User talk:Damorbel

Welcome!

Hello, Damorbel, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

I hope you enjoy editing here and being a Wikipedian! Please sign your messages on discussion pages using four tildes (~~~~); this will automatically insert your username and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or ask your question on this page and then place {{helpme}} before the question. Again, welcome! William M. Connolley (talk) 21:41, 16 June 2008 (UTC)[reply]

GHE

In case you're not aware of how this works: imagine a planet with no atmosphere. R=T^4 sets the sfc temperature. Now assume a layer of atmosphere transparent to solar but opaque to IR. That atmosphere, assuming it starts at 0K, will warm up by absorbing IR from the surface. Whereupon it radiates downwards, and increases the radiation at the sfc, which then warms up. Easy, no? You can even do the maths yourself, I'm sure William M. Connolley (talk) 21:41, 16 June 2008 (UTC)[reply]

Nice to have your contribution William, but I am not completely new here and not at all new to thermodynamics. There is a long history of ideas that depend on breaching the Second law of thermodynamics. Those wacky schemes that promise Perpetual motion, energy out of nothing etc. etc. depend fundamentally on this, If I could arrange for heat to flow from a lower temperature to a higher as in the article describing the Greenhouse effect I would not be discussing the matter with you, I would be busy getting rich! Further, should you find flaws in thermodynamics you should find clarification in Statistical mechanics

When making statements like "Whereupon it radiates downwards, and increases the radiation at the sfc, which then warms up. Easy, no? You can even do the maths" which I adjudge a technical contribution, I would prefer you put them on the relevant talk page. Since you have views about heat flow from a lower temperature to a higher then perhaps they would be more relevant to the article on the Second law of thermodynamics.

The tone of your writing "Easy, no? You can even do the maths" implies that you are not willing to engage the matter which is a shame as I have pointed out flaw which should be resolved not obliterated, I invite you to reverse the undo as a gesture indicating your wish to engage on the matter.--Damorbel (talk) 06:01, 17 June 2008 (UTC)[reply]

This is funny. You've completely run away from the technical issue. If you ever wish to return to it, do let me know. I see DF has given you an answer too. The maths is here William M. Connolley (talk) 06:42, 17 June 2008 (UTC)[reply]

Read your link. Math is rather confused like you say, but you still seem wedded to heat energy going from a cold troposphere to a warm surface. Whatever the math, it's the physics that counts.

This is a fairly simple example of the physics. A cold particle, one with low energy, when colliding elastically in any way, mechanically or electromagnetically with a more energetic (hotter), will always gain energy from the more energetic one; the energetic particle will lose energy to the less energetic one. This is not a new idea!

I asked you not to put this discussion here on my user talk page but on the greenhouse effect talk page. I expect you to agree to me copying it there.

I put the talk here to avoid wasting everyones bandwidth on t:GHE. You may do what you wish with these words, provided that if you move them, you make it clear that its you thats done so, not me. I'm finished with you though - there is nothing more to say William M. Connolley (talk) 20:49, 17 June 2008 (UTC)[reply]

Speedy deletion of Damorbel

A tag has been placed on Damorbel, requesting that it be speedily deleted from Wikipedia. This has been done under section G1 of the criteria for speedy deletion, because the page appears to have no meaningful content or history, and the text is unsalvageably incoherent. If the page you created was a test, please use the sandbox for any other experiments you would like to do. Feel free to leave a message on my talk page if you have any questions about this.

If you think that this notice was placed here in error, you may contest the deletion by adding {{hangon}} to the top of the page that has been nominated for deletion (just below the existing speedy deletion or "db" tag), coupled with adding a note on the talk page explaining your position, but be aware that once tagged for speedy deletion, if the article meets the criterion it may be deleted without delay. Please do not remove the speedy deletion tag yourself, but don't hesitate to add information to the article that would would render it more in conformance with Wikipedia's policies and guidelines. Lastly, please note that if the article does get deleted, you can contact one of these admins to request that a copy be emailed to you. --Cocomonkilla ^{(talk) (contrib)} 21:20, 11 August 2008 (UTC)[reply]

Speedy deletion?

What's all this about speedy deletion? I have made a number of serious contributions notably in Greenhouse Effect http://en.wikipedia.org/wiki/Talk:Greenhouse_effect#Deficiencies and they never saw the light of day, no discussion, no reason possibly deleted by http://en.wikipedia.org/wiki/User:KimDabelsteinPetersen Abusive remarks "we've seen this kind of junk before" by http://en.wikipedia.org/wiki/User:William_M._Connolley

Why is my user name in red?

Your user page was presumably red because you had not yet edited it. This has now happened and you can use it. The "speedy deletion" issue above relates to a new article that you (presumably inadvertantly) created called "Damorbel" and which has been/will be deleted soon as it serves no purpose in "main space". Regards. Ben Mac Dui 21:51, 11 August 2008 (UTC)[reply]

PS Your user page is called "User:Damorbel" as opposed to "Damorbel" (the article you created by mistake). You then put the "hang on" message on "User:Damorbel" - but no-one was trying to delete your user page, just the article. Don't worry, you'll get the hang of it. Ben Mac Dui 21:56, 11 August 2008 (UTC)[reply]

I'm still upset about the abusive stuff from http://en.wikipedia.org/wiki/User:William_M._Connolley Connolley's experise does not appear to extend to thermodynamics, he contradicts funadamental concepts that he would have learnd in a basic course. There is nothing so basic as 2nd law of thermodynamics, he demands "reliable sources".

Here's a tip. All you need to do is this: User:William M. Connolley (with invisible square brackets at each end you can only see in "edit mode") to link to a page in Wikipedia - you don't have to specify the actual web page. Here's another. Users should always behave with civility to one another, (and I know nothing of your dispute), but User:William M. Connolley is quite right about the need for reliable sources. Please read WP:CITE and WP:V. Regards. Ben Mac Dui 19:25, 12 August 2008 (UTC)[reply]

User:Ben MacDui Thanks!

Blackbodies

I have run the blackbody equations (using the Stefan-Boltzmann law) for bodies in Earth's solar orbit. The average temperature for a rotating black body (albedo=0) is 279K, 6°C, 42°F. The average temperature for an Earth without an atmosphere is lower because it reflects (albedo=0.31) some of the incoming energy (light). A more correct temperature would be that for the moon (albedo=0.12) because it has no ice or clouds to reflect energy. It is interesting that the "standard" 254K that everyone uses does not allow for the expected albedo change that loosing an atmosphere would cause.

These are easy equations - E = (1-a) * s * T^4

One reason that you are having trouble with Talk:Greenhouse effect appears to be that you have not solved this equation for various parameters. I suggest using a spreadsheet. For instance, where does your 282K value come from?

incorrect temperature of 255K in place of 282K because it assumes the Earth radiates like a black body

Like I said, my computation gives 279K (which may be wrong) for a perfect blackbody. Good luck. Q Science (talk) 06:39, 13 March 2009 (UTC)[reply]

Urge caution

You have now violated the three-revert rule at Talk:Greenhouse effect. Self-reverting would be a really good idea. Short Brigade Harvester Boris (talk) 22:18, 14 March 2009 (UTC)[reply]

You have now not only broken WP:3RR but done so in full knowledge of the rules. I'm going to ask you what Boris already did: Are you going to self-revert? --Kim D. Petersen (talk) 22:47, 14 March 2009 (UTC)[reply]

You have been blocked from editing for a short time in accordance with Wikipedia's blocking policy for violating the three-revert rule. Please be more careful to discuss controversial changes or seek dispute resolution rather than engaging in an edit war. If you believe this block is unjustified, you may contest the block by adding the text {{unblock|Your reason here}} below.

Vsmith (talk) 22:52, 14 March 2009 (UTC)[reply]

This user's unblock request has been reviewed by an administrator, who declined the request. Other administrators may also review this block, but should not override the decision without good reason (see the blocking policy).

Damorbel (block log • active blocks • global blocks • contribs • deleted contribs • filter log • creation log • change block settings • unblock • checkuser (log))

Request reason:

I wish to be unblocked because my contributions are constantly deleted without explanation or, at best the deleter thinks they are, to put it mildly, irrelevant but gives no further explanation. As regards the policy the three revert rule is for "Since reverting in this context means undoing the actions of another editor" how can I be in breach of the rule when canceling what appears to me as vandalism of my contribution? I have asked the deleter (who I can't always identify, to consider arbitration. It seems a bit daft to ask to ask an automatic deleter for arbitration.I wish to complain about the arbitrary handeling of my contributions by [[1]], [[2]], [[3]] and others unknown to me. --Damorbel (talk) 10:20, 15 March 2009 (UTC)[reply]

Decline reason:

You reverted many times over this issue despite quite a number of editors who agreed with each other that the section of discussion being removed was irrelevant to improving the article, which is the purpose of a talk page. I take this very seriously, because this disruption took place on a talk page. Wikipedia is not a discussion forum, and edit warring, particularly over refactoring discussions, is disruptive. And, just so you don't miss the point: WP:3RR lists "obvious vandalism" as an exception. The fact that you say "appears to me" as vandalism makes it clear that even you know this was not "obvious vandalism", so no, there is no excuse for your behavior. Mango juice^talk 14:00, 15 March 2009 (UTC)[reply]

If you want to make any further unblock requests, please read the guide to appealing blocks first, then use the {{unblock}} template again. If you make too many unconvincing or disruptive unblock requests, you may be prevented from editing this page until your block has expired. Do not remove this unblock review while you are blocked.

May 2009

Welcome to Wikipedia! I am glad to see you are interested in discussing a topic. However, as a general rule, talk pages such as Talk:Global warming controversy are for discussion related to improving the article, not general discussion about the topic. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. Thank you. Kim D. Petersen (talk) 06:19, 12 May 2009 (UTC)[reply]

KimDabelsteinPetersen, would you care to explain how discussion about POV is unrelated to improving the article on Global warming controversy? And also explain why pointing to the unscientific nature of the greenhouse effect is not an important contribution to the Global warming controversy. I suggest you clear (or acknowledge) your defective contribution here yourself, please.--Damorbel (talk) 06:56, 12 May 2009 (UTC)[reply]

Personal comments and argumentum ad editor aren't acceptable on article talk pages, nor is using then as a forum for your pet conspiracy theory on why the Scientific opinion on climate change must be wrong, because you personally believe that the greenhouse effect doesn't exist. If you believe that WMC has done something wrong, in opposition to wikipedia rules and/or guidelines, or broken administrator ethics, then wikipedia contains forums where such complaint is valid and appreciated. Amongst others opening a user conduct RfC or raising it on the administrators noticeboard for incidences. Good luck. --Kim D. Petersen (talk) 07:23, 12 May 2009 (UTC)[reply]

Regarding your comments on Talk:Global warming controversy: Please see Wikipedia's no personal attacks policy. Comment on content, not on contributors. Personal attacks damage the community and deter users. Note that continued personal attacks will lead to blocks for disruption. Please stay cool and keep this in mind while editing. Thank you. Kim D. Petersen (talk) 07:14, 12 May 2009 (UTC)[reply]

Kim D. Petersen, personal attacks, where? Identifying a contribution as POV with an accompanying explanation is not a personal attack; indentifying a contribution as "unsound science" (with an accompanying explanation) is not a personal attack. In what category do you put "deleting contributions without discussion"? The proper classification is "arbitrary deletion", please stop doing it and preferably reverse those you have made. --Damorbel (talk) 07:29, 12 May 2009 (UTC)[reply]

This and this are personal attacks/personal comments (on/about WMC), it has nothing to do with content. Your comments in the same about how the greenhouse effect (apparently) doesn't exist - is original research and unsupported by references. That is also something that talk-pages cannot be used for. Your insistence on these is what gets you warned. Talkpages are not for discussion on general theory or as a soapbox - sorry. --Kim D. Petersen (talk) 11:23, 12 May 2009 (UTC)[reply]

I saw no personal attacks in the referenced comments by Damorbel. If you have something to teach the rest of us regarding this sensitive subject, please be FAR FAR more specific. State exactly where the attack on the person is. You seem to believe that it is sufficient for the personal attack may be inferred by the reader. But let's see it in your own words. blackcloak (talk) 19:15, 23 May 2009 (UTC)[reply]

Good. I will edit these to remove anything to imply that WMC has any POV but be absolutely sure I am only doing this as a gesture of goodwill. But to describe my contribution that the Earth's surface cannot be warmed by radiation from the cold troposphere as original research is far from correct, it is standard science and I will restore it. You may not be familiar with the Second law of thermodynamics, I cannot help that but the second law is not my discovery. --Damorbel (talk) 12:35, 12 May 2009 (UTC)[reply]

Dear Damorbel, Please understand many of us were teaching the Second Law to undergraduates at University before I guess you were born. Can I suggest that you both read it up a little more carefully (in particular try to understand what an isolated system is) and think about other everyday things which would be impossible with your interpretation. I can raise the temperature of a solar tube on my roof from 60C to 90C using only a cold mirror to reflect sunlight. The mirror can easily warm the tube even though the tube is a lot hotter than the mirror. The second law is not violated because the presence of solar flux means it is not an isolated system. This is the same as the case with the troposphere. --BozMo talk 10:06, 14 May 2009 (UTC)[reply]

BozMo, I appreciate your contribution, you certainly have an important point. The radiation striking the mirror does not heat it because a mirror, or any material that reflects or scatters Electromagnetic radiation is not heated by the radiation, the it just redirects it; it is why aVacuum flask has a mirror finish. You are probably aware of this. My argument is that no heat can be transferred from the Troposphere to the Earth's surface by radiation (or any other process) because the troposhere is always a lot colder than the surface, that is why I cite the Second law of thermodynamics; surely as a teacher this is what you taught your students?

I would very much like to know more about the material you used for teaching, we may be able to resolve differences quite quickly if it is available on line. --Damorbel (talk) 10:45, 14 May 2009 (UTC)[reply]

I have had one attempt at explaining to you why you are in error in the statement "no heat can be transferred from the Troposphere to the Earth's surface by radiation (or any other process) because the troposhere is always a lot colder than the surface". This is wrong because there is no isolated system to which the second law could be applied. I have also given you an analogy; heat can be transferred from a cold mirror to a hot pipe. I am not going to spend a lot more time on trying to explain it, because if you are capable of understanding it, the statements are very straightforward. I stopped teaching before the internet existed, so I cannot offer you notes online. However the WIkipedia articles on the second law and on isolated systems were basically correct last time I looked. --BozMo talk 10:54, 14 May 2009 (UTC)[reply]

BozMo, show me one reference that says the Second law of thermodynamics applies only to isolated systems. It is common experience that heat flows from hot to cold in all circumstances unless external work is done by a compressor or some sutch. Do you know of some natural arrangement where heat is taken out of the troposphere and transferred to the Earth? Now think very carefully, You do say you taught Thermodynamics, I asked you for some indication of your material and I don't have it yet, please tell me how you got these ideas, otherwise I might think you are just wasting time. --Damorbel (talk) 11:16, 14 May 2009 (UTC)[reply]

You are rapidly ceasing to look like a good faith enquirer and looking like someone who is being deliberately stupid, or perhaps we are past this point. Have you tried to understand these concepts at all? You ask for "one reference that says the Second law of thermodynamics applies only to isolated systems". I thought you claimed some familiarity with the second law but can you even state what it says? Ok, lets make it easy for you. The correct technical formulations are, as originally formulated in the 19th century by the William Thomson (Lord Kelvin) and Rudolf Clausius, respectively:

A cyclic transformation whose only final result is to transform heat extracted from a source which is at the same temperature throughout into work is impossible.

A cyclic transformation whose only final result is to transfer heat from a body at a given temperature to a body at a higher temperature is impossible.

Now, you can start from these, which you can confirm by dropping the exact phrases into Google and looking at all the places they are quoted. Please derive any statement about a non-isolated system from them? Or perhaps you are thinking about the Rudolf Clausius formulation made popular by song "Heat generally cannot spontaneously flow from a material at lower temperature to a material at higher temperature." Heat can't pass from the cooler to the hotter, you can try it if you want but you'd far better notter? As you have had explained several times heat means "net heat" and the fact that this case is not an isolated system means that "spontaneously" is not satisfied. Of course all the heat in the system in this case derives from the sun (nearly). Of course the heat from the sun is cascading downwards into deep space as per the second law. But the troposphere can influence the cascade in a way which warms the earth without violating the second law. The analogies of a blanket and a mirror are good enough for you to understand them. --BozMo talk 12:05, 14 May 2009 (UTC)[reply]

"The analogies of a blanket and a mirror are good enough". Good enough for what? Often the CO2 in the troposphere is shown as reflecting heat radiation from the Earth as here [4]. Is this what you mean by a mirror? The image shows infrared radiation being returned from the troposphere to the surface. But it is not being reflected as if the CO2 was a mirror. CO2 gas cannot reflect radiation in any significant way, its refractive index is tiny, about 0.00015 and it doesn't conduct electricity. The Greenhouse effect claims that "CO2 absorbs infrared and reradiates it back to the surface" This is true as far as the radiation field goes but because the troposphere is colder than the surface there can be no transfer of heat energy, no "trapping" of heat as it is put sometimes.

