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.
Well Willis, I think you realise that if you put real values in, the current could not be measured.
So as far as Modern Physics is concerned if it cannot be measured you cannot say it exists.
Excuse me? Quite aside from the fact that we are discussing a Gedanken experiment rife with “ideal” things, why in the world do you think that one could be unable to measure the heat flow or electric current? Are you trying to say that the top and the bottom would be so close together in temperature that the difference would not be measurable and therefore could not be said to exist? Or do you think that there is some sort of lower bound to the potential difference a thermoelectric junction can produce?
Or do you have something against dippy ducks?
We can make the air column as large as we like. We can make g as large as we like. It only exists in our mind, after all. We can make it a kilometer high and place it in gravitational field of 100g. For an ideal gas, we can simply scale any lapse rate up to where the temperature difference is 100K. We can use not just one thermocouple but 1000, adding their tiny little voltages in series until it is large enough to measure.
The Carnot Cycle engine is a textbook example of an “ideal” heat engine used in discussions of thermodynamics. One can run an ideal Carnot Cycle heat engine between any two reservoirs at any difference in temperature. It won’t be very efficient, and it will take a long time to go through a cycle (in part because the adiabatic expansion has to be quasi-static to be reversible) and it would be perfectly happy running between the top and bottom of the air column if they were at different temperatures. And then there is the even simpler problem of perpetual heat flow if you place a simple insulated silver rod so that it connects the top and bottom of the reservoir. Then the system never reaches equilibrium — you have heat flow to the top through the silver (surely you don’t want to propose an adiabatic lapse rate in silver that is identical to that of air, do you?) as long as there is a temperature difference, and the heat somehow “flows” back to the bottom due to gravity as fast as it gets there.
A neat trick if you could do it. But you can’t. There ain’t no such thing as a free lunch, and a spontaneous thermal gradient is the potential provider of an eternity of free lunches.
rgb
Robert Brown says at 1/22 8:04am:
“Therefore consider the “jar” argument once again.”
Ok. This may be long, fair warning. Pour a libation of choice, may only interest Robert.
“Grab a jar of air at the bottom of your “equilibrium” room with a DALR. Its pressure is a bit higher and temperature is a bit higher than air at the top.”
Ok. Willis’ cv equilibrium not mine; I haven’t added anything to Willis’ adiabatic GHG-free equilibrated tall air column under presence of inexplicable gravity field. I add for clarification assuming your permission: an open Bjar, b for bottom, and its separate lid actually. Assume here both jar & lid were inside Willis 1 cv and in energy equilibrium (i.e. no “hot” jars or “cold” lids suddenly cross cv.)
“Grab a second jar of air at the top, where pressure and temperature are both a bit lower.”
Ok. Tjar, t for top. With same clarification as Bjar above.
“So far, we know nothing about the density of said air.”
In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).
“Nothing in fluid dynamics requires a fluid to be compressible….I don’t know how you want to idealize your “air”…. Gravity, however, is the same at the top and at the bottom.”
Ok. Lotsa’ thought experiments exist using incompressible fluids. Not my air – I’ll use Willis’ GHG-free atmospheric non-idealized air.
“Moving the jar at the top to the bottom… gently enough that one doesn’t slam the fluid molecules…”
Ok. Before Tjar is moved down to a decreased h, I’ll assume with Robert’s permission, the lid has been closed completely and we have a second body (Tjar + gas) within the cv, nothing across Willis’ 1 cv.
“The fluid in the two jars is not in thermal equilibrium. Surely you agree?”
YES for Tjar moved down (deliberately –w/no time to re-establish equilibrium) and “gently” thus heat flows before re-establishing cv equilibrium since Tjar is touching Willis air.
NO for 3rd body closed Bjar still presumably at the bottom, Robert hasn’t moved it yet. BJar is still in cv energy equilibrium near h=0.
“…the two jars have any sort of thermal pathway” opened between them, they will come to thermal equilibrium at a temperature in between T_b and T_t, one we can actually compute as it will depend on their heat capacities at constant volume which depend on N only.”
Robert didn’t write it, but I presume just a trivial oversight, that closed completely Bjar was then raised, deliberately and gently as Tjar was lowered. So YES, the thermal path way being established between T&Bjars at some h >0, establishes the jars new equilibrium (Caballero’s mean KE) temperature (of h) thru heat flow that ceases eventually.
“Note well — extremely well, if you please — that the heat capacity of the gas gets no contribution from gravity.”
Ok. Noted extremely well.
“From this alone you should be able to see that gravity is decoupled from the problem, because your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect.”
No. Robert & I part views here. Total Tjar energy = its KE+PE & is not ever changed; it is always conserved & constant. No total energy can be harmed or created by moving the jars. Robert is incorrect to write gravity field can be decoupled (i.e. ngh term ignored). Accepting that means no conservation of total energy & then a thermo law is broken.
