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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.

Last updated on 22 October 2010 by TonyWildish.

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Related Arguments

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 901 to 950 out of 1393:

  1. Re #894 KR you wrote:-

    "damorbel - Are you asserting that an individual photon (with a particular energy) can be used to identify the temperature of the object that emitted it, thus affecting it's absorption? "

    Yes I am, that is the basis of all quantum interactions.
    Since 'an object' can be an individual particle, I don't see the diffiiculty.

    Further you wrote:-
    "Or that the possibility of absorption is not a function of photon energy and absorption spectra? If so, you are sadly mistaken."

    Does this mean that photon interactions do not need to be quantised? Well I don't accept that photon interactions can occur in a non quantised way; throughout the whole electromagnetic spectrum photon interactions with massive particles are always quantised; both on the macroscopic level (many particles described statistically) and at the individual particle level.

    NB Photons do not interaction with each other.
  2. darmorbel @897

    Wein's displacement law, and the spectrum you allude to apply only to black bodies. Gases (such as CO2 and H2O) are not black bodies. The photons they emit will correspond to some of the frequencies of a black body, but their distribution of intensities will not. It is therefore not appropriate to talk of "temperatures" of photons.
    The energy of a photon is only related to the quantum states involved in the decay when that photon is emitted. Black bodies are "black" because they have a wide range of quantum states, merging into a continuous band. Otherwise they wouldn't be "black".
  3. Darmorbel @901

    "damorbel - Are you asserting that an individual photon (with a particular energy) can be used to identify the temperature of the object that emitted it, thus affecting it's absorption? "

    Yes I am, that is the basis of all quantum interactions.


    This is incorrect. A molecule of CO2 (for example) has 4 (one doubly degenerate) vibrational modes. These occur at
    2565 cm-1 1480 cm-1 and 526 cm-1 If a molecule is in the 1st level excited bend state it can emit a photon at 526 cm-1 But this reveals nothing about how many molecules are in excited states in the other vibrations or rotational excited states or the translational energy that the molecules have.
  4. damorbel - You have completely misread my question; I have to wonder if it's deliberate. You have two complete strawman arguments misstating what I asked.

    I asked about a photon and "...the temperature of the object that emitted it...". You replied regarding an individual particle, not an object temperature, strawman. An individual photon does not give the temperature of an emitting object. It doesn't even give the temperature of a single molecule, as there are multiple possible emitting down-transitions in electron orbitals for an excited molecule; it doesn't have to drop to ground state.

    Your answer is incorrect!

    "Does this mean that photon interactions do not need to be quantised?" - All photon interactions are quantum interactions, I never stated otherwise, strawman.


    The issue is whether photons emitted by a cooler object can be absorbed by a warmer object (ensemble temperatures), namely the atmosphere and the surface. Since an individual photon does not indicate it's emitting objects temperature, it's absorption depends only upon the absorptivity of the warmer object (determined by quantum interactions, which are off-topic) and the individual photon energies.

    You are really reaching here.
  5. Re #898 KR you wrote:-
    "Actually, the peak of the curve is the mode, not the average; the two are not identical unless the distribution is symmetric. That's a fairly common error. "

    This is a better explanation of mode

    The relation of temperature to the peak is, once more, a critical one in quantum theory.
  6. I referred to the Statistical Mode, not emitting, molecular, orbital, or other modes.

    Mistaking statistical mode for mean (average), or vice versa, is a common error.

    The rest of your post is rather irrelevant to the discussion.
  7. Re #904 KR you wrote:-

    "I asked about a photon and "...the temperature of the object that emitted it...". You replied regarding an individual particle, not an object temperature, strawman."

    I asked you to define 'an object'. Until you do, I don't know what you are talking about. So pleeaaase - do it now!

    For example you wrote:-

    "as there are multiple possible emitting down-transitions in electron orbitals for an excited molecule; it doesn't have to drop to ground state."

    Yes and these transitions can be (almost) anywhere in the EM spectrum where many other kinds of interactions can take place; but please let us stick to thermal matters, i.e. interactions that affect temperature.
    Response: [muoncounter] Pedantic requests, such as the definition of 'an object,' are exactly what have put this thread at over 900 comments, many mere attempt at distraction. If you have difficulty with definitions of everyday language, how can your definitions of scientific terminology be taken seriously?
  8. When the fish is in the boat, it may tend to flop and wiggle violently, desperate to once again command the murky, muddy waters in which it has evolved to thrive.
  9. damorbel - "I asked you to define 'an object'." I cannot find such a request from you. But I'll answer it now that you have asked.

    I am speaking of relevant objects to the radiative greenhouse effect - the surface of the Earth, and the atmosphere. Both are large, ensemble objects, containing temperatures defined by thermal distributions, with thermal emission and absorption spectra defined by their component molecules.

    You cannot determine the temperature of such an object from a single photon. Single photons have a single energy, a single wavelength.

    Absorption likelyhood, the real question at hand, is based upon absorption spectra and individual photon energies. So I will ask again:

    Are you asserting that the possibility of absorption of a photon is not a function of that photon's energy and the potentially absorbing object's absorptivity spectra?

