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During the past several months there have been various, unpublished studies circulating around the blogosphere and elsewhere claiming that the “greenhouse effect” cannot warm the Earth’s atmosphere.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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Joel, to expand on my comment:
The point you are missing is that all that occurs is that the extra IR fails to warm the oceans because evaporation via latent heats of evaporation and condensation transports it quickly upwards and away.
Since that has always happened and is a constant process there is no disturbance of the normal temperature profile of the atmosphere so your comment is an irrelevance.
That feature of the natural world is already reflected in the observations.
A little more CO2 in the air just results in a miniscule unmeasurable change in the speed of upward energy transport and a miniscule unmeasurable adjustment in the air circulation systems.
That also complies with Miscolski’s finding of a constant optical depth for the atmosphere despite the increase in CO2 over the past 61 years.
Joel Shore:
You say “Terry Oldberg: Just when I thought that the claims being made by the posters in this thread could not get any more bizarrre, you prove me wrong! My hat is off to you!” This is an example of an “ad hominem argument.” An ad hominem argument is one that makes one’s opponent the issue rather than the issue that is under discussion. An ad hominem argument is logically illegitimate because it is irrelevant.
The issue under discussion is whether climatology contains the foundational error that I have described. It seems to me that it is beyond dispute that climatology contains this error. If it contains this error then climatology is not a science. Do you have anything to say that relates to this issue?
Terry Oldberg
Andrew W says:
July 25, 2010 at 8:42 pm
Reed Coray, Joel Shore was I think talking about two objects held at a constant temperature by internal heat sources, by bringing the two objects close together each receives additional energy from the other, so the temperature of both objects rises.
Andrew–that’s a good point. However, in the case that each object in isolation is maintained at a constant temperature by an internal heat source, when brought into proximity the question becomes “what is the energy source that is making the temperature of the warmer of the two objects increase?” Some could argue it is the heat being generated internal to the warmer object that is causing the temperature of the warmer object to rise. Without this heat source, won’t the temperature of the warmer object decrease even in the presence of the cooler object. Said another way, suppose that at the instant the two objects are brought into proximity, the warmer object heat source is turned off. Do you still believe the temperature of the warmer object will rise? I don’t see how it can.
On the other hand, if at the instant the two objects are brought into proximity the cooler object heat source is turned off, then provided the temperature of the cooler object is higher than the “background temperature”, the instantenous rate of cooling of the warmer object will decrease because the background temperature the warmer object “sees” in part has increased. This decreased rate of cooling will result in a rise in an temperature of the warmer object (only if its internal rate of energy supply is kept constant). Thus in the case of termination of the warmer object internal energy input, the warmer object does NOT increase in temperature when the cooler object is brought in proximity; and in the case of termination of the cooler object internal energy source, the warmer object increases temperature only as long as its internal energy source continues. Thus it seems to me the source of the warmer object temperature rise is the warmer object’s internal energy source.
BTW. I believe the presence of an internal energy source nullifies application of the Clausius formulation of the sencond law of thermodynamics: “It is impossible to construct a perfect refrigerator”–i.e., it is impossible to move energy from a reservoir at lower temperature to a reservoir at higher temperature without affecting the environment in any other way because I believe the environemnt is being “affected” in some other way–namely a source of energy is being supplied to the system.
Terry Oldberg says:
July 25, 2010 at 2:30 pm
I agree with your analysis, though I presented it differently in a post in this thread http://wattsupwiththat.com/2010/07/23/quantifying-the-greenhouse-effect/#comment-438432
The basic problem is that in climatology different physics frameworks are being used indiscriminately, viz thermodynamics, classical statistical mechanics, and quantum statistical mechanics, leading to paradoxes. This is probably due to the fact that people have not studied physics in enough depth to be able to discriminate, and use the terminologies ad hoc and at taste.
Two additional comments: First, I know it’s boorish for the recipient of largess to ask the doner of that largess for even more; but Anthony is there any reasonable way for WUWT to prvoide a “comment preview capability?” And lacking that, is there any way to increase the size of the “comment section?”
Second, as an avid reader of and occaisional commenter to this blog, I’d like to welcome Ferenc Miskolczi. I have printed out your 2007 paper “Greenhouse effect in semi-transparent planetary atmospheres” and plan to study it. One question I have regarding the virial theorem. As I understand it, when applied to the potential/kinetic energy of physical systems, the model employed is the classic Netwonian model of potential and kinetic energy. Is the theorem still valid in the special relativistic sense when relative to an inertial reference frame the “apparent mass” of an object is a function of its velocity so that there is a coupling between kinetic energy, 0.5*m*v^2, and gravitational potential energy which is proportional to the product of the masses of two objects divided by the square of the distance between them?