Since gravity potential energy cannot be decoupled & conserve TE, the PE of Tjar is reduced (of h), the PE of Bjar is increased (of h). When the established thermal pathway allows Tjar and Bjar to reach energy equilibrium, the KE of the molecules inside each will be the same as the KE of the molecules in Willis’ cv air at h. (After Caballero, we are talking KE mean here, actually each jar has a parcel of molecules.)
After Caballero, each of those jar KE means are the respective jar equilibrium temperatures (of h). Tjar (KE +PE) = constant = Bjar (KE + PE) = Willis’ air (KE + PE) (all of h). In the Tjar case, mean KE reduces as PE increases. In the Bjar case, mean KE decreases as PE increases.
“…your belief that dT/dh (where h is the height of a jar) gets a contribution from gravity at equilibrium is incorrect. dT/dh = 0 (where the derivative should be a partial derivative btw). Adiabatic lapse is a non-equilibrium state.”
In my view, my belief is correct since no thermo law is broken. My view/belief invokes immutable conservation of energy in the 1 cv (as others have stated in this thread). dT/dh is not = 0 which it cannot be and keep total energy = constant. (as an aside, not a partial derivative I think, I could be wrong – trivial matter). Here in presence of inexplicable gravity field, adiabatic lapse IS an equilibrium state. Presence of this g field results in the ngh = PE term necessarily and always materializing. Curiously, this stratification also enables buoyancy – a matter for another day.
“Even if we leave them where they were in the first place and simply run a heat-superconducting wire between them, we expect heat to flow in the wire because they are not in thermal equilibrium and thermal equilibrium does not depend on where you are!”
This gets a little more interesting. Note extremely well here, if Robert runs his heat-superconducting wire between T&Bjars in their original positions (at hugely different h), in my view nothing of interest happens. (NB: don’t need exotic super-conductor wire here, an ordinary works, but maybe SC enables an easier-to-see solution). This (now 4th body) wire is in thermal contact with Willis’ cv gas all the way from top to bottom. At system equilibrium in the cv, that wire will be in thermal contact with the KE of Willis gas at each continuous h so the T of that SC wire will necessarily vary at each h. By immutable thermo law!
The bottom terminal end of the superconducting wire (ending in Bjar) will be at thermal equilibrium of Bjar KE and vice versa all along its length (of h) in Willis air KE to Tjar KE which is also in energy equilibrium. There is never any 1 body in the system cv that is not in equilibrium thus no energy can flow between any of the4 bodies forever – esp. with conservation of energy invoked.
“Once again, you are stuck.”
No, conservation of total energy rescued me. As it has for many of my exploits.
“You have a closed system that — you claim — is in thermal equilibrium with two different temperatures, one at the top and one at the bottom. Yet that means that an ordinary thermometer carried from the bottom to the top would read two values, and, just as would be the case for any two systems whose temperature is measured, means that heat would flow between them if it could.”
Yes, two thermometers will read two different temperatures (KE) even in the equilibrated CV because of conservation of energy in a gravity field. Heat will now NOT flow in Willis’ cv, since all 4 bodies is – at every continuous h in stratified T- in energy equilibrium. No irreversibility. Entropy is constant. No heat flow.
“Air is a conductor of heat and heat will flow from the bottom to the top until the system is in equilibrium.”
Yes initially & stop flowing heat (no energy flow forever here) when balanced – just don’t go across Willis cv, Robert. Same system, different h’s. There are NOT two separate cv.s, here one hot and one cold. Only Willis’ original 1 cv, which in energy equilibrium & touching everywhere.
Yes GHG-free air conducts KE (i.e. T) to the wire from each h at molecular mean KE (Caballero’s temperature defn.), thanks for pointing out. Willis cv can achieve stratified T equilibrium, all 4 bodies touch in energy equilibrium at continuous h’s thru the wire, no Perpetuum Mobile. Nothing crosses Willis’ cv. A drinking bird (got a laugh out of me) could be added but it stops drinking energy at equilibrium.
Courtesy of Robert, I’ve had a fun Sunday afternoon whilst nursing a head cold & refilling hot lemon tea. Now I’ll fill a Bjar w/scotch OTR, turn up the fire, & go re-read Maxwell’s Demon thing and report back (don’t cringe), maybe even w/an elevator speech. Geez, I badly need to employ an editor but this site is inefficient for conversations, my apologies, I gave fair warning.
“Trick, a coffee cup on a table is experiencing the force of gravity, but it is not moving in the table’s frame of reference, therefore it is not accelerating.”
Stephen – The cup IS accelerating at g wrt earth surface/table in their rest frame or cup would have no weight. In this table frame of reference, the coffee cup has mass m & inexplicable gravity generated weight force = mg down on the table at rest wrt surface of earth.
