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.

Regarding N and Z, my understanding of it is this (could be totally wrong, IANAPhysicist):
When a star is accreting, gravitational contraction converts gravitational potential energy into thermal energy. It is this thermal energy which acts as an opposing force to gravity, preventing the star from collapsing completely. As the star gains mass, its gravitational attraction increases as does its thermal energy. At some point thermal energy becomes large enough that fusion occurs and the star becomes stable. What does this have to do with N and Z? As I understand it, they claim that they can estimate atmospheric temperature from the density of the atmosphere. A volume of air that is more dense has more mass than the same volume of air that is less dense. Therefore, shouldn’t the dense volume of air have more gravitational potential energy that can be converted to thermal energy? And, can we assume that any volume of air that is being contracted by gravity will have some thermal energy above the black/gray body energy?
richard verney
The affects of the tides and other orbital parameters will indeed affect the atmosphere. Work is indeed done in the thermodynamic sense of external force acting on a closed system. In fact the Moon and Earth system will experience a loss in Kinetic energy from the “gravitational drag” of tides. These are third or fourth or even higher order effects. Like variations of the trace gas CO2, the affect of these things on “atmospheric temperatures” are extremely small and inconsequential compared to radiative forcing from our Sun.
JJThoms says:
January 20, 2012 at 6:03 am
“Leave this for a few months. Those warm and cool atoms of argon will be wizzing about in the box but they will be loosing no energy. All that wizzing eventually equalizes the temperature throughout the gas – the low pressure gas will have the same temperature as the high pressure gas.”
But according to the Ideal Gas Law, this is not possible. The temperature distribution must follow the pressure distribution so that PV = NkT
where P is the absolute pressure of the gas measured in atmospheres; V is the volume (in this equation the volume is expressed in liters); N is the number of particles in the gas; k is Boltzmann’s constant relating temperature and energy; and T is the absolute temperature.
http://en.wikipedia.org/wiki/Ideal_gas_law
They very fact that there is a vertical pressure gradient results in a temperature gradient.
So in this mythical elevator….
First it has no sides. It has no top. The bottom can vary widely…. Density can be 50% different.
With temperature changes the hight as well as the width of the elevator can grow or shrink.
What is so hard to figure out? As the atmosphere has heat added it grows. This growing increases it’s surface area. This increased surface area. Increases the thermal exposure to the cold of space. It cools.
The atmosphere as well as earths temperature is self regulating.
Jeremy says:
January 19, 2012 at 7:52 pm
Gravity has NO AFFECT ON TEMPERATURE.
How many times must it be said.
You have to do WORK to create a change in temperature – this is basic thermodynamics!!!!
Jeremy please use the standard equation to state your case. simply (Joel I said simply) Q= U+W.
If W=FD and F = mg then W can be done by heated air rising IF volume changes. We have been told there is a diurnal bulge. As W is a path function and W1-2 (air going up) does not have to equal W2-1 (air coming down) then W could be being done via gravity.
KE and PE in U do not change IF the center of gravity does not change.
Bomber_the_Cat :
January 20, 2012 at 3:40 am
In my books an ocean is not a closed system, the way this column of gas is proposed. It is not a good example.
Cosmologists believe in perpetual motion. They believe light travels through the universe ad infinitum without using any energy!
Good luck in this quest to make those who do not understand begin to understand….don Quixote….. er ….. Willis.
For me this is a very simple issue:
1. The Earth is only receiving enough power (or energy flow) from the sun to emit an average of about 240 W/m² from the surface of the Earth. That is derived first, from the solar constant divided by four, the ratio of flat spherical surface area of the Earth to the area of the disk of solar radiation absorbed by the Earth and second, by applying an assumed 30 percent optical reflection factor. Reflected solar energy is treated as never having arrived.
2. According to the Trenberth diagram and the Stefan-Boltzmann equation, the Earth is actually warm enough to be radiating something on the order of 396 W/m² average power from the surface.
3. A transparent, non-greenhouse gas containing atmosphere cannot, by definition, prevent or stop the surface from radiating all this power to outer space. Thus, the surface would be continually losing a net average of 156 W/m² as long as it remained that warm. Fixing this would require a close relative of the perpetual motion machine–the perpetual power creation machine.
