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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 801 to 850 out of 1393:

  1. To moderator: OK, the trolling has indeed gone on long enough. Sorry for adding to it.
  2. LJR #793: So now you are disavowing your own prior arguments? Or wasn't it you who wrote;

    #715: "Low temperature, (lower energy) atmosphere adding radiative heat to the warmer surface is a violation of the 2nd law."

    #758: "See this process (unlike the magic box) abides the 2nd law, Hot to Cold."

    #765: "So regardless of re-radiation and reflection, the atmosphere can NOT warm the warmer Earth...Period."

    You have said all these things. Right there in the thread history. Yet now you deny it to avoid answering my challenge;

    "LJ Ryan, I note that you continue to make nonsense claims about greenhouse gases being unable to redirect energy from a colder area to a warmer one while refusing to answer how a parabolic mirror does so."

    How does the observed reality that EM radiation travels from the surface of a parabolic mirror to the hotter object at the mirror's focal point not disprove your various statements quoted above?
  3. Re 786 Tom Curtis,point by point

    1/ Yes

    2/Not relevant.
    - The spectrum of samples plays no part in heat balsnce (Gustav Kirchhoff 1862 'Ueber das Verhälteniß zwischen dem Emissionensvermögen der Körper für Wärme und Licht'.

    3/"If a photon is absorbed by a surface, the surface gains the energy that was contained in the photon (by conservation of energy)".
    - This would only be true if the surface were at 0K i.e. it is not itself radiating.

    4/"If energy is absorbed by a surface, all else being equal, the temperature of the surface will rise, and the surface is warmed.
    - See 3 above.

    5/"However, the surface of the Earth is typically warmer than the atmosphere, so it itself is radiating energy to the atmosphere, and is radiating more energy than it receives from the atmosphere."
    - Very far from clear. Two surfaces close to each other can both be radiating very strongly but without energy transfer. Energy is only transferred if there is some difference in temperature.

    6/"Therefore, absent any other energy sources, the net effect of the interchange of photons between atmosphere and surface is that the surface cools and the atmosphere warms."
    - Yes. (But see 5/)

    7/" However, if the atmosphere was not there, or did not radiate IR radiation: (a) the total energy emitted by the surface would still be the same, because that energy is solely a function of its temperature and emissivity; but (b) the surface would receive less energy because it would not be absorbing photons emitted by the atmosphere."
    - Regards (a)'Energy' is not emitted (see the answer at 5/)
    - Regards (b)The surface does not get energy from the atmosphere.(see 5/.)

    8/"Therefore, over a given period of time, and ignoring all other energy sources, the Earth will cool quicker without an atmosphere containing GHG than it will with one."
    - This is very different. Now you are talking about 'The Rate of Cooling (Heating). The question of cooling (heating) rate is entirely a function of the albedo (if you take the 'rate' as a % of the temperature). The rate of temperature change (by radiation only) of any body is strongly dependent on its reflectivity, whether it has an atmosphere or not. If a body has a highly reflective surface it means there is very little material to either emit or absorb radiation, so heat transfer is minimal.

    9/ -----> end.
    - No comment.
  4. L.J. Ryan - "Do you agree?"

    No.

    The theoretic black body work (including Kirchoff) is based upon a "white light" excitation, equal across all frequencies, absorption and emission by the black body based upon the absorption/emission spectra.

    The climate, on the other hand, is driven by a band-limited solar input which does not match the thermal emissive spectra, is not greatly affected by greenhouse gases, and hence represents a fixed input, not a match to the thermal spectra at all.

    And, as I stated earlier, given a fixed input power, and a need to radiate that (or change internal energy and hence temperature), emissivity and temperature have an inverse relationship. As effective emissivity of the planet goes down, temperature goes up.

    Asserting that the Earth follows "white light" illumination with interdependent absorption/emission is a complete mistake. It's a fixed input power outside the GHG affected thermal spectra, which is sufficient to radiate the incoming 240 W/m^2. And the black body temperature required to radiate that power is a lower limit on the temperature of a gray body of lesser emissivity.
  5. damorbel

    Please clarify the point you made here:

    "If a photon is absorbed by a surface, the surface gains the energy that was contained in the photon (by conservation of energy)".
    - This would only be true if the surface were at 0K i.e. it is not itself radiating."