    DSL - rotfl.
    Response: [muoncounter] Oddly enough, damorbel used to agree:

    We are all familiar with the Planck spectrum, the amplitude of which is a function of the temperature, But taking one photon (with energy a function of frequency), or even one spectral component, does not represent the entire spectrum thus the temperature is not defined. Although a single photon has energy it does not have a temperature.

    See comment#70, this very thread.

    One could only describe this behavior as 'doublethink.'

  10. damorbel, you seem insistent that the following 'fact' is of great import:

    - the energy of a photon can tell you the 'temperature' of the particle that emitted it

    Firstly, a single particle doesn't have a temperature, which is a statistical measure. It has an energy. This seems to have been explained to you before. However we are discussing gases, liquids and solids that, as has been explained, emit photons with a range of energies. These spectra are clearly the applicable measures when discussing radiative energy transfer.

    Next, from what people have said, it seems you believe that a photon emitted by a cooler object cannot be absorbed by a warmer object. Is this an accurate description of your stance?
  11. damorbel,

    "What % of the heat tranferred to the atmosphere from the ground by radiation:-

    14%?......40%?.......90%?"

    "That is the question I asked you. But I will accept evaporation and convection as 'kinetic'."


    I'm not sure what you're getting at here or why this matters, but I'll give it a stab. If over 90% of the thermal mass of the planet is in the oceans and about 5% is in the land mass, that leaves maybe a percent or two in the atmosphere? Are you looking for an actual number?

    Again though, all the kinetic energy flows from the surface to the atmosphere and back to the surface, relative to the radiative budget, are zero. They have to be because all the energy leaving at the top of the atmosphere is radiative.

    Also, I haven't read this entire thread, but I'm not sure I understand you're fundamental objection here. I don't see how the greenhouse effect violates the second law because it's not about energy going from cold to hot in conduction process. Are you claiming that a photon cannot travel from the colder atmosphere toward the warmer surface?
  12. Damorbel if you keep digging at that rate, you will soon find yourself in molten iron.

    Wein's law says nothing like what you imply. It says what is the most likely frequency of photons emitted by a source according to the temperature of that source, provided it is a blackbody. If there are several sources emitting at the same time with overlapping spectra, all it tells you is how likely it is that one frequency originated with a source rather than another.

    If I'm in a room with some light coming from outside (i.e. from the sun) where a light bulb is on, Wein's law can not allow me to determine what was the source of a given photon. Only the relative probability.

    To go back to the origin of this discussion, it must be made clear that a photon coming from the sun at a given frequency and one coming fron the light bulb at the same frequency will have the same energy. Exactly the same. You previously argued that it was not the case. Wein's law does not allow you to defend that either.

    Photons do not carry ID cards, no matter how badly your confused mind wants it. The energy of a photon depends only on its frequency. There is no way to tell where an individual photon originated only by examining its frequency. You could say where it could not have originated, that's all.

    I must say that throwing the photoelectric effect in the mix was yet one of your funniest moves in that strange display of yours.
  13. muoncounter at 909. I disagree. It can also be just double talk, or talk adapted to the need of the moment when conducting an argument with no other function than sowing confusion.

    Or could it simply be that Damorbel's own confusion is so thorough by now that he is completely lost in the whole thing?
    Response: [muoncounter] Doublethink includes the ability "to forget any fact that has become inconvenient, and then, when it becomes necessary again, to draw it back from oblivion for just so long as it is needed."
  14. 884 KR
    " The "2nd law" objection to the greenhouse effect is based upon a mistaken notion.."

    I don't disagree with you. However I'd like to suggest there is another problem. In the post, the statement of the 2nd law has missed out the phrase: "whose sole result". This is a statement that the 2nd law only applies to a closed system.

    For practical purposes, the system which consists of: the sun, outer space, the solid earth and the earths atmosphere is not a closed system.
    Outer space is, for practical purposes, an infinite sink. The sun is, again for practical purposes, an infinite source of energy. No one can deny that energy from the sun reaches the surface of the earth - at least, nor that radiated energy which isn't reabsorbed somewhere leaves the system...

    This is implicit in the text - body heat is in effect a source of energy external to the heat exchange system which is moderated with blankets.

    So long as some source continues to pump out energy irrespective of the destination of that energy, we're free to build an engine which uses that energy to, for example, concentrate it up to any temperature we can manage. How that engine works (photons, gases, cogs, whatever) is immaterial - the beauty of things like the statistical and termodynmaies is that they are defined for an abstract engine, which applies to all real engines...
  15. damorbel - I think you've been called out; "By thine own words shalt thou be condemned."

    To wit:

    "We are all familiar with the Planck spectrum, the amplitude of which is a function of the temperature, But taking one photon (with energy a function of frequency), or even one spectral component, does not represent the entire spectrum thus the temperature is not defined. Although a single photon has energy it does not have a temperature." - damorbel @70, this thread, 24/11/10 (thanks for pointing this out, muoncounter)

    versus:

    "But there is no need to have a certain number of particles to make a sample, so one particle with the same energy as the average energy of all the particles also has the same temperature as the whole sample." - damorbel @892, this thread, 31/3/11

    Reductio ad absurdum - by contradiction you have disproven your own arguments.

