This is a guest post from ‘Lucy Skywalker’ who has recently returned from a trip to Germany where she attended a seminar given by Roderich Graeff, the engineering concern owner who has been experimenting with equipment he has designed to test the Loschmidt gravito-thermal effect. This line of research is highly relevant to the theoretical work of Hans Jelbring, and also Nikolov and Zeller, who have proposed hypotheses to explain the thermal gradient found in the atmosphere causing the near surface air to be warm relative to higher altitudes.
SECOND LAW CHALLENGED? LOSCHMIDT VINDICATED?
Lucy Skywalker – May 28 2012
Few people know that in the last decade, there have been quite a number of serious challenges (as opposed to perpetual-motion challenges) to the hallowed Second Law of Thermodynamics. Dr Sheehan has organized conferences and written books about all this. He put me in touch with Dr Graeff, the one participant who has been running real experiments. These experiments appear to vindicate the theories of Loschmidt, who 150 years ago challenged his friend Maxwell’s belief that a vertical air column in equilibrium will be the same temperature top and bottom. Loschmidt maintained that gravity would cause the bottom molecules to be warmer than the top ones. But until Graeff, nobody had actually undertaken the experimental research needed to check these theories against measurements.
This work could be extremely important, not least for Climate Science, if it holds up to close scrutiny. After reading Graeff’s paper and his book, I went to Germany to join his seminar, and to examine for myself his apparatus that appears to measure vertical heat gradients in columns of air, water, and other substances in steady, non-convecting equilibrium, and appears to show that in isolation, they are warmer at the bottom than at the top.
On the journey, everything goes wrong. The plane fails to take off and we are abandoned in the airport. The lucky ones find a long queue waiting for help from two junior staff on broken-down laptops. I get new train tickets in Frankfurt, but cannot inform my host because I have the wrong number. The first train is late so it misses the second train. I swap for another ticket but the next train is also late so it misses the third train. Dr Graeff has been driving to the station and back for several hours. I tell him I know it is going to be special – it has to be, after all that.
And it is special indeed. But who am I coming to a seminar whose purpose is to question the hallowed Second Law of Thermodynamics and prove that heat can travel from cold to warm? That’s what my host asks me – and everyone else. Why am I here? and why are you here? and you? and you? Why do you think you can disprove the Second Law then? Who do you think you are to do this? And without a degree in physics too? And do you know what it says? Can you explain it to me please? Ah, but how many versions of the Second Law are there, and which one do you refer to?
I would have been thoroughly disconcerted by now. I never was out to disprove the Second Law – only to examine a modification to a common interpretation of it, that Graeff has spelled out as a result of thousands of hours of close experimental work. But there is a twinkle in mine host’s eye. At 84, Dr Roderich Graeff is as sprightly as the rest of us, listening intently and responding with a mixture of 
sharp challenges and a wicked sense of fun. We’re not going to let him get away with any slipshod experiments – and neither is he going to let us get away with slipshod challenges. When I ask him, how can he be sure the measurements are as accurate as he claims, he turns on me and says, well, what do you think? How are you going to check how accurate they are?
Nullius In Verba.
Beautiful. I will have to work for my understanding, as Graeff had to do, but I love every minute of it.
The visit is a magnificent extension of Graeff’s book which I cannot recommend too highly, if you are open to the challenge. It tells you far more than I can tell here, about Graeff’s story, his life of engineering work, his scientific interests, his surprising discovery, his simple but well-fitting theoretical backing, and enough details to tempt you to replicate his work. He is a genius in classical style, who also comes from a deep place of caring about humankind. After surviving the firebombing of Hamburg (40,000 dead in one night) he wants to end all “Mutually Assured Destruction” systems, proactively build peace, and use his scientific and engineering talents to look for alternative energy sources.
Graeff’s whole house has become a laboratory – and yet it is still the warm home that he designed and built in the beautiful Black Forest many years ago. Empty Thermos flask interiors of all sizes stand around like Russian dolls, waiting to be stacked inside each other, to provide affordable insulation to the vital core experimental columns, whose temperature gradients between top and bottom are going to be measured with homemade thermocouples and fine thermistors, using the finest insulated wire you have ever seen. Sheafs of labelled pairs of wires hang out of large polystyrene boxes holding their precious cargoes of alternate layers of insulating and conducting sheaths that enclose the central mystery: Dewar bottles aka thermos flask interiors containing the core columns of solid, liquid or gas to be tested.
It all looks a mess, but in fact the experiments are well orchestrated. How many has Dr Graeff now done? About 852 1/2, this last decade or so, he tells me. Is this true? Certainly it may be, because he has carefully numbered each one. Why are there so many? Well, he has needed to check out the environment, vertical orientation, effects of size, effects of number of layers, the insulations, thermal equalizers, thermocouples, the instruments’ bias and sellotape fixings, the convectance impeders, substances actually tested, dataloggers, and software. You would think the universities would be fighting each other to have the priviledge of testing Graeff’s amazing work. But no. With a few notable exceptions like Prof Sheehan, the academics refuse to touch anything that challenges the Second Law, even if only to modify its interpretation. No, they cannot fault the experiments. And no, they cannot deny the results either. But still they fear… what? Leprosy? So Graeff goes back to his work, patiently testing over and over again, each tiny detail that might, just might upset the results. But they don’t. The results continue to hold. Tiny, but indisputable, like grit in the mouth.
Calibrate the thermocouples? You have to get the feel of the whole thing, Graeff says. And he is right. This is not 852 separate experiments, this is 852 facets of one basic experiment. Think the thermocouples might give a skewed result? Right, let’s reverse their connections and take the average to eliminate bias. Let’s also run a series of experiments in which the core is inverted. See how long the core needs to settle down… seems to take just under 2 days to invert its gradient if it’s convection-impeded gas or liquid (conductive metals are quicker to readjust), so let’s run a series of 4-day cycles and see how they average out. See how close the spots on the measurement lines are to each other? See what steady lines they make? All that is proof that the measuring devices are measuring true, not wild, and the fact that we can reverse the apparatus and get similar results suggests strongly that it is gravity that realigns the temperature of the inner core to a negative gradient, not a fault in the apparatus.
Still not certain about the thermocouple calibration? But that’s not the point, Graeff says. And he’s right again. The point of the experiments is that they are consistently showing a negative vertical temperature gradient at all in the core axis, even though that is surrounded by layers in which there is no gradient, which are surrounded by layers tending steadily towards the normal interior-of-room positive gradient, warmer at ceiling level than at floor level. Okay. So are all 852 experiments showing a negative temperature gradient at the core, surrounded by a positive gradient, that suggests gravity is, as Loschmidt suggested, affecting the thermodynamic equilibrium of all those core columns?
Well, perhaps a few experiments did not work. Perhaps a Dewar bottle broke. Perhaps a mouse got in and made a nest. Perhaps the cellar door was draughty and letting in the cold outside air. Perhaps the thermocouples were badly attached. Perhaps the Keithley datalogger was playing up. Perhaps the Excel program was faulty. Perhaps… but aren’t we forgetting Einstein’s words, to the effect that just one effective experiment was all that was needed to topple an accepted law? And aren’t 800 varieties of that experiment enough? What hoops has the man got to go through? Hasn’t he earned recognition? 
Why is no university even bothering to check those experimental methods and results, let alone replicate them or study Graeff’s theory which fits the results like a silk glove? How many laboratories are going to kick themselves because they didn’t recognize the telltale evidence of a genius at work, I wonder?
Perhaps it’s easier for me as an outsider to spot genius 🙂 I can feel that frisson a mile off. I recognize the heavy, stuck shape of the doubts of the orthodox when “appeal to authority” is missing from the menu; I recognize the hoops and the highs; I have examined for myself what proofs and patience and precisions are really needed. Still not convinced? Read my Amazon review of Graeff’s book. Get the book itself and read it twice. Then come and meet Graeff and check out his tests for yourself, ask more questions. But by that time, you ought to have realized that he really is on to something important, and it’s something that we too can check and replicate and even publish.
James Clerk Maxwell would be proud.
Next part:
Gathering an Experimental Replication Team. I will describe the experiments in more detail, with background about the 2nd Law controversies. I will describe Graeff’s theoretical underpinning that seems to work and requires omitting a commonly-held assumption about the Second Law, and allows for the effect of gravity as an unavoidable outside influence. Certainly there is no reason to abandon the 2nd Law or even rewrite it in essence. And I’m putting out word to gather a little local team to replicate Graeff’s work.
Tallbloke's Talkshop 
Thanks Rog. My, was that fast. I like the way you’ve put in the pics. Thank you.
The lapse rate he measures for air is higher than theory predicts. Why?
Non-equilibrium thermodynamics rears its head here.
Good question MDGNN.
Graeff tells a whole story around this issue in his book. Briefly, Graeff was puzzled himself as to why measurements were higher than his first predictions by a factor of 5, pretty well precisely 5. Two years later he woke up one morning with a lucid moment and knew he had to amend the specific heat he used in his formulation, by dividing it by the number of “degrees of freedom” of the molecules of the substance in question. Air has basically five degrees of freedom at room temperature, is the short answer. Long answer requires a book or two… at least, Graeff’s two chapters, but more is needed.
Upshot for Graeff was that instead of a theoretical temperature gradient of 0.014K/m, he could now use 0.07K/m… which matched his experimental observations very well indeed. He has also, since then, obtained matches for other substances eg water which has 18 degrees of freedom.
Lucy: thanks for the interesting write-up. As I’ve said elsewhere, I don’t think Graeff is challenging the second law, or at least not the version of it which is formulated in terms of energy distribution. He is pointing out that it is incorrectly interpreted, and that mis-interpretation leads to badly conceived thought experiments which don’t recognise that the presence of gravity invalidates the idea of a column isolated from external forces. Gravity pervades everything.
His degrees of freedom insight could be very important to our understanding of statistical mechanics, which has been an area fraught with assumptions and a lack of solid experimental results. At its foundation it rests on 150 year old ideas about the behaviour of molecules which have never been rigorously tested. There was some discussion around this in the previous thread.
Thanks Lucy. Lapse rate is a dynamic analysis in that a virtual work argument is used to postulate the temperature gradient at which a falling parcel of air would, due to the adiabatic heating from compression, replace a lower parcel with no net change of internal energy.
What Graeff is doing is to establish something else. The practical implication is that there is a mechanism which, because the heating due to gravitational potential energy is greater than that due to lapse rate, always initiates convection, even at night with no heating at the earth’s surface from incident SW energy.
I think this is very important.
MDGNN, spot on. The atmosphere adiabatic lapse rate is of the order of one-tenth of Graeff’s measured gravity gradient. So yes, it appears that convection kicks in to neutralize nine-tenths of the temperature gradient. In mines, with less convection, there is a higher lapse rate – varies from one mine to another.
But there is more. It all fits Nikolov and Zeller’s work of course. And it fits something that to me was mind-boggling when it first hit home: why the Earth’s core is hot, and why we only feel it on the surface on the rare occasions when volcanos erupt. But….. one must also reconcile with the cold ocean floor and a positive temperature gradient in the oceans – although Graeff has clearly and repeatedly measured a negative temperature gradient in water when convection has been impeded.
I have my ideas but I’d rather hear others’ thoughts.
Buoyancy. Check it out. 😉
http://www.teachersdomain.org/asset/phy03_vid_zhot/
Which begs the question – why doesn’t the atmosphere end up warmer up top for the same reason. I think it’s because more heat is being lost to space near the top of the atmosphere, and the bottom of the atmosphere is being warmed by the oceans at the same time.
It wouldn’t hurt to see if we can find some data on the relative buoyancies of warm water and warm air though.
Tallbloke, I think we can explain more with just gravity 🙂 I suggest you think carefully.
Lucy, as you said: “one must also reconcile with the cold ocean floor and a positive temperature gradient in the oceans”. So that’s why I’m taking a look at buoyancy. Then we can look at the force of the gravito thermal effect, incorporating the degrees of freedom hypothesis, and see what drops out.
At the global average SST of 17C, and a salinity of 39,000ppm, water weighs 1028.616kg/m^3
at an average seafloor temperature of 2C and the same salinity, water weighs 1031.205kg/m^-3
=2.595kg of buoyancy/m^-3
rho of Air at gobal average 15C is 1.2250
rho of Air at -25C is 1.4224
I’ll do some sums in a bit. 🙂
Correct TB: you get at the ocean data from the equation of state. The really interesting bit [paper to be published when it gets past pal review] is that the end of ice ages is caused by massive melting of the Antarctic ice pack due to the tsi increase and a reduction of local cloud albedo, no CO2 involved
The melt water sinks and restarts the ocean currents. You can see this in the transects N-S of the Pacific Ocean showing isotherms. Only at the Antarctic does it get near 0 °C at the ocean base. Elsewhere it’s more saline and levels off at 1.8°C. When the ice ages end the deep Southern ocwean rises by 2 K due to the massive contraction of the ice area!
Another picture you could add:
I apologise if this is a daft question, but it has been ’bugging me’. The air is never just pure air, it is ‘contaminated’ with aerosol, liquid and solid particles, and as such they would be under the influence of gravity, just like the molecules in air.
Aerosol are everywhere and in huge quantities. Take a deep breath. Even if the air looks clear, it’s nearly certain that you’ll inhale tens of millions of solid particles and liquid droplets.
http://earthobservatory.nasa.gov/Features/Aerosols/page1.php
So if the buoyant aerosol are concentrated by gravity wouldn’t they concentrate the air temperature?
TB
Graeff’s theoretical calcs for water (and salinity and near-zero behaviour don’t really matter for my point here) show 0.04K/m. Experiments with convection-impeded water back this up with 0.045K/m.
Now multiply that up. You get 40-45K/km. Whoa. Where did it go?
Steveta
Wicked 🙂
Pardon my ignorance of physics, but does not gravity cause pressure? With gasses, this pressure means that molecules are closer together which means that vibration per molecule may be uniform for the column of gas, but the increased number of molecules mean a higher temperature is detected at the bottom compared to the top of the column. Is there any other mechanism that needs to operate to give the results obtained?
I am quite surprised that this topic is even controversial. Of course the kinetic energy of molecules increase as they move lower in a gravity well. This must increase their temperature. As the molecules try and escape the gravity well, their kinetic energy is reduced and so is their temperature.
We see the same effect with light. Light is red-shifted as it tries to escape a gravity well, which indicates a reduction in energy (temperature). Light is blue-shifted as it approaches a gravity well, which indicates an increase in energy (temperature).
This is very simple to simulate with perfectly elastic balls on a computer model (which is a valid model for air molecules). The balls at the bottom of the simulation (nearest the gravity source) will over time come to have the highest kinetic energy per unit of volume and thus the highest temperature. The balls at the top of the simulation (furthest away from the gravity source) have the lowest kinetic energy per unit of volume (and thus the lowest temperature).
Intuitively you can see this. As the air molecules try and escape gravity they are slowed until they reverse direction and fall back towards the surface. Otherwise they would escape to space. At that point, when reverse direction and have lost almost all kinetic energy, they are effectively near absolute zero.
In opposition to the effect is convection. As the molecules near the surface gain kinetic energy their volume expands, making the air lighter, which then rises. Thus in dry air we should see a lapse rate equal to the kinetic energy imparted by gravity. We will not see this in moist air, because of the added energy from condensation/evaporation of water (10 kkcal/mole as I recall).
Added to this is the compressibility of air as compared to liquid. This should also have an effect, because it allows for more energy to compressed into a smaller area with gasses as compared to liquids. This could well explain why gasses have an opposite direction (sign) for lapse rate as compared to liquids.
First, thanks to Dr. Graeff, Lucy, TB, and MDGNN for covering most of the critical questions above and I agree with most of the points raised so I don’t want to spin wheels over the same topics. But speaking of those topics, the atmosphere and ocean temperature gradients, let’s assume Dr. Graeff is completely correct in the measurements. This would mean there must be some missing differences across these substances that seems to be hiding in the physics. For instance, if ocean water does naturally have such a large gradient then the density gradient is fighting over the natural gradient and in fact override in the end, is the more powerful of the two effects, so in the end you still end up with colder water deeper by the unique density of water. We need some numbers here.
The atmosphere would be somewhat the opposite with the gradient explaining most of the warmer-lower lapse but you have a ~288K to 216K, or even 3K, temperature gradient opposing this gradient. Does the subtraction match? Seems we need some detailed calculations here also to explain both cases. This seems to lead to the point where radiation in the lower atmosphere has indeed a very tiny influence, less than 66 Wm-2 at the tropics and not too far above zero at the poles, leaving all heat flows to the matter’s mass itself as N&Z implies.
Seems just a subtraction problem between two opposing heat tendencies from different parameters.
Seems Dr. Graeff has covered most of my suspect calibrations so I give him much more weight shown in his reports. Pick up the calculators everyone and help answer these.
Truthseeker says:
May 30, 2012 at 2:50 am
Pardon my ignorance of physics, but does not gravity cause pressure?
Gravity acting on mass produces a radial pressure gradient from the centre of gravity outwards.
With gasses, this pressure means that molecules are closer together which means that vibration per molecule may be uniform for the column of gas, but the increased number of molecules mean a higher temperature is detected at the bottom compared to the top of the column. Is there any other mechanism that needs to operate to give the results obtained?
We need to consider the difference in compressibility of gases and liquids, and the ‘degrees of freedom’ in which the molecules are able to oscillate. Also the degree to which that oscillation becomes restricted as pressure increases. However, the low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume
Which leads me to wonder what role electricity might play in the opposite gradient found in the ocean. Salt water is a fairly good conductor, much better than air, which is a fairly good electrical insulator. Could it be that as water molecules are forced closer together their proximate oscillations generate charges which cause energy to flow in the form of electrical current to less dense regions of the ocean (nearer the surface)?
The 104.45o angle between the hydrogen atoms in water molecules is variable, and I would expect it to change under pressure as water molecules become more densely packed. I think that might cause an electrical potential difference (voltage) to arise between deeper denser water and less dense water above it.
Sounds a bit ‘far out’ I admit, but is it worth considering?
Possibly relevant reading:
http://en.wikipedia.org/wiki/Van_der_Waals_force
“Van der Waals forces are relatively weak compared to covalent bonds, but play a fundamental role in fields as diverse as supramolecular chemistry, structural biology, polymer science, nanotechnology, surface science, and condensed matter physics. Van der Waals forces define many properties of organic compounds, including their solubility in polar and non-polar media.”
http://www.lsbu.ac.uk/water/molecule.html
“The charge distribution depends significantly on the atomic geometry and the method for its calculation but is likely to be about -0.7e on the O-atom (with the equal but opposite positive charge equally divided between the H-atoms) for the isolated molecule [778].d The experimental values for gaseous water molecule are O-H length 0.95718 Å, H-O-H angle 104.474° [64].e
These values are not maintained in liquid water, where ab initio (O-H length 0.991 Å, H-O-H angle 105.5° [90]) and diffraction studies (O-H length 1.01 Å, O-D length 0.98 Å [1485]; O-D length 0.970 Å, D-O-D angle 106° [91])f suggest slightly greater values, which are caused by the hydrogen bonding weakening the covalent bonding and reducing the repulsion between the electron orbitals. These bond lengths and angles are likely to change, due to polarization shifts, in different hydrogen-bonded environments and when the water molecules are bound to solutes and ions. Commonly used molecular models use O-H lengths of between 0.957 Å and 1.00 Å and H-O-H angles of 104.52° to 109.5°.”
http://www.lsbu.ac.uk/water/vibrat.html
“Increasing the pressure on water decreases the O····O distances (graphed elsewhere) so increasing the covalent O-H distances and lowering their stretch frequency [804]. Raised pressure also causes a reduction in long, weak or broken bonds and an increase in bent and short, strong hydrogen bonds”
Lucy: Did Graeff find any difference between pure water and salt water? We might get some clues if he has tested them both under similar conditions.
