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All IPCC definitions taken from Climate Change 2007: The Physical Science Basis. Working Group I Contribution to the Fourth Assessment Report of the Intergovernmental Panel on Climate Change, Annex I, Glossary, pp. 941-954. Cambridge University Press.

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The greenhouse effect and the 2nd law of thermodynamics

What the science says...

Select a level... Basic Intermediate

The 2nd law of thermodynamics is consistent with the greenhouse effect which is directly observed.

Climate Myth...

2nd law of thermodynamics contradicts greenhouse theory

 

"The atmospheric greenhouse effect, an idea that many authors trace back to the traditional works of Fourier 1824, Tyndall 1861, and Arrhenius 1896, and which is still supported in global climatology, essentially describes a fictitious mechanism, in which a planetary atmosphere acts as a heat pump driven by an environment that is radiatively interacting with but radiatively equilibrated to the atmospheric system. According to the second law of thermodynamics such a planetary machine can never exist." (Gerhard Gerlich)

 

Skeptics sometimes claim that the explanation for global warming contradicts the second law of thermodynamics. But does it? To answer that, first, we need to know how global warming works. Then, we need to know what the second law of thermodynamics is, and how it applies to global warming. Global warming, in a nutshell, works like this:

The sun warms the Earth. The Earth and its atmosphere radiate heat away into space. They radiate most of the heat that is received from the sun, so the average temperature of the Earth stays more or less constant. Greenhouse gases trap some of the escaping heat closer to the Earth's surface, making it harder for it to shed that heat, so the Earth warms up in order to radiate the heat more effectively. So the greenhouse gases make the Earth warmer - like a blanket conserving body heat - and voila, you have global warming. See What is Global Warming and the Greenhouse Effect for a more detailed explanation.

The second law of thermodynamics has been stated in many ways. For us, Rudolf Clausius said it best:

"Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature."

So if you put something hot next to something cold, the hot thing won't get hotter, and the cold thing won't get colder. That's so obvious that it hardly needs a scientist to say it, we know this from our daily lives. If you put an ice-cube into your drink, the drink doesn't boil!

The skeptic tells us that, because the air, including the greenhouse gasses, is cooler than the surface of the Earth, it cannot warm the Earth. If it did, they say, that means heat would have to flow from cold to hot, in apparent violation of the second law of thermodynamics.

So have climate scientists made an elementary mistake? Of course not! The skeptic is ignoring the fact that the Earth is being warmed by the sun, which makes all the difference.

To see why, consider that blanket that keeps you warm. If your skin feels cold, wrapping yourself in a blanket can make you warmer. Why? Because your body is generating heat, and that heat is escaping from your body into the environment. When you wrap yourself in a blanket, the loss of heat is reduced, some is retained at the surface of your body, and you warm up. You get warmer because the heat that your body is generating cannot escape as fast as before.

If you put the blanket on a tailors dummy, which does not generate heat, it will have no effect. The dummy will not spontaneously get warmer. That's obvious too!

Is using a blanket an accurate model for global warming by greenhouse gases? Certainly there are differences in how the heat is created and lost, and our body can produce varying amounts of heat, unlike the near-constant heat we receive from the sun. But as far as the second law of thermodynamics goes, where we are only talking about the flow of heat, the comparison is good. The second law says nothing about how the heat is produced, only about how it flows between things.

To summarise: Heat from the sun warms the Earth, as heat from your body keeps you warm. The Earth loses heat to space, and your body loses heat to the environment. Greenhouse gases slow down the rate of heat-loss from the surface of the Earth, like a blanket that slows down the rate at which your body loses heat. The result is the same in both cases, the surface of the Earth, or of your body, gets warmer.

So global warming does not violate the second law of thermodynamics. And if someone tells you otherwise, just remember that you're a warm human being, and certainly nobody's dummy.

Basic rebuttal written by Tony Wildish


Update July 2015:

Here is the relevant lecture-video from Denial101x - Making Sense of Climate Science Denial

 


Update October 2017:

Here is a walk-through explanation of the Greenhouse Effect for bunnies, by none other than Eli, over at Rabbit Run.

