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.
IMPOSSIBLE ….. Willis in a nutshell…..
It’s amazing, but if we ignore the Puzzler shells that fall short of a given altitude, and calculate the mean kinetic energy of the Puzzler shells that *do* reach a given altitude, then we find that these Puzzler shells have a greater mean kinetic energy (at altitude) than the mean energy of the shells leaving the gun barrel.
In fact, in the limit of very large altitude, the vanishing small fraction of Puzzler shells that reach these far-higher-than-average altitudes turn out to retain twice mean the kinetic energy at-altitude as the shells leaving the gun barrel (showing that the energy-gain is a factor of two requires a tedious calculation).
So in this statistical sense, the Puzzler shells become “hotter” as they climb!
Joe Born says:
January 21, 2012 at 3:26 pm
I just read Velasco. Unless I missed something, Velasco agrees with Robert Brown and myself.
First, they say:
So Velasco et al say that Coombes and Laue conclude correctly that the temperature is the same throughout, i.e. isothermal.
Then they analyze the same question for an infinite rather than a finite system. Their conclusion is:
So they are distinguishing between finite and infinite systems, and they clearly say that the equilibrium situation in both is not choice (2) above, but choice (1), that temperature is the same throughout because the system is in equilibrium.
So I’m unclear why you think Velasco disagrees with me. If I understand it, which is always an open question, it seem they agree that the final distribution is isothermal. What am I missing?
w.
William, that’s why I constructed the WUWT Beach House story … to show in concrete, practical terms that Willis Eschenbach is entirely correct in asserting that “gravito-thermal” theories do violate thermodynamical principles.
tallbloke says: January 21, 2012 at 1:17 pm
“This is not a perpetual motion machine. All you have done is exploit the Gibbs free energy between the two dissimilar gases until it’s gone.”
OK, what’s the end state? Suppose you connect heat engines across both top and bottom. If there is a temperature difference, heat flows and work is done. If not, then both top and bottom are at the same temperature, and no heat is flowing. This recreates the original conditions – sealed cylinders. But they can’t both be at their lapse rate. What will the gradient be?
Willis:
Bless you; you’ve finally engaged. And, as I said, I’m happy to be set straight; I don’t contend I know the answer. I was just trying to attract some eyes to those papers to see if others see what I thought I did.
The part you quote is highly ambiguous; I couldn’t be sure what they were saying. But I think the key is Equation 8, which I read as the mean single-molecule kinetic energy as a function of altitude. In my last post I applied it, with f=3, to an N=2 system for which E=2mgz_mid. The result appealed to my intuition.
When you put realistically large values of N to it, you get a lapse rate that, as I said above, is negligible for practical purposes (to the extent that our hypothetical gas column is practical) but nonetheless non-zero. That is why over the last couple of weeks I came to question the equilibrium-implies-isothermal result everyone swears by.
“It’s already been described up thread, but I’ll do it again. Take two vertical cylinders 100 m tall. Fill one with helium and the other with xenon at 1 atmosphere. Thermally couple the bottom of each cylinder to the other and insulate the rest of the cylinders. The DALR for helium is 0.001888 K/m and 0.061 K/m for xenon. If gravity causes the DALR to be established in both cylinders, the temperature difference at the top of the cylinders would be 6K. If we then run a heat engine from that temperature difference, the xenon will warm and the helium cool until there is no temperature difference at the top.”
There minor things like, your lapse rate will affected by having pressure greater and less then the atmosphere- bottom of both will have more than 1 atm. Top of helium will be greater then ambient air pressure, and Xenon one at top will have less then ambient air.
One could have your lapse by having helium at or below ambient pressure at top- this also means less pressure at the bottom- so all of helium cylinder will have slight vacuum. With Xenon just have pressure at bottom 1 atm and top will have slight vacuum.
This wouldn’t make much energy. A sunny sidewalk and shaded sidewalk has far more temperature difference.
And insulating these cylinders would difficult- gases do not conduct heat well. Air is great insulator if you inhibit convection. This machine seems to generally inhibit convection- on purpose, you are shielding it from the energy of sun.
So this would make some tiny amount of energy powered by gravity.
There other ways of powering it by gravity. You use a balloon using buoyancy of helium. Inflate balloon from bottom of cylinder. Lift water with balloon. Deflate balloon at top into lower pressured helium, float balloon down with less inflated balloon, fill balloon up again, lift water, etc.
Probably get more energy. But still a lousy way to make energy.
Hydro dams are better way to make energy from gravity.
No gravity- no energy from hydro dams. Less gravity, less power from a hydro dam. More gravity and more energy from hydro dam.