Hi Rog and Lucy,
I am a bit unsure as to how to approach this, I may have some info that may impact on the experiment that you wish to clone. I am not willing to post this info. You have my email. Regards Wayne.
Ray C says:
May 29, 2012 at 3:11 pm
I apologise if this is a daft question, but it has been ’bugging me’. The air is never just pure air, it is ‘contaminated’ with aerosol, liquid and solid particles, and as such they would be under the influence of gravity, just like the molecules in air…
Air contains mainly nitrogen and oxygen, but also CO2, A, N, H2O, as well as.. the aerosols. Graeff’s calculations are an approximation for N2 and O2 because all the rest have different temperature gradient factors. Degrees of freedom – too complicated to explain here and the approximation is good enough.
Apologies for slow responses. My main computer is down 😦
ferd berple says:
“This is very simple to simulate with perfectly elastic balls on a computer model (which is a valid model for air molecules). The balls at the bottom of the simulation (nearest the gravity source) will over time come to have the highest kinetic energy per unit of volume and thus the highest temperature.”
Do you have a link to this program?
Your line of reasoning is most persuasive and yet the br1 simulation gives the isothermal solution.
Tallblokes Hans Jelbring Post.
An Alternative Derivation of the Static Dry Adiabatic Lapse Rate
It would be most instructive to compare both simulations.
Truthseeker says:
May 30, 2012 at 2:50 am
What I did was try thinking slowly into exactly what gravity does to the individual molecules. This leads to molecules not only packed closer but also moving faster (on average of course).
ferd berple says:
May 30, 2012 at 3:31 am
I am quite surprised that this topic is even controversial…
Indeed. More is the pity. But it’s understandable, given a) the stature of those who really formulated the laws of thermodynamics b) really it would have been impossible to measure at lab scale until pretty recently. And it took Graeff’s engineer’s mind to realize that though the official standards of accuracy for thermocouples is too low to catch the temperature gradient at lab scale, he could utilize the thermocouple for gradient measurement which is ten times more accurate than absolute measure.
Lucy,
It is quite the mind tease if you try to follow all the laws and theories to generate a viable planetary model. Many areas were not considered in experimentation which just cluttered up science results with bad thesis conclusions which we have learned over centuries.
Technology has changed but our consensus scientists ignore this to protect the past from changes.
Bias? Dam right there is.
From gravity to relativity is just a guess.
Doing an experiment on land to the ocean can have quite a different conclusion due to many factors from land heights, reflected energy, velocity differences, angles of solar heat at that particular moment, water loss, salt changes, incorrect pressure measuring, etc.
Once you understand the mistakes, how they were generated and what bias there is, you will understand that science should have been constantly reviewed to new discoveries or technologies which show vast areas of human mistakes. Statistical gathering on on orb with parameters of averaging has generated another error of laziness upon our science community if they want to go farther to understand our planets past and it’s future. Water loss and planetary slowdown generate huge changes over the billions of years.
Bryan:
“Your line of reasoning is most persuasive and yet the br1 simulation gives the isothermal solution. ”
Glad someone was paying attention!
I wrote that simulation specifically to see how it would apply to Graeff’s results. There are many papers deriving an isothermal distribution of a gas under gravity, but I didn’t really believe it until I checked it out.
I must admit I have no plausible explanation for his results, and everyone I have spoken to about it are stumped as to why it could be true or where it is wrong. I’d be very happy to advise Lucy on a replication, I hope it works out!
I even tried replication using a home-made centrifuge, but only managed to measure wind-chill – it needs a vacuum or a lot better insulation than I had 😦
Lucy, TB, Mydogs et al – FYI this remarkably clear and simple video by Carl Brehmer has re-appeared on the web – http://myweb.cableone.net/carlallen/Greenhouse_Effect_Research/Greenhouse%20In%20A%20Bottle%20Reconsidered.html he put it up at the end of last year but only a few noticed it. Keep plugging away, sooner or later the truth will be out and they will have to admit that CO2 has feck all to do with it.
Lapogus: I have derived the replacement for CO2-GW. But in so doing I am being forced [joke!] to change everything!
These are : no direct thermalisation, needed to explain the mechanism by which Kirchhoff’s Law of Radiation does not apply at TOA so emissivity DOWN = zero. [Yes TB, it’s the only logical outcome.]
Also of course, the ludicrous boundary condition at the other end, IR UP = S-B bb AND recycles S-B bb ‘back radiation’.
Correct these and radiation is but a minor component of the heat transfer. The side effect is that Tyndall was wrong [easily proved by Nahle’s Mylar balloon experiment] and collateral damage is Lindzen and all the others who believe in direct thermalisation.
You also overturn Sagan’s incorrect aerosol optical physics and Houghton’s treatise has major errors although he was on the right track, just a lack of statistical thermodynamics’ understanding as with most other scientists.
And the real culprit is – biofeedback via change of average cloud albedo convolved with other natural effects.
The indirect thermalisation of IR at cloud droplets is the physical cause of Miskolczi’s experimental proof of constant IR optical depth independently of [CO2].
PS Brehmer has got the wrong end of the stick. The CTE effect is minor compared with adiabatic temperature rise at constant volume plus the increase of absorptivity of CO2 with T.
Brian (and evidently br1), on slide 4 in that presentation (http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw) is where he made the mistake it seems.
He says:
“If you normalize those distributions to have an area equal to ‘1’, and plot the ideal 1D Maxwell-Boltzmann distribution (black dashed line), then you can see they are all VERY similar.
A little bit noisy though.”
It is the normalization itself that gives the equal temperatures at any height. Each of the four (red, green, blue, and magenta) curves on the left plot all do have similar curve and therefore they all have a Maxwell-Boltzmann distribution at each height and the normalization does just that, it show all four are M-B.
But, he then takes the normalized sets and computes the temperatures from the artificially normalized sets and of course, they are all have a temperature very close to each other. He should have used his four sets from the left plot and each would have in fact had a decrease in temperature with height and the proper MB for each of the four temperatures.
That simulation would not be too hard to reproduce correctly.
Or that is what I see in that presentation.
wayne:
“It is the normalization itself that gives the equal temperatures at any height. ”
Afraid not.
The temperature of each curve is related to the mean velocity, which for a 1D Maxwell-Boltzmann distribution goes as:
v_mean=sqrt(2*kB*T/(pi*m))
so all I did was take the mean velocity of the velocity histogram and then solve for temperature.
Scaling all the velocities in the distribution by a constant factor will not change the mean, hence will not change the estimated temperature.
“Scaling all the velocities ”
that is to say, scaling the *count* of the velocities will not change the distribution. This is only related to particle number, not velocity distribution.
Hi br1. I see now see what you are up-scaling to normalize, the counts, I misinterpreted that plot, you are right there, but I still have questions with your results even though that is one neat way to approach it.
What threw me is you have the maximum counts in all four cases with a velocity of zero and I must have mentally switched the axes. So what gives there? The max count at zero velocity.
Think I’ll take a stab at reproducing your results in 3d with axis component separation. Takes a bit more execute time but I just don’t want to leave any leaf unturned, for I have also simulated balls in a column as Brian mentioned and the top balls were always found at a decreased mean velocity over time… but you have tweaked my curiosity.
Sorry Wayne, BR1’s presentation is spot on and you are mistaken. The DENSITY of particles does not (directly) determine the temperature of a gas. A high density gas and a low density gas can have exactly the same temperature, as long as the average KE is the same. The simulation shows that the average KE is (very nearly) the same [and still follows the MB distribution], so the temperature is (very nearly) the same at every altitude.
Thanks, BR1, for such a clear, simple refutation of the “adiabatic lapse rate = thermal equilibrium” hypothesis.
*****************************************************************
I must say, I have rarely seen so much unsupported wishful thinking as shows up in the comments in this thread. From oceans to mines to statistical mechanics to the shape of water molecules, there is no end to wild speculation about what is “obvious”. (There is an old joke among mathematicians that if you want to find the error in a proof, look to the place where the author says “it is obvious that …” ) Yet the one bit that is actually supported with calculations and graphs and theory is the one bit that seems to be most strongly questioned.
Opinions about science are easy. If you want to convince people, then back up the opinion with number and equations. Graeff has attempted to do that (moderately convincingly). BR1 did it quite well. But until you can refute standard answers (like the center of the earth is warm because of heat from infalling material/radioactive decay that has not cooled yet because of 4000 miles of pretty good insulation) then accepting other unconfirmed causes (like the Graeff effect) is premature.
Well Tim F, you didn’t respond to my
2b or not 2b ?
question concerning the Verkely et al paper highlighted by ‘Trick’ in the last discussion on this
“Exner pointed out that the confusion arose from defining the problem in an inconsistent way…(Maxwell) discusses the classical formulation of the problem and its answer (the profile will be isothermal) but then he argues that convective motions…will be important…. Bohren and Albrecht…consider an ideal gas in a gravitational field, and seek the state of maximal entropy… result in an isentropic profile. This can be regarded as a confirmation of Maxwell’s idea… This brings us outside the domain of classical thermodynamics, and hence one can expect that the temperature profile will no longer be isothermal; we will derive below what profile forms the outcome.”
In part 2b, Verkley paper derives “what profile forms the outcome” showing the common misconception of an isothermal profile must be replaced by a non-isothermal, isentropic profile.
so I must say your editorialising looks a bit silly to me. This is an open question we are investigating, and the Loschmidt conundrum isn’t so easily settled.
Graeff is measuring an effect, but what is it? How does it relate to the open atmosphere and oceans?
Please contribute with ideas rather than rhetoric.
“Well Tim F, you didn’t respond to my
2b or not 2b ?”
I am pretty sure I already addressed all this Tallbloke, but one more time …
1) The accepted standard answer for an isolated, insulated column of air was already given by Gibbs & Boltzmann over 100 years ago. Classical thermodynamics (and now statistical mechanics) predicts an isothermal gas with a pressure gradient and with no convection.
2) The actual atmosphere is NOT isolated nor insulated. Specifically, it is heated quite strongly from the bottom and cooled quite strongly from the top (as is a mine). This means the idealized answer from Gibbs and Boltzmann should not be expected to apply to the atmosphere. The heating in the actual atmosphere leads to convection, which leads to a temperature gradient. NO ONE questions that there should be (and that there actually is) a gradient due to the heating at the bottom and cooling at the top. The strong convection limits the importance of heat flow by conduction, so the adiabatic approximation becomes more accurate.
All that the paper you cite concludes is that:
a) a column of air perfectly insulated from the surroundings but allowed to conduct within itself should be isothermal (in accordance with classical thermodynamics).
b) a variety of idealized heated/convecting/non-equilibrium conditions should lead to the isentropic conditions and hence to the adiabatic lapse rate (in accordance with classical thermodynamics).
c) the actual atmosphere is somewhere between these extremes.
So when “(Maxwell) discusses the classical formulation of the problem and its answer (the profile will be isothermal) but then he argues that convective motions…will be important…”all he is saying is that (a) is the ideal answer, while is acknowledging that (c) is the real-world answer. “Confirmation of Maxwell’s idea” is simply confirmation of sunshine from the sun and IR radiation to space.
You can’t say “the expected answer for a heated/cooled atmosphere is a temperature gradient (ie case (c)), therefore case (a) (with completely different conditions) should also be a temperature gradient.” Neither Maxwell nor Gibbs nor Boltzmann nor Verkley make this error.
PS EVEN IF Graeff is right, his results would have little to do with the actual atmosphere.
[Reply] Well, you still fail to address ‘Trick’s points about Verkely et al section 2b. And similarly, bottles of co2 in the lab have little to do with the actual atmosphere.
Try using the standard atmosphere calculator.
http://www.digitaldutch.com/atmoscalc/
Input the altitude, get back these *variables*: temperature/pressure/density/speed of sound
Wayne:
“So what gives there? The max count at zero velocity.”
It’s a 1d MB distribution, not a 3d distribution. In a simulation with no collisions, the dimensions are independent, so the horizontal components won’t make any difference, and they won’t change as there are no horizontal forces.
Still, I would be very interested in your 3d sim. How long do you reckon it might be before you can post a result?
You wrote one sentence I would like clarification on, about ‘energy per volume’. I imagine the energy per volume at low altitude is indeed greater than at high altitude due to the increased number of particles per volume, but it is the average velocity per particle that gives temperature. What did you mean exactly?
br1: ” it is the average velocity per particle that gives temperature.”
Ya think? How do we measure temperature? If you use a bulb thermometer, the heat capacity of the surrounding medium is going to have a significant effect on the reading. That will be strongly affected by the density of the medium.
Maybe this is where misunderstanding arises. The bulk temperature of a gas is a different matter from the temperature of an individual particle.
On your 1D simulation, that is why I feel it really needs to be 3d for completeness. I also really need to also account for the x-y planes velocity components per layer, also to be complete. On when I can post it, as soon as I can, if I can. I will not complete it and then not post the results just because it verifies your results. Just proper science. Within the equations it does seem a 3D should exactly match the 1D when merely solving in 3D and isolating the z component.
Also br1, please point to the mention of ‘energy per volume’. I don’t remember saying anything close to that, of course, the lower volumes will always have more total energy per volume due to the density by the gravity. I could not even find it in a search of this post. Maybe it was Wayne Job?
Do you think opposing high velocity air streams have friction? Seems that is one key parameter. Or are we going to stay out of reality and in pure theory? If there is a real gradient that is the culprit.
Tallbloke: “Ya think? How do we measure temperature?”
I agree temperature is not as simple as my statement implies, which was just for a pure ideal gas. I’m always amazed at how subtleties and complications in practice can have unexpected effects, and I certainly haven’t seen the end of them. On a positive note, sometimes I think I’m making progress on the basics.
I should have typed “that might be the culprit”, paying too much attention to spelling instead of the words themselves. 🙂
Wayne: “Also br1, please point to the mention of ‘energy per volume’.”
Sincere apologies, it was ferd berple.
About friction between winds, there probably is something like that, so long as one doesn’t mind using the word ‘friction’. The effect of finite gas particle size could be very important. I can almost prove that finite particle size should not affect the isothermal argument, but I can’t quite prove it in 3d. In 1d I can prove it (I think). I haven’t had time to write a full 3d simulation of a gas in a box yet.
Interesting topic folks. Haven’t posted on this subject here before or much of anywhere else for that matter outside the WUWT threads.
br1 says at 3:04pm:
“The temperature of each curve is related to the mean velocity, which for a 1D Maxwell-Boltzmann distribution goes as:
v_mean=sqrt(2*kB*T/(pi*m))”
Note in the M-B v_mean formula there is no gravity force term ~g. M-B distribution is derived with the assumption of no external forces on the gas particles. This classic M-B solution does obtain isothermal for no gravity field in an enclosed adiabatic non-GHG ideal gas column (no other outside forces).
Add gravity and there is an external force ~mass on each particle – which changes things from M-B distribution, so can’t invoke classic M-B. The Verkley paper 2a shows exactly how to derive the classical isothermal solution in the presence of gravity where the non-GHG gas column is allowed to do work (across the control volume) on the gas column above & below.
When the non-GHG gas column in the presence of gravity is idealized perfectly (i.e. adiabatically) isolated so no work is allowed across the control volume (as Graeff seems to be experimentally striving for), Verkley et. al. paper 2b shows exactly how the ideal gas temperature profile with height differs from the classical isothermal solution to a non-isothermal, isentropic (max. entropy point) solution at LTE; btw also consistent with 0th, 1st & 2nd Thermo Laws which are used in derivations.
This paper is fairly short, condensed & ideal solutions are related to real standard atmosphere so one has to read it thoroughly, slowly or a couple times to get all the nuances. Really got to look close at the constraints as Exner wrote (and Graeff appears to be doing) as confusion results if you don’t.
The paper points out in discussion how to think about mixing. Call mixing as convection, conduction and/or turbulent. For me, it is helpful to just think: of course there is vigorous mixing of the non-GHG ideal gas column constituents under normal standard troposphere (80% of earth’s atmosphere) initial conditions (i.e. no p~0 hPa, no T~0K).
>Well, you still fail to address ‘Trick’s points about Verkely et al section 2b
You apparently missed this part above:
This IS section 2b of the paper! If the conditions preclude effective conduction within the column, but allow for vertical motion (eg natural convection or upslope/downslope winds), then a temperature gradient WILL occur (the adiabatic lapse rate). There is no contradiction. No breaking of thermodynamic principles. No startling new science.
>And similarly, bottles of co2 in the lab have little to do with the actual atmosphere.
Where in any of this discussion did we start talking about bottles of CO2?
[snip]
[Reply] That was my response to your comment regarding the relationship of Graeff’s controlled experiment with the real atmosphere. I’m sure you can see my point.
Trick: Welcome, and thanks for pitching in. I’m having trouble deciding whether Tim Folkerts has tacitly conceded the point or not. Maybe you can help.
“The DENSITY of particles does not (directly) determine the temperature of a gas. ”
Tim, I never said any such thing in this post, and you know it. Trying to make me look foolish just does the same to you. Can you stop the out-of-context or mangled cheap shots?
I have no idea what point you think I have conceded, Tallbloke. If you could clarify, I would appreciate it.
But to make more clear what my position is, let me quote the abstract:
A few things jump out.
1) “Using an elementary variational calculation”. So if you don’t know calculus of variations, you can’t really comment on the math. Many engineers would know this, but I suspect most would be quite rusty on the topic.
2) “an isothermal profile is obtained in accordance with classical thermodynamics”. The author confirms the standard result.
3) “appropriate in the case of convective mixing”. So the solution with a temperature gradient is for a case with active convection, which is clearly not the case for Graeff’s experiments.
As to Trick’s comments …
>Add gravity and there is an external force ~mass on each particle –
>which changes things from M-B distribution, so can’t invoke classic M-B.
Sorry, but Wikipedia and I both disagree. “The Maxwell–Boltzmann distribution applies to any system composed of atoms, and assumes only a canonical ensemble, specifically, that the kinetic energies are distributed according to their Boltzmann factor at a temperature T” Unless you are prepared to show either than 1) this is an incorrect statement of the law or 2) that this is not a canonical ensemble, you are are making an unsupported (and incorrect) hypothesis. Gravity adds additional degrees of freedom, but those degrees of freedom would also have 1/2 kT of average thermal energy. The velocity distribution of of the MB distribution still applies.
>When the non-GHG gas column in the presence of gravity is idealized perfectly
>(i.e. adiabatically) isolated so no work is allowed across the control volume
First of all, “adiabatic” means “no heat”, not “no work”. For 2b and the “adiabatic lapse rate”, we are talking about parcels of air WITHIN the system being adiabatic with respect to each other, ie the thermal conductivity is negligibly small. So for example, if you fill an insulated piston with air, that air is adiabatically separated from the rest of he atmosphere. If you take that cylinder to a higher altitude, the cylinder will expand (doing work) and cool. This type of process is the core idea behind the adiabatic lapse rate (and the idea of “potential temperature”). I could do this in a column of air and create a temperature gradient.
But I have to KEEP doing this to maintain the gradient (according to both classical thermodynamics and Verklay). When there is active convection, conductivity is indeed negligible, but if convection stops, then conduction is the only game in town, and it will bring the temperature back to a uniform profile (according to both classical thermodynamics and Verklay).
