Guest Post by Willis Eschenbach
Since at least the days of Da Vinci, people have been fascinated by perpetual motion machines. One such “perpetuum mobile” designed around the time of the civil war is shown below. It wasn’t until the development of the science of thermodynamics that it could be proven that all such mechanisms are impossible. For such machines to work, they’d have to create energy, and energy cannot be either created or destroyed, only transformed.
I bring this up for a curious reason. I was reading the Jelbring hypothesis this afternoon, which claims that greenhouse gases (GHGs) are not the cause of the warming of the earth above the theoretical temperature it would have without an atmosphere. Jelbring’s hypothesis is one of several “gravito-thermal” theories which say the heating of the planet comes from gravity rather than (or in some theories in addition to) the greenhouse effect. His thought experiment is a planet with an atmosphere. The planet is isolated from the universe by an impervious thermally insulating shell that completely surrounds it, and which prevents any energy exchange with the universe outside. Inside the shell, Jelbring says that gravity makes the upper atmosphere colder and the lower atmosphere warmer. Back around 2004, I had a long discussion on the “climateskeptics” mailing list with Hans Jelbring. I said then that his theory was nothing but a perpetual motion machine, but at the time I didn’t understand why his theory was wrong. Now I do.
Dr. Robert Brown has an fascinating post on WUWT called “Earth’s baseline black-body model – a damn hard problem“. On that thread, I had said that I thought that if there was air in a tall container in a gravity field, the temperature of the air would be highest at the bottom, and lowest at the top. I said that I thought it would follow the “dry adiabatic lapse rate”, the rate at which the temperature of dry air drops with altitude in the earth’s atmosphere.
Dr. Brown said no. He said that at equilibrium, a tall container of air in a gravity field would be the same temperature everywhere—in other words, isothermal.
I couldn’t understand why. I asked Dr. Brown the following question:
Thanks, Robert, With great trepidation, I must disagree with you.
Consider a gas in a kilometre-tall sealed container. You say it will have no lapse rate, so suppose (per your assumption) that it starts out at an even temperature top to bottom.
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
I say no. I say after a million collisions the molecules will sort themselves so that the TOTAL energy at the top and bottom of the container will be the same. In other words, it is the action of gravity on the molecules themselves that creates the lapse rate.
Dr. Brown gave an answer that I couldn’t wrap my head around, and he recommended that I study the excellent paper of Caballero for further insight. Caballero discusses the question in Section 2.17. Thanks to Dr. Browns answer plus Caballero, I finally got the answer to my question. I wrote to Dr. Brown on his thread as follows:
Dr. Brown, thank you so much. After following your suggestion and after much beating of my head against Caballero, I finally got it.
At equilibrium, as you stated, the temperature is indeed uniform. I was totally wrong to state it followed the dry adiabatic lapse rate.
I had asked the following question:
Now, consider a collision between two of the gas molecules that knocks one molecule straight upwards, and the other straight downwards. The molecule going downwards will accelerate due to gravity, while the one going upwards will slow due to gravity. So the upper one will have less kinetic energy, and the lower one will have more kinetic energy.
After a million such collisions, are you really claiming that the average kinetic energy of the molecules at the top and the bottom of the tall container are going to be the same?
What I failed to consider is that there are fewer molecules at altitude because the pressure is lower. When the temperature is uniform from top to bottom, the individual molecules at the top have more total energy (KE + PE) than those at the bottom. I said that led to an uneven distribution in the total energy.
But by exactly the same measure, there are fewer molecules at the top than at the bottom. As a result, the isothermal situation does in fact have the energy evenly distributed. More total energy per molecules times fewer molecules at the top exactly equals less energy per molecule times more molecules at the bottom. Very neat.
Finally, before I posted my reply, Dr. Brown had answered a second time and I hadn’t seen it. His answer follows a very different (and interesting) logical argument to arrive at the same answer. He said in part:
Imagine a plane surface in the gas. In a thin slice of the gas right above the surface, the molecules have some temperature. Right below it, they have some other temperature. Let’s imagine the gas to be monoatomic (no loss of generality) and ideal (ditto). In each layer, the gravitational potential energy is constant. Bear in mind that only changes in potential energy are associated with changes in kinetic energy (work energy theorem), and that temperature only describes the average internal kinetic energy in the gas.