Equivalently in the cup rest reference frame, the table would be seen to be accelerating at a (=g) and the cup has no way of knowing whether (F=weight=mg=ma) is from table accelerating or from an inexplicable gravity field.
Wait for my maybe imperfect language to be corrected below…but that’s generally ok I think.
If this is an honest plea for correction, say no more.
Forget the fact that Earth is rotating and the small non-inertial correction (Coriolis force) as it is quite negligible.
The cup sits on the table. Gravity exerts a force down of mg. The table exerts a “normal force” up on it of N. The cup is at rest. F = N – mg = ma = 0. Therefore the correct statement is that the forces acting on the cup are balanced, with gravity pulling it down and a column of electrostatic and pauli exclusion forces that stretch from the center of the earth to the top of the table pushing it up. It is indeed not accelerating.
The “weight” of the cup is mg, and of course this doesn’t change whether or not the cup is sitting at rest or falling. If we put the cup on a scale, a scale measures N, the normal force applied across the scale, and at rest N = mg.
Put the cup in an elevator (resting on a scale) and as the elevator accelerates up
N – mg = ma > 0
so that N > mg, the cup “weighs” more (not really, but in the accelerating frame it appears to). Accelerate it down and N – mg = ma < 0 and it weighs less. Drop it (so it is in free fall) and mg = ma and N = 0, so as far as the scale is concerned it is "weightless", although the weight of course has been mg throughout.
All of this is in my online intro physics book, with pictures and considerable explanatory text, if you care to learn it.
http://www.phy.duke.edu/~rgb/Class/intro_physics_1.php
rgb
@Trick 1:37 pm The cup IS accelerating at g wrt earth surface/table in their rest frame or cup would have no weight.
No, there is a FORCE due to gravity Fg = G(m1*m2)/r^2 There is no acceleration in this formula.
As I said, this is a statis problem. there is an equal and opposite force exerted from the table. No net force. F=(mass)(acceleration) if net force = 0, mass > 0, then acceleration = 0.
If the table broke, The upwards force from the table would drop and Fg would then cause the cup to accelerate downward.
Robert Brown says
“Excuse me? Quite aside from the fact that we are discussing a Gedanken experiment rife with “ideal” things, why in the world do you think that one could be unable to measure the heat flow or electric current?”
I said as far as I know there has never been an EXPERIMENT to test which of these conjectures are true for a thermally isolated column of gas in a gravitational field.
1. That is has an isothermal distribution
2. That is has an adiabatic distribution
Someone tries the experiment thermocouple with very long copper/silver leads and the most sensitive ammeter but finds no measurable current.
After consulting the physical tables and numbers he finds that the apparatus was not sensitive enough to detect a current.
I think we are at that stage with this particular method.
The two tubes with gases of widely differing Cp seems a better route but there are problems there too.
Joel Shore,
I am referring back to your post of 1/21/12 here:
http://wattsupwiththat.com/2012/01/19/perpetuum-mobile/#comment-872192
You say:
” 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.”
You are confusing two different things (and insulting good scientists). You have either not read or do not understand Jelbring’s paper. (I have no final opinion on the N&Z paper since I have not yet seen their Part 2 which explains their thermodynamic reasoning). You are confusing the adiabatic lapse rate derivation and the total heat content of the atmosphere. They are separate thermodynamic entities.
For a “dry” atmosphere in an electromagnetic and gravitational field, the “first law” can be written:
dU = CpdT + gdz (1)
This equation is also known in meteorology as “dry static energy” and is closely related to “potential temperature”.
For a system (atmosphere) in steady state, dynamic equilibrium, the internal energy (U) of the system is constant (conservation of energy) and dU = 0. This means that at any point in the system (atmosphere):
CpdT + gdz = 0 (2)
And for a system in static equilibrium
CpT + gz = constant (3)
This is the focal point of the Jelbring paper. As you ascend vertically in the atmosphere the thermal energy (and therefore temperature) decreases and the gravitational potential energy increases. Energy is conserved. Thus there is a vertical thermal gradient in an atmosphere that is under the influence of both an electromagnetic and gravitational field. The gradient can be represented by rearranging equation (2) to give:
dT/dz = -g/Cp (4)
Thus the vertical temperature profile in an ideal steady state atmosphere is solely a function of the gravitational acceleration and the specific heat capacity (at constant pressure) of the atmospheric gas. It does not matter if the gases are GHG’s or not, it does not matter what the down welling radiation is, it does not matter what the pressure is (as long as it is above 0.2 atm.), it does not matter what the density is, it does not matter what the absolute temperature is at any given point. The dry adiabatic temperature gradient is a constant, is a function of gravity and heat capacity only, and is a direct result of the first law and the conservation of energy (yes, there is a law defining the DALR).