>btw also consistent with 0th, 1st & 2nd Thermo Laws..
Sorry, but no. The 0th law states “if two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other.” Suppose my “third system” is a thermometer. If the top of the column (system 1) is in thermal equilibrium with the bottom (system 2)< then the thermometer must be in equilibrium with both (ie read the same temperature for both).
PS. Were referring to Gore's CO2-in-a-jar experiment? Surely you don't think that because you disagree with me, and that you disagree with Gore, that somehow I must support Gore misguided scientific understandings.
And now I have more important things to work on, so I won't be checking back as often.
Wayne,
When you said
it seemed that you thought that the normalization process itself would lead to incorrect temperatures, when in fact all it does is to make the graphs easier to see and to compare (precisely because the density IS different). That is what i was commenting on when I stated that density does not matter, since either the original data or the normailzed data give the same result.
Your later comments do seem to recognize that the normalization is not a problem, so I apologize for misinterpreting your position.
It’s great to know Tim F has more important things to work on than the laws of thermodynamics, we can be expecting some truly mould breaking science from him. 🙂
I’ll let Trick answer for himself regarding most of Tim’s responses, but I’ll just make the following two points in response.
1) Graeff still gets a (reduced) gradient in freely convecting air.
2) Tim F has fallen for the same fallacy Robert Brown did wrt the ‘zeroth law’. It is perfectly possible for a temperature gradient to exist through contacting systems in thermal equilibrium, because the bottom of the upper system will be at the same temperature as the top of the lower system where they touch each other. So no heat will flow, and there will be thermal equilibrium, even though the average temperature of the upper system is different to that of the lower. I’ll do a back to basics post on this since it seems to confuse high-and-mighty physicists such as Robert Brown and Tim Folkerts.
It doesn’t confuse lowly nuts-and-bolts engineers like me. I’m kind of glad Tim F has decided to ride his high horse off elsewhere for now, he was confusing the discussion with his misunderstandings of thermodynamics.
Does anyone know if Graeff took into account the fact that thermocouples are pressure sensitive?
http://www.tandfonline.com/doi/abs/10.1080/0895795031000088269
“because the bottom of the upper system will be at the same temperature as the top of the lower system where they touch each other. ”
Huh? you seem to be suggesting that an object can have a temperature profile across its lenght, but that any two adjacent microscopically thin layers of the object must be at the same temperature.
How does that work?
Trick:
“Add gravity and there is an external force ~mass on each particle – which changes things from M-B distribution, so can’t invoke classic M-B.”
subtlety here: note that in my simulation I didn’t invoke the classic MB – it was an *output* of the simulation. This was actually the question I was trying to answer, and what the temperature was of whatever distribution the simulation came up with.
The only place the MB distribution was invoked was thermal contact with the ground. This was treating the ground as a ‘thermal wall’, as described in this paper: http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf . The interaction between the wall and the particle can’t be affected by gravity (surely?). After that, the particle was on its own under gravity with a force F=-mg acting on it. Throw enough particles into the air, and one *ends up with* a MB distribution at each height, and each distribution has the same temperature.
p.s. For those interested in writing their own simulations for this, the phi1 distribution in the thermal walls paper is not actually a MB distribution. One can generate velocities in the phi1 distribution by using its inverse cumulative density function
V_phi1=sqrt((-2*kB*T/m)*log(1-rand))
where ‘rand’ is an evenly weighted random number selected between 0 and 1.
Steveta_uk: the anvil compression of the T/Cs was to extraordinarily high pressures. 5 GPa is about 50,000 atmospheres!
Steve, OK, sloppy formulation. You can split hairs as fine as you like, the gradient has continuity where it passes through the contact point.
ferd berple says:
May 30, 2012 at 3:31 am
…In opposition to the effect is convection. As the molecules near the surface gain kinetic energy their volume expands, making the air lighter, which then rises. Thus in dry air we should see a lapse rate equal to the kinetic energy imparted by gravity…
Added to this is the compressibility of air as compared to liquid. This should also have an effect, because it allows for more energy to compressed into a smaller area with gasses as compared to liquids. This could well explain why gasses have an opposite direction (sign) for lapse rate as compared to liquids.
& Tallbloke considers buoyancy…
I’ve really had to think what is is like at molecular level. I think Ferd and Rog are on to something, this helps make sense of the difference between free water (oceans cold at bottom) and free air (atmosphere cold at top). I’d like to add further:
Take one cubic metre of water. Turn it into vapour and it will extend to a height of a kilometre, if covering the same ground area. Gravity has a huge part to play with gas / air / vapour, compared with liquid / oceans, if we look at it this way. It is perhaps then less surprising that gravity only just wins out with air (nine-tenths of Graeff’s direct gravity effect gets neutralized) and loses with water.
Tallbloke: re testing pure water and salt water, I don’t know. Probably, knowing the man. Electrical effects: an interesting thought, particularly as it’s deep sea fishes that actually utilize electrical energy for light. But I suspect it can still be explained by gravity plus a little quantum stuff maybe. I’ll email you Roderich’s email, you can ask him yourself.
Wayne, br1, thanks for interest, thumbs-up, and followup studies.
I cannot watch the presentation as my main PC is down. However, I note that it was Boltzmann and Maxwell that Loschmidt challenged. And personally I have absolutely no problem with non-isothermal theory, now I’ve studied Graeff’s experiments, grasped his theoretical backing (from his paper and book) and met the man and seen the lab work for myself. Personally, I prefer to start from the evidence of the experiments and not worry about any theory that seems to suggest things should be otherwise, unless the theory is very robust.
In this case, the isothermal theory is indeed upheld by a venerable line of theoreticians, starting with Willard Gibbs I think but continuing right up to the present time. But heck, one careful experiment upsets the lot! And I do think that Graeff’s theoretical backup is brilliant: simple, matches data, uses good maths and thermodynamic basics. What’s the problem?
Admittedly, it does read more easily in Graeff’s book, when he explains delightfully how he came to the theory in two profound intuitive “aha!” moments, one at New Year 1999, one a couple of years later. I had trouble originally grasping the “degrees of freedom” stuff until I poked my nose into the quantum mechanics involved – which helped a lot though I still find it a bit of a mystery. But it is actually quite logical, that the direction of gravity only affects one of the degrees of freedom of the substance in question, and that the equipartition theory explains why the standard specific heat has to be divided therefore by the number of degrees of freedom.
What is involved is energy, work done: gravity acts on each molecule, and work done can be easily converted to temperature by any engineer with standard thermodynamic knowledge, Graeff says. The only difference between apples and molecules is the need to include the degrees of freedom for molecules. If I have understood it correctly.
Mydogsgotnonose says:
May 30, 2012 at 12:28 pm
Lapogus: I have derived the replacement for CO2-GW. But in so doing I am being forced [joke!] to change everything!
These are : no direct thermalisation, needed to explain the mechanism by which Kirchhoff’s Law of Radiation does not apply at TOA so emissivity DOWN = zero. [Yes TB, it’s the only logical outcome.]
Also of course, the ludicrous boundary condition at the other end, IR UP = S-B bb AND recycles S-B bb ‘back radiation’.
Correct these and radiation is but a minor component of the heat transfer. The side effect is that Tyndall was wrong [easily proved by Nahle’s Mylar balloon experiment] and collateral damage is Lindzen and all the others who believe in direct thermalisation.
You also overturn Sagan’s incorrect aerosol optical physics and Houghton’s treatise has major errors although he was on the right track, just a lack of statistical thermodynamics’ understanding as with most other scientists.
And the real culprit is – biofeedback via change of average cloud albedo convolved with other natural effects.
The indirect thermalisation of IR at cloud droplets is the physical cause of Miskolczi’s experimental proof of constant IR optical depth independently of [CO2].
Well MDGNN, your dog may not have a nose but I think your own nose beats any normal dog’s nose. I had to admit I cannot understand every word you write, but the “scent” is heavenly and I think you really do understand things.
“TOA IR emission effectively zero” – this seems right on to me. Heck, if that’s what TOA did,it would be visible like a halo round other planets. And we’d be deafened by IPCC pointing to such evidence. No, there’s nowt but theory. OTOH, gravitational warmth gradient (tempered by convection) just fits the evidence like the proverbial silk glove… or something.
Tim Folkerts says:
May 30, 2012 at 4:36 pm
Sorry Wayne, BR1’s presentation is spot on and you are mistaken. The DENSITY of particles does not (directly) determine the temperature of a gas…
Thanks, BR1, for such a clear, simple refutation of the “adiabatic lapse rate = thermal equilibrium” hypothesis….
*****************************************************************
I must say, I have rarely seen so much unsupported wishful thinking as shows up in the comments in this thread. From oceans to mines to statistical mechanics to the shape of water molecules, there is no end to wild speculation about what is “obvious”…
I did a bit of real science with Tim’s comments.
Method
I tested for the presence of the word “obvious” and it showed up only twice… both times in Tim’s comments. I then considered whether I had implied anything like “obvious” in my post. I am puzzled because it seems missing to me… its presence is not at all “obvious” to me.
I put Graeff’s work in the context of current serious challenges to the Second Law. “Serious” does not mean “obvious” but perhaps this conflation is obvious to Tim.
I said right at the beginning “This work could be extremely important, not least for Climate Science, if it holds up to close scrutiny.” I don’t see anything about “the obvious” there either.
I said “Graeff goes back to his work, patiently testing over and over again, each tiny detail that might, just might upset the results. But they don’t. The results continue to hold. Tiny, but indisputable, like grit in the mouth.” Still no sign of either the word “obvious” or, I think, the spirit or flavour of that word.
wayne says:
May 31, 2012 at 2:01 am
“The DENSITY of particles does not (directly) determine the temperature of a gas. ”
Tim, I never said any such thing in this post, and you know it….
My observations in this little experiment is that Tim is not observing this thread at all carefully, let alone the experimental work which this piece is actually about.
My conclusions:
1) This is a rather rude time-waster. Therefore, although Tim has written a lot more, I am ignoring it because I want to pay attention to the matter in hand, namely Graeff’s demonstration of a consistently negative temperature gradient in the thermally-shielded core, even when surrounded by a positive one.
2) This is a shame. For all I know, Tim Folkerts is a Doctor of Science. He – you – really should pay more attention and learn more civility. I hate to see talents misused or go to waste.
Tim Folkerts says at 3:14am:
————————————————————–
“As to Trick’s comments …
>Add gravity and there is an external force ~mass on each particle –
>which changes things from M-B distribution, so can’t invoke classic M-B.”
Tim continues:
“Sorry, but Wikipedia and I both disagree. “The Maxwell–Boltzmann distribution applies to any system composed of atoms, and assumes only a canonical ensemble, specifically, that the kinetic energies are distributed according to their Boltzmann factor at a temperature T” Unless you are prepared to show either than 1) this is an incorrect statement of the law…”
————————————————————–
I am prepared to show that the quote above has confused Tim in 2 ways that cause disagreement:
1) M-B is not a law – it is a probability density function.
2) Tim quotes from the wiki page on the equipartition theorem. Tim should quote more clearly w/less confusion from the M-B distribution wiki page which starts out right at the top:
“…the Maxwell–Boltzmann distribution describes particle speeds in gases, where the particles move freely without interacting with one another, except for very brief elastic collisions in which they may exchange momentum and kinetic energy…”
A gravity field causes the gas particles to no longer “move freely” thus M-B distribution can’t be invoked per the M-B wiki page et. al. Apparently the wiki page on equipartition theorem could be improved to reduce confusion and disagreement in at least one poster on the internet.
Exner would cry & not be happy since Tim’s “…confusion arose from defining the problem in an inconsistent way.”
br1 says at 10:37am:
“The only place the MB distribution was invoked was thermal contact with the ground. This was treating the ground as a ‘thermal wall’,…”
Once again, Exner would not be happy. You are adding confusion by apparently modeling different constraints than Graeff’s experiments & the constraints in 2b of the Verkley paper which theory shows obtain an ideal non-isothermal, isentropic profile. Your simulation would be properly non-isothermal, isentropic given the same exact constraints. Try it!
The paper you cite does not mention gravity or show any gravity terms. It states: “The distribution of velocities of the molecules leaving this surface will be determined by the temperature of the wall.”
Thus there is no gravity field affecting the velocities of the molecules leaving the thermal wall surface in the paper. M-B can be invoked as in the paper at the wall but then turning M-B off: “.. the particle was on its own under gravity with a force F=-mg acting on it….” is mixing constraints.
Exner is happy since apparently Graeff’s experiments (still striving to approach constraint perfection) and Verkley 2b (perfect) utilize the same exact constraints in a gravity field both showing non-isothermal profiles.
Trick:
“Once again, Exner would not be happy.”
Sorry, I’m not familiar with Exner (I’m relatively new to the climate debate). Could you provide a link?
“You are adding confusion by apparently modeling different constraints than Graeff’s experiments & the constraints in 2b of the Verkley paper which theory shows obtain an ideal non-isothermal, isentropic profile. Your simulation would be properly non-isothermal, isentropic given the same exact constraints. Try it! ”
umm, how does convection apply to an isolated+insulated+static gas?
“Thus there is no gravity field affecting the velocities of the molecules leaving the thermal wall surface in the paper.”
of course, I’m glad we can agree on this.
“M-B can be invoked as in the paper at the wall but then turning M-B off: “.. the particle was on its own under gravity with a force F=-mg acting on it….” is mixing constraints.”
what is this all about? Gravity is not ignored, it simply has no effect on the thermalisation of the particle with the wall. The velocities with which the particles are emitted from the wall are *instantaneously* determined – it is a boundary condition. Gravity acts at all times.
For example, imagine the simpler case of a perfectly reflective wall in the presence of gravity. The velocity of the particle the moment after colliding with the wall would be the negative of the velocity the moment before it collided with the wall. The presence of gravity has no effect on the reflective boundary condition, but that doesn’t make it inconsistent or ‘mixed’.
“Exner is happy since apparently Graeff’s experiments (still striving to approach constraint perfection) and Verkley 2b (perfect) utilize the same exact constraints in a gravity field both showing non-isothermal profiles.”
But Graeff’s experiment is not meant to have convection. If it does have it, then there is something screwy going on anyway – where can it come from?
br1:”But Graeff’s experiment is not meant to have convection. If it does have it, then there is something screwy going on anyway – where can it come from?”
From the fact that warmer air rises to displace cooler air. Graeff found a gradient of around 0.04oC/m when he prevented convection by introducing a glass powder medium. But he also found a gradient of around 0.02oC/m when allowing the air to freely convect.
Lucy has more details.
br1 4:34pm – tallbloke posted the link above 5/29 9:45am which leads to Verkley paper which contains the note:
“Exner pointed out that the confusion arose from defining the problem in an inconsistent way…textbooks by Exner (1925, 60–62)…”
br1 continues:
“But umm, how does convection apply to an isolated+insulated+static gas?…Graeff’s experiment is not meant to have convection. If it does have it, then there is something screwy going on anyway – where can it come from?”
Comes from nature allowing forced convection and free convection. Graeff’s baseline experiment does not allow forced convection (as in the atmosphere w/a surface heat source and heat sink at top to deep space) but his base experiment and Verkley 2b do allow free convection to obtain non-isothermal profile.
br1 also writes:
“…the simpler case of a perfectly reflective wall in the presence of gravity. The velocity of the particle the moment after colliding with the wall would be the negative of the velocity the moment before it collided with the wall. The presence of gravity has no effect on the reflective boundary condition, but that doesn’t make it inconsistent or ‘mixed’.”
No. The reasonably continuous moment before and the moment after particle colliding with any wall velocities ARE differently affected by gravity. There IS a gravity effect on the incident and reflective vert. velocities b/c there is an h difference (small but present at any observable moment time slice). There also is a boundary condition which ideally is ignored but can’t be in Graeff’s non-ideal experiments.
Trick: “M-B can be invoked as in the paper at the wall but then turning M-B off: “.. the particle was on its own under gravity with a force F=-mg acting on it….” is mixing constraints.”
Br1 asks: “what is this all about? Gravity is not ignored, it simply has no effect on the thermalisation of the particle with the wall.”
Geez, gravity having no effect means gravity is ignored as in the paper cited. You wrote your simulation invoked no-gravity M-B at the thermal wall (as in the paper cited) then turned on gravity away from the walls. This is mixing gravity & no-gravity constraints. This constraint switch is not part of Graeff’s experiments or Verkley paper at all. If your simulation gets a different answer, it is not necess. a surprise & as above Exner is not happy with confusion from inconsistency.
To make Exner happier, keep gravity having a continuous effect in the simulation as then it will be consistent and introduce less confusion w/Graeff experiments and Verkley paper.
br1
I think your problem lies with the assumption that the molecule leaves with the same velocity as it arrives. This contradicts commonsense. Think into it. Be a little molecule. What you have to look for is conservation of momentum / energy. The falling molecule is energized by gravity, thwacks the bottom wall which says ouch you’ve given me a red cheek. Bottom wall surface molecule is slightly energized in this way, and it manifests as temperature increase. Bouncing air molecule loses a tiny bit of energy from the impact, rises to top wall which says ouch you’ve given me a pink cheek (balance of energy / temperature is transferred from top wall to molecule here, because said molecule is flagging under gravity).
OK?
Again I say, read the book especially Graeff’s chapters on his theoretical insights arriving, plonk! plonk! and his calculations to convert kinetic energy into thermal energy. Did you read my last note to you? Do you think molecules are real? Are they less affected by gravity, individually as air molecules, than collectively as Newton’s apple? Surely not!
Correction: the top wall has to say “you’ve given me a blue cheek” because the impacting molecule here draws net energy away from the wall.
Tim Folkerts, I noticed your gracious apology. Thanks Tim, that was very acceptable. Let’s just stay to this interesting topic, it does need to be correctly answered one way or the other, don’t you agree?
br1’s simulation is a lesson in itself. I have created the distributions, one dimension and three, and am going to carry it a bit further to see if his re-scaling can apply in the manner that he is using. One thing that does pop out is that in 3D, two of the random velocities, x and y, are not affected in any pure manner whereas the z velocity component is warped by the gravitational field and it initially appears that the re-scaling he is using no longer works in a real 3D world but I have more writing to do over the weekend.
Look at the distribution terms. The left terms are all constant. The velocity squared term linearly scales with gravity’s mostly constant acceleration as the height varies. The right term also includes a velocity squared term as you would expect. If you are really a scientist, look at this. With the x-y unvarying and one linearly decreasing it seems that even though br1’s stab may hold mathematically in one D, it seems no longer the same in 3D. This might be where some of the confusion lies. But, of course, I’m not finished yet.
TimF, ouf course that is speaking of the v = √(vx^2+vy^2+vz^2) in the 3D equation as compared to the 1D M-B distribution. (just so you can see where that I spoke above applies) Now rerun br1’s with those terms included and you might see what I am speak of.
Trick, sorry, seems much I was saying to Tim is the same topic you are presenting to br1. I should have read all of the recent comments. Didn’t mean to just duplicate you, in so many words.
Excellent stuff everyone. I think this thread is demonstrating the virtue of continuing a polite discourse on a subject over a period of months rather than having a two day shootout and burying the corpses quickly without a post mortem.
🙂
I have finally identified a couple of quirky inferences and non-sequiteurs in Maxwell’s 1872 tome on heat and will be writing a post elucidating a couple of consequent logical errors which may have a strong bearing on the reasons why paradoxical situations have arisen in statistical mechanics. Basically, the ‘zeroth law’ has been extended beyond it’s remit by later scientists.