Phil. says:
July 25, 2010 at 12:06 pm
Look up radiative shielding on thermocouples on Google you’ll find plenty of examples.
I would suggest you read a basic textbook on radiational heat transfer since your knowledge is sadly lacking.
I went there, Phil, and the few places which describe the matter, show you to be —once again— both a charlatan and snake oil salesman.
The facts are just these: The shields were used to prevent undue influence of the surrounding atmosphere. That’s it, and no thing else.
Since the thermocouple itself is a temperature sensing device, the discussion regarding a colder body causing a warmer body to increase in temperature and the discussion of shielding of thermocouples to prevent undue influence are entire unconnected.
Why does everyone keep banging on about thin shells and plates.
Imagine a planet with little or no atmosphere which is heated by it’s own sun. It receives 250 watts/m2. The planet will warm until it has an average temperature of ~258K when it will radiate 250 w/m2 back to space. So we have
Incoming solar radiation = Outgoing IR radiation, i.e. thermal equilibrium.
Now give the planet a layer of atmosphere which includes IR absorbing gases (ghgs). The gases absorb 50 w/m2 of the planets outgoing IR so now we have the situation where 200 watts/m2 are emitted directly through the ‘IR window’ but 50 w/m2 is absorbed and emitted by the ghgs.
The planet-atmosphere is still in equilibrium, i.e. it receives 250 w/m2 and gets rid of 250 w/m2. But……
…. the atmosphere radiates in both (every) directions – up and down – so while 50 w/m2 is radiated out to space 50 w/m2 is radiated back to the surface. The surface of the planet now receives 300 watts/m2 (250 from it’s sun + 50 from the atmosphere) which means it will warm up to ~270K.
However you want to word it, the ghgs make the planet warmer than it would otherwise be without them. Whether you choose to think of it as reduced cooling (or reduced radiation loss) or heating – it’s your choice.
Ok – this example is massively oversimplified but the general concept is correct.
Our earth receives ~235 w/m2 from the sun but the earth’s surface temperature is ~288K (or 15 deg C). This equates to an energy emission level of ~390 w/m2. We must be getting the extra energy from somewhere and, since other heat transport processes such as convection and latent heat tend to remove heat from the surface, it must be coming from atmospheric radiation.
A lot of posters seem to think that the extra heat is somehow magically appearing from thin air. It’s not – we have a constant enrgy source (the sun). I note Roy Spencer had to make this point at least half a dozen times. If the energy gained from the sun is greater than the (net) energy being lost by radiation we will warm up. I’ve just boiled a kettle of water to make a cup of tea (we’re terribly civilised here in the UK). After 1 minute the rate of energy input was the same as it was after 10 seconds but, because the energy gain was far greater than any energy loss, the water heated up quickly.
Hi John, a couple of obs.
1) If the atmosphere absorbed 50W, it would re-emit a bit more than 25 up and a bit less than 25 down, due to curvature of the planet. How do you get your 50 down 50 up?
2) The GHG’s wouldn’t only absorb IR emitted by the planet, but some of the incoming IR from the sun too. That would reduce the amount of solar radiation arriving at the surface wouldn’t it?
Re: My comment
John Finn says:
July 26, 2010 at 2:25 am
I’ve noticed a slight error in my reasoning/numbers in the above post. I’ll leave it as an exercise for anyone who is till reading this thread.
However the general result remains correct.
Heh, beat you to it.
tallbloke says:
July 26, 2010 at 2:36 am
Hi John, a couple of obs.
1) If the atmosphere absorbed 50W, it would re-emit a bit more than 25 up and a bit less than 25 down, due to curvature of the planet. How do you get your 50 down 50 up?
You were quick off the mark. I noticed the error pretty much as soon as I clicked the “post comment” button. That’s what comes of trying to use easy numbers. There is also a more subtle problem in that the amount of atmospheric absorption is a function of surface emission which means that using a constant absorption figure of 50 or 100 watts/m2 is a bit silly. Basically I over-simplified the process. All in all not my best effort. I think I’ll go back to bed.
2) The GHG’s wouldn’t only absorb IR emitted by the planet, but some of the incoming IR from the sun too. That would reduce the amount of solar radiation arriving at the surface wouldn’t it?
Although this is a factor I think it’s reasonable to ignore it in this illustration.
tallbloke says:
July 26, 2010 at 3:15 am
Heh, beat you to it.
Well done. You’ve won 10000 quatloos which you are free to spend at Lucia’s blackboard blog. Be warned, though, the currency may now be out of circulation now that betting on the monthly UAH temperature anomaly seems to have stopped.