I simply do not understand about the "blanket", I have never seen a scientific explanation of how CO2 acts like a blanket. How can it be different from the atmosphere of O2 & N2 that protects the Earth from a lot of "nasties"? Can you give me a link on the blanket effect, please? --Damorbel (talk) 18:33, 14 May 2009 (UTC)[reply]

Almost every night, the surface of the Earth is colder than the air a few thousand feet above the surface. Therefore, at night, heat flows from the warmer atmosphere back to the colder surface. For examples, see this description of the atmosphere's temperature profile (including the related Flash graphs and a Lapse Rate Animation). Because the bulk of the troposphere does not change temperature between day and night (easily seen in these graphs), it does not play a direct role in the greenhouse effect. Q Science (talk) 22:10, 14 May 2009 (UTC)[reply]

Damorbel, The blanket and mirror were analogies used because you seemed to be making fairly basic mistakes suggestive that you did not understand accurate scientific explanation. This is confirmed when we tried to work off accurate scientific description you didn't get it. There was no suggestion of a mirror or woolly blanket space but simply the warning that you were applying a law you did not understand in a way which would preclude the existence of mirrors and blankets, therefore you needed to understand the laws better. I think by now the onus is rather on you to explain what you do understand, since direct approaches and approaches by analogy seem to be failing. What is your understanding of the second law of which you claim some knowledge? Please, not just some vague notion overheard in a taxi that it says something should not happen, but an accurate statement which you believe you understand. Also explain where do you get that statement from? Perhaps explain where you think it applies and does not apply (with as much generality as you can muster so please don't say "except where there is a heat pump", explain what about a heat pump makes it an exception). Do not worry if you make mistakes, just explain what you have been told. You can use the language in the relevant Wikipedia articles if you understand them. Perhaps if we can find any solid foundation in this we can (re)build on it?--BozMo talk 11:39, 15 May 2009 (UTC)[reply]

"An accurate statement which you believe you understand", I gave you a link to the Second law of thermodynamics, I have the impression you have not read it; otherwise, what is wrong with it? "What about a heat pump makes it an exception", I suggest you read this article Heat pump. The article explains how heat can be extracted from a cool body and tranferring it to a body hotter than the cool one (cooling the cool body further). Heat pumps are (thermally) efficient ways of heating your house, they work like air conditioners in reverse. In the States "air conditioners in reverse", using the heat pump principle, are used widely to warm houses.

I am not sure which of your "accurate scientific descriptions" I did not get. Which law do I "need to understand better"? Q Science points out (above) that the the atmosphere retains heat from the Sun at night keeping the surface warm, much warmer than, let us say, the Moon. But CO2 and the other greenhouse gases play a negligible role in this because they are such a small proportion of the atmosphere; it is the bulk of the tropospheric O2 & N2 that retains heat, stopping the temperature plummeting at night. But this is not how a blanket works, a blanket retains an amount of air near a warm body, preventing the air from drifting away, quite different to the troposphere where there is plenty of "drifting away" (particularly during gales).--Damorbel (talk) 09:52, 29 May 2009 (UTC)[reply]

O2 & N2 retain heat, but can not get rid of it. H2O transfers that heat back to the ground, CO2 cools the stratosphere. (Yes, I know this is OR, but it is true.)

Not quite; H2O (latent heat flux) is a net input of heat to the atmosphere. Short Brigade Harvester Boris (talk) 15:48, 29 May 2009 (UTC)[reply]

Evaporation adds heat to the atmosphere without producing a temperature change. (However, it does cool the surface.) When clouds form, most (perhaps 80% to 90%) of the latent heat is released toward space. (Some also creates wind.) I have not been able to find a reference, but I suspect that the released heat has a blackbody spectrum rather than a gas emission spectrum. When condensation (dew, fog, frost) occurs, most of the latent heat is returned to the surface, again without changing the temperature. As a result, the temperature quits dropping. Q Science (talk) 17:19, 29 May 2009 (UTC)[reply]

Heat pumps work because energy (electricity) is used to force the heat to go the wrong way.
A cold atmosphere can add heat to a warmer surface because the atmosphere is warmer than the background microwave radiation. Search this blog for "Let's examine two objects in isolation" (because the page is very long). If Barton Paul Levenson agrees, his excellent description should be made into a wiki page. Q Science (talk) 12:31, 29 May 2009 (UTC)[reply]

"O2 & N2 retain heat, but can not get rid of it" if the air is at a stable temperature molecules in a given small volume have the same temperature thus when GHGs are radiating they are kept warm by the O2 & N2. All the so-called greenhouse gases cool the Earth by radiation, you can see them doing it in the IR band here [5].

Unfortunately, that won't work in my browser, the images are not downloaded. However, most of the IR satellite images I've seen elsewhere show the temperature of the cloud tops. I think this represents blackbody radiation from liquid water and not IR released from atmospheric gases. Q Science (talk) 17:19, 29 May 2009 (UTC)[reply]

Satellite images show the radiation in different IR bands, the lighter images are generally of the colder parts, but cold is relative here, most of the radiation from the atmosphere is from material above 220K, very much warmer than the 2K-3K of deep space.--Damorbel (talk) 23:53, 29 May 2009 (UTC)[reply]

2nd law of thermodynamics says the heat goes from the warm surface to the cool troposphere then to very cold deep space, it is very simple. The GH effect claim that as you say "A cold atmosphere can add heat to a warmer surface because the atmosphere is warmer than the background microwave radiation" i.e. that somehow heat can go from the troposphere to the surface (against the thermal gradient) is bizarre in the extreme. I know "thousands of scientist believe it", I also know that thousands of scientists know nothing of thermodynamics. --Damorbel (talk) 15:32, 29 May 2009 (UTC)[reply]

Did you read the description I referenced above? Stated another way, Stefan's equation does not use temperature to the fourth power, it actually uses a difference in temperature. Q Science (talk) 17:19, 29 May 2009 (UTC)[reply]

Sorry, I did not find the reference to anything saying "Stefan's equation does not use temperature to the fourth power" could you show me better? Thanks.--Damorbel (talk) 23:53, 29 May 2009 (UTC)[reply]

To locate the correct section in the reference you must search for the string provided, but without the double quotes..

In Black_body#Radiation_emitted_by_a_human_body, P = A s e (T^4 - To^4) shows how to use the Stefan–Boltzmann law where the surface is at one temperature and the atmosphere is at another. This indicates that a cold atmosphere will reduce the amount of energy lost by a warmer surface. The mistake most people make is to simply set To=0. On the Earth, even this equation will give the wrong result because some radiation is simply reflected from the cloud bottoms without affecting the temperature of the atmosphere. Q Science (talk) 01:53, 30 May 2009 (UTC)[reply]

Do not use talkpages as a forum

Please do not use talk pages such as talk:Global warming controversy for general discussion of the topic. They are for discussion related to improving the article. They are not to be used as a forum or chat room. If you have specific questions about certain topics, consider visiting our reference desk and asking them there instead of on article talk pages. See here for more information. Thank you. --Kim D. Petersen (talk) 15:50, 29 May 2009 (UTC)[reply]

Kim D. Petersen I'm vastly curious about this idea of our reference desk, is this where a group of "our owners" hand out advice? Can I become one of "our owners" too? Do you have to pay money to take part in the work of our reference desk? --Damorbel (talk) 14:37, 16 June 2009 (UTC)[reply]

Radiative Heat Transfer

Continued from KimDabelstein's talk page.

Hello Mr./Ms./Dr. D. We've spoken before. Now I happen to spend almost none of my time thinking of radiative heat transfer; however, I have a few simple thoughts that I hope Kim won't mind cluttering his talk page (he of course has the right to kick us off :) ).

You seem to assert that there can not only be no net radiative heat transfer from a cold body to a warm body, but that there can be no transfer period. By this, do you mean to imply that radiation given off by a 500K body may never reach a 1000K body? I really doubt this.
If I'm wrong, and you do accept that a cold body's radiation can be received by a warmer body (though net transfer is always toward the colder), then continue: Imagine a warm body, next to a vacuum. The warm body will radiate into the vacuum, and after some period of time, the temperature will decrease until the rate of heat produced by the body equals the rate lost. Now imagine a warm body, surrounded by a cooler body, surrounded by vacuum. Would it not make sense that, if the cooler body could radiate some heat toward the warm body, that the warm body would then have a higher equilibrium temperature?

Granted, these are off the top of my head, but they represent my fundamental concerns with what you say (as myself, not citing any sources). Awickert (talk) 08:39, 30 May 2009 (UTC)[reply]

Perhaps I am not clear. It is essential to distinguish between radiation and heat transfer, the presence of a radiation field does not define heat transfer, no more than two bodies in contact defines heat transfer. Heat tranfer is dependent on temperature difference, always from hotter to colder, be it by contact (conduction) or radiation. Two bodies remote from each other and above 0K have both an electromagnetic radiation field, the magnitude of which can be calculated by the Stefan-Boltzmann law well known for the T⁴ dependency. If these two bodies have the same temperature then the fields are equal and no heat is tranfered. When they have different temperatures the heat transfer is proportional to the difference of the 4th powers, see here [6].

This is not ground breaking science, it has been around since the 19th century, why it is not more generally known I do not know, even those claiming to know about quantum theory seem not to realise that quantum theory emerged from this 19th century science.--Damorbel (talk) 09:35, 30 May 2009 (UTC)[reply]

I'll leave this here for one more round, and then move to your talk if necessary. So it seems you (perhaps by a convention I don't know about) define heat transfer as net transfer of heat. That makes sense then. If I extrapolate your statement about heat transfer being proportional to the difference of T⁴ between two bodies, I would say that you would seem to agree with my previous statement that a body with a temperature intermediate between a warm body and a vacuum would thereby slow the rate of net heat loss from that warm body. I assume my conjecture of your opinion is wrong, because it goes against your argument. But my conjecture of what you say seems to work with my argument that a cool body between a warm body and a cold body may slow the rate of heat loss from the warm body and thereby cause the warm body to have a higher steady-state temperature. Awickert (talk) 09:49, 30 May 2009 (UTC)[reply]

What is the difference between heat tranfer and "net heat tranfer"? --Damorbel (talk) 10:18, 30 May 2009 (UTC)[reply]

In conduction, for example, heat transfer is diffusive, which is based on an idea of random motion of packets of heat. Although these packets move randomly, the hotter ones move faster and the overall mixing of hot and cold heat packets is due simply to their motion. So although net heat flows from a hot body to a cold body, an individual packet may move back from the cold body to the hot body. So that's what I mean. In radiative transfer, a single wave or whatever. And this is why it's the temperature gradient that is important, and heat flows more slowly from a 255 K material to a 254 K one than from a 1200K material to a 273K material. I'm fine using your convention, and as a matter of fact, that is how I almost always use "heat transfer" - but it seemed useful to talk about the individual packets this time around.

So, to get to the main point (again), it seems your problem is that you say that the cold troposphere can not warm the (warmer) Earth's surface. This is correct - but it can prevent it from getting colder. Reiterating: warm body - cool body - cold vacuum, or warm body - vacuum: the second example has a higher temperature gradient. Awickert (talk) 17:53, 30 May 2009 (UTC)[reply]

What you describe is an interesting but new theory of heat. Conventionally heat is the kinetic energy of particles, the temperature of material is a measure of the translational, i.e. in the x y z axes, energy. Thermal energy is transferred between particles by collisions which lead to the exchange of momentum between them. These particles do not, because of the way momentum is exchanged (think of billiard balls exchanging momentum) have the same momentum but there is a distribution of momentum which leads to a distribution of energy described by the Maxwell-Boltzmann statistics. This may bear comparison with your idea of "heat packets". It is not possible to measure the energy of the individual molecules and thus decide on their temperature, so temperature is not defined this way. The only way you can measure temperature is by observing the average (translational) energy of an ensemble of particles.

Thus because temperature is a statistical average measure defined over a period when all energy flow from one part of the ensemble to another has disappeared (i.e there is no average energy flow internal to an oject - it is in local thermodynamic equilibrium LTE) and it now has a temperature.

In a new situation, two objects each in LTE but with different temperatures are brought into contact (or perhaps a "radiation exchange situation") will make a new ensemble that is not in LTE (because of the temperature difference). These two objects will now exchange energy in such a way that the energy of the cooler object is increased, thus its temperature rises, and the hotter body loses the same amount of energy. After a time energy flow between the two objects slows to, let us say, zero; we now have a new thermodynamic equilibrium local to a new system that comprises both objects (because they both have the same temperature). Because the two objects are now thermally in contact they will continue to be a system (system = "going together") because, even if heat is added to one part only, making the system temperature undefined, soon the (individual) temperatures will be equal and the system (made up of two connected parts) will have a new temperature.--Damorbel (talk) 10:59, 2 June 2009 (UTC)[reply]

I just do not understand how you can claim that, if I interpret you correctly - I assume you mean the greenhouse gases, "can prevent" (the Earth) "from getting colder". I noted above that particles in gases exchange thermal energy by collision, in solids it is similar but by vibrations along the bonds that hold the solid together. When the particles making up some (or all) of a given volume of gas have a dipole moment accessible to thermal energy these molecules will be able to exchange energy by radiation in addition to the collision process. This "radiative energy exchange" cannot introduce a temperature gradient that is different from the collision process, the LTE condition I described above is equally valid for radiation. There is a difference betwen the two, gas under pressure (in a pressure vessel) maintains its pressure by exchanging momentum with the vessel walls (normally thought as having the same temperature as the gas, if they had another temperature they the gas would heat or cool accordingly). If the walls were transparent, radiating gas would exchange energy with the environment external to the container; if the walls reflect then there would be little or no energy exchange (c.f. Thermos flask). There is little reflection from gases in the troposphere, as I noted above radiating gases transfer heat to colder places in the environment, they also absorb heat from warmer places, they do not generate a thermal gradient when doing this because the collision process quickly re-establishes LTE. The radiation process itself works to restore LTE but without any reflecting effect and, given the long mean free path of photons in the atmosphere, much of the radiation eascapes so that the heat getting in to the atmosphere and is freely dispersed into deep space.--Damorbel (talk) 10:59, 2 June 2009 (UTC)[reply]

Thank you for your lengthy response. I will try to answer it in two parts, and keep it as brief as possible (I have largely failed at this, because I'm still having trouble seeing how we're reaching such different conclusions):

First, I did not intend to create a "new theory" of heat. What I was trying to do is to illustrate why heat conduction (e.g., via Fourier's law), when coupled with a conservation equation, gives heat transfer to be a diffusional process. My illustration was through an analog for temperature-driven heat flow in another diffusional process, Brownian motion of molecules. This works less well in thinking about conduction, but (as I'll get to below) works somewhat better in thinking about radiation (at least insofar as random orientations are concerned). I also always understood temperature (not heat) to be related to the average kinetic energy of the system, as it (and not heat) is the state variable, with heat being the thing that is transferred.

I am interleaving my response to make it easier to relate point and counterpoint. I wrote "new theory of heat" which was a short way of saying the idea of "random motion of packets of heat" is not applicable to conduction, which I agree is a diffusion process. My problem with your explanation is that it seems very different from the (conventional) Kinetic theory of heat which doesn't have "packets of heat". The flow of heat by conduction is best illustrated in solids where the molecular parts are held in a rigid structure by intermolecular forces called bonds, these bonds have some elasticity (think of springs) and heat is stored as the vibrational energy of molecules and bonds. Heat diffuses because the molecules transmit energy from one to the next through the bonds. In metals heat is partially transported by the motion of conduction electrons. Heat conduction is not useful when considering an atmosphere because convection is far more powerful heat transfer method; convection can result in temperature inversion, a cup of water is cold at the bottom and warm at the top, similarly the air in a room is warmest near the ceiling. Heat flow is no longer the simple matter of "hotter to colder" of the second law of thermodynamics, fluids expand and float upwards in the gravitational field thus some of the heat energy is used to overcome gravity and resultant change in gravitational energy becomes very relevant. (I can only answer in bits, I'm busy and it is not a simple matter).--Damorbel (talk) 20:42, 4 June 2009 (UTC)[reply]

Thanks for the answer, and sorry for the late response: also busy. Yes, I understand all this - I was using conduction as a simple example, and clearly failing to use an appropriate analogy. Awickert (talk) 05:19, 12 June 2009 (UTC)[reply]

As to the convection, my response below deals with a radiative balance only, so let's save that for later unless necessary in order to stay on topic. Also, as to not waste your time in explaining things, I'll point out that I'm not wholly uneducated: feel free to state what you think without needing an accompanying explanation of the basic physics. Awickert (talk) 05:26, 12 June 2009 (UTC)[reply]

I also do not see how your statement about the atmosphere not preventing cooling would work. My only point of total confusion in your paragraph is your "There is little reflection..." sentence due to (probably) a few accidental grammar issues when you typed: when you say "do not generate a thermal gradient", what is "doing this"? You speak of collisions, but you're mostly talking about radiation. In any case, I assume you are talking about a very localized thermal equilibrium, because there is a definite gradient between here and outer space. My point of a little confusion is that radiation can not create a different thermal gradient than collision alone: if I put a heat lamp at the bottom of a jar, the radiation will cause the molecules to become excited and re-radiate (sending energy up the jar) and will cause the molecules to move faster (propagating the heat signal through collision). The two certainly work in tandem in the jar, and I believe which one dominates depends on the speed and intensity of the radiation and particle motions, with respect to the size of the system. Overall, due to limitations in particle speeds and their net-random-walk (ideal gas) motions, radiation dominates on longer length-scales.

For my reasoning: Consider two bodies radiating towards each other in a conservative full-space. Radiation intensity increases with temperature. Therefore, more energy will be sent from the body with higher temperature towards the body with lower temperature than vice versa. As time progresses, this process will cause the hot body to lose heat and the cold body to gain heat until, after time and exponential-decay-wise, they reach a thermal equilibrium. I'd like to know if you accept this.

Next, consider that the hot body is being heated from some source, but not the second, or that the same source is heating the first body more than the second. In either of these cases, equilibrium will not imply equal temperature, since there are new heat input terms that will also need to be balanced, and the first body will be able to be hotter than the second. This OK with you? (I'm trying to figure out where we lose each other.)

Next, consider that the cold body is vacuum. Let's say that it is very cold and that its radiation is near-negligible. If the volume of the vacuum is much greater than that of the first body, then the whole thing will end up at the temperature of the vacuum. Otherwise, the heating term on the first body has to be pretty big. But this is indeed what happens in our Solar System: the Sun heats the planets and produces a large temperature gradient between them and outer space. Still OK?

After this, let's add a third body between the planet and outer space. This is the atmosphere. The atmosphere receives heat energy from the sun, preventing it from directly hitting the ground. It also receives heat radiating off the planet's surface. Because the atmosphere is finite in volume, it is comprehendible that it reaches an equilibrium temperature that is given by its temperature inputs (from the planet and star, for example) and its loss of heat due to radiation to the big heat sink of outer space. Still OK?

