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.
Trick says:
January 25, 2012 at 8:08 pm
Circular logic or begging the question. The condition that the sum of potential and kinetic energy is constant is the definition of the adiabatic lapse rate. You have assumed your conclusion. The adiabatic lapse rate happens because the air packet is allowed to expand and do work against the surrounding pressure but is not allowed to reach thermodynamic equilibrium by heat conduction from the surrounding air or any other means for energy to enter or leave the parcel. There is nothing in thermodynamics or meteorology that says an air column can only have the adiabatic lapse rate. The adiabatic lapse rate is the upper limit for a stable atmosphere. If the lapse rate were greater than the adiabatic rate, convection would occur. But lapse rates less than the adiabatic rate are observed all the time. In fact, temperature increases with altitude in the stratosphere. Temperature increasing with altitude near the surface happens all the time at night. That’s how dew and frost form. In the Arctic over the sea ice or in the Antarctic during winter, the temperature inversions reach high into the atmosphere. That’s because the ground cools by radiation faster than the air can conduct energy to the ground. The important word in that sentence is conduct, which you seem to think never occurs.
DeWitt Payne,
Good luck with your experiment. You didn’t respond to the issues I raised [except for the cotton insulation, which I explained]. As I made clear, my view is strictly limited to questioning whether a doubling of CO2 would have any significant effect. If we are in agreement, there’s nothing to discuss, because as I said, I’m not doing the Wood experiment. Planet Earth is doing a fine experiment, and showing that CO2 is at best a bit player. And my oft-repeated hypothesis has never been falsified:
At current and projected levels, CO2 is harmless, and beneficial to the biosphere.
If we are in agreement with that, fine. If not, you need to explain why the rise in CO2 has no apparent effect on global temperatures. As you can see, it’s still not clear to me where you’re coming from.
Trick says:
January 25, 2012 at 8:08 pm
Trick: “The gas molecule(s) will have the same total energy throughout by 1st Law.”
Paul: “No, they won’t. Molecules of higher total energy go higher; this selection effect has been pointed out repeatedly. “
Trick: “Paul Birch misinterprets or misunderstands Caballero.”
I did not refer to Caballero. The selection effect is really easy to understand. Forget molecules. Think tennis balls, fired straight up by a row of spring loaded catapults. Some of the catapults are stronger, so some of the tennis balls shoot out faster. The faster they are (that is, the higher their total energy KE+PE) the higher they go. If you stand on a tall ladder, the only balls you will see passing you are the fast ones (the ones that had the highest muzzle velocity, the ones with the highest total energy). The slow ones never get that high. Your position up the ladder selects tennis balls of higher total energy.
————-
“Trick: “There is no law AFAIK that says the molecule(s) mean kinetic energy or temperature (as defined by Caballero in 2.1 top post link) has to be the same thoughout one thermodynamic system in a gravity field.”
Paul: “Yes, there is. The zeroth law that says that in thermal equilibrium the temperature is everywhere the same, and the second law that says you can’t extract net work from a system in thermal equilibrium, but can from any macroscopic temperature difference. Except for a negligible relativistic correction (of order 1E-12 for the Earth), gravity is irrelevant to this.”
Trick: “Here Paul can’t extract work since the entropy is constant due to the process being reversible, the molecules gain just as much kinetic energy falling to the bottom as they lose in rising to the top & vice versa: voila reversible. ”
The second law does not say that you can’t extract work from reversible processes; on the contrary, adiabatic expansion is the archetypical way to extract work from a gas reversibly (ie. isentropically). In this case the random motion of the particles in the gas is “reversible”, but cannot be used to extract work, because the gas is in thermal equilibrium, which is to say, all at the same temperature, as required by the zeroth law.
DeWitt Payne says at 1/26 7:03am:
“Circular logic..”
DeWitt – this experiment can really be performed in the non- ideal gas, near-adiabatic gravity infested real world. Given apparently no one has yet cited a ref. to it having been done appropriately & thus dismissing the demon(s), then I take that to mean experiment is difficult but logic is not circular.
Proof: insert the thermometers ideally in thought experiment – take a reading. They read the same or different. Circle ends.
Extending to atmosphere makes a difficult thought experiment compound the difficulty. Makes me ask, if we cannot yet dismiss the demons from the simpler thought experiment (a gas column in gravity field), then on what basis do we move on to the more difficult atmosphere?
I think the answer lies in – you need to operate with a view on the thermometer readings in ideal experiment, that enough physical laws (0th, 1st, 2nd) are reliable enough to allow moving on to the real atmosphere to draw law-consistent conclusions in agreement with stuff experimentally observed in the real atmosphere.
Among things we’ve learned can’t do so far to prove any conclusion physical: divide by 0, construct a Perpetuum Mobile esp. one that allows energy extraction, and decrease entropy.
