Guest post by Ben Herman and Roger Pielke Sr.
We have received a further question on our post:“The Greenhouse Effect” by Ben Herman and Roger Pielke Sr.
The question is summarized by the following text
Anyway my question refers to the common example of taking away the atmosphere and observing a cold surface. But as I understand it, the mean daytime surface temperature on the moon is over 100 C, with no greenhouse effect. The mean nighttime temp drops to -150 C. http://www.solarviews.com/eng/moon.htm
This is important to note, because encouraging a popular picture in which the presence of the atmosphere only warms the surface takes all the convection and fluid dynamics out of the discussion, and that’s where all the important complexities are.
Isn’t it more the case that the atmosphere both warms and cools the surface, depending on circumstances? The IR absorption of H2O and other GHG’s warms the surface relative to what it would otherwise be, but as the lunar case shows, convection and turbulent mixing cools the surface relative to what would happen without an atmosphere. Take away the atmosphere and you take away both warming and cooling mechanisms.
We have reproduced the substance of our follow up answer below.
Predicting the surface temperature indeed involves the interaction of the atmospheric and ocean turbulent sensible and latent fluxes, long- and short- wave radiative fluxes and interfacial fluxes between the surface and the atmosphere. I have been urging for years to move away from the surface temperature to characterize global warming and cooling (and replace with ocean heat content changes in Joules) because the surface temperature is such a limited sample of the heat content changes of the climate system as well as involving these complicated feedbacks.
On the Moon, there is, of course, no atmosphere, so its surface temperature results from the difference between the surface long wave radiative emissions, the amount of solar radiation absorbed and reflected, and the conduction of heat into and out of the surface. The effect of the atmosphere on Earth is to mute the diurnal (and seasonal) temperature range as a result of the turbulent fluxes, and other effects (such as clouds and precipitation). These atmospheric effects, for example, result in lower daytime and higher nighttime temperatures from what they otherwise would be. I presume this is the cooling and warming effects that you refer to. However, even with these effects, the surface is clearly warmer than it would be without the CO2 and water vapor IR absorption bands.
But the reasons are that the atmosphere scatters back to space some sunlight, and takes up some of the surface heating through conduction, and mixes it it by convection and turbulence. Also, the relatively rapid rotation of the earth on its axis does not permit the daytime side to reach equilibrium before it starts nighttime cooling. As a result, daytime temperatures on earth are cooler than they would be with no atmosphere, and warmer at night than with no atmosphere.
Of course, the Moon, with no atmosphere, still has to have basically the same effective radiating temperature as does the Earth. This should be
[sigma *Tmd**4 + sigma* Tmn**4]/2 = sigma*Te**4 where Tmd is the daytime temperature of of the Moon, Tmn is the night time temperature of the Moon, and Te is the effective radiating temperature of the Earth.
The fact that the daytime time temperature is warmer than the Earth’s temp is simply a result of the fact that the Moon is not in an equilibrium state – it warms up during the daytime and cools down at night, just as does the Earth. However the warming during day and cooling at night must balance each other or the Moon ( and the Earth) would be steadily heating up or cooling down over time. The daytime warming occurs because the outgoing IR cannot balance the absorbed solar during the day. The nighttime cooling occurs because the outgoing IR is greater than the non-existing solar at night. The existence of a partially absorbing atmosphere does, as you stated, keep days cooler and nights warmer.
Also, the length of a day on the Moon is 29.5 earth days, almost a full Earth month. Therefore the daylight side of the Moon heats due to solar radiation, for half a month. Then when it’s night, it cools for another half month. Thus the daytime and nighttime temperatures are much more extreme. There is no greenhouse effect on the Moon, of course, and if the Moon’s day was the same 24 hours as an Earth day, its day and night temperatures would not vary as much but its radiative equilibrium temperature would be the same.