Now, I know we're not OK with at least some of the following part. But here I go. If we assume an isotropic medium (or a well-mixed anisotropic one such that it appears isotropic), radiation is given off in every direction. Solar radiation that hits the atmosphere can be re-radiated upwards (towards space) or downwards (towards the Earth's surface). Heat on the Earth's surface can then be conducted or convected downwards, or re-radiated upwards. The portion of the heat that is re-radiated upwards may be either intercepted by the atmosphere or go through to space. But if it is intercepted by the atmosphere, it may be re-radiated in any direction: in our case, the important directions are up and down. This is why having an atmosphere capable of absorbing Earth's thermal emission fingerprint is important for the thermal balance. Because hotter bodies radiate more than colder bodies, net heat will flow to the coldest body. (I know you do not like "net heat", but radiation is quantized, by electron-hopping in molecular orbitals, in our case, so it is important to distinguish it from quanta of energy sent out in each individual radiation event.) However, because radiation can be sent in any direction regardless of temperature gradient, some heat may be sent from a cold body to a warm body. This results in the observed warmer thermal equilibrium between a cool body and a warm body than between a cold one and a warm one, as well as the fact that heat transfer rate is a function of temperature gradient.

I would very much appreciate learning at which points you dissent: as far as I can see, my above explanation (which took me way too long to write) fits the current state of knowledge of radiative heat transfer and accurately surmises the physical system. Awickert (talk) 08:38, 3 June 2009 (UTC)[reply]

Anthony Watts

You might find this interesting. The POV pushers are at it again. Q Science (talk) 20:48, 16 June 2009 (UTC)[reply]

Is Anthony Watts pushing a POV? Can you summarise? A POV can be helpful when seeking to resolve some scientific matter but for me the weaknesses of the AGW "science" are so obvious that they need only quite ordinary physics to illustrate them. There is nothing in any physics that shows the temperature of the Earth is in any way dependent on the albedo, nor is there any mechanism (such as absorption/emission) to show that GHGs, like H2O or CO2, can change the average surface temperature over the whole planet. A simple lab experiment will show that, although these gases absorb and emit IR radiation, they have exactly the same thermal characteristics as other gases, their ability to absorb/emit is just another heat transfer mechanism. Photons transfer heat in a way similar to molecular collisions, a fact recognised by A Einstein in his 1916 paper. More recently this has been endorsed by describing a "photon gas".--Damorbel (talk) 01:21, 17 June 2009 (UTC)[reply]

Sorry, I thought you were familiar with his work. You should really read his book. (It is short, be sure to look at the pictures.) At any rate, the pro-Global Warming crowd are attempting to discredit Watts by using various ad hominem attacks. (Attack the person, not the message.) Q Science (talk) 07:46, 17 June 2009 (UTC)[reply]

Nice to know but I think that researching global warming by examining local records could only work if you were able to correlate temp., press., wind etc. on a grid based on cells 1m cube, i.e. never.--Damorbel (talk) 15:09, 17 June 2009 (UTC)[reply]

KimDabelstein removed this from the dicussion 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).

That is not correct - thermal radiation is photons with a frequency distribution based on the temperature of the object. The number of photons is based on the emissivity of the surface convolved with the distribution. "Amplitude" and "electric field" are not terms associated with this process. The energy of each photon is based on its frequency (or wavelength) only. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

You should be aware that for the analysis of thermal radiation the electromagnetic model is substantially error free because there are no relevant quantum effects that would change the answer. (If you know of some, please say what they are - thanks in advance.)--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

"Propagation of an electro-magnetic wave" is not the same phenomena as an "electric field". Old TV's use electric fields to accelerate electrons to produce images. Q Science (talk) 00:21, 29 June 2009 (UTC)[reply]

Agreed! An electric charge has an electric field (measured in V/m). When it moves it causes a current (i=dQ/dt) that in turn generates a magnetic field. When a charge accelerates (d²Q/dt²)the electric and magnetic fields combine to launch an electromagnetic wave that travels at the speed of light. This wave is entirely general; "a wave" is not necessarily a sinusoid, it can be a pulse or a photon, an electromagnetic wave has only to have orthogonal electric and magnetic fields and propagate at the speed of light in a vacuum. --Damorbel (talk) 08:38, 29 June 2009 (UTC)[reply]

"Amplitude" is wrong term

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.

Again, "amplitude" is not the correct term. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

What is the "correct" term then?--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

I am not really sure. Both intensity and flux are normally used, but they both have multiple meanings and, in this case, both have problems because the amount of energy per photon varies a lot over the spectrum we are interested in. I have been using IR radiation, IR energy, or just energy. (Not a good answer, sorry.) Q Science (talk) 00:21, 29 June 2009 (UTC)[reply]

Maybe Radiative flux is best. Q Science (talk) 00:28, 29 June 2009 (UTC)[reply]

As noted above an EM wave is very general, the quantum concept comes from the interaction of EM waves with atoms where an interaction (absorption/emission) can only take place with energies equal to those of the orbiting electrons. If an electron in orbit is to receive energy the wave has to have enough energy to push it to a higher orbit, it also has to have the same frequency, phase and timing i.e. be a photon. Should the orbiting electron be above the ground state i.e. already have excess energy, an incident wave with insufficient energy to boost the orbit it may well cause the orbiting electron to drop to a lower orbit simultaneously emitting a photon; this is Einstein's great discovery of stimulated emission. You can easily see from this that, for the stimulated emission of a photon, the phase of the stimulating wave (photon) and the (stimulation) emitted photon must be the same i.e. their amplitudes must add. This is the principle of the Laser. --Damorbel (talk) 08:38, 29 June 2009 (UTC)[reply]

Great description for UV light, not applicable for IR which affects molecular vibrations, not electrons. Q Science (talk) 02:29, 30 June 2009 (UTC)[reply]

I have never argued that the thermal interactions of IR with molecules require quantum effects to describe them, you introduced the photon but it is not needed.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

If both objects have the same temperature, the EM waves have the same amplitude and there is no heat transfer. It must be realized 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.

"Electric field" is not the correct term. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

If electric field is not the correct term what is? We are concerned with Electromagnetic radiation after all.--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

Back Radiation

The "back radiation" concept is rather well illustrated by this diagram used by the IPCC [[7]] 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).

The energy from an object has nothing to do with the temperature of a surface that receives that radiation. For instance, energy (light) takes 8 minutes to get here from the sun. How could the temperature of the Earth affect the energy the sun emits. In addition, the Moon and the Earth have different surface temperatures, but they both get the same energy per square meter from the Sun. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

"The energy from an object has nothing to do with the temperature of a surface that receives that radiation" It most certainly does, two adjacent bodies both emit radiation to and absorb radiation from each other, this is summed up by Stefan's law. Above 0K both bodies emit and absorb radiation, if there is a temperature difference between them, this difference tends to zero with time as energy passes between them by radiation (always from the hotter to the colder). --Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

You keep confusing energy emitted with net energy transferred. Two objects at the same temperature can both be emitting a lot of energy, but still have a net energy transfer of zero.

Objects above 0K produce Electromagnetic radiation because of the acceleration of electric charge (see above). In no way is this the "emitting of energy", it is "generating an electromagnetic (photon) field". By means of this electromagnetic field, energy may be transferred (or not) between two or more objects, dependent on their relative temperature, relative distance etc.--Damorbel (talk) 08:38, 29 June 2009 (UTC)[reply]

Things cool off because they loose energy. If the energy is not emitted, what happens to it? Q Science (talk) 02:29, 30 June 2009 (UTC)[reply]

If it is isolated (no other way like to lose energy e.g. conduction) then it stays at the same temperature. But it also has no way to acquire energy, and thus get a temperature, in the first place.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

The Earth rotates, therefore a specific location gains energy during the day and looses it at night. Again, If the energy is not emitted, what happens to it? Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

Try this, put the 2 objects a light year apart. Does the temperature of one object affect the energy emitted by the other? Now, place the same 2 objects a foot apart. How does that change the amount of energy emitted by either? Why? Q Science (talk) 00:21, 29 June 2009 (UTC)[reply]

Terminology - "volts per meter"

"volts per meter" describes an electric field, not radiation. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

The full name for the matter under discussion is Electromagnetic radiation, it is called this because it has an electric (and a magnetic) field. Photons are a more advanced way of describing Electromagnetic radiation, they are really only necessary for subatomic scale effects. To learn more about this I reccommend this Canonical quantization, it's tough stuff, much of it due to Paul Dirac. --Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

Please show me where Electromagnetic radiation is measured in volts per meter. Q Science (talk) 00:21, 29 June 2009 (UTC)[reply]

[8]--Damorbel (talk) 08:38, 29 June 2009 (UTC)[reply]

Wow. You know he is talking about RF transmission, right? He is not real clear on what volts per meter means, but it is probably the peak voltage difference from one end of the dipole to the other. Sort of like, the voltage from one side of a photon to the other side of the same photon. In the case of an RF survey, that would be the peak to peak voltage at the base of a receiving antenna at some distance from the transmitter. (I think) Basically, the purpose of the survey was to see if cell phones (or some other transmitter) cook your brain. Q Science (talk) 02:29, 30 June 2009 (UTC)[reply]

"About RF transmission, right" Why then does he write "Many people think of them simply as radio waves, but EM waves cover a much broader frequency spectrum. EM waves extend from the very lowest frequency (Hz) to frequencies beyond radio waves, light waves, X-rays, and gamma rays."?

I sometimes have the impression you do not pay much attention to what I write.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

I would still like a reference where light is measured in "volts per meter". Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

More on Back Radiation

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 [[9]]. 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 "back radiation" has a different frequency distribution than solar radiation. It is also emitted below the clouds. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

So what? Does this change the answer?--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

You asked (paraphrased) "why isn't the back radiation scattered by the clouds in the same way as the solar radiation?" My answer was "the backradiation is emitted below the clouds. Therefore, the clouds can not reflect this back to space." Q Science (talk) 00:21, 29 June 2009 (UTC)[reply]

When the "back radiation" hits the ground some of it should be reflected rather than absorbed, as with the incoming sunlight. But the "back radiation" is shown as 324W/m² which is "energy per second", energy per second is not a measure of radiation, should this supposed radiation be reflected there is no energy transfer. --Damorbel (talk) 08:38, 29 June 2009 (UTC)[reply]

About 30% of UV and visible light are reflected, but only 2% of IR radiation. This is just they way things are. On the other hand, clouds "reflect" lots of heat. They absorb the energy without changing temperature by evaporating water. This creates a little turbulence, the water vapor is mixed with cooler air, recondenses, and then reradiates heat (IR radiation). All without changing the local temperature very much. Thus, clouds "reflect" IR radiation very well.

The idea that reflection of radiation effects the equilibrium temperature of any object in any solar system is a serious flaw in the greenhouse effect thesis, it completely falls to bits under Kirchhoff's analysis of thermal radiation.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

BTW, the diagram shows "324W/m2 absorbed", and simply does not show the additional energy that is reflected since it does not affect the surface energy balance. For energy from the Sun, reflected energy must be accounted for.

"Simply does not show" Excuse me but the 324W/m² is from the source i.e. the troposphere, it is a fraction of this that should be reflected rather than absorbed.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

You should read the original paper, the 67 W/m2 absorbed by the atmosphere could be as high as 85 W/m2.

"You should read the original paper" I have, and many others. "high as 85 W/m2" what difference does this make?--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

Trenberth is one of the main IPCC authors and editors. Therefore, any issues with his papers are important. In this case, it shows the significant lack of data used to support the IPCC position. In addition, some of the numbers in the IPCC documents don't agree with the paper they come from. One interpretation of that specific paper is that it indicates that the basic Greenhouse model is extremely wrong. You might try a google search on Trenberth alternate greenhouse model. Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

Also 67 W/m2 actually means 67 joules/(second meter*meter steradian). (I am not clear on the steradians, an angular measurement. The values given may already assume integration over the whole sky.) The way it is expressed is just a shorthand method of representing energy transferred. Actual measurements are usually made with a cosine integrator (whatever that actually means). Q Science (talk) 02:29, 30 June 2009 (UTC)[reply]

Speaking of radio waves (above), I am sure that you know that some radio waves are reflected by the ionosphere, depending on their frequency. It seems that it would also make sense that some IR radiation is reflected by other parts of the atmosphere. Therefore, the question becomes, what percent is reflected and what percent is emitted? And how do we know? If a significant part of the Greenhouse effect is based on reflection, then what part, if any, will Greenhouse gases play in that reflection? Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

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 (UTC)[reply]

Basically, I agree with you that the diagram is nonsense. If you read the original source paper, you will see that most of the numbers are highly disputed or simply guessed at. It was presented as a framework to help understand the physics, not as the basis for supporting Anthropogenic Global Warming. (The IPCC did that.) But you need to learn the terminology. Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

This would not matter except the whole greenhouse effect argument (as put forward by the IPCC) is based on it.--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

(Please pardon my out-denting the original text. Restore if you disagree.) Q Science (talk) 19:04, 26 June 2009 (UTC)[reply]

No problem, perfectly clear to me. Thanks for taking the trouble.--Damorbel (talk) 17:38, 28 June 2009 (UTC)[reply]

Basic Theory

The truth is that we have a lot of equations that describe light very well. But we still don't know what it is, how it is absorbed, or how it is emitted. Comparisons with radio waves are interesting, but no one has proved that they are the same phenomena. I once read some information explaining why scientists think that x-rays and light are different versions (frequencies) of the same phenomena. It included a number of experiments to support their arguments. However, none of this is final. We don't even know how gravity works, but the equations are good.

"Don't know what it is"! What on earth do you mean? This is in danger of becoming a waste of time, the matter under discussion is electromagnetic waves that are produced by accelerating electric charge, this is the link from classical mechanics made to electrical effects by James Clerk Maxwell the critical link is Coulomb's law which describes the mechanical force between electric charges. I should come as no surprise that the exact functioning of this needed the developement of quantum theory. --Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

We can produce synchrotron radiation (x-rays) by accelerating electrons, and radio waves, but I don't know of any experiment where light is produced that way. Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

Again, why synchrotrons, why electrons? Acceleration of charge arises all over the place. The acceleration of electrons in your TV CRT produces some EM radiation, there is bremstrahlung and Cherenkov radiation, dipole vibrations in CO2 and H2O molecules can both absorb and emit IR. The list is very long.--Damorbel (talk) 13:48, 30 June 2009 (UTC)[reply]

Using the term photon is an admission that we don't know what a photon is, only that we can describe some of its properties. Newton thought it was a particle. Einstein proved that the energy is quantized. Yet diffraction patters exist. The science is anything but settled. Q Science (talk) 02:29, 30 June 2009 (UTC)[reply]

"We don't know what a photon is". If you do not accept the scientific method you could say the same about an electron or any other scientific concept. There is a very old idea that if you cannot see something it doesn't exist, is this the "Q Science standpoint"? If so I am not the slightest bit surprised at your responses to my really rather basic physics.--Damorbel (talk) 06:57, 30 June 2009 (UTC)[reply]

The whole idea behind string theory is that none of the basic "particles" is understood. Even quarks are a bit of a guess. I am not sure what that has to do with "scientific method". Q Science (talk) 10:03, 30 June 2009 (UTC)[reply]

"The whole idea behind string theory is that none of the basic "particles" is understood" What do you mean by this? Is string theory itself included in this general dismissal? This is and old story, I've pointed out before developements like relativity (special and general), quantum theory etc. did not all "disprove" previous theories such as Newtonian gravity, wave theory of light etc., they extended them and made them more powerful.

I think you are well beyond the matter in hand. The evidence for quantum effects is vast; many, many observations are explained by the various quantum theories, many predictions are made sucessfully on the basis of quantum theories, in short modern physics would not work without quantum effects. Conversely string theory has predicted nothing, it is an attempt to explain an apparent shortfall in the mass of galaxies [10].

You keep introducing irrelevant concepts such as photons, string theory etc. I have spent a lot of effort showing you that conventional physics explains planetary temperature etc. without exotic inventions such as the greenhouse effect or even quantum theory; you clearly have not been following what I have written; you seem to support other editors who delete my contributions, I seriously wonder what you think you are doing. --Damorbel (talk) 12:02, 30 June 2009 (UTC)[reply]

Spencer on Second Law

Damorbel,

Roy Spencer has a blog post that you should read:

http://www.drroyspencer.com/2009/04/in-defense-of-the-greenhouse-effect/

In particular, the section on the Second Law of Thermodynamics is relevant to some of your concerns. 66.159.87.108 (talk) 14:35, 30 June 2009 (UTC)[reply]

Seen it! Check the date.--Damorbel (talk) 15:10, 30 June 2009 (UTC)[reply]

Who cares if you seen it, and when. Have you read it, understood it, and do you agree with it? If not, what specifically do you disagree with, and why? blackcloak (talk) 03:24, 4 July 2009 (UTC)[reply]

Check the date. Yes, I know he posted it on April 1st, but it's not an April Fool's post — this one (from the same day) is: http://www.drroyspencer.com/2009/04/mr-gore-recants/ If you don't think Spencer knows what he's talking about, then I'm afraid I don't think you know what you're talking about. 66.159.87.108 (talk) 20:01, 5 July 2009 (UTC)[reply]

I'm not the slightest bit interested in your opinons on what I write about thermodynamics. If you had reason to think I was writing tripe you would be explaining why it was tripe, not just saying "I'm afraid I don't think you know what you're talking about". Get busy and work out for yourself what's wrong, then we'd both know something--Damorbel (talk) 20:21, 5 July 2009 (UTC)[reply]

Here goes

To briefly review...

There is nothing in thermodynamics that allows an absorbing/emitting component to induce a temperature gradient in a mixture of gases. The nearest you can get to this is convection which allows the hotter gas to rise to the top of the tropopause, the complete opposite to the so-called greenhouse effect. The fact that the Earth's surface is warmer than the tropopause is completely explained by the pressure gradient arising from the Earth's gravitational field.

