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William M. Connolley (
talk) 21:41, 16 June 2008 (UTC)
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)
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Cocomonkilla (
talk) (
contrib) 21:20, 11 August 2008 (UTC)
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)
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)
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)
User:Ben MacDui Thanks!
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?
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)
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)
{{
unblock|Your reason here}}
below. Vsmith ( talk) 22:52, 14 March 2009 (UTC)
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)
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)
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.
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)
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)
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)
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)
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)
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 a Vacuum 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)
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)
"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)
"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)
"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].
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)
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)
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)
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)
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)
Continued from KimDabelstein's talk page.
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)
What is the difference between heat tranfer and "net heat tranfer"? -- Damorbel ( talk) 10:18, 30 May 2009 (UTC)
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)
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)
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)
You might find this interesting. The POV pushers are at it again. Q Science ( talk) 20:48, 16 June 2009 (UTC)
Blackcloak, the rigorous approach is to treat this radiation as Electromagnetic radiation (EMR). The important matter is how heat is transferred between two places by this radiation. You no doubt know that thermal radiation comes from the accelerating of electrical charge by thermal vibrations. To be rigorous, thermal radiation is not heat, it is an electric field with an amplitude related to the temperature of the source object (no radiation at 0K).
Two objects exchange heat when one is hotter than the other because the hotter one has an EM wave with a higher amplitude. The amount of heat transferred depends on the difference in the amplitudes between the hotter and the colder.
If both objects have the same temperature, the EM waves have the same amplitude and there is no heat transfer. It must be 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.
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).
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"?
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)
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)
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.
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)
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)
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.
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.
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.
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 1st item in the article. Common experience shows that all objects, micro to mega, tend to equalise temperature both locally and globally.
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.
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 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.
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].
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.
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)
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)
This section has now become unmanageable, please put your responses in the new section below.--
Damorbel (
talk) 08:16, 10 July 2009 (UTC)
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)
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)
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)
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)
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)
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)
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)
I saw this post by you
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)
PS: You might find this interesting - AIRS - Atmospheric Infrared Sounder currently flying on the NASA Aqua Mission
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:
Let F be the total energy emitted by the earth as a black body
What is presently done is to say that a fraction of the incident radiation E is reflected, the rest absorbed, and re-radiated by the earth as F:
and then solve for Earth's temperature . 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 , then a better equation is:
If you solve that for the Earth's temperature you get:
Solving you get =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)
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 , and emits where 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:
or . This is always true, at every wavelength, even in non-equilibrium, because and 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 for a white body is 1, emissivity 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:
The present analysis on the black body page is wrong, because it assumes =1 while is not. If you do it right, you get
and you get your equation above:
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 but in the ir range we have and . Then conservation of energy says incident=reflected+emitted or:
or
As noted in my original analysis, . 99.9 percent of the sun's radiation is vuv, so we can say , so now
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)
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)
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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)
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)
"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)
...do this kind of stuff [23] William M. Connolley ( talk) 21:08, 10 May 2010 (UTC)
Heald http://en.wikipedia.org/?title=Wikipedia_talk:WikiProject_Physics&diff=prev&oldid=458540639
Pratt - Headbom http://en.wikipedia.org/?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458754024&oldid=458750956
Heald
http://en.wikipedia.org/?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?
Pratt http://en.wikipedia.org/?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)
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)
Heald [25]
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)
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
[26] William M. Connolley ( talk) 16:58, 22 March 2012 (UTC)
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)
http://en.wikipedia.org/wiki/User_talk:William_M._Connolley#request_for_advice
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)
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)
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)
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)
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)
. . . 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)
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:-
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)
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)
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)
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)
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)
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)
http://en.wikipedia.org/wiki/Talk:Heat/Archive_14#Needs_Revision._Contains_a_fundamental_blunder.
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
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)
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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)
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.
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Cocomonkilla (
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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)
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)
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)
User:Ben MacDui Thanks!
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?
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)
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)
{{
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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)
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)
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.
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)
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)
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)
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)
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)
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 a Vacuum 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)
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)
"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)
"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)
"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].
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)
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)
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)
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)
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)
Continued from KimDabelstein's talk page.
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)
What is the difference between heat tranfer and "net heat tranfer"? -- Damorbel ( talk) 10:18, 30 May 2009 (UTC)
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)
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)
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)
You might find this interesting. The POV pushers are at it again. Q Science ( talk) 20:48, 16 June 2009 (UTC)
Blackcloak, the rigorous approach is to treat this radiation as Electromagnetic radiation (EMR). The important matter is how heat is transferred between two places by this radiation. You no doubt know that thermal radiation comes from the accelerating of electrical charge by thermal vibrations. To be rigorous, thermal radiation is not heat, it is an electric field with an amplitude related to the temperature of the source object (no radiation at 0K).
Two objects exchange heat when one is hotter than the other because the hotter one has an EM wave with a higher amplitude. The amount of heat transferred depends on the difference in the amplitudes between the hotter and the colder.
If both objects have the same temperature, the EM waves have the same amplitude and there is no heat transfer. It must be 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.
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).
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"?
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)
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)
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.
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)
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)
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.
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.
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.
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 1st item in the article. Common experience shows that all objects, micro to mega, tend to equalise temperature both locally and globally.
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.
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 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.
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].
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.
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)
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)
This section has now become unmanageable, please put your responses in the new section below.--
Damorbel (
talk) 08:16, 10 July 2009 (UTC)
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)
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)
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)
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)
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)
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)
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)
I saw this post by you
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)
PS: You might find this interesting - AIRS - Atmospheric Infrared Sounder currently flying on the NASA Aqua Mission
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:
Let F be the total energy emitted by the earth as a black body
What is presently done is to say that a fraction of the incident radiation E is reflected, the rest absorbed, and re-radiated by the earth as F:
and then solve for Earth's temperature . 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 , then a better equation is:
If you solve that for the Earth's temperature you get:
Solving you get =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)
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 , and emits where 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:
or . This is always true, at every wavelength, even in non-equilibrium, because and 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 for a white body is 1, emissivity 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:
The present analysis on the black body page is wrong, because it assumes =1 while is not. If you do it right, you get
and you get your equation above:
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 but in the ir range we have and . Then conservation of energy says incident=reflected+emitted or:
or
As noted in my original analysis, . 99.9 percent of the sun's radiation is vuv, so we can say , so now
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)
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)
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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)
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)
"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)
...do this kind of stuff [23] William M. Connolley ( talk) 21:08, 10 May 2010 (UTC)
Heald http://en.wikipedia.org/?title=Wikipedia_talk:WikiProject_Physics&diff=prev&oldid=458540639
Pratt - Headbom http://en.wikipedia.org/?title=Wikipedia_talk%3AWikiProject_Physics&action=historysubmit&diff=458754024&oldid=458750956
Heald
http://en.wikipedia.org/?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?
Pratt http://en.wikipedia.org/?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)
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)
Heald [25]
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)
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
[26] William M. Connolley ( talk) 16:58, 22 March 2012 (UTC)
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)
http://en.wikipedia.org/wiki/User_talk:William_M._Connolley#request_for_advice
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)
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)
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)
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)
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)
. . . 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)
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:-
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)
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)
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)
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)
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)
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)
http://en.wikipedia.org/wiki/Talk:Heat/Archive_14#Needs_Revision._Contains_a_fundamental_blunder.
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
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