Here’s the tricky part. In equilibrium, the density of the upper and lower layers, while not equal, cannot vary. Right? Which means that however many molecules move from the lower slice to the upper slice, exactly the same number of molecules must move from the upper slice to the lower slice. They have to have exactly the same velocity distribution moving in either direction. If the molecules below had a higher temperature, they’d have a different MB [Maxwell-Boltzmann] distribution, with more molecules moving faster. Some of those faster moving molecules would have the right trajectory to rise to the interface (slowing, sure) and carry energy from the lower slice to the upper. The upper slice (lower temperature) has fewer molecules moving faster — the entire MB distribution is shifted to the left a bit. There are therefore fewer molecules that move the other way at the speeds that the molecules from the lower slice deliver (allowing for gravity). This increases the number of fast moving molecules in the upper slice and decreases it in the lower slice until the MB distributions are the same in the two slices and one accomplishes detailed balance across the interface. On average, just as many molecules move up, with exactly the same velocity/kinetic energy profile, as move down, with zero energy transport, zero mass transport, and zero alteration of the MB profiles above and below, only when the two slices have the same temperature. Otherwise heat will flow from the hotter (right-shifted MB distribution) to the colder (left-shifted MB distribution) slice until the temperatures are equal.
It’s an interesting argument. Here’s my elevator speech version.
• Suppose we have an isolated container of air which is warmer at the bottom and cooler at the top. Any random movement of air from above to below a horizontal slice through the container must be matched by an equal amount going the other way.
• On average, that exchange equalizes temperature, moving slightly warmer air up and slightly cooler air down.
• Eventually this gradual exchange must lead to an isothermal condition.
I encourage people to read the rest of his comment.
Now, I see where I went wrong. Following the logic of my question to Dr. Brown, I incorrectly thought the final equilibrium arrangement would be where the average energy per molecule was evenly spread out from top to bottom, with the molecules having the same average total energy everywhere. This leads to warmer temperature at the bottom and colder temperature at elevation. Instead, at thermal equilibrium, the average energy per volume is the same from top to bottom, with every cubic metre having the same total energy. To do that, the gas needs to be isothermal, with the same temperature in every part.
Yesterday, I read the Jelbring hypothesis again. As I was reading it, I wondered by what logic Jelbring had come to the conclusion that the atmosphere would not be isothermal. I noticed the following sentence in Section 2.2 C (emphasis mine):
The energy content in the model atmosphere is fixed and constant since no energy can enter or leave the closed space. Nature will redistribute the contained atmospheric energy (using both convective and radiative processes) until each molecule, in an average sense, will have the same total energy. In this situation the atmosphere has reached energetic equilibrium.
He goes on to describe the atmosphere in that situation as taking up the dry adiabatic lapse rate temperature profile, warm on the bottom, cold on top. I had to laugh. Jelbring made the exact same dang mistake I made. He thinks total energy evenly distributed per molecule is the final state of energetic equilibrium, whereas the equilibrium state is when the energy is evenly distributed per volume and not per molecule. This is the isothermal state. In Jelbrings thought experiment, contrary to what he claims, the entire atmosphere of the planet would end up at the same temperature.
In any case, there’s another way to show that the Jelbring hypothesis violates conservation of energy. Again it is a proof by contradiction, and it is the same argument that I presented to Jelbring years ago. At that time, I couldn’t say why his “gravito-thermal” hypothesis didn’t work … but I knew that it couldn’t work. Now, I can see why, for the reasons adduced above. In addition, in his thread Dr. Brown independently used the same argument in his discussion of the Jelbring hypothesis. The proof by contradiction goes like this:
Suppose Jelbring is right, and the temperature in the atmosphere inside the shell is warmer at the bottom and cooler at the top. Then the people living in the stygian darkness inside that impervious shell could use that temperature difference to drive a heat engine. Power from the heat engine could light up the dark, and provide electricity for cities and farms. The good news for perpetual motion fans is that as fast as the operation of the heat engine would warm the upper atmosphere and cool the lower atmosphere, gravity would re-arrange the molecules once again so the prior temperature profile would be restored, warm on the bottom and cold on the top, and the machine would produce light for the good citizens of Stygia … forever.
As this is a clear violation of conservation of energy, the proof by contradiction that the Jelbring hypothesis violates the conservation of energy is complete.
Let me close by giving my elevator speech about the Jelbring hypothesis. Hans vigorously argues that no such speech is possible, saying
There certainly are no “Elevator version” of my paper which is based on first principal physics. It means that what I have written is either true or false. There is nothing inbetween.
Another “gravito-thermal” theorist, Ned Nikolov, says the same thing:
About the ‘elevator speech’ – that was given in our first paper! However, you apparently did not get it. So, it will take far more explanation to convey the basic idea, which we will try to do in Part 2 of our reply.
I don’t have an elevator speech for the Nikolov & Zeller theory (here, rebuttal here) yet, because I can’t understand it. My elevator speech for the Jelbring hypothesis, however, goes like this:
• If left undisturbed in a gravity field, a tall container of air will stratify vertically, with the coolest air at the top and the warmest air at the bottom.