4. An average surface power radiation of 396 W/m² can only continue with an atmosphere that can absorb a net 156 W/m² from the outgoing radiated power and return it to the surface.
All technical details of exactly how that happens are irrelevant to this power balance requirement.
Kevin Vaughan,
No they don’t, light is red shifted if it moves opposite a gravitational field, and blue shifted if moving with the field. The frequency shift shows a change in energy due to the gravitational field. Gravity influences everything, even electromagnetic waves.
Willis, or anyone else.
First, the gravity theory is a non-starter. Willis is correct and people seem to keep missing some very fundamental properties here. Heat is not temperature. Heat is temperature at volume. Keep molecular temperature the same, but increase density and you have more heat. Willis is exactly correct in saying that the upper less dense molecules can have the same temp as the lower more dense molecules and it all be at equilibrium.The PE and KE will vary over height (as related to the gravity point source) so total energy (TE) of any specific given volume of gas can and will come to complete equilibrium over time as the temperature of each given molecule reaches the same temperature. It’s a volume issue. His proof is also correct. If this didn’t happen, you would and could have a perpetual motion issue to deal with. Case close, can we move on to something new please? This doesn’t prove GHG theory correct, it just says the alternative theory is incorrect and people are looking like buffoons. (Engineer by schooling)
I do have a question though. As has been said over and over, our atmosphere volume has to increase if it warms (Total heat). Why are we trying to measure average temperatures, which is meaningless and damn near impossible, instead of not just measuring the height of the atmosphere instead? You would think that an average height of the atmosphere would be much easier to do as a function of time and any trends would then be easy to spot.
AJ
Tim Folkerts says @ur momisugly January 20, 2012 at 6:06 am
Tim, please refer to this page and specifically to the sentence that reads “Ignoring tiny corrections for gravity, the gas will be distributed uniformly in the container, so the only unknown is the velocity distribution function”.
OzWizard says:
January 20, 2012 at 6:43 am
Try to understand it Willis, please. All your thought experiments are irrelevant unless you can demonstrate where their math, or their data, or their ‘grey body’ model, or their regression is wrong. To do that you need to understand what they have done and, by your own admission, you do not yet understand what they have done.
Their math, etc can be completely correct AND Willis can also be correct. All it takes is understanding that the GHE has a maximum value determined by exactly the parameters K&Z use. So, you have a GHE, just as Willis’ thought experiment requires, to warm the surface. You also have many planets that all have reached their own maximum GHE based on atmospheric mass, gravity and irradiance.
“For such machines to work, they’d have to create energy”
I certainly don’t believe in perpetual motion machines, but the above statement is dead wrong. For such machines to work they would have to reduce entropy, not create energy.
One of my imaginary friend lives on a spherical planet that is not rotating, has an atmosphere consisting of one type of gas and a spherically uniform gravitational field.
There was a time when the planet was uniformly lit from a light source in the surface fed by geothermal energy and some of that geothermal energy uniformly warmed the atmosphere from the surface.
Unfortunately the geothermal fuel ran out and now the planet has no light or heat since.
Before this event my imaginary friend built a tall tower of regularly spaced temperature guages.
They were fortunate that the heat source was not strong enough to cause atmospheric molecules to achieve escape velocity. They found a magic point where the atmosphere reached a maximum height at which the molecules had zero kinetic energy and the temperature read 0 deg K. But the temperature below was none zero because of the geothermal heat source and heat was lost to space through radiation from the atmosphere to maintain equilibrium.
Since the heat source died it has been found that the height at which the molecules of atmosphere have zero velocity is decreasing and the molecules of atmosphere below are cooling.
They have a theory of global cooling that predicts that unless a heat source is found, one day in the future the whole atmosphere will have a uniform temperature of 0 deg K but until then the surface will be warmer than the top of the atmosphere because it is always 0 deg K at the top.
anna v says:
January 20, 2012 at 7:24 am
Bomber_the_Cat :
January 20, 2012 at 3:40 am
In my books an ocean is not a closed system, the way this column of gas is proposed. It is not a good example.
Not only that but water is incompressible.
From the “lay” sidelines:
I know of a perpetual motion machine: a priori scientific debate. Generates its own energy. Of course that’s debatable.