That is true in optically thin gases, but not when the gases are nearly opaque in the frequencies of concern (optically thick). Remember, the stratosphere gets warmer with height and, therefore, has no convection. My (possibly incorrect) interpretation of this is that the stratosphere (and everything below it) is therefore opaque in the CO2 absorption bands. Q Science (talk) 17:09, 4 July 2009 (UTC)[reply]

It really makes no difference if it is O2/N2 or pure CO2. 1/ Do you not remember Kirchhoff's law of thermal radiation? The temperature of material in a radiation field (sunlight) is only dependent on the intensity of the field (= distance and temperature of the star) and not the composition of the material, it does not depend on the % of radiation reflected (albedo). I gave a link to Opacity (optics) with an explanation why opacity is irrelevant because it has a term for scattering that has nothing to do with the absorption and emission of radiation but is to do with reflection, a non-thermal process.--Damorbel (talk) 18:35, 4 July 2009 (UTC)[reply]

If that was true, then white objects and black objects, when placed in direct sun light, would have the same temperature. Q Science (talk) 16:14, 5 July 2009 (UTC)[reply]

Yes they would, as would transparent ones and highly reflecting ones. Look at it this way, some of the material absorbs and emits radiation, some of it reflects it; your white object absorbs less than it reflects, the black one the other way round but for each case the amount of material active in absorbing and emission is the same, a large % for the black and a small % for the white thus the outgoing heat always balances the incoming.

It is important to remember the "thermal equilibrium" requirements in Kirchhoff's law of thermal radiation. This means that heat must not come from (or go to) another place e.g.it has been suggested that Jupiter has an internal heat source. Internal heat would raise Jupiter's temperature in a way that depends on the planet's emissivity. For a the same amount of internal heat a planet with low emmissivity would become hotter than one with a higher emissivity. Put another way, the outgoing heat is no longer equal to the incoming heat.

Which is true for Jupiter, the outgoing heat is greater than the heat it gets from the Sun. Q Science (talk) 00:16, 6 July 2009 (UTC)[reply]

For objects that do not have uniform colour or are not rotating etc. the surface temperature is different from place to place but the average remains the same, it has to, otherwise where would the excess/shortfall of heat come from/go to?

One final point, looking from a distance the black body might appear to be hotter than the white one because it radiates more in infrared (because it absorbs more...)--Damorbel (talk) 18:08, 5 July 2009 (UTC)[reply]

If you use a thermometer, the black object actually is hotter, ie, it stores more heat. Q Science (talk) 00:16, 6 July 2009 (UTC)[reply]

??? I don't understand. At a given temperature a black object (high emissivity radiates more heat than a white one, it also absorbs more at whatever temperature. With no other heat source the temperature is stable and the outgoing heat equals the absorbed heat. Since the absorptivity and emissivity are equal then the colour is irrelevant (Kirchhoff's law of thermal radiation)--Damorbel (talk) 08:27, 6 July 2009 (UTC)[reply]

The absorptivity is in one frequency band. Here, black and white are not equal (0.9 vs 0.2). However, they emit in a different frequency band where the emissivities are almost equal (0.81 vs 0.86). This is because the Sun is hotter than the objects. (At the same temperature, black and white objects emit about the same amount of heat.) (See Satellite thermal control for systems engineers, Robert D. Karam, for very explicit details. Its a book, so the on line text is missing pages. You will like pages 16, 157, Table 6.1.) Q Science (talk) 15:43, 6 July 2009 (UTC)[reply]

I have already mentioned that for an object that is not uniform then the local temperatures generally are not the same. The satellite problem is much bigger, I know, I used to work on satellites. Different parts of a satellite produce heat so the thermal equilibrium goes out the window. Satellites are very far from uniform in shape. Take flat solar panels, if during manoeuvres they go edge on to the Sun their temperature drops very quickly, easily to -150C. Thermal control of a satellite is very important if a long service life (or any life at all) is required. There are programs to simulate thermal performance but the simulation generally has to be checked is an expensive device called a "Thermal vacuum chamber", atleast it used to be.--Damorbel (talk) 20:34, 6 July 2009 (UTC)[reply]

How quickly will it drop to -150C? I understand it takes 3 months to reach 45K (-228C). Q Science (talk) 04:04, 7 July 2009 (UTC)[reply]

There can be no general answer to this. The kind of solar arrays I had in mind are mainly silicon cells about 1mm thick mounted on 0.1mm kapton with copper interconnects, I imagine this has a very short thermal time constant in the order of minutes, far less than three months. Think of radiative cooling from your car roof, it is really quite quick. Satellite thermal control is very interesting but equilibrium conditions are not generally the problem. Heat transfer is the main difficulty, keeping heat sources from getting too hot and extremeties from getting too cold. For Earth orbiting satellites there are problems from night and day, this depends very much on the orbit, in synchronous orbit they only have "night" at the equinoxes and then a maximum of 2hr. None the less this causes thermal cycling that may be very stressful. Deep space missions have to have special provision, you can find a quick summary here [11], note the reference to the "traditional radio-isotope thermal generators". --Damorbel (talk) 06:28, 7 July 2009 (UTC)[reply]

Well, assuming that it is less than 8 hours, the question becomes, "Why doesn't the Earth's surface get that cold every night?" By definition, the answer is "the Greenhouse Effect". So the next question is, "How? What is it about the atmosphere that keeps the surface warmer than a satellite?" Q Science (talk) 18:11, 7 July 2009 (UTC)[reply]

According to this [12] the Earth loses heat at 235W/m², the atmosphere has a mass about 10tons/m². At night this 10 tons is cooling at 235W/m². 10 tons of air at about 270K contains about 270 x 10 x 1000 x 1000 = 2.7x10⁹J of energy. At 235W/m² during 12hr it will lose 235 x 12 x 3600 = 10.152.000J (10⁷J) meaning about 1 deg. temperature drop overnight. Now there must be a mistake here because common experience indicates that the atmosphere cools rather more than this. But the calculation is not out by orders of magnitude; one factor not included is that convection arising from the Sun's input will continue into the night because of accumulated heat from the Sun. Sufficient to say the surface temperature doesn't drop like on the moon because of the thermal inertia of the atmosphere, the ground and sea water, an inertia that is not affected by the presence of CO2. You must check what I wrote above about frost on calm, clear nights for another aspect of surface temperature. --Damorbel (talk) 20:46, 7 July 2009 (UTC)[reply]

But the surface cools by a lot more than that. The question is, Why does the surface lose only 10C instead of the expected 150C? To remove the question of thermal inertia, assume an ice chest (insulated) with the top off. Why doesn't the bottom get really cold? (Let's assume a desert to avoid dew and frost, both of which are importent.) Q Science (talk) 21:17, 7 July 2009 (UTC)[reply]

I really do not understand your expectation that a 10T mass of air can cool to -150C in a short time. No material changes temperature without a change of energy, see this table [13], air has a specific heat capacity of just over 1000J/kg/deg. that means that for a change of 1C, 1kg must lose or gain 1000J, the basis of my calculation. Now, when temperatures are changing, the atmosphere and the surface are a very complex interacting set of effects that depend very much on the weather, when considering global climate effects weather, as a local effect, is irrelevant, don't you think? I suspect you are back wasting time again.--Damorbel (talk) 05:32, 8 July 2009 (UTC)[reply]

I am sure you are not suggesting that the surface does not cool at night. And we both know that it cools a lot more than the atmosphere above it for precisely the reason you just gave. All I'm asking is, Why doesn't the inside of an insulated container get as cold as a satellite in orbit would get if the Sun was blocked for 8 hours? (And I agree with you, CO2 has nothing to do with it.) BTW, the heat of vaporization of water is 2,261.11 J/g (2,261,110 J/kg) and that completely explains why dew keeps the temperature from falling, but in a desert, there is no dew. Q Science (talk) 08:01, 8 July 2009 (UTC)[reply]

Your question is far too vague. What orbit is the satellite in? Is there an orbit with eight hours shadow? What are the precise thermal properties of this container? A few seconds thought should reveal that a sufficiently insulated container could be made with almost any thermal time constant, well over eight hours.--Damorbel (talk) 06:44, 9 July 2009 (UTC)[reply]

Sorry, I was thinking you could turn radiators edgewise to the Sun for a few hours and they would get cold. After all, you said that they would "quickly" drop to -150C. Since you did not provide a number for "quickly", I assumed 8 hours might do it. I was also thinking that objects at the surface surrounded with an insulator on all sides, except for the top pointing toward space, should lose heat pretty fast, just like the satellite. Something like placing a small black object in a Styrofoam ice chest without a lid. It seems like this should lose heat (and reduce temperature) just as fast as the satellite. For fun, let's just assume that they are both made from the same material so they would have the same thermal properties. (Sorry, I didn't mean to imply an orbit with 8 hours of shadow.) Q Science (talk) 07:28, 9 July 2009 (UTC)[reply]

This link [14] is to a book on satellite technology you will find a graph (Fig 10-5a) showing the thermal response of a solar array to a sudden change in sunlight, it shows the initial temperature change of about 50C in 10 minutes which is about what I would expect. Such a temperature change may well cause buckling and failure in such a large structure unless care is to avoid it. The second graph (Fig 10-5b) is for GSO (geostationary orbit), this shows a much faster and bigger change of temperature because thermal radiation from the Earth has little effect. The first graph is for LEO (low Earth orbit) where radiation from the Earth keeps the satellite warm even when it is in the shade.--Damorbel (talk) 09:35, 9 July 2009 (UTC)[reply]

Wow, great book. The temperature drops about 50C in 10 minutes, 100C in 35 minutes, and in high orbit about 240C in 70 minutes. Since those graphs prove that the greenhouse effect is real, the next question is, How does it work? Q Science (talk) 16:17, 9 July 2009 (UTC)[reply]

"Since those graphs prove..." Oh yes? May I ask how, please?--Damorbel (talk) 17:25, 9 July 2009 (UTC)[reply]

Well, why doesn't the roof of my car experience a similar drop in temperature? There must be some reason. The nonsense with the insulated cooler is just to stop wind and the thermal mass of the planet from making things complicated. Q Science (talk) 18:02, 9 July 2009 (UTC)[reply]

"Blanket"

Blankets, duvets, fibre insulation greenhouses etc. all do their job by confining air (gas) to the proximity of an object. If cool air has free access to a warm surface it will cool it by convection unless the circulation associated with convection is impeded; if the circulation is impeded the heat transport is by conduction. Blankets and double glazing both work because the conductivity of air is very low, this is what "creates a separation" i.e. the temperature gradient from inside to out side the blanket, greenhouse etc. Spencer states "A blanket – real or greenhouse — doesn’t actually create the separation between hot and cold…it just reduces the rate at which energy is lost by the hot, and gained by the cold." But time and again even GHE aficionados say that the effect is not the same as a real GH, real GHs do "actually create the separation between hot and cold" due to the confining effect of the glass on the air.

IT’S NOT A REAL GREENHOUSE

Spencer remarks on the difference between GHE and reality but then claims "the infrared atmospheric greenhouse effect instead slows the rate at which the atmosphere cools radiatively, not convectively." How do the GHGs "slow down the rate"? The only way you can get a temperature gradient in gas is to suppress circulation completely. But even that doesn't work for long. Temperature gradients in gases due to radiative heat transport virtually do not arise; any temperature gradient induced by radiation external to a given volume of gas is quickly removed by gas-to-gas radiation (within the volume itself) and also by diffusion and convection.

Without an absorbing emitting atmosphere the atmosphere would be very hot away from the surface

However even if the atmosphere was 600C the radiative temperature of the sky would be the extreme cold of space. Because heat rises you would get frosts even during the day in sheltered places

GH gases cool the atmosphere from what it would otherwise be. Therefore they are receiving conduction and radiation and emitting radiation in all directions. And therefore the sky temperature is not near absolute zero.

All you then have to plug in is Stephan-Boltzman net radiation heat loss using the temperature of the surface and temperature of the atmosphere to see that you are unlikely to get frosts during the day in sheltered places apart from when the air temperature is already quite low.

So called backradiation is just following on from prevosts 1791 exchange theory that all hot objects emit radiation based on temperature without reference to the temperature of other objects and just involves a reduction in cooling when the atmosphere object is being warmed by a nearby hotter earth surface object, as decribed by net radiation heat loss curves. Andrewedwardjudd (talk) 14:00, 18 March 2012 (UTC)andrewedwardjudd[reply]

2^nd Law of Thermodynamics

A moment's thought will reveal that Spencer's ideas on this are strictly his own:- "First of all, the 2nd Law applies to the behavior of whole systems, not to every part within a system", Oh really? This was, for me, the real April 1^st item in the article. Common experience shows that all objects, micro to mega, tend to equalise temperature both locally and globally.

Net flow?

Is there any other kind? If, as Spencer claims "A hot star out in space will still receive, and absorb, radiant energy from a cooler nearby star" then, since it is gaining energy it would get even hotter. Classical thermodynamics says that this can only happen by the application of external work, generally in the form of a heat-pump.

Again Stefan-boltzman says that the hot object emits more radiation than the cold object but the cold object also emits radiation but less, with no laws saying that radiation coming towards an object only gets absorbed due to the temperature of the emitter. Therefore there is a net transmission of an amount of radiation from the hotter object to the colder object.

The hot star emits far more radiation than the cold planet that emits radiation absorbed by the hot star. The warmed cold planet however slows down the rate of cooling of the hot star.

Andrewedwardjudd (talk) 14:06, 18 March 2012 (UTC)andrewedwardjudd[reply]

Further

Oft repeated, nonsense "In other words, a photon being emitted by the cooler star doesn’t stick its finger out to see how warm the surroundings are before it decides to leave." But the hotter one does emit more and more energetic photons and the temperature effects are, according to Quantum electrodynamics (QED), dependent only on the probability of these photons being aborbed, thus one object cools and the other gets warmer.

The hot star cools while keeping the cold planet from being the temperature of space. Therefore the hot star does not see the temperature of space in the direction of the warmed but cold planet. Net radiation heat loss curves do the rest. Andrewedwardjudd (talk) 14:09, 18 March 2012 (UTC)andrewedwardjudd[reply]

And the jars?

The jars have no special influence on each other; take away one jar and you take away its heat without waiting for it to cool naturally. If the jars are thermally connected their temperature may equalise before they cool completely, there is nothing unusual about that.

Kirchoff's Law

He refers to "a misunderstanding of Kirchoff's Law" (Kirchoff's law of thermal radiation) and goes on to say "the infrared opacity of a layer makes that layer’s ability to absorb and emit IR the same". Presumably he means Opacity (optics) where it says (correctly) "it describes the absorption and scattering of radiation" thus his analysis adds in scattering. Scattering, like reflection, is a process that redirects radiation without absorbing it, it is not a thermal process. This is one of the GHE's received wisdoms and it is an error that permeates the whole matter, not least in calculating a planetary temperature [15].

Frost and dew

In a clear sky the absorption and emission by the atmosphere takes place at high altitude where the troposphere is very cold. Since, in calm conditions, heat transport accross the surface is poor there can be a local drop in temperature since the radiated heat is not replaced quickly by the normal surface transport effects i.e. wind and waterflow.

THE GREENHOUSE EFFECT WORKS…FOR NOW

He claims:- "The greenhouse effect is supported by laboratory measurements". How does he simulate the effects of gravity that gives the atmosphere it's temperature gradient? How does he simulate the convection currents and water evaporation that transport surface heat into the upper troposphere. Prof. Spencer may be sceptical on doom scenarios but his physics is not too hot!--Damorbel (talk) 14:23, 4 July 2009 (UTC)[reply]

Pressure and temperature

The last sentence, here quoted, has no basis in science: "The fact that the Earth's surface is warmer than the tropopause is completely explained by the pressure gradient arising from the Earth's gravitational field." The existence/presence of a pressure gradient does not imply anything about temperature (ref. to his use of the word warmer) or the establishment of a temperature gradient. For one thing, if such a relationship existed, we'd have an equation that describes it. In fact the heat equation helps us show that, in the limit (time), 1) for the case of no boundary conditions, the temperature of a body must reach a uniform and constant value, and 2) for the case (one dimensional) of two boundary conditions (with different boundary temperatures), the temperature must be first order (linear) in distance between the two boundaries. If you need a longer explanation (like how to squeeze out valuable information from differential equations without actually solving them), just ask a specific question. Not until Damorbel can get this point straight in his head is there any reason to proceed to lower level (i.e. all the other) issues. When two people disagree, you have to find and resolve the differences in the underlying basis (assumptions) of that disagreement. My comment above identifies the problem assumption. In my past experience with Damorbel, when we get to this point, he just freezes up, or off we go into lala land. Nothing responsive to the core idea is forthcoming. Perhaps we'll get some strange indecipherable, tangential comment. He simply does not obey normal rules of discourse when faced with irresolvable contradiction. This is not meant to be critical of him; it's just my perception of his behavior. blackcloak (talk) 07:17, 9 July 2009 (UTC)[reply]

Tell me Blackcloak, what is the effect on the temperature of the change of atmospheric pressure with altitude? It is well known from the Combined gas law that pressure, temperature and density of a gas are all interrelated, the concept of a stable atmosphere having a uniform temperature (on a global scale) is simply not sustainable--Damorbel (talk) 09:35, 9 July 2009 (UTC)[reply]

I am sure that you are aware that the stratosphere gets warmer with height. While the tropopause (at 11km) is between -45C and -70C, the stratopause (at 50km) is about -3C. In winter, in the north where the lakes freeze, the stratopause temperature is higher than the surface temperature. Q Science (talk) 16:17, 9 July 2009 (UTC)[reply]

"The stratopause temperature is higher than the surface temperature." Yes, and the temperature of the Thermosphere gets above 2000K, so what? The key to this is the fact that convection stops at the tropopause. Most relevant also: 90% of the atmosphere is below the tropopause, it is this 90% that completely dominates atmospheric effects on the surface temperature.--Damorbel (talk) 17:25, 9 July 2009 (UTC)[reply]

Your question was "what is the effect on the temperature of the change of atmospheric pressure with altitude?" The fact that the stratosphere temperature increases with a decrease in pressure indicates that there is NO correlation between temperature and pressure. Therefore, something else is causing these effects. The decrease in pressure explains why air cools as it rises (via the Gas law), but it does not explain why there is a cooler place to rise to. Q Science (talk) 18:02, 9 July 2009 (UTC)[reply]

My mistake, I should have written "change of tropospheric pressure with altitude", the gas laws only apply in the troposphere, above the tropopause the pressure is too low, there is no convection and little heat so the stratosphere has very very little effect on the surface.--Damorbel (talk) 21:00, 9 July 2009 (UTC)[reply]