• This also is happening with the Earth’s atmosphere.
• Since the top of the atmosphere cannot be below a certain temperature, and the lower atmosphere must be a certain amount warmer than the upper, this warms the lower atmosphere and thus the planetary surface to a much higher temperature than it would be in the absence of the atmosphere.
• This is the cause of what we erroneously refer to as the “greenhouse effect”
Now, was that so hard? It may not be the best, I’m happy to have someone improve on it, but it covers all the main points. The claim that “gravito-thermal” theories are too complex for a simple “elevator speech” explanation doesn’t hold water.
But you can see why such an elevator speech is like garlic to a vampire, it is anathema to the “gravito-thermal” theorists—it makes spotting their mistakes far too easy.
w.
KevinK says:
January 20, 2012 at 7:28 pm
“We can do this but we use diodes (aka rectifiers) to accomplish this task. the GHE does not behave as a rectifier does (i.e. energy only flows one way).”
Actually, that is exactly what is happening, so this might help you comprehend. The energy comes in shortwave. It goes out longwave. So-called GHGs are transparent to shortwave, opaque to longwave.
KevinK says:
January 20, 2012 at 7:28 pm
For the other people with whom I have been discussing my hypothesis of “GHG” cooling: Anyway, that’s the standard greenhouse argument. I am not saying GHGs necessarily cause the ground to heat more than it would otherwise. But, what they are doing has the same steady state effect as if they were, and the interpretation in the steady state is merely ambiguous. Kevin has simply gone off the rails, and I am trying to help him understand the GHG argument, and that he has done nothing to refute it.
Ok, I now declare this an OFFICIAL FARCE…….
We clearly have two camps, the believers in the GHE and those that have sound technical reasons to DOUBT it.
I suggest we all shovel the snow from our driveways and reconvene next summer to discuss it again……………..
Cheers, Kevin (see you after the thaw)
KevinK;
Well said. May I offer a simple analogy and then launch into an extension of your explanation that ties back to N&Z? Thanks!
Analogy:
I dam a river and form a lake. At equilibrium, the river flowing in has a flow rate of 240 m3/min. Outflow at the dam = 240 m3/min. I raise the dam one meter. After a short period of time, the flow rate in is… 240 m3/min and the outflow is 240 m3/min. The depth of the water increases by 1 meter, and the potential energy stored as a consequence of that can be calculated, but the flow rate at equilibrium changes not one bit. The same is true of doubling C02. Temporary fluctuation in the system, but one equilibrium is established again, the incoming w/m2 and the outgoing w/m2 are exactly what they were before. There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt. Since everyone wants to average w/m2 and convert that to degrees, I say let ’em. If they cannot show that the incoming watts absorbed has changed, then at equilibrium the outgoing watts have to be the same too. Since the outgoing watts are the same, temperature of the planet is the same. Go ahead folks, use SB Law to contradict that.
Onto N&Z building on what KevinK explained.
Imagine that there are upward bound photons being released from earth surface and from every layer of the atmosphere at every elevation. Some escape, and some don’t. The most photons emerge from the earth surface, BUT… they have the smallest chance of getting out to space because they have the greatest chance of being absorbed on the way up. As we increase in altitude, there are less upward bound photons released from a given layer because a) the temperature is lower and b) the density is lower. BUT…the percentage chance that any upward bound photon will escape to space increases with altitude. Let’s repeat that. The higher the altitude, the greater the percentage chance that any given upward bound photon will escape to space is.
As KevinK explained, once you hit equilibrium, every photon rising from earth surface cools the surface. If it is intercepted somewhere in the atmosphere, then that spot in the atmosphere warms. If the photon is re-emitted, then that spot in the atmosphere cools. If the photon goes downward, it must inevitably be absorbed again, either in the atmosphere or the earth surface, and where ever that happens, that spot warms and re-emitts. At some point though, the photon escapes, taking with it the exact same amount of energy as it brought when it entered the system. The net is zero.
So…as seen from space, let us assume we can “see” an average (ignorant term and wrong but for sake of argument I’ll live with it for this example) we can “see” 240 w/m2 exiting the system. How many watts came from where? Some of the photons would in fact be emitted from earth surface and go straight to space uninterrupted. How many? Not many. Even though the most upward bound photons in the system are being released from the warmest surface (earth surface) not many escape. The frigid TOA though is pouring the photons into space. It might be cold at TOA, but every upward bound photon emitted at TOA has a 100% chance of escaping to space.
Hence….