Trick: Do I remember correctly that you went to the trouble of obtaining a copy of Exner’s textbook? Is there any chance you could photograph relevant pages for us?
Wayne: Keep going, it’s great to see independent minds converging towards consistent solutions.
tallbloke 7:35pm – I did borrow a copy of Bohren&Albrecht 1998 text to learn the detailed constraints of Verkley 2004 2b theory if that’s what you mean. B&A text is easy to obtain, read & reinforces Exner 1925: get the constraints consistent or stand confused. True everywhere – telling that Exner had to write this down & still usefully remembered to this day.
I recall my thermo Prof. once teaching (thankfully over & over & over) that a control volume is student’s biggest friend in the course; better learn to draw one to pass. An equivalent in statics is the free body diagram. I’ve put those concepts to good practical use over the years. Can you even think of integrating a differential equation in practice w/o specific boundary conditions? Remember to thank a teacher who made a difference.
Trick: Ah, that was it. Here’s one I’m going to get out of the library tomorrow:
The tragicomical history of thermodynamics, 1822-1854 / C. Truesdell.
Lucy Skywalker’s says: I’m putting out word to gather a little local team to replicate Graeff’s work.
Well said! In my opinion the only real issue to be addressed right now is not whether this or that theoretical speculation is right or wrong but WHETHER GRAEFF’S EXPERIMENT IS REPRODUCIBLE.
I fully support your proposal to repeat his experiment and hereby volunteer to lead and manage a team that does so. Until that work is completed, I personally will refrain from entering into this fascinating debate-from-theory, however attractive and addictive it is becoming.
Oh…and roll out your Part 2 Graeff article soon, Lucy!
Ok, thx, ordered up that one too. The B&A text is also complete w/the thermo tragicomical history. Some really great anecdotes (so much that some complain, ha). I ended up reading B&A (mostly) cover to cover.
David/Lucy: I’m thinking a borosilicate glass vacuum tube of the type used for solar hot water panels might make a good test rig core.
Overlay of recreated data plot and figure 2 from PDF
Data as XLS is here.
If anyone wants to look in more detail, including using more expansive scaling, there it is.
Would be useful if someone definitely identified the traces, perhaps mark them on the apparatus drawing, although this omits the locations, how the connections are routed, etc., a lot missing.
The gx numbers are derived from intermediate filenames here.
I’m not in contact with these people so asking is not something I wanted to do, especially when I am taken as hostile.
TB Re. Replicating the Graeff experiment:
I had in mind a simple plastic tube about 5m high and surrounded by a colossal amount of polystyrene foam (I really don’t understand the rationale behind Graeff’s use of multiple layers of different insulation materials. Keep it simple!) Add high spec, high sensitivity electronic temperature sensors at top and bottom. The starting temperature profile inside the tube is completely irrelevant but would be around ambient. Just wait for it to settle down over time and record the top-to-bottom differential temperature. I see no point in having the elaboration of two parallel tubes. Just repeat the experiment with different contents and measure the differential top-to-bottom temperatures in each case.
By the way, Graeff’s idea of using the difference in temperature between the tops of his two different columns to drive a heat engine is NOT a violation of any of the Laws of Thermodynamics as he claims. Any work thus generated would be as a result of heat flowing from the hotter to the colder column top, not (as he unaccountably claims) the reverse. And the heat thus flowing would either be limited to the surplus heat contained in the hotter column (assuming perfect insulation) in which case it would soon all be used up – so no perpetual motion there. Alternatively, if the insulation is imperfect, the ambient heat leaking in (however small) would be what is powering the heat engine – so no perpetual motion there either. (Actually, in reality the heat engine would not work anyway unless its discharge temperature was less than ambient!) .
However this lack of violation of the Laws of Thermodynamics is not at all bad news for Graeff because it means that if his discovery of a temperature differential between bottom and top of column due to gravity is proven true, it must be possible to incorporate this observed phenomenon within the existing Laws – a far less tough challenge than trying to buck them.
Tallbloke, don’t forget that at the sort of pressures you see in the deep ocean a cubic meter of water will hold more molecules than it will at the surface (ie be denser due to compression) as well as density changes due to temperature.
Water is generally considered uncompressible at “normal” pressures but it does compress slightly at very high pressures. I am not sure you included that density change in your calculations.
http://www.newton.dep.anl.gov/askasci/eng99/eng99607.htm
If you did account for the pressure change density my apology.
Larry
Has anyone worked out the thermal time constants associated with this column of air? On my calc, a metre of stationary air has a time constant of about 5 million seconds, or two months. That’s not the time to reach equilibrium – it’s something like a half-life. Equilibrium would take about a year. And the aim of the experiment seem to be to keep the air stationary.
The time constant goes with length squared. So a 2m column would have a time constant of eight months, etc.
In contact with the thermal mass of glass and the container?
Well, the glass and container are supposed to be super insulators. So they should have an even longer time constant. Otherwise the experiment is in trouble. There’s no way you can find a temp gradient in air if the boundaries can conduct it away.
David: Glass would minimise electrostatic effects wouldn’t it? The problem with a 5m high rig is that you need a 5m high temperature controlled room to put it in. I’m wondering if it might not be easier to control the temperature of a warmer than ambient room rather then one which is cooler. In general, it’s easier to control additional heat input and escape than maintain constant T in a room cooler than ambient.
Nick: As you can see from the plot Tim C has replicated (well done Tim!), Graeff did run the experiment for months. However it seems that when he inverts the water filled rig, the gradient re-establishes itself over a couple of days.
Larry: I think I mentioned earlier that the compression is around 1.8% at the sea floor.
Tim, I agree we need better experimental description.
I may have got the air time constant wrong – it now looks to me to be less than a day. However, water is long. For 1 m the time constant is
density*SH/k = 1000*4200/0.58=7240000 sec=84 days
So I see the experiment with water ran for a little over one time constant, which doesn’t really get to steady state.
Nick, IIRC the test rig was less than a metre, about 700mm?
TB, the water column was 850mm. That brings it down to about 60 days.
We all agree that the density gradient is quickly established.
The molecules in the air column are all approximately ‘in the right place’.
So the time to rearrange between an isothermal or non -isothermal equilibrium may not be as great as a timescale of months.
Is it possible that part of the difference between the results obtained by br1 and Hans Jelbring is due to the fact that br1 is considering individual particles whereas Hans Jelbring is considering ensembles of particles with ‘gaps’ in between?
A rising parcel of air expands due to the lowering of surrounding pressure, but also contracts due to cooling as its kinetic energy is exchanged for potential energy. In an adiabatic process at equilibrium, these opposing tendencies should cancel each other. If the gravito thermal theory of Loschmidt is correct, the net result should be a temperature gradient at equilibrium equal to the gravitational lapse rate (for dry air).
I’ll be interested to see what difference keeping gravity in play throughout will make to br1’s calcs, along with the 3D analysis Wayne has undertaken. This issue is shaping up to be a blockbuster storyline, with twists and turns and reversals along the way. 🙂
Lucy:
“I think your problem lies with the assumption that the molecule leaves with the same velocity as it arrives.”
Sorry, but I don’t make that assumption. I presume you got this from the ‘reflective wall’ comment to Trick? That was only an illustration to show how gravity has no effect at a boundary.
“The falling molecule is energized by gravity, thwacks the bottom wall which says ouch you’ve given me a red cheek. Bottom wall surface molecule is slightly energized in this way, and it manifests as temperature increase. Bouncing air molecule loses a tiny bit of energy from the impact, rises to top wall which says ouch you’ve given me a pink cheek (balance of energy / temperature is transferred from top wall to molecule here, because said molecule is flagging under gravity).”
This was the first simulation I wrote on the subject. Here is a presentation I made to Roderich at his meeting over a year ago:
(apologies if you can’t read it due to your PC being down, but it will be a nice motivation to get your PC back to life! I can email it if that would help?).
Note in the last slide I ask a big “why?” as to how the ‘random mean’ behaviour (which does give a gradient!) can be different from the long term random behaviour (which doesn’t give a gradient!). To answer that question I wrote the second simulation, which was the http://www.slideshare.net/brslides/maxwellboltzmann-particle-throw presentation , which shows that a column of gas under gravity does not support a temperature gradient when only driven by a thermal wall with no further energy inputs or losses.
I then tried inter-particle attraction in a multi-particle gas in a box, but that also gave no gradient. I show theoretically why that is the case in http://www.slideshare.net/brslides/walton-applied-tovanderwaalsgas
Yesterday I wrote a simulation which swaps energy between vz and hidden energy reservoirs (degrees of freedom!), and that DOES give a gradient. However, I have to do a lot of sanity checking to see what assumptions are in that, and while the result is intriguing I don’t want to rest my hopes in it yet.
Apart from that I’m a bit stuck for an explanation. I have no motivation to ‘disprove’ Roderich’s results – I would much rather they are what they seem to be, than some experimental mistake!
tallbloke:
” Graeff found a gradient of around 0.04oC/m when he prevented convection by introducing a glass powder medium. But he also found a gradient of around 0.02oC/m when allowing the air to freely convect.”
which shows that the effect we are looking for is not due to convection!
David Socrates says:
May 31, 2012 at 9:02 pm
Lucy Skywalker’s says: I’m putting out word to gather a little local team to replicate Graeff’s work… I fully support your proposal to repeat his experiment and hereby volunteer to lead and manage a team that does so…
Thanks David that sounds fantastic. But first, wait for my second article where the constraints will hopefully be spelled out a little more. What I would most like to do is take a little group of really interested people to visit Graeff, where a number of things are likely to become quickly clear, thereby saving us from inventing wheels unnecessarily. For instance, the layers… but I will discuss this in the next piece.
David S: By the way, Graeff’s idea of using the difference in temperature between the tops of his two different columns to drive a heat engine is NOT a violation of any of the Laws of Thermodynamics as he claims.
David, actually Graeff Tallbloke and I are aware that there is no essential violation of the Second Law. But coming in cold and seeing the results, one may wonder about a claim of violation because one has been taught that a column of air in equilibrium is the same temperature top and bottom. But in reality, that was no more than the surmise of Maxwell and all others except Loschmidt (backed with impressive-looking maths from Willard Gibbs etc), and until Graeff it had never actually been tested.
So Graeff jokes at his expense.
tchannon says:
May 31, 2012 at 10:27 pm
Overlay of recreated data plot and figure 2 from PDF
Data as XLS is here.
If anyone wants to look in more detail, including using more expansive scaling, there it is.
Would be useful if someone definitely identified the traces, perhaps mark them on the apparatus drawing, although this omits the locations, how the connections are routed, etc., a lot missing.
The gx numbers are derived from intermediate filenames here.
I’m not in contact with these people so asking is not something I wanted to do, especially when I am taken as hostile.
Tim, thank you, that is an amazing re-creation. It’s possible / likely that Graeff has the original data. Yes, that graph of Graeff’s is very badly labelled, and next article I hope to produce a better version so you can see easily what is what, because in fact, when one unravels it, all details are telling.
And thank you for joining in and doing this work when I realize you have some problem with it. I don’t see you as hostile, just with some problem but I do not know what it is. Email me privately if you would rather – because I would like to try and understand your sticking-point. Then either I could explain what I now understand, or I might agree with you.
Trick:
““Exner pointed out that the confusion arose from defining the problem in an inconsistent way…textbooks by Exner (1925, 60–62)…”
This was a complaint against those who said the equilibrium condition should have a gradient!
Which is somewhat ironic given the present discussion 🙂
“Graeff’s baseline experiment does not allow forced convection (as in the atmosphere w/a surface heat source and heat sink at top to deep space) but his base experiment and Verkley 2b do allow free convection to obtain non-isothermal profile.”
Verkley 2b *assumes* convection. Read the first sentence of section 2b: “Now, in Ball [1956, Eq. (4)] it is argued that for convective motions…” and he goes on from there.
But where does Graeff’s convection come from??? There is not meant to be an external heating differential, and there are not meant to be winds. So where can it come from? And to avoid glib answers, please explain the why and how of how convection gets established?
“The reasonably continuous moment before and the moment after particle colliding with any wall velocities ARE differently affected by gravity.”
you can’t be serious? Take the reflective wall again. Let’s make it ‘one level more real’ and say that the wall is composed of springs that have a bit of ‘give’ in them, but keep the idealisation that they are massless and lossless (hey, let’s not take too much reality in one go!). Even though the impinging particle may sink in to the surface a little, it will be thrown out with the same velocity it came in with. This is true whether gravity is present or not. So there is still no inconsistency or ignoring going on.
Add one more level of reality – an atom size is about 10^-10 m. I’m not sure how much it sinks in to a surface under STP (not much), but there are also EM interactions that may be on the order of 10^-6 m. Now consider that in the simulation I took the height of gas up to 4.10^3 m. Are you trying to imply that the effect of gravity over 10^-6 m will dominate what happens up to a height of 10^3 m? You are off by over 9 orders of magnitude.
“Geez, gravity having no effect means gravity is ignored as in the paper cited.”
Not at all. Your argument is like saying that because gravity acts in the Z direction and does not act in the X direction, that having a simulation with both X and Z components is inconsistent.
“You wrote your simulation invoked no-gravity M-B at the thermal wall ”
No, that is not the case. To be precise, MB is not invoked at the wall at all. What is invoked is a phi1 emission velocity distribution (which is *not* a MB distribution!). The presence of gravity *has no effect* on this condition, it is basically a thermal continuity condition.
“To make Exner happier, keep gravity having a continuous effect in the simulation”
It does – there is no time where gravity is not present in the simulation.
Also note the main point of the simulation – it is to see how the distribution changes with altitude. Note that *even though* the particles are slowed down due to gravity, the *distribution of velocities* does not change. This result stands on its own and is quite independent of the lower boundary condition. It is also what the old farts have been going on about for the last 150 years. Even Verkley and Exner say the same thing!
As to leaving out bits of reality in a simplified simulation – of course that is true. Roderich measures a gradient where my simulation has none, but the question is *why*? It is not due to the naive thought that because gravity slows down molecules with height, therefore the gas cools. There is something else in there, but I don’t know what it is.
br1:”tallbloke:
” Graeff found a gradient of around 0.04oC/m when he prevented convection by introducing a glass powder medium. But he also found a gradient of around 0.02oC/m when allowing the air to freely convect.”
which shows that the effect we are looking for is not due to convection!”
It shows the effect we are looking for is reduced by convection. The important point is that the fact that there is convection shows that there is an effect.
” I presume you got this from the ‘reflective wall’ comment to Trick? That was only an illustration to show how gravity has no effect at a boundary.!”
Gravity always has an effect. It doesn’t suddenly stop acting on particles just because they are near boundaries.
Lucy, have you got my emails about the Keithley datalogger?
Nick Stokes says:
June 1, 2012 at 3:23 am
Has anyone worked out the thermal time constants associated with this column of air? On my calc, a metre of stationary air has a time constant of about 5 million seconds, or two months. That’s not the time to reach equilibrium – it’s something like a half-life. Equilibrium would take about a year. And the aim of the experiment seem to be to keep the air stationary.
Wikipedia: Thermal time constant
Time constants are a feature of the lumped system analysis (lumped capacity analysis method) for thermal systems, used when objects cool or warm uniformly under the influence of convective cooling or warming. In this case, the heat transfer from the body to the ambient at a given time is proportional to the temperature difference between the body and the ambient…
The whole point of the experiment is to impede convection, in a thermally-neutral environment ie temperature top and bottom being seen/measured as equal. And even more, the whole point of this work is to look closely at the actual data obtained, in the context of apparatus produced with an excellent engineering awareness of things like boundary conditions where unwanted heat exchanges might happen. I am setting aside theoretical work in general, in favour of simple direct observations at this point.
So for now, I don’t see that your “thermal time constant” is relevant. If you think it is relevant, please explain why in simple practical layman’s terms, relating to the apparatus described.
My next article will try to explain the experimental rationale in more detail.
Tallbloke, short answer, yes. Am penning a reply but it’s not a simple yes/no.
Lucy,
The time constant comes directly from the heat equation which governs simple conduction – the mechanism that these experiments aspire to. It governs the approach to equilibrium. You can think of it as a measure of how long it takes for a heat differential at one end of the column to have an effect at the other.
What it means is that if you started out with a temperature gradient, with or without gravity, it would take a time interval of something like that long before it would be substantially reduced. The 100 or so days might be long enough – I just wondered if anyone had checked.
Nick, sorry, I see you did talk a bit more, and Tim too, about potential boundary problems. I will address these, but the short answer is that they are not bad enough to upset the primary message of data indicating a negative temperature gradient while surrounded by a positive one.
Nick, another short answer is that the “consensus” of many experiments Graeff has done, is that it takes typically about two days for extraneous effects to level out to the limits of instrument detection and actual experimental needs. I will also talk about the averaging techniques Graeff uses in order to bypass certain biases and wobbles.
David Socrates:
“Alternatively, if the insulation is imperfect, the ambient heat leaking in (however small) would be what is powering the heat engine – so no perpetual motion there either.”
but this violates 2LoT, as this could be done continually and it should not be possible to get continual work from a heat bath.
Note the statement about heat travelling from cold to hot only applies in the vertical direction (upper surface cools down, lower surface heats up, away from isothermal). Heat would flow from hot to cold across the horizontal gradient between the tops of two columns of dissimilar gases.
br1: Roderich measures a gradient where my simulation has none, but the question is *why*? It is not due to the naive thought that because gravity slows down molecules with height, therefore the gas cools. There is something else in there, but I don’t know what it is.
Please explain why you consider this a “naive thought”.
To me, this is the very essence of the matter, that air molecules do behave rather like apples under gravity, they do fall and gain kinetic energy in consequence and this kinetic energy can be translated directly into thermal energy. One major difference is that molecules move in all directions. Related to this is another major difference between molecules and apples falling, that you have to reckon with the degrees of freedom, because vertical energy gets translated into energy in all directions, after a few collisions. The three axes xyz are not enough, you also must include the xyz factors of rotational energy, also vibrational energy between atoms in molecules, and degrees missing because quantum theory says so at room temperatures. Only monoatomic gases, helium, argon etc, have just three degrees of freedom corresponding to xyz. All this needs to be factored into the models imho.
Again, I recommend people to read Graeff’s own account of how he arrived somewhat intuitively at his theoretical / mathematical backups to his experimental evidence. After that, I leave you to theorize away because I think you are more at home in this dimension than I am, and you may yet attain the definitive theory. Like Faraday I will stick with the experiments, and leave you Maxwells to do the maths.
tallbloke:
“It shows the effect we are looking for is reduced by convection.”
so let’s say there are two effects, one which produces a gradient and one which reduces it. Convection effects reduce the gradient, so what effect produces it? How does that relate to Verkley 2b, which deals with convection?
As I see it, the Verkley paper has two main sections:
2a, no convection which gives an isothermal distribution
2b, convection which gives a gradient
As an isothermal trend would obviously reduce Graeff’s gradient, then it seems that Graeff has a third effect, and there is no mention of what that could be in any paper I have read. Hence the interesting puzzle. That my simulation doesn’t have this ‘third effect’ is not surprising, and I don’t take it as disproof of Graeff’s results, but I do take it as proof against the idea that gravity must produce a temperature gradient.
“Gravity always has an effect. It doesn’t suddenly stop acting on particles just because they are near boundaries.”
Gravity doesn’t stop acting, but it doesn’t always have an effect.
Take the case where gravity acts only in the Z direction, then what effect does it have in the X direction?