I am pretty sure that the gas law still applies at the mesopause. Once the pressure is really low, then molecules travel many feet before bumping into other molecules. Even at that point, the gas law still applies. (Or please provide a reference that says otherwise.) In fact, while I agree that there is no convection in the stratosphere, I see no reason that there would not be convection in the mesosphere where the pressure is even lower. Q Science (talk) 06:09, 10 July 2009 (UTC)[reply]

No the gas laws do not apply at the mesopause or even above the tropopause basically because the thermal energy density is dominated by other energy sources, see here [16] the postulates of kinetic theory no longer apply above the tropopause.--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

First, you used the word "effect," which implies an underlying cause. Effect is the wrong word, if you want to be clear. It seems you are asking what is the change in temperature as a function of time,

I certainly am not, --Damorbel (talk) 21:00, 9 July 2009 (UTC)[reply]

Do you now see how your imprecise language leads to confusion? blackcloak (talk) 00:16, 10 July 2009 (UTC)[reply]

If you think "what is the effect on the temperature of the change of atmospheric pressure with altitude?" is imprecise then you must have a hard time following this discussion. The variation of atmospheric pressure with height is often approximated by meteorologists using a concept called Scale height, there you will find discussion about the temperature also.--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

Let me guess. You don't. blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

I don't what? (Please respect my request to post in the new section http://en.wikipedia.org/wiki/User_talk:Damorbel#Pressure_and_temperature_.2F2)--Damorbel (talk) 19:54, 10 July 2009 (UTC)[reply]

Eliza? Is that you in there? blackcloak (talk) 04:39, 11 July 2009 (UTC)[reply]

but this is purely informed speculation. Next, you specify pressure with altitude, which is pretty clearly dP/dz. So, with these interpretations, you are asking what is the equation that relates dT/dt to dP/dz. Well, first of all, you can not use the combined gas law directly because it does not recognize the continuous change from one state to another; it tells you about the end result after a change. The gas law also requires keeping one of the three P,T,or V constant.

only if you integration by parts which gives the wrong answer since P, V, and T vary jointly, it is quite impossible to hold one steady while varying either of the others.--Damorbel (talk) 21:00, 9 July 2009 (UTC)[reply]

You'll have to explain how you think integration by parts allows you to create a differential from a gas law equation that does not have time as in independent variable. Actually, within the realm anticipated by the gas equation, you DO hold one quantity "steady." If it is not "steady," you wait until to return to the original value. For instance, in Boyles law, a change in pressure will lead to a change in temperature, but before you determine the ending volume, you have to wait for the temperature to reach, or drive it to, the initial temperature. blackcloak (talk) 00:16, 10 July 2009 (UTC)[reply]

When you say about Boyle's law "you have to wait for the temperature to reach, or drive it to, the initial temperature" this waiting is to allow for a change in energy (cooling/heating) as a result of the work done changing the pressure. In the atmosphere there is no work involved (no change in energy) in moving a mass of gas up or down in the atmosphere other than that needed to overcome inertia and friction, never the less the temperature of the mass drops as it goes up and rises as it descends. When considering a mass of gas in the atmosphere you must not forget that in addition to the P V T energy it also has potential energy (mgh) proportional to its altitude--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

Whatever the reason for having to add or remove thermal energy is, the final temperature must reach the initial temperature for the results of Boyles law to apply. Boyles law says nothing about why there may be a temperature change following a rapid change in pressure. blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

Regarding the sentence that states in part "... there is no work involved in moving a mass of gas up ..." is wrong. When you add potential energy to a mass by lifting it against gravity, you must do work. That work (energy) associated with lifting the mass can not contribute to the total thermal energy content of the lifted mass. blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

"you must do work". No, it is self evident that a volume of air moving up and down does not need any force to lift it. If you add heat to a volume of air it will rise steadily to the tropopause, if you had to do work then it would only rise while you apply the force. It is quite comparable to a submarine, with neutral buoyancy the sub stays at the same depth, make the buoyancy just a tiny bit positive and the sub will rise all the way to the surface--Damorbel (talk) 19:54, 10 July 2009 (UTC)[reply]

Moving up and down means that there is an acceleration, however slight, and since F=ma, there must be a force to accomplish this. Basic Newtonian mechanics. What you find self evident is in direct conflict with Sir Newton's now accepted laws of motion. That dates your understanding of certain things to somewhere preceeding 1664 (don't hold me to the date, I did not bother to look it up). blackcloak (talk) 04:31, 11 July 2009 (UTC)[reply]

Regarding the sentence which states in part "never the less the temperature of the mass drops as it goes up and rises as it decends" appears to be wrong. Just by moving it up can not change its temperature, provided you are measuring the temperature at the same relative point in the mass. By that I mean that if you make an initial measurement of temperature at 10km above the surface, and then raise the entire mass of air from the surface up by 1km, and then make the temperature measurement at 11km (the same relative point) above the surface, there will be no change in temperature. Now you may be thinking that there is a change (decrease) in pressure that then leads to a change (lower) in temperature (volume increase also). This too is wrong because the pressure at any point is proportional to the mass above that point. By moving the entire mass up 1km, and then making a pressure measurement at the 11km point, the mass above that point is the same as it was initially. (I am of course assuming that the force of gravity is constant over the 11km, which is a good approximation, and not an effect that you considering in your argument since you have not mentioned it.) blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

If I read you correctly your concept of "moving a volume up 1km" means lifting the volume + the gas above it 1km then that is quite unrealistic, it would leave a vacuum underneath the moved volume. No, when you move a volume up 1km you convert some of its pressure energy (as it expands) and some of its thermal energy (as it cools) into gravitational potential energy i.e. the energy it has due to its height above the surface.--Damorbel (talk) 19:54, 10 July 2009 (UTC)[reply]

Thought experiments are always unrealistic. No, there would not be a vacuum; the lifting would be done by air release at the surface (if you want to know how the thought experiment would be conducted) for the purpose of lifing and doing work on the atmosphere above it (I thought that was obvious.). And no, the concept is not mine. The concept and the words are yours. See above. Of course you'll argue that I misinterpreted your words, which we both know is an all too common occurrence. But to paraphrase my earlier remark, the communication errors are a direct result of your imprecise language. blackcloak (talk) 04:31, 11 July 2009 (UTC)[reply]

It also has has a constraint described as a closed system, a difficulty that takes some degree of careful argument to overcome. Now the heat equation offers some help if you can convert the question to one where the relationship between dT/dt and dT/dz is asked. Here is where the heat equation offers an answer provided you can (en)close the system by arguing that no mass or energy leaves or enters the volume of interest. But even if you can not provide a resonable argument for ignoring, substantially, external effects, you still know what the trends over time must be- a movement towards the mean. As for your comment about the typical timeframe of changes in temperature of the atmosphere, you are correct. The external forces (outside the closed system) serve to constantly change the conditions; mathematically this becomes a differential equation with a forcing function.

I am not asking for a forcing function since the change of pressure with height does not involve any change of energy in (i.e. work done on/by) the troposphere--Damorbel (talk) 21:00, 9 July 2009 (UTC)[reply]

The thermal energy provided by the Sun is the forcing function. Perhaps you are now only considering what happens at night, when the external source of energy is essentially zero. Again, you're not being perfectly clear. blackcloak (talk) 00:16, 10 July 2009 (UTC)[reply]

The Forcing function (differential equations) of the Sun's energy gives a day/night variation around an average temperature which has no effect on the pressure/temperature gradient in the troposphere, even the most ardent GHE enthusiast recognises this, you are just wasting time pushing it.--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

I agree, we need not pursue this. Afterall it was an afterthought in my comment. You never responded to the central point, that of a closed system (a tactic I predicted earlier). blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

A good place to start learning how this is handled mathematically is Kreyszig, Advanced Engineering Mathematics, 7th edition, pg 99, nonhomogeneous equations. The key point, for the present purposes, it to understand that the external forces (energy from the sun causes changes in the temperature of the earth's surface and the atmosphere) can act over a time period that is short compared with typical time periods associated with heating and cooling processes (dT/dt) in the atmosphere and oceans. If you understand electrical circuits, this is like a full wave rectifier driving an RC circuit to produce a (mostly) constant output voltage, and then considering what happens at the output when the AC supply voltage fluctuates dramatically and quickly (relative to the RC time constant). blackcloak (talk) 19:25, 9 July 2009 (UTC)[reply]

I think there may be confusion between two matters, Q Science was discussing the day/night change in temperature of a satellite; my question to you was about change of tropospheric temperature and pressure with altitude only because you do not seem to realise that the global (average) the tropospheric temperature and pressure profiles of a given planet are defined by the gas laws only.--Damorbel (talk) 21:00, 9 July 2009 (UTC)[reply]

They're only "defined by the gas laws" in your imagination (mostly because you are not taking the time to define your closed system). You have not learned the difference between correlation and causality. When, and if, you read the gas laws carefully enough (and btw the wikipedia entries are not particularly thorough is describing the constraints) perhaps you'll understand why you are not applying them properly. Now, to try to understand what you mean specifically, when you use the term "temperature and pressure profiles" are you referring to the instantaneous values or the long term averaged (with averaging over a year) profile? Or are you talking about the spatial average over all points on the globe (or given planet, in your words)? Again, you are not being precise enough for unambiguous communication. Please, add all the extra words you need in order to describe exactly what you are saying. Reread your contributions while trying to anticipate how your reader may misinterpret what you are saying. Then fix any problems you foresee. I spend a fair amount of time responding to you. I get the feeling you are just rattling off quick, poorly thought through, responses. blackcloak (talk) 00:16, 10 July 2009 (UTC)[reply]

What I have in mind is common to all planetary atmospheres, see here [17] and here [[18]]

The gas laws define the P V T relationship on the basis of kinetic theory, kinetic theory is very successful but does not apply at low pressures e.g. above the tropopause, this limitation applies because the kinetic theory postulates

This section has now become unmanageable, please put your responses in the new section below.--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

Pressure and temperature /2

New contributions here, please

(Repeated from above) No the gas laws do not apply at the mesopause or even above the tropopause basically because the thermal energy density is dominated by other energy sources, see here [19] the postulates of kinetic theory no longer apply above the tropopause.--Damorbel (talk) 08:16, 10 July 2009 (UTC)[reply]

Neither of those appears to support your statement. Specifically, all the postulates of kinetic theory still apply. I tried to find a definition for the pressure where the Knudsen regime begins, but was not able to. Q Science (talk) 14:00, 10 July 2009 (UTC)[reply]

Follow your link the kinetic theory postulates to system and explain to me how your arguments conform to this requirement. (Move this part below as you see fit.) blackcloak (talk) 19:15, 10 July 2009 (UTC)[reply]

See below.--Damorbel (talk) 19:54, 10 July 2009 (UTC)[reply]

If the kinetic theory still applied then the temperture would continue to drop and convection would still take place in the stratosphere The kinetic theory fails because, in the statosphere, the external input of energy to it exceeds that from the troposphere, that is the postulate "The average kinetic energy of the gas particles depends only on the temperature of the system" is no longer met, "the system" being the transport of heat in a system comprising the Sun's energy input to the troposphere and the Earth's surface and the transport of that energy through the troposphere by convection, water vapour and radiation. As the stratosphere article explains the main source of heat for the stratosphere is direct input of ultraviolet radiation from the Sun. The stratosphere is really quite different from the troposphere, if you look here[20] you will see that heat transport is done by quite dissimilar processes.--Damorbel (talk) 19:03, 10 July 2009 (UTC)[reply]

I did. Nothing applies. blackcloak (talk) 04:31, 11 July 2009 (UTC)[reply]

A "nothing applies" response tend to close the discussion, doesn't it? I am afraid you will have to explain what you think does explain the workings of the troposphere and the stratosphere, and explain the relevance (if any) of the gas laws; if no relevance is acceptable to you, do you have another explanation?--Damorbel (talk) 06:47, 11 July 2009 (UTC)[reply]

I'm simply not interested in being redirected to a conversation about minutia when the basics are not resolved. Why don't you tell us what it would take to convince you that the gas laws do not predict a temperature profile (as a function of altitude) based on the (reasonably easy to predict) pressure profile (as a function of altitude)? If you think the gas laws provide the means for computing the temperature profile, then give us the equation. (Make the system as simple as you have to (and tell us exactly where you are making the simplifications) in order to create the equation.) Remember, the gas laws have no time dependencies built into them, i.e. the equations do not have time as an independent variable, so you can only hope to predict a very long time average when an external source of thermal energy is, essentially, pulsing. blackcloak (talk) 06:56, 12 July 2009 (UTC)[reply]

Is there something you don't get about the combined gas law that I gave (09:35, 9 July 2009)? Below the tropopause the product PV is constant so, knowing the pressure P at a given height, you can calculate T. You should also know the surface temperature, if you look at page 4 of this pdf you can see just how it works.--Damorbel (talk) 08:47, 12 July 2009 (UTC)[reply]

Maybe we can get somewhere if I just play dumb. You say PV is a constant below the tropopause. Let's call that constant K=PV. Now the gas law says PV=NrT, which now means that K=NrT, or solving for T we have T=K/(Nr), which is a constant (no altitude dependence). Not too complicated, is it? blackcloak (talk) 02:24, 13 July 2009 (UTC)[reply]

It depends what you mean by "dumb". The altitude comes from stacking up volumes of gas against the force of gravity; V=NrT/P (at the surface P is high and V is low; at the tropopause it is the other way round.)--Damorbel (talk) 10:30, 13 July 2009 (UTC)[reply]

So the product PV is not a constant? I don't get what you're saying if you're actually objecting to the conclusion T=K/(Nr). And, if "the altitude comes from stacking up volumes ..." then you should be able to write an equation that gives T as a function of altitude. What is that equation? (We'll come back to the time dependency part after we straighten out what happens in the static case (i.e. equilibrium/steady state case) later.) blackcloak (talk) 04:47, 14 July 2009 (UTC)[reply]

I may have said PV is constant, it is PV/T that is constant i.e. the energy; this comes from the relation PV=RT Joules/mole. This is true in a container but in the atmosphere it is PV/T + potential energy (pe) that is constant (PV/T=R-pe), from this you can see that with increasing altitude some of the thermal energy is transformed into potential energy. In the atmosphere a unit mass of air at the surface has no potential energy (h=0). At the tropopause, let us say 10,000m, it has potential energy (pe)= mass x g x 10,000J. Thus a stable condition exists in the atmosphere where the temperature drops with increasing altitude. This is not easy to solve with a simple equation, a fairly easy way is to divide the mass of the atmosphere into equal mass parts and solve for equal energy density (J/kg). You know that each part has the same total energy but it is split between thermal and potential energy, you can do this is with a spread sheet. I'm sorry if this is not very clear but it is not an easy matter, the difficulty arises because, with increasing altitude, both temperature and pressure change, they change together and with a fixed energy ratio i.e. the thermal energy (te) and potential energy (pe) of a mass of gas change with height in a fixed ratio, pe = 2te, this is explained by the virial theorem.--Damorbel (talk) 09:24, 14 July 2009 (UTC)[reply]

I don't think the virial theorem should be used here because it relates potential energy between objects, such as the gravitational potential between stars. In your argument, you are not interested in the potential energy between air molecules (where it would apply), but instead between the surface and some part of the atmosphere. Think of a rock on the surface, it has a potential energy with respect to the center of the Earth, but its temperature has nothing to do with how far it is from the center of the planet or its mass. In the case of the atmosphere, the pressure makes the equation fail to be useful.

The virial theorem applies to all materials that comprise particles, the physics of particles depends in no way on their size and number, it matters not if they are stars or molecules, it merely shows the relationship between potential and kinetic energy in a thermodynamic system.--Damorbel (talk) 09:56, 15 July 2009 (UTC)[reply]

Also, PV/T=R-pe does not make sense because pe depends on mass and none of the other terms care about mass. I think that the number of molecules (n) and their density need to be added for the equation to work. Q Science (talk) 15:47, 14 July 2009 (UTC)[reply]

R is in joules per mole as is PV, in a gravitational field the mass of a mole becomes pressure.--Damorbel (talk) 09:56, 15 July 2009 (UTC)[reply]

So let's see if this makes any sense. You are now say that PV/T is energy. P is force per unit area, so the units are mass times acceleration divided by distance squared. This is mass times distance divided by the quantity seconds squared times distance squared. So P has units of mass divided by the quantity seconds squared times distance. Therefore PV has units of distance cubed times mass divided by the quantity seconds squared times distance. This means PV has units of distance squared times mass divided by seconds squared, which is the equivalent of units of mass times velocity squared, which means PV has units of energy. When we ask what the units of PV/T are, we must conclude that they are energy divided by degrees, which happens to be the units of entropy. So are you talking about energy or entropy? blackcloak (talk) 04:17, 15 July 2009 (UTC)[reply]

Entropy (S) is measured in Joules, as is PV, as is RT. When referring to gases i.e. PV and RT, it is usually defined in J/mol. --Damorbel (talk) 09:56, 15 July 2009 (UTC)[reply]

Can you cite a source for your contention that Entropy has units of joules? Or, do you think temperature is a unitless measure, like N? blackcloak (talk) 16:22, 15 July 2009 (UTC)[reply]

The following is from Entropy. Q Science (talk) 19:11, 15 July 2009 (UTC)[reply]

It can be seen that the dimensions of entropy are energy divided by temperature ... "joule per kelvin" (JK−1).