    You are a troll - willing to say anything, even contradict yourself, in order to prolong an argument. Nothing you have written can be taken seriously, as you are not engaged in a scientific discussion.

    I have no idea as to your motivations. Perhaps you just like to argue - in that case I consider you a ( -snip- ). Perhaps you are arguing points you don't believe in for ideological reasons - in that case I consider you an ( -snip- ). Or perhaps you do this because it's your job? I'm familiar with that last case; my brother spent years as a denialist of second hand smoke dangers for a major tobacco company. In that case I would ask you the question I asked him - "How much does a soul go for these days?"

    Overall, I'm disgusted.

    Everyone - I would encourage you to consider this demonstrated behavior when evaluating anything that damorbel writes, whether here, or on his multiple attempts to redefine the Wiki page on thermodynamics.
    Response: [DB] I completely agree with you, word for word, but I have a role to fulfill. Sorry for the snips.
  16. A bit of summary here, then. I'll put it in the form of a proof for clarity.


    - Individual photons have energies, but these do not represent the temperatures of the objects that emitted them. You can say what objects could not have emitted that photon based on temperature, but not which one has.

    - Absorption of a photon by an object (warmer or colder than the emitting object) has a likelyhood based upon the absorption spectra and the energy of the individual photon; not the temperature of the emitting object.

    - The Earth's surface has an emissivity and absorptivity of ~0.98 in the IR spectra, so 98% of those photons impinging will be absorbed.

    - 98% of surface impinging atmospheric thermal radiation (aka "backradiation") will be absorbed by the Earth, as per the Earth absorptivity and atmospheric emissivity spectras.

    - Each photon absorbed, by the first law of thermodynamics, adds to the internal energy and hence temperature of the absorbing object.

    - The emitting mass of the atmosphere (due to the lapse rate) is colder than the Earth's surface.

    - Hence a colder object raises the temperature of a warmer object by it's presence.

    - Therefore: The assertion by Gerlich and Tscheuschner that a cooler object heating a warmer object violates the 2nd law of thermodynamics is categorically false.

    Q.E.D - Quod erat demonstrandum.

    ---

    I don't think that I need to say anything more on this topic. Adieu.
  17. DB - I understand the moderation role, not a problem.

    I would encourage everyone to form their own opinions of each poster's contributions in light of their content, and act accordingly.
  18. I would, KR, but if the light is coming from a cooler object, I won't be able to see the contribution.
  19. Perhaps, then, DSL, 'the cool shall rule' by virtue of superior vision?
  20. DSL and KR @ 918, 919.
    ROFL

    Thus the troll is reduced to its initial insignificance. Reality and the reality-based have prevailed. Yeah!
  21. Re #900 KR You wrote:

    "and that this absorption (by the 1st law of thermodynamics, conservation of energy) affects and slows the total, net energy transfer to the atmosphere and hence to space."

    You write about the 1st Law and radiation as if these were the only two energy processes involved - you take into account radiation only. But for a thermodynamic analysis you must include all forms of energy involved in the whole thermodynamic system that comprises the atmosphere.

    By confining your consideration to radiation only you may well get the answer you are seeking but that is hardly science!

    As I have mentioned before, you must also account for the gravitational energy of the gas that makes up the atmosphere; it is, after all, the gravitaional component that gives the troposphere its temperature profile (lapse rate) of -6.5K/km.

    Any attempt to may an 'energy balance' that doesn't include gravitational energy is not going to give an accurate picture.
  22. damorbel - I'm quite surprised to see you back. Have you read my posting here?

    Evaporation, convection, and the adiabatic lapse rate have all been covered in tedious detail on this thread; if you're interested, look it up.

    But (personal opinion) I do not consider it worthwhile to debate with someone who (like you) is willing to contradict your own posts in order to prolong an argument - that is trolling, not science.
    Response: While these topics have indirect relevance to the 2nd law and its relationship to the GHE, this thread is not intended as a substitute for a college level physics course. As you pointed out, these topics have already been covered here in excruciating detail. Future off-topic or repetitive comments will be deleted.
  23. Damorbel, you didn't need anyone's help to cover yourself with ridicule. It is glaringly obvious that, not only you haven't the fuzziest idea about these matters, but you are willing to contradict yourself for the sake of argument.

    This is rather amusing: in the instance noted above, you later adopted an incorrect position, opposite to your originally stated one, which was correct. And you defended the latter with all your rethorical might. Really, that is comical. Cut your losses.
  24. KR

    When investigating complicated system it's important to approach the mechanism front-wise and linearly. Starting at the start (solar input) avoids confusion as introduction of secondary input/variables (forcing) manifest. If in fact, all the tenets of GHG theory are valid then such a stepwise approach will only hone their formulation.

