By raypierre , with the gratefully acknowledged assistance of Spencer Weart
In Part I the long struggle to get beyond the fallacious saturation argument was recounted in historical terms. In Part II, I will provide a more detailed analysis for the reader interested in the technical nitty-gritty of how the absorption of infrared really depends on CO2 concentration. At the end, I will discuss Herr Koch’s experiment in the light of modern observations.
The discussion here is based on CO2 absorption data found in the HITRAN spectroscopic archive. This is the main infrared database used by atmospheric radiation modellers. This database is a legacy of the military work on infrared described in Part I , and descends from a spectroscopic archive compiled by the Air Force Geophysics Laboratory at Hanscom Field, MA (referred to in some early editions of radiative transfer textbooks as the "AFGL Tape").
Suppose we were to sit at sea level and shine an infrared flashlight with an output of one Watt upward into the sky. If all the light from the beam were then collected by an orbiting astronaut with a sufficiently large lens, what fraction of a Watt would that be? The question of saturation amounts to the following question: How would that fraction change if we increased the amount of CO2 in the atmosphere? Saturation refers to the condition where increasing the amount of CO2 fails to increase the absorption, because the CO2 was already absorbing essentially all there is to absorb at the wavelengths where it absorbs at all. Think of a conveyor belt with red, blue and green M&M candies going past. You have one fussy child sitting at the belt who only eats red M&M’s, and he can eat them fast enough to eat half of the M&M’s going past him. Thus, he reduces the M&M flux by half. If you put another equally fussy kid next to him who can eat at the same rate, she’ll eat all the remaining red M&M’s. Then, if you put a third kid in the line, it won’t result in any further decrease in the M&M flux, because all the M&M’s that they like to eat are already gone. (It will probably result in howls of disappointment, though!) You’d need an eater of green or blue M&M’s to make further reductions in the flux.
Ångström and his followers believed that the situation with CO2 and infrared was like the situation with the red M&M’s. To understand how wrong they were, we need to look at modern measurements of the rate of absorption of infrared light by CO2 . The rate of absorption is a very intricately varying function of the wavelength of the light. At any given wavelength, the amount of light surviving goes down like the exponential of the number of molecules of CO2 encountered by the beam of light. The rate of exponential decay is the absorption factor.
When the product of the absorption factor times the amount of CO2 encountered equals one, then the amount of light is reduced by a factor of 1/e, i.e. 1/2.71282… . For this, or larger, amounts of CO2,the atmosphere is optically thick at the corresponding wavelength. If you double the amount of CO2, you reduce the proportion of surviving light by an additional factor of 1/e, reducing the proportion surviving to about a tenth; if you instead halve the amount of CO2, the proportion surviving is the reciprocal of the square root of e , or about 60% , and the atmosphere is optically thin. Precisely where we draw the line between "thick" and "thin" is somewhat arbitrary, given that the absorption shades smoothly from small values to large values as the product of absorption factor with amount of CO2 increases.
The units of absorption factor depend on the units we use to measure the amount of CO2 in the column of the atmosphere encountered by the beam of light. Let’s measure our units relative to the amount of CO2 in an atmospheric column of base one square meter, present when the concentration of CO2 is 300 parts per million (about the pre-industrial value). In such units, an atmosphere with the present amount of CO2 is optically thick where the absorption coefficient is one or greater, and optically thin where the absorption coefficient is less than one. If we double the amount of CO2 in the atmosphere, then the absorption coefficient only needs to be 1/2 or greater in order to make the atmosphere optically thick.
The absorption factor, so defined, is given in the following figure, based on the thousands of measurements in the HITRAN spectroscopic archive. The "fuzz" on this graph is because the absorption actually takes the form of thousands of closely spaced partially overlapping spikes. If one were to zoom in on a very small portion of the wavelength axis, one would see the fuzz resolve into discrete spikes, like the pickets on a fence. At the coarse resolution of the graph, one only sees a dark band marking out the maximum and minimum values swept out by the spike. These absorption results were computed for typical laboratory conditions, at sea level pressure and a temperature of 20 Celsius. At lower pressures, the peaks of the spikes get higher and the valleys between them get deeper, leading to a broader "fuzzy band" on absorption curves like that shown below.
We see that for the pre-industrial CO2 concentration, it is only the wavelength range between about 13.5 and 17 microns (millionths of a meter) that can be considered to be saturated. Within this range, it is indeed true that adding more CO2 would not significantly increase the amount of absorption. All the red M&M’s are already eaten. But waiting in the wings, outside this wavelength region, there’s more goodies to be had. In fact, noting that the graph is on a logarithmic axis, the atmosphere still wouldn’t be saturated even if we increased the CO2 to ten thousand times the present level. What happens to the absorption if we quadruple the amount of CO2? That story is told in the next graph:
The horizontal blue lines give the threshold CO2 needed to make the atmosphere optically thick at 1x the preindustrial CO2 level and 4x that level. Quadrupling the CO2 makes the portions of the spectrum in the yellow bands optically thick, essentially adding new absorption there and reducing the transmission of infrared through the layer. One can relate this increase in the width of the optically thick region to the "thinning and cooling" argument determining infrared loss to space as follows. Roughly speaking, in the part of the spectrum where the atmosphere is optically thick, the radiation to space occurs at the temperature of the high, cold parts of the atmosphere. That’s practically zero compared to the radiation flux at temperatures comparable to the surface temperature; in the part of the spectrum which is optically thin, the planet radiates at near the surface temperature. Increasing CO2 then increases the width of the spectral region where the atmosphere is optically thick, which replaces more of the high-intensity surface radiation with low-intensity upper-atmosphere radiation, and thus reduces the rate of radiation loss to space.
Now let’s use the absorption properties described above to determine what we’d see in a typical laboratory experiment. Imagine that our experimenter fills a tube with pure CO2 at a pressure of one atmosphere and a temperature of 20C. She then shines a beam of infrared light in one end of the tube. To keep things simple, let’s assume that the beam of light has uniform intensity at all wavelengths shown in the absorption graph. She then measures the amount of light coming out the other end of the tube, and divides it by the amount of light being shone in. The ratio is the transmission. How does the transmission change as we make the tube longer?
To put the results in perspective, it is useful to keep in mind that at a CO2 concentration of 300ppm, the amount of CO2 in a column of the Earth’s atmosphere having cross section area equal to that of the tube is equal to the amount of CO2 in a tube of pure CO2 of length 2.5 meters, if the tube is at sea level pressure and a temperature of 20C. Thus a two and a half meter tube of pure CO2 in lab conditions is, loosely speaking, like "one atmosphere" of greenhouse effect. The following graph shows how the proportion of light transmitted through the tube goes down as the tube is made longer.
The transmission decays extremely rapidly for short tubes (under a centimeter or so), because when light first encounters CO2, it’s the easy pickings near the peak of the absorption spectrum that are eaten up first. At larger tube lengths, because of shape of the curve of absorption vs. wavelength, the transmission decreases rather slowly with the amount of CO2. And it’s a good thing it does. You can show that if the transmission decayed exponentially, as it would if the absorption factor were independent of wavelength, then doubling CO2 would warm the Earth by about 50 degrees C instead of 2 to 4 degrees (which is plenty bad enough, once you factor in that warming is greater over land vs. ocean and at high Northern latitudes).