You need to learn not to spring traps. blackcloak (talk) 04:35, 21 July 2009 (UTC)[reply]

You conclude that "... the thermal energy (te) and potential energy (pe) of a mass of gas change with height in a fixed ratio, pe = 2te ..." But you also say that at the surface, the pe is zero, so at the surface the te must also be zero. With both te and pe zero, can the temperature be anything other than absolute zero (which I think you must know is impossible)? blackcloak (talk) 04:17, 15 July 2009 (UTC)[reply]

As you noted it is the change of thermal energy of a mass of gas as it gains potential energy when rising in a gravitational field. For a mass of gas held together by gravitational self attraction the whole energy is split 1:2/pe:te. The distribution of gravitational pe of a rocky mass is different from that of a gas because solids (and liquids) are scarcely compressible and they stratify, gases form easily predictable density rofiles in a gravitational field.--Damorbel (talk) 09:56, 15 July 2009 (UTC)[reply]

Please note that my only usage of the word change was when I quoted you. The way you used the term change implies you believe te is a function of height. If you had wanted to suggest that there was a differential relationship, you would have said that the change in te wrt height is a fixed ratio (whatever you mean by a fixed ratio). If you had simply wanted to say that, at any altitude (below the tropopause), the te and pe maintain a fixed relationship, you would have said the te and the pe always maintain a fixed ratio. What you have not said, specifically, if I am not mistaken, is what governs the temperature. Is just the thermal energy responsible for establishing the temperature of a mass of gas? blackcloak (talk) 16:22, 15 July 2009 (UTC)[reply]

Despite my stumblings it remains true that the so-called blanket effect of greenhouse gases is a complete misunderstanding of atmospheric physics, the warmer surface is actually the effect of gravity compression of O₂, N₂ CO₂ or any other gas and nothing at all to do with absorption or emission.--Damorbel (talk) 09:56, 15 July 2009 (UTC)[reply]

And I thought I was stubborn. You live and you learn. Well, that's the expression anyway. blackcloak (talk) 16:34, 15 July 2009 (UTC)[reply]

I don't claim to understand the matter completely but when a group of people (scientists?) claim that the Earth "radiates in the infrared like a blackbody" then even those with just a little knowledge must surely have their curiosity roused, I mean it's just like claiming the Earth is flat.--Damorbel (talk) 05:33, 17 July 2009 (UTC)[reply]

Just think of statements like the one you quote as over simplifications that are not intended to be taken too literally. Most simple minded explanations of the greenhouse effect suggest that heat is "trapped" by greenhouse gases. That one word leads to a terrible misunderstanding of what happens in reality. Most simple minded explanations of the greenhouse effect talk about the heating of the atmosphere and the surface due to greenhouse gases. This over simplification gives uninformed readers the idea that the earth is predominately heating up. 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, in addition to faster heating. To make things worse, the IPCC contributes to the confusion, and wikipedia editors contribute to the misunderstandings by not accepting edits that correct errors promulgated by IPCC (and other POV) material designed to communicate in overly simplified language. I suggest you not try to see scientific truth in newspaper article level descriptions of what you know to be a very complicated set of interactions among a variety of physical processes. Get over constructing mental models that require you (and ask us) to abandon accepted physical laws. And find some physicist you trust (obviously we haven't achieved that level of respect) to help guide you through your re-learning journey (as I mentioned oh so many months ago now). blackcloak (talk) 06:54, 20 July 2009 (UTC)[reply]

Damorbel, I think I finally see what you are trying to say. You appear to be describing the Dry Adiabatic Lapse Rate (DALR) which is a function of the gravitational constant and the average atomic mass of the atmosphere. This value is constant and does not change with height. It determines how much a parcel of air will warm or cool when it is moved from one altitude to another. However, the DALR describes a change in temperature, not the actual temperature. When the air over a hot parking lot rises, it cools at the DALR, and quits rising when it has cooled to the same temperature as the surrounding air. When air flows from central Antarctica (high) to the coast (at sea level), the air warms by gravity compression at the DALR. As a result, the Antarctic coast will be 20C (or more) warmer than the central plateau. The Environmental Lapse Rate (ELR) is the actual change in temperature with height, and this varies all the time. The different layers of the atmosphere are defined (in part) because they each have an ELR different than the layers above and below. However, the DALR is the same in all layers of the atmosphere. Q Science (talk) 19:11, 15 July 2009 (UTC)[reply]

Lapse rate is only an approximation, if the formula gives a constant value with height then it is fundamentally inaccurate since the basic change of temperature with height is linked to pressure which declines exponentially with height. The actual temperature is a function of the insolation as well as height, but the value of the measured lapse rate is the indication it gives of the changes taking place due to the movement of air, quasi static conditions such as inversion etc. all of which are powerful factors in weather forecasting but have little to do with the basic, gravity generated temperature profile. --Damorbel (talk) 12:11, 17 July 2009 (UTC)[reply]

The logarithmic change in pressure causes a logarithmic change in density. As a result, the DALR is constant. The error (approximation) is due to the fact that g (the gravitational constant) is not constant, but decreases slightly with altitude. There is also a factor related to specific heat being a function of pressure. (DALR=g/cp page 77) The actual temperature is not a function of height, but is instead a function of exactly the other items you mentioned above. To be clear, there is no "gravity generated temperature profile", but gravity does define the DALR which affects the maximum possible surface temperature. In general, the daily maximum surface temperature is more than 70°C below the DALR limit. Q Science (talk) 21:26, 17 July 2009 (UTC)[reply]

Objects in orbit

I saw this post by you

"Two bodies in the same orbit, one of black carbon (albedo 0.05) and one gold plated (albedo 0.95) will have the same actual temperature but wildly different infrared signatures, giving apparently equally wildly different temperatures."

Actually, they will have the same temperature only if the albedo is the same as the emissivity. (Actually, if the solar absorptivity is the same as the IR emissivity.) However, because that is seldom the case, the actual temperature of objects in orbit very much depends on the surface material. (See Satellite thermal control for systems engineers, Robert D. Karam, page 158, table 6.1) Q Science (talk) 22:35, 19 November 2009 (UTC)[reply]

PS: You might find this interesting - AIRS - Atmospheric Infrared Sounder currently flying on the NASA Aqua Mission

Thanks for your comment. Having worked on satellite solar panel design, perhaps I can use this experience to expand my explanation. First of all the statement :"Two bodies in the same orbit..." applies only to the average temperature of spinning spherical bodies such as planets. One of the problems with flat solar panels is their orientation towards the Sun, if they are in the desirable "Sun facing" attitude they will be warm but hopefully not too warm because their output voltage (and thus efficiency) would drop. However if, during manoeuvres, they should go into an edge-to-the-Sun attitude all solar heat input is lost and the temperature drops drastically. A flat solar pannel has a low thermal time constant and may be damaged by thermal shock since its temperature can suddenly go below 100K. I know some solar cell designs that were tested by plunging them in liquid nitrogen to see if they cracked.

A body larger than a point (unless it conducts heat perfectly) will have different temperatures at different places, even the spinning sphere will have a warm equator and cole podes.

Many satellites are far from spherical, fortunately the ones I dealt with were spinning cylinders, making the thermal calculations fairly easy. Three-axis stabilised boxes with solar "paddles" that cast shadows are a pig and sometimes need active thermal control. Heat dissipation is not uniform throughout any satellite and this, together with the Sun being eclipsed at times, keeps a satellite thermal design engineer quite busy.

The simple science behind the statement your question is that only material (such as CO₂) that absorbs heat can radiate it. Where thermal equilibrium exists this absorbing material radiates what it receives, otherwise its temperature would rise (or fall) until it reaches an equilibrium temperature. Material that does not absorb/radiate radiation may reflect it (albedo) or transmit it (transparency). Material that reflects or transmits radiation is neither heated nor cooled by the incedence of radiation. Material that is composite (but perfectly heat conducting!) i.e. both absorb/radiates and also reflects/transmits radiation has a temperature that is only dependent on the mean intensity of the radiation is receives. The material that reflects/transmits has it temperature defined by the material that absorb/radiates. In the case of solar systems the mean intensity is arrived at by measuring the distance, the temperature and the size of the heat source(s) i.e. the star; note that the distance is the only variable in this calculation.

One final point, if the body is heated by internal energy its temperature now is partially determined by its thermal emissivity, high for black carbon and low for gold plating. Similarly if a body is heated transiently e.g. a nuclear explosion, the black one will quickly radiate its excess heat and cool to its equilibrium temperature, the gold plated one will do this much more slowly because it is such a poor radiator (that's why shiny teapots are the best!)--Damorbel (talk).

A rough greenhouse model

Hello - I became interested in the discussion on black body and the temperature of the earth. Heres a rough model of the greenhouse effect: Let E be the total energy incident on the earth from the black body sun:

E(T_{S})=\left(\sigma T_{S}^{4}\right)\left(4\pi R_{S}^{2}\right)\left({\frac {\pi R_{E}^{2}}{4\pi D^{2}}}\right)

Let F be the total energy emitted by the earth as a black body

F(T_{E})=\left(\sigma T_{E}^{4}\right)\left(4\pi R_{E}^{2}\right)

What is presently done is to say that a fraction $\alpha$ of the incident radiation E is reflected, the rest absorbed, and re-radiated by the earth as F:

(1-\alpha )\,E(T_{S})=F(T_{E})

and then solve for Earth's temperature $T_{E}$ . The problem is that the infrared portion of the earth's radiation does not make it out, it gets reflected back by the greenhouse effect. Infrared radiation starts at wavelengths around 700 to 2500 nanometers and goes higher in wavelength. A VERY ROUGH model is to take the average and say that everything below about 1600 nanometers gets out, the rest does not. If you do the calculation, you get that only about 57 percent of the Earth's black body radiation actually escapes. So if you say $\gamma =0.57$ , then a better equation is:

(1-\alpha )\,E(T_{S})=\gamma \,F(T_{E})

If you solve that for the Earth's temperature you get:

T_{E}=T_{S}{\sqrt {\frac {R_{S}{\sqrt {\frac {1-\alpha }{\gamma }}}}{D}}}

Solving you get $T_{E}$ =286.0 Kelvin or about 12.9 degrees Celsius or about 55.5 degrees Fahrenheit. Pretty good guess. (But no references) PAR (talk) 04:09, 9 January 2010 (UTC)[reply]

Let me better understand your estimate for gamma of .57. You take the average, in wavelength, of the 700 and 2500 nanometer limits .5*(700+2500), then compute the relative power in (long?) one of the two portions (700 to 1600, 1600 to 2500) relative to the whole (700 to 2500) using a T^4 power vs temp relationship (or some suitable rough estimate over the two wavelength ranges), yielding the 57/100 ratio? It would be nice to see the details of this critical part of your calculation. blackcloak (talk) 04:10, 4 March 2010 (UTC)[reply]

Thank you for taking the trouble to post this. The question of surface temperature the Earth is far from simple. We all know that it is heated by radiation from the Sun and cools itself by radiating into deep space. It is the same for all objects in orbit, with or without an atmosphere.

To understand the factors governing the surface temperature of a planet it is best to start with a very simple model a model that has many shortcomings when trying to represent the Earth but has the merit of seperating the variables, so we propose a rotating spherical body that conducts heat well and is small enough to have a uniform temperature. Such a body has a surface area 4 times its projected area ("its disc"). Let us suppose that it is a black body according to Kirchhoff's definition i.e. it absorbs all radiation falling on it, it is also opaque (transmits nothing) and it reflects nothing; further it radiates according to its temperature. Such a body absorbs solar radiation which heats it, its temperature will change until it is radiating the same energy as it receives. This simple relationship is detailed by the formula:

{\frac {T_{E}}{T_{\odot }}}={\sqrt {\frac {R_{S}}{2D}}}

Now suppose we modify our black body by covering a percentage of its surface evenly with small, perfect mirrors so that this percentage of incoming radiation is reflected, it is no longer a black body. The light reflected by the mirrors does not contribute to the heating of our non-black body, also these mirrors do not radiate heat (this is the logic of shiny metal tea pots, thermos flasks and multi-layer insulation). You will realise of course that the mirrors also reduce the surface area available for radiating heat because they do not radiate heat at any temperature. The consequence is that the temperature of our now non-black body is exactly the same as that of the black body, the reduced absorbing area being exactly compensated by the reduced emission area.

This argument is equally valid for all possible non-black bodies, coloured or partially transparent. Further, real planets are not good conductors of heat and may be watery and have significant gassy atmospheres but the argument is still valid when considering "global" effects. When considering a real planet it is obvious that there is no characteristic single temperature but that is a different matter, the distribution of temperature over the surface depends on how heat is transported over the surface, this was excluded by our initial assumptions.

There are further consequences for our non-black model. If we observe our black body model from a distance using an infrared thermometer we will get an accurate measure of its temperature, this does not apply in the least to a non-black body. First of all the IR radiation from a non-black body is less than from a black body at the same temperature so our thermometer will always register a lower temperature, for a gold plated object this can be 95% less. The non-black body also reflects (or transmits) incident radiation, how will we focus our IR thermometer so that it does not now include some of the reflected radiation from our non-black body in its computation?

I am interested in how my arguments strike you because the consequences for suggested planetary climate change are very severe, a substantial part of the argument for climate sensitivity depends on the equilibrium temperature (for that is what we are talking about) being sensitive to albedo, (α in your formulas.) --Damorbel (talk) 11:12, 9 January 2010 (UTC)[reply]

YES! My analysis above is wrong because I forgot Kirchhoff's law of thermal radiation and you did not. You get Kirchoff's law by supposing that there is a grey body enclosed by a surface, lets say a sphere, and the inner surface of the sphere is a perfect mirror. At equilibrium, the grey body will be at temperature T and it will be immersed in a "photon bath" at temperature T. (Everything has the same temperature at equilibrium) A photon bath is just radiation in all directions with a black body distribution and intensity, because the black body distribution is to photons what the Maxwell distribution is to massive particles. By conservation of energy, the energy of black body radiation falling on the grey body must equal the amount reflected plus the amount emitted. The grey body reflects $\alpha B(T)$ , and emits $\epsilon B(T)$ where $\epsilon$ is the emissivity, 0 for a white body, 1 for a black body and B(T) is the emission spectrum of a black body. That means incident=reflected+emitted or:

B(T)=\alpha B(T)+\epsilon B(T)

or $1-\alpha =\epsilon$ . This is always true, at every wavelength, even in non-equilibrium, because $\alpha$ and $\epsilon$ are properties of the material. This is the same as when you say the mirrored (white) part of a body doesn't emit, but the black parts do. Reflectivity $\alpha$ for a white body is 1, emissivity $\epsilon$ is therefore zero. Now if you go to the case where some energy from a black body like the sun falls on the grey body, you get by conservation of energy incident=reflected+emitted:

E(T_{S})=\alpha E(T_{S})+\epsilon B(T_{E})

The present analysis on the black body page is wrong, because it assumes $\epsilon$ =1 while $1-\alpha$ is not. If you do it right, you get

E(T_{S})=B(T_{E})

and you get your equation above:

{\frac {T_{E}}{T_{\odot }}}={\sqrt {\frac {R_{S}}{2D}}}

Which gives an earth temperature of about 42 degrees fahrenheit. If you again do the greenhouse analysis that I tried above, you say that there are two wavelength regimes, "vuv" (visible and ultraviolet) and "ir" (infrared) with separate emissivities and alphas. Lets say in the vuv range we have $\epsilon =(1-\alpha )$ but in the ir range we have $\epsilon =0$ and $\alpha =1$ . Then conservation of energy says incident=reflected+emitted or:

E_{vuv}(T_{S})+E_{ir}(T_{S})=\alpha E_{vuv}(T_{S})+(1)E_{ir}(T_{S})+\epsilon B_{vuv}(T_{E})+(0)B_{ir}(T_{E})

or

E_{vuv}(T_{S})=B_{vuv}(T_{E})

As noted in my original analysis, $B_{vuv}(T_{E})=\gamma \,B(T_{E})$ . 99.9 percent of the sun's radiation is vuv, so we can say $E_{vuv}(T_{S})=E(T_{S})$ , so now

{\frac {T_{E}}{T_{\odot }}}={\sqrt {\frac {R_{S}}{2D{\sqrt {\gamma }}}}}

If we make the cutoff at 1600 nm, we get gamma=0.57 and an earth temperature of 117 degrees, so thats wrong. If we make the cutoff at 2500 nm, we get 68 degrees F, if we make it at 3500 nm, we get 53 degrees F. Anyway, the average Earth temperature is around 50 degrees, so this tells me that the greenhouse effect is responsible for about 7 degrees Fahrenheit difference. Check out http://fuse.pha.jhu.edu/~wpb/spectroscopy/atm_trans.html to see the atmospheric absorption as a function of wavelenth. 1 um is 1000 nm, 10 um in the middle is 10,000 nm. PAR (talk) 16:27, 9 January 2010 (UTC)[reply]

Once more thanks for your attention. I would like to look carefully before commenting further.--Damorbel (talk) 17:02, 9 January 2010 (UTC)[reply]

Good. Also, alpha should be reflectance, not albedo. PAR (talk)

Too right! Albedo is very ill defined, that anybody should predict anything with albedo as a parameter is difficult to believe.--Damorbel (talk) 17:41, 9 January 2010 (UTC)[reply]

There seems still to be some wavelength dependency from the variable γ, apparently due to infrared reflection by greenhouse gases. When modeling the working of GHGs you must stick strictly with what they do, they don't reflect (or even scatter) radiation, they absorb and emit it, how strongly they do it is a function of temperature; reflection and scattering are independent of temperature.

We are getting into deep water here, we should no longer be thinking of radiation for heat transport (warming) due to GHGs. Because of the thermal structure of the atmosphere (the temperature declines at 6.5C/km altitude in the troposphere) there can be no radiative transfer of heat from the troposphere to the water/rock surface. Note I said transfer of heat, this would be needed if there was to be a warming effect at the surface. Only exceptional tropospheric conditions called inversion give rise to temperature inversion (temperature rising with altitude), these conditions do not happen on a global scale in the troposphere.

The thermal structure of the atmosphere (troposphere, stratosphere and mesosphere is far from simple because, as in all thermodynamic problems, all significant sources and sinks of energy (not just heat energy but gravitational energy also) have to be accounted. For example the stratosphere itself is a truly global temperature inversion starting at thre tropopause, because ultraviolet energy from the Sun is absorbed there heating it [21] almost to the surface temperature 50km below.