Now let’s clear up your confusion as to how this applies to the surface temperature. The temperature at any given point is a function of the total heat content of the atmosphere. The heat content is a function of the incoming solar radiation and the outgoing long wave radiation. The total heat content can change (due to a change in the effective emission height for instance), and therefore the absolute temperature at any point can change, but once steady state is reached, the temperature gradient remains a constant. In all cases, the temperature at the surface will be warmer than the temperature at any altitude.
The “green house effect” is simply the difference between the surface temperature and the effective temperature at the effective emission altitude (z) (Jelbring’s outer sphere and diameter “d”). It is controlled by the adiabatic lapse rate, not by “back radiation”. “Back radiation” is a function of the surface temperature (and the lapse rate), not the other way around.
In Summary, 1) the GHE is simply a function of the effective emission height (i.e., effective emission temperature) and the lapse rate. 2) Surface temperature is simply a function of the total heat content of the atmosphere.
This is the Jelbring hypothesis and it is very straight forward. Where is the “utter nonsense”?
By the way, in your hypothetical, “isothermal” atmosphere, there would be no GHE (Ts = Te). This result would be true no matter the GHG concentration, whether it is zero or >1000 ppm. It would also be very cold!
I must also emphasize that this system (Jelbring) does not represent a perpetuum mobile – it is strictly a result of elementary thermodynamics. If this is a perpetuum mobile, then the rotation of the planets around the sun in the solar system is a perpetuum mobile – I don’t think Newton would approve! He understood gravity.
Bill
P.S., You also commented on my Thomas Kuhn quote in your post:
”“Because the unit of scientific achievement is the solved problem and because the group knows well which problems have been solved, few scientists will easily be persuaded to adopt a viewpoint that again opens to question many problems that have previously been solved.” (Page 169)
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!]”
You miss the point that Kuhn is trying to make. There would be many more paradigm shifts in science if the group practicing “normal” science were not so protective of their territory and utterly resistant to new ideas. Science would advance at a much faster rate. That 10,000 number contains many real paradigm shifts that were denied. Here is another Kuhn quote for you and Donald Brown:
“When examining normal science….we shall want finally to describe that research as a strenuous and devoted attempt to force nature into the conceptual boxes supplied by professional education.” (Page 5)
Bill
By the way, I have said it before, but I will repeat it here:
“I am confident that the complete Water-cycle is important to the surface- and near surface- temperatures. Therefore Water Vapor (WV) does also have a better capability than other “dry” gases of keeping heat in. The “Moist” or “Saturated Adiabatic Lapse Rate” is proof of that. None of the “lapse-rates” however, dry or moist, have changed since they were first established back in the 19th century.
Yet – CO2 has increased by 40 – 45 % since that time and if a pocket of ascending air contains 40 to 45 % more of a gas that is constantly being warmed by radiation from the surface below – and we are certain that must be the case because Kiehl & Trenberth (K&T) say that 40 watts per square meter are escaping from the surface out through “The Atmospheric Window.” – Therefore all radiation from the surface must be capable of reaching out to space – and must, by definition warm all gases in its path that absorb it.”
So, why does it not show up in the lapse rates?
In my view this is incorrect Robert. Within Willis cv in gravity field, can define h as 0 at bottom of cv. We know PV=nRT (for all of h) and thus rearranging CAN find density (of h) = n/V = P/RT (all of h).
I don’t know whether to do an itemized reply all at once or reply as you make mistakes. I’m also not sure what “cv in gravity field” stands for. What is cv? Also, if you look at the explicit context of my remark, it was fluids, and I later said that if one presumes an ideal gas… The point being that real gases have compressibilities that are not idea.
But either way, if you look at section 8.1.9 in my online physics textbook (the section titled “variation of pressure in compressible fluids” you will see the derivation of the density of an ideal gas (equal 711). So I think I’m familiar with this and you are quite correct — for an ideal gas;-)
Indeed, in the space immediately below this I derive the standard textbook isothermal variation in pressure that is a self-consistent solution to the buoyancy problem:
P(z) = P_0 exp(- Mg/RT)
where M is the molar mass of the ideal gas, T is the temperature, and R is the ideal gas constant (or factor in N_A to turn R into k if you prefer).
Note that this solution:
a) certainly exists, for an ideal gas;
b) is clearly in equilibrium. T is the temperature throughout the air column in question. No second law violations here.
This is a column that is clearly in static force equilibrium. The gas perfectly balances weight, buoyancy, and density throughout. That was the point of the \rho g z term used in the derivation (which if you read back in the text is the basis of fluid pressure in a static fluid). It is also in thermal equilibrium. Heat content will not be conducted up or down the column because it is all at the same temperature. This column does not violate the second law.