    Let's start with what we undoubtedly agree:

    (e1)emissivity + reflectivity =1 earth's albedo =.3

    (e2)emissivity = absorbed energy/ incident energy---or stated continuously--- emissivity = absorbed power/
    incident power

    (e3) σTe4= S/4 * (1-A) flux density emitted via blackbody earth = flux density absorbed via blackbody earth (note to muoncounter, blackbody equivalent) So σTe4= 240 W/m2, represents the theoretical maximum power emitted and adsorbed, via SW, by the surface.


    This theoretical max flux can be used to calculate actual max flux absorbed by earths surface. Intuitively it make sense, a surface must absorb energy before radiating said energy. So properly, this must be calculated prior to surface to atmosphere emissivity consideration. Also, immediately jumping to actual surface temp and backing out flux, will as said earlier lead to erroneous conclusions.


    (e4) Because 240 W/m2 is max blackbody absorption it is equal to max incident power.

    (e5) Earths ε = .98. Using (e2) earths actual absorbed power =.98 * incident power = .98*240 W/ m2 = 235 W/ m2

    (e6)Because 235 W/ m2 represents the true absorbed SW radiation value it also represents earth's maximum gray body emissions due to SW.

    (e7) When surface LW emission flux is equal to SW solar absorption the earth's system is in equilibrium; 235 W/ m2 equates to 254K. Because the 1st law must be upheld this represents the temp maximum via solar radiation. Any additional temperature increase must come from non-radiative energy input. I suspect you are screaming "What about forcing". Ok, by adding radiative reflection/re-radiation flux to solar input, with white, black and gray atmosphere emissivities, LW forcing is easily evaluated.


    The following section (e8),(e9), (e10), demonstrates GHG physics. Specifically, by adding flux regardless of quantitative magnitude and/or vector magnitude.

    (e8) Atmosphere ε=0 Teq=time to equilibrium (e7) (235 W/m2)

    White
    Note: As Time approaches 2Teqthe Surface flux approaches infinity. Also, as Time approaches 2Teq the atmosphere becomes transparent to visible surface emission...when visible emissions = 240 W/m2 TOA equilibrium is at hand. According blackbody emissions this equates to ~1200K.


    (e9) Atmosphere ε=1 Teq=time to equilibrium as defined by (e8)

    Blackbody

    Note: As time approaches 4Teqthe Surface flux approaches infinity. Also, at 3Teq the atmosphere radiates 235 W/m2 plus the 5 W/m2 originally reflected by the surface (e5) give the required 240 W/m2 TOA...302K.

    (e10) Because(e8) confers the maximum temperature (~1200 K) for TOA equilibrium and (e9) confers the minimum temperature (302K)for TOA, radiative forcing is shown to be a false mechanism. That is, since actual temperature (288 K) is well below the minimum temperature established by blackbody atmosphere, ε < 1, will generate a temperature higher then 302K.

    (e11) If gray body ε=.612 then, according to (e1), gray body reflectivity = .388

    (e12) If gray body ε=.612 then, according to (e2), gray body absorbed = incident flux* ε.

    (e13) Atmosphere ε=.612 Teq=time to equilibrium as defined by (e8)

    Gray
    Note: TOA is achieved when surface radiates 768 W/m2...341K.

    (e14)As demonstrated, a body's emission can not be increased by it's own reflection,re-radiation, or insulation. As demonstrate lower energy does not increase higher energy, low light does not make more luminous a brighter surface. As demonstrated atmospheric forcing, GHG physics is a false mechanism which in fact violates the 2nd Law. Choosing to ignore this fundamental law leads to fallacious results. Fallacious result such as 341K with an atmosphere emissivity of .612. But just as fallacious misapplication of physics which leads to ε=.612 equating to 288K. Notice I did not say atmospheric radiation does not exist...I did not say the downward radiation does not exist. Atmospheric radiation is isotropic however, lower energy atmospheric radiation can not increase the higher energy surface. Solar input, assuming .3 albedo, can only account for 240 W/m2 flux and therefore delta T between solar input and actual temperature must be a result of non-radiative input.
  25. 924 L.J. Ryan.

    Nice embedded link to the lecture notes of some Professor Jin-Yi Yu - who has some other nice explanatory material for those finding this all a bit hard.
    http://www.ess.uci.edu/~yu/ess55.html

    Where did the rest of the material come from (it is only polite to reference sources, after all)?
  26. les 925

    "Where did the rest of the material come from"

    Calculated base upon GHG physics. For example, an atmosphere with LW ε=0, will reflect all terrestrial emissions. That is, until accumulated energy is sufficient for visible spectra emissions.

    By the way surface energy for 1.998046785(Teq) should be 120422 W/m^2.
  27. L.J. Ryan
    that's just your interpretation.
  28. Riccardo 927

    you said:
    "that's just your interpretation"

    Be a bit more specific please. What is "that"?
  29. LJRyan @924:

    1) The emissivity/absorptivity of a body varies with wave length. Therefore there is no absolute emissivity, only emissivity/absorptivity relative to a certain range of frequencies of emissions/absorptions. Hence we have:

    (Eq 1a') reflectivity(sw) = earth's albedo(sw) = 0.3

    Eq 1a") 1 - reflectivity(sw) = emissivity(sw) = 0.7

    (Eq 1b') reflectivity(lw) = earth's albedo(lw) = 0.02

    (Eq 1b") 1 - reflectivity(lw) = emissivity(lw) = 0.98

    Where (sw) indicates the range of wavelengths at which the Sun's radiation is most intense, and (lw) indicates the ranges of wavelengths at which the Earth's thermal radiation is most intense, and where equation (1a'and ") consider the whole Earth system, while equations (1b'and ") consider only the surface of the Earth, and do not include the effects of the atmosphere.