Smokey says:
January 26, 2012 at 7:03 am
Obviously, if I believe that Wood’s results were wrong then I must believe that, all other things being equal, doubling the CO2 partial pressure will increase the average surface temperature. But, of course, all other things are never equal in a non-linear possibly chaotic system like the Earth and its atmosphere. I believe I fall into the lukewarmer camp because I think the overall climate sensitivity is at the low end of the IPCC range, on the order of 1.5 C/doubling. As far as policy, I think it’s pointless and possibly immoral to spend vast amounts of money to prevent a small increase in risk from things like extreme weather events when there are people at risk now from the extreme weather events that happen without an increase in temperature.
Skeptics should be concentrating their efforts not on the fundamental science, much of which is correct, and concentrate on the really flaky stuff like the IPCC Working Groups 2 and 3 reports on effects of warming and costs of mitigation. The first seem to be overestimated and the second wildly underestimated, not to mention unlikely to happen.
Smokey says:
January 26, 2012 at 7:03 am
Smokey, the corona (atmosphere if you will) of the sun is aprox 1-3 million K. Does the corona of the sun heat the sun?
DeWitt,
Thanks, looks like we’re pretty much in agreement: there is no looming disaster.
• • •
mkelly,
I’m not sure I understand your comment. Hotter radiating objects heat cooler objects, don’t they?
Bryan,
It’s winter in case you haven’t noticed. I need at least four hours of cloud free sunshine between 9:30AM and 3:30PM (otherwise the sun is too close to the horizon and I have a large hill on the western horizon as well) with little wind to get good results. I’m lucky to get that happening one day/week.
Vaughan Pratt refusing to get involved in a nit-picking session at Climate, Etc. is not the same as not defending his results. He hasn’t pulled the web page, and as someone once said about Wood, the results still stand. Look at the level of criticism here. I’ve been told by two different people that I need to put cotton inside the box to replicate Wood.
Right now I’m modifying one of my boxes so I can measure the air temperature profile from the exposed rear surface to the window. I’ll also switch the boxes between the different insulating boxes. But even if I can convince you that Wood was wrong, I’m betting that somehow you’ll manage to convince yourself that it doesn’t matter.
You also keep bringing up the Penn State experiment as if it somehow proves Wood was correct. It doesn’t. The important finding with regard to Wood from PSU is that the surface temperature at night with the IR transparent film can be lower than for the IR absorbing film. The reverse is never true. But this effect is small because the net radiation from ground to space on a clear, cool night with low humidity (which also doesn’t happen every night) is only ~100 W/m² compared to ~1,000 W/m² from direct normal sunlight on a clear day at local noon. Also, if there’s dew or frost, that makes both films IR absorbing. The geometry is such that cold air that forms near the IR absorbing film falls to the surface causing convective mixing while there is a temperature inversion in the IR transparent tunnels with no convection. The end result was that the IR absorbing film was not cost effective, not that it didn’t work.
DeWitt Payne says
“But even if I can convince you that Wood was wrong, I’m betting that somehow you’ll manage to convince yourself that it doesn’t matter. ”
On the other hand I have no doubt that you will carry out a careful and honest experiment.
After scrutinising your work I will try to replicate your results.
But I need to see a finished report before starting.
I have no hidden agenda.
The experiment tops any theory.
Roll on better weather
Smokey the answer is that it does not. It is not dense enough compared to the sun. If something that hot has no effect then how exactly does a few CO2 molecules.
mkelly,
It seems to me that if the Sun’s corona is at 3 million degrees, and the Sun’s surface is less than one one-hundreth of that, that the corona would heat the chromosphere. It just wouldn’t heat it very much, because the corona has so few molecules.
Paul Birch says at 1/26 7:08pm:
“The second law does not say that you can’t extract work from reversible processes…”
Paul ! – Yes it does. Double check. 2nd law forbids entropy to go down. You extract the work from a reversible process, drive entropy down. Can’t. Then it is irreversible, some energy left town.
Paul Birch continues:
“…the higher their total energy KE+PE the higher they go.”
Yes. But here none of the catapults are stronger by definition KE + PE = constant. 1st law. If the molecules all have the same constant total energy in the closed system, they all go just as high. Just like when your tennis balls are launched with the same total energy. Same height. 1st law.
Trick,
The First Law only requires that the total energy of a closed system remain constant. It does not require that the total energy at every location within, say, an insulated column of air, be constant. You can’t change the gravitational potential energy distribution without changing g, but the temperature profile can be anything within the constraints that the total energy be constant. It can increase or decrease with altitude. However, the system is only stable, i.e. in thermodynamic equilibrium, when it’s isothermal.
Trick says:
January 26, 2012 at 7:23 pm
Paul Birch: “The second law does not say that you can’t extract work from reversible processes…”
Trick: “Paul ! – Yes it does. Double check. 2nd law forbids entropy to go down. You extract the work from a reversible process, drive entropy down. Can’t. Then it is irreversible, some energy left town.”
Oh dear. Reversible processes are isentropic. Adiabatic expansion is one of them. The entropy in the gas stays the same, even though its enthalpy goes down.