Last updated on 7 October 2017 by skeptickev. View Archives

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Further reading

  • Most textbooks on climate or atmospheric physics describe the greenhouse effect, and you can easily find these in a university library. Some examples include:
  • The Greenhouse Effect, part of a module on "Cycles of the Earth and Atmosphere" provided for teachers by the University Corporation for Atmospheric Research (UCAR).
  • What is the greenhouse effect?, part of a FAQ provided by the European Environment Agency.

References

Comments

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Comments 826 to 850 out of 1393:

  1. Re #824 "damorbel #820: You attribute a quotation from message #804 to me. It was actually written by KR." Very sorry!
  2. damorbel #821: "The Sun is at about 5780K ... The Sun occupies only a small part of the sky so we do not get a the full 5780K here, only 279K." someone; mods, NATO, the UN ... anyone please make it stop! The abuse of physics here is right up there with any human rights violations! (just so as this isn't pure rant: you do see the full temperature give or take attenuation - just not integrated over the whole of our 'aperture' (the sky), nor do we see it's full irradiance; but that does not affect the light spectrum and, therefor the temperature we see, as such)
    Response:

    [DB] You have my sympathies, FWIW. I would reiterate an earlier suggestion I made: DNFTT. And we all know by now the players in this drama. If you see recycling of earlier arguments, please point them out for deletion and possible stronger action. Thanks!

  3. LJ, it's the same old question: where does the energy being emitted by the atmosphere go? It cannot choose its path. If a molecule of CO2 3 inches above a fallow field in Idaho emits a photon downward at 3:00 in the afternoon on July 29th, and that photon is not impeded before striking the sun-warmed molecules that make up the soil, what happens to the photon? Does it slam on the brakes and say to itself "Damn, I almost violated the alleged 2nd Law of Thermodynamics!"? Or does it hit the soil and "bounce off"? Can it be absorbed? More home experiments! Take two pots of boiling water, both with a constant heat source of 90C. Place a 50C heat source ten feet (so there can be no question of convective interference) above pot no. 2. Will the temperature of pot no. 2 increase at all? Will the 50C heat source add its energy to the 90C source and make the water hotter than for pot no. 1? Yah, ok, DB. I'm done--and I barely got started.
  4. Re #814 your comment (as mod) was:- "Response: [muoncounter] Perhaps you should have checked the link in my response to #784. Unless you are a different damorbel, you gave us the parable of your disdain for textbooks some months ago. Do try to keep track of your own words; they are there for all to see." If you look at #784 carefully. I wrote :- "But I don't know which textbook I am suppose to read or whether it is a requirement for scientists to read text books. Personally I recommend original works, textbook contents are at least 2nd hand if not much more; at university my tutors always advised original texts, they had a low opinion of published textbooks." - I was responding to DB's comment in my #783 where he links to scaddenp #753. I responded to scaddenp's remark in 753 where he wrote:- "I asked if the experiment didn't go your way, whether you would be prepared to abandon your view and read the textbook. (ie, behave like a scientist)." This is of course a personal attack on me and I usually avoid responding to them. But, since you are in a special position as a moderator, I thought it would be a good idea to let you know the origin of these remarks that I, for one, see as highly irrelevant.
    Response:

    [DB] The comment is not an attack on you personally; that would be ad hominem and would be disallowed. The remark in question was directed to your very own words. Please be consistent and do not feign ignorance. The definition of "is" has already been debated.

    [muoncounter] You repeated the substance of a prior comment, which adds nothing to the current discussion. You've done the same thing a number of times. If you find the instruction to stop that particular behavior a personal affront, so be it.

  5. damorbel - the gambit is a way to end the argument. If you arent philosphically prepared to accept experimental evidence as the arbiter, then yes it is a well-deserved attack on you and meant to expose you to other reader of this. On the other hand, if you do accept that reality is the arbiter, then then the game is played like this: An experiment is proposed: (you can propose it). You calculate by any means you like, the outcome of the experiment. I am sure you mean to do within your understanding of physics. Someone else (not you), calculates the experiment via the relevant textbook physics. If you are right, then time for us to help you polish a paper. If textbook is right, then time for you to go back to school and stop complaining the climate scientists dont understand physics.
  6. What's this talk of photons having temperatures in Kelvins? A photon's energy depends on its frequency or wavelength, not on the temperature of its source, which is what Damorbel seems to imply. The better informed here correct me please, as this is the way I see it: the temperature of an EM radiation source affects the spectrum of the radiation and that's about it. An individual photon at a given frequency couldn't care less whether it came from a 5 gazillion degrees source or a light bulb, does it? If it does, how exactly does that manifest? A different spin angular momentum? Or what?
    Response: There is a common misconception that all photon sources output photons of only a single frequency that is determined by the temperature of the source. In fact, the blackbody radiation curve is a distribution of photons of multiple frequencies, with an increase in temperature causing a shift in that distribution of emitted photons so that more of the higher frequency/energy photons are emitted relative to the lower frequency/energy photons, but there still is emission of photons of multiple frequencies.