Note also that this variation of pressure with height is very, very close — within around 3% — to what is observed in the atmosphere, even though the atmosphere is better described by an adiabatic lapse rate than an isothermal one because it is not in equilibrium. And even the adiabatic lapse rate isn’t a very good approximation to the actual variation. It certainly doesn’t describe the air column from top to bottom, with troposphere, stratosphere, thermosphere and so on. It doesn’t even work all that well only within the troposphere. It’s just a useful baseline for a gas system heated at the bottom and cooled at the top — e.g. the atmosphere.
rgb
Robert Brown says at 1/22 8:04am:
“b) If you persist in arguing that it cannot, then you have established gravity as a Maxwell’s Demon for air. Please read about Maxwell’s Demon as it would greatly improve your understanding of detailed balance and why this argument ultimately microscopically fails.”
Ok, I’m back quickly (admittedly scotch OTR is only half gone) from reading Maxwell’s demon for air: this applies to a container of air not in a gravity field where the demon decreases the entropy by opening & closing a door selectively letting thru certain molecules. Here, Willis’ cv keeps entropy constant so gravity & energy conservation solution can stand w/o establishing gravity as a demon.
Could apply Maxwell’s demon to Willis’ premise of GHG-free air container in a gravity field for some additional fun I suppose. May even discover a PE demon that arises since Maxwell didn’t have gravity in his container. BTW – some think the demon when added to the cv adds enough entropy to keep it constant or increasing. Don’t let any demons cross the cv!
I will not persist in arguing same ground or hijacking a thread, ceptin’ what sneaks in later. My view stands on energy conservation, Robert’s view stands on writing “gravity is decoupled from the problem” which necessarily introduces ignoring some of molecule’s total energy in a gravity field. My view can certainly change but only in keeping w/conservation of energy in an inexplicable gravity field.
“c) Given the stable thermal gradient in this case, one can trivially construct a heat engine that does perpetual work moving heat from the bottom to the top and then “re-using” this energy after gravity sends it back to the bottom. That’s again a simple fact.”
No, one cannot construct a perpetual motion machine in Willis’ cv + gravity: the jars, lids and wires, all reach energy equilibrium exactly & since they are all touching, heat stops flowing eventually. I’m at end of Robert’s 8:04am post finally.
Robert Brown says at 1/22 3:03pm:
“What is cv?”
Control volume.
Joel Shore;
Yes…You can (modulo the point that you need to replace “GHGs” with “greenhouse effect” since technically clouds are not a GHG since they are not a gas but they are part of the greenhouse effect). Because in the absence of any greenhouse effect, the effective radiating layer is at the surface and the highest blackbody surface temperature that can support the emission of an amount of power equal to 24W/m^2 times the earth’s surface area is a temperature of 255 K.
So, in the absence of the greenhouse effect, the Earth’s surface temperature has to be at least ~33 K less.>>>>
1. To be correct your statement would have to say that in the absence of GHE, conduction, convection, heat capacity, oceanic and atmospheric currents, the earth’s surface temperature has to be at least ~33K less.
2. To be correct, you would also have to note that ~33k less could only occurr in one scenario, and one scenario only, which would be an earth with a uniform temperature across all latitudes, from day to night, season to season, and exposed to uniform radiance. All other instances result in a differential that exceeds ~33K and my simple math model demonstrates that this number is CONSIDERABLY higher than 33K.
Robert didn’t write it, but I presume just a trivial oversight, that closed completely Bjar was then raised, deliberately and gently as Tjar was lowered. So YES, the thermal path way being established between T&Bjars at some h >0, establishes the jars new equilibrium (Caballero’s mean KE) temperature (of h) thru heat flow that ceases eventually.
You still don’t understand. Yes, by “jar” I mean that you adiabatically isolate the fluid samples at the top and the bottom, in jars with lids that do not conduct heat. The jars contain fluid at some pressure and density. I’ve constrained them so that their pressure and density profile will not change — it supported their weight before, it still supports their weight. Molecules that hit the walls bounce elastically off of them, precisely. Note that this elastic collision process is precisely similar in ever respect to having actual parcels of fluid at all of the walls that instead transfer molecules in equilibrium back into the “jar” at the same rate that random motion transfers them out. After all, equilibrium is adiabatic, right? No net transfer of energy, momentum, or particle number across the imagined boundaries of parcels of fluid in equilibrium. A perfectly elastic boundary that reflects particles back into the volume has the exact same effect, within the usual thermal noise that vanishes in the large particle number limit.
Now that the fluid is in the jars, however, we can see the irrelevance of gravity.
We can move the jars anywhere and the temperature inside of them doesn’t change. You seem to agree, although you completely miss the point about “thermal pathway” and how the invariance of the temperature inside of those jars as they are moved up and down matters.