Update #2 John Nielsen-Gamon has alerted us to more information on the Moon’s radiative temperature. John e-mailed
I read your blog post on Greenhouse Part 2. I also recently came across the Science of Doom web site; it seems to be of very high quality. You might want to link to http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/ [on] your post to direct the reader to further details on the radiative temperature of the Moon.
Update – corrected text (underlined) h/t to Gerald E. Quindry
David Springer said “If you look down at earth from space at night you won’t see any CO2 spectral (emissive) lines. You will see CO2 absorption lines. ”
The behavior is symmetric. If you look down from space you see CO2 absorption lines against a warm background. If you look up from the surface, you see CO2 emission lines against a cold background. It all depends on the CO2 temperature relative to the background.
Water vapor concentration.
With H2O you have to be careful about numbers as it varies so much. Near the surface it can reach 3-4% in warm air. The atmospheric average is more like 0.4% (4000 ppm), because so much of the atmosphere is quite cold.
I’m a little astounded here. I’m at the end of this thread, and it appears that both sides are on board that the same side of the moon faces the earth. This is beyond ludicrous!
The same side of the moon always faces the Sun! A la “DARK Side of the Moon”. Anybody ever heard of a penumbra or Earthshine? How did we map the dark side before the space age? I think there’s a lot of CO2-worry goin’ on in both camps! WUWT?! Let’s get this correct.
Phil. says:
July 31, 2010 at 11:56 am
Dave Springer says:
“This is not how it works in a cold dense gas. When CO2 absorbs energy it its absorption bands it causes nearly instantaneous collision with another molecule, the vast majority of which are nitrogen, oxygen, and water vapor. The kinetic energy in the collisions is radiated in continous blackbody spectrum with the peak frequency determined by the sensible temperature of the gases.”
Phil: The collision part is true but nitrogen, oxygen, and water vapor do not radiate as blackbody spectra.
Just finished my Sunday chores and came to this thread.
Phil, you have to take this back. I am sure it is reflex reaction to saying no whatever Dave is sayin.
Every type of molecular ensemble radiates according to a modified for that type ” black” body temperature . It is elementary statistical mechanics from which, temperature comes out as a statistical adjunct of the kinetic energy of each molecule, and from which the black body formula was guessed at.
http://en.wikipedia.org/wiki/Statistical_mechanics
Your statement would mean that O2 and N2 are at 0 Kelvin, which is the classical limit of temperature when there is no kinetic energy. If there is kinetic energy, there is temperature and there is gray body radiation.
tallbloke says:
July 29, 2010 at 3:25 am
you are misspelling your contributor’s name:It seems he writes it Eggert, NOT Eggart.
It is an interesting analysis. He should submit it to an engineering journal and get it peer reviewed by engineers.
Ralph Dwyer says:
You might want to read this: http://www.straightdope.com/columns/read/429/why-does-the-same-side-of-the-moon-always-face-the-earth (and http://answers.yahoo.com/question/index?qid=20081026193951AARQoVE for good measure).
BZZZZZZZZZZZZZT! Wrong.
Greenhouse gases don’t regulate how much heat escapes. They slow down the rate of escape. This results in an increase in temperature on the warmer side of the insulating layer of gas. The increased temperture differential raises the rate of thermal transfer.
All thermal energy at the earth’s surface escapes out into space. Every last iota of it.
Write that down.
@JimD
No, you don’t see emission lines in the clear night sky looking upward. You see absorption lines. The following article discusses it in detail and if you ever took Astronomy 101 in college (I did) it is taught there as well.
http://www.gemini.edu/sciops/telescopes-and-sites/observing-condition-constraints/ir-background-spectra?q=node/10789
There are lovely graphs of infrared spectrum covering 15 micrometers and longer. You will note there is no energy near 15um. A portion of the spectrum is absent. That is an absorption line not an emission line.
What part of that don’t you understand?