Paul Birch: “…the higher their total energy KE+PE the higher they go.”
Trick: “Yes. But here none of the catapults are stronger by definition KE + PE = constant. 1st law. If the molecules all have the same constant total energy in the closed system, they all go just as high. Just like when your tennis balls are launched with the same total energy. Same height. 1st law.”
The molecules do not all have the same energy. They have a wide range of energies (or momenta) randomly drawn from a Maxwell distribution, from zero all the way up. All of the catapults are different.
It’s all about the energy. When a molecule collides with something (a photon, another molecule) there is an exchange of energy. The molecule can go faster or slower after the exchange.
If the energy is random kinetic, it is called heat. Temperature is the RMS average random kinetic energy. Note not all kinetic energy is random. The non-random part can do work. So can a difference in temperature, but not a uniform temperature.
The part of the energy that is not heat includes latent energy of evaporation, energy of position, and energy of motion – aka wind. These correspond to various degrees of freedom. In a real atmosphere, all of these are going on at the same time. Because of gravity, part of the motion (the non-random up-and-down part) gets converted from speed to energy of position. You can get an approximation of this by calculating the ratio of density to pressure. If this ratio stays the same, temperatures approximate isothermal..
But when they don’t, you get a change of temperature with altitude. Since in the real atmosphere we do get a lapse rate, which is less in a wet atmosphere due to part of the energy being in latent form. The details are very heavy math, which is why it’s easy to overlook some part.
Ocean heat: IR goes to evaporation until the air is saturated with water. Visible light goes to depth. In the North Atlantic, this leads to the Gulf Stream, and the heat radiates to the atmosphere and space over thousands of miles. In the Equatorial Pacific, the surface current is mostly East/West, so it gets warmer over time, until El Nino dumps the excess heat.
Ice Ages: should be caused by an increase in albedo from 30% to about 39%. This could happen from volcanoes, or from the heat engine slowing down. If the Polar areas remain frozen, while the tropics and temperate areas have more clouds, especially high level clouds, temperatures slowly drop. Or, in the case of the Younger Dryads, rapidly drop. We know that at the end of Ice Ages,, temperatures rapidly increase. The mechanism is so far unknown, but should have positive feedbacks until it gets close to current temperatures. So the rapid increase in temperatures from 1850 to 2000 is probably not CO2 related, and may have just ended around 1998. My guess is ENSO, and that could be a 1000 year cyclical pattern.
Repeat: temperature is the average random kinetic energy per molecule. If you double the number of molecules and double the random kinetic energy, the temperature remains the same. The lapse rate is due to the fact that a greater proportion of the total energy at altitude is in the form of positional and non-random kinetic energy. Jet streams at 8 km are often hundreds of km per hour, and since this is non-random, results in lower temperature. Winds are lowest at the surface, and this is part of the reason surface temperatures are higher. Energy, changes in the form of energy, and energy flows – including but not limited to radiation – explain the weather, and over time, the climate. Emissions are random, absorption is non-random due to the lapse rate, so H2O and CO2 concentrations play a role, but the IR path is only one of several paths that determine energy flows.
This is a nomination for a new top level post. It is a follow up to the comment above from Stephen Rasey above on Jan 21, 2012.
Two Sides of the Albedo; The Problem with 240 W/m^2.
by Stephen Rasey
240 Watts/m^2 is an important number in climate science[4]. People use it to calculate the black body temperature of the Earth ( Tbb=255 K = -15 C) and compare it to some measured average temperature, (Tm = 285K = +15C) which converts back to a blackbody heat flow of 390 W/m^2. The difference between 240 W/m^2 and 390 W/m^2 is explained as the Green House Effect.
There is a big problem here. Along the pathway from the sun to the surface of the Earth in this algebra, we have violated the laws of Thermodynamics at least twice.
Let me be clear at the beginning. I believe in a Greenhouse Effect (GHE). I am questioning the physical model and the arithmetic used to calculate the size of the GHE. In the end, I hope to convince you that a more useful estimate for solar heating of the Earth’s surface is 341 W/m^2, equivalent to about 279 K, not the 240 W/m^2 and 255 deg K generally assumed.
How people get to 240 W/m^2 is that take a solar constant (So = 1364 +/- 3 W/m^2) as the energy received full on at the distance of the earth. They then divide by 4 to spread that energy evenly over a sphere representing the earth’s top of atmosphere, yielding 341 W/m2.[4] THIS is the ORIGINAL SIN! Dividing by 4 leads to a dead planet; it is an isothermal planet without net heat flow, anywhere. Next, this 341 W/m^2 is passed through an albedo mechanism, reflecting 30% and transmitting a net 240 W/m^2 to the ground. At this point, it is equivalent to a Blackbody at 255K and thus the second sin against thermodynamics is committed.