If you think that is too idealised, and say that the real world is more complicated, then please think about the orders of magnitude involved. There is a gradient of 0.07 K/m to be explained. The simulation I wrote went to a height of 4000 m, which should give a temperature drop of 280 K, yet showed no trend to about +-0.3 K.
br1 says:
June 1, 2012 at 11:57 am
David Socrates:
“Alternatively, if the insulation is imperfect, the ambient heat leaking in (however small) would be what is powering the heat engine – so no perpetual motion there either.”
but this violates 2LoT, as this could be done continually and it should not be possible to get continual work from a heat bath.
br1, I suggest you re-read Graeff’s reformulation of the Second Law which is at the end of his paper (link in article at top). The problem is that the Second Law really applies to closed systems where there is no input of external forces. But when we remember to include gravity, most systems are not really closed.
Whether this system is potentially useful in terms of usable electric energy, is another matter. Currently the electric output is minuscule, and does not draw enough on the system to really put it out of equilibrium. And establishing the gradient currently takes a day or two… except with metals which are much quicker but also seem to have a lower gradient.
Lucy:
“Please explain why you consider this a “naive thought”. ”
Basically, it was this thought that brought me to Roderich in the first place. I was wondering what the effect is, and presumed it must have been dealt with before. After a quick Google, I came across Roderich, and contacted him from there.
The only problem is that the gradient goes away once you go into the details. That is explained in the three slideshare presentations. I emailed them on to you, hoping you can still read pdf’s. I admit I have been struggling since.
However, my daily work is experimentally based, and I would be delighted if you could replicate the result.
br1If you think that is too idealised, and say that the real world is more complicated, then please think about the orders of magnitude involved. There is a gradient of 0.07 K/m to be explained. The simulation I wrote went to a height of 4000 m, which should give a temperature drop of 280 K, yet showed no trend to about +-0.3 K.
Indeed, this gradient is to be explained.
As stated above, I think that what happens in the free air is that convection neutralizes about nine-tenths of gravitational warming in air, and wins out in water – the results being the known temperature gradients and adiabatic lapse rates.
I think this dual action – gravity versus convection – is the reason for the high wind speeds on Jupiter etc. as well as the heat at depth.
Now apply this to the Sun 🙂
br1, thank you. I see I will have to try and follow your presentation. I’ll have a quick peek but give me time – as right now I would rather get on with the second writeup! Personally I have a strong feeling that something must have been omitted from the factors you use in your presentation – likewise with Willard Gibbs. I trust we are in agreement that experimental data trumps model “data”…?
TB,
Interesting discussion.
We are programmed to think a certain way due to the laws and theories that were imposed on us by others as THEIR concepts of gravity, relativity, etc.
Velocity is very, very interesting…
We are mostly made of water which is a different density of our atmosphere.
When we are laying down on our backs, from our nose to our backs are in a slightly different velocity. Stay there too long(dead) blood(mostly water) pools to the lowest area.
When you stand up, From your toes to the top of your head is in many velocities of the atmosphere.
We currently measure pressure incorrectly by what effects it has on a different density such as water.
Getting into centrifugal force or lack of changes on this orb as at the 48 degree latude is where water flow differently.
Now, is that by pressure differences or the theory that our dense core is trying to suck us down and why are we NOT attracted to denser materials like rocks?
br1: That my simulation doesn’t have this ‘third effect’ is not surprising, and I don’t take it as disproof of Graeff’s results, but I do take it as proof against the idea that gravity must produce a temperature gradient.
Not so fast. 🙂
You haven’t addressed the issue of dealing with singleton particles as opposed to gaseous ensembles such as are present in the real atmosphere. I’m not sure your simulation is relevant to the problem.
Take the case where gravity acts only in the Z direction, then what effect does it have in the X direction?
It tends to change the X motion into a Z motion.
Lucy:
” Personally I have a strong feeling that something must have been omitted from the factors you use in your presentation – likewise with Willard Gibbs.”
Seems like it. I’ll have fun with adding ‘hidden’ degrees of freedom. At the moment, that gives me all sorts of gradients, either positive or negative, depending on how I implement it. The sim I’m writing at the moment conserves energy, but may break momentum laws – how does energy from a vibrational mode give a translation? I think I know what to try next…
“I trust we are in agreement that experimental data trumps model “data”…?”
sort of. The only problem here is that so long as the effect is energetically small, there is the possibility that energy leaked in somehow in a way that you didn’t expect.
So a replication would be great, as would a way to increase the energies involved.
tallbloke:
“You haven’t addressed the issue of dealing with singleton particles as opposed to gaseous ensembles such as are present in the real atmosphere. I’m not sure your simulation is relevant to the problem.”
well, there is always more to be included. But if you read my sentence with emphasis on the *must*, I think it stands:
“I do take it as proof against the idea that gravity *must* produce a temperature gradient”.
I certainly don’t consider it the case that I have ‘disproved’ Roderich’s results, or that there is *no* theory behind them.
The only problem here is that so long as the effect is energetically small, there is the possibility that energy leaked in somehow in a way that you didn’t expect.
Well, in that case you have a real problem trying to imagine how energy could leak from the positive gradient that surrounds the core of the experiment, to a negative gradient at the core. To me, THAT spells out no because of 2LoT!!
I see Occam’s Razor: the “leaking” effect of gravity produces the central surprise negative gradient.
Folks – I see a target rich environment since my last post. I’ll pick off a couple and spend some time with the rest.
Lucy 10:48am: ” The whole point of (Graeff’s) experiment is to impede convection”
br1 12:27pm: “As I see it, the Verkley paper has two main sections: 2a, No convection…2b, convection.”
No, Verkley paper theory 2a and 2b allow free convection as does Graeff exp.
Really need to grok the difference between forced convection and free convection & be specific. Suggest google them. Free convection will be operating in Graeff’s base experiment and in both Verkley 2a and 2b. Forced convection is not part of the experiment (thus impeded – by insulation). Forced convection is not considered in Verkley at all.
—————————————————–
Nick Stokes 11:48am: “The time constant comes directly from the heat equation which governs simple conduction – the mechanism that these experiments aspire to.”
Seems to me the time constant calculated on conduction only would result in way too long a time to LTE b/c air is such a good insulator. Vigorous mixing, free convection, turbulence, conduction, maybe more are also allowed to act in Graeff experiment and Verkley theory. I seem to remember reading (maybe B&A text) that LTE thermo theory can tell us precious little about time.
—————————————————-
br1 10:27am: “What is invoked is a phi1 emission velocity distribution (which is *not* a MB distribution!). The presence of gravity *has no effect* on this condition, it is basically a thermal continuity condition.”
Ok phi1 dist. is not M-B, it is something else. BUT as a particle approaches the wall from below its velocity will be reducing; as it leaves the wall still moving up the velocity will be reduced even more & the whole point of Graeff experiments and Verkley 2b is to show this velocity change must be “thermalized” to use your word. Gravity must affect the thermal continuity through the wall collision.
Since in the phi1 distribution the presence of gravity *has no effect*, the “thermalization” process near the wall in your simulation is a different constraint than Graeff experiments and Verkley 2b.
It occurs to me that the incident angle and reflection angle will be the same with an elastic wall & elastic particle in the presence of gravity, but the velocity (hence T) will always vary with a vary in height – except!! in phi1 velocity dist. where there is no gravity.
Exner is not happy with a simulation using phi1 dist. and gravity at the same time. Exner may also say I badly need an editor, ha.
br1: “I certainly don’t consider it the case that I have ‘disproved’ Roderich’s results, or that there is *no* theory behind them.”
Okay, I’ll relax. I still don’t think you’ve disproved Loschmidt either by the way. 🙂
Let’s see what Trick has to say later. I’ve got TSI fish to fry now.
Trick
I’m puzzled by your remarks about free and forced convection – and I did check by looking them up. Graeff’s experiment detailed in his linked paper has two tubes filled with water, one of which also has ultra-fine glass spherules, which impedes convection. The result is a substantial difference in temperature gradients between the two. The convection-impeded tube shows a temperature gradient of -0.05K/m whereas the free-convection tube shows a gradient of only -0.01K/m.
Do you regard “impedance” as a form of “forced convection”?? because if so, I would regard the use of “forced” here as misleading, as “forced” generally means fans, pumps, ie moving objects, not static blocks.
Take a simple case of parcel of air with two forces acting on it.
Gravity acting down, buoyancy force acting up.
If the forces are equal the parcel will ;
a) Remain at rest
b) Move up or down at constant speed
Case b is often mistakenly thought of as convection but it is not.
This situation is called the Neutral Atmosphere by Meteorologists.
If the the forces acting on the parcel of air are not balanced the parcel will accelerate in the direction on the greater force.
This is convection.
If a third external force acts on the parcel this gives rise to forced convection.
Bryan: Thank you,that’s simply and clearly stated. So in a neutral atmosphere, why would an air parcel move at all? I would guess the atmosphere is never really neutral, but the local forces which give an impetus to an air parcel are (by definition) not going to continue acting once the air parcel has moved away, and so it will ‘coast’ under its own momentum at a reducing velocity until friction stops it.
David Socrates
Please would you email me! Your emails are on my kaput computer 😦
Lucy at 4:47pm: “..ultra-fine glass spherules, which impedes convection.”
Sure, I suppose anything that impedes free convection would involve some sort of added force hence forced convection is invoked simply b/c convection is no longer free. Here the mass/viscous/et. al. effects of the spherules are added forces in any random system perturbation & the LTE results apparently change. I wrote Graeff’s base experiments have free convection to avoid confusion (make Exner happy) with any modified constraint experiments.
Bryan 5:10pm: “If the forces acting on the parcel of air are not balanced the parcel will accelerate in the direction of the greater force.”
Agreed. With no 3rd force there is no forced convection (so left w/free convection) and random perturbations might for a time unbalance the parcel’s weight & buoyant forces which eventually re-balance.
br1, in your three simulations you keep speaking of M-B distributions but your velocities are coming from “phi1” distributions in all cases, or that is the way I read it. Why? Have you ever just used the M-B distribution for the random velocities? Ffor the true distribution is M-B, not phi1, right?
Carrying that thought on, and this would be one big coincidence, if a comparison of MB to phi1 does show different results and a gradient, it could be a possibility that this is exactly the effect, the differential between MB & phi1, that Graeff may be showing in his results. (your phi1-instead-of-MB exactly cancelling the very effect we are speaking of here). That would be weird coinidence mathematically (but of huge importance).
That was just a thought.
Also it sure would help if you would expand on your equation sequences used in your simulations, especially the random picking of the velocities, for an attempt to duplicate your work leaves many subtle places where I, or someone else, might not parallel your exact algorithm or relationships. That is, are the velocities randomized linearly? Squared or root to pick even energy levels? Most of that detail is not in the words and left out of the explanation.
Also, you do realize that at each vertical level in your 4000m column, at ie. 1000m, 2000m, etc., the distribution at each level, horizontally, will be a MB distribution, there is no reason this would not be so. But the constant term in the MB dist. is M/2.R.T, or m/2.kB.T, and by you “re-scaling” your results, it seems to me, your are merely re-adjusting the ‘T’ terms as your very last adjustment to be equal again and then making that the point in your summary that the ‘T’s are all equal (with random variance). Maybe I am wrong there so could you please explain why this is not so.
I have part of this duplication in 3d coded, but this is going to take a while, getting a little tedious and memory intensive. (but my first assumption was to use MB as the random velocities and then I homed in on exactly what the “phi1” was, and that threw a new kink in my logic trying to parallel you)
David Socrates 5/31 9:02pm: “I personally will refrain from entering into this fascinating debate-from-theory, however attractive and addictive.”
Right on David. I am just not strong enough, ha.
Wayne 8:22pm: “…the true distribution is M-B, not phi1, right?”
No. Neither M-B nor phi1 have gravity turned on. Any simulation will be incomplete using either M-B or phi1. What is needed is M-B distribution formula modified for gravity effect. I wonder if anyone has ever done that? Someone here may know; it is good thesis material if not.
The other hard part of the simulation is how to model conduction? 0th law consistent? The experiment as well as Verkley 2b follow 0th Law.
The impossible part of Graeff’s experiment is keeping his furnace from being a forced convection if apparatus located in a house and if in the back yard, then impossible to keep the sun from doing the forced convection thing.
Nick Stokes is right to think about time as it is not obvious at what time the experiment can be called finished. If left long enough then the experiment will eventually reach room temperature if inside or if outside then ~standard atmosphere temperature profile.
Trick: “No. Neither M-B nor phi1 have gravity turned on.”
Trick, I do realize that. I was wanting clarification of the reason br1 was using phi1, even in the case of no gravity at all, instead of the distribution from simple M-B for the random initial velocities, so in the first moment at zero height the M-B is satisfied. That’s all. Maybe I’m overlooking something. Was it because the results matched zero gradient? I can’t seem to get it from the words in his simulations.
“What is needed is M-B distribution formula modified for gravity effect. I wonder if anyone has ever done that?”
That’s pretty much were I am trying to head, I’m already getting a feeling that my results and br1’s are going to diverge in the end when I go 3d but I do want to start parallel to his logic if possible.
I’ve been following Graeff’s work with interest for a long time. His results seem at least plausible to me.
In my experience, when one speaks of “Breaking the Second Law”, it immediately triggers a hostile response in a large percentage of people. That does not strike me as effective communications. The general threshold of proof for breaking the Second Law is much higher than that.
It seems more effective to me to describe his work in terms of demonstrating a case where the Thermodynamic Limit does not apply. Read the link. It very much applies to Graeff’s work.
That is less likely to run into resistance. Getting Graeff’s work to be generally accepted will suffice in and of itself.
Lucy: the glass beads do more than inhibit convection, they will also change the physical properties of the water by the hydrogen bonding interactions with the surface.
You may also get leaching of Na, essentially ion exchange, and this will order the water.molecules to a further degree. This chemistry is very subtle.
Hi wayne,
Glad you are proceeding with the simulation, I’ll help out where I can.
” in your three simulations you keep speaking of M-B distributions but your velocities are coming from “phi1″ distributions in all cases, or that is the way I read it. Why?”
The two are very closely related. I was somewhat sloppy in my text, as I sometimes interchanged the wording, but the relation between them, and what you get when use one or the other is discussed in: http://polymer.chph.ras.ru/asavin/teplopr/ttk98pre.pdf
The way I think of it is that if you consider a volume with gas particles in it (thermodynamic equilibrium, no gravity!), and count what the velocities of all the particles are (binning the velocities in finite steps, in other words, counting how many particles have velocity v1<v<v2, then stepping along the v1, v2 limits to cover all the velocities of interest), you get a 3D MB distribution. If you consider a volume of gas and bin just one component of velocity (v1<vx<v2), then you get a 1D MB distribution (which is just a Gaussian). BUT if you take an infinitely thin slice through the volume (as opposed to an actual volume), and count the normal component of velocity (v1<vz<v2, infinitely thin slice), then you get a phi1 distribution. This is slightly problematic mathematically, as the density of particles in an infinitely thin slice is zero, and yet we know that particles will pass through that slice, so that is circumvented by using the normal velocity applied to a thin volume to give the flux impinging on the wall. So phi1 is the distribution of velocities needed to regenerate a MB distribution in the neighbourhood of the wall (thermodynamic equilibrium, no gravity).
Now, what happens to the phi1 distribution at a wall if we turn on gravity?
Absolutely nothing!
Because it represents thermal continuity through an infinitely thin plane, and we know that the wall has a temperature and is in equilibrium with the gas neighbouring it, gravity makes no difference whatsoever AT THE WALL. I wish Trick could see this, as it would avoid his claims of confusion. What the distribution is *away* from the wall actually makes no difference, and can be modified how you like, the phi1 distribution at the wall will remain unchanged.
"Also it sure would help if you would expand on your equation sequences used in your simulations, especially the random picking of the velocities"
The phi1 distribution given in the paper is a Probability Density Function. To select random velocities in the distribution, first you need to integrate it to get the Cumulative Density Function (you can use Mathematica's "The Integrator" online) with limits from -infinity to v. As PDFs are normalised to unity area, the CDF ranges from 0 to 1. Now you can invent a random number between 0 and 1 (standard random number generator), set that as the CDF value, and then find out what the corresponding velocity is. If you re-arrange the CDF to solve for velocity you get the Inverse Cumulative Density Function. The velocities you pick will have the distribution of the PDF and you are done.
For a phi1 distribution, the ICDF is
v_phi1=sqrt((-2*kB*T/m)*log(1-rand))
For a 1D MB (=Gaussian), it is
v_phi2=sqrt(2*kB*T0/m)*erfinv(rand)
where erfinv is the inverse error function.
About rescaling, T is not scaled. If it was, then there would be no noise in it???
br1, think you got all of the gray areas i was having a bit of a problem seeing through. 🙂 (also it’s starting to feel like if I had taken a course in statistical td it probably wouldn’t have hurt either) Thanks.
Q Daniels
Good to see you here. I’ve tried to go very carefully with the wording. My title says “challenged” not “broken” and has a question mark following. I’ve tried to show awareness of people’s sticking-points, and this will be picked up in my next piece. Graeff himself uses quite provocative language sometimes but those who pay attention to what he says, realize his starting-point was complete surprise with the evidence in his own tests that would not go away, and that he quickly realized appeared to challenge at least a common interpretation of the Second Law.
MDGNN, yes indeed. Possibly significant near-surface effects as well as convection inhibition – to say nothing of the combined gravity column effect of water plus glass plus air (glass microbubbles I believe). Probably Graeff is aware of these factors but I shall ask. Still, the most important thing is not so much the size of the negative temperature gradient (which depends on the substance tested), as its very existence.
TB,
A great deal of discussion is convention on our planet surface, but…
Parameters are different in space with no pressure or surrounding gases.
Experiments with plastic bottles of one in the freezer and the other in the heat of the sun show gases change compression. Warmer gases vibrate in containment while colder gases contract in density.
Let us take how ice forms. It is NOT on top of each other(unless you have some precipitation involved). But it forms under each other in layering.
Each layer is also at a different velocity and different pressure keeping the freezing water to the surface with the lack of energy being absorbed by water. A great deal of air is trapped in water to keep the buoyancy and pressure is greater while velocity is less and heavier as the deeper you go.
Snowflakes form with freezing and rotation to give these interesting patterns of flakes but they do not form as a ball but on a flat plane.
br1,
I’ve read all your slides. There’s nothing in there that surprises me.
Particularly, in the Walton paper, if you have a spherically symmetric MB distribution at all points, you get an isothermal result.
Also of note, the Gravitational Lapse Rate does not occur at the Thermodynamic Limit, and I think (without checking the math carefully) that it wouldn’t occur at n=1, either.
Presumably the claimed effect will scale with length and cross-sectional area? For a much larger signal, why not just build a large thermopile array and then insulate it, rather than have to deal with other columns of difficult-to-control materials?
Such a solid-state device would be more mobile, allowing easier results arising from rotating the device relative to the gravitational vector and moving to different locations. Apologies if this has already been discussed, I’ve not caught all of this topic.
br1,
I’m working on your ‘why’.
There’s a subtle assumption in using the MB Distribution. I don’t think the assumption is correct.
The MB Distribution is a result of equipartition, which is fine. The assumption is that particles retain the velocities they had at the last randomization. This is correct for systems for which there is no acceleration between collisions. It is not correct for systems where particles accelerate between collisions. In effect, using the MB distribution assumes there is no gravity acting at the particle level.
It’s a very subtle assumption. I haven’t checked, but I believe it traces back to Maxwell.