    Consequently, when somebody receives a photon of a given frequency, that person can state only the relative probabilities of the temperature of that photon's source. That photon could have come from a source of any temperature. As a commenter said a bit ago, photons do not carry ID cards.
  7. Ryan, your assertion that the earth, if receiving energy faster than it looses it, will not warm beyond 255K here violates the first law of thermodynamics - what you propose does not conserve energy. Either you are wrong or the Laws of Thermodynamics are wrong - take your pick.
  8. Re #831 You wrote:- "What's this talk of photons having temperatures in Kelvins? A photon's energy depends on its frequency or wavelength, not on the temperature of its source" The temperature of a particle (in an ideal gas) is measured by the amount of energy (Joules) in the particle. The Boltzmann constant relates the energy to the temperature in Kelvins, the formula is E = 3/2 kT where k is the Boltzmann constant 1.3806504×10^−23J/K Photons are considered to be energetic particles, the term photon gas is used frequently. As energetic particles photon energy can be given as temperature or e/v (electron volts) Photon energy is also a function of the oscillation frequency of the electron that originates the photon, so photon energy is given by the formula E = hv where h is the Planck constant = 6.62606896×10^−34 J/s and v the frequency You wrote:- "...the temperature of an EM radiation source affects the spectrum of the radiation and that's about it." Not just the spectrum but the energy also. You wrote:- "An individual photon at a given frequency couldn't care less whether it came from a 5 gazillion degrees source or a light bulb, does it? If it does, how exactly does that manifest? A different spin angular momentum? Or what? " Each photon is created by an individual electron that gets its energy from the particle that where the electron is found. A photons energy is directly related to the temperature of the particle emitting it. When a photon is emitted it carries momentum, there is a recoil reaction on the particle emitting the photon which means the emitting particle loses the amount of momentum taken away by the photon. This is just the same but on a smaller scale, as a bullet leaving a gun. Thus photon energy is directly related to the source temperature and the photon 'knows' this because the frequency v is a direct function of the temperature.
  9. Re #831 (in the grey area) someone wrote:- "That photon could have come from a source of any temperature. As a commenter said a bit ago, photons do not carry ID cards." If they wrote that, then it is not correct. Photons are emitted (and absorbed) by individual (accelerating) charged particles. The source of a photon characterises it by the energy the photon has. The photon keeps this energy (unless it changes energy in a gravitational field) until it is absorbed by another charged particle, even if it has to cross the universe before this happens.
    Response:

    [DB] I am simply gobstoppered. Please think about what you just wrote some more. As written, does-not-parse.