In the past month on WUWT, there have been several highly enlightening posts by Brown, and Eschenbach [3] that have made clear that when any non-greenhouse gas[1] is in a gravity field of a planet, and the surface of the planet is isothermal, then the atmosphere must be isothermal at the temperature of the ground. The purpose of those posts was to refute the theory that a gravitational field can cause the lower part of the atmosphere to be warmer than the upper part. Until the temperature lapse rate goes to zero, work can be extracted from the temperature difference. Therefore, the atmosphere at equilibrium must be isothermal if the heating of the ground is uniform. There is no net heat flow. There is no convection, no net conduction, no latent heat exchange, no heat of vaporization or of fusion because there is no temperature differences to drive the heat flow. It’s a dead planet.
For this next argument, I ask you to imagine 4 concentric spherical shells, from inner to outer, call them G, B, T, S. G is the Earth Ground, a solid sphere with a gravitational field.. B is the Albedo layer. T is the Top of the Atmosphere with an ideal gas between T and G. S is the surface of solar irradiance. These shells are all spherically symmetrical with a common center. Why do we have S as a sphere? Because we divided solar insolation by 4 !! The only way to uniformly heat G is to have a spherically uniform radiator. When we do that, using the inverse square laws, we see that in order to irradiate G with X W/m2, surface S must also radiate at X W/m^2. S and G must be at equal Black Body temperatures for any X! We have not specified radius of S or G or T, except by “concentric” Rg .LT. Rb .LT. Rt .LT. Rs. Therefore, the temperature at T must also be equal to S. The Atmosphere between G and T is isothermal with everything at the same temperature. The Atmosphere has a pressure gradient, but is without a temperature lapse rate.
Wait a minute! What about the Albedo layer B? Don’t we have to multiply the Energy flow from S through T to G by the albedo fraction “A”, call it 30%. So, to plug in some numbers, S is radiating by 341 W/m2 at 279 K. T is at the same temperature. 341 W/m2 encounters the Albedo B, and only (1-A) or 240 W/m2 get through to the Ground. Therefore Ground must be bathed in 240 W/m2 that implies a black-body temperature of 255 K. Oops! Suddenly, we are confronted with a non-isothermal system, with the ground colder than the top of the atmosphere. Thermodynamic alarm bells should be going off in your heads.
We can plug Dr. Brown’s silver wire into the atmosphere near T and the other end into the Ground G and extract work from the temperature difference, warming Ground G until it warms back to near temp of T. We then unplug the wire. Ground G is again only receiving 240 W/m2 again. But who here believes that Ground G will drop in temperature again back to 255 deg K? How can it get colder than its surroundings? Of course, it will not.
At thermal equilibrium the Temp of G must be the same as T and S. Otherwise we could use the silver wire to extract work. The Albedo fraction (A) of Albedo B must make no difference in the temperature of G! Remember, this is a DEAD PLANET we are considering here. We made the Original Sin of dividing by 4 long ago.
Something is missing. There is an albedo layer. It is blocking about 100 W/m2 of downward solar irradiance heat flow. But, the albedo has two sides. It also blocks 100 W/m2 of heat flow upwards. It is serving as an insulator, a device that reduces the transmission of heat in both directions, but cannot by itself change the ultimate temperature on its two sides. An albedo is not a Maxwell’s Demon.
Don’t believe me? Try this on your next campout or hike. In the morning, fill your water bottle. Wrap it in a “space blanket” with an albedo at least 90%. So little of the sun’s heat will get to the bottle that the temperature of the water should plummet. By noon, you should have a welcome block of ice to cool you down. Read the directions of the space blanket carefully! If you put it on wrong side out, you’ll heat the water to boiling instead of freezing it. — No. Space Blankets don’t work that way. They have two equal sides. They cannot refrigerate their interiors; they can only retard the heat flow.
Albedo’s have two sides, too. They reflect heat away. They reflect heat back to the ground. They insulate. They retard the heat flow, not change the temperature of the end state. They do this without GHG’s (2). But an Albedo, whatever it reflectivity, cannot change the temperature of the ground —- ON A DEAD PLANET.
Put an Albedo on a rotating planet, illuminated by a near-point-source sun, possessing a night and a day, equator and poles, and make the albedo time varying by the minute and hour – it will make all the difference in the world! Solar insolation is anywhere from 1365 W/m2 to zero and all manner of heat capacity mechanisms come into play. But on that world, you don’t divide the solar irradiance by 4.
Stephen Rasey
Notes:
(1) The same is true if the atmosphere contains GHGs for the same reasoning, but I don’t want to go there, yet.
(2) But Albedo is from clouds and ice, and H2O is a powerful GHG! I’ll concede the point, but maintain we are talking about two separate properties of the same compound. Either way, a GHG driven albedo cannot act as a Maxwell’s Demon and keep the concentric spheres at different temperatures.