Incidentally, that’s why I think your centrifuge test won’t work. In a centrifuge, the particles are not accelerated between collisions, at least until you’ve got a relativistic (frame-dragging) centrifuge.
Q. Daniels 6/3 10:36pm: “…you get an isothermal result.”
br1 slides 6/1 9:41am:
p. 6 “…gas will certainly be isothermal.”
p.7 “…the equilibrium condition, and have shown that this will be isothermal.”
No, Verkley 2b shows the correct non-isothermal LTE solution.
—————————————————————————–
Q. Daniels continues 11:58pm: “I’m working on your “why”. There’s a subtle assumption in using the MB Distribution. I don’t think the assumption is correct.”
Right, br1 is showing the classical thermo grandmaster isothermal solution circular logic. br1’s slides show two pages of Walton; on p.186 find “Thermal equilibrium demands equality of (10) and (11)…”
This “demand” is the classical assumption which leads to the barometric eqn. The “demand” assumption is required to let the integration of Walton p. 186 eqn. (11) proceed holding T constant so there should be no surprise finding classic result isothermal (T=constant) solution.
Why? The Verkley 2004 paper 2b theory progresses past the earlier classical work because no such assumption or “demand” need be made anymore. The paper’s integration (detail shown in B&A text) is general, employs the thermo Laws and is very elegant in showing ideal non-isothermal, isentropic solution result is closer to earth’s standard atmosphere profile than the classical isothermal result.
br1 should use generalized physics as in Verkley 2b if simulation seeks to result in the non-isothermal solution Graeff is striving to find experimentally & Lucy et. al. want to replicate – at least at some time constant before the apparatus reaches LTE with the surroundings.
Trick,
I don’t see any disagreement between the things we’re saying. At this point, my primary interest is in seeing a clear, acceptable presentation of the physics, as per my first comment in this thread. The hard part is getting past the word ‘acceptable’, which has nothing to do wth Physics.
I believe the following hypothetical statement is narrowly true, subject to the validity of the condition.
… if you have a spherically symmetric MB distribution at all points, you get an isothermal result.
I do not believe the statement to be applicable to a gas under gravity, for the reason I later stated.
Put differently, I accept that the math involved in the statement is correct, but the physical assumption is not, and therefore the math is irrelevant. The Stat. Mech. ‘proof’ of isothermal in fact proves nothing because it is based on a faulty premise.
Lucy May 31st 1.25 pm. Don’t get the wrong end of the stick. What I am saying is that Kirchhoff’s Law of Radiation does not apply at TOA because it is extreme non-equiibrium physics, convection changes to radiation as the gas thins.The [Met Office?] modellers with whom I had a run-in on BH have too little basic physics’ education to understand that there is a reason for the fine print in the text books.
What they do is assume Kichhoff’s Law applies so DOWN emissivity = UP emissivity thus injecting 240 W/m^2 DOWN [TB knows about this], and it has to be balanced by the back radiation cycling UP at the bottom of the atmosphere.
I wish these people had brought in a professional engineer to tell them they were wrong 25 years’ ago. As it is, they’re busily blasting anyone who has the temerity to point out that they have invented a perpetual motion machine, the cause of the imaginary predicted positive feedback.
My present model is a bit of a change in that i am deducing that Everything is wrong. It’s not the first time I have been an iconoclast! You don’t get rewarded for it.
Q. Daniels:
“I’ve read all your slides. There’s nothing in there that surprises me.”
they only contain standard results, so may not be surprising. However, these can be non-intuitive to some.
“The assumption is that particles retain the velocities they had at the last randomization.”
No, that assumption is not anywhere. In particular, for the gravitational throw simulations, F=-mg was acting at all times, and the particles slowed down, came to a stop, then reversed direction before accelerating back to the ground. Still the distribution was isothermal.
“In effect, using the MB distribution assumes there is no gravity acting at the particle level.”
For the particle throw simulation, the MB distributions were *outputs* of the simulation, and were not assumed.
“Incidentally, that’s why I think your centrifuge test won’t work.”
The centrifuge experiment has been done, see http://www.worldscinet.com/ijmpb/23/2322/S0217979209052893.html (though this is subscription. I have a copy). He derives and measures a temperature difference in a centrifuge of
DeltaT = (omega^2.r^2)/(2*Cp)
I disagree with his derivation (as does Graeff!), but it is definitely worth replicating.
Trick:
“No, Verkley 2b shows the correct non-isothermal LTE solution.”
Graeff’s experiment is for a static, insulated gas, in what is meant to be an constant temperature bath. Such an undisturbed setup should have maximum entropy. As Verkley showed, the maximum entropy result is isothermal (Verkley 2a).
“Right, br1 is showing the classical thermo grandmaster isothermal solution circular logic. br1’s slides show two pages of Walton; on p.186 find “Thermal equilibrium demands equality of (10) and (11)…” This “demand” is the classical assumption which leads to the barometric eqn.”
This demand is that in equilibrium the energy flowing up equals the energy flowing down. How on earth else can you define equilibrium???
“The “demand” assumption is required to let the integration of Walton p. 186 eqn. (11) proceed holding T constant so there should be no surprise finding classic result isothermal (T=constant) solution.”
T is not held constant! It is *found* that T is constant. In other words, the *output* of this model is the isothermal result, it is not a condition of the model.
“Why? The Verkley 2004 paper 2b theory progresses past the earlier classical work because no such assumption or “demand” need be made anymore. The paper’s integration (detail shown in B&A text) is general, employs the thermo Laws and is very elegant in showing ideal non-isothermal, isentropic solution result is closer to earth’s standard atmosphere profile than the classical isothermal result.”
Sure, which is explained by convection. Read another snippet of Verkley2b:
“convective turbulent motions are now taken into account, albeit implicitly. Their role is to mix the potential temperature field, to strive to homogenize it.”
This also leads to a result that is not the maximum entropy result, as he continually points out.
“br1 should use generalized physics as in Verkley 2b if simulation seeks to result in the non-isothermal solution Graeff is striving to find experimentally & Lucy et. al. want to replicate – at least at some time constant before the apparatus reaches LTE with the surroundings.”
If you are saying that Graeff’s results are only temporary, then they are no longer of interest. The whole point is that they *are* (meant to be) in equilibrium.
br1 11:58am – Fun discussion.
“Graeff’s experiment is for a static, insulated gas…”
Yes, I agree to my knowledge.
br1 continues: “…Verkley showed, the maximum entropy result is isothermal (Verkley 2a).”
No, an insulated gas column with work allowed to be done on the gas column above or below is NOT static in Verkley 2a and is shown isothermal. My understanding is that Graeff’s experiment is fully enclosed thus his static experiment allows no work done on the column of air above or below. Check Verkley Fig. 1 which part 2a terms “…a midair atmospheric column…”. Graeff’s experiment is not simulating this part 2a midair column to my knowledge so Exner would be unhappy comparing part 2a results to those of Graeff exp., i.e. different constraints.
In part 2b, Verkley subtly changes the constraints to those shown in the B&A text which are NOT a mid-air column. The B&A column IS static & the perfectly insulated air column constraints show no work allowed on the air column above and below which are the exact same as Graeff’s AFAIK. Unfortunately the Verkley et. al. paper is not very clear on this, you have to get a copy of B&A to see their figures & read about their constraints. The static, perfectly insulated 2b result changes to non-isothermal, isentropic & Exner is happy comparing Verkley 2b and Graeff.
br1 continues: “This demand is that in equilibrium the energy flowing up equals the energy flowing down. How on earth else can you define equilibrium???”
This is energy conservation & is not the demand I am pointing out. The demand in Walton eqn. (11) is that the temperature be assumed constant over the integrated pressure&height of the column to let the math dept. take T outside the integral as a constant. Look and you will see.
br1 continuing: “T is not held constant! It is *found* that T is constant.”
No & no. Look close. This Verkley 2004 points out explicitly in 2c (2a is “former”, 2b “latter”):
“The principal difference between the two is as follows: in the former, the vertically integrated absolute temperature is kept constant (classical thermodynamic approach); in the latter, the vertically integrated potential temperature is kept constant (Bohren and Albrecht 1998).”
So, classically, the absolute temperature is held & found constant (isothermal) just because this assumption is “demanded” to perform the Walton (11) math integral. The harder math integration (T assumed not constant during the integration but energy/enthalpy is conserved) & entropy maximization is performed in the B&A text.
br1: “If you are saying that Graeff’s results are only temporary, then they are no longer of interest.”
Yes & no. Yes, I am saying Graeff’s results ARE temporary. Only because there is no perfect insulation; the apparatus WILL reach the surrounding temperature eventually. But as Nick Stokes points out it IS interesting what experimentally happens for a short enough time constant.
Trick:
“Graeff’s experiment is not simulating this part 2a midair column to my knowledge”
Incorrect! (Isn’t this fun 🙂 )
The thermocouples in Graeff’s container are slightly away from the top and bottom surfaces of the container (he showed me a typical insert). Hence they are sampling the temperature of a mid-air column, and Verkley 2a applies. That the container is closed is irrelevant, because the variational constraints apply at the thermocouples. That the container may exchange heat with the surroundings (even though it is intended to be insulated) may not meet the constraint of 2a, and this may yet be a flaw in both the experiment and the analysis (in comparison to *both* 2a and 2b).
“The B&A column IS static & the perfectly insulated air column constraints show no work allowed on the air column above and below which are the exact same as Graeff’s AFAIK.”
Verkley says repeatedly in 2b that the air column has convection, and this convection is needed to redistribute theta and make L constant, Eqn(13). He even uses the words ‘turbulent’ and ‘heated’ several times, hence this column isn’t static in the thermodynamic sense. If you are saying the constraints of 2b contradict Fig 1, and that the column is now pinned at heights z1 and z2 (which seems to me unlikely), even that makes no difference, as these are only tools used in a variational analysis. That Verkley Eqn(13) is a constant is not determined by putting a lid on the column!
I admit I haven’t read B&A, not sure I’ll get the chance, so I may not be able to form my own opinion about what they have to say. However, I’ll have a quick check to see what I can find. There may indeed be some subtleties I am missing.
“This is energy conservation & is not the demand I am pointing out.”
OK, sorry.
“The demand in Walton eqn. (11) is that the temperature be assumed constant over the integrated pressure&height of the column to let the math dept. take T outside the integral as a constant. Look and you will see.”
Err, I looked and no, absolutely not!
To see why, start with Walton Eqn(10). The integral is over *velocity in region 2* which has temperature T2. The fact that T2 can be taken outside the integral in this case only means that the region *has* a temperature, which of course doesn’t change with velocity, and doesn’t change with time as it is assumed to be in equilibrium.
Then go to Eqn(11). This describes the energy of the molecules thrown up from region 1, which has an unknown temperature T1. Please note that like Eqn(10) the integral is again over the *velocities of the molecules in region 1*, it is not over height! T1 is the temperature of region 1 (which of course is not changing as it is in equilibrium), and only some molecules leaving region 1 can reach region 2, due to the potential energy needed to get there. Only molecules with velocities greater than sqrt(2gh) can get there, and when they get there they have lost mgh potential energy, so they carry (0.5mv^2-mgh) energy each. The T1 in the integral is that of region one, because that is the sample of molecules they are coming from! T1 is of course independent of the velocities of the particles in region 1, it is only saying that they *have* a (unknown as yet) temperature, and it says nothing about what temperature they will have when they reach region 2. The assumption is that T1=/=T2, otherwise he could have just used T and be done with it.
Then he invokes equilibrium energy flux, and *finds* T1=T2.
“Only because there is no perfect insulation; the apparatus WILL reach the surrounding temperature eventually.”
Well, this contradicts Graeff’s claims, and is what makes the claim interesting. He has run the setup for well over 6 months without disturbance, and found that it comes to equilibrium away from ambient. This is quite a controversial claim!
p.s. apologies for all the emphasis and exclamation marks, but as we seem to be able to read the same text and come to opposite conclusions, I feel that I must do something to distinguish the important points.
br1, Trick,
Please have some patience with my poor communications skills. Sorting through what I’m trying to say usually takes a fair amount of effort. My thoughts on this subject aren’t even ready for my tech writers.
I’m trying to work backwards on this thing. I believe there are fundamental errors in the derivation of the isothermal profile.
Particularly, in the Walton paper, if you have a spherically symmetric MB distribution at all points, you get an isothermal result.
This comment applies to the Walton paper, as well as other stat. mech derivations of isothermal. It does not directly apply to you molecule throws.
In using calculus, there is a fundamental assumption (obscured by the calculus notation) that continuous functions are close enough to the behavior of discontinuous functions (atomic gases), and that errors do not scale with increasing values of n. This is only exactly correct at n=1 (like your throws) and n=∞ (the Thermodynamic Limit). At n=1, there is no discontinuity beecause there is no separation between quanta. At the Thermodynamic Limit, there is no discontinuity because there are no particles to be discontinuous (quantized).
Put differently, attempting to apply the Thermodynamic Limit with gravity means you either have a non-zero particle mass and a Black Hole, or zero particle mass and therefore a non-atomic (non-quantized) gas. Do you see a different physical explanation?
I suppose what I really should do is take one of the interim equations in either Verkley 2a or Watson, and replace the Calculus notation with Limit notation.
br1 5:58pm says:
“To see why, start with Walton Eqn(10). The integral is over *velocity in region 2* which has temperature T2. The fact that T2 can be taken outside the integral in this case only means that the region *has* a temperature, which of course doesn’t change with velocity, and doesn’t change with time as it is assumed to be in equilibrium.”
Temp. doesn’t change with velocity??? I’m gobsmacked. Hmmmm…well let’s see…
Ok start w/eqn. (10). Exact Walton clip: “..region 2 so that the net transfer of energy upward through P for molecules of all velocities will be…(insert Eqn. 11 integrated over dw as if T(w)=const. in the region)”
Emphasis on “all velocities”.
The mean of a parcel of particles KE (i.e. velocities as particle m = const.) anywhere in region 2 is defined as the temperature of that parcel. Walton writes region 2 has particles of velocity between w and (w+dw). Hence any parcel of these particles in region 2 will have temperature T varying (with vel. thus with KE). In this manner T is really a function T(w) and cannot come outside as a constant when Walton performs the integration over dw. N’est ce pas?
UNLESS if a demand of equilibrium is made to just assume T constant everywhere within region 2 then of course the result is isothermal; that demand just makes it so. The integration by Walton of eqn. (7) and eqn. (10,11) proceeds when this demand is made. Result T1=T2 isothermal by demand.
I predict you would enjoy the B&A 1998 text. Worth an effort to get it. Try your local librarian; some univ. library in your reasonable area might have it & deliver for free interlibrary exchange. I’ve done that before & found librarians live for that kind of request challenge, ha.
Enjoy reading thru the 12 or so detailed & discussed steps to get the general non-isothermal answer where no demand of equilibrium need be made for the more general integration of eqn. (7,10,11) – only demand is that no energies/enthalpys be harmed on the way to max. entropy. I liked the B&A text so much, read pretty much the whole thing.
Geez, wish B&A was the text of my thermo courses instead of the one written by the prof., lol.
Hat tip to TB for all the bandwidth on this arcana…
Trick -“Only because there is no perfect insulation; the apparatus WILL reach the surrounding temperature eventually.”
br1 responds – “Well, this contradicts Graeff’s claims, and is what makes the claim interesting. He has run the setup for well over 6 months without disturbance, and found that it comes to equilibrium away from ambient. This is quite a controversial claim!”
Nothing can come to LTE away from ambient absent energy input & I haven’t seen that in any claims, will take more than 6 months then in this case. There is no perfect insulation. Only in theory. Theory can have a system with constant entropy (no friction, adiabatic, inelastic, all that stuff) whereas all practical systems have to increase entropy to LTE.
Graeff’s experiment is still interesting transiently on the long road to LTE as he strives toward better measurement and insulation. But in the long run, his T will be at ambient absent any trickle of energy input. What guess do you make for the curves posted by tchannon 5/31 10:27pm? Looks like something trends down for couple weeks then back up over several months.
————————————————————–
br1 5:58pm – “The thermocouples in Graeff’s container are slightly away from the top and bottom surfaces of the container (he showed me a typical insert). Hence they are sampling the temperature of a mid-air column, and Verkley 2a applies.”
(Wiping the spit scotch off my monitor..) no, no, no – this just moves the thermocouples up off the bottom it doesn’t change the system at all. Geez. Graeff’s apparatus has a bottom & top. Thus 2b. Verkley 2a midair column has no bottom or top for work pass thru, and that makes all the difference.
Why? I am not sure Verkley even noticed the difference in constraints that B&A applies for part 2b (NOT a midair column) – b/c they do not discuss it. My guess is that the 2a column has no top or bottom so the air columns above and below can do just the right amount of work on the particles btw Z1, Z2 so that they slow to isothermal below half way and speed up to isothermal above halfway. Magic applies in theory but that is what the theory seems to say. I dunno really, maybe someone else has a better comment.
—————————————–
Q. Daniels 10:19pm – Oh man, need to visit the math dept.
Trick says: June 5, 2012 at 12:27 am
…What guess do you make for the curves posted by tchannon 5/31 10:27pm? Looks like something trends down for couple weeks then back up over several months.
This curve is the jewel in the crown, but Graeff has labelled it really badly and I’ve just done an improved version to show its riches in part 2. That sheaf of parallel lines trending down then up is the absolute temperature measured by thermistors and scaled on the right hand side of the graph, in Graeff’s paper on the water column (ref at top). The sawtooth lines are the temperature gradients, measured directly as such by thermocouples.
I’m generally staying off further comments now while doing part 2.
The 3D symmetrical distribution implicit in MB despite gravity only acting in vertical direction is causing most of the disagreement.
Why not sidestep it?
Start with simple kinetic theory of gases computer simulation without gravity with each molecule given the same initial kinetic energy .
Elastic collisions of molecules with ‘walls’ and each other.
Each molecule(particle) will move away from their initial velocity value by collision.
No heat loss or gain through walls of column.
A random distribution occurs with similar volumes containing about the same number of molecules if the total molecular count is high.
Now add gravity.
An ordered pressure and density gradient will quickly form from the no gradient start.
If Graeff is correct then over a much longer time a temperature gradient will also form.
Interesting mindset to laboratory testing and actual parameters.
Gravity is still a guess but many parameters were never considered or included.
What is being missed by all sides currently is the volume differences of the atmosphere to an actual container.
Because of our planets shape, the container would have to be pie wedge shape in order to contain the same volume as to what laboratories use. The velocity difference from on our planet surface to the outer atmosphere has much less gases due to compression differences and volume space.
br1, I said I would let you know my results… and I’ll be darn, you seem to have converted me and have been right all along. My calculations (a spreadsheet model) always put the molecular distributions, one at ground level, the other at any given altitude (within limits), right over the top of each other. It doesn’t matter if it’s one, two, or three dimensions, any atm. molar mass, initial temperature, or if any x-y tangent velocities is included, or not, always the same, plotted dead over the top of each other.
Seems you have convinced me of something I have been searching for a quit a while now! Now I really wonder why Dr. Graeff is getting the results he reports, for now it doesn’t seem to be gravity… and I was so sure it the potential energy’s influence.
TB, want a copy of that sheet? It would behoove you to take a look. It’s really simple, just eight columns and a graph. Maybe write a post with it so other can critique or learn from to for so many have stumbled on this topic for at least six month’s now. (or maybe someone can point me where this is wrong, for I can’t find it)
Hi Wayne, Trick, br1 Lucy and everyone. I’ve been away for a few days and haven’t caught up with this excellent thread yet. Did anyone address my issue concerning the difference between considering single particles and ensembles of particles in gases?