Ben Herman and Dr. Pielke, Sr., address the following question:
“Anyway my question refers to the common example of taking away the atmosphere and observing a cold surface. But as I understand it, the mean daytime surface temperature on the moon is over 100 C, with no greenhouse effect. The mean nighttime temp drops to -150 C. http://www.solarviews.com/eng/moon.htm
This is important to note, because encouraging a popular picture in which the presence of the atmosphere only warms the surface takes all the convection and fluid dynamics out of the discussion, and that’s where all the important complexities are.”
What they [Herman and Pielke] do in addressing the question is reassert the primacy of the assumption that Earth, the Moon, and all bodies should be treated as black bodies. This assertion is plain as day in the following from Herman and Pielke.
“However the warming during day and cooling at night must balance each other or the Moon ( and the Earth) would be steadily heating up or cooling down over time. The daytime warming occurs because the outgoing IR cannot balance the absorbed solar during the day. The nighttime cooling occurs because the outgoing IR is greater than the non-existing solar at night.”
Then Herman and Pielke refer us to a website that does the same thing in Spades. The reference is in the following:
“[One of their correspondents says:] You might want to link to http://scienceofdoom.com/2010/06/03/lunar-madness-and-physics-basics/ [on] your post to direct the reader to further details on the radiative temperature of the Moon.”
The conclusion of the article that they reference is the following:
“Conclusion
So the reason that the moon – with a surface with a real heat capacity – appears to have a warmer climate “than predicted” is just a mathematical error. A trap for the unwary.
The right way to calculate a planet’s average radiation is to calculate it for each and every location and average the results. The wrong way is to calculate the average temperature and then convert that to a radiation. In the case of the earth’s surface, it’s not such a noticeable problem.
In the case of the moon, because of the wide variation in temperature, the incorrect method produces a large error.
So there’s no “lunar explanation” for the inappropriately-named “greenhouse” effect.”
In case it is not obvious from this “Conclusion,” what the author has done is construct an argument that is a perfect circle. Before revealing the circle, let me set the stage. He is responding to “A Greenhouse Effect on the Moon?” by Martin Hertzberg, Hans Schreuder, and Alan Siddons. It is published at http://climaterealists.com/index.php?id=5770. The thesis of that paper is that the black body assumption is highly problematic in climate science. They, Hertzberg and other, use a method that does not make the black body assumption, a method that they say NASA used, and they arrive at the result that there is an Greenhouse Effect on the moon even though there is no atmosphere. The criticism of their argument takes their calculations, shows that those calculations differ from the ordinary calculations used with black bodies, reformulates those calculations as black body calculations, and reveals that their error is a simple failure to understand the math involved. This argument is a perfect circle. The argument is no less circular than the argument that the Bible is the revealed word of God because the Bible says that it is the revealed word of God.
Hertzberg, Schreuder, and Siddons wrote their article as a criticism of the assumption that the Moon could be treated as a black body. To respond to that argument by showing that its calculations differ from the black body calculations is perfectly circular. It simply says that you are not permitted to criticize the black body assumption.
This attitude seems common to the vast majority of climate scientists. They are going to stick to the black body calculations regardless. Yet that is to cleave to a system of mathematics that might be inadequate to the task at hand and whose use might discourage a closer look at physical processes. If scientists are going to retreat, yes defensively, to their mathematics anytime someone wants to look at actual physical processes then their mathematics has become the number one roadblock to progress.
Now I think I understand why no one will answer my question: “Why do we assume that Earth must radiate into space all the radiation that it receives from the sun?” The answer is that our mathematics, embodied in the black body assumption, demands that we do so. Yet if we follow that assumption, we will never understand actual physical processes on Earth. We will not understand them because we will insist that they conform to the black body assumption, yet they cannot do so. Earth is a dynamic creature that is moving and changing all the time. It needs huge amounts of energy to do this. Part of that energy comes from solar radiation and that part of solar radiation is not radiated to space. My solar panel heats my water in Spring, Summer, and Fall but not in winter. However, at all times it is taking radiation from the sun and not returning it.