AJB says: January 20, 2012 at 7:51 am
“Ignoring tiny corrections for gravity, the gas will be distributed uniformly in the container, so the only unknown is the velocity distribution function”.
I agree completely. If you can ignore gravity, then gases are distributed uniformly in a container. This is a very good approximation for car engines or steam turbines, and so engineers can reasonably ignore the effect of gravity that makes the pressure at the top of a steam vessel 0.0001 atm less than the pressure at the bottom.
However, the thought experiment here specifically removes ALL OTHER EFFECTS BESIDES GRAVITY. Now we are ONLY dealing with those “tiny corrections”. And for a column 10 km high (sort of like the atmosphere), then those gravitational effects become quite important. Density and pressure are known to drop with altitude due to those effect. The question remaining is “does temperature also drop with elevation in a perfectly insulated container?” I say “no”.
Tim,
Have a think about why you say no. Then let the rest of us know. And please don’t just throw names around. Describe the physical phenomenon that leads you to the conclusion.
Thanks
“Otherwise heat will flow from the hotter (right-shifted MB distribution) to the colder (left-shifted MB distribution) slice until the temperatures are equal.”
Robert, your mistake seems right there in the last statement. The molecule members of the upper hotter right-shifted MB distribution cannot equalize (they are actually already in equilibrium) with the lower cooler left-shifted MB for the acceleration that changes the molecules vertical velocity. I think I am right on that, would you reconsider?
You keep want to say the molecules at various levels must be at the same mean velocity but by your own same example, they can’t in the gravity well case. Maybe the MB derivation doesn’t have a gravitation terms when applied vertically. I’ll stop and check it now.
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.
Hi Willis,
Permit me to make a small change to this (most of the rest of what I’ve read so far of your post is fine). The final equilibrium arrangement is one where the average kinetic energy per molecule is the same everywhere in the gas column. There are a lot more molecules in the denser gas at the bottom, so there is a lot more internal kinetic energy per unit volume down there, even though the kinetic energy per molecule does not change.
One has to be very careful about how one combines internal kinetic energy and work-based energy. The usual approach is to turn them into the enthalpy of the gas, especially when the gas is in thermal equilibrium with a non-uniform density profile. The usual formula for enthalpy of an ideal gas, E = U + PV, assumes a closed volume, thermal equilibrium, and more or less uniform pressure. Thus one can look at how the enthalpy changes as one e.g. compresses the gas or adds heat. This again becomes complicated when one takes into account variation in density and is the sort of thing that goes into estimating the DALR.
The simplest way to view this is in terms of the heat capacity. The heat capacity of the gas in any volume large enough to hold “many” molecules and be at thermal equilibrium is proportional to the number of molecules in the volume. The gradient in the density that gravity generates in the compressible fluid is accompanied by a gradient in the heat capacity, so that in equilibrium (at the same temperature) the gas at the bottom does have more energy per unit volume than the gas at the top.
The air temperature in the thermosphere can be well over 1000C — in principle hot enough to melt metals. But the atmosphere out there is very thin and the heat capacity is miniscule. The stratosphere — a layer named because of the lack of vertical transport and turbulence — gets warmer from the bottom to the top. That doesn’t necessarily mean that the energy per unit volume increases though, as the density still decreases. It just means that air molecules at the top of the stratosphere are moving, on average, faster than the molecules of air at the top of the troposphere, exactly the opposite of what one would expect from a naive “greater density equals greater temperature” model.
We now return to the regularly scheduled discussion (I haven’t finished reading your post:-).
rgb
Tim Folkerts says:
January 20, 2012 at 6:06 am
Tallbloke says: “No, as we’ve been saying all along, as have other people on this thread, at the lowest energy state, molecules at the top of the atmosphere have the same total energy as those at the bottom, but less of the total is available as kinetic energy …”
–As has been stated many times in many way, science is not decided by consensus, and science is certainly not by consensus of non-experts. Of the people on this list who seem to have formal training in physics, the agreement seems quite strong that isothermal is the the equilibrium condition for the given thought experiment.
Among OTHER trained physicists I have consulted, that is ALSO the agreement for isothermal rather than a lapse rate.