There are a few finer points we need to take into account in order to relate this experiment to the absorption by CO2 in the actual atmosphere. The first is the effect of pressure broadening. Because absorption lines become narrower as pressure goes down, and because more of the spectrum is "between" lines rather than "on" line centers, the absorption coefficient on the whole tends to go down linearly with pressure. Therefore, by computing (or measuring) the absorption at sea level pressure, we are overestimating the absorption of the CO2 actually in place in the higher, lower-pressure parts of the atmosphere. It turns out that when this is properly taken into account, you have to reduce the column length at sea level pressure by a factor of 2 to have the equivalent absorption effect of the same amount of CO2 in the real atmosphere. Thus, you’d measure absorption in a 1.25 meter column in the laboratory to get something more representative of the real atmosphere. The second effect comes from the fact that CO2 colliding with itself in a tube of pure CO2 broadens the lines about 30% more than does CO2 colliding with N2 or O2 in air, which results in an additional slight overestimate of the absorption in the laboratory experiment. Neither of these effects would significantly affect the impression of saturation obtained in a laboratory experiment, though. CO2 is not much less saturated for a 1 meter column than it is for a 2.5 meter column.
So what went wrong in the experiment of poor Herr Koch? There are two changes that need to be made in order to bring our calculations in line with Herr Koch’s experimental setup. First, he used a blackbody at 100C (basically, a pot of boiling water) as the source for his infrared radiation, and measured the transmission relative to the full blackbody emission of the source. By suitably weighting the incoming radiation, it is a simple matter to recompute the transmission through a tube in a way compatible to Koch’s definition. The second difference is that Herr Koch didn’t actually perform his experiment by varying the length of the tube. He did the control case at a pressure of 1 atmosphere in a tube of length 30cm. His reduced-CO2 case was not done with a shorter tube, but rather by keeping the same tube and reducing the pressure to 2/3 atmosphere (666mb, or 520 mm of Mercury in his units). Rather than displaying the absorption as a function of pressure, we have used modern results on pressure scaling to rephrase Herr Koch’s measurement in terms of what he would have seen if he had done the experiment with a shortened tube instead. This allows us to plot his experiment on a graph of transmission vs. tube length similar to what was shown above. The result is shown here:
Over the range of CO2 amounts covered in the experiment, one doesn’t actually expect much variation in the absorption — only about a percent. Herr Koch’s measurements are very close to the correct absorption for the 30cm control case, but he told his boss that the radiation that got through at lower pressure increased by no more than 0.4%. Well, he wouldn’t be the only lab assistant who was over-optimistic in reporting his accuracy. Even if the experiment had been done accurately, it’s unclear whether the investigators would have considered the one percent change in transmission "significant," since they already regarded their measured half percent change as "insignificant."
It seems that Ångström was all too eager to conclude that CO2 absorption was saturated based on the "insignificance" of the change, whereas the real problem was that they were looking at changes over a far too small range of CO2 amounts. If Koch and Ångström had examined the changes over the range between a 10cm and 1 meter tube, they probably would have been able to determine the correct law for increase of absorption with amount, despite the primitive instruments available at the time.
It’s worth noting that Ångström’s erroneous conclusion regarding saturation did not arise from his failure to understand how pressure affects absorption lines. That would at least have been forgivable, since the phenomenon of pressure broadening was not to be discovered for many years to come. In reality, though Ångström would have come to the same erroneous conclusion even if the experiment had been done with the same amounts of CO2 at low pressure rather than at near-sea-level pressures. A calculation like that done above shows that, using the same amounts of CO2 in the high vs. low CO2 cases as in the original experiment, the magnitude of the absorption change the investigators were trying to measure is almost exactly the same — about 1 percent — regardless of whether the experiment is carried out at near 1000mb (sea level pressure) or near 100mb (the pressure about 16 km up in the atmosphere).
Ray, There is a difference between blackbody radiation and line radiation. An iron bar glows red when heated. The lines which are emitted by iron vapour are green.
Blackbody radiation is thermal radiation, which depends on temperature. Line radiation is produced by electronic relaxation of atoms, or vibrational and rotational relaxation of liquid and gas molecules. These three are all produced in a different manner. Blackbody radiation is produced in yet a different manner. “Thermal radiation is generated when heat from the movement of charged particles within atoms is converted to electromagnetic radiation.” See http://en.wikipedia.org/wiki/Thermal_radiation
There are other types of radiation too such as fluorescence, phosphorescence, and Bremsstrahlung, X-rays and Gamma-rays. They are all electromagnetic radiation but they are all produced in different ways and have different characteristics.
Alastair McDonald (#345) wrote:
Alastair, physicists seem to believe that they have a solution to the solar coronal heating problem – without throwing out everything we know about the interaction between matter and electromagnetic energy. If you read the Wikipedia article a little further you will see this. Electronic excitation in the corona is important given the fact that the corona is hot enough that matter exists as a plasma. With regard to temperature measurements in the corona, there would however be a difference inasmuch as the corona is no doubt distant from a local thermodynamic equilibrium, but not to the degree that you imagine.
Electronic excitation is not an issue in the earth’s atmosphere because neither the radiation no matter are hot enough that electronic excitation would be involved.
As has been pointed out to you on numerous occasions (without any acknowledgement on your part, either in terms of agreement or disagreement), the thermal infrared radiation of which we speak is not due to electronic excitation, that is, it is not due electrons jumping from one orbital to another within an atom. It is due to the molecules themselves assuming different quantum rotational and vibrational states, and at higher temperatures it would involve stretching.
Think about it: how do we know that carbon dioxide absorbs infrared at various parts of the infrared band? At a basic level, without recourse to quantum mechanics, we can measure it in a lab. Why do we claim to know the same thing with regard to its emission at the same wavelengths? Because we are able to deal with emission the same way: measurements in a lab.
The various quantum molecular states involve energies far lower than what is involved in the corona of the sun. The longwave radiation relevant to the greenhouse effect as it exists within our climate system isn’t frozen out except under rather specific circumstances – because it is within the range of the energies associated with the quantum molecular states.
This isn’t electronic, it is molecular.
You claim that carbon dioxide is unable to participate in the greenhouse effect except near the surface of the earth. You claim that the reason is that the thermal radiation gets frozen out higher up in the earth’s atmosphere, or on other occasions, you claim that the energy gets carried off by non-greenhouse gases such as oxygen.
With regard to the latter, what every energy may be lost as the result of collisions may be gained by collisions, and as long as the collision rate exceeds the emission rate by a couple of orders of magnitude, a local equilibrium will be established between thermal radiation and the thermal state of matter – at above ~20 mb. Over all frequencies and wavelengths? No, but over the relevant frequencies and wavelengths.
Now with to the former, your argument is that the energy gets frozen out because the temperature of the atmosphere is too low to achieve electronic excitation. But the same would no doubt hold at the surface. It is able to achieve the state of excitation because the excitation is molecular, not pertaining to electrons shifting from one quantum orbital state to another, but to molecules shifting from one quantum rovibrational state to another. Genuine scientists studying the interaction of infrared thermal radiation and various gases know this as a result of lab experiments. Genuine scientists also know this as the result of quantum statistical mechanics, both in terms of how quantum mechanics applies to molecules and the equipartion theorem applies to both radiation and matter.
However, we also know this as the result of satellite imaging of the atmosphere.
I pointed out as much in 343:
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When you argue against the established science, you are making ad hoc arguments which are typically grounded in a lack of understanding and armchair-theorizing. You are doing so without recourse to empirical observation, systematic empirical laboratory experimentation, and both the relevant conceptual and mathematical framework. You seem to think that you are justified in doing this in part because the people who you are speaking with don’t know everything there is to know.
Well, neither do you.