  10. #831: "As a commenter said a bit ago, photons do not carry ID cards." #834: "If they wrote that, then it is not correct. Photons do carry ID cards? Will the madness never cease? 'Get the new EZ-photon identification card! Never get held up by those pesky laws of physics again! With EZ-photon you too can make your own decisions about what forms of matter you choose to interact with. EZ-photon! Because reality is just so passe!'
  11. Damorbel, you are digging yourself into a bottomless pit of nonsense. This sentence makes no sense, and wouldn't even if the syntax was correct: "Each photon is created by an individual electron that gets its energy from the particle that where the electron is found." You say this: "photon energy is given by the formula E = hv" That is, in fact, correct. Where in that formula is the temperature of the source hidden? In other words, what distinguishes the energy of a photon at a given frequency emitted by a source at a certain temperature from the energy of a photon at the same frequency coming from a source at a different temperature? There are only 2 terms to the energy of a photon, one is a constant. You are saying that, if the other is also kept constant, the product of the 2 can nonetheless be different according to a factor that is not part of the equation. Do you realize how idiotic that is?
  12. Damorbel @831 Each photon is created by an individual electron that gets its energy from the particle that where the electron is found. A photons energy is directly related to the temperature of the particle emitting it. This is incorrect. Molecules emit photons when they transition from one quantum state to a less energetic one. The frequency of the photon is determined by the difference in energy between the two states, as related by E = hv. The energies of the quantum states are fixed, determined by the atomic makeup of the molecule and the strength of the bonds between those atoms. Thus the frequency (and hence energy) of the photon is not determined by its temperature. Temperature will control the intensity of the radiation at a given frequency, since that will determine the proportion of molecular in excited quantum states that can decay. IR radiation is emitted by vibration of atomic nuclei within a molecule, Microwave by rotation of the molecule as a whole, and visible/UV radiation is emitted by electrons. Two photons of identical frequency are not "tagged" by their emitting source. However the spectrum (plot of frequency versus intensity) of a given molecule (especially a gas) is a sufficient finger-print to identify the substance uniquely, and the relative intensities of the various frequency bands can often be used to infer temperature of the emitting substance.
  13. Re #835 & #386 If you find what I wrote in #833 unclear check this link and find - when a cosmologist talks - . when a cosmologist talks about the 'temperature' of a photon Then tell me what the problem is with 'the temperature of a photon'.
    Response: [Dikran Marsupial] I suspect there is a good reason the article talks of an "equivalent temperature" rather than simply a "temperature".
  14. damorbel, That article is talking about curve matching an observed wavelength spectrum to a known emission profile at a given, not the temperature of an individual photon. Individual photons of equal wavelength are identical regardless of source temperature.
  15. Re #839 Bibliovermis you wrote :- "Individual photons of equal wavelength are identical regardless of source temperature." If photons of 'equal wavelength' are 'identical' then they have the same energy also, which according to the link means they have the same 'temperature'. The only difference betwen a photon and a particle moving at less than c is that a photon must be absorbed to give up its energy. When you talk about 'curve matching' and 'spectrum' you are no longer talking about individual particles (including 'photons' as particles). These terms form part of statistical mechanics, the science of large collections of particles. But the concept of temperature is not confined to 'large collections of particles', temperature is an intensive property, meaning individual particles have a temperature also.
    Response: [Dikran Marsupial] No, it means they have the same "equivalent temperature", it does not imply they were emitted by bodies of the same temperature.
  16. Re #838 in Response: [Dikran Marsupial] you wrote :- "I suspect there is a good reason the article talks of an "equivalent temperature" rather than simply a "temperature"." There is. Photonic energy is regarded as electromagnetic, although when they are created they take mechanical momentum from the emitting particle and give it (the momentum ) up when absorbed. But photons are not mechanical 'objects'; they have no mass so can't collide. Collision is how mechanical particles exchange momentum (thus energy), according to kinetic theory. Temperature is essentially a mechanical concept, that is why the energy of a photon gives it an 'equivalent' temperature.
  17. Discussing an individual photon's 'temperature' is a bit of semantic play, which is why quotes are used. Individual photons with equal 'temperature' (energy / wavelength / frequency) are identical regardless of source temperature.
  18. Re #840 in Response: [Dikran Marsupial] you wrote :- "No, it means they have the same "equivalent temperature", it does not imply they were emitted by bodies of the same temperature." Photons are generated in different ways, but when they are generated by molecular motions they have energy directly related to the temperature of the particles. Einstein wrote a paper about this in 1916 "Zur Quantentheorie der Strahlung" and I've never seen it contradicted. Photonic energy is regarded as electromagnetic, although when they are created they take mechanical momentum from the emitting particle and give it (the momentum ) up when absorbed. But photons are not mechanical 'objects'; they have no mass, so they can't collide. Collision is how mechanical particles exchange momentum (thus energy), according to kinetic theory. Temperature is essentially a mechanical concept, that is why the energy of a photon gives it an 'equivalent' temperature.
    Response: [Dikran Marsupial] You are still not making the distinction between the "effective temperature" of a photon/emitting particle and the temperature of the emitting body. Oh well, you can lead a horse to water...
  19. This has gone on long enough. I would like to encourage all to abide by this principle we keep on talking about yet keep on ignoring: DNFTT. Of course, every time we try, they spew another humongous piece of absurdity and we can't help but point it out. We have to stop doing that. They have all done has done an excellent job of demonstrating the extent of their confusion and no amount of redirecting can reconcile them with reality. At some, point, one's mind must be acknowledged as having declared itself. We're long past that.
  20. Re #842 Bibliovermis you wrote :- "Individual photons with equal 'temperature' (energy / wavelength / frequency) are identical regardless of source temperature." This is precisely what Einstein's 1916 paper is about, he shows how the electromagnetic 'Planck black body spectrum' is equivalent to the Maxwell-Boltzmann energy distribution in an ideal gas.
  21. Re #843 Response: '[Dikran Marsupial] You are still not making the distinction between the "effective temperature" of a photon/emitting particle and the temperature of the emitting body.' Sorry, I must have missed something; "effective temperature"? I had not realised this matter had been commented on. Can you help me?
    Response: [Dikran Marsupial] I think you need to read the article that you introduced to the discussion here a little more carefully. Note the author talks about the 'temperature' of a photon, the quotes imply that the meaning of temperature was not the usual meaning of the word.

