July 2009 (UTC)[reply]

Oh yes, I forgot to include in the list that greenhouse gases do not "cause" global warming. The energy of the sun that reaches the earth does. Greenhouse gases are (well, one of) the enabling agents, not the cause, of thermal energy entering and leaving the atmosphere. blackcloak (talk) 15:55, 22 July 2009 (UTC)[reply]

Semantics. The atmosphere is responsible for the climate being different than that of the moon. The question is - How? One of the answers I found indicates that Trenberth (one of the IPCC writers) is a moron. Though, perhaps I am the one who doesn't have a clue. At any rate, discussions with Damorbel (and others) have helped me clarify the arguments. Any good physics theory is full of holes. The trick is to find and understand them. Q Science (talk) 18:17, 22 July 2009 (UTC)[reply]

In the political world, and the entire subject of global warming is now dominated by politics, words matter. Why? Because words carry baggage, and in this case inferences which not too subtly imply impending doom (ok, that's a slight exaggeration). In fact, the same gases that "cause" warming also "cause" cooling. What actually warms up and cools down (and what we feel) is predominately (99%) the non-greenhouse gases, even though those non-greenhouse gases do not (mostly) accept or release electromagnetic energy in the IR. I use words pretty carefully, and find your dismissal as semantics evidence of a mind too willing to compromise. And btw the moon does not have a climate, unless you're thinking of meteorites, UV, solar flares and cosmic rays. blackcloak (talk) 20:47, 22 July 2009 (UTC)[reply]

Whether it is "the sun" or "greenhouse gases" that cause "global warming" is semantics. The way those terms are currently used is propaganda (ie not supported by the science). And both sides of the argument are guilty of using those terms incorrectly. The important thing is what is actually happening. And yes, I am willing to ignore the "propaganda terminology" to try and understand the physics. With respect to the Moon, I was referring to the fact that its surface gets both hotter and colder than the Earth's surface. Sorry that wasn't more clear. Q Science (talk) 22:22, 22 July 2009 (UTC)[reply]

Dismissing as semantics using "the sun" or "greenhouse gases" means you don't distinguish between a source and an enabling agent. Without understanding such fundamental differences, you'll never understand the physics, much less be able to communicate clearly what you've learned. In effect you're asking your reader to translate, inevitably inaccurately, your words before trying to understand them. blackcloak (talk) 06:59, 20 March 2012 (UTC)[reply]

Talk:Greenhouse gas

Thank you for your contributions to the encyclopedia! In case you are not already aware, an article to which you have recently contributed, Greenhouse gas, is on article probation. A detailed description of the terms of article probation may be found at Wikipedia:General sanctions/Climate change probation. Also note that the terms of some article probations extend to related articles and their associated talk pages.

The above is a templated message. Please accept it as a routine friendly notice, not as a claim that there is any problem with your edits. Thank you. -- TS 15:28, 25 February 2010 (UTC)[reply]

I am aware of the probation on the Greenhouse Gas article. My recent contributions have been to the discussion page of the Greenhouse Gas article. I am not aware of any probation on the Greenhouse Gas discussion page. If you look back in the record, you will see that I have been refusing to make changes to the article as I consider it a big waste of time. And we all know that many potential editors are completely disgusted with a wikipedia system that allows poorly qualified individuals to make content decisions. blackcloak (talk) 18:42, 25 February 2010 (UTC)[reply]

I'm sure you didn't mean it to, but interspersing your comments throughout a complex comment of mine has made mine all but incomprehensible. I didn't just write it to you, but for you and others to read and discuss. Please refactor your contributions to the end of the discussion. --Nigelj (talk) 22:42, 11 March 2010 (UTC)[reply]

I disagree with your conclusion about rendering your comments incomprehensible. Those of us who indent properly appreciate the advantage of being able to track responding comments via physical proximity. blackcloak (talk) 22:55, 11 March 2010 (UTC)[reply]

I have fixed the problem. The solution was to replace the first colon with a pound sign. Q Science (talk) 23:09, 11 March 2010 (UTC)[reply]