Is time for scientists to step back from their work and look for another system to replace the math of the black body. Some have suggested that quantum theory can do the job, but quantum math is pretty complicated. My preferred approach is to focus on the physical processes and develop the math that is needed to describe each of them. La Nina would be a good start. Develop a good description of the natural regularities that make up the La Nina phenomenon. Do not subjugate that description to black body assumptions. It seems to me that doing the contrary will inevitably lead to agreement with AGW proponents. Why? Because as long as we stick with the black body we are discussing nothing but the characteristics of the CO2 molecule, in abstraction from physical processes, and a mathematical version of Earth that is far too idealized to accommodate actual physical processes.
Joel Shore says:
August 1, 2010 at 4:41 am
Thanks Joel. I needed that. Drinkin’ without thinkin’ and then dinkin’makes one ripe for plinkin’. I am living proof. My dad kept telling me life is twice as difficult when you’re half-witted. I just have to try twice as hard to compensate.
David Springer,
It is very different for astronomers who need to know how what wavelengths are transmitted when looking at spectra from stars or background IR signals. At wavelengths like 14 microns where the transmission goes to zero, what do you think you see? You see the atmospheric CO2 emission at its temperature, which completely replaces the transmitted astronomical signal, and makes that wavelength useless for astronomy. What you don’t see is zero flux at that wavelength.
Theo Goodwin says:
August 1, 2010 at 12:53 pm
Now I think I understand why no one will answer my question: “Why do we assume that Earth must radiate into space all the radiation that it receives from the sun?” The answer is that our mathematics, embodied in the black body assumption, demands that we do so.
That the earth within a year or so must radiate back into space all the radiation it receives has nothing to do with black body radiation. It has everything to do with simple arithmetic and energy conservation, two statements you would find hard to contradict.
If the earth on average did not radiate away what it received, energy conservation implies that it will be getting hotter and hotter and hotter, and when time=infinity, i.e. very very long, it would reach the temperature of the sun. It would not go higher because it would then violate the second law of thermodynamics.
This is the same old argument, hand waving but effective, of why the universe as observed is not infinite. If it were infinite, it would have an infinite number of stars. The radiation from that infinity of stars would be so great that every body in the universe would have the temperature of the stars, planets included. Since planets do not have the temperature of the stars, the universe is not infinite.
The black body radiation formula is a useful tool for predicting the radiation from a body and is not taken as is by engineers, but is modified for each substance as the “gray body formula”. That climatologists misuse it is another story.
I have to laugh at the quote:
The right way to calculate a planet’s average radiation is to calculate it for each and every location and average the results. The wrong way is to calculate the average temperature and then convert that to a radiation. In the case of the earth’s surface, it’s not such a noticeable problem.
In the case of the moon, because of the wide variation in temperature, the incorrect method produces a large error.
The earth, being a much more complex system than the moon, and having a large variety of emissivities (gray body constant) and enormous changes of temperature too, day, night and seasonal and latitudinal , has not problem with the way the radiation is calculated!! and presumably the errors are small !!!!
To compound the hubris, the 2m air temperature is taken as the radiating body temperature, when it is well known that it is the ground and oceans that radiate the bulk of energy, that radiation goes as T^4, and the ground can be sizzling or freezing while the air temperature is temperate. One can cook eggs on rock and sand in the summer and the arctic gets 15C anomalies in the winter while the ice is at -40K, and they talk of small errors for earth!!!
Continued:
Is time for scientists to step back from their work and look for another system to replace the math of the black body. Some have suggested that quantum theory can do the job, but quantum math is pretty complicated.
Black body comes out of quantum statistical mechanics, and actually it is one of the reasons quanta were considered a necessity.
http://en.wikipedia.org/wiki/Planck%27s_law
Ultimately, Planck’s assumption of energy quantization and Einstein’s photon hypothesis became the fundamental basis for the development of quantum mechanics.