    The question does not ask whether the energy absorption is balanced out by emissions, just whether an absorbed photon leads to an energy gain. Are you saying that it does not for an object above 0K? If not, what happens to the energy of the photon? Where did it go?

    As for how the energy is emitted, keep in mind the Stefan-Boltzman Law. The amount of energy radiated is dependent on the temperature of the material, not whether it has absorbed an extra photon recently.

    That means if the surface is constantly emitting and absorbing photons, and we increase the number of photons being absorbed, the only way for the surface to emit that extra energy is to get warmer. Do you agree with this? If not, what happens to the energy? How can the extra energy be emitted without the object getting warmer (again keeping in mind Stefan-Boltzmann)? If it isn't emitted, and it doesn't raise the temperature, what happens to this energy?
  6. e, damorbel is consistently and repeatedly avoiding the question as to whether he will even philosophically accept the idea of an experiment as the way to settle a question. (asked here. This to me implies someone only interested in arguing with no intent to resolve anything, perhaps even a paid troll; or someone who prefers a faith-based position. I suggest there is no point discussing with him at all.
  7. Philippe Chantreau 797

    "I am talking about a spherical blackbody receiving solar radiation at a constant rate in w/sq m, in the solar spectrum. If, by any means, that blackbody's ability to radiate energy out is impaired, what will happen to its temperature?"

    As long as the emissivity remains constant, i.e. remains a blackbody, the temperature remains unchanged.

    Got it Philippe. The spherical blackbody temperature will remain the same.
  8. LJR: "As long as the emissivity remains constant, i.e. remains a blackbody, the temperature remains unchanged."

    So let's stop talking about theoreticals; what happens when the emissivity of a real planet decreases? By your statement, if solar input remains the same and emissivity decreases, temperature must increase. Or do you now wish to change that as well?
  9. muoncounter 799

    "Now that's a radical change of heart, as you specifically described earth's blackbody temperature here. And you've totally ignored the complete defenestration of your argument by KR (#773-775) and CBD (#785)."

    If you did not understand the context, sorry. Earth blackbody temperature, is a generally (I think) accepted naming convention for earths blackbody temperature equivalent.
  10. muoncounter 808

    "So let's stop talking about theoreticals; what happens when the emissivity of a real planet decreases?"

    If emissivity decreases, the "real planet" will absorb less solar flux.

    "By your statement, if solar input remains the same and emissivity decreases, temperature must increase."

    Input would not remain the same (see above). Lower input results in a lower temp. NOT an increase.
  11. L.J. Ryan - "If emissivity decreases, the "real planet" will absorb less solar flux. "

    Incorrect - please read my post here, and recognize that input power is not affected by the greenhouse gases and their resulting IR emissivity.

    You have so far ignored my post.

    The real world climate receives band-limited solar energy unaffected by IR emissivity, hence a fixed input power. Output power (which must, at dynamic equilibrium, match input power) is determined by IR emissivity and the dependent variable, temperature.

    Please respond or acknowledge.
  12. KR 811

    Without addressing the specifics of your post 804, which I will get later this evening or tomorrow, muoncounter 808 made no reference "IR emissivity". Therefore, I stand by 810.
    Response: [muoncounter] Note the use of the words 'real planet;' this is specifically about solar input and how a planet with a greenhouse atmosphere responds. Don't bother replying if you are going to haggle over this nonsense; especially since you claim the right to define 'blackbody' however it suits you.
  13. DSL 800

    "Do I have this right?"

    I'm not sure I understand your entire concept.

    "But, of course, all atmospheric layers do radiate, and some of this radiation is absorbed by warmer surfaces and warmer atmospheric layers."---To what end. The warmer layers/surface does not get warmer do to this absorption.

    "Here's the key, though: the current temperature of any atmospheric layer or material within the system is that specific temperature because the atmosphere is already adding its radiated energy. The system is dynamic. if we take away the atmosphere, the surface eventually (quickly) reaches a lower equilibrium temperature. "---Again, I'm not sure I understand...but the lower temperature atmosphere can not increase the temp. of warmer surface.