Don't forget with a proper calculation of the equilibrium temperature (282K instead of 255K) there is no need of a "Greenhouse effect" to account for the Earth's observed temperature and futhermore the planetary temperature is no longer a dependent on the albedo (sorry - reflectance!) so it makes no difference to the mean planetary temperature if we plant black or white flowers. --Damorbel (talk) 21:15, 9 January 2010 (UTC)[reply]

Please check out what I wrote in talk:black body. I get 278.7 for the Earth temperature, not 282, but the point is valid. I would not go so far as to say that albedo, reflectance, etc. does not have an effect, but the effect given in the black body article is wrong. PAR (talk) 21:33, 9 January 2010 (UTC)[reply]

My 282K figure is dreadfully unscientific, I merely added the deep space (CMB) temperature of 2.7K to the Sun/Earth value. I have made a number of attempts, as have others, to get some sense into this black body question but since it is the basis of the whole "planetary doom" industry, progress will inevitably be slow. Would you care to estimate how many people earn their living directly from this nonsense? Many times greater is the number of people whose whole belief system is built round it, let alone the political vultures who perceive the rotten logic with multiple escape routes as rich pickings for their short term populist solutions. It isn't remotely patronising to say that they don't actually know what they are talking about!--Damorbel (talk) 10:17, 10 January 2010 (UTC)[reply]

Well, when it comes to science, I am fanatically uninterested in the political implications one way or the other. The truth is the truth and I don't care whose ox gets gored. I understand that there are plenty of people on both sides whose income, political power, and self esteem depend on the facts bearing out the theory that they have invested in, and that their scientific arguments will be colored by this. On the other hand, this does not constitute a proof that they are wrong. I'm happy to apply my bullshit detector to either side, so don't count me as a loyal supporter of any side. PAR (talk) 15:32, 10 January 2010 (UTC)[reply]

"I am fanatically uninterested..." Truly great attitude, I share it largely but very few do. "so don't count me as a loyal supporter of any side", I seem to have struck a raw nerve and I'm sorry, but I don't think you will find a single support plea in any of my postings. I like to think I am on a learning curve that has someway to go, my logic is: either I have not understood the matter properly or my explanations are inadequate. This latter has various sides, I never expect to get our dog to understand physics, he just wags his tail and wants to go for a walk, in the meanwhile I may have found a better explanation! However I still get irritated by those who delete my contributions, even to talk pages, without any discussion. --Damorbel (talk) 15:59, 10 January 2010 (UTC)[reply]

Where you got it wrong

D, I think your analysis is correct for your model. But that's not the model used in the articles or their sources. You say "covering a percentage of its surface evenly with small, perfect mirrors so that this percentage of incoming radiation is reflected, it is no longer a black body....The consequence is that the temperature of our now non-black body is exactly the same as that of the black body, the reduced absorbing area being exactly compensated by the reduced emission area." So far so good. But then "This argument is equally valid for all possible non-black bodies, coloured or partially transparent." This is wrong. Think about what "coloured" means: the absorptivity or emissity is wavelength dependent. Now re-read the articles with this in mind, and stop turning them to nonsense. Dicklyon (talk) 04:03, 20 December 2010 (UTC)[reply]

I would very much like to know why you think it is wrong. It happens to be very similar to Kirchhoff's position. Planck's radiation law confirmed Kirchhoff's position in detail by successfully explaining the relationship between his radiation formula and experimental observation, the great goal of all scientists in every discipline. Planck's radiation formula places limits on radiation at high frequency that were seen to be the same as those by imposing quantum restrictions on the energy in high frequencies; that is why it is a valid argument to see Kirchhoff, with his 'black body' and Planck with his 'harmonic resonators' as the true precursors of quantum theory. Challenge this if you like, but do it properly, please don't put your challenge forward as current (encyclopedia) science without careful consideration; I understand Wikipedia has rules about this sort of thing. --Damorbel (talk) 09:37, 20 December 2010 (UTC)[reply]

It's wrong because it's not the model that the sourced calculations are based on; as to whether it's a good model or not, that depends on how well it matches the physics being model. I don't think Kirchhoff looked at the question of whether the earth's emissivity in the long wavelengths was close to its absorptivity in short wavelengths. Nobody is disputing that Planck and Kirchhoff had great and correct results; but they don't imply what you keep saying they do, since they only equate absorptivity and emissivity at each wavelength individually, not across different wavelengths. Dicklyon (talk) 10:54, 20 December 2010 (UTC)[reply]

"they only equate absorptivity and emissivity at each wavelength individually, not across different wavelengths." That changes nothing. The transfer of thermal energy by radiation is essentially a thermal matter which requires, if a source has a temperature its spectrum is the Planck distribution. And temperature is the matter in hand. When considering the 'mechanical' temperature, the distribution of velocities is the Maxwell-Boltzmann distribution. Both of these distributions are for equilibrium conditions, for Planck it is the spectrum plus the plus the temperature (or total energy; a distant source may well have the spectrum, indicating the temperature, but the energy is reduced by the inverse square law and the actual temperature achieved by a body intreceptig the radiation is correspondingly reduced). So, if the Planck spectrum is modified by a wavelength function, the energy transfer rate can only be less than with a 'black body'. In the case of 'mechanical heat, a body not in equilibrium has a non-uniform temperature, so the velocity (or energy) distribution of the particles no longer corresponds to the M-B distribution.

Kirchoff's discovery was to realise that, in a closed situation like a cavity, the rate of (internal) heat tranfer arising from disequilibrium was a maximum if the parts were black, if they were not black i.e. possessed a (any) wavelength function, then the rate of heat transfer would be reduced, equilibrium would take longer to happen but the final temperature would be the same throughout, just as if all the parts were in 'mechanical' thermal contact.--Damorbel (talk) 12:20, 20 December 2010 (UTC)[reply]

Yes, and Kischhoff was correct about that last part. But you're wrong to say "The transfer of thermal energy by radiation is essentially a thermal matter which requires, if a source has a temperature its spectrum is the Planck distribution." Dicklyon (talk) 15:19, 20 December 2010 (UTC)[reply]

"But you're wrong to say "The transfer of thermal energy by radiation is essentially a thermal ..." When I say thermal it is in the sense of 'sometthing with a temperature....' strictly speaking the concept of a (single) temperature can only be used for something in (thermal) equilibrium, that is to say there is no temperature gradient to be found in it thus all parts have the same temperature - 'the entropy has reached a maximum'. The same applies to (thermal) radiation; a radiation field that does not have a planck spectrum, and most do not, is not 'in equilibrium', that is to say the shape of the spectrum is not that of a Planck function; an (extreme) example is a laser beam. A laser beam may contain a lot of energy but you cannot assign a temperature to it because the single frequency (more than one in most lasers) is not the result of random processes but a carfully synchronised and phase locked (coherent) emissions. A laser beam is absorbed (or not) just like any other light, it will warm up the material absorbing it but you can never say the laser source has a 'temperature' because the laser spectrum means it is 'out of equilibrium', thus transfer of energy by means of laser light 'is not (a) thermal (processs)'. --Damorbel (talk) 19:41, 20 December 2010 (UTC)[reply]

Right, your analysis is inapplicable, since the Sun is hotter than the Earth and the whole thing is far from equilibrium. Dicklyon (talk) 22:29, 20 December 2010 (UTC)[reply]

"the whole thing is far from equilibrium". Yes it is. But it is in a steady state, the outflow of heat (LW IR) is equal to the insolation. Kirchhoff's 1862 paper states that the temperature of a body is independent for equilibrium, for the Earth it would then have the average temperature but the disequilibrium arises because 1/ the Sun is very hot, 2/ Earth only intercepts a small portion of the Sun's output 3/ the Earth does not have a uniform surface temperature (i.e. it isn't in thermal equilibrium either).

None of these things stop us estimating the surface temperature of the planet, we must take into account the distance of the Sun, the fact that Earth is a good approximation to a sphere, it revolves on its axis, it is in an orbit round the Sun, the axis is tilted wrt the plane of its orbit round the Sun etc. etc. Kirchhoff's law still applies in this steady state condition, changes in albedo only affect the rate at which the steady state returns after a disturbance. NB this does not apply if the Earth had an internal heat source which would mean there was a permanent disequilibrium wrt the Sun. If you wish you can do the sums for two heat sources, the Sun and some internal heat source. --Damorbel (talk) 08:26, 21 December 2010 (UTC)[reply]

True until you said "Kirchhoff's law still applies in this steady state condition, changes in albedo only affect the rate at which the steady state returns after a disturbance." If changes in the albedo don't come with proportionate changes in emissivitiy in the far IR band where the planet radiates, then the albedo can matter a lot in the equilibrium temperature. You keep trying to say that Kirchhoff said more than what he said. You haven't looked into wavelength dependence as I suggested, which is where your confusion will be resolved. Dicklyon (talk) 19:32, 21 December 2010 (UTC)[reply]

"You keep trying to say that Kirchhoff said more than what he said." Since he said "equilibrium will be reached for wavelength independent (or 'arbitrary') bodies I absolutely do not see how you can say "You haven't looked into wavelength dependence". Kirchhoff even wrote that "if the emissivity and absorptivity were different you would have an impossible condition where the temperature could never be stable". Do you have a copy of his 1862 paper? What have you seen in it?

Its not just Kirchhoff's paper, Planck's radiation law comes from various attemps to justify the observed value of the black body radiation function. Planck's formula opened the door to quantum theory. I respect your interest but you have a lot of work to do if you want to establish the idea of albedo sensitivity; there is so much of modern physics dependent on this black body concept of Kirchhoff.

There is more than one aspect to the albedo, you mention wavelength but there can be albedo without a wavelength function, a partially reflective surface may well have no wavelength function but still reflect a fraction of the incoming radiation, the moon and Mercury come close to this; the Moon has an albedo of 0.136 and has no particular spectral characteristic; likewise Mercury, albedo = 0.142.

I thought we already agreed that we're not talking about thermodynamic equilibrium. What Kirchhoff said about that, which you're quoting, is correct. But you keep trying to apply it to a different situation, in which the different temperatures of the sun (and its radiation) and the Earth is a major effect that you keep ignoring. It's explained in the article and elsewhere that the absorptivity about 0.7 and emissivity around 0.95 for these different spectra are the result of the wavelength dependence of the Earth's surface. Dicklyon (talk) 22:36, 21 December 2010 (UTC)[reply]

Talk:Global warming controversy

Thank you for your contributions to the encyclopedia! In case you are not already aware, an article to which you have recently contributed, Global warming controversy, 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 15:39, 6 February 2010 (UTC)[reply]

DAB

[22]

Not a forum, not a battleground

A few of your recent edits indicate that you are sometimes treating article talk pages as a forum for general discussion of the topic rather than discussion of improvements to the associated article. More worrisomely, you seem to be personalizing your misunderstandings, turning discussion into a battleground. Please do not do this. Have you considered posting your questions at Wikipedia:Reference desk/Science? - 2/0 (cont.) 20:25, 26 March 2010 (UTC)[reply]

I have no idea who posted this (unsigned item), (some kind of spook?) I suggest you link to what to are commenting about or stop bothering me. This kind of anonymous stuff is really freaky.

"you seem to be personalizing your misunderstandings something about me being an ignoramous is it? Alright I confess, I am completely ignorant of what you are on about.--Damorbel (talk) 20:42, 26 March 2010 (UTC)[reply]

Please be more mindful of the civility policy or you may be blocked from editing. Thank you, - 2/0 (cont.) 05:48, 12 May 2010 (UTC)[reply]

Please explain who you are and what prompted you to post this note. The elements of civility in communication are not observed by anonymous comments and threats of blocking.--Damorbel (talk) 10:16, 12 May 2010 (UTC)[reply]

You see the "2/0" above - that is a link to a username, so your poster is not, in fact, anonymous. "2/0" is an example of what wikipedia calls a "username"; you may have noticed that unlike in Real Life not all usernames follow the common first-name then surname pattern. For example, "Damorbel" William M. Connolley (talk) 10:27, 12 May 2010 (UTC)[reply]

Thus no explanation. You seem interested William, may I ask why?--Damorbel (talk) 11:31, 12 May 2010 (UTC)[reply]

Certainly you may. This page is on my watch list, and you seemed puzzled, William M. Connolley (talk) 11:35, 12 May 2010 (UTC)[reply]

"so I tried to help?" I don't understand. You regularly delete my contributions to talk pages with, if I'm lucky, an offensive remark. You seem to have a restricted view of the applicability of thermodynamics to climate processes yet you regularly delete my contributions on these matters. The idea that you are trying to help is not one I would deduce from what you do.--Damorbel (talk) 12:27, 12 May 2010 (UTC)[reply]

We can (if you like) argue about contributions elsewhere, though I doubt that would be profitable. I'm explaining my motivations here, on this page William M. Connolley (talk) 12:47, 12 May 2010 (UTC)[reply]

"I'm explaining my motivations" I think I can guess your motivations but that is scarcely relevant. It is your editing that I object to. You belong to a category of editors who mainttain that they are know the truth about scientific matters, what 'the consensus is' and regard themselves as having a responsibility to put their view in place of others. William, you don't have this exalted status, your version of various thermodynamic matters is frankly hilarious, have you ever studied the subject in depth? I see not the slightest talent in your contributions on matters with thermodynamic content, yet you freely delete other peoples contributions, even in the discussion pages. What do you want, to eliminate discussion? Seems to me you have issues other than thermodynamics motivating your actions.

I am writing this while considering making a financial contribution to Wikipedia and one of the considerations is how much time I should devote to it. My understanding is that your style of editing, particularly towards the contributions of other, breaks a number of the rules that are in place to ensure proper consideration is given to the articles in Wikipedia. So my question to you is, do you feel that you should have the right to interfere directly, i.e. with minimal, if any, discussion, in the contributions of others and do you expect to continue editing in your current manner? --Damorbel (talk) 11:01, 12 December 2010 (UTC)[reply]

Don't...

...do this kind of stuff [23] William M. Connolley (talk) 21:08, 10 May 2010 (UTC)[reply]

Links

Heald http://en.wikipedia.org/w/index.php?title=Talk%3ABoltzmann_constant&action=historysubmit&diff=458513047&oldid=458497334

Heald http://en.wikipedia.org/w/index.php?title=Wikipedia_talk:WikiProject_Physics&diff=prev&oldid=458540639

Iblis http://en.wikipedia.org/w/index.php?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458544897&oldid=458540639

Heald http://en.wikipedia.org/w/index.php?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458540639&oldid=458540482

Pratt - Headbom http://en.wikipedia.org/w/index.php?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458754024&oldid=458750956

Heald http://en.wikipedia.org/w/index.php?title=Talk:Boltzmann_constant&diff=prev&oldid=458513047

Really? Surely you know that personal rmarks and opinions about the other contributors are a no-no in Wikipedia?

Dick http://en.wikipedia.org/w/index.php?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458716646&oldid=458715969

Pratt http://en.wikipedia.org/w/index.php?title=Wikipedia_talk:WikiProject_Physics&diff=next&oldid=458608860

So you approve of personal comments on article talk pages, do you? --Damorbel (talk) 06:16, 4 November 2011 (UTC)[reply]

At the very least personal remarks are 'off topic', such remarks as "If anyone can get through to User Damorbel and help him out" is laughable if not pathetic. The writer of this comment did not respond in any detail to the points I raise, my impression is that a contributor with an understanding of thermodynamics would have responded with a detailed argument, you can see the kind of thing I mean here [24]--Damorbel (talk) 07:23, 5 November 2011 (UTC)[reply]

Heald [25]

Reminders

As a new editor in 2008, you were alerted to the need for verifiable sources and the proper way to conduct a content dispute

In this 2011 wikietiquette assistance thread, it was pointed out to you that "talk pages are not the place for theoretical or scholarly discussions about the topic"

Out of curiousity, was this you as well? It is a rhetorical question. If so, please do not bring the same sort of prevarication to wikipedia, and if not, then I apologize. I just couldn't help noticing the apparent coincidence between subject matter and user name.

As for collapsing the thread at Greenhouse effect, talk pages are not for theoretical debates of the subject matter. If you wish to post some draft article text, complete with what you think are properly cited (according to wiki) and verifiable (according to wiki) sources, then by all means please do. I will continue to collapse general topic debate and invite you to seek dispute resolution if you find that objectionable. Though I expect you will be told the same thing you were told in the link above. NewsAndEventsGuy (talk) 16:52, 19 January 2012 (UTC)[reply]

Wiki seems to be a total mess on heat

I think part of the problem comes from an American text book writer who was publishing from 1932 until fairly recently

"Heat. Heat, like work, is a measure of the amount of energy transferred from one body to another because of the temperature difference between those bodies. Heat is not energy possessed by a body. We should not speak of the “heat in a body.” The energy a body possesses due to its temperature is a different thing, called internal thermal energy. The misuse of this word probably dates back to the 18th century when it was still thought that bodies undergoing thermal processes exchanged a substance, called caloric or phlogiston, a substance later called heat. We now know that heat is not a substance. Reference: Zemansky, Mark W. The Use and Misuse of the Word “Heat” in Physics Teaching” The Physics Teacher, 8, 6 (Sept 1970) p. 295-300. "

The problem with Wiki is that no matter whatever herculean efforts you make to correct 'somebody who is wrong on the internet' there are always a small army of other individuals who will ensure that wiki is never going to be a source of knowledge you can rely on. Andrewedwardjudd (talk) 16:40, 14 March 2012 (UTC)andrewedwardjudd[reply]

It seems to be much more than 'one text book writer'! As far as heat is concerned there seems to be a large number of people who have a mental fog when trying to distinguish between heat and thermal energy; they then edit at full speed without checking for consistancy; that is one of the causes of the mess. --Damorbel (talk) 17:01, 14 March 2012 (UTC)[reply]

my point about the text book writer is that he was producing undergraduate text books from 1932 till recently and might have slowly poisoned the minds of nearly all students and text books writers. There must be a reason that this change has happened.

By the way, the way wiki works is that you have to produce good quality references to support your idea. If there are no modern textbooks or other authoritative sources that support your view then you have no case to alter wiki. The text books can be total garbage but if you cannot find reliable sources you have no chance on wikiAndrewedwardjudd (talk) 18:08, 15 March 2012 (UTC)andrewedwardjudd[reply]

You write "you have no case to alter wiki" So in this matter the Wiki principal appears to break down. May I ask what book you are thinking of? Perhaps you don't want to put it in public, perhaps we should be careful but I'm told the dead can't be libelled!