(3) Other relevant WUWT posts:
Brown, Jan 24, 2012: http://wattsupwiththat.com/2012/01/24/refutation-of-stable-thermal-equilibrium-lapse-rates/
Eschenback: Jan 19, 2012: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/
Eschenback: Jan 13, 2012: http://wattsupwiththat.com/2012/01/13/a-matter-of-some-gravity/
Brown, Jan 12, 2012: http://wattsupwiththat.com/2012/01/12/earths-baseline-black-body-model-a-damn-hard-problem/
Rasey, Jan 21, 2012: http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872239
[4] Just a small sample of references:
http://data.giss.nasa.gov/gistemp/2007/
Earth’s Climate History, Antón Uriarte
Image: Uriarte: Appendix 1 [Note Thermals and transpiration on an isothermal uniformly-insolated dead planet.]
http://www.soest.hawaii.edu/GG/FACULTY/POPP/Lecture2.ppt [same dynamic weather systems on a 342 W/m2 isothermal dead planet.]
Alley, Earth: The Operator’s Manual Page 352 #9
Anderson, Decadal Climate variability: dynamics and predictability. Pg. 296-7. The Figure 2 on 297 with “Surface temperature without Greenhouse effect” – Venus = (minus 46 deg C)
Wendisch-2012, “Theory of Atmospheric Radiative Transfer: A Comprehensive Introduction. Fig. 1.2 after Trenberth et al 2009. Same diagram as Uriarte above.
Stephen Rasey,
You seem to be laboring under two misconceptions.
1) “AVERAGING IS A SIN”
If an actuary say the average life-span in the US is 78.1 years, no one should interpret this to mean that everyone will always live exactly 78.1 years.
If an economist says that the US average income is $39,945, no one should interpret this to mean that everyone will always earn exactly $39,945.
If a scientist says the average incoming solar radiation is 340 Wm^2, no one should interpret this to mean that every square meter always will receive 340 W/m^2.
Any scientist who reports an average is well aware that averages do not tell you the value at every time for every location. The whole point in finding an average is to have a simple number that best summarizes the situation. Now, if a climate model actually assumed that the insolation was and even 340 W/m^2 everywhere, that would be Bad Science™. Fortunately, no climate model I know of makes such a stupid assumption.
2) “340 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM”
You explicitly use this assumption when you say “The only way to uniformly heat G is to have a spherically uniform radiator…. S and G must be at equal Black Body temperatures for any X!”
I could easily divide the outer shell (which you called “S”) into 1 m^2 regions, and then have 1/2 of each square meter surface radiating 2 * 340 W/m^2 and the other 1/2 radiating 0 W/m^2. Or have 1/10 radiating 10 * 340 W/m^2 and have 9/10 radiating 0 W/m^2. Or have 1/200,000 of the surface radiating at 200,000 * 340 W/m^2 and 199,999/200,000 radiating 0 W/m^2 (which would ~ correspond to the temperature of the sun & the faction of the sky covered by the sun). Or put another way, I could place a small, hot 340 W light bulb above every square meter of earth (with all of the light focused downward, of course) and produce a uniform 340 W/m^2 “insolation” at the TOA.
IF you HAD to make a model with globally uniform insolation, having1/200,000 of the shell “S” radiate at 200,000 * 340 W/m^2 would be the best option (for a clearly bad model that ignores night/day and poles/tropics). (By the way, this model WOULD show effects of albedo and of GHGs, and the surface could be above 255 K.)
Other than these two points, I agree with the rest of your analysis and conclusion. Surrounding the entire earth with a uniform 255 K shell (with any albedo) shining light onto the ground (with any albedo) with an intermediate layer (with any albedo) and any atmosphere (with or without GHGs) will cause the surface to be 255 K. (Actually, we would have to further assume no geothermal energy).
This would indeed be a very dead world, with no visiblelight for photosynthesis.
Tim, First your point 2:
240 W/m^2 is what is equated to a 255 K blackbody.
340 W/m^2 equates to a 279 K blackbody.
That difference is key to my central point that 279 K should be the ground temperature in an average insolation model. So the greenhouse effect is the increase from a base tempature of 279 K to some agreed upon average 288K.
Therefore the Greenhouse effect on Earth does not
add 33 deg K to a base like of 255 K,
but only 9 deg K to a base line of 279 K.
As to your point 1:
We are agreed that it is a stupid assumption. But lots of people, use 240 W/m^2. Each of the references I noted in [4] above made use of 240 W/m^2.
Not counting my own uses, there are 49 uses of it in this one thread.
I cannot speak to what GCMs use. But a lot of literature is littered with 240 W/m^2 and 255 K as the energy and temperature of the Earth without the greenhouse effect.
Google “240 W/m albedo” and look at the first couple of pages. Or Goggle ‘albedo “255 K”
Or better yet, Google ‘ “240 W/m” “255 K” ‘ and look at the list of hits. Look at the ones ending in “.edu”
Pull up the WUWT post: Earth’s baseline black-body, A damn hard problem. Do a find on “240”. A lot of people have fallen into the conceptual trap that the albedo itself, in a uniformly insolated planet, will change the temperature of the ground. Using Dr. Brown’s silver wire, you could pull power from the temperature difference across the albedo. It cannot be so.