The greater the mass of the atmosphere, the lower the chance of any energy packet escaping to space. At earth surface, conduction and convection will massively dominate radiance, and photons released via radiance have a near zero chance of escaping directly to space anyway. At high elevations, while the temperature is must lower than earth surface, conduction and convection are vastly reduced. While the number of photons is also vastly reduced due to the lower temperature, the percentage chance that any given one of them will escape is vastly higher.
Hence…
The greater the mass of the atmosphere, the greater the surface pressure and the greater the dominance of conduction and convection and the less impact on temperature of radiance. The greater the mass of the atmosphere, the more skewed the ratio of photons released at high altitudes will be. And if one “averages” the upward bound flux that actually escapes to space from earth surface to TOA, what will one get?
One will get exactly the total flux being absorbed.
If 240 w/m2 is being absorbed, then 240 w/m2 is existing. The greater the mass of the atmosphere, the more likely that any given energy packet will be returned from the surface straight back to the surface, and thus the higher the surface temperature. The greater the atmospheric mass (and hence surface pressure) the more skewed to higher (colder) altitudes the percentage of escaping photons will be. BUT… the average will always equal exactly the mount being absorbed.
Hence, as per N&Z…
The greater the mass of the atmosphere, the higher the surface pressure and so the higher the surface temperature. Calculate insolation absorbed, and you now have two variables, insolation and surface pressure, that define the surface temperature. Double or quintuple CO2 and you know how much the surface temperature changes? It changes by the amount that the mass of the atmopshere (and hence surface pressure) changes. But the incoming and outgoing watts remain (at equilibrium) unchanged.
Just like the water going over the dam.
Dammit
Myrrh says:
January 20, 2012 at 5:38 pm
Stephen Wilde says:
January 20, 2012 at 3:25 pm
Gravity does not separate by temperature. It separates by mass.
This is what I meant, but here said more succinctly..
======
er.., I meant weight…
http://www.exploratorium.edu/ronh/weight/
“jae says:
January 19, 2012 at 7:05 pm
DAMMIT, WILLIS:
PLEASE ADDRESS THE EMPIRICAL EVIDENCE, WHICH WILL ULTIMATELY RESOLVE THIS ARGUMENT! WHY DO YOU REFUSE TO DO THIS?
[Moderator’s suggestion: If you didn’t YELL at him maybe more would be accomplished? Maybe? -REP]”
It seems that some of Anthony’s “moderators” need some moderation. Is it really the moderator’s place to tell me just how to express myself on a blog? I think NOT! Just what the hell do you think you are?
Bart says:
January 20, 2012 at 6:17 pm
Of course there’s an upper limit, but there’s not very much mass between that level, where the atmosphere consists almost entirely of hydrogen and helium, and the level where LTE applies, which would be isothermal at the surface temperature under the conditions specified by Willis in his original Gravity post.
Out of curiosity I calculated the pressure a function of altitude for a purely adiabatic atmosphere. For a surface temperature of 255K, it would reach a pressure of zero and a temperature of 0K at an altitude of about 26.2 km. But there’s no way it would stay that way as the thermal diffusivity gets quite large as the pressure goes to zero.
ferd berple says:
January 20, 2012 at 7:12 pm
“Why do you think the oceans are colder on the bottom and hotter on the top? Gravity separates the oceans by temperature.”
Yes, but only accounts for a small portion of the temperature gradient in the upper ocean.
KevinK says:
Speaking of sanity check, you may want to try one on your argument. For example, can you extend it to show that there’s no advantage to wearing a jacket when you go out on the cold or putting any insulation in the attic of your house? (Basically any case where you have the steady-state temperature of an object determined by the balance of what energy receives and what energy emits.)
It is amazing how difficult basic concepts become for people, even “engineering professionals” when the concepts conflict with what they strongly want to be the case!
Bart says:
January 20, 2012 at 6:50 pm
What useful conclusions can be drawn? A surface that can absorb radiation but not emit it at any wavelength at any temperature while having zero heat capacity goes far beyond postulating a perfectly transparent atmosphere or a surface with an absorptivity identically equal to 1 as limiting cases in thought experiments. It’s so far removed from reality as to be worthy of the term unphysical and the results are completely irrelevant to anything real. It says nothing whatsoever about the atmospheric greenhouse effect on a real planet. Especially it does not say that addition of greenhouse gases cause the surface of a rotating spherical planet with an IR absorbing atmosphere in orbit around a single star to cool compared to an atmosphere that is more transparent in the IR.
“Taking the most simplistic approach possible:
A uniformly irradiated world, no rotation, no poles, two gas atmosphere (no water/condensation)
Atmosphere is uniformly dense – no pressure gradient:
Assuming there is only 1 GHG gas present in trace amounts and only that GHG can absorb IR energy. Assuming the ‘one other gas’ TG is totally transparent to all incoming and outgoing radiation.