Energy from gravity is commonly used.
Now, back radiation is something like 200 watts per square meter? If so this would dwarf all hydro dams or all nuclear energy if you get 5% of it. The only source of energy greater that back radiation is solar energy, but alas no one has figured out how to harvest this huge source of energy.
Robert Brown says:
Exactly. There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave. It is not. Whenever one finds the actual lapse rate to be close to the adiabatic lapse rate, the reason likely is that the lapse rate would even be higher given how the atmosphere is being differentially heated by radiation (or conduction from the Earth’s surface)…But a lapse rate above the adiabatic lapse rate is unstable to convection, which then transports heat up into the atmosphere until the lapse rate drops down to the adiabatic lapse rate. The adiabatic lapse rate is just a stability limit…Think of it as a ceiling on the lapse rate. If the lapse rate is naturally below that ceiling (as it is in the stratosphere and, as Robert has noted, in certain cases even in the troposphere), then such a lapse rate is perfectly acceptable. If the lapse rate exceeds that ceiling, then it is pushed down to the adiabatic lapse rate.
It is really no more magical than that.
Tallbloke to Willis:
“the rules for falsifying a proposition by appeal to the consequent theoretical constructability of a perpetual motion machine of the second kind are very clear. You have to specify the machine and demonstrate that it will produce work.”
Willis may have considered this unworthy of a response. Or he responded and I missed it. But one simple answer is to place an infra-red sensitive photocell at the top of the atmosphere and generate power with the i-r emitted from the surface due to the thermo-gravitational effect.
And the reason, in case it has been forgotten, why the supposed Jelbring effect makes a sunless planet luminous is, as stated by Willis with abbreviation, as follows:
“…assume that we have the airless [sunless] blackbody planet … The temperature of this theoretical planet is, of course, the theoretical S-B temperature.
Now suppose we add an atmosphere to the planet, a transparent GHG-free atmosphere. If the theories of N&K and Jelbring are correct, the temperature of the planet will rise.
But when the temperature of a perfect blackbody planet rises … the surface radiation of that planet must rise as well.
And because the atmosphere is transparent, this means that the planet is radiating to space more energy [i.e., micro-wave background] than it receives. This is an obvious violation of conservation of energy, so any theories proposing such a warming must be incorrect.”
I think this is worth repeating, since it seems to eliminate all need for further discussion.
Joel Shore says:
January 21, 2012 at 7:39 pm
You are right, upon re-reading my explanation doesn’t track at all. Let me try again in a simpler manner.
The value of the Nte(Ps) is given in equation 7 as:
exp(0.233001 Ps ^0.0651203 + 0.0015393 Ps ^ 0.385232)
This has four numbers in it that were used to fit the value of Nte to the surface pressures of the various planets. These are the four “free parameters”. The fact that they are fits is evidenced by the number of decimals. This kind of fitting can be easily done in Excel using the “Solve…” menu item.
There are also variables which are somewhat arbitrarily chosen although as you say this is the lesser problem. One issue is that on earth the surface pressure is defined as being at sea level. But where is the reference level on a planet with no ocean? At the lowest point? At the average elevation? RMS average elevation?
In any case, the point remains. They have used a very suspect equation (a single variable appears twice) to fit the surface pressure of eight planets and moons to the “Nte”, which in turn is the amount the planet is above the “greybody” temperature. This kind of 4-free-parameter fit, particularly with only 8 datapoints, and pressures that vary by 15 orders of magnitude, carries absolutely no weight with me. With apologies to those who have heard it, the apposite story is as follows, by Freeman Dyson:
Four free parameters, eight data points, that adds up to nothing at all in my world. Those who think that this is “evidence” of anything other than rampant curve-fitting are fooling themselves.
w.
William Gilbert says:
The problem with using Thomas Kuhn to support your notions is that people quoting Thomas Kuhn have correctly predicted about 10,000 of the last 10 paradigm shifts (although that first number is probably too conservative an estimate). [This sort of makes economists, for which this sort of quip was originally applied, look good by comparison!]
And, the the supporters of the notion that the surface of the Earth can emit 390 W/m^2 without some of this radiation being absorbed by greenhouse gases in the atmosphere have still not produced a simple explanation of how this can be done using CORRECT physics principles.
Besides which, we know for a fact that the reason why the Earth’s surface is emitting 390 W/m^2 is not because there is some magical new source of energy but because there is absorption occurring in the atmosphere such that only ~240 W/m^2 is escaping to space.