What I hold is that the whole of the scientific community knows a great deal more than any one individual, or for that matter, what any one individual would ever be capable of knowing. There is a division of cognitive labor which that exists within a community which is ultimately all engaged in the same thing: the empirical study of the unified nature of reality. For any given well-established scientific knowledge, there are generally mulitiple independent lines of investigation which provide it with justification, and a conclusion justified by multiple independent lines of investigation is generally justified to a far greater degree than it would be if it were justified by any one line of investigation considered in isolation.
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Consider an ant colony consisting of 10,000,000 individuals.
Each individual has perhaps 10,000 brain cells. Individually an ant brain isn’t capable of very much. But an ant colony has as collectively as many brain cells as a human being. These brain cells are by no means as tightly integrated as a human brain. Nevertheless, collectively an ant colony is capable of learning at a level which far exceeds that which any one ant brain is capable of in isolation, such that an older, larger colony knows better than a younger smaller colony than to attack a colony larger than itself.
The same principle applies to human beings.
Both genetically and in terms of the innate capacity of my brain, there is very little difference between me and someone who lived 800,000 years ago in a clan consisting of 100 individuals. However, as a human being living in today’s modern society, I have the benefit of a legacy stretching back to perhaps 10,000 years, and I have the benefit of living in a community consisting of nearly 7,000,000,000 individuals. Collectively we are capable of knowing a great deal more than any one individual is capable of in isolation.
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The same principle applies even in terms of the temporal integration of your understanding from one moment to the next.
Engaging in extreme cartesian doubt where one would be unwilling to accept anything which is not immediately “self-evident,” including whatever memories one might have of a moment ago, if one were systematic about it, one would not even be able to take recourse to even the most basic distinctions between dreaming and perception, identification and evaluation, cognition and emotion, or even the subject who is aware and the object which the subject is aware of. At a slightly higher level, one might choose to accept perception, but regard any evidence much beyond this, including memories of more than a few seconds ago as objects which were placed there to create the mere appearance of an enduring and ancient world which was created ten seconds or ten thousand years ago.
But we are capable of moving beyond such extreme skepticism because we are able to recognize the principle that a conclusion which receives justification from multiple independent lines of empirical investigation is generally justified to a far greater degree than it would be if it only received justification from any one line of investigation considered in isolation. We realize that any argument for such extreme skepticism must presuppose the very sort of knowledge which it would deny us recourse, and as such is ultimately (self-referentially) incoherent. Thus for example a radical skeptic might choose to undercut all claims to knowledge by arguing that all claims to knowledge might ultimately be grounded in nothing more than a dream.
But what does the skeptic mean by “dream”?
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At a certain level, you are making the same sort of mistake as the young earth creationist or radical skeptic.
You arbitrarily accept some of the conclusions of science, those which fit your personal “theory,” but not others. But how do you know the conclusions of science which you accept? What justifies them? The conclusions of science exist within a vast network of human knowledge which is far greater than that which any one individual is capable of understanding in isolation from a human community.
Right here on this blog we form a community which exists within a far vaster community. We are capable of learning from each other because we participate in a dialogue.
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Consider this: if you have one individual with only three insights, he is capable of making only three connections between any two insights.
But if you have two individuals with three insights each, by learning from one another, they will have six insights and they will be capable of making fifteen connections between any two insights. Between five individuals, the community is capable of making one hundred and five connections. But if a given individual refuses to learn from the others, if he refuses to participate in the dialogue, then he is once again capable of making only three connections.
In terms of the principle involved, this is the position you place yourself in by refusing to acknowledge and learn from the insights of others.
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This is important enough to point out and to stress both for your sake and for the sake of the communities which you are a member of.
When you insist that you have some alternative to modern empirical science, you know just enough to sound authoritative to some. By insisting that your own personal and largely ad hoc theory is an alternative to well-established science, you find it necessary to cast doubt on well-established science. As such you are undercutting the purpose of this community and inadvertently serving to further the interests of those who claim that there is no threat to humanity posed by global warming because global warming does not exist. Whether you choose to recognize it or not, your ad hoc skepticism serves the denialists.
I would ask that you reconsider your approach, for the sake of your understanding, this community, the human community and out of the recognition of the threat posed by climate change which you acknowledge exists.
Rod/Ray/Hank/Timothy/Alastair: I have gained from your discussions and so would attempt to clarify some things as I see them. Heat radiation is that radiation exchanged between bodies at equilibrium whatever the nature of the transitions. The black body has an infinite continuum of allowable frequencies. A cavity can approximate a black body arbitrarily well provided any port approaches zero size, and there is at least one region which absorbs at all frequencies inside. Black body radiation describes heat exchanged between black bodies.
Line or narrow band emitters can be at equilibrium, but parts of the spectrum will be empty. Similarly, line or narrow band absorbers can be themselves at equilibrium while letting higher energy radiation pass through.
Interacting quantum oscillators with the same transition frequency create higher and lower energy transitions. A large number of identical oscillators form bands, so a collection of atoms can form an approximation to a black body. Isolated atoms can (mostly) only emit single quanta. Interacting ideal gas molecules at our pressures show line broadening from occasional collisions, but spend too much time separated from neighbors to form bands. The exception is water, which molecules have a relatively strong attraction and a permanent dipole that makes them interact over much greater distances than they would otherwise, thereby forming a continuum at moderate concentration.
Since temperature is only defined at equilibrium, we rapidly get into difficulties in non equilibrium situations, which is most of the time. We can talk about apparent temperature, calculated from the ratio of vibrational states, which could be different from that using the distribution of velocities, and all these could be different from the apparent temperature of radiation passing through from a black body. This is the LTE state. Different thermometers will read differently in non-equilibrium situations. Go buy a dozen and some paint and try it.
Coherence is the result when atoms with inverted population distributions are placed in a cavity and are “stimulated” to emit in concert with a standing mode in the cavity – continuous mode lasers. Coherent radiation is monochromatic, and vice versa. I think this is what RodB was talking about.
Finally, the Einstein coefficients do not require equilibrium, so we can write a general NLTE transfer equation if we have the numbers, and as far as I can find we do.
Allan Ames (#353) wrote:
This is why we speak of higher temperatures and pressures. But notice – speaking of a band should not be taken to imply that the individual lines no longer exist as distinct entities but only that they consist of an identifiable natural grouping. This will exist independently of the pressure or temperature. But even in the case of the bleeding between lines or bands which I mentioned above, the lines and bands will be distinguishable – its just that the boundaries between the lines or the bands will not be as well-defined due to spreading.
Allan Ames continues:
Water has some interesting properties including a fairly impressive ability to amplify parity violation which may have been a significant factor in the origin of the chirality of life. However, there natural broadening of lines due to the uncertainty relationship which exists between time and energy, line broadening, band broadening, etc. All of these involve to varying degrees a blending where while a line or band remains distinct from another, they are no longer entirely separate. They never were.
Allan Ames continues:
Under LTE conditions, temperatures measured by different approaches will be identical since LTE is a local thermodynamic equilibrium.
With respect to thermometers which are painted different colors, well, that is why God created albedos. Temperatures are measured in the shade. When you speak of a thermometer which is painted a different color registering a different temperature, you are speaking of a thermometer which has been placed in the sunlight and/or which hasn’t had the opportunity to achieve equilibrium with its environment.
This is in 237. I wrote:
The question at this point is, “How do you carve things?” You could define temperature in non-standard ways so that you are measuring temperatures in the sunlight using different colored thermometers just like you could define a color which consists of either blue or red, but it would simply anything. Indeed, it would make things more complex. Or we might choose not to distinguish between the thermal radiation of the sun and the local thermal radiation of either the atmosphere or the surface, as I suggested in 237. But we make these distinctions because they simplify matters – much like the choice of coordinate systems when attempting to solve Einstein’s field equations in General Relativity. Put your thermometers in the shade and let them achieve an equilibrium with their surroundings.