I don’t care where the jars are when we establish a thermal pathway between them. I didn’t say that we have to bring them together and make them touch. It doesn’t, after all, matter where they are or how the pathway is established. They now just jars containing gases at two different temperatures and pressures that do not vary no matter where you carry them to — they are inside perfect thermos containers with infinitely rigid elastic walls. Drill a hole in the thermos and put a perfectly insulated silver wire so that it runs in between them, and heat will flow from the hotter to the colder, whether or not the jars have been moved at all.
The point is that there is no difference between the perfectly adiabatic walls of the jar and the gas that otherwise surrounds and uplifts the jarful of the gas. Remove the jar, put the silver wire into the air column, and heat will flow because the parcels of air at the top and bottom are not in equilibrium.
There is, as previously shown, nothing at all contradictory about a column of ideal gas supporting its own weight isothermally. In fact, I provide a reference for where I derive its description literally as a textbook example. This is the only static distribution of gas in which heat cannot flow due to conduction. There is no bulk transport or convection, either. It is not only a self-consistent solution, it is the only self-consistent solution to the problem of evaluating pressure and density in which heat will not flow from hotter to colder regions, because you don’t need a silver rod — the gas conducts heat all by itself.
rgb
So, why does it not show up in the lapse rates?
Because CO_2 is 0.03% of the atmosphere? And because the atmosphere is already optically “thick” from the CO_2 that is already there? And because the things that determine the troposphere-stratosphere boundary height probably have nothing serious to do with CO_2 per se?
Not disagreeing with your argument, by the way. I’m just pointing out that it is quite reasonable to expect the GHE to vary extremely slowly with CO_2 concentration on all of these accounts. And it does — without positive feedback, CO_2 forcing due to the additional CO_2 is probably 0.3-1.0C, although some might argue that it should be a bit higher than that — still certainly less than 2C and most of the estimates I’ve read are around 0.6C.
The argument is over the feedback. The GHE is quite real and almost certainly is responsible for a large chunk of the warming of the earth compared to an ideal (if not terribly realistic) blackbody. In fact, it is probably responsible for part of more than the 33C or whatever that is estimated compared to this, because the initial state for a rotating Earth with a heat capacity would be cooler than the ideal blackbody in the first place.
However, it is largely a saturated effect straight up. Doubling it doesn’t give us 66K warming — nobody but maybe Hansen with his “boiling seas” asserts that. The question is whether or not there is additional forcing, in particular from increased water vapor that will presumably be added to the atmosphere as it heats. But that in turn neglects the many negative feedbacks that self-organize into things like the global oscillations and increased albedo, that all increase cooling as water vapor and temperature rise.
Although the data are still inconclusive, the evidence is mounting that a) climate variation from non-GH sources is at least as large as GH related variation; b) overall feedback is somewhere in the weakly negative to weakly positive range, excluding the “catastrophic” part of CAGW already in the sense that it is already more likely than not that there will be no catastrophic warming, and the boundaries are actively being lowered by every year in the current stretch where the Earth seems to be failing to follow the “catastrophic” prediction curves.
This is skepticism at its best, quite frankly. It isn’t crank science, openly violating laws of thermodynamics. It is simply observational evidence that isn’t agreeing with the predictions of the more extreme models with high feedback. The other place where really good skeptical science is being done is in the examination of the sun and the various oscillations. Solar-albedo feedback is quite plausible and represents an additional driver that can easily erase almost any amount of GHE warming at \Delta T \approx 91 \Delta \alpha — a variation of 0.011 in albedo is enough to completely cancel — or explain — the warming of the twentieth century. Not just some of it, all of it. The oscillations are all self-organized structures that exist because they improve long term heat loss and keep the Earth in balance. Their variation is again on the same order of \Delta T as that observed, at least in any short run 30-50 year stretch. Both of these have to be better understood and incorporated before making rash conclusions about “catastrophe” at our doorstep. The science isn’t really settled, not at all, however much the greenhouse effect per se is reasonably well understood and observationally supported.
rgb
No, one cannot construct a perpetual motion machine in Willis’ cv + gravity: the jars, lids and wires, all reach energy equilibrium exactly & since they are all touching, heat stops flowing eventually. I’m at end of Robert’s 8:04am post finally.
In other words, they weren’t in thermal equilibrium before. Exactly. Because the air is all touching and heat will flow until everything is the same temperature, because the “jars” are a convenient fiction to help you isolate your point of view, something you are having a hard time doing.