Much the same situation arises with 'Climate Change'. The IPCC explanations are just thermodynamic rubbish, but who understands thermodynamics these days? --Damorbel (talk) 21:05, 15 March 2012 (UTC)[reply]

The text book writer is Zemansky

Wiki is not a source of knowledge as such. It is a source of referenced material. If you have the correct view it has to be published before it can be cited. So arguing you are right if nobody is listening and you have no references is pointless on Wiki. And presumably on a topic like Heat there are not multiple government agencies and corporations at work to ensure deleting editors are always in the majority. Presumably! ??? !!! :-) So it should be possible to get a concensus if you are prepared to do the hard work knowing that in time another group of editors will arrive to delete all the hard work because you were 'oh so wrong and oh so poorly educated'.

What i have learnt is that when you get difficult editors like those on the heat page you can develope a sub paragraph as a minority view, and keep adding references to it and they cannot delete the very high quality references. Eventually the sub para causes them to explode and they start making it smaller but at least you get the references into the article with huge extensive verbatim citations of the point you wish to see in wiki. So write something like some writers like Kelvin1,2,3,4,5,6 Maxwell,78,91, Joule 111,212, Clausius,234,234 and others 99,100,102 have argued that heat is a dynamic force or living force and that heat is contained in the internal moving force or kinetic energy of an object. And then in the references you quote verbatim huge screeds of text saying exactly that and eventually your references are more or less bigger than the page itself. And there is sweet FA they can do about it.

But if you add text any other way or try to talk about it on the talk page you waste your efforts where the wiki pages are hugely hard to work with when some twit or agent is constantly removing your work. You can spend months on wiki achieving zero otherwise.

There is already such a paragraph on the heat page so it can be altered slightly with references slowly added till it tends to make the 'strict' definition look a bit daft.

often the references are already there and you only have to add a bit more text to get it to more accurately reflect the writers view. As was the case for Joule on sensible and latent heat and the living force.

On climate change if you were referring to backradiation that is thermodynamically supportedAndrewedwardjudd (talk)andrewedwardjudd

Enabling topic evasion

[26] William M. Connolley (talk) 16:58, 22 March 2012 (UTC)[reply]

You seem puzzled. But it seems clear enough: why are you restoring the comments of a topic-banned editor? William M. Connolley (talk) 18:23, 22 March 2012 (UTC)[reply]

You write "But it seems clear enough" Why should it appear clear enough? To be perfectly clear, I haven't got a clue what you are on about, I have searched Help and found nothing about 'topic evasion', is it your idea? If it isn't in Help it cannot be that important.

I undid the deletion in question because I wanted to read it, OK? I have been discussing 'Heat' with this guy because I am well qualified to do so and the disscussion was about important aspects of 'heat' which is a fairly obscure matter, OK?

Can you explain your reason for banning the guy? Recently you have made no significant contribution to the matter in question so I have no idea why you want to ban anybody. --Damorbel (talk) 18:38, 22 March 2012 (UTC)[reply]

because I wanted to read it, OK? - no, not even vaguely believable.

Can you explain your reason for banning the guy? - no, because I haven't. How could I? Read the diff I provided William M. Connolley (talk) 18:47, 22 March 2012 (UTC)[reply]

"no, not even vaguely believable." Go on William, give it a go! If you can believe in CO2 caussed AGW you should be able to believe this! HI! HI! --Damorbel (talk) 18:53, 22 March 2012 (UTC)[reply]

Your comments were moved to the article talk page

Incidentally, you probably deserve a conduct warning for an edit commentary of "blatant vandalism" when an editor was restoring a deal of content which you appeared to have deleted. Do try harder please. --BozMo talk 07:26, 25 August 2012 (UTC)[reply]

!

http://en.wikipedia.org/wiki/User_talk:William_M._Connolley#request_for_advice

Cheese

Discussion moved to page Talk:Wensleydale cheese#Creamery, since it became long and really belongs to article page now. Staszek Lem (talk) 17:25, 20 November 2012 (UTC)[reply]

You wrote: "suggest you try Witgenstein 'Of what you know not - speak not'." Colleague, you are thoroughly confused here. I am not writing anything about this cheese. I am editing what is already written, in order to comply with wikipedia rules. And that latter thing I believe I do know better than Witgenstein. If you have anything to add to the article, citing reputable sources, I will be more than happy to help you to accomodate your addition to fit wikipedia. But unfortunately unreferenced opinions have no place in wikipedia articles. Staszek Lem (talk) 20:02, 20 November 2012 (UTC)[reply]

Um, user:Staszek, you appear to be editing the article quite successfully without introducing any knowledge, relevant material or references. Please carry on editing a subject you clearly are sans connaisance, it is so amusing to watch you struggling! --Damorbel (talk) 20:22, 20 November 2012 (UTC)[reply]

This comment (and lack thereof in the talk page of the article in question) demonstrates that you have yet to improve your understanding of how wikipedia works. In particular, there is a huge amount of work here than can be done without French. I am glad you are amused. Most people become mad when disagreed with. Staszek Lem (talk) 01:18, 27 November 2012 (UTC)[reply]

please stop posting on my talk page

Damorbel, your comments on my talk page belong in the talk page of the relevant article. Please stop posting on my talk page. Thank you. Waleswatcher (talk) 21:27, 23 November 2012 (UTC)[reply]

Saved

Indeed "planetary equilibrium temperature" is not immediately relevant, that is something for advanced students.

But you seem not to be fully aware of the meaning of "intensive quantities"? Or the meaning of temperature. A temperature can only be defined for a system in equilibrium, a system not in equilibrium has two or more temperatures, the requirement for a single temperature being that the particle energy has a Maxwell Boltzmann distribution

Of course, as I remarked before, the temperature of a thermodynamic system is independent of the size of the system; rather intuitive don't you think? This drives one to the logical conclusion that the smallest system that can have a temperature is a system comprising only one particle!

The meaning of the Maxwell Boltzmann distribution is statistical of course; (possibly from statistical mechanics - do you think?) In a Maxwell Boltzmann distribution the particles have an equal probability of accessing the system states. --Damorbel (talk) 19:58, 30 November 2012 (UTC)[reply]

Community topic ban requested

Your last edit at Talk:Boltzmann constant [27] was completely out of line with the standards of collegiality and civility expected here, as well as demonstrating apparently insoluble lack of competence and understanding. I have made a request at WT:PHYSICS for a community topic-ban to bar you from all further editing of articles and talk pages related to thermodynamics. Jheald (talk) 21:57, 8 December 2012 (UTC)[reply]

Hello Damorbel. If you wish, you can respond in the thread at Wikipedia talk:WikiProject Physics#Proposal: for User:Damorbel to be community-banned from all articles and talk pages on thermodynamics. Thank you, EdJohnston (talk) 02:43, 10 December 2012 (UTC)[reply]

The discussion at WT:PHYS gives you a chance to tell your side of the story. If you pass this up, people may assume that your position has no defence. People do seem to be very unhappy with your editing. Possibly a deal can be worked out. EdJohnston (talk) 13:49, 11 December 2012 (UTC)[reply]

approx

\approx

Heat flow

Okay, I'll bite: what is wrong with using the word "flow" to describe the transport of thermal energy from one place to another? I've seen you're objections on other talk pages, but I've never understood what goes wrong if one thinks of heat as flowing. Does that lead to a false prediction? Spiel496 (talk) 00:46, 27 January 2013 (UTC)[reply]

Using any term is, of course, a matter of ' usage ' and . Flow is transport of a "fluid". At one time heat was regarded as a fluid 'caloric'; this failed, basically because none of the experimental observations could be made compatible with with anything that could reasonably be accepted as a fluid. The difficulties were very great, was caloric weightless? Was caloric created by friction?

Fluid mechanics generally require the flowing substance to be conserved and the analogy is used in physics in the term 'flux' one speaks of magnetic flux. Particle flow is generally spoken of in similar terms e.g. neutron flux. Of course these are not fluids but generally they do not appear (or disappear), if at all, without some very powerful reaction.

On the other hand, heat, being the kinetic energy of particles, can change into another form almost without being noticed. A good example of this is found in the troposphere where the temperature drops with altitude; almost nobody will explain that this 'fall in temperature with height' is due to thermal energy being converted to gravitational potential energy, but that doesn't stop it being so.

Good science (and good encyclopedias) rely on precise terminology. In the case of heat, in the first instance it must not allow confusion with the 'caloric' theory and secondly thermal energy may well be transported by fluid flow, as in convection, which may well cause further confusion if the words chosen are not precise.

The idea of 'heat flowing' is so firmly embedded in popular culture it may never be displaced, which is rather unfortunate because currently it is just another brick in the wall that divides science and popular culture. --Damorbel (talk) 10:44, 27 January 2013 (UTC)[reply]

You use the 'caloric' theory to make your point. Let me ask you this: If the caloric theory had never existed, would you still object to the term "flow" to describe the transport of heat?

If your answer is Yes, then my advice would be: Don't muddle your argument by bringing up this discredited theory. People hear it like "Don't brush your teeth; the Nazi's did that." The natural response is "I don't care what some bogus theory says; I still think heat 'flows' ". To show that the "flow" concept really is misleading, then skip caloric and go right to some of the "experimental observations (that couldn't) be made compatible with anything that could reasonably be accepted as a fluid".

The fact heat can be created and destroyed is a real tangible difference from a fluid like water. But would you say that it is still wrong to think of a "heat flow" in a passive system, like a chunk of metal, where there energy is conserved? If Yes then, again, don't muddle your argument with whether a fluid is conserved. Besides, one can imagine a system where a fluid is effectively created/destroyed, for example through a chemical reaction or phase change, so I'm not convinced people really associate "fluid" with "absolutely conserved". Spiel496 (talk) 18:55, 28 January 2013 (UTC)[reply]

I don't understand your problem. Heat doesn't flow because it isn't a fluid, heat transfer doesn't follow any of the rules of fluid flow equations, for that you need the Navier–Stokes equations. Your example of 'a chunk of metal' is good only in so far as heat diffuses in solids, that is what Fourier's discovery was all about. Fluid flow may well transport relatively hot (or cold) material but that cannot be described as heat flow, it is mass transport. --Damorbel (talk) 21:27, 28 January 2013 (UTC)[reply]

The Navier-Stokes equations: Yes, that is a good point. (Which you could have brought up in the first place.) It would definitely be a mistake to suggest that heat transfer is described by the Navier-Stokes equations. That may be a more narrow view of the word "flow" than most others have. When I hear the phrase "heat flows" I think of it more like a gas flowing in a porous medium. I would think the equation for this type of flow would be fairly similar to the heat equation.

. . . similar to the heat equation. Why not? If diffusion fits what is happening then use it! Diffusion happens to fit mass transport in semiconductor processes, both in manufacturing and conduction by electric charges; as also in the mixing of gases. The trick is not to use flow where it doesn't apply because you will get the wrong answer! --Damorbel (talk) 07:39, 29 January 2013 (UTC)[reply]

I would argue that one gets the wrong answer not by using the word "flow", but rather by using the wrong formulas. But I get your point. So does it fix everything if one replaces "heat flows from a warmer region to a cooler one" with the statement "heat diffuses from a warmer region to a cooler one"? Spiel496 (talk) 19:03, 29 January 2013 (UTC)[reply]

If you think accurately you should be able to express yourself just as well. You will find heat related matters in Wikipedia are a total shambles. Look at the Heat article, the second sentence has:-

Heat is not a property of a system or body, but instead is always associated with a process of some kind, and is synonymous with heat flow and heat transfer.

Which is rubbish. Heat is due to the motion of the particles making up a 'body' (solid, liquid and/or gas), to be strictly accurate it is due to the momentum of the particles, but more often it becomes confused by calling it 'the translational kinetic energy'. It has to be momentum because momentum is a vector quantity.

I could go on but why on my own talk page? --Damorbel (talk) 21:22, 29 January 2013 (UTC)[reply]

Warning, we will seek to have you blocked if you keep re-inserting your anti-consensus views in the Heat article

This is a copy of what I posted on Talk:Heat:

Damorbel, I will try not to make any personal comments here. If you don't stop re-inserting your views against the consensus here, we're going to have to ask admins to take action against you to prevent this. You keep re-inserting your incorrect statements about heat being kinetic energy, etc. It's good that you're interested in this subject but you should take thermodynamics course or read textbk if you want to find out how the scientific community defines these terms rather than how you are defining them. Or, please try reading the Heat article in another language as linked on the left side, using Google Translate to read it in English. Otherwise, there are other venues where you can write essays about why these terms should be (re)defined in the way you describe, but not in an encyclopedia, please. In case it helps you at all, let me point out that thermal energy is stored in part in the oscillations of molecules and atoms, and oscillations involve a continual transformation between kinetic energy (reaching a max when the particle is moving fastests) and potential energy (when the particle is at its most extreme displacement). Look up oscillations of springs for this concept. Thermal energy is a term used to describe a property of a system or body, while the term heat is defined as only a property of a specific process that transfers a given amount of energy in certain ways (which represents a change in energy of one system and opposite change in another). Also, Joules/second simply aren't units of energy (nor of heat), they're units of power (pls. look it up), which is energy per unit time. DavRosen (talk) 14:30, 18 July 2013 (UTC)[reply]

Wikipedia operates by consensus, and we use a WP:BRD cycle: Bold edit (any edit that is undiscussed), Revert, Discuss. There is no second B after the first Revert. No matter how much someone thinks the article should say something, that has to be articulated on the talk page during the Discuss cycle so that a consensus can be obtained on what the article should say. Heat is a technical subject that requires the review of an expert on the subject to make sure that what the article says makes sense. We welcome experts, but sometimes it can be difficult for an expert to get their point across to the rest of the editors working on the article. I recommend patience, lay out what you see as needed changes, and if they are not agreed to, there is not much that can be done other than count to ten, work on some other articles for a while, and bring it up again in a few months. Apteva (talk) 02:37, 19 July 2013 (UTC)[reply]

Discussion at Administrator's Noticeboard.

This is to inform you that in view of your latest activity at Talk:Heat, I have made a request at WP:AN for you to be community topic-banned from all articles and talk pages on thermodynamics.

The discussion can be found here. Jheald (talk) 21:40, 18 July 2013 (UTC)[reply]

In accord with a notice that I should inform you about it, I have made an entry here.Chjoaygame (talk) 05:06, 20 July 2013 (UTC)[reply]

They only need to be notified about the thread, not every post to the thread. Apteva (talk) 02:39, 21 July 2013 (UTC)[reply]

Given the allegation (diff) made by Cyclopia (talk · contribs) connecting you with banned user GabrielVelasquez (talk · contribs), I have made a request at WP:SPI for a sock-puppet investigation to try to resolve the allegation one way or the other. Jheald (talk) 20:18, 22 July 2013 (UTC)[reply]

Gabriel Velasquez? That'l be the painter, init?

E's dead, dontcha know?

Naw, s' Diego Velázquez 'oos ded!

User:Jheald, be careful you don't loose too badly. Have a nice day.

--Damorbel (talk) 20:38, 22 July 2013 (UTC)[reply]

Topic Ban

Per the consensus at WP:AN, you are hereby indefinitely topic banned from all edits related to thermodynamics, broadly construed. Note that topic bans apply to all spaces on Wikipedia--article space, talk space, user space, etc. Any violation of this ban will result in your account being blocked. You may appeal this ban to either the Arbitration committee or the Administrator's Noticeboard (see WP:UNBAN), but you may not appeal for at least six months.

And just to address one criticism I assume you will level, as you mentioned it at AN, the whole point of me closing this and issuing the ban is that I have not investigated the matter in great detail. While I looked at the broad outline and some of the evidence, I did not make this decision make the ban on my own; rather, my purpose here is simply to enact the very clear consensus of both involved and uninvolved editors in that discussion. That, in fact, is how Wikipedia works: by consensus. Qwyrxian (talk) 12:44, 24 July 2013 (UTC)[reply]

Heat Archive 14

http://en.wikipedia.org/wiki/Talk:Heat/Archive_14#Needs_Revision._Contains_a_fundamental_blunder.

Review discussion

I inserted the link for you [28], which is technically a violation of the rules on editing other's comments, so please feel free to revert if you find it unhelpful. NE Ent

Discretionary sanctions 2013 review: Draft v3

Hi. You have commented on Draft v1 or v2 in the Arbitration Committee's 2013 review of the discretionary sanctions system. I thought you'd like to know Draft v3 has now been posted to the main review page. You are very welcome to comment on it on the review talk page. Regards, AGK [•] 00:16, 16 March 2014 (UTC)[reply]

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Courtesy notice: discretionary sanctions for the topic of climate change

This is a standard message to notify contributors about an administrative ruling in effect. It does not imply that there are any issues with your contributions to date.

You have shown interest in climate change. Due to past disruption in this topic area, a more stringent set of rules called discretionary sanctions is in effect. Any administrator may impose sanctions on editors who do not strictly follow Wikipedia's policies, or the page-specific restrictions, when making edits related to the topic.

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I would have simply removed your comments at Talk:Climate change denial as off-topic. --Ronz (talk) 16:57, 30 April 2019 (UTC)[reply]

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GHE

Speedy deletion of Damorbel

Speedy deletion?

Blackbodies

Urge caution

May 2009

Do not use talkpages as a forum

Radiative Heat Transfer

Anthony Watts

KimDabelstein removed this from the dicussion on Greenhouse effect

"Amplitude" is wrong term

Back Radiation

Terminology - "volts per meter"

More on Back Radiation

Basic Theory

Spencer on Second Law

Here goes

To briefly review...

"Blanket"

IT’S NOT A REAL GREENHOUSE

2nd Law of Thermodynamics

Net flow?

Further

And the jars?

Kirchoff's Law

Frost and dew

THE GREENHOUSE EFFECT WORKS…FOR NOW

Pressure and temperature

Pressure and temperature /2

Objects in orbit

A rough greenhouse model

Where you got it wrong

Talk:Global warming controversy

DAB

Not a forum, not a battleground

Don't...

Links

Reminders

Wiki seems to be a total mess on heat

Enabling topic evasion

Your comments were moved to the article talk page

!

Cheese

please stop posting on my talk page

Saved

Community topic ban requested

approx

Heat flow

Warning, we will seek to have you blocked if you keep re-inserting your anti-consensus views in the Heat article

Discussion at Administrator's Noticeboard.

Topic Ban

Heat Archive 14

Review discussion

Discretionary sanctions 2013 review: Draft v3

ArbCom elections are now open!

ArbCom Elections 2016: Voting now open!

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2^nd Law of Thermodynamics