Guest post By Ben Herman and Roger A. Pielke Sr.
During the past several months there have been various, unpublished studies circulating around the blogosphere and elsewhere claiming that the “greenhouse effect” cannot warm the Earth’s atmosphere. We would like to briefly explain the arguments that have been put forth and why they are incorrect. Two of the primary arguments that have been used are
- By virtue of the second law of Thermodynamics, heat cannot be transferred from a colder to a warmer body, and
- Since solar energy is the basic source of all energy on Earth, if we do not change the amount of solar energy absorbed, we cannot change the effective radiating temperature of the Earth.
Both of the above statements are certainly true, but as we will show, the so-called “greenhouse theory” does not violate either of these two statements. (we use quotation marks around the words “greenhouse theory” to indicate that while this terminology has been generally adopted to explain the predicted warming with the addition of absorbing gases into the atmosphere, the actual process is quite a bit different from how a greenhouse heats).
With regards to the violation of the second law, what actually happens when absorbing gases are added to the atmosphere is that the cooling is slowed down. Equilibrium with the incoming absorbed sunlight is maintained by the emission of infrared radiation to space. When absorbing gases are added to the atmosphere, more of emitted radiation from the ground is absorbed by the atmosphere. This results in increased downward radiation toward the surface, so that the rate of escape of IR radiation to space is decreased, i.e., the rate of infrared cooling is decreased. This results in warming of the lower atmosphere and thus the second law is not violated. Thus, the warming is a result of decreased cooling rates.
Going to the second statement above, it is true that in equilibrium, if the amount of solar energy absorbed is not changed, then the amount of IR energy escaping out of the top of the atmosphere also cannot change. Therefore the effective radiating temperature of the atmosphere cannot change. But, the effective radiating temperature of the atmosphere is different from the vertical profile of temperature in the atmosphere. The effective radiating temperature is that T that will give the proper value of upward IR radiation at the top of the atmosphere such that it equals the solar radiation absorbed by the Earth-atmosphere system.
In other words, it is the temperature such that 4 pi x Sigma T4 equals pi Re2 Fso, where Re is the Earth’s radius, and Fso is the solar constant. Now, when we add more CO2, the absorption per unit distance increases, and this warms the atmosphere. But the increased absorption also means that less radiation from lower, warmer levels of the atmosphere can escape to space. Thus, more of the escaping IR radiation originates from higher, cooler levels of the atmosphere. Thus, the same effective radiating temperature can exist, but the atmospheric column has warmed.
These arguments, of course, do not take into account feedbacks which will kick in as soon as a warming (or cooling) begins.
The bottom line here is that when you add IR absorbing gases to the atmosphere, you slow down the loss of energy from the ground and the ground must warm up. The rest of the processes, including convection, conduction, feedbacks, etc. are too complicated to discuss here and are not completely understood anyway. But the radiational forcing due to the addition of greenhouse gases must result in a warming contribution to the atmosphere. By itself, this will not result in a change of the effective radiation temperature of the atmosphere, but it will result in changes in the vertical profile of temperature.
The so-called “greenhouse effect” is real. The question is how much will this effect be, and this is not a simple question. There are also questions being raised as to the very sign of some of the larger feedbacks to add to the confusion. Our purpose here was to merely point out that the addition of absorbing gases into the atmosphere must result in warming, contrary to some research currently circulating that says to the contrary.
For those that might still question this conclusion, consider taking away the atmosphere from the Earth, but change nothing else, i.e., keep the solar albedo the same (the lack of clouds would of course change this), and calculate the equilibrium temperature of the Earth’s surface. If you’ve done your arithmetic correctly, you should have come up with something like 255 K. But with the atmosphere, it is about 288 K, 33 degrees warmer. This is the greenhouse effect of the atmosphere.
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Jim D says:
“Smokey, some people here were questioning whether CO2 effects can even be seen or measured, which was amazing to me.”
OK. Simply state the climate sensitivity to CO2. It’s my turn to be amazed. ☺
I won’t be too hard on you. Just make it to one decimal place.
The sensitivity number is key to the entire debate. Provide it, or admit you don’t know what it is.