    2) (Eq 2) is ok as it stands, provided it is indexed for wavelengths as in equation 1.

    3) Equation 3 is false in that you interpret σTe^4 as dealing with the Earth's surface only, while S/4*(1-A) definitely deals with the total incoming radiation at the top of the atmosphere. Consequently, equation 3 should read:

    (Eq 3) σTe^4 = S/4*(1-A) where Te is the effective temperature of the outgoing radiation at the top of the atmosphere, S is the total incoming solar radiation at the top of the atmoshere and A is the Earth's Albedo.

    Because equation 3 is valid only for the TOA, it places no limit on surface temperature of the Earth by itself, and no limit on the maximum energy radiated by the surface per second.

    4) Equation 4 is false. The maximum black body absorption at the TOA is S/4, ie, the case where the Earth's albedo is zero. In this case, it is approximately 340 w/m^2.

    5) Equation 5 is false. You had already applied the shortwave emissivity by compensating for Earth's albedo. There is no need, and it is contradictory to apply a second and different emissivity value.

    6) Adjusting for errors to date, equation 6 is true. Stated correctly it is that the maximum out going radiation from the Earth at the top of the atmosphere, averaged over time, is S/4 or approx 340 w/m^2. Note, this is a TOA equality, not a surface equality, so as yet it tells us nothing about the Earth's surface temperature.

    7) Equation seven and comments are irrelevant because of the preceding errors. We are not screaming about forcings. We are wondering why you can't even get the simple things right.

    I will not comment further at this time because:

    a) You have not established the appropriate groundwork, and are instead working on a host of demonstrably false assumptions.

    b) Your tables which carry your argument have unclear symbols, and are derived by an unexplained method. In other words, they are simply bare assertions.
  30. L.J. Ryan - Fascinating post, you've obviously put a lot of work into it.

    It is, unfortunately, completely incorrect. Tom Curtis has shown that far more rigorously than I could.

    As les noted, you have an embedded link to a Professor Jin-Yi Yu's lecture on the greenhouse effect. It's a nice presentation - I suggest you actually read it, and it's conclusions, rather than mining it for equations.
  31. LJR, I have to thank you for the link to Pr Jin-Yi Yu's lecture, which indeed could not be recommended enough. It certainly would be a nice addition to the site, or to the scientific guide.
  32. Tom Curtis 929, KR, les

    Did you actually comprehend my post or just offer a knee jerk retort.

    re 1)Your Eq1 equations confirm mine. I'll prove it to you...what is the SW flux incident on the earths SURFACE?

    What is the SW absorptivity of the earth SURFACE?


    re 3) My (e3) is "mined" directly from Jin-Yi Yu lecture...KR, les and Philippe Chantreau seem to think highly of his presentation. So if I have got it wrong so does he. Your added commentary regarding TOA blackbody temperature is specious. Two reasons: first, the quote "Energy emitted by Earth = Energy absorbed by Earth" are not my words...look it up. Notice no mention of TOA. Second, the emission as defined by Stefan–Boltzmann law are from a blackbody's surface, not some arbitrary distance in order manufacture a energy gain... your definition is complete nonsense.

    re 4) Ok, let's eliminated earth's albedo and re-calculate blackbody maximum....T= (340 W/ m2 *1/σ).25 = 278K.
    Any interested readers should note, the absolute maximum blackbody temperature, with absolute maximum possible SW input is still, 10o colder, then actual temperature. Let me repeat that, GHG physics fabricates energy sufficient enough to confer temperature 10o above that which the sun provides!

    Tom Curtis do you deny this fact?

    re 5) You want to make SW absorption by the earth SURFACE 240 W/m2, fine. Shall I recalculate with this slightly higher flux, SURFACE absorption. As I suspect you know, there will be little change to the results.

    re 6) Ok, lets correct the record, please provide the actual SW absorption for the earths SURFACE.

    re 7) Since you obfuscate GHG physics mechanics invoking earths albedo, lets remove albedo and test this falsehood. No albedo, maximizes SW input, 340 W/m2 incident to the SURFACE. Given this most fantastic crutch, do you, Tom Curtis, KR, les, Philippe Chantreau, DSL, RW1, Stu, Phil, scaddenp, e, muoncounter, John Cook, DB, any of the other { -snip- } have confidence radiative forcing will work?

    Any interested readers should note, I will spot GHG physics the full solar input 340 W/m2 to the SURFACE, no albedo, and I contend it can not rectify the required energy to achieve earth's mean temperature of 288K.
    Response: [muoncounter] Please read the Comments Policy more carefully; accusations such as that removed from your comment are unnecessary and unacceptable. Do not overuse the emphasis; it is tantamount to shouting.
  33. Ryan.