Hence, what N&Z and Jelbring have exposed is how many otherwise intelligent people will apparently believe utter nonsense if it supports what they really strongly want to believe. I keep thinking that one of these guys is going to step up and admit that their hypothesis is all a Sokal-style hoax ( http://en.wikipedia.org/wiki/Sokal_affair ). These “theories” really have revealed some very embarrassing things about the AGW-skeptic movement…which, to their credit, some skeptical folks like Willis, Robert Brown, and Roy Spencer are trying desperately to correct, although it makes it all the more amazing that such crazy nonsense continues in light of their best efforts to squelch it.
As an example of the arbitrary nature of the N&Z choice of values for surface pressure in their paper, they say the pressure of the earth is 98,888.20 pascals. I particularly liked the fact that it is allegedly accurate to the nearest hundredth of a pascal.
The oddity is that standard sea level pressure for the earth is generally taken to be 101,325 pascals (no tenths or decimals) … so their number is off by a couple of percent. They describe their number as “mean atmospheric surface pressure” and “mean total surface pressure”, so it is unclear just what they think they are measuring. I don’t even know what “total surface pressure” might be.
Finally, whenever you see parameters given to seven significant figures, and the earth’s surface pressure given to two decimals, you know you are dealing with amateurs. That kind of fake precision is a big red flag at any time.
w.
The best discussion of the perpetual motion paradox that I have seen on this thread is the one by Robany here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-871446
But I did not see any responses to him. Anyone have any comments?
Personally, I am having a lot of trouble with the whole PM concept for the Jelbring system. Gravity is a time invariant externally imposed field of force acting on mass. When did Newton’s second law of motion become a perpetual mobile of any kind?
Bill
Willis Eschenbach says:
January 19, 2012 at 10:32 pm
You are right and I was wrong. If the g force were constant, it would be isothermic. The g force is inverse square with distance, so the temperatures are a little different, but >99% of the difference from 0-30 km is pressure, and <1% is temperature. I don't have the exact equations, but I can do orders of magnitude.
“…assume that we have the airless [sunless] blackbody planet … The temperature of this theoretical planet is, of course, the theoretical S-B temperature.”
Which is what temperature? 2 K
“Now suppose we add an atmosphere to the planet, a transparent GHG-free atmosphere. If the theories of N&K and Jelbring are correct, the temperature of the planet will rise.”
So atmosphere would certainly be 2 K.
N&K and Jelbring are saying the atmosphere warms the planet?
“But when the temperature of a perfect blackbody planet rises … the surface radiation of that planet must rise as well.”
Well, a blackbody is imaginary.
And it’s suppose absorb all radiant energy of whatever wavelength and emit all energy at all wavelengths.
So blackbody which similar to earth wouldn’t have internal heat, because blackbody is suppose to be a perfect conductor of all energy- unlike rock. So if black surface is 2 K, it’s core planetary temperature is also 2 K. And so now a dinky atmosphere is suppose to warm the entire planet. A million hydrogen bomb would not warm that planet- does that mean nuclear bombs don’t create heat?
William Gilbert says: 7:42 pm
It seems to me that you have given insufficient consideration to the mechanical energy changes in Pressure P as a function of z and T. And doesn’t this formula apply to a constant mass? Therefore the volume is a function of pressure which is a function of z.
Some other people like Joel Shore have commented: There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave. It is not. I agree. The Adiabatic Lapse Rate only defines the critical threshold above which convection becomes possible.
The ALR is far more like the Critical Angle of Repose of a Pile. This defines the maximum angle (slope) of a pile before it becomes unstable and sloughs off.
ALR does not require that the temperatures reach this lapse rate. They only define a maximum of the lapse rate before the system will be unstable. A column of gas can exist at ½ the ALR and not convect, but it will still thermally conduct. So what will happen to this column if no energy is applied and it is thermally insulated from the outside? Will the temperature gradient increase making the bottom hotter and the top colder? No. It will become isothermal and its pressure distribution will change accordingly.
A physicist says:
January 21, 2012 at 2:36 pm
Tiring of of the Sysyphean task with the copper blocks, you lay a pair of insulated pipes up the mountain between your houses. You fill the pipes with a carefully selected refrigerant. At the top, the gaseous refrigerant is exposed to the cool mountain air in a condenser. This condenses the gas, yielding liquid refrigerant, with the waste heat warming her house.
The liquid refrigerant then flows by gravity down to the bottom of the hill in the first pipe. At the bottom it is evaporated in a standard air conditioner and cools your house.
At that point it is again a gas, and so it flows up the second pipe to the condenser at the top, driven by the counterbalancing weight of the liquid refrigerant in the other pipe that is pressurizing the expansion valve.