This is where the entire alarmist contingent collapses into disarray.
anna v says:
July 28, 2010 at 7:13 am
Somebody check me, because I think that even in the naive picture of a single photon hitting a single CO2 ( or H2O) the assumption that 50% of redistributed energy goes outwards and 50% downwells violates conservation of momentum.
The photon gives up energy equal to h*nu and transmits momentum by h*nu/c. Conservation of momentum means that the total of the redistributed energy in thermal photons/ collisions will have also a component of that momentum, which is pointing up and out. Otherwise there would be radiation pressure pushing all those CO2 and H2O molecules up and away.
Since the number of molecules is limited but the number of long wave ground radiation photons unlimited we would end having all H2O and Co2 in the stratosphere :). A bit tongue in cheek but am waiting for corrections.
You appreciation of the paradox is correct, Anna. The magnitude of the loss of energy during photon-molecules collisions is given by their total absorptivity and total emissivity. The redistribution of the energy proposed by the AGW idea means the creation of wave energy (and consequently of force and work) from nothingness. Besides, the AGW idea makes an imaginary molecule that is a perfect absorber-emitter, if as it was a blackbody, which is not the situation of the carbon dioxide, which is a weak absorber and a weak emitter.
I don’t know if you have read that, when referring to the carbon dioxide absorptivity, AGW proponents always say “[the carbon dioxide]… which is a strong absorber at 14 μm waveband…” The assertion is incorrect. In physics, we always avoid to mention the term “strong” because it gives the impression that the subject we are talking on is exerting its properties at a maximum level of the scale. Rephrasing the AGW assertion, physicists would say, “[the carbon dioxide]… exhibits its maximum absorptive power at 14 μm…” The difference between the correct phrase (physicists’ one) and the twisted one (AGW’s one) is from Earth to Pluto…
Nasif Nahle says:
July 28, 2010 at 8:43 am
Dear John,
The example about inflating a balloon by deflation is an analogy . . . . . . [edit]
————
Nasif,
Appreciate your reply. It will take me some time to think about it. Thanks.
John
I believe you are correct in that light pressure from upwelling radiation must push the molecule out towards space but it isn’t much of a push compared to the force of gravity pulling back in the opposite direction. I’m not sure that holds true for hydrogen and helium atoms at the top of the atmosphere though but certainly for heavier molecules.
But now that you bring it up I believe that’s why it takes the best laser interferometer and digital signal processor money can buy to dig the 15um signature spectral line for downwelling radiation from CO2 molecules out of the continuous blackbody spectrum. In a cold dense gas the molecules are so close together that when it gets a kick in the butt from a photon in its absorption band it causes an almost instant collision with another molecule instead of raising an electron into higher orbit which decays with the reemission of another 15um photon. The collision instead raises the blackbody temperature of the gas and a portion of that that downwells in continuous blackbody spectrum with the peak power frequency set only by the temperature.
Part of the blackbody downwell will of course also do some molecular butt kicking of its own but those will be pushing molecules back towards the earth not away from it.
899 (July 27, 2010 at 10:44 pm):
Not withstanding the definition in the American Heritage English Dictionary which you cite, the “projection” of the IPCC models is a mathematical function that maps the time to the global average temperature anomaly. In contrast, a “prediction” is an extrapolation to the outcome of a statistical event. The complete set of these events form the statistical population for the testing of the associated model. A population of this kind does not exist for the IPCC’s models. That it does not exist has the significances that the IPCC’s models: a) are not falsifiable and b) cannot be statistically validated.
At http://icecap.us/images/uploads/SPINNING_THE_CLIMATE08.pdf, the IPCC expert reviewer Vincent Gray reports that when he raised the issue of how the models could be validated, the IPCC failed to answer his question. Instead, they changed the language of the assessment reports such that “validation” was changed to “evaluation” and “prediction” was changed to “projection.”
In the “evaluation” of a model, the computed global average temperature anomaly is compared to the measured global average temperature anomaly. This comparison supports computation of the error. It does not support falsifiability of the model. Few people with interests in climatology seem to understand that an ability to compute the error is not synonymous with the falsifiability of the model.