899 says:
July 25, 2010 at 8:05 pm
“So, before you start blowing snot in my direction, you ought get your own crap in one sack!!”
My post was meant as a support to you, 899. So the “snot” was meant to go in the CAGW direction. Sorry if some of it ended up in your lap!
hehe.
As others have mentioned; Those two elements; If both has current running throught them, and is in effect two separate active heating elements, and you bring them toghether, all you get is a bigger oven. All you have proven, is that if you turn on more and more heating elements in an oven, it gets hotter.
This setup is in my opinion irrelevant to the real “problem”. It would be much more interesting if element number two had no heating element.
It occurs to me that perhaps there is a better analogy.
A bucket of water. With a in-pipe and a out-pipe.
The in-pipe is sunlight. The out-pipe is radiation to space. The water-level is the temperature. Certainly we are now back to that, yes, the equilibrium can be adjusted. And is adjusted. There are valves at both the in and the out-pipe.
But not much new is explained.
We are back to the fact that there is nothing magical about CO2. That the effect is very small. Water vapour is much much more interesting, and is created all the time.
There is no tipping point. On the contrary, the system is very stable.
John Finn says:
July 26, 2010 at 2:25 am
You have once more described Peden’s oven, the paradoxical chicken that cooks by itself because of the reflected radiation:
http://i480.photobucket.com/albums/rr165/magellansc24/hansen_oven.jpg
I am appalled at the way people wade in, hand wave and mix classical concepts with quantum concepts and feel so satisfied they congratulate themselves. They should apply to the Delphi oracle.
The whole back scattering picture is a mish mash, that is why it leads to Peden’s paradoxical oven.
I am curious, has somebody an explanation why Peden’s oven is a paradox and not a reality? Hand waving says one has free energy .
Hint. Look at cbs’s post
chicken and reflective surface get fast in equilibrium at low temperature, long before the feedback starts 🙂 assuming there is infrared reflection.
That is why one needs shells in real calculations of physics problems, and integration tools.
Wow, what a discussion.
One group say that the radiant heat, lets say, from the moon is adding heat to the sun. While the other group say it doesn’t mount to a hill of beans. If it radiant heat from the moon is adding heat to the sun then we have a runway sun situation.
kwik,
the most interesting thing about h2o is that it has an effect on albedo when in the form of clouds that affects the amount of incoming solar power that is often assumed as a constant.
The reason that O2 and N2 don’t normally absorb or emit IR is that they’re symmetric molecules and thus can’t interact with an electromagnetic field. But while they’re in the process of collision, there will be some asymmetry set up which could lead to emission. The question is just how likely this is?
.
It is likely with a probability of 100% .
Actually it happens during each collision and there are billions of them in every cubic mm of air .
This phenomenon which makes N2 and O2 absorb and radiate is called “collisionaly induced dipolar momentum” and plays a role which increases with density and temperature .
This is a quantum mechanical effect which is extremely difficult to compute but experimentally proven .
However at usual densities and temperatures the induced dipolar momentum is small so that the emission of N2 &O2 is relatively small too .
See f.ex http://www.chem.ualberta.ca/~abrown/research/resother.html
.
AnnaV
.
You are hundred times right but you are shouting in the desert .
Clearly most people here and elsewhere keep mixing microscopical quantum mechanical conceptual frames with macroscopical thermodynamical frames what leads to all kinds of paradoxes .
.
Btw don’t fall for this “thermalization” garbage either .
If it is true that excited CO2 gives energy away to N2 by collisions , it is also true that N2 excites unexcited CO2 by collisions .
And the two rates are exactly equal in LTE !
So it is trivial QM textbook knowledge that CO2 absorbing IR does not “heat” the atmosphere in any usual sense of the word “heat”
And even if like Phil is saying the emission time is much longer than than the mean time between collisions , it is absurd to say like he does that there is a net energy transfer from CO2 to N2 by collisions .
Think 10 seconds and you will find easily why saying that there is a net energy transfer between 2 molecular species in LTE is ridiculous 🙂
tallbloke says:
July 26, 2010 at 2:36 am
Hi John, a couple of obs.
1) If the atmosphere absorbed 50W, it would re-emit a bit more than 25 up and a bit less than 25 down, due to curvature of the planet. How do you get your 50 down 50 up?
The atmosphere for these purposes is approx 10 km thick compared to the Earth’s radius of approx 6,400 km, to a very good approximation there is no curvature.