For instance, they say:
“all of the following are true:
— Boltzmann distribution of kinetic energy
— Boltzmann distribution of potential energy
— Boltzmann distribution of total energy.”
This is in contrast to Tallbloke’s claim that ONLY total energy follows the Boltzmann distribution. —
Hello Tim,
I am just intrigued by your logic. You state about the Tallbloke´s statement: “science is not decided by consensus, and science is certainly not by consensus.” This is absolutely true regarding scientists that are not corrupt or ingorant.
Then you state:
“Among OTHER trained physicists I have consulted, that is ALSO the agreement for isothermal rather than a lapse rate.”
Then you use the consensus argument to “disprove” what Tallbloke claimed.
Do you recognize your fallacy? I am making the comment since you pretend to be scientific.
I can ssure you that I didn´t bother about consensus when I wrote my E&E article. If I had done it would never had been written.
Best Hans Jelbring
Why not place a 10metre double insulated tube on a long arm centrifuge, fill tube with dry gas at surface pressure, allow temperatures to settle, the temp should be the same at both ends of the tube.
Rotate centrifuge to give 20g.
We should have multiplied the gravity effect by 20, the short length of tube 10m not 1km, would reduce the effect by 1/1000 so the temperature difference measured should be in the order of 180 millidegrees C assuming 9degrees/Km.
Any temperature difference other than zero would be worth investigating…..
Tallbloke says:
“Loschmidt, Lagrange, Laplace, Jelbring and me vs Willis, you, Jeremy, Boltzmann and Maxwell
Outside, now!”
Thanks for the laugh. ☺
Let me make one purely semantic argument. The adiabatic lapse rate is ~ 10 K/km. “Adiabatic” means “no energy flow”. So if you can make the approximation that there is little energy flow compared to other energies involved, then the adiabatic lapse rate would be a good estimate of the situation ion the atmosphere. Conduction of heat from one part of the column to another is by definition non-adiabatic. Therefore if conduction is the primary means of energy transport, the situation would not be expected to follow the adiabatic lapse rate.
And that lead me to one slight variation of the though experiment. Consider our perfectly insulated column. I will modify this SLIGHTLY by putting a heat reservoir at the bottom that holds at exactly 300 K (this would be similar Willis’s uniformly lit planet where the SB temperature is 300K at the surface). Now I force the air in the column to be well beyond the adiabatic lapse rate (perhaps I add temporary barriers every 100 m and cool each section by 2 K from the one below ie twice the adiabatic lapse rate). If the barriers are removed, two things will happen.
1) the column of air will be unstable to convection, and the air will start mixing like crazy.
2) there will be heat conduction P/A = k * (Delta_t).
Process 1 will continue until the lapse drops to ~ 10 K/km, at which point it will stop. Process 2 does not need to stop when Process 1 stops. In fact, Process 2 will continue as long as there is any temperature gradient.
NOTE 1: Process 1 will be MUCH quicker than Process 2. I guesstimate Process 1 would be mostly completed within a few hours. But even after convection stops, Process 2 would drive the whole column toward a uniform temperature of 300 K. I guesstimate this would take several months due to the low thermal conductivity of air and the great distances involved.
NOTE 2: If there the air is not perfectly isolated from the rest of the universe causing even a TINY heat sink at the top of the column (even a few mW/m^2), then the rate of conduction will not be enough to erase the lapse rate. GHGs (and even tiny bits of dust) in the upper stratosphere are enough to radiate thousands of mW/m^2 (ie several W/m^2), which means any real atmosphere will have a lapse rate. This tiny leak from the top is enough to maintain the lapse rate, and hence explain why the thickness of the atmosphere plays a major role in surface temperature.
A physicist says:
January 20, 2012 at 6:18 am
“Here is a Car-Talk Puzzler-type question that (hopefully) will illuminate why Luke and other posters are mistaken.”
“Alice has a cannon that shoots vertically, with a random initial vertical velocity, whose root-mean-square initial value is 100 meters/second, whose mean value is zero, and which is normally distributed (a Bell-shape curve).”
Is there any relevance about your shooting of she is all the time shooting fron the same altititude?
Will she shoot with the same initial root-mean square 100M/s from any altitude?
I am not a physicist so excuse me if misunderstanding you.