Cordially,
Terry Oldberg
I still say trying to figure this out from QM view is a waste of time. We know that insulators insulate. The fiberglass batting in my walls and attic block every last bit of infrared radiation from crossing the barrier. What it does is causes a conversion of radiative power into a higher temperature on the impinged surface which must then cross the barrier by conduction alone before it can radiate out the opposite side. The conductive transfer of energy is slow as molasses in January through 9 inches of fluffy fiberglass.
CO2 does exactly the same thing except it only blocks a narrow band of radiated power.
Y’all are making a simple thing so damn complicated no one can understand it. Stop.
Let me amend that. My fiberglass insulation has a layer of craft paper that forms a vapor barrier to block convective energy transport. CO2 doesn’t block convection at all which why it’s pretty silly to call it a greenhouse effect as greenhousess work by blocking convection. Apt terms would have been “the insulation effect” and “insulation gases”.
Dave Springer says:
July 28, 2010 at 10:34 am
[–snip for brevity–] I believe you are correct in that light pressure from upwelling radiation must push the molecule out towards space but it isn’t much of a push compared to the force of gravity pulling back in the opposite direction. I’m not sure that holds true for hydrogen and helium atoms at the top of the atmosphere though but certainly for heavier molecules. [–snip rest–]
I think you might be incorrect in thinking that ‘upwelling’ happens.
More it is —I will think— that the energy is passed on to adjacent atoms/molecules, rather than the individuals themselves actually moving.
anna v
I very much sympathize with your comments, I think that the entire construction of the Greenhouse Effect rests on a mish-mash of concepts from classical thermodynamics, statistical mechanics and other fields. However, I think it would be valuable to scrutinize the GE models in the simplests possible way and demonstrate the paradoxes with reference to the axioms.
First and foremost, the word “equilibrium” is used extensively in the debate without any attention payed to the fact that, by definition, in statistical mechanics the temperature is the same everywhere within an ensemble at equilibrium. This is quite important since the celebrated Stefan Bolzmanns law ultimately rests on stat. mech. This might however be challenged since the incoming radiation is of another quality than the outgoing radiation, however, the 2nd law is quite commanding on this point: Heat flows spontaneously from higher to lower temperatures. In any case, the way in which we treat the incoming solar radiation ultimately defines which kind of thermodynamic system we are dealing with: Is it an equilibrium system that should be treated by stat. mech? Or is it a non-equilibrium system for which other thermodynamic laws apply? In the latter case it would be very strange to imagine the atmosphere being isothermal in the absence of GHG (see Spencers articles), since it would correspond to an infinite heat conductivity of the atmosphere.
Secondly. The radiation diagrams used in the popular descriptions of the GE cannot possibly depict heat flows (if they depict anything). This is because the radiation depicted has a specified direction and therefore it should be treated as a radiation pressure. However, I think climatologists would have a very hard time to harmonize the ENERGY- and MOMENTUM-balance equations that these diagrams imply.
anna v says:
July 28, 2010 at 7:13 am
Somebody check me, because I think that even in the naive picture of a single photon hitting a single CO2 ( or H2O) the assumption that 50% of redistributed energy goes outwards and 50% downwells violates conservation of momentum.
The photon gives up energy equal to h*nu and transmits momentum by h*nu/c. Conservation of momentum means that the total of the redistributed energy in thermal photons/ collisions will have also a component of that momentum , which is pointing up and out. Otherwise there would be radiation pressure pushing all those CO2 and H2O molecules up and away.
Incorrect Anna because you seem to be regarding the receiving molecule as a solid ball. In the case of a molecule of CO2 absorbing at ~15μm the bond has been excited to vibrate, consequently the atoms are vibrating faster, i.e. the momentum has increased. The translational momentum is not effected. When the molecule emits a photon the vibrational momentum of the molecule drops consistent with the emission of a photon with momentum. Most of the absorbed energy is given up to surrounding molecules via collisions in the lower atmosphere so the total translational momentum of the surrounding molecules is increased
(i.e. temperature goes up). In macroscopic terms it’s like setting a spring vibrating in a tray full of marbles, the marbles will acquire momentum at the expense of the spring.
@899
There are at least three kinds of ways that the ground cools off at night. Conductive, convective, and radiative.