    Thanks for including me in your rant.
    But you will note that I have not participated in your calculation thread of argumentation. The reason I have not us that as a physicist, I know thus is not how one actually thinks about this kind if problem.
    I've made my remark about the blog post. Respond to that or not; but I'm not being sucked into this particular bit of kinda-garden-proifiness.
  34. les 933

    I'm not sure what you are saying/asking...are you referencing your post 925 where you asked: "Where did the rest of the material come from (it is only polite to reference sources, after all)?"

    Assuming yes, these formula are straightforward applications of GHG physics. I know that sounds obligatory, yet this is a very true answer. It is only this "physics" charade which permits such misapplication of know laws.

    Do you wish to see the actual equations implicit in within data tables? If so, no problem. Please reply affirmatively and I will post later this evening.
  35. Ryan 934

    Assume No.

    HTH.
  36. L.J. Ryan - The most important issues I see with your postings are:

    - Absorptivity/albedo vary with wavelength; your initial problem statement did not incorporate that.

    - 240 W/m^2 enters the climate at TOA (boundary condition), including all SW albedo effects. The appropriate satisfaction of that boundary condition is that 240 W/m^2 leave the TOA. However, you make the (false) assumption that this power level is a boundary condition on the surface. That boundary condition misapplication is a huge error, and leads you in the wrong directions, and to ridiculous results.

    As stated before, given a known amount of outgoing radiation, the black body temperature is an absolute minimum on the temperature of an equivalently radiating graybody, due to the relationship of emissivity and temperature.

    The following is a rough zero dimensional calculation, but actually is quite close to measured effects:

    Without GHG's (ignoring affects on SW albedo), 240 W/m^2 of SW would enter the climate, and the only exit would be LW radiation from the surface through LW transparent atmosphere, average surface LW emissivity of 0.98, hence a temperature of -16.7C.

    With GHG's and an effective LW emissivity of 0.612 (that's measured, L.J., not made up, from the LW spectra to low orbit), the temperature is:

    T = [ P / ( ε * σ ) ]^0.25 K (Stefan-Boltzmann equation)

    or

    [ 240/(0.612 * 5.6704*10^-8 ]^0.25 - 273.15 = 15.2C

    Which, not surprisingly, matches our experience; ~15C surface temperature. Those are the measurements and the math, with an appropriately applied boundary condition of 240 in/240 out.

    If your hypothesis does not match the measurements, it's probably time for a new hypothesis!
  37. - 240 W/m^2 enters the climate at TOA (boundary condition), including all SW albedo effects. The appropriate satisfaction of that boundary condition is that 240 W/m^2 leave the TOA. However, you make the (false) assumption that this power level is a boundary condition on the surface. That boundary condition misapplication is a huge error, and leads you in the wrong directions, and to ridiculous results.

    Surely this assumption that LJR makes is equivalent to assuming that there >is no Greenhouse effect. In other words LJR's argument is circular: he's assuming what he's trying to prove.
  38. LJR is banging the table based on his notion that 'the blackbody temp at a given flux represents the maximum temperature.' Putting this unsubstantiated phrase into the google machine, there are no occurrences of it other than in these SkS comments. There don't even appear to be any denier blogs sourcing this sentence. However, there are multiple occurrences of statements which say, in essence, 'the wavelength of maximum flux represents (in this case via an inverse proportion) the blackbody temperature.'

    Are we simply witnessing a case of reversed word order? If so, this is truly much ado about nothing.
  39. I am sorry not to have responded before this, damorbel (872). Sadly, in the interval, we seem to have left basic thermodynamics (and G and T) behind.

    Heat is not what is measured by temperature. Internal energy is what is measured by temperature.

    Before internal energy can do anything (create work or raise temperature elsewhere) it must be transferred from a higher to a lower temperature, from a source to a sink. It will then become heat, which is the net transfer of energy, and the work it generates, or the warming it produces, will conform to the second law. That is why we must use the net transfer of energy to the atmosphere, e 873, from the surface, and not the back transfer (the negative term in Stefan Bolzmann) to calculate any possible GHG effects.

    From the source, surface, to the sink, atmosphere, most of the energy transfer is via conduction, convection and evaporation. From the source, atmosphere, to the final sink, space, all the transfer is radiative. However, the first and second laws are not concerned with transfer mechanisms. The final outgoing energy, atmosphere to space, must balance the incoming solar energy (first law). If we make the simplifying assumption of an effective emission level in the atmosphere, this fixes the outgoing radiative temperature (at 255K as it happens). Anything which increased that temperature would increase outgoing radiation, and the atmosphere would cool down. Anything which reduced that temperature would have the opposite effect.

    So why is the surface temperature about 33 degrees C higher? Because of the lapse rate. As damorbel has pointed out, gravity compresses the atmosphere and the work done in the compression warms it up.

    This has nothing to with radiative effects. It is a function of gravity and specific heat (page 45 of Elementary Climate Physics by Taylor), and is about 6K per kilometre of altitude.