It is not the lack of theoretical tools but their misuse that is the problem in climatology.
anna v says:
July 31, 2010 at 11:36 pm
Phil. says:
July 31, 2010 at 11:56 am
Dave Springer says:
“This is not how it works in a cold dense gas. When CO2 absorbs energy it its absorption bands it causes nearly instantaneous collision with another molecule, the vast majority of which are nitrogen, oxygen, and water vapor. The kinetic energy in the collisions is radiated in continous blackbody spectrum with the peak frequency determined by the sensible temperature of the gases.”
Phil: The collision part is true but nitrogen, oxygen, and water vapor do not radiate as blackbody spectra.
Just finished my Sunday chores and came to this thread.
Phil, you have to take this back. I am sure it is reflex reaction to saying no whatever Dave is sayin.
Every type of molecular ensemble radiates according to a modified for that type ” black” body temperature . It is elementary statistical mechanics from which, temperature comes out as a statistical adjunct of the kinetic energy of each molecule, and from which the black body formula was guessed at.
http://en.wikipedia.org/wiki/Statistical_mechanics
Your statement would mean that O2 and N2 are at 0 Kelvin, which is the classical limit of temperature when there is no kinetic energy. If there is kinetic energy, there is temperature and there is gray body radiation.
Anna it’s time to revisit your Physics of Gases textbook, gas molecules do not radiate as blackbody spectra, they radiate at certain lines characteristic of transitions between various molecular motions, for the IR region typically between vibrational and rotational modes. The quantity emitted at any frequency cannot exceed the quantity emitted by a backbody at the same temperature. Oxygen and Nitrogen molecules possess no radiatively active vibrational or rotational modes and so do not radiate. You are attempting to apply the physics of solids to gases and in this case it doesn’t work.
Phil. says:
August 2, 2010 at 7:34 am
Every matter radiates according to its temperature in a gray body formula, i.e. emissivity and spectrum changes from black body.
Phil , you cannot but be wrong that O2 N2 and H2O do not have thermal properties that will lead to a gray body radiation . They are gasses with kinetic energy independent of the vibrational and rotational modes, and in collision even if symmetric, will be deformed and radiate in the appropriate thermal frequencies. Otherwise they should have 0 kinetic energy which means 0K Temperature, classically.
Anna V. writes:
“That the earth within a year or so must radiate back into space all the radiation it receives has nothing to do with black body radiation. It has everything to do with simple arithmetic and energy conservation, two statements you would find hard to contradict.”
This I agree with. I agree because of the phrase “within a year or so.” See, that permits me to ask my question, though I guess I reformulated it incorrectly in the original. My reformulated question, adding the word “daily” reads as follows:
Now I think I understand why no one will answer my question: “Why do we assume that Earth must radiate into space all the radiation that it receives from the sun DAILY?” The answer is that our mathematics, embodied in the black body assumption, demands that we do so.
If we could make the simple assumption that Earth sends out whatever it receives from the sun DAILY, then the AGW folks have a simple calculation to make when they are calculating the heating caused by CO2 in the atmosphere. Take away the regular period, the DAILY, and what period do you work with? So, I ask you: What is the period that the calculation is based on? If there is no regular period, such as daily, then how is the calculation based? It has to become process based. It has to be based on physcial hypotheses about the way the radiation passes through the myriad winding paths of Earth. If you are going to make those kinds of calculations, I do not see why they would assume the mathematics of black bodies. That system of mathematics is just too simple. On the other hand, perhaps a definite period is not actually needed for black body calculations. If so, why do all these articles assume some definite period?
The work of Siddons and others that I quoted was suggesting a calculation based on treating the Moon as a three dimensional object. You might find their attempt laughable, most new ideas are. I took them to be suggesting an alternative to the standard mathematics. To respond by saying that they have naive ideas about the standard calculations is irrelevant, isn’t it?
anna v, I just read up on this and it turns out that symmetric molecules like O2 and N2 are radiatively inactive because they have no dipole moment, so they can’t emit or absorb photons to change vibration states. This is different from H2O and CO2 that are “IR active” due to their electric field asymmetries. Therefore collisions of O2 and N2 can only change their translational and rotational kinetic energy and vibration states with no photons involved, and mostly they remain in the ground state for vibration leaving five energy modes to share (“equipartition”) their kinetic energy (three translational and two rotational). These gases therefore have no absorption or emission spectra.