Assuming all energy absorbed by the surface from the sun is then emitted at one IR wavelength which is only (and totally absorbed) by the one GHG gas
Assuming all IR energy (100%) is absorbed by that GHG in the first 10 meters above the surface.”
Let’s stop here.
Let’s look at this world. Unless one can see this IR energy [humans without devices can’t], from space this planet would be utterly black.
Next problem is nothing absorbs all the sun’s radiation.
If we assume you mean that a limited amount of the sun’s radiation is absorbed and have the rest reflecting, we fix the problem of having a utterly black world and it’s conceivable that the GHG is absorbing this limited band width. It’s quite simple actually, you have CO2 ocean and a CO2 GHG gas. The CO2 ocean only absorbs and emits what the CO2 gas can absorb and emit.
Venus is almost this- but Venus ocean is not a liquid, but instead it’s a hot dense gas. Though don’t think matters whether it’s gas or liquid. On Venus visible light reaches the surface but we can assume any wavelength the sun emits which CO2 can absorb isn’t shining from the sun to the surface- it’s being absorbed and emitted by the CO2. The CO2 is not a “surface” [liquid or solid] but what difference does it make?
So it seems you are basically talking about Venus. If you don’t mean the planet is utterly dark from space [or on the planet].
Continuing:
“Assuming that energy is transferred rapidly be collision to TG molecules near the surface
Assuming it is largely the TG molecules which carry the energy to the upper atmosphere by convection
Assuming the energy can only be emitted to space by being passed back to a GHG in the upper atmosphere”
Again still Venus. Though difference is CO2 not trace gas, and TG is actually minor constituent
of atmosphere [around 3% but that is about 3 earth atmospheres of N2- so a lot of N2 but dwarf by massive amount of CO2]
“Double the quantity of GHG, energy is totally absorbed and passed to TG twice as rapidly.
After the system equilibrates there is a trace amount of extra energy retained in the atmosphere at any point in time. ”
We could 1/2 the amount of CO2 on Venus and thereby double the amount of N2 relative to the CO2:)
I think Venus would cool if lost half it’s CO2.
But replaced the half of CO2 with same amount of
N2- took out 46 atm of CO2 and replaced with 46 atm of N2, then I would suppose that the Venus surface would become much dimmer- not allowing as much diffused sunlight from reaching the surface. But don’t think that would change the planet’s temperature.
“(related to that held by the extra GHG molecules).
There are now twice as many GHG molecules in the upper atmosphere with the opportunity to emit energy to space, but there is also twice as much opportunity they will intercept each-others IR emission – average emission to space remains the same, and atmospheric temperature remains the same.
“Double the quantity of TG. There is twice the opportunity of collision between GGH and TG at all levels of atmosphere, energy near the surface is transferred more quickly to the TG.”
As said don’t think there would any difference.
Move Venus to earth orbit and Venus would change quite significantly- though require a long time to radiate the energy- so could a million years to change significantly.
“Repeat the exercise with graduated layers of from greater to lesser density atmosphere. Do the effects remain the same layer by layer?”
As saying above, if you half the amount CO2 in the atmosphere, made a 46 atm instead of 92 atm, this would radically change the temperature- lower by say 100 K. Half it again, and I would guess the temperature would lower again significantly and because Venus does not rotate, one could get liquid CO2 on nite side and cascading lowering of atmosphere pressure, and further lowering of temperature. Or have similar effect of moving Venus out to Earth orbit.
Venus would develop very small ocean of liquid CO2- possibly seeping under ground or making lakes in lowest elevations.
thepompousgit says:
January 20, 2012 at 7:28 pm
He did that (those) lecture(s) after I graduated, as I remember. I do have a copy on CD of the lecture on deriving the 1/r² force relationship from Kepler’s Laws, though. It’s too bad there weren’t pictures of the blackboards he used. Newton used conic sections. Feynman used a more readily comprehensible geometric approach. As I remember, Feynman did it that way because he couldn’t follow Newton’s proof. Conic sections were big in Newton’s day.
Cheers, Kevin.
Joel Shore says:
January 20, 2012 at 7:15 pm
thepompousgit says (in response to Hans Jelbring):
I get my suntan from UV radiation. Where is this fairytale land that enables one to get a suntan from IR radiation? Or are you just saying whatever pops into your head because you know the ignoramuses will believe whatever you say? Sorry if this is rude, but the difference between UV & IR is really basic stuff!