    So, we if we can accept:

    1) that there is sufficient water vapour in the atmosphere to absorb all, or nearly all, of the net outgoing energy, and

    2) the effective radiative temperature of the atmosphere to space is 255K at an effective altitude of about 5 kilometres,

    there is no need to pursue greenhouse theory into the realms of quantum electrodynamics. G and T are right, and the simplistic “back-radiation” theories of AGW are wrong.

    However, there is one more explanation that is more difficult to refute: the “higher is colder” theory which suggests that increased absorption in the atmosphere will raise the effective radiative altitude to a higher and colder level via the lapse rate.

    This will reduce outgoing radiation, and the atmosphere and surface will warm to compensate. Effectively, the lapse rate will shift to the right.

    Does the evidence support this theory? The radio-sonde and satellite data should show this effect over the past 40 years when CO2 has been increasing relatively rapidly.

    I will down-load the data to try to find out. Please don’t expect anything conclusive.
    Response: Not gravity again :( See, among other rebuttals, Tamino's little gem, the detailed response (be sure to read the comments, too) by Chris Colose, and the lengthier series of posts on Science of Doom. If that's not enough, start searching the intertubes for "Steven Goddard Venus."
  40. LJR, your shouting is not impressive. As for this question " have confidence radiative forcing will work?"
    Answer is yes. Muoncounter above summarizes well some of your confusion. You should look again at Pr Jin-Yi Yu's lecture.
    This statement:"So if I have got it wrong so does he" does not follow from logic at all.
  41. Fred Staples, "Please don’t expect anything conclusive."

    Fear not.
  42. Fred

    Is the atmosphere being actively compressed by gravity? Is it more dense here at the surface than it was yesterday? The answer, of course, is no. Therefore no work is being done.

    Indeed, we can experience this in everyday life. Pump up your bicyle tyres quickly to 60psi. You've done work on them and they've gotten hotter due to compression.

    Leave the bike alone for a bit and come back later. The tyres are now at the ambient temperature. How did that happen? They're still at 60psi (the force analagous to gravity here is the tension of the inner tube) but the fact that they're at significantly higher pressure than their surroundings doesn't make them hotter*.

    Gravity, like the tension of an inner tube, is no substitute for thermodynamics.

    *PS I had one 'sceptic' argue back that when you cycle around the tires get hot. Apparently not familiar with friction!
  43. A tiny Excel exercise for L.J.Ryan and others:

    Create a new spreadsheet. Enter the values below:

    A1: 1.0
    A2: A1-0.01


    Copy A2, paste from A3 to A50.

    B1: =(240/(A1 * 5.6704*10^-8 ))^0.25- 273.15

    Copy B1, paste from B2 to B50.

    Explanation: With 240 W/m^2 (fixed) radiated as power, and varying emissivities, what gray body temperature (in degrees C) is required to radiate that 240 W/m^2, using the Stefan-Boltzmann relationship? An emissivity of 1.0 represents a perfect black-body, 0.98 represents the Earth's surface with no GHE, and an emissivity of 0.61 is quite interesting.
  44. Fred Staples @939, first let me surprise everyone and congratulate you on being almost entirely correct.

    I notice that you indicate that:

    1) The effective temperature of the outgoing radiation is 255 K;

    2) That the effective altitude of radiation to space is about 5 km.

    From these it follows that

    3) The (average) temperature of the atmosphere at about 5 km is about 255 K.

    You also note that:

    4) The lapse rate (in the troposphere) is entirely determined by the local gravitational accelerationg, g, and the specific heat of the atmosphere; and that

    5) The lapse rate is approximately 6 degrees per km (6.5 is more accurate).

    Therefore:

    6) The average surface temperature is (5 * 6.5) + 255 = 288 K.

    That is the greenhouse effect in a nutshell.

    To see this, consider an example in which the atmosphere absorbs (and radiates) no IR radiation. In that case the effective altitude of radiation to space would be 0 km, and hence the surface temperature would be (0 * 6.5) + 255 = 255 K.

    It is not the lapse rate, therefore, which is responsible for the elevated surface temperatures, for it is (near) constant in both scenarios. Rather it is the presence of IR absorbing and radiating gases in the atmosphere.

    We can consider the case where the concentration of GHG in the atmosphere decreases. This will lower the effective altitude of radiation to space, and consequently lower the surface temperature (but not the effective temperature of radiation to space). Conversely, if we increase the concentration of GHG, that will raise the effective altitude of radiation to space, the surface temperature will rise, all else being equal.

    Hence, from principles you have just espoused, the green house effect follows by simple logic.

    Back radiation does come into the picture. Without the return of some energy from the atmosphere to the surface, thus reducing the net transfer of energy from surface to atmosphere, an increase of the Earth's surface temperature would be impossible. But that transfer (in keeping with the 2nd law of thermodynamics) can never exceed the energy transfer from the surface to the atmosphere (except locally and temporarily). If it rises to a level which would warm the surface by more than the amount indicated by the lapse rate and the effective altitude of radiation to space, the result is simply an increase of of convection and evapo/transpiration, thus nullifying the effect.
  45. KR

    You said :

    However, you make the (false) assumption that this power level is a boundary condition on the surface.

    and

    T = [ P / ( ε * σ ) ]^0.25 K

    [ 240/(0.612 * 5.6704*10^-8 ]^0.25 - 273.15 = 15.2C


    from where does the P (240W/m^2) LW originate....the surface? This very flux is the sole source of LW...without the 240 W/m^2 of SW input the there will be no LW. So save changes to albedo or solar radiation, this IS a boundary condition for radiative energy. Said otherwise, because P input to the surface must equal P output at equilibrium...you can not get more LW flux out then SW flux in. Re-radiating Pout can not increase Pin. Delta T must be a function a non-radiative input.