Thermal equilibrium is not “energy equilibrium” — it is detailed balance, not equal energy density everywhere. Gravity cannot act as a heat source in a static unmoving fluid, it only does work on fluid that is rising or falling in bulk. There is no fluid rising or falling in a static column of air. That was also the bit of how raising and lowering the jars has no effect on the temperature of the air inside. If you want to talk about the adiabatic expansion of parcels of air, they have to move up or down. In equilibrium they do neither. Even if an adiabatic thermal profile is reached first as you take air in a sealed container and “turn on” gravity, as it sits there it will quickly become isothermal because air conducts heat and heat flows from hot to cold until they are the same temperature. Your proposed thermal gradient is not static thermal equilibrium.
rgb
Okay, I give up trying to elicit a serious discussion from the many physicists here about why the Velasco et al. conclusion is or is not valid. But just in case there exists among them a spark of intellectual curiosity none has so far betrayed, I leave my layman’s (no doubt inaccurate) description of the technique the Velasco et al. and Román et al. papers employ to demonstrate that such a system would exhibit a non-zero temperature lapse rate.
Román et al. deal with a vessel that is disposed in a gravitational field and contains N identical-mass monatomic molecules (i.e., molecules whose kinetic energy is all translational; none of that energy resides in tumbling or vibration). The vessel so isolates the molecules that their aggregate total (kinetic plus potential) energy E is constant. To arrive at the molecular-velocity (and therefore temperature) distribution as a function of height, the authors use a technique they characterize as “counting microstates.” Rather than count discrete states, though, they integrate through a continuous “phase space.” That is, since each molecule’s state can be completely described by its (three-dimensional) position and (three-dimensional) momentum, the total system state can be described by an ordered set of 6N scalars, and the system state at any instance can therefore be thought of as a position in a 6N-space, a “phase space.”
Their technique is based on the assumption, which I take it is well accepted in statistical mechanics, that the probability of the system state’s falling within any region of that phase space is proportional to the volume of that phase-space region. Now, the constraint that energy is fixed restricts the system to a (zero-volume) hyperparaboloid in that phase space, so the probability that the (energy-constrained) system will fall within a given range of states having that energy becomes the ratio of that range’s area on the hyperparaboloid to the total hyperparaboloid area. Equation 2 adumbrates this approach, giving a function that is zero at all places in the phase space except those that have an energy E, where its Dirac delta function gives it an infinite value that results in a finite integral of that function.
.
To find the position and momentum distribution for a single one of those molecules, Román et al.’s Equation 9 integrates all the other molecules’ positions and momenta out of that function and restates the result into a form that lends itself to use of a magical identity they introduced in Equation 5 as proved by a pay-walled paper. This is how they perform the integration that leads to their Equations 12, 14, and 15, which Velasco et al. adopt as their paper’s Equations 5, 6, and 7 for the distributions of molecular position and velocity.
It is from the thus-obtained Velasco et al. Equations 5 and 6 that they claim to obtain their Equation 8 for temperature as a function of altitude. It is that equation that I interpret as saying that, for any finite number of molecules, there would be a non-zero (but for significant number of molecules very small) temperature lapse rate.
Bryan says on January 22, 2012 at 9:42 am:
“O H Dahlsveen
Your interesting passage from Fourier did not deal with a thermally isolated gas .
Which means no heat enters or leaves the gas sample.”
==========
Sorry Bryan, I have obviously misunderstood your statement – but I have thought a lot about it and – I have come to the conclusion that my answer should have been:
“Hmmmm, maybe it is because it is impossible to thermally isolate a gas, at least I have never heard of a way to do it. – You can isolate against light and electromagnetic energy radiation. But heat I am afraid only conducts and will therefore find its way through any container. – Just look at a thermos flask.” —
Actually lots of research, that we never hear of, is going on to find the “Perfect Gas/Liquid Isolator” So if you can improve the thermos flask to achieve 0 heat loss I shall be sending you begging letters.” However if you are successful my guess is that you will find the gas inside your “Super Thermos” will never change its temperature.
William Gilbert said @ur momisugly January 22, 2012 at 2:53 pm
It seems to take a “strenuous & devoted attempt to force” the Received View (currently accepted paradigm) into minds here. How much more difficult to teach if there were no Received View and every institution taught its own version of Physics. I would agree that if the group practicing “normal” science were not so protective of their territory and utterly resistant to new ideas we would have more progress. OTOH, if they were utterly unprotective of their territory and accepted every new idea willy-nilly, there would be close to nil, or at best, glacial progress. No standing on the shoulders of giants allowed as it were. Just continual reinvention of the wheel (perpetuum mobile?) Kuhn was cognisant of this.
Oddly, The Character of Scientific Revolutions was the first book I purchased for my new Kindle. The second is a book called The Book of Lilith highly recommended by a friend who knows I rarely read fiction. You’ve gotta laugh 🙂
Forgot to put the link in to The Book of Lilith:
http://www.phy.duke.edu/~rgb/Lilith/Lilith.php
Stephen Rasey says at 1/22 2:23pm:
“No,… There is no acceleration…”
Merely trying to point out (me – poorly) whether the cup is accelerating or not depends on the reference frame. Cannot merely say the cup is not accelerating. Consider all ref. frames. Not just the familiar one.