    I don't agree with core of your last paragraph. Lower temp IR can not increase the temp. of a warmer surface.
  14. CBDunkerson 802

    "LJ Ryan, I note that you continue to make nonsense claims about greenhouse gases being unable to redirect energy from a colder area to a warmer one while refusing to answer how a parabolic mirror does so."

    NONE of the listed post in 802 do I say "unable to redirect energy from a colder area to a warmer". What I do say: lower temp IR can not increase the temp of a warmer surface.

    You said:
    "How does the observed reality that EM radiation travels from the surface of a parabolic mirror to the hotter object at the mirror's focal point not disprove your various statements quoted above?"

    EM Radiation is reflected, not absorbed and re-radiated, therefore the dish temp is irrelevant.
  15. Re 805 e you write:-

    "The question does not ask whether the energy absorption is balanced out by emissions, just whether an absorbed photon leads to an energy gain."

    It takes time, I agree not very much, to absorb a photon, in this time only a body at 0K will not emit a photon.

    Perhaps it seems trivial to consider individual photons, infinitesimal time periods etc., these are normally handled at macro level by statistics but the bottom line is what happens at the individual photon, particle etc.

    You wrote further:-
    "As for how the energy is emitted, keep in mind the Stefan-Boltzman Law. The amount of energy radiated is dependent on the temperature of the material, not whether it has absorbed an extra photon recently"

    The S-B law is about power (W/m^2) not energy; the energy absorbed can only be found by integrating the power WRT time, during which a body above 0K is also emitting (power) according to S-B law. The equilibrium temperature is where the emitted power equals the absorbed power.

    It is worth noting that the equilibrium temperature requires only a power balance. A black body is the most efficient emitter but if the body is not black e.g. coloured, a gas or with a refractive index >1 etc., then its emissivity is less than 1. and will not emit so much power at a given temperature; conversley if the emissivity <1 a body will be hotter than a black body, this is why it is a great mistake to assume the Earth 'radiates like a black body'.
  16. LJR #814: So, your latest nonsensical dodge is that EM radiation behaves differently if it has been absorbed and re-emitted than if it 'travels directly' and/or is reflected? Ignoring the ridiculousness of that claim for the moment;

    How does the EMR emitted by a remote control travel from the couch to the warmer receiver in the heat generating electronic equipment it controls?

    How does the EMR emitted by a microwave oven travel from the cold walls of the oven to the warmer food being cooked?

    How does the EMR carrying radio and television broadcasts travel from the transmitter to warmer locations around it - rather than radio and television being routinely interrupted by minor local temperature variations?

    Heck, how does the EMR of 'sunlight' travel from the cold of space to the warmer upper atmosphere to the warmer still lower atmosphere? By your claims we should all live in perpetual darkness because sunlight cannot approach the warmer surface of the Earth.

    The 2nd law of thermodynamics does not say that energy can not flow from cold to hot (regardless of whether it has been absorbed and re-emitted somewhere along the way). That is a ridiculous lie which violates thousands of observations from everyday life. What the 2nd law of thermodynamics actually says (in this context) is that the net flow of energy between objects in a closed system will always be from cold to hot... that is, energy flows from the cold objects to the hot ones and vice versa, but since the hot objects are giving off more energy than they receive the net flow is from hot to cold.

    BTW, your version of events violates the 1st law of thermodynamics... energy can neither be created nor destroyed. You argue that 'colder EM radiation' (setting aside that radiation has no temperature) cannot raise the temperature of a warmer surface... so what happens to it? You've got energy hitting a surface and not making if warmer. It simply ceases to exist. Violating the first law of thermodynamics.
  17. Re #784 Re:-
    "Response: [muoncounter] We've heard the one about textbooks before. No need recycling your old ideas when they didn't work first time around"

    I have never raised the matter of text books, it would not occur to me to do so. It has been raised multiple times by other contributors and they received no warning about it I have responded about the reliability of text books. Are there matters to which I may not respond?