    To claim effective LW emissivity of 0.612 is measured, is specious. How do measure “effective emissivity”?

    Emissivity with the qualifier effective, only aids in obfuscation. “Don't bother looking a the real mechanics of GHG physics, it can all be explained by effective emissivity"
  46. Wow.

    Comment 945 and counting!

    Not to intrude too much on everyones fun, but a few questions for Fred:

    "From the source, surface, to the sink, atmosphere, most of the energy transfer is via conduction, convection and evaporation." Source please! Radiation transfer from an object at temerature x is determined only by its temperature, not other parallel heat transfer mechanisms.

    "and the simplistic “back-radiation” theories of AGW are wrong. ". Oops. Small problem here Fred. Back Radiation has been observed for decades and has been increasing in the radiation frequencies of the GH gases. Damn pesky thing observations aren't they. A beautiful theory derailed by a mere observation.

    "That is why we must use the net transfer of energy to the atmosphere, e 873, from the surface, and not the back transfer (the negative term in Stefan Bolzmann) to calculate any possible GHG effects" Wrong Fred. Yes the net of the two flows will be what determines (is) the actual energy flux. However this is made up of several different phenomena that occur because of different mechanisms. Radiation flux from the surface is driven solely by surface temperature. Absorption by the atmosphere depends of the absorption properties the GH gases alone. Back Radiation depends on the temperature of the lower atmosphere at an altitude where the path back to the surface is not 'optically thick'. Several phenomena coming together to create the GH Effect, rather than the GH Effect being caused by the net of the heat flux.
  47. Fred

    "If we make the simplifying assumption of an effective emission level in the atmosphere, this fixes the outgoing radiative temperature (at 255K as it happens). Anything which increased that temperature would increase outgoing radiation, and the atmosphere would cool down. Anything which reduced that temperature would have the opposite effect."

    You are missing several factors here Fred. As GH Gas levels increase, then the altitude at which emitted photons have a clear path out to Space increases. With altitude Temperature decreases so the emission temperature decreases so less radiated because the altitude of radiation increases. To use a simple analogy, its like a Cloud Bank. At the bottom of the 'cloud' a photon can 'see' the ground so can reach it. At the top of the 'cloud' a photon can 'see' Space and can reach it. One of the two effects of increased GHGas levels is to increase the upper altitude of the 'cloud'. So when a photon can reach Space it will be emitted from air that is colder.

    There is an excellent article on the radiative physics of the GH Effect by Prof Ray PierreHumbert in Physics Today here http://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf

    Have a look at his figure 3a, modelled and measured spctrum of radiation leaving the Earth. At around wavenumber 670, right in the highest absorption region for CO2, the amount of energy reaching space spikes up.

    Right in the middle of the highest absorption part of the CO2 band the emission temperature spikes back up! Why? Because the altitude at which the path to Space at that wavenumber is clear because of the high levels of absorption is SO high that it is above the altitude where lapse rate causes a temp drop with altitude and is high enough that atmospheric temps are climbing in the upper stratosphere. The models of radiative transfer are so accurate that they capture this altitude dependent temperature behaviour superbly.

    That graph is a thing of transcendent beauty.
  48. 947 - Glenn.
    Nice article link.

    As an aside, and of only the slightest relevance to this blog post, I note his statement "The planetary warming resulting from the greenhouse effect is consistent with the second law of thermodynamics because a planet is not a close system"... as noted above and to which 934 Ryan should have assumed I was alluding.

    Now, I suggest that as, as noted by the Economist, this is the 150th anniversary of some relevant and great physics (the unification of Electric and Magnetic fields) - we pay respect to this great moment by giving up on this assault on the nodal discipline of physics by attempting to prove/disprove another of it's great achievements (the laws of thermodynamics) through bean-counting and pseudo-modelling.
  49. les 948

    Let me see if understand your revised position on the GHG physics and the 2nd law.

    I paraphrase you (maybe): Because the earth's radiative energy system is not a "closed loop system" and the 2nd law applies only to "closed loop systems", GHG physics therefore does not follow nor is required to follow the 2nd law.

    Do I have this right?
  50. L.J. Ryan, you 'quote' the phrase "closed loop system" twice in your comment above, yet I cannot find it anywhere in the preceding comments.

    I assume this is meant to be a reference to comments about over-simplified versions of the 2nd law of thermodynamics (e.g. 'energy can only flow from hot to cold') only being applicable to 'closed systems'.

    The 2nd law of thermodynamics is universal (so far as we know)... however, poorly stated versions of it require qualification.

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