Air is a poor conductor of heat due mostly to having very little heat capacity. A vacuum is the ultimate in poor conductors as it has zero heat capacity. Storm windows still commonly used in temperate climates operate on that principle. They go on in the fall and come off in the spring. By adding a second layer of glass with a few inches of air between them significantly reduces conductive heat loss through a single pane. The ultimate is newer windows with a vacuum between two panes which virtually eliminates all conductive heat loss. Conductive heat transfer is static in that no atoms or molecules are displaced. The just vibrate in place against each other and the energy is mechanically transmitted.
Convective transfer is where atoms and molecules physically move from one place to another. A single pane of glass in a window stops all convective heat transfer. Opening the window more or less lets you adjust convective heat transfer to suit your needs. A fan will make it happen even faster.
Radiative transfer is the only one that doesn’t require a medium. Light at most frequencies zip right through a double pane vacuum window. Curtains across the window allow you to adjust radiative heat transfer.
Another largely overlooked heat transfer path is a heat pump which is how air conditioners work. A fluid, preferably one with high latent (non-sensible) heat capacity and easily compressible (like freon and water vapor) is compressed at one location which makes it give up latent heat to environment (typically by conduction and convection in a radiator with a fan forcing air across it), is piped to another location, and there expanded which sucks in heat from the environment through conduction and convection (with the same aid as the compression side).
The heat pump mechanism gets short shrift in global warming. There’s a naturally occuring heat pump in the water cycle. Water at the surface is expanded (evaporation) which sucks in heat from the environment, rises high in the air, and is there compressed (condensation) where it gives up the latent heat far from the source. In thunderstorms it works so well that water evaporated on the ground falls back to earth as ice instead of rain even on days when surface air temperature is 100F+.
Anyhow, upwelling definitely happens. Confirm with just about any of these passive infrared night vision products:
http://www.infrared1.com/
which convert infrared radiation “welling” from anything you point them at to a visible visible wavelength that your eye can see. These can also be found in homes commonly in the form of intrusion detection where the infrared viewing device is tuned to detect infrared blackbody radiation at and slightly around 98.6F (body temperature).
There is no need to be a physicist to understand how these things work. Virtually all warming and cooling modes of the earth’s surface can be explained in terms relating them to the common household items.
The unnecessary complications in this thread have gone far over the top – the lingua franca has become Pedantry Plus Plus.
Phil. says:
July 28, 2010 at 1:32 pm
Incorrect Anna because you seem to be regarding the receiving molecule as a solid ball. In the case of a molecule of CO2 absorbing at ~15μm the bond has been excited to vibrate, consequently the atoms are vibrating faster, i.e. the momentum has increased. The translational momentum is not effected. When the molecule emits a photon the vibrational momentum of the molecule drops consistent with the emission of a photon with momentum. Most of the absorbed energy is given up to surrounding molecules via collisions in the lower atmosphere so the total translational momentum of the surrounding molecules is increased
(i.e. temperature goes up). In macroscopic terms it’s like setting a spring vibrating in a tray full of marbles, the marbles will acquire momentum at the expense of the spring.
Anna is correct because if the molecule of CO2 is not a perfect absorbent, as you are mentioning, the “bounced” EM force makes the Prad in the same direction of the photon stream doubles, so the force of the electromagnetic radiation is applied as 2U, instead U alone. The work exerted by the wave colliding with a molecule of CO2 would double in the same direction of the photon stream, that is, to the upper layer of the Earth’s atmosphere.
Dave Springer says:
July 28, 2010 at 2:45 pm
Well said Dave. I’m with you 100%.
Too much information with too little knowledge is the problem for many contributors here and in the wider climate community 🙂
The hydrological cycle is the key. If it can change in response to other changes (solar, oceanic AND composition of the air) as it clearly does then AGW is dead as a Dodo and Miskolczi’s findings are all the proof one needs.
Stephen Wilde says:
July 28, 2010 at 3:30 pm
Dave Springer says:
July 28, 2010 at 2:45 pm
Well said Dave. I’m with you 100%.
Too much information with too little knowledge is the problem for many contributors here and in the wider climate community 🙂
The hydrological cycle is the key. If it can change in response to other changes (solar, oceanic AND composition of the air) as it clearly does then AGW is dead as a Dodo and Miskolczi’s findings are all the proof one needs.