Allan Ames continues:
Granted. Monochromatic, in phase, etc. But this is the result of a population inversion, not the sort of thing that we are dealing with here.
Allan Ames continues:
We have known of these effects for quite some time now.
Here is one of the earlier abstracts I have been able to find from 1989, although there are earlier versions of this paper:
Additionally, you will notice that the incorporation of non-LTE effects into radiation codes in use goes back at least as far as 1996.
Please see:
You will notice that this is the very same radiation research that modern climatology was based on at the same time in the 1980s. Military research – as has been stressed on numerous occasions. Nothing new.
In any case, I have pointed out just recently the existence of NLTE in the atmosphere:
One take-away point if you look at the graph: the difference between local and non-local thermodynamic equilibria is a matter of degree not kind. At some level of accuracy or another there will be some non-local effects however insignificant. Lets face it: reality can get rather messy. But it will be a matter of degree. Viewing much of the atmosphere strictly from inside a framework of either LTE or partial-LTE works quite well under most circumstances less well in others. But even any
In any case, climate models have incorporated the very effects which you believe are important.
Finally, if at times it seems that I am unfamiliar with some of this, it is because I haven’t started looking into climatology until the past couple of months and my background in physics is largely informal. But the research is there, the knowledge is there. And if you want, you can look it up – as I have done.
CORRECTION to #354
Where I wrote:
… using blockquotes, that is the extent of what I quote at that point, and the rest should be outside of the quote.
We I continue with,
… that should be outside of what I am quoting. Sorry for the confusion on my part.
Hank (344, again), I really couldn’t get much out of the discourse in PhysicsForum that you referenced. Maybe it’s me. But thanks for the effort and assistance.
All, I scanned the latest posts. They seemed informative, but I didn’t detect a direct answer to my very simple(???) 3rd question of #337, which was (with some clarification): Assume the temperature of a mole of gas (say CO2) that has a boltzman distribution of translation kinetic energy and every molecule is in its ground vibration and rotation states. (Is that possible?) Now add infrared radiation so that a percentage of the molecules (any % you want) has energy absorbed into its rotation and vibration energy levels, and assume no other energy transfer ala equipartition, collision, radiant emission, or from any other source, and that it is given enough time to settle (though given my assumptions I don’t know what “settle” would be…). Did the temperature of the mole, as measured by a thermometer, go up? down? or stay the same? Put another way (but still the same question), did the induced kinetic energy of the atomic vibrations and the molecular rotation add to the kinetic energy of the translation molecular movement?
A second side question (sorry): Assuming no other energy transfer, would this mole of CO2, if sufficiently compressed (though that might mess me up as it is itself an energy transfer…) emit black/graybody radiation per Planck’s function. And if so where would that emitted energy come from: translation, vibration, rotation, or some combo (I’m assuming not from electron, chemical or nuclear, etc.)?
Although it doesn’t seem quite as critical to me as before, I think this (and related questions on molecular energy transfer) has an impact on the base process of global warming — what gets warmed and how. I guessed it would have been straight-forward, but it’s looking far from that, including a bunch not even well known.
I need to go back and study the posts. They look good, and I appreciate yourall’s assistance and input.
> assume no other energy transfer …. Did the temperature of the mole, as measured by a thermometer, go up?
Not as long as your assumption holds true, while no energy was transferred to the thermometer — the thermometer won’t detect anything until energy is transferred to or from it.
What I am saying is that cavity radiation, blackbody radiation, and thermal radiation are all the same thing, but it is different from line radiation of which there are many types. I prefer to use the term blackbody because it describes the prefect absorber, which through Kirchhoff’s law of emission equalling absorption, emits with the shape of Planck’s function.
I agree that a blackbody is an ideal which does not really exist. However, it is a good approximation to reality and Planck’s function can be used when modeling the radiation emitted by the Earth’s surface etc. The reason Planck’s function is written as B(T) is because it is the Blackbody function.
which describes the ideal
Re #357
That is a good question. And the answer is that if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.
In fact what happens is that the vibrational energy can be re-emitted, or it can be degraded into rotational energy or translational energy. Water vapour can emit the rotational energy, but for carbon dioxide the rotational energy can only degrade to translational energy i.e. kinetic energy.
The answer to the second question is that CO2 does not emit blackbody radiation. It only emit lines related to it vibrational energy. Nitrogen does not even emit lines due to vibrational energy, whereas water vapour radiates vibrational lines, rotational lines, and continuum (blackbody?) radiation.
The energy for the lines comes from the excitation which may have arisen in collisions, making its original source translational energy. In other words as the gas radiates it cools down (kinetic energy gets less so registers a lower reading on a thermometer.)
HTH.
Cheers, Alastair.
Re #357
Rod, that is a good question. And the answer is that if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.
In fact what happens is that the vibrational energy can be re-emitted, or it can be degraded into rotational energy or translational energy. Water vapour can emit the rotational energy, but for carbon dioxide the rotational energy can only degrade to translational energy i.e. kinetic energy.
The answer to the second question is that CO2 does not emit blackbody radiation. It only emits lines related to its vibrational energy. Nitrogen cannot even emit lines from its vibrational energy, whereas water vapour radiates vibrational lines, rotational lines, and continuum (blackbody?) radiation.
The energy for the lines comes from excitation, which may have arisen from absorption, but also from collisions so the original source is translational energy. In other words as CO2 radiates, it cools down (the kinetic energy gets less so it registers a lower reading on a thermometer.)
If you consider the centre of a large mass of CO2, the radiation emitted by the molecules will equal that absorbed from other molecules. The same equality will happen with the collisions sharing out the translational energy.
The conventional idea is that the translational and vibrational energy will also reach a stable balance, and so the vibrational emissions will depend on the translational energy, ie the temperature of the gas. This leads to a state of local thermodynamic equilibrium (LTE.)
However, LTE only applies to the centre of a mass of gas. In the atmosphere, scientists are already aware that the top of the atmosphere is in non-LTE. Where they are going wrong is not realising that at the other edge, the base of the atmosphere, LTE does not apply either. Here there is a net inflow of energy: radiation, sensible and latent heat, whereas at the top of the atmosphere there is a net loss.
HTH.
Cheers, Alastair.
> CO2 does not emit blackbody radiation. It only emit lines related to it vibrational energy.
So does this modify your idea, and you no longer think the stratosphere is dark in the infrared band?
It sounds like you now do believe that in the upper stratosphere the CO2 is radiating in the infrared (as observed by the IR astronomers and the climatologists), and perhaps your argument is about how quantum mechanics works but not about what’s observed? But it also sounds like this eliminates the possibility of falsifying your idea by looking for IR radiation in the stratosphere.
Any other way to distinguish your idea from the commonly held physics? Must be able to falsify it if there’s a real difference.
Before the thread is cut, may I make a final plea for answers to queries I raised in post 204. Some but not all of the questions were answered. I would still appreciate enlightenment on the following:
In starting this debate, Ray Pierrehumbert explained that, at certain wavelengths, there is already saturation by CO2 (atmosphere optically thick and little or no outgoing OLR radiation to space from these wavelengths). However, increasing atmospheric CO2 would bring contiguous wavelengths in the wings into play such that they, too, would become optically thick. My question relates to whether modellers are considering these wing areas in isolation with respect to CO2. Is it not the case that they are already fairly blocked by water vapour such that extra CO2 would have much less greenhouse effect than would have been the case had no water vapour been present. It is my understanding that the Hitran data relate to single gases and not to greenhouse gas mixtures. Have any experiments of the type described by Raypierre been conducted anywhere with such mixtures?