Theo Goodwin and anna v,
The balance between incoming and outgoing radiation is the long-term limit. Even annual averages don’t balance out exactly. Note that adding greenhouse gases changes the net balance at the top of the atmosphere by reducing the outgoing part. This balance can be restored by atmospheric warming to a new equilibrium, or, as some would have it, more cloud cover or aerosols which may come from natural processes or man-made. Obviously the debate is on which of these will dominate in the restoration of equilibrium, and I don’t need to say what the consensus is.
anna v says:
August 2, 2010 at 11:34 am
Phil. says:
August 2, 2010 at 7:34 am
Every matter radiates according to its temperature in a gray body formula, i.e. emissivity and spectrum changes from black body.
Phil , you cannot but be wrong that O2 N2 and H2O do not have thermal properties that will lead to a gray body radiation .
But I’m right, I thought you were a physicist, as I said before go back and read some physics of gases!
Jim D says:
August 2, 2010 at 7:10 pm
anna v, I just read up on this and it turns out that symmetric molecules like O2 and N2 are radiatively inactive because they have no dipole moment, so they can’t emit or absorb photons to change vibration states. This is different from H2O and CO2 that are “IR active” due to their electric field asymmetries. Therefore collisions of O2 and N2 can only change their translational and rotational kinetic energy and vibration states with no photons involved, and mostly they remain in the ground state for vibration leaving five energy modes to share (“equipartition”) their kinetic energy (three translational and two rotational). These gases therefore have no absorption or emission spectra.
This is a given. But a much greater given is that gases to first order follow classical statistical mechanics and therefore thermodynamics. Classical statistical mechanics does not know about vibrational and rotational levels. It knows about the kinetic energy carried by each and every molecule and gives its statistical distribution which defines temperature entering as a constant in the formula of the statistical distribution. Every molecular ensemble that has a temperature, i.e. its molecules are not frozen at 0K radiates according to the black body radiation, which classically needed the correction that Planck introduced to fit the data.
Now how, when one looks at the microstate does a symmetric molecule radiate? A neutral symmetric molecule when hit by another neutral symmetric molecule gets deformed and higher moments ( quadruple, octuple) appear and it radiates away, so as not to destroy the smooth transition from classical to quantum statistical mechanics, and thus thermodynamics :). It is the kinetic energy that is transformed into radiation, the average kinetic energy falls, and the gas cools.
This is elementary statistical mechanics and quantum statistical mechanics.
Think a bit. If symmetrical molecules would not cool or heat with collisions, they would not only be the perfect insulator but considering that more than 90 percent of our atmosphere is N2 and O2 we would be either cooking or freezing. Collisions at the microstate level mean absorption and emission of soft photons turned into and out of the kinetic energy of the total molecule.
Theo Goodwin says:
August 2, 2010 at 5:51 pm
My reformulated question, adding the word “daily” reads as follows:
Now I think I understand why no one will answer my question: “Why do we assume that Earth must radiate into space all the radiation that it receives from the sun DAILY?” The answer is that our mathematics, embodied in the black body assumption, demands that we do so.
Well, without entering into the discussion of whether an average global whatever exists, ( temperature humidity heat capacity etc) the numbers given are given as an average over the globe over the year. From the evident heat capacity of the earth it is clear that this is not an instantaneous or a daily or a weekly sum. Actually if we look at the PDO, the ENSO, the AO and all those acronyms that have long periods it is evident that yearly is too short also. So no, our mathematics does not demand that we balance the budget in a specific time interval as tight as daily. Energy conservation does give constraints but certainly not daily ones.
Phil. says:
August 2, 2010 at 8:58 pm
anna v says:
August 2, 2010 at 11:34 am
Phil. says:
August 2, 2010 at 7:34 am
Every matter radiates according to its temperature in a gray body formula, i.e. emissivity and spectrum changes from black body.