And I’m with you. However, it would be unnecessary to talk in those (QM) terms if AGW proponents would not have resorted to those terms.
By the way, the heat transfer by conduction-convection at the boundary layer surface-atmosphere is the fundamental combination of mechanisms that takes the energy from the surface for being convected away by the atmosphere. For example, two days ago, the net energy exchange between the surface and the atmosphere by conduction-convection was 200 J.
Therefore, Sun, the water vapor, the oceans, the land and the subsurface materials are the systems and subsystems that maintain our Earth nicely warm.
To me, the case is closed.
Phil. says:
July 28, 2010 at 1:32 pm
Net momentum is the vector sum of all linear moments in a closed system. The net momentum of a vibration is precisely zero. Try again to find the missing momentum.
Photons carry a linear moment. Conservation of linear momentum is a fundamental law of nature. Momentum is not conserved unless the molecule as a whole has a net momentum increase equal to that carried by the supposedly annihilated photon. Either that or the photon lives on with no change in net momentum.
Light sails are composed of atoms, none of which are “solid little balls” as you put it. How do photons manage to transfer momentum to the light sail?
Since the photon imparts a change in momentum to the sail how is this reflected (pun intended) in the net momentum of the photon given that by law the velocity of photon cannot change and by law the momentum must be conserved?
Hopefully after thinking these things through you’ll come to realize that the photon isn’t absorbed and unless the molecule was put in linear motion there was no exchange of momentum that must be conserved in the closed system comprising molecule and photon. The photon is reflected, it imparts kinetic energy to the molecule (the vibration) and, since energy has to be conserved too, for the books to balance and in light of the fact that the photon’s velocity cannot change, the m part of mv must change by an amount corresponding to mass/energy equivalency (E=MC2). If measured the photon will have a lowered frequency.
Ya gotta make the conservation books balance somehow, Phil. No if ands or buts about it without violating fundamental physical laws or inventing new ones.
P.S. @Phil
You should have taken to heart Anna’s admonition about trying to treat EMR as quantum units in contexts where it is properly treated as waves obeying the laws of electrodynamics which are analog not quantum. Wave/particle duality. Write that down.
Phil. says:
July 28, 2010 at 1:32 pm
anna v says:
July 28, 2010 at 7:13 am
Somebody check me, because I think that even in the naive picture of a single photon hitting a single CO2 ( or H2O) the assumption that 50% of redistributed energy goes outwards and 50% downwells violates conservation of momentum.
The photon gives up energy equal to h*nu and transmits momentum by h*nu/c. Conservation of momentum means that the total of the redistributed energy in thermal photons/ collisions will have also a component of that momentum , which is pointing up and out. Otherwise there would be radiation pressure pushing all those CO2 and H2O molecules up and away.
Incorrect Anna because you seem to be regarding the receiving molecule as a solid ball. In the case of a molecule of CO2 absorbing at ~15μm the bond has been excited to vibrate, consequently the atoms are vibrating faster, i.e. the momentum has increased. The translational momentum is not effected. When the molecule emits a photon the vibrational momentum of the molecule drops consistent with the emission of a photon with momentum. Most of the absorbed energy is given up to surrounding molecules via collisions in the lower atmosphere so the total translational momentum of the surrounding molecules is increased
(i.e. temperature goes up). In macroscopic terms it’s like setting a spring vibrating in a tray full of marbles, the marbles will acquire momentum at the expense of the spring.
I believe it is incorrect to say that the momentum (translational) of a molecule made up of multiple atoms is a function of the rate of vibration of those atoms. Isn’t it true that in a reference frame in which the center of mass of a multi-atom molecule is stationary, the momentum of the molecule is zero independent of either the number of vibrations per unit time or the amplitude of those vibrations? If such a molecule “absorbs” a photon, then conservation of energy says that the energy prior to and after the collision must be conserved (some, probably most, of this energy will be manifested in vibrations, but some will be manifested as translational motion of the molecule’s center of mass), and conservation of momentum says the the momentum (both translational and rotational) must be conserved. Since in the reference frame at rest with respect to the center of mass of a molecule prior to a collision with a photon the photon has momentum, after the collision/absorption of the photon, the center of mass of the molecule will have motion in that reference frame. Vibrations, no matter how fast, of the atoms about the center of mass do not “represent/contain/or-whateever-other-verb you want to use” this momentum.