One other query. CO2 is heavier than air and I understood that it was mixed in the lower atmosphere as a result of convection. Why, then, does it not form a thin layer immediately above the influence of convection? From what I’ve read it seems to be invading the stratosphere.
Sorry to drag you’all away from quantum mechanics but I’m only a simple soul looking for simple answers that I may be able to understand. Thank you in advance.
re Alastair, 360-1: I need to be picky to insure I understand your point exactly. You say,
“…if the increase in the vibrational energy and rotational energy does not change, then the kinetic temperature (measured with a thermometer) will remain the same because the translational energy has not changed.”
I assume your “increase” is what I posed in my question — the blasting of my CO2 with some IR, and not anything that happens later. Then your answer says NO, the temperature of the gas does not increase. Which implies that the kinetic energy of the vibrations and rotations DO NOT go into determining the temperature, only translation does. Is this your contention?
An anomaly: later you talk of the molecules emitting IR radiation and “cooling the gas.” Problem: IR line emission comes from vibration and rotation energy; if that doesn’t go into temperature re above, how can the gas cool when the ‘rotation’ energy is decreased (through IR radiation emission)?
2) If vibration energy transfers to translation (among others) does it then indirectly affect the temperature?
3) my understanding is that CO2 picks up rotation only from its vibrational levels. (CO2 does not have a dipole moment so can not absorb into rotation levels directly from IR, but only after vibration absorption makes the molecule a quasi dipole. (I don’t know if after the initial absorption the molecule can now absorb into rotational energy directly or not…) This would imply that rotation can transfer to translation or emission but only after going through vibration first. I think this is what you said; is it?
4) Why is it that CO2, if properly dense and reasonably heated, can not emit “continuum” blackbody radiation, but water vapor can??
5) Is Hank (362) correct? Are you changing your stance that GHG don’t re-emit? Or is it that you contend they can emit but lose their vibration/rotation energy so quickly through transfer via collision or otherwise to translation, that they don’t?
Hank Roberts (#362) wrote:
Actually Alastair (#361) wrote:
… so CO2 radiates energy gained from collisions. As for Co2 not emitting rotational, I believe that is right as well.
Finally, with respect to the difference between line thermal radiation and blackbody thermal radiation, I think they are as different as black and white. But the thing of it is that there is a continuem of grays between black and white. And as I have suggested, a similar continuem appears to exist between the two archetypal forms of thermal radiation. However, I personally still have a fair number of holes to fill in on this.
Re 357 Rod B:
Some questions as posed are difficult to answer in a few lines, so we answer hopefully related questions. Sometimes there are just too many questions and we get writers cramp.
1) At any finite temperature some fraction will be in higher states.
2) Adding net IR – heat – to a system will increase the total heat in the system by E = N h nu.
3) Energy is partitioned among the various modes if equilibruim is reached. Assuming you add energy to one mode, interaction with other atoms is usually necessary to achieve equilibrium.
4) BB radiation? If compressed enough probably. What upwells from Venus?
5) The new material made from compressed gas would have its own modes, not necessarily directly related to those of the molecules from which it was made. The energy of compressing the gas would “squeeze out” most of the lower energy motions. To bring OCO molecules close enough to allow quantum mechanical interactions says you put a lot of PV energy into it. Energy levels in closely coupled systems become properties of the group, not any individual atoms. Part of what distinguishes solids from gases is that there are different modes of motion of the atoms or electrons in the phase. Remember that all energy levels are derived from interactions between positively charged nuclei and negative electrons. Sometimes these levels are derivable from simple mechanical models like vibrating springs, sometimes not, like the electronic levels which are purely quantum mechanical.
re 361, 362 McDonald & Roberts:
I keep missing the point of your debate. I think I have it, than all of a sudden the same point comes back from the opposite direction. Could I ask if any of the following align? If not, what, or if so, what else is missing?
My understanding is that molecules which are IR absorbers in spectrophotometers interact with thermal radiation at their characteristic wavelengths, ww. This interaction can have at least three effects. One is to absorb ww photons and re-radiate (almost) ww photons. Another is to remove ww photons and increase kinetic energies. A third is to remove kinetic energy and radiate ww photons.
I’m trying to sort out which of Alastair’s statements are from his unique theory and which are from generally accepted physics, myself.
re 361, 362 McDonald & Roberts
Allan Ames (#367) wrote:
If I understand things correctly, the three effects exist and are all important. That seems to be something everyone agrees upon at this point. Once a local thermodynamic equilibrium over certain parts of the longwave spectra exists, the amount of energy being lost to kinetic energy after absorption is equal to the amount of kinetic energy being lost to reemission. And above 20 mb, the collision rate greatly exceeds the rate at which radiation is reemitted, so as an approximation, one may regard this as a local thermodynamic equilibrium.
However, as both you and Alastair have pointed out, the atmosphere isn’t strictly a local thermodynamic equilibrium even above 20 mb. I don’t know all of the details as of yet, but there are partial local thermodynamic equilibria and non-local thermodynamic equilibria – at different wavelengths. You probably know more about this part of it than I do. In any case, the deviation from a local or partial thermodynamic equilibrium is a matter of degree, and as the pressure increases, the deviation will decrease until for nearly all parts of the relevant spectra it becomes negligible at 50 km and below.
Finally, unless I am mistaken, the only real disagreement at this point is over the extent to which blackbody thermal radiation and line thermal radiation are different. I at least don’t think that is especially important. We will probably sort it out later – and learn more in the process.
Douglas Wise (#363) wrote:
Well, it will certainly be blocked where water vapor is dominant, but generally water vapor doesn’t make it much above 4 km. Interestingly, though, even at 4 km it isn’t that evenly distributed. I know of some mpgs which show this as the result of infrared sounder readings. Similar mpgs exist for aerosols and I believe even for carbon dioxide. I will have to check. However, as you point out, CO2 goes well above 4 km. We know this from both gas measurements and soundings. For example, here is an infrared image of CO2 longwave reemission at 8 km:
NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/
… so while it would be blocked below 4 km, it will absorb upwelling (and downwelling) radiation from different altitudes, and there will be reemission – principally from the far wings at lower altitudes and closer to the peaks at higher altitudes.
Douglas Wise (#363) wrote:
At a certain level, this is actually what the essay is about. But they have been repeated under different conditions with different mixtures since. And then of course there are the measurements we take of the atmosphere itself.
Douglas Wise (#363) wrote:
Good questions.
The answer seems to be due to diffusion – which is driven by thermal-kinetic motion. It is a slow process – which is part of the reason why there is such a lag between the the anthropogenic emission of carbon dioxide and the achievement of equilibrium in terms of the greenhouse effect.
Anyway, I will look up the sounder-vids a little later, but if anyone wants to find them a little sooner, I believe they are at the same site as the link.
CORRECTION to 369
I had written:
This is false.
The amount of energy being lost by absorption to kinetic energy will (typically) be less than the amount of energy being lost by kinetic energy to reemission, at least in the case of both water vapor and carbon dioxide. What is equal is the capacity to absorb or reemit. However, energy gained by convection will add kinetic energy, and on the whole, this kinetic energy is lost to radiation – which cools the atmosphere.
In the case of ozone, it is more split, but with radiation being lost to kinetic in the lower part, and less kinetic being lost to radiation above, so than on the whole ozone tends to warm the atmosphere. The reason for this difference is, I believe, due to the fact that it is absorbing ultraviolet radiation from sunlight rather than longwave thermal radiation from the earth.