So now you have a modern refrigeration cycle operating without any effort necessary on your part, no need to move those 19th century copper blocks … and folks, that’s just one of many 21st century technology advances brought to you by TGC USA, a wholly owned subsidiary of ThermoGravitiCorp Ltd, a Cayman Islands Corporation. Invest now in the energy source of the future! Pick up the phone and dial 1-(800)-472-8489, that’s 1-(800)-GRAVITY, call right now for full details on where to send your checks, and remember, this opportunity won’t last!
w.
pegged that one, willis.
now we can see that there is a difference between static and dynamic.
tallbloke says:
January 21, 2012 at 3:59 pm
I partly agree. There is also rotational and vibrational energy.
That is true for a block of atmosphere that changes altitude without gaining or loosing energy.
However, thermal conduction from warm to cold means that energy will move from one block of atmosphere to another. As a result, the actual lapse rate will be much closer to zero than it is to the DALR. The question is – How much closer?
Let’s assume that the average speed of a molecule is extremely high. When a molecule moves either toward or away from the center of the Earth, it will undergo an acceleration for some time (dt). If the change in velocity due to gravity is a significant part of the average speed, then there should be a measurable induced lapse rate. However, I suspect (I have not computed this) that the gravitationally induced change in velocity will be at least nine orders of magnitude less than the average velocity (mainly because dt is extremely small). As a result, the gravitationally induced lapse rate will not be significant with respect to the current models. In other words, it would be indistinguishable from zero. (Of course, when the pressure gets low enough, dt increases and the lapse rate should become noticeable.)
More to the point, the DALR and the gravitationally induced lapse rate are created via totally different mechanisms. As a result, their magnitudes are not even close.
Stephen Rasey says: January 21, 2012 at 10:06 pm
“ALR does not require that the temperatures reach this lapse rate. They only define a maximum of the lapse rate before the system will be unstable. A column of gas can exist at ½ the ALR and not convect, but it will still thermally conduct.”
I agree with almost everything you said, but not this. I explained here how forced vertical motion in the stable regime acts as a heat pump, driving heat downward, and the gradient toward the ALR. And there’s plenty of wind to do the forcing.
Nick Stokes, hold on there. There is nothing forcing a vertical motion in my very short example. It is a closed, insolated system, a simple sealed column that we heat to 1/2 the ALR rate, seal it up, and then see what happens.
I agree with you, if you pump any system, especially with a day-night cycle, you will set up a non-isothermal atmosphere with a non-zero lapse rate. That is all the more reason I think “divide by 4” climate models are worse than useless; they lead you down a blind alley.
Anthony and Willis, I would love to see a post that looks into matter of how the suns radiation budget is used up from space to ground to space.
The title of the post can be: Two sides of the Albedo debate
Re: Joel Shore 9:25 pm
The question that has been bugging me for a year is that discussions seem to universally take the suns energy 1367 W/m^2, divide by 4 to remove night and day and make a thermally dead planet (which we now know has an isothermal atmosphere !! ) and get 342 W/m^2. We give the earth a 30% albedo, so only 70% make it way to the ground, about 240 W/m^2.
No, Willis, the planet Earth receives 342 W/m^2. Some of it bounces immediately off the Albedo. Most of it goes through and some of it bounces around a bit – because the Albedo layer has two sides: a top and a bottom.
I asked this question in a Jan 12 “Earth’s Baseline…” comment. Why do we account for albedo only on the direct-from-sun energy, but everyone seems to ignore it on surface-to-sky path? I set up a general formula where the top Albedo was A= 30% as accepted, but I allowed for an unknown Albedo reflectivity from the bottom as “a”. Using infinite series of decaying reflections…
So if A=a, then the heat at the ground would be 340 W/m^2 with 240W/m^2 of heat leaking through the layer. Add to that the 30% originally reflected and you have the original 340 W/m^2. Maybe “a” isn’t equal to A. But where is the proof that “a = 0”?
There is really no such thing as a “one-way mirror”. It really only a partially reflective two-way mirror. The Albedo layer over the earth acts the same way. But I fear many people treat it as a one-way valve, a diode, a mirror with only one side. There are two sides.
If I am wrong, then I want to establish prior claim to a new invention: “The Portable Backpackers Liquid Oxygen Generator.” Get a cardboard box, cover it in 99% reflecting Mylar (one way of course). Let’s see, 340 W/m^2 * (1 %) leaves 3.4 W/m2 inside the box. Yep, 3.4 W/m^2 equates to 88 deg K, and Oxygen liquefies at 90 deg K. (Oh, that was easy! Why hasn’t anyone thought of this before…)
[Blockquoting fixed. —w.]