    I do appreciate the work done by mods.
    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.
  18. Following damorbel's logic @815, I can never pay of my debt to the bank. After all, it takes time to complete a transaction, and during the time, the Bank will make many transactions in which they credit other accounts. Therefore, if I make a payment on the debt, they will never have received it, for in the time it takes for them to receive it, a larger amount will have been paid out by them. Therefore, they should be under no obligation to credit my account with the amount paid.

    If that sounds like casuistry, it is only because the argument mirrors damorbel's.

    Transparently for those not inclined to casuistry, if in damorbel's book keeping, the energy of the incoming photon is immediately credited to the outgoing radiation, then the amount of energy lost by the absorbing body is reduced by the amount of energy gained from incoming photons. Because less energy is lost, the body will therefore have more energy (and hence be warmer) than an equivalent body that started at the same temperature and emissivity but did not have the incoming radiation.

    In fact, damorbel is really trying to run two sets of books here, and hoping that we do not notice. In one set of books he credits the incoming energy to the simultaneous outgoing radiation so that he does not have to account for the absorbed energy in discussing the temperature change of the absorbing body. In the second set of books he debits all outgoing energy from the emitting body. Only by keeping both books separate can he pretend that a cool body interacting with a warm body can change the equilibrium temperature, ie, the temperature at which incoming energy matches outgoing energy, for that warm body.
  19. Re #785 CBD you wrote:-
    "LJ Ryan, I note that you continue to make nonsense claims about greenhouse gases being unable to redirect energy from a colder area to a warmer one while refusing to answer how a parabolic mirror does so."

    The absorption of radiation and re-emitting is not what is meant by 'redirect(ion)'.

    Redirection is what a mirror does, it does not change the wavelength (colour) of the (redirected) radiation; the energy is not absorbed by a mirror (or any other reflective process e.g. scattering, as with fog.) thus the temperature of the mirror is not changed.

    The converse is also true, the redirection of light by a mirror is, at first order, independent of the temperature of the mirror - a hot mirror works much the same as a cold one.
  20. Re #804 CBD you wrote:-
    "The theoretic black body work (including Kirchoff) is based upon a "white light" excitation, equal across all frequencies, absorption and emission by the black body based upon the absorption/emission spectra."

    Arguing Kirchhoff's contribution was confined to 'black bodies' is not correct.

    Kirchhoff was the first to consider 'arbitrary' bodies, ones that reflect, refract, are (partially) transparent etc., e.g. mirrors, coloured bodies, gases etc., thus all radiation, including that with a narrow spectral range.

    Subsequent developments in atomic and quantum theory have not invalidated his work, which would have been rather unlikely because his work was the inspiration for it!
  21. Re #816 CBDunkerson you wrote:-

    1/"How does the EMR emitted by a remote control travel from the couch to the warmer receiver in the heat generating electronic equipment it controls?"
    - Remote controls emit high brightness narrow band IR that is focussed; the detector at the receiver can distinguish the RC bright spot against yhe background radiation coming fron your cup of tea and othe objects near you. Also the detectot has a filter to pass only the narrow band radiation emitted by the RC so that the effect of broadband (thermal) IR from your tea cup, the room etc., is reduced.

    2/"How does the EMR emitted by a microwave oven travel from the cold walls of the oven to the warmer food being cooked?"
    - The microwave energy is repeatedly reflected by the walls until, sooner or later, it encounters the food to be cooked, where it is (mostly) absorbed. If there is no food in the oven the microwave energy builds up (there are warnings not to run the thing at full power when it's empty) and you will damage the device, possibly bustin the magnetron.

    3/ "How does the EMR carrying radio and television broadcasts travel from the transmitter to warmer locations around it - rather than radio and television being routinely interrupted by minor local temperature variations?"
    - Same as 1/, except the transmitter is not focussed (much).

    4/ "Heck, how does the EMR of 'sunlight' travel from the cold of space to the warmer upper atmosphere to the warmer still lower atmosphere? By your claims we should all live in perpetual darkness because sunlight cannot approach the warmer surface of the Earth."
    - The Sun is at about 5780K and that has the spectrum we see (well, our eyes do not see the infrared (IR) and ultraviolet (UV) but it is stil there). The Sun occupies only a small part of the sky so we do not get a the full 5780K here, only 279K. But sunlight still behaves with some of the properties of a 5780K source, it ionises O2 in the stratosphere to make ozone and it tans your skin, burning it if you are not careful.