@Tim Folkerts 9:38am,
First, the temperature would be 279 K, not 255 K. That is my essential point. I’m not giving ground on that.
Secondly, Divide solar insolation by 4 to average the insolation — EQUALLY — over the planet Earth. I didn’t make that up. A lot of climate scientists do that every day as a “Toy” model. It’s in the books. Its in the lecture notes. Equally over the planet means it is Spherically symmetrical. Therefore, the source of illumination must be spherically symmetrical. That is a direct consequence of dividing by 4.
Thirdly, by your 1/200,000 of the shell in radiating at 200,000 * 340/W ^2 and the rest at 0K, you basically make my point. Double the size of the shell. Each of those radiating areas have to spread their radiation out by the inverse square law so each only provides 1/4 of the energy per square meter from S to G. But each radiating point has expanded by a factor of 4. So four times the surface area providing only 1/4 the energy per meter of S to the Ground G equals no net change in radiation to G. G, or at least the top of the albedo layer B, receives 340 W/m^2, regardless of the size of S.
Fourth, because the ground is uniformly insolated, no point different than another, there would be an isothermal atmosphere, even in a gravitational field (See Brown: Refutation…), and there can be no temperature difference across any albedo layer, elsewise a silver wire across the layer would be able to extract work.
Albedos can change the temperature of the ground only when they are time-varying and/or there is a night and a day – i.e. non-uniform insolation.
That is why dividing by four is the Original Sin in climate science. Dividing by 4 creates an isothermal planet model. Any illustration (such as Uriarte) that shows 341 +/- 1 W/m^2 as incoming solar energy, and also shows almost 100 W/m^2 in “Thermals”, “Convection”, “Evapo-Transpiration” heat flow is inconsistent in its core assumptions.
Stephen, I thought I had responded earlier, but it looks like it got lost. So let me repeat the main points.
1) Yes, I should have said 2) “240 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM” (not 340 W/m^2). I got ahead of myself and wasn’t paying enough attention to the specific number. But this is really a secondary point.
The main point had to do with the SPECTRUM of the incoming light and this idea of yours: “A lot of people have fallen into the conceptual trap that the albedo itself, in a uniformly insolated planet, will change the temperature of the ground. ”
* If the incoming light is from a shell with a uniform temperature completely surrounding the earth, then everything within that shell will become that temperature independent of albedo, emissivity, atmospheric pressure, or GHGs. In this case you are right that worrying about albedo is “a trap”. The “sunlight” and the “earthlight” have the same spectrum and everything will be symmetric, so changing the interior cannot affect the balance of energy. You cannot have something on this world like clouds that reflect sunlight and absorb earthlight, because sunlight and earthlight are the same.
* If the incoming light is from “a million tiny stars”, then the other factors DO matter. There is no longer symmetry between the incoming sunlight (mostly 0.1-4 um) and the outgoing “earthlight” (mostly 4-100 um). Now albedo can and will matter, cooling the earth by reflecting away the incoming light. Now GHGs can and do warm the earth by letting sunlight in quite well, but not letting the earthling out as easily. Now, ignoring albedo is wrong, and your thinking would be “a trap”.
Since “a million tiny stars” is a closer analogy for uniform sunlight, then the effects of albedo & GHGs will matter, even in a uniformly illuminated world. Albedo WILL lower the effective radiating temperature from 279 K (340 W/m^2) to 255 K (240 W/m^2). GHGs WILL raise the surface temperature by preventing earthlight from escaping directly to space (which apparently amounts to ~ 33 K).
Stephen,
I did type too quickly earlier and got the wrong number. I should have typed
2) “240 W/m^2 OF RADIATION MUST HAVE A 255 K BLACKBODY SPECTRUM”,
not 340 W/m^2, so we are in agreement there.
But that is not my key point. The key point is what the incoming “uniform radiation” is like. There are two main possibilities to consider.
1) If the incoming “sunlight” is from a shell with uniform shell with a constant temperature completely surrounding the earth (as you postulate), then the temperature anywhere inside will be the same temperature as the shell independent of albedo, emissivity, atmospheric pressure, or GHGs. The “earthlight” and “sunlight” will be identical, and anything inside (like clouds or space blankets or GHGs) will have an identical effect on energy flows up and down, and cannot change the equilibrium temperature.
If the shell is 340 W/m^2, the earth will be ~ 279 K.
If the shell is 240 W/m^2, the earth will be ~ 255 K.
Worrying about albedo would indeed be a “trap”, since it won’t affect this world with this “sunlight”.
2) If the incoming light is from “a million tiny stars” then the conditions of the earth can and will matter. The “sunlight” will be mostly 0.1-4 um, while the “earthlight” will be mostly 4-100 um. This means that clouds or GHGs can have a different effect on upward and downward light. Clouds can and will reflect away “sunlight” and absorb “earthlight” (lowering the effective blackbody temperature of the the earth from 279K to 255K). GHGs can and will absorb “earthlight” better than “sunlight”, which has the effect of warming the surface (apparently from ~ 255K to ~ 288 K based on actual observations).