Timothy, CO2 can absorb into rotation but not emit from it????
Allan (366), I appreciate the helpful explanation why a sufficiently compressed gas acn emit BB radiation. I assume it would follow that the top of the atmosphere’s radiation is line/spectral only, and not BlackBody — not near warm or dense enough. (GHG emission only; not including the atmospheric albedo and such)
I don’t accept (or understand) your statement, “…all energy levels are derived from interactions between positively charged nuclei and negative electrons.” What about straight translation energy, rotation, or even vibration, which is movement of the entire atom?
Rod B (#372) wrote:
As I understand it, since carbon dioxide is a linear molecule which is only transiently dipolar as the result of bending, and therefore it would neither be able to absorb or emit in a strictly rotational mode. However, given the fact that it is transiently dipolar, vibration may be coupled with rotation, with the combined existing in quantized energy states. Those states can both absorb and emit.
Rates of diffusion are illustrated with a nice applet here: http://chem.salve.edu/chemistry/diffusion.asp#model
I vaguely recall CO2 is well mixed after several years’ time (the peaks from seasonal growth in different regions can be tracked as they move around the world– the variations can be followed until they get diffused.
[[It is my understanding that the Hitran data relate to single gases and not to greenhouse gas mixtures. Have any experiments of the type described by Raypierre been conducted anywhere with such mixtures?]]
In brief, yes. Combustion engineers did a lot of it in the ’40s to understand what happened in various engines and industrial processes. Carl Sagan used their data in his first model of the Venus atmosphere (1960).
Hottel, H. C.; Egbert. R. B. 1941. Trans. Am. Soc. Mech. Engrs. 63, 297.
[[One other query. CO2 is heavier than air and I understood that it was mixed in the lower atmosphere as a result of convection. Why, then, does it not form a thin layer immediately above the influence of convection?]]
Interesting question. I have no idea what the answer is. Possibly it’s related to the fact that CO2 is such a small fraction of the air by volume.
Re #364 where Rod wrote “I assume your “increase” is what I posed in my question — the blasting of my CO2 with some IR, and not anything that happens later. Then your answer says NO, the temperature of the gas does not increase. Which implies that the kinetic energy of the vibrations and rotations DO NOT go into determining the temperature, only translation does. Is this your contention?”
The answer to that question is Yes. The vibrational and rotational energy do not go into determining the (Maxwellian) temperature. But the energy of vibrations and rotations are not considered ‘kinetic’ energy. They should be thought of as vibrational energy and rotational energy, or as internal energy. It is true that during a vibration the energy of an atom change from pure potential energy, when the atom reverses direction , and pure kinetic energy but that kinetic energy is not what we are considering. The kinetic energy that is measured is the translational movement of the molecule.
2) If vibration energy transfers to translation (among others) does it then indirectly affect the temperature?
Yes again. The vibrational energy and the translation energy are interchanged in collisions, and as such will reach a stable balance. Thus before your blast of radiation the molecules would have been rotationally and vibrationally excited as the result of collisions, and after the blast they would have returned to a stable state. In the new stable state, the excess excitation would have been shared with the translational energy to raise the temperature of the gas.
3) my understanding is that CO2 picks up rotation only from its vibrational levels. (CO2 does not have a dipole moment so can not absorb into rotation levels directly from IR, but only after vibration absorption makes the molecule a quasi dipole. (I don’t know if after the initial absorption the molecule can now absorb into rotational energy directly or not…) This would imply that rotation can transfer to translation or emission but only after going through vibration first. I think this is what you said; is it?
No, what I have not said is that you actually get mixtures of vibration and rotation. The reason there are so many lines making up a CO2 vibration band is because the energy level depends on the vibration and rotation taken together. So CO2 can absorb vibrotional lines but, unlike water vapour, not pure rotational lines.
Vibrational energy is greater than rotational energy, so if a molecule loses some of its vibrational energy but not all, then it can go from being vibrationally excited to rotationally excited.
4) Why is it that CO2, if properly dense and reasonably heated, can not emit “continuum” blackbody radiation, but water vapor can?
That is a problem that still puzzles scientists. It may be because H2O can form dimers, or it may be that water vapour is condensing into water which does emit BBR, then evaporating again.
5) Is Hank (362) correct? Are you changing your stance that GHG don’t re-emit? Or is it that you contend they can emit but lose their vibration/rotation energy so quickly through transfer via collision or otherwise to translation, that they don’t?
Yes, I can’t make up my mind how the spectrum seen from space is formed. Both for Mars and for the Earth, the CO2 band is at a fixed brightness temperature which is independent of the surface temperature.
But currently I think that at the base of the atmosphere CO2 does not re-emit because it it losing its energy through collisions too quickly, but at the top of he atmosphere the air is too thin and so CO2 does emit from the non-LTE region.
re 372 Rod B: This is going way beyond what we need for the current discussion, but there have been traces of the issue in other comments. Translational motion (of the center of mass in the absence of a field) is not quantized. As for the rest, please do some searches on molecular structure & bonding. Look at the bonding or valence electrons. These electrons are contributed by the atoms in the molecule and shared within the molecule. What we think of as vibrations are alternate solutions for the distribution of electrons around the atoms with energies differing by an IR quantum. The notion of mechanical analogues for electronic distributions comes from molecular mechanics (MM), which finds such analogues empirically. MM is the work horse in statistical mechanics, particularly solution properties of drugs. Spectroscopically, the mechanical analogues are useful, and even quantitative, but do not tell the whole story though they cover most of the IR where we are concerned. Look up triplet state and flouescence to see more on molecular spectroscopy.
On the matter of line broadening, note that even though molecules have no net charge or dipole, there are local variations in charge. One O might have a touch more + than the other, say, or there could be a small gradient across an O. These local charge fluctuations exist in all molecules and allow iteractions of the electrons when the atoms are close enough. Look up electric quadropole, and note that there are sextapoles, octapoles, etc, which increasingly come into play as the molecules get closer and closer.
Remember, for RC purposes, there is still a water continuum to work out.
re Tomothy Chase, various. I find Alastair’s thoughts provocative, and would like to understand well enough to agree or disagree.
As for the origin of spectral lines in CO2, the earlier references on non LTE discuss them in detail. I would note the symmetry between absorption and emission.
I have been dismissive of non-equlibrium thermo, and now confess to having had an epiphany, also known as a cessation of stupidity from the following (just read the first few sections): http://home.earthlink.net/~dckennedy/pubs/nonEqThermoRad.pdf
Allan Ames (#380) wrote:
I would recommend studying studying the science. I myself will be getting into it in more detail in the next few weeks. However, I can already see that our understanding of absorption and reemission both in the labs and the atmosphere is fairly well developed. And we have been taking measurements of longwave emissions from CO2 at mid-troposphere for some time. At this point they are becoming rather detailed.
Please see:
NASA AIRS Mid-Tropospheric (8km) Carbon Dioxide
July 2003
http://www-airs.jpl.nasa.gov/Products/CarbonDioxide/
I have been interest in nonequilibrium for a while for a number of reasons and I will be reading the article with considerable interest. However, I noticed he is dealing with entropy production principles. This shouldn’t be a problem with regions near local thermodynamic equilibria, but if you are dealing with far from equilibrium it becomes more problematic. From what I can tell, investigations in that region are still fairly tentative.
To get a sense of the state of the field, I would recommend:
Classification and discussion of macroscopic entropy production principles
arXiv:cond-mat/0604482v3 [cond-mat.stat-mech] 2 May 2007
Stijn Bruers
http://arxiv.org/PS_cache/cond-mat/pdf/0604/0604482v3.pdf
Thanks to those who replied to my queries (#363). If I may, I would like to push Timothy Chase (#370) – or anyone else – into providing further information.