Loschmidt originally considered a constant volume process, but Andreas Trupp thought it should have been cast as a constant pressure problem.
NASA says Cp = Cv + R (the gas constant)
http://www.grc.nasa.gov/WWW/k-12/airplane/specheat.html
They say on that page:
“Thermodynamics is a branch of physics which deals with the energy and work of a system. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiments.”
In other words, when you try to apply thermodynamics theory to individual molecules or ‘particles’, you run into trouble.
Gravity acting on atmospheric mass creates a pressure gradient which in turn creates a density gradient. The number of particles in a given volume will affect the heat capacity of the ensemble. This is not considered in br1’s analysis so far as I can tell.
Now heat capacity doesn’t of itself affect ‘instantaneous’ temperature necessarily, but it sure makes a big difference to the amount of heat that a given air parcel can transfer to anything else. SO I think we need to be considering not only the particles, but the size of the spaces in between them.
tallbloke 2:15pm: “ …try to apply thermodynamics theory to individual molecules or ‘particles’, you run into trouble…”
Yeah. Big time. I would suggest troposphere gas pressures/temp.s are the ones of reasonable interest (~80% of earth’s atmosphere). Apparently br1 is simulating and Graeff is experimenting to learn more about troposphere gas physics.
So reasonably about 1000 hPa to 250 hPa and T=315K to 225K.
Fig. 2 graph in Verkley shows the range quite nicely. Lower than p ~225 hPa and other high altitude physics become important to consider/discuss since the standard dots fall off the standard & theory curves below that pressure & above that altitude.
Don’t know whether this contributes to the discussion, but here’s what two acknowledged experts in the field – Michael Flanders and Donald Swann – think about the first and second laws.
😉
Trick:
“Temp. doesn’t change with velocity??? I’m gobsmacked.”
Sure it doesn’t.
Consider a box of gas, equilibrium, no gravity. The velocities of molecules are described by a distribution with a characteristic temperature. The SINGLE value of temperature describes the *distribution* of velocities (note the plural ‘velocities’). Hence an integral over those velocities must use a single value of temperature.
“UNLESS if a demand of equilibrium is made to just assume T constant everywhere within region 2”
Yes, that demand is made.
If region 2 (which is a thin slice at a fixed altitude) has a temperature, then T2 is the temperature of the slice (not really a difficult concept!). Anything else is implying that region 2 is not in equilibrium. This says nothing of the temperature of region 1, or whether the two regions have the same temperature as eachother.
“then of course the result is isothermal; that demand just makes it so.”
No! That region 2 *has* a temperature says nothing of the temperature of region 1. Indeed, the assumption is that region 2 has a different temperature than region 1.
“Nothing can come to LTE away from ambient absent energy input & I haven’t seen that in any claims”
Yet this is Graeff’s claim. Once Lucy posts the relabelled version of the figure, you will see that there is an aluminium shell in one of the layers which has a temperature gradient very close to zero. This is because aluminium is a good thermal conductor. Within that shell, and within some more insulation is the central chamber where all the fun stuff happens. It has a temperature gradient which is persistent in time, and which re-establishes itself when the setup is inverted so that it is always cold on top and warm on bottom. The hypothesis is that this gradient is due to gravity’s influence on particle kinetic energy.
Furthermore, the external environment (the basement) has a gradient which is warm on top and cold on bottom. So the equilibrium internal temperature gradient is the opposite of that of the environment. What do you make of such a claim?
“(Wiping the spit scotch off my monitor..) no, no, no – this just moves the thermocouples up off the bottom it doesn’t change the system at all. Geez.”
Yes, I agree!!! Glad we can agree on something 😉
Pardon me for putting words into your mouth, but you seem to be of the opinion that Verkley2a is for an open column, while 2b is for a closed column.
What Verkley says is in the abstract
“If one assumes that there is no net heat exchange between the column and its surroundings—implying that the vertical integral of the absolute temperature remains constant—an isothermal profile is obtained in accordance with classical thermodynamics and the kinetic theory of gases. If instead the vertical integral of the potential temperature is kept fixed—as argued by several authors to be appropriate in the case of convective mixing—an isentropic profile results.”
What happens at the edges of the column in Fig 1 is only a mathematical tool for *calculating* what he states in the abstract. The true constraints on the system are as restated in the text:
Verkley2a: “The classical variational problem for a midair atmospheric column is to maximize S, the mass M and the enthalpy H being kept fixed.”
Verkley2b: “maximize S for fixed M and L”
Whether the system is open-ended or closed-ended is irrelevant. Maybe one case or the other is handier for calculational purposes, but it doesn’t affect the physics at all.
To see what is appropriate to Graeff, one has to ask: “Does he have fixed M and fixed H?” in which case 2a applies, or “Does he have fixed M and fixed L?” in which case 2b applies. I presume we can agree that he has fixed M, but does he have fixed H or fixed L? Or neither? Do you think that fixed L is achieved by putting a lid on the container???
You are right to be concerned about insulation, but we still need to answer why Graeff gets a gradient which is cold on top and warm on bottom.
What B&A say, I don’t know.
Lucy:
“I’m generally staying off further comments now while doing part 2.”
Wise choice!
Personally, I am all in favour of replication.
Note that for the centrifuge measurement by Liao, he gets a 2 K temperature difference from hub to outer radius. Depending on how many people you get on board (and count me in), one should also consider replicating this (though feel free to stick with one implementation for better focus).
wayne:
“Now I really wonder why Dr. Graeff is getting the results he reports”
which is where I’m at as well. That gravity is involved seems clear, due to his setup inversion experiments, but what is it affecting?
Be wary of a centrifuge result, reason: it needs a highly specialised device more suitable for nuclear isotope enhancement, if without the extreme speed. The drum/rotor has to operate in a very hard vacuum and above isotope requirements would need other measures to reduce thermal disturbance to near zero. As a reminder, until the 1930s no-one could get a result in a centrifuge, then someone twigged the thermal problem.
I reckon the answer already exists from that field but published data is I assume unobtanium.
br1 5:19pm: “Consider a box of gas, equilibrium, no gravity.”
NO! I refuse. Unless Graeff moves the subject of this thread to space.
Rest of my answers will point to “0 g”. Hint why: Of course, that box above is isothermal, T=const. at ideal LTE. No debate. Walton region 2, classical guys & Verkley/B&A and Graeff all have g turned on though and that makes all the difference. Velocities vary, parcel T varies w/vert. velocity component , non-isothermal. Can’t do the Walton integration straightforward.
The Verkley 2a & 2b constraints do matter b/c 2a and 2b get different results at max. entropy! Why? Constraints ARE different & matter, see the Exner comment. Too, 2b constraints get a profile much closer to real standard atmosphere.
Cold on top and warm on bottom will be transient on the way to real non-ideal LTE. Can’t wait for Lucy’s part 2 post.
Is my understanding correct:
There is a cylinder of material. This is filled with porous material and that in turn is filled with a non-solid.
The core of the cylinder is in good thermal contact with the body of the cylinder via the porous material. The body of the cylinder has the same temperature profile as the core.
The supposition is the cylinder will have a gradient.
br1 says: June 5, 2012 at 5:27 pm
Note that for the centrifuge measurement by Liao, he gets a 2 K temperature difference from hub to outer radius. Depending on how many people you get on board (and count me in), one should also consider replicating this (though feel free to stick with one implementation for better focus).
Look up the Hilsch vortex tube. Here we see what I think is same or related effect for real, big time. TimC is right about centrifuge problems.
tchannon says: June 5, 2012 at 8:17 pm
will answer you in next article.
I’ve another wrench to throw in the works.
Let’s suppose for a moment (as I do) that Graeff is correct, and the Gravitational Lapse Rate is faster than the Dry Adiabatic Lapse Rate. What are the implications?
1) Any equilibrium would be at a balance between the GLR and DALR.
2) The tendency towards the DALR would be through convection.
3) Convection is a non-linear dynamic process. Another name for this is “Fluid Dynamics”.
4) Therefore there is no LTE.
5) Therefore there is no general solution for a gas in gravity.
That would go a long way towards explaining the difficulties in deriving such a solution.
Alternatively, the contrapositive form of 4 is:
Assuming the existence of LTE also assumes the GLR is <= to the DALR.
As I explained above, the ALR dominates because it’s based on a virtual work argument and absolute. However, the GLR, if it exists, will ensure there can only rarely be an unstable equilibrium wrt ALR. This is temperature inversions.which counter the effects of the GLR thus ensuring that unstable equilibrium.
Trick:
“Walton region 2, classical guys & Verkley/B&A and Graeff all have g turned on though and that makes all the difference. Velocities vary, parcel T varies w/vert. velocity component , non-isothermal. Can’t do the Walton integration straightforward.”
In one sense I sympathise with this view. Walton has assumed that region 1 velocities are described by a MB distribution with T1, and that region 2 velocities are described by a MB distribution with T2. Hence there is no assumption of being isothermal, but there is a challengeable assumption that the MB distribution correctly describes the velocity distribution under gravity at each height.
That this assumption is in fact correct is not detailed in the paper. Fortunately (for Walton) it is correct, as my and wayne’s simulations show, but I agree it could do with an extra line or two. I’ll see what I can manage 🙂
“The Verkley 2a & 2b constraints do matter b/c 2a and 2b get different results at max. entropy! Why? Constraints ARE different & matter”
Sure the constraints matter, but I think we disagree on that the constraints actually are. Verkley states that 2a is under the constraints of constant M and constant H, while 2b is under the constraints of constant M and constant L. To look at this closer, please refer to Verkley Table 1 where he calculates H, L and S for the different constraints.
The first line in the table reads:
Observed, H=2.0032, L=2.3150, S=4.3764
He has calculated M already, so he keeps that for all his calculations. To apply the constraints of 2a, he sets H equal to the observed H, allows L to be free, and maximises S. He finds:
Isothermal, H=2.0032 (the same as Observed, because it is constrained to be!), L=wrong, S=way too high.
Next he applies the constraints of 2b, which is that L is fixed. He then allows H to be free, and maximises S. He gets:
Isentropic, H=wrong, L=2.3150 (the same as Observed, because it is constrained to be!), S=a bit too high
As he still doesn’t get the right answer, he does the whole point of the paper, which is to combine Iosthermal and Isentropic while maximising S to get
Intermediate, H=2.0032 (fitted to data!), L=2.3150 (fitted to data!), S=just right
The constraints are what he is doing with H and L, not what he is doing with Figure 1, which is just a calculational tool!
Or do you really think the atmosphere is better described by a closed container, rather than a free column???
“Cold on top and warm on bottom will be transient on the way to real non-ideal LTE.”
So you don’t believe in a temperature gradient due to gravity, or how am I to interpret this?
tchannon:
” As a reminder, until the 1930s no-one could get a result in a centrifuge, then someone twigged the thermal problem.”
Could you expand on this a bit please? What exactly is the thermal problem you refer to?
“The core of the cylinder is in good thermal contact with the body of the cylinder via the porous material. The body of the cylinder has the same temperature profile as the core. The supposition is the cylinder will have a gradient.”
Graeff finds a negative gradient (but a smaller one) on the outside of the cylinder. As you step through the layers of insulation, one finds the largest negative gradient in the middle, then lesser gradients until one reaches the aluminium shell(s) which are close to isothermal in height, then positive gradients further out until one reaches the room temperature gradient at the outside of all the layers. These gradients seem to be well ordered, so again unlikely to be just some random data error.
br1 11:43am: “..2a is under the constraints of constant M and constant H…”
Yeah. Finally found a poster can see the inner details of Verkley but unfortunately can’t see B&A. No time right now but may work on a more hearty post next coupla’ days. While waiting for Lucy…hint, hint.
In the meantime, think about what const. H means. This enthalpy H is the sum of the system’s internal energy plus (the pressure at the boundary of the system and its environment times the volume).
Here in 2a delta H = 0 = delta (that sum right above). The sum can’t change but each component of it can change equally +/- with the other.
The pressure can be transmitted across the open ends to the column above and below in 2a but not in 2b. Very interesting.
Then think about PV = nRT. T=const (isothermal) in 2a as well as V and n are constant. Then how in the world does P become const. in the face of gravity? The resolution is work can cross the top & bottom in 2a. This is unnecessary in 2b so to me, 2b is the more elegant solution. May take reading B&A to come over to this side of the force.
Back later.
br1,
Averaging an orb is currently science’s mother or all screw-ups.
This then ASSUMES every point of our planet MUST follow the exact same parameters even though many factors are occurring at the same instance.
That single calculation MUST be enhanced a million fold for EVERY single molecule and point on this planet.
Laboratory science and actual parameters of our planet are very much two different areas of fiction and science fiction.
Water is far denser to the atmosphere and timeframes of energy has to be much different in rotation and velocities.
Remember, we ourselves are denser in material with mobility in our atmosphere so naturally we would be heavier to the atmospheric gases no matter which point you are on this planet. Gases are denser and heavier on our planet surface and get less so as you go higher.
Tallbloke, you are missing a key idea when you say “It is perfectly possible for a temperature gradient to exist through contacting systems in thermal equilibrium, because the bottom of the upper system will be at the same temperature as the top of the lower system where they touch each other.”
The zeroth laws tells us that
* if A is in thermal equilibrium with B, and
* if B is in thermal equilibrium with C, and
* if C is in thermal equilibrium with D,
then A and D must be in thermal equilibrium.
Let “B” be the bottom of one column of gas (and “A” is the top). Let “C” be the bottom of the other column filled with a different gas. Let me try some crude ASCII art.
______
|.A.||.D.|
|….||…..|
|….||…..|
|….||…..|
|.B.=.C.|
———-
Each column is insulated, except at the bottom of the columns, where they are in contact with each other. By putting these two in contact (at the same altitude of course), they will eventually come into equilibrium — ie T_B = T_C. It is well known, however, that different gases have different lapse rates — ie at the top we have T_A ≠ T_D if a gradient equal to the lapse rate has been created.
Now, you could try to argue that A doesn’t have to have the same temperature as B when in thermal equilibrium (ie that a lapse rate is in fact the condition for thermal equilibrium) because of the difference in altitude. The same goes for C & D.
But now we have reached a contraction with the zeroeth law. You claim
* A is in thermal equilibrium with B, and
* B is in thermal equilibrium with C, and
* C is in thermal equilibrium with D,
* A is NOT in thermal equilibrium with D, (ie they are at the same altitude, but have different temperatures).
So yes, there is a very clear contradiction with the zeroth law. We are left with three possibilities:
1) my logic is flawed
2) your interpretation is flawed.
3) the zeroeth law is in fact broken in this situation.
I think (2) is the source of the problem. You are welcome to argue why either (1) or (3) is instead the problem.
[Reply] Thanks, I’ll let you know why 1) is the answer later, must dash for my train. 😉
The second law of thermodynamics is trivial – simply define the boundary of the system and account for all energy crossing the boundary.
There are no free lunches.
Of course, if you learned your thermodynamics from wikipedia or from climatology you’re screwed – the Boltzmann model was buried a long time ago along with Boltzmann.
Since there are no statistical mechanics calculations, where is the equation of state?
Even the ideal gas law is multiple variable problem, i.e., you need both a pressure and a temperature model.
When reality breaks, change reality to fit the model.
“t’s great to know Tim F has more important things to work on than the laws of thermodynamics, we can be expecting some truly mould breaking science from him. ” 🙂
Ah, but rewards have to be balanced with the odds of success. On a scale of 0 – 1, I would estimate…
* importance of overturning laws of thermodynamics: 0.95
* odds that this discussion and my contributions will lead to that goal: 0.000000001
* relative importance: 0.000000095
There are plenty of things in my life that rank higher. 😉
Right then Tim F:
First of all, the zeroth law doesn’t say what you say it does. It talks about three systems not four. So your gedanken experiment is interesting, but the zeroth law doesn’t rest on it.
According to wikipedia:
“The zeroth law states that systems are said to be in thermal equilibrium if they are able to transfer heat between each other (for example by conduction or radiation) but do not do so. If two systems are in thermal equilibrium with a third system, they are also in thermal equilibrium with each other. ”
Now, right there we have a contradiction. If there is a gradient running through three systems in sequential thermal contact they will not transfer heat. The average temperature of the three systems will all be different and so the outermost systems would not be in thermal equilibrium if they have heat capacity and were suddenly brought into contact. However, if the outermost systems are given time to equilibriate with the gradient before they come into contact then they will indeed be in thermal equilibrium.
More when I get home off the train.
“First of all, the zeroth law doesn’t say what you say it does. ”
Yes, it does.Think!
As you quoted
We just showed “A” & “C” are in thermal equilibrium. We already agreed that “C” and “D” at different altitudes in the same column are in thermal equilibrium. (NOTE: These are NOT sequentially arranged. A is above B; B is next to C, D is above C (and D is next to A) )
A D
| |
B=C
So we can conclude
So “A” & “D” must be in thermal equilibrium.
Elementary logic!
**********************************************************************************************
You already agreed that “The zeroth law states that systems are said to be in thermal equilibrium if they are able to transfer heat between each other (for example by conduction or radiation) but do not do so.” So you are saying that if A & D are brought into thermal contact (“they are able to transfer heat between each other”) AND they purportedly have different temperatures, yet they will not transfer energy (“but do not do so”). Remember, A & D are different gases at the same altitude. They are different temperatures, yet they must not be able to exchange heat to come to the same temperature because they are in equilibrium already!
Doesn’t it make your head hurt trying to believe that??
br1 June 6, 2012 at 11:58 am,
Gleaned this awhile ago. My expertise is elsewhere as might eventually become apparent.
Context is isotope enrichment using artificial gravity, an obvious idea first tried, 1920s, perhaps earlier (don’t recall). No-one could get it to work. Someone had a bright idea early 1930s that the problem was thermal, the heat exchange with the air outside of the container, so they put the container in a vacuum, and got results.
Mechanically this was disastrous because the drive and a gas flow (for enrichment is necessary) has to pass through a stationary vacuum chamber, horrendous seal problems. I don’t know how hard a vacuum but to stop conduction and convention needs maybe 5 torr or preferably better, making matters even worse because there are few materials suitable for high vacuum, and sliding seals?
I recall coming across a very weird thing about vacuum based insulation, apparently it also needs narrow gaps, counter intuitive, something about free path length. Also it is only under high vacuum conditions that it is worth worrying about radiation, why for all the faffing with shiny surfaces, nothing much changes. Seems to be an awful lot of misunderstandings and nonesense out there, which also is stated by some.
Hi Tim F,
In your first comment (since you ran off to evade your last beating 😉 ), you said:
“The zeroth laws tells us that
* if A is in thermal equilibrium with B, and
* if B is in thermal equilibrium with C, and
* if C is in thermal equilibrium with D,
then A and D must be in thermal equilibrium.”
Where in any of that is there anything about altitude? Where is the gravitational field?
Your gedanken experiment fails because A and D, when brought into contact will produce a flow of heat which will deplete the heat content of one column or the other until the thermal equilibrium is restored. At which point energy will on average be evenly distributed (although there would still be a temperature gradient). The same setup was proposed by Peter Berenyi, Robert Brown, and several others trying to inventorize a perpetual motion machine of the second kind from Jelbring’s thesis. They all failed.
tallbloke 5:29pm – “First of all, the zeroth law doesn’t say what you say it does. It talks about three systems…”
Yes. And I would like to point out the air column theory and Graeff’s experiment is just ONE system so strictly the 3 system zeroth Law is not applicable. Sure, some have said just subdivide the one system into 2 or more, but then you simply have 2 * 1 ideal gas systems so both will equilibrate at the contact.