The surface will be a uniform temperature, but the atmosphere will be cooler as you go up (due to IR radiation emitted from the atmosphere to space).
Worrying about albedo is now critical, and your thinking has now become the “trap”.
Both of these worlds are “uniformly illuminated”, but the second is a much better approximation of a uniformly illuminated earth with 340 W/m^2 of “sunlight”. The second has the correct spectrum of “sunlight” to match real sunlight. The second will have effects of GHGs and albedo, even on a uniformly illuminated world.
Tim, you bring up an excellent point that the spectrum of the energy received is significantly different than the energy emitted. Thank you.
I thought about this, too in a comment from Jan 12. (quote shortened, full comment here)
It is the value of “a=0” people use to get to 255 K if they assume A=30%
In the next comment, I showed that the total energy leaking through the albedo is the same (1-A) regardless of the value of “a”.
You discuss in Point 2 the issue of spectrum conversion very well. It is a serious issue, but I don’t think you can just assume the 255 K value is correct.
On the one hand, you argue that an Albedo of clouds lowers the blackbody temperature of the Earth, but then GHG’s ride to the rescue to raise its temperature. Wouldn’t make more sense to net out the change? “a = A” might not be right, but “a = 0” is quite likely wrong. For one thing, even at night, we can look up and see the bottom of the clouds from scattered light from the ground. There has to be some return.
Four thoughts:
1. There still exists the argument of the Isothermal atmosphere in a gravitational field and Dr. Brown silver wire which we bridge across the albedo layer. If “a” does not equal “A”, then we have a perpetual temperature difference across the albedo with which the silver wire can extract work. That is a good reason to say “a” must equal “A”. But it is not proof. Well, maybe it is, and I just don’t grok why, yet.
2. In your million star suns, the total energy may average 340 W/m^2, but spectrum will not be a 340 W/m^2 = 279 K blackbody, but a black body of much higher temperature and heat flux attenuated by distance (or spacing of the stars). So now we have 340 W/m^2 hitting the TOA, but not blackbody spectrum. How does that change arithmetic or Brown’s isothermal atmosphere? Does the albedo become a Maxwell’s Demon?
3. The literature doesn’t seem to spend much time on this spectrum conversion. Yes, solar energy is converted to heat. But at the same time, they talk of “thermals”, “latent heat”, “convection” in a model that needs to have an isothermal ground. At equilibrium there must be no net energy flow. There is no day and night, everywhere it is the same grey illumination. No wind, no weather. A significant piece of the literature is inconsistent in this regard.
4. The whole “divide by 4” average insolation is a bad idea to try to explain climate mechanisms. Might as well divide by zero. It is better to scrap it rather than fix it. Climate is a dynamic equilibrium, not a static one. Day and night cannot be averaged. Eppur si muove.
Two errata to my post from 9:59 pm above:
On the one hand, you argue that an Albedo of clouds [ made of GHGs ] lowers the blackbody temperature of the Earth, but then GHG’s ride to the rescue to [eliminate the drop and further] raise its temperature. Is the lowering of the blackbody temperature a real thing or only an accounting gimick?
3. The literature doesn’t seem to spend much time on this spectrum conversion. Obviously incorrect. Spectrum conversion is a big part of the literature. It is just not addressed much brief path from sun to ground. (1364 / 4 = 340) * 70% = 240 W/ m^2 implying 255 K black body.
Stephen,
This is getting a bit off-topic from the original post, but I will make a couple quick comments (or perhaps not so quick — we will see).
>Why do we account for albedo only on the direct-from-sun energy,
>but everyone seems to ignore it on surface-to-sky path?
For your analysis, there are not just two parameters for reflection (albedo for sunlight and earthlight). First, you need to consider reflection, transmission and absorption –> R+T+A = 1 (all light will either get reflected by, transmitted through, or absorbed by an object it interacts with) (albedo = 1 – reflectivity). You need to consider how these relate to both sunlight and earthlight (actually, you should consider it for each individual wavelength, but since sunlight (0.1-4 um) and earthlight (4-100 um) basically don’t overlap, you can pretty well treat them as two distinct ranges with two separate values) . You need to consider how these differ for the ground and the atmosphere.
So you would have a matrix of parameters like R_s_g (reflectivity of sunlight by the the ground) or T_e_a (transmission of earthlight through the atmosphere).
You say “a fraction “a” is reflected back downward”, so this would be R_e_a in my notation. But the atmosphere does not reflect much IR at all, so “a” would be very close to zero. And the ground does not reflect much earthlight (or “atmospherelight”, but “atmospherelight” would be the same basic wavelengths as earthlight).