    For the rest - no comment.
  22. Re #818 Tom Curtis you wrote:-

    "Following damorbel's logic @815"

    - no comment.
  23. LJR, by definition, if the ability to radiate is decreased, the emissivity does change and the blackbody becomes grey. How else can it possibly manifest? It has been pointed to you earlier that this is exactly what the atmosphere does.

    And what was this thing again with the cooker cooler below ambient but in fact staying at the same temperature as the air?

    This is a waste of time.
  24. damorbel #819: We already covered this nonsense... you claim EM radiation behaves differently when reflected than it does when absorbed and re-emitted. Again, irrelevant even if it weren't completely false. Real world examples of each;

    Photographer taking a picture of a 'penguins on ice' exhibit in a zoo. The ice is colder than the mirror in the camera and the film yet the image is carried by the EM radiation from cold to hot. The ice does not 'disappear' from the picture. EMR flowing from a cooler object to a warmer one while being reflected along the way.

    Two identical pots of water on an electric stove. Heat them both to the maximum temperature of the stove for an extended period and then turn one off while putting the other at half heat. If energy were unable to travel from cold to hot then the pot which was warmed up to the maximum heat could not absorb and re-emit heat from the now half as hot burner and would therefor cool down at exactly the same rate as the pot whose burner was turned off (until it reached half heat and stopped cooling). Yet this does not happen. The pot with half heat cools down slower / remains boiling much longer than the pot with no heat... because even though it is (initially) hotter than the burner it is still receiving additional heat... heat flowing from a cooler object to a warmer one by absorption and re-emission.

    One example involves reflection, the other absorption and re-emission. Yet both show energy flowing to a cooler area from a warmer one. Ergo, the reflection vs absorption and re-emission distinction you keep making is meaningless. It is observed reality that energy can and does flow from cold to hot in either case.

    Your arguments continue to be complete nonsense... and you continue to avoid any attempt to address the countless real world examples proving that.

    damorbel #820: You attribute a quotation from message #804 to me. It was actually written by KR.
  25. damorbel #821:

    "Remote controls emit high brightness narrow band IR that is focussed; the detector at the receiver can distinguish the RC bright spot against yhe background radiation coming fron your cup of tea and othe objects near you."

    The question isn't how the receiver distinguishes the signal from the remote from ambient energy (which you get completely wrong BTW). It is how it receives the signal at all if, as you claim, energy cannot flow from the cold remote to the warmer receiver.

    "The microwave energy is repeatedly reflected by the walls until, sooner or later, it encounters the food to be cooked, where it is (mostly) absorbed."

    But how? Are you amending the 2nd law of thermo-ridiculousness to; Energy cannot flow from cold to hot unless it is repeatedly reflected first? Still doesn't explain the remote control.

    "Same as 1/, except the transmitter is not focussed (much)."

    The IR remote isn't focused much either... which is why you can bounce the signal off a wall behind you and still have it work on the television (or whatever) in front of you. However, that still doesn't explain how either travels from cold to hot.

    "But sunlight still behaves with some of the properties of a 5780K source, it ionises O2 in the stratosphere to make ozone and it tans your skin, burning it if you are not careful."

    So now you are claiming that only the temperature of the 'origin point' of the EMR matters... it can go through cooler and warmer areas so long as none is warmer than the origin point. Yet the remote, microwave, and radio examples all show EMR traveling to areas warmer than their origin points.
  26. Re #824

    "damorbel #820: You attribute a quotation from message #804 to me. It was actually written by KR."

    Very sorry!
  27. 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!

  28. 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.
  29. 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.

  30. 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.
  31. 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.
  32. 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.
  33. 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.
  34. 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.

  35. #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!'
  36. 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?
  37. 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.
  38. 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".
  39. 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.
  40. 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.
  41. 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.
  42. 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.
  43. 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...
  44. 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.
  45. 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.
  46. 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)

  47. 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).
  48. 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.
  49. 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.
  50. 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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