Anthony and Willis, I would love to see a post that looks into matter of how the suns radiation budget is used up from space to ground to space.
The title of the post can be: Two sides of the Albedo debate
Re: Joel Shore 9:25 pm
The question that has been bugging me for a year is that discussions seem to universally take the suns energy 1367 W/m^2, divide by 4 to remove night and day and make a thermally dead planet (which we now know has an isothermal atmosphere !! ) and get 342 W/m^2. We give the earth a 30% albedo, so only 70% make it way to the ground, about 240 W/m^2.
No, Willis, the planet Earth receives 342 W/m^2. Some of it bounces immediately off the Albedo. Most of it goes through and some of it bounces around a bit – because the Albedo layer has two sides: a top and a bottom.
I asked this question in a Jan 12 “Earth’s Baseline…” comment. Why do we account for albedo only on the direct-from-sun energy, but everyone seems to ignore it on surface-to-sky path? I set up a general formula where the top Albedo was A= 30% as accepted, but I allowed for an unknown Albedo reflectivity from the bottom as “a”. Using infinite series of decaying reflections…
So if A=a, then the heat at the ground would be 340 W/m^2 with 240W/m^2 of heat leaking through the layer. Add to that the 30% originally reflected and you have the original 340 W/m^2. Maybe “a” isn’t equal to A. But where is the proof that “a = 0”?
There is really no such thing as a “one-way mirror”. It is really only a partially reflective two-way mirror. The Albedo layer over the earth acts the same way. But I fear many people treat it as a one-way valve, a diode, a mirror with only one side. There are two sides.
If I am wrong, then I want to establish claim to a new invention: “The Portable Backpackers Liquid Oxygen Generator.” Get a cardboard box, cover it in 99% reflecting Mylar (one way of course). Let’s see, 340 W/m^2 * (1 %) leaves 3.4 W/m2 inside the box. Yep, 3.4 W/m^2 equates to 88 deg K, and Oxygen liquefies at 90 deg K. (Oh, that was easy! Why hasn’t anyone thought of this before…)
A physicist says: January 21, 2012 at 2:36 pm
On the planet Stygian …
Next morning, exchange the now-warmed beach block for the now-cooled mountain block, and do it again. Which amounts to this: free heating, air-conditioning, and electrical power for you and your sister, forever, with zero work input! Thank you, thermo-gravitic theory!
Well, of course, you could do this on Earth. Which emphasises my point that it takes work to maintain the ALR. Energy that comes from the wind.
I should also respond to Joel Shore’s
“There seems to be a common fallacy around here that the adiabatic lapse rate is some sort of law of how gases in a gravitational field behave.
…
If the lapse rate is naturally below that ceiling …, then such a lapse rate is perfectly acceptable. If the lapse rate exceeds that ceiling, then it is pushed down to the adiabatic lapse rate.”
No, there is a wind driven heat pump, which by forced motion in the stable regime, pushes the gradient up toward the ALR, exactly analogous to the convective engine operating when the lapse rate exceeds the ALR. It’s not a fallacy.
Stephen Rasey says:
January 21, 2012 at 11:40 pm
Stephen, the measured albedo of the planet is about 0.3. Because this is a measured figure, perforce it includes all “double reflections”. It includes reflections from the surface plus reflections from the clouds and aerosols minus any double reflections of the type you have noted.
As you point out, double reflectors will be present whenever there are clouds. As a seaman, I have seen this in the form of what we used to call “atoll blink”. This is where from over the horizon you can see the reflection of unseen lagoons on the underside of the clouds. The sun strikes the shallow water in the blue lagoon. That lovely turquoise blue light makes a curious glow on the underside of any clouds hanging over the lagoon. That light can guide the canny mariner home when the atoll can’t be seen.
But to answer your question, yes, it’s included.
Having said that, there’s a couple kinds of albedo. One is what is reflected back to the eyes of a viewer between the earth and the sun. Another, the bond albedo is what is reflected not just back into the observers eyes but in all directions. The bond albedo cannot be practically measured, you’d need a swarm of hundreds of satellites in a sphere around the earth to measure scattering in all possible directions. So it’s done with a big pile of computer modeling of the reflection. Do they include double reflections? I’d be shocked if they didn’t. I say this because to calculate bond albedo you have to calculate what’s happening at low grazing angles to the planet. At those angles, tall clouds shade large areas, including other clouds. In addition much more light is reflected from the surface, and has to fight through a lot of cloud to get out.
So it’s not a simple problem, and I suspect the scientists have not ignored the issue. I don’t have the references to hand, but if you google something like “double reflection” “bond albedo” “cloud” on google scholar and look for PDF papers, you’ll find something.
w.