As I understand it, wavelength saturation implies almost zero transmission such that almost no infra red photons of the wavelength in question can reach the top of the atmosphere and escape as OLR. (If photons of this wavelength are both absorbed and re-emitted rather than transmuted to another energy form, then I take it that the density of greenhouse gases in a vertical column is sufficient, where saturation occurs, to richochet them up and down to the extent that hardly any make it to the top).
It is suggested that, as atmospheric CO2 increases, wavelengths in the wings also become optically thick and add to the greenhouse effect. I can fully appreciate this argument in a theoretical sense and if one ignores water vapour . However, if the relevant wing areas are already saturated or nearly saturated by a combination of CO2 and water vapour, then, clearly, there will be less upside potential for extra greenhouse gas warming than would have been the case had no water vapour been present.
Timothy Chase correctly points out that atmospheric water vapour distribution is patchy. In the atmosphere over deserts, for example, one might expect that extra CO2 could have its full extra warming potential. Over oceans, however, the potential would not be maximised. Timothy states that water vapour typically remains below 4 km but that there will be CO2 above that level still capable of acting as a second blanket. This is where I get into difficulty. If there is saturation or near saturation from a greenhouse gas mixture in the first 4km of the atmosphere, it seems to me that the CO2 above would be largely unemployed (redundant). I accept that I started the previous sentence with an “if”.
The question that I would like an answer to is this: When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of saturation or near saturation by a combination of water vapour and CO2 in the overlapping wavelengths? Alternatively, in producing a formula for radiative forcing by CO2 which is applicable in the range 180-280 ppm, have they addressed the possibility that the formula would have to change as CO2 levels significantly exceed those experienced in the previous one million years and possible saturation by a combination of water vapour and CO2 kicked in?
re 381 Tomothy Chase. Thank you for the references. I can summarize my epiphany as the notion that both fermions and bosons are part of thermodynamics, but it does not go much deeper than the statement — more a matter of tolerance than understanding. It does allow me to accept that having gas and radiation at different apparent temperatures is a thermodynamically consistent state, though not one described by the Schwartzschild equation as far as I can see.
Douglas Wise — note the heat in the lower atmosphere isn’t going away just because there’s a lot of water vapor, it’s still in the system and is going to keep bouncing around until it gets transferred to the upper atmosphere.
Where there is much less water vapor, at the top of the atmosphere level where the heat can escape, there’s no question of saturation.
Try a pure poetry analogy — bubbles rising from the bottom of a pool full of water move up and out. Say the bottom of the pool fills up with mud — the bubbles rising up will move much slower while rising up through the lower few feet of mud, but once they get above the mud they still rise up to the top.
re 382 Douglas Wise. Are you familiar with LBL model at the HITRAN site or the transfer calculators? I am not sure but these may help. I hope the GCM modelers will say they would like to reproduce what the LBL models do.
http://cfa-www.harvard.edu/hitran
http://snowdog.larc.nasa.gov/
http://snowdog.larc.nasa.gov/cgi-bin/rose/flp200503/flp200503.cgi
http://geosci.uchicago.edu/~archer/cgimodels/radiation.html
re 384. Hank, I’m probably better at poetry than physics but you either haven’t really helped or Raypierre got this thread off to a false start as regards explaining saturation to laymen. I liked the poetry concerning the red M and M sweets (candies) and the two fat kids who ate the lot (saturation). Your mud analogy contradicts this. The mud merely delays the photons on their journey to space but, nevertheless, they eventually get out (not saturation). Seems to me you can’t both be right. (Raypierre said nothing came out of the end of his female experimenter’s tube.) Am I being too literal with the poetry?
re 385. Thanks Alan, you may well have given me the information I need. It’s getting late in the UK and there’s no good my looking at your links till tomorrow when the whisky loses its influence. My fear is that, even tomorrow, I may not understand what’s in them. However, I’ll try and then, if necessary, come back to you for a translation that a geriatric biologist might be able to grasp.
Douglas Wise (#382) wrote:
Saturation for a given layer of atmosphere means essentially that the emissivity at a particular wavelength is 1, or alternatively, that it is opaque to that wavelength. All the radiation is absorbed, but it is also re-emitted. However, it is re-emitted isotropically, that is, in all directions.
Eventually, all of the thermal radiation which enters the system will leave the system just as the sunlight which is absorbed at the surface gets reemitted as thermal radiation. However, the big question is how much the temperature must rise before the amount of thermal radiation leaving the system is equal to the amount of thermal radiation which is entering the system?
Infrared radiation which is re-emitted towards the surface is radiation which is not immediately escaping to space. Likewise, radiation which is absorbed by the surface will raise the temperature of the surface, increasing the amount of thermal radiation it emits. The feedback of absorption and reemission between the surface and the atmosphere will continue until the temperature of the surface has increased sufficiently for the amount of thermal radiation leaving the system to equal the amount entering the system.
A good read on this is at:
Learning from a Simple Model
https://www.realclimate.org/index.php/archives/2007/04/learning-from-a-simple-model
However, you might also want to check out the spreadsheet at:
htt://www.editgrid.com/user/timothychase/Greenhouse
In any case, since the radiation gets re-emitted and must pass through the layer of the atmosphere which is extremely dry and where carbon dioxide is dominant, the carbon dioxide can and will absorb the radiation even if it has been absorbed before, re-emitting roughly half of what it absorbs back towards the surface – just as the water vapor below does.
With line-by-line analysis, they take this into account throughout the atmosphere, although models typically analyze things in terms of perhaps fifteen layers of atmosphere. However, climate sensitivity has been most accurately determined not by the models, but by paleoclimate analysis over the past 400,000 years. It would appear to be roughly 2.8 degrees Celsius.
PS to 387
To play around with the spreadsheet, you can do “File->Export As->Excel” and change the figures once you have the spreadsheet on your computer. Like the “simple model” itself, it is extremely oversimplified, but at least it should give you some sense for how the feedback process underlying the greenhouse effect works.
Alastair, good post (378). One thing still haunts me. I will assert (and might be wrong…) that vibrational energy is kinetic. It consists of particles of mass with velocity. The fact that it oscillates should not make any difference (other than affecting the average velocity. Translation oscillates of a sort, too (hits another molecule head on, comes to a stop, shoots out in another direction). Also, it is vibration that is the primary generator of temperature in liquids and solids. What you say sounds sensible; but I just can’t resolve this (for one) quandary.
Allan (379), maybe it’s just me (so sorry if I’m slogging down the thread), but I don’t think the thrust of this thread, which I’m (very slooooowly) working toward, can be comprehended without a pretty solid and detailed knowledge of inter- and intramolecular energy absorption, emission, storage, transfer. And it’s not totally clear to me, at least.
I did not follow the line about electron distribution being the alternate description of vibration energy. The vibration is of the atoms oscillating with another atom. Is it the electrons moving with the nucleus )or not???) that your referring to? Neither did I follow “just a touch more + charge”; O can get one more proton (then it’s no longer O) or lose one electron to ionization, e.g. Is the latter a “touch more +”?
A quick glance at your reference (380) looked promising. Thanks.
Re #389
Rod, can I put it this way “When we talk about the kinetic energy of a molecule, we are referring to the the translational energy of the whole molecule, not any kinetic energy of its parts. We tend to use the terms ‘kinetic energy’ and ‘translational energy’ interchangeably.
But free molecules with translational energy only exist as gases, so the heat energy of liquids and solids is not translational energy. It is in the form of molecular and atomic vibrations.