Yeah…The way I was going to put it is that infrared radiation doesn’t give you a sun tan for the same reason that cell phones don’t give you brain cancer: The energy of the photons are way too small to cause this sort of cell damage to occur.
Einstein figured this out about a century ago.
=============
Go out in the midday sun in the tropics, photon for photon UV won’t be able to penetrate further than the epidermis, the first layer of skin, to give you a tan while thermal infrared, heat, will penetrate some inches and its energy absorbed by the water in you, you’re mostly that and water is the great absorber of heat, will move the water into vibration, kinetic energy, heat. UV will damage you from the outside in and you won’t know it’s happening until you can feel the effects; thermal infrared, if you’re unable to sweat it out, will boil you from the inside out. You’ll boil quicker than you’ll get skin damaged to death… Put on a shirt, you stop UV in its tracks, no suntan, that’s how strong an energy it is.. it’s tiny, really tiny, compared with thermal infrared.
UV doesn’t give you a tan because it burns you, you can’t feel UV, it doesn’t burn you, it gives you a tan because the melanin in your body is doing its best to neutralise excess of it scrambling your skin on a DNA level. It is essential for the production of vitamin D. Thermal ir is bigger and packs more punch, stand in front of a fire, that’s thermal ir you can feel burning you directly. Compare with sunbed.
There’s a world of difference between UV photo-damage and thermal infrared heat damage.
http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2671032/
Thermal infrared is the size of a pin head, near ir is microscopic, how big is UV compared with them?
“8) Once our bundle of energy is absorbed by the GHG it ceases to exist as IR light and is converted to heat (Warming Event #2). This is our little bundle’s second warming event within a few milliseconds, boy am I proud. Note that with any flavor of the speed of light our bundle can make it to the top of the atmosphere (TOA) in a few milliseconds at most. ”
What is meant by warming event. Has the GHG gas molecule velocity increased, or is it excited molecule which may soon emit the same wavelength or same different wavelength that it is capable of emitting.
The isentropic state is the state of maximum entropy. If you start with an isothermal atmosphere and thoroughly mix it without adding energy, you end up with an isentropic state which has the adiabatic lapse rate. Once in this state, the air can’t be unmixed (because entropy is maximized), so in this sense it is a final state for a given energy content. This doesn’t mean that an isothermal state or any stable lapse rate cannot last an infinite amount of time. I don’t think the problem is well posed because many states can last infinitely long if only diffusion is acting, but only the isentropic state cannot be transformed to another state by mixing, in my opinion.
gbaikie says:
January 20, 2012 at 8:47 pm
I’ve actually run some numbers on Venus using the line-by-line program at spectralcalc.com . They have a database of high pressure CO2 absorption lines. Because the surface temperature is so much higher, the 15 μm band of CO2 doesn’t contribute much. Even at 92 bar, there are still ranges of relatively low absorption for CO2 alone. However, those gaps are nearly filled by SO2 and H2O, even though the partial pressure of H2O is small. Replacing half the CO2 with nitrogen does cool the surface, but not by much. Another interesting point is that it isn’t completely dark in the visible spectrum at the surface when the sun is above the horizon. One of the Russian probes that made it to the surface managed to transmit some visible light photographs before it failed. You do have to lower the total pressure to get major a major reduction in temperature. We had a big discussion of this over at scienceofdoom.com .
Joel Shore;
Speaking of sanity check, you may want to try one on your argument. For example, can you extend it to show that there’s no advantage to wearing a jacket when you go out on the cold or putting any insulation in the attic of your house? >>>
Good lord man, what brand of cereal did your physics degree come in anyway? The person wearing the jacket and the home with the attic both have INTERNAL heat sources. The increased layers of insulation change the temperature gradient such that is is more pronounced closer to the heat source, hence warmer “inside” the jacket and “inside” the house. Which has what to do with the topic at hand?
While you’re trying to come up with an answer for that:
The earth absorbs 240 w/m2. At equilibrium, the earth emitts 240 w/m2. Temperature (accoding to you and your wrongly applied SB Law) is 255K
CO2 quintuples. The system is perturbed until equilibrium is once again established in which event:
The earth absorbs 240 w/m2. At equilibrium, the earth emitts 240 w/m2. Temperature (according to you and your wrongly applied SB Law) is 255K.
Change from quintupling CO2 = 0.
Bill Hunter says:
January 20, 2012 at 8:35 pm
“Yes, but only accounts for a small portion of the temperature gradient in the upper ocean.”
Any can disagree. Tell us why you disagree. What accounts for the other portion?