I believe your string/marble analogy is flawed. A vibrating string will when striking a marble cause the marble to move away from the string; but the string will be displaced in the opposite direction in such a way as to conserve momentum of the string/marble system. If the ends of the string are rigidly attached to the tray, this momentum will be transfered to the tray and the tray will move in the direction opposite to the marble’s motion.
Rule of thumb. If the wavelength of the EMR stretches across a large number of number of molecules then classical electrodynamics apply. Trying make sense of that context in quantum dynamics or quantum mechanics is an exercise in futility (although it’s really fun to try). I’m not going to attempt to express 15 micrometers in an exact number of air molecule widths but it’s safe to say it’s a large number of them in any air pressure much short of a perfect vacuum.
near the width of the molecule in question quantum dynamics apply. greater than the width of the molecule in question classical electrodynamics apply not quantum dynamics. I’m not going to attempt to compute an exact wavelength
The debate has expanded beyond making individual replies.
1. I see that the idea that surface downward radiation has an emission spectrum is somehow controversial here despite the fact that this spectrum is regularly measured, and, yes, CO2 lines are visible. I know that for anti-AGWers admitting this leads to a slippery slope where you have to admit CO2 does something, so I can see why people resist even this basic fact, or would prefer not to know.
2. How is radiation modeled? If you are at some level in the atmosphere, you can integrate downwards to get the upward flux, for example. What do you integrate? Simply put, for each spectral line due to CO2 or H2O you find its upward flux from the temperature profile and corresponding gas concentration profile as you integrate down, and eventually you may reach a level where the line saturates, or you reach the ground which emits at that wavelength according to its temperature. Once you have upward and downward radiative flux profiles at each level for all wavelengths, the flux convergence and atmospheric heat capacity give the radiative heating rate via the normal rules of thermodynamics. This is a physics model that works and can be verified by measuring spectra at various levels. Look up line-by-line radiative transfer models for more information on this. There is a pressure-broadening (or collision) effect that marginally blurs the lines, but no more than that, and can be accounted for too.
3. How does terrestrial radiation warm the atmosphere? Several people have already explained this, and I just repeat them. The IR photon is absorbed by a CO2 (or H2O) molecule, and the energy is transferred to other atmospheric molecules by collision, which is therefore seen as a temperature change due to this increase in energy that would not have been there had the CO2 not absorbed that photon. With no GHGs, those IR photons just go through and impart no energy to the O2 and N2. With more GHGs, more energy is absorbed and shared with the O2 and N2. Sort of obvious, really.
4. Climate sensitivity to CO2? I say 2.8 degrees per CO2 doubling, but that is just my opinion. It can’t be proven now, and we can only wait for the next few decades to prove it by measurement. This corresponds to 0.2 C for each 20 ppm, which is the current rate per decade (see World Climate Widget), so I just go with that.
Phil. says:
July 28, 2010 at 1:32 pm
Incorrect Anna because you seem to be regarding the receiving molecule as a solid ball.
No, I am not. I am treating it quantum mechanically. A molecule absorbs the photon by the excitation of a line. It gains an energy of h*nu and a momentum of h*nu/c.
Conservation laws hold in quantum mechanics and yes, the translational motion of the molecule will get a push in the direction the photon was traveling.
When it emits photons, it looses translational energy of h*nu and momentum from the direction of emission of the photon h*nu/c.
If it gives up the energy in collisions, the same holds, momentum is a conserved constant of motion in all physical theories/systems.
@jimd
I already conceded at least for the sake of argument that water vapor accounts for 2/3 of down emission at the south pole while 1/3 is due to CO2.
Time to move on. South pole has lowest absolute humidity of anywhere on the planet at 0.03%. Temperate climes go 1% to 5% and oceans tend to the higher end of that range.
If almost no water vapor produces twice the effect of CO2 then 100 times that much water vapor is going to make CO2 insignificant comparison.
The tail doesn’t wag the dog, Phil. I think Max Planck told me that when I met him in a previous life.