If you google Maxwell Temperature you will find this page ‘Kinetic Energy’ http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/kintem.html where it says:
It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration.
HTH,
Cheers, Alastair.
Re #382 where Douglas asked:
When climate modellers determined climate sensitivity for a doubling of CO2, did they take into account the possibility of saturation or near saturation by a combination of water vapour and CO2 in the overlapping wavelengths?
The climate modellers are denying that saturation occurs. See Raypierre’s essay that heads this thread. They then claim that the increase in CO2 will cause a warming which will be amplified by the subsequent increase in water vapour. They do not know the amplification, so climate sensitivity is unknown. However they guess that for doubling CO2 temperatures will rise by 3 C.
FWIIW, I am saying that CO2 is saturated, but increasing CO2 will still increase global temperatures. That is because the saturated absorption happens closer to the ground and so it melts the ice at high altitudes and latitudes. The resulting change in albedo means that the surface absorbs more solar heat, and so there is a large rise in temperature, much more than the 3 C they predict.
re 389 Rod B: In the overall discussion, it is a small point that molecules are not mechanical systems, though they sometimes act the same way. Check out “quantum oscillator” and “coupled quantum oscillator” and the “correspondence principle”. It is worthy to discuss mostly to note that interactions between molecular energy state can occur in many ways, with a variety of distance dependences. We are fortunate that these can be reasonably well described with Lorentzian average curves with pressure or concentration dependent widths, but we should keep in mind that the Lorentzian is an approximation, and each interaction will have its own statistics.
re 391 Alastair: If one could run a GCM with LBL calculation in, say, 1KM intervals near the surface, with NLTE inclusion and consistent handling of energy exchanges, as would this accomplish what you wish? (For me this would be a good start, but just a start.)
Alastair McDonald (#391) wrote:
They are claiming that saturation occurs – at particular wavelengthes, but that there is both line broadening and band broadening as the result of higher temperatures and pressures – effects that are easily measured in the laboratories.
We know that it will be amplified by water vapor in large part due to the Clausius-Clapeyron relation. However, other factors are involved, in part due to amplification as the result of the albedo flip (melting ice will expose more dark soil and ocean which will increase the absorption of sunlight and thus thermal energy within the system), the distribution of the continents, etc..
We know that the climate sensitivity given the current distribution of the continents is roughly 2.8 degrees as the result of of the paleoclimate record for the past 400,000 years.
We know that CO2 is not saturated at most of the relevant wavelengths at higher altitudes as the result of laboratory experiments at different pressures and temperatures, the laws of quantum mechanics, and satellite imaging of radiation being reemitted at different altitudes which has the signature of carbon dioxide. The last of the involves our ability to measure emissions at more than 2000 different wavelengths.
For more info, including videos based on the data we are getting from the satellite measurements, please see:
AIRS – Multimedia: Videos: Animations
http://airs.jpl.nasa.gov/Multimedia/VideosAnimations/
Incidently AIRS stands for atmospheric infrared sounder – the name of the satellite which is getting the data which makes possible the imaging. In essence, CAT scans for the atmosphere, rendering visible that which is invisible to our eyes – given the limitations of human vision.
If its saturated at groundlevel, then this is precisely where CO2 will nott be effective. Moreover, absorption by water vapor at the same wavelengths will completely swap the effects of carbon dioxide. Pesky lab experiments again.
Please see:
RealClimate » A Saturated Gassy Argument
https://www.realclimate.org/index.php/archives/2007/06/a-saturated-gassy-argument/
… and the paleoclimate record says otherwise. 400,000 years of it.
Alastair, you could do much better than this.
You could learn the science rather than fantasy role-playing the unappreciated super genius. Think about it: something genuinely productive, maybe even helping us out. But what you just did is counterproductive.
Alastair writes:
> I can’t make up my mind how the spectrum seen from space is formed. Both for Mars and for the Earth,
> the CO2 band is at a fixed brightness temperature which is independent of the surface temperature.
Alastair, aren’t you here contradicting the basic idea of the whole process, that when CO2 increases, Earth warms up, and over several centuries Earth comes back to radiative equilibrium eventually by increasing the outgoing energy from the top of the atmosphere?
Or did you mean something different?
Timothy, a quibble, maybe. Do (can?) models do a line by line analysis?? The lines must number in the tens of thousands, all with different absorption constants that vary by molecule and by temperature. Then do 15 atmospheric layers, then do it in 100km by side sections, then project it with all other effects and forcings by some time interval through 50-100 years… and where most integrations have (ought?) to be done numerically. I’m no model nor supercomputer expert, but that sounds like a humongous number of flop calculations.
Alastair, I’m sorry for being a tiresome tenacious guy, and I know I’m being picky. Everything gets close but still kinda dances around the question. You say (390),…. “When we talk about the kinetic energy of a molecule, we are referring to the translational energy of the whole molecule, not any kinetic energy of its parts. We tend to use the terms ‘kinetic energy’ and ‘translational energy’ interchangeably….” (emphasis mine). This sounds like it could be for convenience of analyzing, e.g. It comes close but it does not quite say, “Only a gas’ translation kinetic energy determines temperature. (Period) With a gas we use [not “tend to”] the terms kinetic energy and translation interchangeably because they are identical.
Even your source quote (which I have seen), … “It is important to note that the average kinetic energy used here is limited to the translational kinetic energy of the molecules. That is, they are treated as point masses and no account is made of internal degrees of freedom such as molecular rotation and vibration.” precisely says (implies) “for this analysis we will only look at the translation part of kinetic energy…” It does not, again, state definitively that translation energy in a gas is the only kinetic energy that determines the temperature.
I’m beginning to be convinced that the above is in fact true. I haven’t done the math yet, but playing with 3/2kT (translation only) going to 5/2kT to 3kT to 7/2kT (added vibration/rotation energies), it’s not obvious how adding energy to virotation can increase the T. (Though it gets messy when one changes the n of nkT and also changes the masses and average velocities of the equivalent 1/2mv2). And maybe the temperature of solids ands liquids coming only from vibration I can write off as just another mystery of physics (really, not a smart-a comment). Given this, then absorbed radiation can increase the temperature of a gas only after it gets transferred from virotation to translation, either in the same molecule or another via a collision. Secondly, emission decreases the temperature of a gas only if the source energy of virotation was just transferred from some molecule’s translation. Radiation being absorbed and immediately re-emitted (with whatever probability that happens) without any intervening energy transfer does not affect the temperature of a gas what-so-ever. Is this correct?
Rod B (#396) wrote:
They were doing a trillion bits per second a few years back (2002) with the NEC Earth Simulator. I don’t know what they are doing now – but you know how computers have been advancing. They keep improving the resolution, and where they find that it makes a significant difference compared to previous calculations they improve it some more. Likewise, the algorithms are streamlined, made as efficient as they can make them – so that every flop counts. But yes, its a lot of calculations – and I believe it takes weeks to do a single simulation, although someone will hopefully correct me if I am wrong about that.
Timothy (et al) says, “We know that the climate sensitivity given the current distribution of the continents is roughly 2.8 degrees as the result of of the paleoclimate record for the past 400,000 years.” I understand this but, for what it’s worth, it’s one of the thing that makes me vaguely nervous and enhances skepticism. Sounds like measuring with an eyeball glance, marking with a paintbrush, and cutting with a diamond cutter.
PS to 398
If I remember correctly the temporal resolution is at the level of a complete calculation for every simulated minute of every simulated day. The one exception to this is in the polar regions where due to wind speeds a higher temporal resolution is required.