Phil , you cannot but be wrong that O2 N2 and H2O do not have thermal properties that will lead to a gray body radiation .
But I’m right, I thought you were a physicist, as I said before go back and read some physics of gases!
The physics of gases is much broader than the obsessively focused upon quantum mechanical radiation levels by climatology. It so happens that nature has also a continuum spectrum that has no need of absorption and emission levels which climatologists seem to ignore.
Yes, I am a physicist and I have a physicists pov and knowledge in enough depth to know that you really do not know what you are talking about but are parroting the mantras on emission and absorption of radiation necessary for the famous greenhouse model.
Gases do not cool by emitting and absorbing radiation in limited lines only. There is also direct transformation of kinetic energy into radiation from the fields that the molecules have and their distortions in collisional interactions.
Think the following experiment:
Take a ball of O2 at room temperature and leave it in the vacuum of space in the dark. It would need no container as it would keep its shape by gravity. What do you think will happen? Will it keep its temperature or will it cool to the background microwave temperatures?
anna v says:
August 3, 2010 at 12:49 am
Phil. says:
August 2, 2010 at 8:58 pm
anna v says:
August 2, 2010 at 11:34 am
Phil. says:
August 2, 2010 at 7:34 am
Every matter radiates according to its temperature in a gray body formula, i.e. emissivity and spectrum changes from black body.
Phil , you cannot but be wrong that O2 N2 and H2O do not have thermal properties that will lead to a gray body radiation .
But I’m right, I thought you were a physicist, as I said before go back and read some physics of gases!
The physics of gases is much broader than the obsessively focused upon quantum mechanical radiation levels by climatology. It so happens that nature has also a continuum spectrum that has no need of absorption and emission levels which climatologists seem to ignore.
Yes, I am a physicist and I have a physicists pov and knowledge in enough depth to know that you really do not know what you are talking about but are parroting the mantras on emission and absorption of radiation necessary for the famous greenhouse model.
Regrettably you don’t, until you educate yourself further it’s a complete waste of time talking to you.
anna v,
I see Phil has lost patience. Anyway, I will try it a different way. Does a perfect gas radiate and cool or does it maintain its temperature when left alone? Statistical mechanics is about perfect gases that do not lose energy except through collisions with cooler gases or with moving cylinders, etc. Nowhere in statistical mechanics is there anything about the perfect gas radiating energy. Gases are not perfect, but very close to it, otherwise it would be useless to learn about perfect gases which explain hydrostatic pressure, expansion by cooling, heat capacities, and all kinds of basic phenomena and conservation rules. O2 and N2 are for all intents and purposes perfect gases.
The other part of my answer is about temperature. The mean molecular kinetic energy defines the temperature according to a simple proportionality rule. Collisions transfer kinetic energy and hence temperature leading to thermal equilibrium when the kinetic energy is distributed evenly among the molecules in the atmosphere.
Jim D says:
August 3, 2010 at 6:21 pm
I have lost patience with Phil also, and believe that a person who can spout what he does and asks for others to be educated is not worth the effort.
Consider this:
How does isolated in space matter lose energy?
Only by radiation.
There is no other way.
The black body radiation formula started as the classic formula of explaining radiation and classically nothing is known about emission and absorption lines and symmetric or asymmetric molecules. Have a look at http://en.wikipedia.org/wiki/Black_body to see that in statistical mechanics there are no exceptions to the black body loss of heat by radiation. Check and see that there is no word “line” in the whole article.
This statement is wrong then:
Nowhere in statistical mechanics is there anything about the perfect gas radiating energy.
All matter radiates and absorbs energy. An ideal gas is matter.
To say that matter composed of molecules that in isolation each have no dipole moment and therefore even in the aggregate when collisions happen continuously and energy is transferred between the molecules are still symmetric and cannot radiate ignores basic physics. Which I am sorry to say is the habit of people playing with explaining climate.