    [muoncounter] The article uses 'equivalent temperature,' rather than 'effective temperature'.

    [Dikran Marsupial] You are quite right, mea culpa - oh the irony! ;o)

  22. Re #846 Response: [Dikran Marsupial] I think you need to read the article that you introduced to the discussion here a little more carefully. Note the author talks about the 'temperature' of a photon, the quotes imply that the meaning of temperature was not the usual meaning of the word. I think the last sentence of Paul Walorski's article sums up the matter quite well:- "So it's not so much that the photons are all at a temperature of 2.7K but rather that they appear as if they were emitted by a single blackbody which was itself at a temperature of 2.7K." That would only be true if the Planck spectrum and the Maxwell-Boltzmann distribution were equivalent. My use of the word 'equivalent' is deliberate. Re #844 Philippe Chantreau. I'm sorry you have this reaction but the 2nd Law is what is in question. The 2nd Law of Thermodynamics it is very well established, it is not easy to grasp all its implications and failure to take them into account has brought many a beautiful hypothesis crashing down. I'm afraid the concept of temperature is just about as close to the heart of the 2nd Law as you can get.
    Response: [Dikran Marsupial] The "they" in the sentence is the key there, you can't tell from a single photon the temperature of the emitting body, you need to look at the distribution of energies of a large number of photons and do some curve-fitting (and make an assumption or two).
  23. Re #847 Response: [Dikran Marsupial] The "they" in.... you can't tell from a single photon the temperature of the emitting body," Indeed you can't. But what you do know is the amount of energy the emitting particle has (or more accurately 'had') and that is (was) its temperature. Further: "you need to look at the distribution of energies of a large number of photons and do some curve-fitting (and make an assumption or two" Only true if you have a large number of particles. If you have a large number of particles ('real' particles - not photons) they are continually colliding and thus exchanging energy. Because of this they all have different energies and thus different temperatures but the critical point is they have an averge energy that corresponds to the measured (average) temperature.
  24. An object, at 20C, has an 80% absorptivity for 6 micron photons. Absorptivity is unchanged by temperature - the temperature is for later reference. (q) The object is struck by a 6 micron photon from a hotter object (40C) which includes 6 microns in it's emission spectra. What is the probability of absorbing the photon? (a) 80%. (q) The object is struck by a 6 micron photon from a cooler object (0C) which includes 6 microns in it's emission spectra. What is the probability of absorbing the photon? (a) 80%. Photons do not carry ID cards (the earlier quote was originally from me, I believe) indicating the temperature of the emitting object. The temperature of the emitting object is not encoded in the energy of an individual photon. Absorption depends only upon the individual photon energy and the (separate) object absorptivity spectra. You cannot refuse that 6 micron photon because you somehow "know" that it came from something colder. A spectra of photons can be statistically analyzed to determine the temperature required to emit that spectra (given some idea of the emission spectra of the object), but individual photons have energies, not temperatures. And each individual photon adds to the energy of the absorbing object. --- All of these 2nd law objections are based upon one or more such fundamental misunderstandings of physics, and are hence incorrect. The radiative greenhouse theory is entirely supported by thermodynamics.
  25. Damorbel, you've got nothing to say. Cut the BS and answer the substantive questions: Energy of a photon E=h.v Where is the temperature of the source? You have not the slightest clue of what you are babbling about and neither does LJR. "Trolling" is the only accurate way to describe what both of you did on this thread.

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