As always, thanks for playing. There’s a lovely consolation prize waiting as you exit stage left – an autographed picture of Niel’s Bohr sleeping in pajamas covered in Bohr diagrams of carbon dioxide molecules. We thought that especially apt given your undying interest in them.
@jimd
One last question for you and and apology for calling you phil in my last missive.
I focused the interferometers in my orbital sockets upward at the bottom of a leaf and noticed bright green spectral line. Is that because the leaf is a source of green photons or is it merely a reflection of that portion of the continuous blackbody spectrum illuminating it from the ground beneath?
Please try to give your answer in domains mixed willy nilly between quantum and classic domains. They’re much more fun to read that way!
RE: Phil.: (July 24, 2010 at 9:38 am) “Based on the theory you’d expect a transition from log(CO2) to sqrt(CO2) at higher concentration.”
My MODTRAN data does seem to have a sqrt(CO2) form as it rises off the log curve, however I have found that the seventh power of the Log of the zero-corrected CO2 concentration seems to best model the transition region *below* 100,000 ppm. My Excel Solver optimized calculations seem to match the results returned by the MODTRAN tool with less than 0.05 deg K error for MODTRAN CO2 concentration settings from 0.0 to 100,000 ppm.
Dave Springer says:
July 28, 2010 at 11:15 pm
Dave,
A quick run of some numbers yielded the following: 1976 std atmosphere has 64 w/m^2 due to h2o presence and 29 w/m^2 due to co2 presence out of 107 w/m^2 ghg contribution.
Substituting some crude h2o content for poles at winter (1/5 that of the 0.77% of the Std Atm) and tropics (3.3x that of the Std Atm), the results came out with 46w/m^2 due to h2o and 80w/m^2 in the tropics. There was not much change in the relative effect of the co2 doubling in w/m^2, perhaps 10% more at the pole. Also, the fractions chosen were based upon the surface concentration differences rather than the entire column content and are probably significantly overstated. Another over simplification was that the numbers were left at the mean surface temperature which does affect the spectral content.
The net results were that at the poles a co2 doubling without feedback nets a 4 w/m^2 out of 88 w/m^2 total ghg forcing and cloud forcings of ???? while at the tropics the ghg forcing was 122w/m^2 compared with the 1976 standard atm 107w/m^2 due to ghgs and 43 w/m^2 due to cloud effects. The co2 doubling does have more effect at the poles 4 out of 88 vs 4 out of 107 when it comes to fractions and that might even be versus 4 out of 150 if there are practically no clouds at the pole.
While there’s nothing much guaranteed here, it is interesting to see that the co2 is stronger but not dominant at the poles with reduced h2o and that the h2o is still dominant, despite the over 2 halvings reduction. I’ve noticed that many people’s minds switch over to linear mode in their interpretation even though like co2, we’re still dealing with an exponential attenuation with h2o.
Terry Oldberg says:
And yet, your claiming that I am making an ad hominem argument is every bit as much of an ad hominem argument as my claiming that you are employing a “red herring”. (See, everyone can play these fun logic games!)
The models make predictions for various different forcing scenarios…and one of these scenarios will come the closest to being the actual forcing scenario we follow. Therefore, the models under this forcing scenario can be compared to the actual global temperature response. Alternatively, the models can be re-run using the actual measured greenhouse gas concentrations. Ergo, the models can be tested…and their conclusions falsified or not falsified. (Of course, like any complex scientific problem, it isn’t really a black-and-white answer of whether the models are correct or not, but rather an issue of how well they do.)
One can also look at many other things, such as the observed vs predicted pattern of the warming (both in the horizontal and the vertical), etc., etc.
The fact that virtually all reputable scientific organizations on the planet disagree with you ought to give you a hint that you have in fact gone astray. Yes, I know that “appeals to authority” do not logically show something to be true or false. However, they are a good way of gauging the likelihood that your argument is good or misguided…what one might call “a reality check”. And, since your logical argument is so simple, do you really think all of these scientific authorities could have missed something so obvious? Of course, above I have told you some reasons why it is misguided; the fact that scientific authorities don’t subscribe to it is further evidence of its misguidedness.