My point: Einstein’s master insight was that the familiar “pull” of the Earth’s gravitational field is fundamentally the same as constant acceleration. The star position deviated w/sun gravity.
Table ref. frame: Force on Table = +F, cup mass m summation of forces F-ma = 0. F = ma. Same cup, same table, same situation, so same cup mass has to be accelerating at a (=g). Wish I could have written that simple deal 100 yrs ago. Nothing here about force of earth’s attraction. Now: Is the table accelerating?
This applies to what happens to the molecules in Robert Brown’s Bjar – so on topic, loosely. Need to account for total energy in a gravity field or the equiv. acceleration.
Robert Brown says:
January 22, 2012 at 3:48 pm
Agreed
Indeed
Let me suggest that you are mistaking the successful operation of a governing system for the absence of feedback. If the active cloud-based tropical thermo-regulatory system were to fail for only a few weeks, temperatures would soar to unimaginable heights.
This implies, not a “weakly negative to weakly positive” feedback, but the active application of both strong positive and strong negative feedbacks, in amounts that balance each other to maintain a temperature range..
The catastrophic part has always been much more spin than science.
The regulation of incoming radiation through changes in albedo (primarily in the tropics) is consistently neglected in climate science. It is the single largest factor. When the clouds form in the late tropical mornings around the ITCZ, incoming solar is cut by 40 W/m2. This implies that the variation of the time of tropical onset is a major regulatory factor, overwhelming any small changes from
Thank you very kindly, I have been one of few voices crying in the wilderness on this question, saying both that the greenhouse effect is real and that it doesn’t run the variations at our current temperatures.
All the best to you, thank you for your patience and perseverance, Oh, and thanks for the link to your online textbook. I can’t imagine what a pleasure it must be to say “see equation number seven hundred and freakin’ eleven in the physics textbook I wrote” …
w.
w.
Bart,
Even I don’t spend all my time reading this blog, which is one reason I haven’t commented on ‘Bart’s Law’. The other reason is that it makes no physical sense to me. That makes it hard to offer constructive criticism. One thing I will say is that you may be right about one minor point. Making the atmosphere over an isothermal planet absorb and emit in the thermal IR, may indeed lower the average temperature of the atmosphere compared to a transparent isothermal atmosphere. It would cause the surface to warm and the upper atmosphere to cool. Off the top of my head, I suspect that the warming of the near surface atmosphere might not be enough to offset the cooling of the upper atmosphere.
Joe Born says:
January 22, 2012 at 4:02 pm
Sirrah, that is vile mendacity. I read the paper at your behest, and I responded to you in detail. My opinion was, you didn’t understand Velasco. You can take your ball and go home with my blessing, but by gosh, you can’t blame me for it after I did exactly what you are now saying no-ne has done.
w.
Willis, don’t you not have anything to say about the empirical evidence? Why do other planetoids with atmospheres show about the same amount of “warming, DESPITE the amounts of GHGs present? “Why is it not getting warmer these days, when CO2 emissions are increasing drastically? The “bottom line” is always empirical evidence, but you just seem to shake it off!
I keep wondering just WHY you will not address this issue. You have no problem addressing other issues. Come on!
Could you have a case of confirmation bias?
William Gilbert,
Your argument that the atmosphere must show an adiabatic lapse rate is circular. Your requirement that KE + PE be constant with altitude is the definition of adiabatic expansion. Obviously you will get an adiabatic lapse rate with that condition. But there is, in fact, no requirement that KE+PE be constant with altitude. In the real atmosphere, it almost never is. Potential temperature, and thus KE+PE usually increases with altitude.
davidmhoffer says:
The point is that the greenhouse effect alone accounts for about 33 K or more…and it is the only thing that explains the important part of the warming…i.e., the part associated with the radiation emitted from the surface being 390 W/m^2 instead of 240 W/m^2. The rest is mainly what I would call “illusionary warming”. I mean, do we really want to base an estimate of the average temperature of the moon that depends radically on whether we average the temperature over the first micron, first mm, or first meter of its surface?
As for convection, it is actually what prevents the greenhouse effect from being about twice as large as it is.
That is because your model is too simple. It is probably not at all realistic for the Earth…even an Earth absent a greenhouse effect but otherwise similar.
I don’t think either of us would describe the current Earth as being having “a uniform temperature across all latitudes, from day to night, season to season, and exposed to uniform radiance” and, yet, for all intents and purposes, the error in assuming this in regards to distinguishing between averaging T^4 (and taking the 4th root) and averaging T is small.”