Also, it is wrong to say “and (1-a) continues into space” since you did not include the energy absorbed by the atmosphere. With a cloud cover, nearly all of the earthlight is absorbed by the cliouds. This means “a” is very close to zero, but the fraction that escapes to space is ALSO very close to zero.
>If “a” does not equal “A”, then we have a perpetual temperature difference
>across the albedo with which the silver wire can extract work.
>That is a good reason to say “a” must equal “A”
The only problem comes if you try to make “a” and “A” different for the same wavelength. This would violate fundamental rules of physics, but you are not doing that There is no problem with perpetual temperature differences and perpetual extraction of work if there are “connections” to objects of different temperature (ie the sun and space). If you remove those connections (eg by putting an isothermal shell around the earth or enclosing the system in insulated walls), then perpetual temperature differences and perpetual extraction of work would be a problem.
>At equilibrium there must be no net energy flow.
It is important to distinguish between “steady-state” and “equilibrium” (which is much more restrictive than steady-state). A bathtub with water running in at the same rate it drains is a steady-state situation, but it is not an equilibrium situation. When the earth is gaining 240 W/m^2 of sunlight and losing 240 W/m^2 of earthlight, that is a steady-state situation, not an equilibrium situation. (This assumes an average over large areas and long times. For shorter times or areas, steady-state is not a good approximation.)
The earth is NOT in equilibrium — not even close. The earth IS, however, very close to steady-state (as far as net energy is concerned). Deviations from steady-state over periods of decades are pretty much synonymous with “climate change”.
4. The whole “divide by 4” average insolation is a bad idea
>to try to explain climate mechanisms.
Again, I don’t think anyone is using an average this to “explain climate mechanisms”. Averaging the incoming sunlight (or averaging evaporation or averaging outgoing thermal IR or … ) is simply a handy way to summarize a very complex situation. It is a way to present the basics to a wide audience. Actual analysis of details must (and does) worry about details. An audience like you or me or Willis or Dr Brown or Anthony or Dr Mann enjoys pondering the details, but IMHO that does not make the summary “a bad idea” for a different audience.
Tim,
Let me refocus our discussion to one point very much on topic to this thread of Perpentuum Mobile.
Back on Feb. 7, my long post was for the explicit purpose to show that climate scientists take average solar insolation (1367 / 4 ) W/m^2, then multiply by (1 – 30% albedo) to achieve a 240 W/m^2 and they equate that to a 255 K black body. I showed via the 4 concentric shells, that there would have to be a temperature of 279 K above the albedo, and 255 K below the albedo at equilibrium. This creates a Perpentuum Mobile, a violation of Thermodynamics. Therefore the model used to defend the 255 K black body temperature of the Earth is fatally flawed.
In your most recent post of Feb. 11, you said, I don’t think anyone is using an average this to “explain climate mechanisms”. “Anyone” is a pretty universal word. I only have to show a someone using it to invalidate the statement. My references in [4] of the Feb. 7 are sufficient for that. But I’m a sport. Show me ONE reference that derives a 255 K Earth ground blackbody average temperature that DOES NOT arrive at it via “average solar insolation”,
Your arguments concerning spectrum conversion and how the albedo is a function of wavelenght are very astute. They cannot be easily dismissed. But neither can Thermodynamic arguments be dismissed so as to create a Perpentuum Mobile.
The Earth is a system in dynamic equilibrium. No argument. But that is not the system under discussion.
“Average Solar Insolation” describes a planet with a uniform temperature. It is a “Toy” to “summarize a complex situation”, but summarize it so badly it leads to erroneous conclusions. Like the “Surface Temperature without Greenhouse Effect:”
Earth = -15 deg C, Venus = -46 deg C.
(From page 297, Anderson, Decadal Climate Variability: Dynamics and Predictability.)
Stephen, you are missing an important point, I think.
The average EMITTED radiation is often described as 255K blackbody radiation.
The average ABSORBED radiation is never described as 255K blackbody radiation.
They are equal in the sense that emitted and absorbed radiation are both ~ 240 W/m^2, but they are most certainly not the same spectrum, and never will be. This means they can (and do) have two very different effects on earth. It is kind of like saying a $1 bill and 100 pennies are the same. Yes, they have the same official value, but they don’t act the same in my pocket. I can’t put the pennies in a vending machine.
Giving me 100 pennies is distinct from giving me a $1 bill, even though in some sense they are the same.
Giving the earth 240 W/m^2 of sunlight is distinct from giving the earth 240 W/m^2 of thermal IR, even though in some sense they are the same.
PS. The outgoing radiation is, of course, modified by GHGs, so even it is not strictly 255 K BB radiation if GHGs are present.
PPS You say “I showed via the 4 concentric shells, that there would have to be a temperature of 279 K above the albedo, and 255 K below the albedo at equilibrium. ” but this is incorrect in your scenario. If the earth was surrounded by 279 K BB radiation, then the temperature everywhere inside would indeed be 279 everywhere, including below the “albedo layer”.