[reply] So to a very good approximation there will be slightly more lost to space than re-readiated to Earth. RT-mod
Chad Woodburn says:
Because of the very different temperatures of earth and the sun, the terrestrial and solar radiation occur at very different wavelengths (see http://scienceofdoom.com/2010/06/01/the-sun-and-max-planck-agree/ ). CO2 does not absorb very much in the visible, UV, and near-IR wavelength of the solar radiation (see http://www.globalwarmingart.com/wiki/File:Atmospheric_Transmission_png ).
Stephen Wilde says:
Sorry, but this is still a nonsensical statement. An increase in CO2 causes an imbalance between the amount of energy the earth is receiving and the amount that it is emitting back out into space. The only way to remedy that imbalance is to increase the amount being emitted to space (or to decrease the amount received through an increase in albedo or something like that). And, the only way to increase the amount being emitted to space is to warm up the atmospheric layers responsible for that emission because the only way that the earth can emit energy to space is via radiation and the amount emitted is specified by the Steffan-Boltzmann Equation.
So, you can’t just wave your hands and say “Upward energy transport”. That transport won’t do a darn thing to change the emission back out into space unless it changes the temperature structure in such a way that the emitting layers warm enough to increase the emission back up to where it balances the absorption again.
Can back radiation warm the oceans?
A simple experiment shows scientific proof (if that could ever be such a word) that it can’t warm a simple container of water nether mind the oceans.
Two identical containers of water (same volume), one in the shade and one in sun left during the day. The local temperature of the air during the day doesn’t reach higher than 21c with very light winds. The starting temperature of the initial water was 15c for both containers.
At the end of the day when just as the sun was no longer reaching the container, the temperature from both was recorded.
The temperature of water in the shade was 15c.
The temperature of water in the sun was 35c.
Therefore during a whole day exposed to the atmosphere and back radiation there was no change in temperature in the shade.
When placed in the sun there was a 20c increase in temperature, caused by about 250w/m2 short radiation from the sun. If back radiation contributed on it’s own and was a value around 25w/m2 (just made up for simple calculation) there would have been an expected 2c rise in temperature from the shaded water.
Thanks to evaporation and therefore latent heat there was no temperature increase in the volume of water in the shade and easily neutralised the miniscule skin effect of long wave radiation.
This included all the atmospheric gases with no noticeable change, so what does just all CO2 contribute, never mind just the CO2 that is anthropogenic.
Back radiation can’t heat a volume of water and latter relies only on short waves, convection and conduction.
Terry:
First of all, the reason there is a contradiction between G&T and everyone else is not due to anything except the fact that G&T are wrong and to say otherwise is just to obfuscate the issue.
Second of all, there is no field of science that I know of where you can’t find examples of people using terminology is slightly different ways and this does not lead people to conclude that the field has foundational errors and is non-falsifiable and all that other stuff.
********
Andrew W says:
July 24, 2010 at 5:55 pm
I don’t understand the logic that people are using to suggest the oceanic absorption of back IR radiation can be ignored or minimised, back radiation hits the ocean rather than land, what happens to it? It is absorbed, then what? It either goes into warming the water or it causes evaporation. If the latter the, energy doesn’t disappear, it’s still trapped in the troposphere.
********
I’d have to agree w/this particular point (and even w/Joel Shore above). Even if one assumes the extra “back (IR) radiation” causes no temp increase but converts completely to surface evaporation, the extra “heat” from IR can’t vanish, it’s now in the form of increased water vapor in the atmosphere (more latent heat). If all of it doesn’t convert to evaporation, what’s left will increase the temp of the water surface.
TomVonk says:
July 26, 2010 at 6:07 am
“[…]Think 10 seconds and you will find easily why saying that there is a net energy transfer between 2 molecular species in LTE is ridiculous :)”
When LWIR from outside your LTE is received by CO2, the equilibrium is disturbed. A new equilibrium is achieved through collisions.
Look at the 14.2 mikron picture in this series posted by James Gibbons :
http://mas.arc.nasa.gov/gallery/comparison.html
It looks foggy but the details are not completely washed out; this indicates that partial re-emission takes place but also that part of the energy is redistributed locally and not re-emitted in this frequency band.
Phil. says:
July 26, 2010 at 6:51 am
tallbloke says:
July 26, 2010 at 2:36 am
Hi John, a couple of obs.
1) If the atmosphere absorbed 50W, it would re-emit a bit more than 25 up and a bit less than 25 down, due to curvature of the planet. How do you get your 50 down 50 up?
The atmosphere for these purposes is approx 10 km thick compared to the Earth’s radius of approx 6,400 km, to a very good approximation there is no curvature.
[reply] So to a very good approximation there will be slightly more lost to space than re-readiated to Earth. RT-mod
Yes but not at the 2 sig fig level used in that example.