Been thinking about an increase over the SB average. It would be higher because of the thermal lag in a purely mechanical atmosphere system without GHG’s. In a GHG free atmosphere the molecules cannot shed the energy gained mechanically from the ground during the day. I do think that GHG’s would raise the temperature further, in line with the equations in Loudon, but not according to Hansen’s empirical relation.
davidmhoffer says:
January 20, 2012 at 8:10 pm
“There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt.
The water retained is the analogy to energy retained. And, if heat capacity is unchanged, that means the temperature must go up in the GHG paradigm.
However, here’s what I believe is wrong about the GHG paradigm: the example is inverted. The dam is already there. Having the dam a particular height means the water behind it is at a particular height. In introducing IR emitting gases, you are, in fact, opening up new floodgates in the dam (more avenues for energy to radiate away), and the water will go down.
DeWitt Payne says:
January 20, 2012 at 8:23 pm
Of course there’s an upper limit,…
I was speaking of a temperature limit. Your objection was that there was a 2nd law violation because the temperature could increase forever. It couldn’t for two reasons: 1) all real atmospheres will radiate somewhere 2) in the worst case, the atmosphere simply escapes and stops heating.
“Out of curiosity I calculated the pressure a function of altitude for a purely adiabatic atmosphere.”
You can’t get adiabatic conditions when there is no heat sink in the upper atmosphere. This is a basic assumption in all the text books when they calculate adiabatic lapse rates, and one of the items which led me to my hypothesis, which is that IR gases in the Earth’s atmosphere are, in fact, heat sinks.
DeWitt Payne says:
January 20, 2012 at 8:46 pm
“…”
I’d appreciate it if you try that again. You’re not talking about anything I’m talking about. I am saying that, on this hypothetical planet, the surface is a blackbody, but its emissions are suppressed while massive amounts of heat are being drawn out of it to feed an ever hungrier atmosphere. At least, that’s my current running hypothesis. It could change to accomodate new facts as I gather them. But, so far, I do not see any evidence that says this scenario is unphysical or impossible, and it is very appealing from the point of view of stability.
ferd berple says:
January 20, 2012 at 7:19 pm
Willis says:
“IF GRAVITY ACTUALLY SEPARATES TEMPERATURE, WE HAVE FREE ENERGY FOREVER!”
If Willis says it, then he is pulling your leg. Gravity can affect a tall column of gas and will separate temperature, but there is no work being done in that process and there is no pump involved, just compression and expansion. As much material goes up as down. Just because there is a temperature gradient does not mean you can extract energy from it indefinitely because extraction will cool the whole system.
This is straightforward. A system at any given temperature above 0 K can be cooled. A system with a temperature gradient can be cooled using a heat engine that exploits the temperature gradient. But that is nothing like ‘free enery forever’ That is ‘stored energy being converted to a non-thermal form’.
A spherical planet with some gravity and an atmosphere isolated from the universe will have a temperature gradient from top to bottom, cold to hot, because of the universal gas law and Brownian motion. Energy can indeed be extracted from the temperature difference between the top and bottom of the system provided that device is not entirely made of gas – it has to not respond to the universal gas law. It will produce, say, electricity. That is a conversion of heat into electrical energy. The whole system cools in response. That is not free energy and ‘gravity’ will not make up for what is missing. That is a thermoelectric generator. Over time the bottom of atmosphere will cool until, in theory, the atmosphere condenses. Gravity does not ‘heat the atmosphere’ is causes the heat in the atmosphere to be unequally distributed according to the pressure gradient.
Bart says:
January 20, 2012 at 11:00 pm
davidmhoffer says:
January 20, 2012 at 8:10 pm
“There’s been an increase in heat stored in the system, but in terms of the average w/m2 exiting the system as a whole, they are exactly the same as they were before down to the last watt.”
I just realized your and Kevin’s thought experiment supports my hypothesis. As you say, all those photons backradiated into the Earth should just be spit back out again. So, why should there be gaps in the emissions at TOA in those bands?
Are they re-radiated back from the surface at even lower wavenumber? (There’s no way the Earth is re-radiating the backradiation at higher energy.) If so, shouldn’t we see a hump at lower wavenumbers in the emissions spectra? In fact, we see just the opposite: the emissions spectrum at TOA over Guam, for example (see here in Fig. 2), in the low wavenumber range appears to fit an isocline near 275K, while that at higher wavenumbers appears to follow a 300K or so isocline.
Curiouser and curiouser…
Now, consider my mechanism in which the IR absorbing gases actually pin down the spectrum in their radiative range, while the rest of the spectrum is trying to push itself up around it. That is a much better fit with what is actually measured!
