Perpetuum Mobile

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.

Image Source

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.

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Tim Folkerts
February 14, 2012 6:32 am

Instead of saying “The average ABSORBED radiation is never described as 255K blackbody radiation.”
I should have added” … except in thought experiments like yours or Jelbring’s.” Certainly we can postulate such a system, but it is so far from reality that it should not be equated in any way with statements about earth’s energy balance. A uniform shell around the earth is a useful alternate reality for sharpening our understanding, but it is definitely NOT what people are generally talking about.

February 14, 2012 7:09 am

Tim, we are talking past each other and If I am not clear, then I am sorry.
When I say 279 K above the albedo and 255 K at the ground, I am using the model that climate scientists use to achieve the figure of 255 K. I am showing that that model violates Thermodynamics and the 255 K must be wrong. You agree that 279 K must be the temperature below the albedo. Yes, it is a thought experiment. It is a simplied model to help determine the strength of the Greenhouse effect (255 K to 288K) and it is a Thermodynamic blunder. It is not my model; it is in many books and papers.
Average Insolation is a steady-state, static model. Jelbring is wrong. The atmosphere is isothermal with the ground to the TOA and the albedo cannot change its temperature. When we are in a steady state condition, there can be no net absorbtion. Your spectrum shifted example is one worth exploring, but since we cannot have any net absorption by the albedo layer and we must not commit the error of the albedo shell making a Perpetuum Mobile, we have to explore more what happens. Because whatever happens, it cannot result in 255K.

Tim Folkerts
February 14, 2012 10:13 am

Stepehen,
I think we can agree that:
* it is NOT a violation of thermodynamics to run a machine perpetually using the hot sun (5770 K) and cold outer space (3 K) as the two thermal reservoirs.
* it IS a violation of thermodynamics to run a machine perpetually using the warm shell (279 K) and warm surface (279 K) as the two thermal reservoirs.
The “million tiny suns” earth is using the hot sun (albeit inefficiently) and cold outer space (albeit inefficiently) as the two thermal reservoirs. The various thermal connections cause the ground to be ~ 288 K (primarily warming from the sparse sunlight coming in, but also IR coming in from the atmosphere and IR going out to space). The various thermal connections cause the atmosphere to be ~ 220 K near the tropopause (continuously cooled by radiation to outer space). Two of the absolutely key factors are albedo (limiting the warming of the ground somewhat) and GHGs (allowing cooling of the atmosphere). Given these two temperatures (which are maintained by the hot sun and the cold space), we can indeed run a heat engine perpetually without any violations of thermodynamics.
You say “Your spectrum shifted example is one worth exploring, but since we cannot have any net absorption by the albedo layer”. But we can have a perpetual net absorption of energy from the hot reservoir (at least as long as the sun shines), and a perpetual net emission of energy to a cold reservoir, which is indeed exactly what is needed to run a heat engine perpetually.

Tim Folkerts
February 14, 2012 6:53 pm

PS Stephen, you might want to read up on Kirchhoff’s law of thermal radiation. http://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation
Specifically note that “a corollary of Kirchhoff’s law is that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.”
So the “albedo layer” must emit radiation just as well as it absorbs that same radiation. If this was NOT true, then your violation of thermodynamic laws could indeed happen. In a sense, you have rediscovered the reasoning that lead Kirchhoff to this law.

February 14, 2012 11:24 pm

Very well argued, Tim. I’ve learned something. Thank you.

* it is NOT a violation of thermodynamics to run a machine perpetually using the hot sun (5770 K) and cold outer space (3 K) as the two thermal reservoirs. …..
The “million tiny suns” earth is using the hot sun (albeit inefficiently) and cold outer space (albeit inefficiently) as the two thermal reservoirs.

I concede that with the million starry sun model it is possible to drive an average insolated planet into a non-isothermal state if it has an albedo that varies with wavelength. The efficiency, as you say, is low, however the temperature difference across the albedo is likely small, but can be non-zero.
It provides a good explanation why in the formula: Normalized Energy to the ground = (1-A)/(1-a), that “a” does not have to equal “A” where “A” is top of albedo reflection and “a” is bottom of albedo reflection (or now modified to “absorption –> emission downward”). I’m still a long way from accepting “a = 0”, for as you note: the “albedo layer” must emit radiation just as well as it absorbs that same radiation. “ For that reason, I’m a long way from believing in the 255 K Earth “Surface temperature without Greenhouse Gas”. I’ll admit, it doesn’t have to be 279 K; might be a little more, or a little less, but I don’t see 255 K as reasonable since the albedo has two sides.
Here’s an interesting visual about that starry-sun-sky.
Since there are 41253 square degrees in a sphere and the sun takes up about 0.20 square degrees, there is room for about 200,000 suns in the celestial sphere. Let’s cut the sun up into 200,000 stars of 6000 deg K. Each will then be 12.6 square arc seconds or less than 4 arc seconds across. Jupiter, viewed from earth is 32 arc seconds or over 64 times bigger in area (though much dimmer!) than one of our 200,000 starry suns. One brilliant 4 arc-second spec for every spot in the sky the size of the sun. …. Hmmm…. too bad we cannot marvel at the spectacle because that uniform albedo is in the way. Just as well, it would hurt the eyes.

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