DeWitt Payne says:
January 20, 2012 at 9:15 pm
“Because the surface temperature is so much higher, the 15 μm band of CO2 doesn’t contribute much. Even at 92 bar, there are still ranges of relatively low absorption for CO2 alone. However, those gaps are nearly filled by SO2 and H2O, even though the partial pressure of H2O is small. Replacing half the CO2 with nitrogen does cool the surface, but not by much.”
In light of my comment above, I note that the composite emissions spectrum of Venus is “pinned down” in the H2SO4 band, while tiny bits are taken out in the CO2 and other bands. How about giving it a go? Suppose, for the sake of argument, that I am right. Where does it lead you in interpreting Venus’ emissions spectrum?
In response to
Jordan says:
January 20, 2012 at 2:28 pm
Hans Jelbring says:
January 20, 2012 at 5:21 pm
Come on guys, you are both missing the point. The basic question is – Does increasing the amount of greenhouse gases in the atmosphere cause the atmosphere to get warmer (the consensus) or colder? In my paper, I suggest a atmosphere made from a hypothetical gas that does not absorb or emit IR radiation. This avoids the very valid exceptions that both of you have made. In this very hypothetical case, the atmosphere has no way to loose energy to space and, therefore, becomes isothermal.
At that point, it does not matter if you add salt particles, O2, or CO2, there is now a mechanism for heat to escape the atmosphere. (However, with O2 the loss is so small it might not be measurable.) Since some of that heat returns to the surface, it gets warmer as the atmosphere cools.
As long as the atmosphere is optically thin, the entire thickness of the atmosphere will cool about the same amount and an adiabatic lapse rate will form near the surface. If the atmosphere becomes optically thick (nearly opaque), then only the top will cool significantly and the lapse rate will not be very close to the adiabatic rate.
On the real Earth, the dry adiabatic lapse rate is about 9.8 K/km, the typical measured troposphere lapse rate is about 6.5 K/km. With an atmosphere lacking any IR emitters, the lapse rate will be 0 K/km (isothermal). Since the actual lapse rate is between 0 and 9.8, if follows that the atmosphere is optically thick in the frequencies that matter.
Beside the mechanisms you have suggested, there are other cooling scenarios caused by pressure gradients produced by the surface temperature variations between the day and night sides of the planet, as well as the pole vs equator differences. But remember, the point of this exercise is not to describe some hypothetical planet, but to determine the function of greenhouse gases.
DeWitt Payne says:
January 20, 2012 at 9:59 am
People, please pay attention to this interchange. Hans Jelbring asserted categorically that “Any surface radiation power exceeding 100 W/m^2 is bull.”
In response, DeWitt posted up an actual measurement from a surfrad station. It shows a 24 hour plot. The MEASURED, OBSERVED surface radiation swings between 300 (night) and 400 (day) watts per square metre. So what is “bull” is Hans’s crazy claim.
I want you all to consider the level of willful blindness about climate science that it would take to get to the year 2012 and still be making Hans’s claim that nowhere on the earth is there surface radiation exceeding 100 W/m2. I mean, to do that you would have to avert your eyes from half the scientific papers published about the climate. You’d have to never, ever google “upwelling longwave” to see if your 100 W/m2 claim is correct. You’d have to think that the Stefan-Boltzmann equation didn’t work. You’d have to never look at a global energy budget.
Tallbloke, upstream you claimed that I was making an “ad hominem” attack when I said:
Now … perhaps given Hans Jelbring’s demonstrated ignorance regarding upwelling radiation in this interchange, you might be willing to rethink your comment?
w.
PS—Please be clear why what I said is not an ad hominem attack. I am not saying that Hans’s scientific theories are wrong because Hans is ignorant of basic information about radiation. That would indeed be an ad hom attack, where you try to bring down the man’s scientific claims by attacking the man.
I am saying that Hans’s scientific theory are wrong, and that Hans is ignorant of basic information about radiation. The two are not linked. Hans’s theory fails on its own merits, with no reference to the author of the theory.
The same is true about my comment about how people who are generally ignorant of basic science believe in theories like Jelbrings. That’s not an ad-hom argument either. Saying that someone is ignorant of basic science does not mean that any given belief of theirs is wrong, that would also be an ad hominem attack.
You see, an ad hominem attack means that you are attacking the man rather than attacking the argument. I’m not saying that those folks arguments are all wrong, or are wrong because they don’t know science.
I’m just saying that a lot of them don’t understand basic science, which is a simple statement and not an ad hominem in any form.
Marc77 says:
January 20, 2012 at 12:00 pm
Once again, we are not discussing whether we can draw power from the Earth’s atmosphere. Everyone agrees we can do that. We are discussing whether we can draw power from either the atmosphere in Jelbring’s thought experiment, or (equivalently) from a tall, thermally insulated cylinder.
w.