I think that is what got the classical guys in trouble, they invoked the 0th Law to assume then find equilibrium in T (isothermal) but that is not truly how the 0th law was justified by the grandmasters – they formulated it with 3 systems not 1 system.
Exact Verkley 2b theory shows the 1 system completely enclosed, perfectly insulated ideal gas in the presence of gravity to become non-isothermal at LTE max. entropy point.
tchannon says, May 31, 2012 at 10:27 pm: Overlay of recreated data plot and figure 2 from PDF
Hi Tim. First of all, belated congratulations for brilliantly reverse engineering the Graeff data and so quickly. I am intrigued to know how the heck you did it!
Second I think there is a more fundamental theoretical problem with using a centrifuge than the bearing issue you mention. A centrifuge does not really emulate the situation in the real atmosphere where, from ground to (say) 15km, the force due to gravity varies hardly at all (only by 0.25% at the equator or by -0.17% at the poles, coriolis forces accounting for the difference).
What does vary in the real atmosphere over that 15km rise is the pressure due to the weight of the air above each point. But are we convinced that a linear variation in artificial g would be a valid proxy for a linear variation in atmospheric pressure?
I would say, not. Others may disagree.
I wrote:
“That this assumption is in fact correct is not detailed in the paper. Fortunately (for Walton) it is correct, as my and wayne’s simulations show, but I agree it could do with an extra line or two. I’ll see what I can manage”
Didn’t take too long, and I should have done this a year ago. Posted here for your delectation:
All these proofs start to look similar after a while, but I don’t think I saw anyone else use this exact argument. Yet it seems too simple not to have been done before.
Anyway, enjoy!
Tallbloke says: “Where in any of that is there anything about altitude? Where is the gravitational field?”
That’s the beauty. I address simply the zeroth law. You posit that a thermal gradient for the thermal equilibrium situation. Unless you posit the same thermal gradient for all gases (which no one seems to do) then your position is at odds with the zeroth law.
” … until the thermal equilibrium is restored .. ”
Which it the very definition of “not in thermal equilibrium” ! You are categorically saying “A” and “D” are not initially in equilibrium, but need to exchange energy (and change temperature) until they reach equilibrium. Only after exchanging energy will A & D be in equilibrium. But if A & D exchange energy and change temperatures, then they will no longer be in equilibrium with B & C. So A & D have to exchange energy with B & C ….
It is an endless loop of contradictions, of systems exchanging energy in a desperate attempt to restore equilibrium. It is a perpetual motion machine. It is a violation of the 0th and 2nd Laws.
“They all failed.”
Sorry my money is still on physics professors and physics textbooks, not on random bloggers with little or no formal background in thermodynamics. They may have failed to convince *you*, but they got the physics right.
Trick says: “And I would like to point out the air column theory and Graeff’s experiment is just ONE system so strictly the 3 system zeroth Law is not applicable. ”
The 0th law is always applicable. 🙂
In this case, we could consider the thermocouples at the top and bottom to be two of the systems, and the column to be the third. If the thermocouples are in equilibrium with the column of air (and the column of air is itself in equilibrium), then the 0th law requires that the two thermocouples be in equilibrium with each other. So if the two thermocouples were brought into contact, then they should not exchange energy, and should not change temperatures.
The only way I can see out of this is for there to be a universal gradient that should appear in any material (air, water, steel, glass … ) in a gravity field. If *every* system changed at the same rate dT/dy, then the two columns I postulated would always have the same temperature at the tops AND bottoms, so they would always be in equilibrium. Then the thermocouples would change as they were lifted from the bottom to the top, so that the lower one would cool as it rose so that it matched the temperature of the top thermocouple, with no energy transfer required when they touched.
But what is that universal rate? The only choice consistent with classical thermodynamics or the computer model of bouncing balls discussed above is for dT/dy = 0.
Tim F says:
(and the column of air is itself in equilibrium)
Lol. 🙂
Talk about begging the question! 😉
The whole Loschmidt debate is about whether energetic equilibrium is the same thing as thermal equilibrium or not.
Tim F says:
“Sorry my money is still on physics professors and physics textbooks,” not on random bloggers with little or no formal background in thermodynamics.
We’ve got some heavyweights on our side too.
Tim F says
not on random bloggers with little or no formal background in thermodynamics.
I bet I have more of a background in both formal and hands on thermodynamics and fluid mechanics than you do.
Tim Folkerts 8:35pm: “…consider the thermocouples at the top and bottom to be two of the systems..”
Thermocouple systems made of ideal gas? Geez.
TIM! We are just considering ideal GAS systems here. There is only 1 ideal gas system, not 3; 0th not applicable.
0th law only applicable to 3 systems, not 2, not 1. So not always applicable – many authors have trouble calling it a “law” for that reason.
Tim Folkerts wrote:
Trick says: “And I would like to point out the air column theory and Graeff’s experiment is just ONE system so strictly the 3 system zeroth Law is not applicable. ”
The 0th law is always applicable.
I agree. To the extent that it is true, the 0th law is always applicable.
One of the corollaries of the 0th law is that in the thermodynamic limit, a system comes to equilibrium at some some uniform temperature.
Graeff’s work appears to directly contradict this.
In a conflict between reproducible empirical evidence and a law of physics, the law loses. Going the other direction is circular logic.
We used to believe in conservation of energy, before it was updated to conservation of mass-energy. Even now, after the clear demonstration of Trinity, the First Law of Thermodynamics is generally written in terms of conservation of energy.
Graeff’s work does not violate the Second Law directly. It does, however, expose a vulnerability. That vulnerability was separated from the Second Law and called the Zeroth Law.
I believe the problem with the 0th law is relates to the words “in the thermodynamic limit”. The thermodynamic limit does not apply to gravitational system because there is no physically valid meaning for n=∞.
Can anyone offer a valid physical interpretation of the meaning of n=∞ in a gravitational system? What I come up with is that for any finite system, it means a non-quantized (non-atomic) gas.
br1 11:43am: “Verkley states that 2a is under the constraints of constant M and constant H, while 2b is under the constraints of constant M and constant L.”
Ok I’m b-a-a-ck after beginning to work thru inner details of Verkley 2a at 1:42pm with above as a start.
2a: Constant M. Constant enthalpy H = gas internal energy + (Pressure on boundary and environment * volume).
So, if H is given constant, then H doesn’t change and…
Change in enthalpy H = 0 = change in internal energy + change in (Pressure on boundary and environment * volume) thus:
Change in internal energy = – change in (Pressure on boundary and environment * volume).
Consider the mass of a parcel of particles of mean velocity v, the mean KE of which is the isothermal T, (at height h above bottom ref. h=0). The sum of all m would be M.
Change in IE is sum of all parcels in column = change in sum (½ * m * v^2 + mgh)) = – change in (Pressure on boundary and environment * volume).
Now can see better if turn g off and on. See the difference? If g=0, isothermal means no change in (Pressure on boundary and environment * volume). Good, this happens in deep space. (Just invoke star trek force fields at top & bottom of air column out there.)
This last eqn. with g .NE. 0 on earth is why earth’s air column in Verkley 2a has to be open on top & bottom. As the air column moves up (increase in h) with isothermal const. T, the sum (mgh) term increases (the KE = T = constant term has no change) and the change in – (Pressure on boundary and environment * volume) must increase (top p goes down less than bottom p goes down, V const.) and vice versa.
The work of moving column mass M a distance h up is equal & opposite to the work in changing the (pressure on boundary & environment) when T = isothermal. This is the physical reason Verkley 2a leaves the air column open to environment on top & bottom to get T=isothermal mathematically by proper general integration. IIRC B&A discusses this reasoning, w/any typo.s of mine improved, ha.
Verkley 2b closes the top & bottom of the air column and the resulting T profile changes to non-isothermal. From assuming constant M constant L. On the way to non-isothermal max. entropy B&A invokes conservation of energy which for a gas means constant H.
Very elegant. AND these 2b assumptions ARE being tested experimentally by Graeff, so he will find non-isothermal result as trump card if replicable. 2a is harder to test I would think (until star trek force fields are invented).
Verkley Fig.2 then shows bottom 80% real standard atmosphere is mostly in between 2a & 2b solution. The classical isothermal solution is just about as close but the resulting const. T profile is far different from reality.
DS,
Actually it was relatively simple to reverse engineer, still took a day, as a challenge.
Essentially we struck lucky in this instance. If you use a PDF or Postscript viewer, change the magnification, if the plot scales smoothly it was originally plotted vector format, not bitmap. To confirm this was the case I exported the PDF page as plain Postscript and opened it with an editor. Identifying the plot was easy, most commands
a b c d s (means stroke but not the only way it might be done)
By hand I copied the traces, grids etc. to files named oddly enough, g1, g2, g3 etc., A bit of hand cleaning up and then the fun starts. Machines are consistent but tend to make a mess of things, was thus.
Finding a short section of four points I had the rectangular plot area, hence page offset, width and height. Here it is, file g2 which I didn’t show
162.72 850.99 480 0 S
642.72 850.99 0 -258.56 S
642.72 592.43 -480 0 S
162.72 592.43 0 258.56 S
This draws two horizontal and one vertical, as a closed figure a rectangle.
At that point I started writing a decode program. The a and b above are essentially a pen x,y, c and d the do something relative. Lots of mess such as do nothing at all. In the end I was able to condense it down to x,y points running 4.2 to whatever (notice there is a blank region around the plot)
It was then a matter of looking at the plot scales, figure out how to move from page x,y to plot scales, at the same time I automated for a list of files (adage, engineers are lazy, we do things so we can do nothing). On seeing results which plotted sanely I needed to either interpolate or as it turned out average when it collapse to a single set of x numbers exactly, nice. Finally a small prog took each file and created a giant text array. Paste result into spreadsheet and plot.
Nothing hard or fancy in there, in reality most fancy looking things are boring hard work.
The overarching lesson from this is simple: PDF to numbers can be done successfully.
Your point about a wild force gradient in a centrifuge, I like it, yep, hadn’t thought of that one.
Q. Daniels 10:14pm: “Can anyone offer a valid physical interpretation of the meaning of n=∞ in a gravitational system?”
You mean a black hole? Thermodynamics near the speed of light. Ha. Still the Laws must hold. Somehow.
“… the 0th law is always applicable.”
Need a valid application to 1 gas system then, don’t think you can theorize one even, the 0th Law states 3 systems. Sure can apply 0th to 3 ideal gas systems.
“Graeff’s work appears to directly contradict this.”
No, only in very long LTE, this is not a physical application of 0th law. Transiently before the insulation breaks down (i.e. let’s in enough external heat flow), Graeff will find non-isothermal gas T profile. Leave it alone for long enough and thermocouples will synchronize with basement T profile & oscillate with the basement furnace clicking on & off, with an insulation delay.
I don’t see any contradiction, good to replicate though.
>> TIM: and the column of air is itself in equilibrium
>TALLBLOKE: Talk about begging the question!
Tallbloke,
You are missing the point. I am NOT claiming in this sentence that the column are isothermal, merely that there are no changes occurring. The question is “what sort of temperature profile exists in an air column that is in thermal equilibrium?” (Or perhaps you are implicitly agreeing that “in equilibrium” = “uniform temperature”, )
You seemed to have been claiming throughout the discussion that when a system (a column of air) is in equilibrium, there will be a temperature gradient.
I am claiming that when a system (a column of air) is in equilibrium, there will be a uniform temperature.
We are BOTH talking about equilibrium conditions (eg Graeff’s columns are given sufficient time so that there is no further energy flow and no further temperature changes).
Or are you now saying that even after days or weeks, that the gas in Graeff’s columns are never in equilibrium, and so they never have to obey the 0th Law?
br1 wrote:
Yesterday I wrote a simulation which swaps energy between vz and hidden energy reservoirs (degrees of freedom!), and that DOES give a gradient. However, I have to do a lot of sanity checking to see what assumptions are in that, and while the result is intriguing I don’t want to rest my hopes in it yet.
I’d love to see that. How can I help?
br1 wrote:
Posted here for your delectation:
http://www.slideshare.net/brslides/analytic-velocity-distribution-under-gravity
I looked at it. It’s very easy to make a subtle assumption.
ncd = nab (8)
This is only valid for collisionless gases. It seems like such an obvious equation.
If collisions occur, then some of the particles in the range will collide and not reach plane 2, while other particles slower than va will collide and change direction sufficiently (even if they don’t change kinetic energy) to reach plane 2. Further, the probability of a collision is itself a function of the total velocity, including the vz term you are varying.
All these proofs start to look similar after a while…
I suspect the assumption in (8) is pervasive, and not original on your part. 😉
Trick says: “Thermocouple systems made of ideal gas? Geez.”
Seriously? This is what you offer as a rebuttal?
We are addressing universal, fundamental issues of thermodynamics. We are looking at systems involving REAL gases (and thermocouples and insulation and glass beads and water). Ideal gases are fun to study, but they are hardly the core of this discussion.
The 0th Law is what allows us to define “equal temperature”. Without it, the whole concept of temperature is meaningless, since a thermometer could measure different temperatures for two systems that have been allowed to come to equilibrium. Any time you are using a thermometer, you are implicitly applying the 0th Law to compare the temperature of the . The readings of the thermocouples are essentially meaningless if the 0th Law is not correct. So yes, the 0th Law is indeed important here, and does indeed apply to the thermocouples!
See Wikipedia (among countless other sources) for more info:
Tim F:
“Or are you now saying that even after days or weeks, that the gas in Graeff’s columns are never in equilibrium, and so they never have to obey the 0th Law?”
Graeff’s empirical experimental measurements are getting a gradient significantly bigger than the dry lapse rate which we would expect from the gravity pervading the column.
In order to find out why, we are going to begin by replicating the experiment.
That’s how real science used to proceed before Kevin Trenberth came along and said from behind his computer screen
“The data are surely wrong.”
Q Daniels says “In a conflict between reproducible empirical evidence and a law of physics, the law loses. ”
That is so true. But there is also the idea that “extraordinary claims require extraordinary evidence”. Overturning the 0th and/or 2nd Law of Thermodynamics is indeed an extraordinary claim.
I have not yet seen what I consider extraordinary evidence.
Graeff has produced some evidence. The set-up needs to be independently examined for any stray energy inputs; any stray voltages; …. Then it needs to be independently confirmed, hopeful with a very different set-up and very different instrumentation to confirm the same gradient appears. Graeff as done some impressive work, but it is only a start. It has not reached the point where thermodynamic textbooks need to be completely re-written.
Quantum mechanics caused textbooks to be rewritten. Relativity caused textbooks to be rewritten. But both of these were supported by numerous independent experiments before they gained acceptance.
Trick wrote:
Transiently before the insulation breaks down (i.e. let’s in enough external heat flow), Graeff will find non-isothermal gas T profile. Leave it alone for long enough and thermocouples will synchronize with basement T profile & oscillate with the basement furnace clicking on & off, with an insulation delay.
By my read, Graeff’s apparatus started in approximately the condition you describe, and then diverged from that. That is to say, the pattern asserted itself against the the breakdown. That goes to my third point of this thread, that the concept of equilibrium is not necessarily applicable.
I’ll summarize my points:
1) The Thermodynamic Limit does not apply to gravity. Care must be taken here.
2) Quantization matters. System behavior for finite values of n is potentially different from behavior with n=1 or n=∞.
3) The existence of an equilibrium condition is not necessarily a valid assumption.
Tim F 11:17pm: “Seriously? This is what you offer as a rebuttal?”
Yes Tim, that and Verkley 2b. There is only one ideal enclosed gas system being non-isothermal per Verkley 2b, you just can’t throw in the thermocouples to make 3 systems and invoke equilibrium to make the enclosed column in the presence of gravity isothermal. 0th law doesn’t work that way.
In this case, the thermo laws ARE correctly applied to this 1 system in Verkley 2b. Heck, yes, you could toss in the thermocouples at different heights to get 3 systems, then invoke 0th law as you clip for measurement. All your clip says is thermocouples WILL equilibrate with gas system in contact w/gas at each height. The clip does not and cannot say anything about the T inside the 1 gas system.
Meaning 0th Law doesn’t say the thermocouples have to read the same at different enclosed heights; THAT is a result of the proper math application of 1st and 2nd Laws and the ideal gas Law to the 1 enclosed gas system w/gravity. Graeff can then invoke the 0th law to do the measurement as you clip.
Here is MY wiki 0th law clip, often expressed:
“If two systems are in thermal equilibrium with a third system, then they are in thermal equilibrium with each other”
3 systems in the law not 1.
Q. Daniels 12:15am: “Graeff’s apparatus started in approximately the condition you describe, and then diverged from that…”
Yeah. The real enclosed, insulated gas column experiment has to start with some initial condition thermal profile. Lucy may tell us in part 2, I haven’t had the time/interest to read up on Graeff exp. other than here.
So you have a dynamic thermal system starting from a set of say unbalanced initial T conditions acting w/o other energy input in the presence of gravity for some time. The system will then arrange T in time to comply & balance with natural forces. My intuition from experience and initial Graeff results look like the time for the rebalancing w/o other input energy is short enough that it happens before the longer time due to GOOD insulation that it takes outside energy to reach the system (and it WILL: no perfect thermal insulation) to force a rearrange in T profile.
As above somewhere by br1, the usual basement experience, mine included, is that it is colder at bottom and warmer at top. This is opposite to the 1st balance the close enough to ideal enclosed gas system experiment will find. Eventually though, as Tim F of the 0th law police writes, the heat will travel warmer to colder through the multiple systems of various layers of insulation to the 1 enclosed gas system of interest.
The basement is found to be in whatever thermal condition once you put a control volume around it and account for energy flow in & out. Nat. gas or oil or coal to the furnace, ground water & earth through the walls, entry and exit doors, windows if any, the famous gravity gradient. Humans standing around, lights. Get ‘em all et. al. right and compute the basement T profile.
Trick,
I think you’ve mostly captured why I think this is a Zeroth Law experiment, rather than a Second Law experiment. Looking at the data, it looks to me like the run time is sufficient that the insulation leakage is already incorporated into the results, and not sufficient to prevent the gradient from establishing itself.
The device has two thermocouples, each of which are in equilibrium with the interior. The Zeroth Law says they should be in equilibrium with each other. The evidence seems to indicate otherwise.
The Zeroth Law is very intuitive, which is why it wasn’t even written as a law until later. It’s easy to think that it must be so. That doesn’t mean that it actually is that way, though.
br1 6/6 11:43am: “Or do you really think the atmosphere is better described by a closed container, rather than a free column???”
I do. Walton writes: “…one considers a vertical column of gas in earth’s atmosphere…isolated from all disturbances…” i.e Verkley 2a so Walton math is just the classical work out from his earlier references. All of this classic work, in whatever formulation, runs up against an integral that contains T(z) or T(h) or here T(w). Every one of them & I mean EVERY, passes this step to easily by invoking equilibrium to make T constant (isothermal) and proceed to solve the integral to get the classic answer of isothermality and the barometric eqn. in the open ended gas column. Verkley’s 2a work & constraints are condensed to show this. Equilibrium is a town where thermodynamics lives, equilibrium is not itself the law.
Verkley Fig. 2 straight vertical line (no lapse) shows the isothermal solution. It is different from the real standard atmosphere in two ways: 1) actual T, 2) T profile.
The difference in actual T & predicted isothermal T is around the same (higher AND lower) as the predicted non-isothermal closed end system Graeff is experimenting with.
Notice the non-isothermal solution T profile “corresponds remarkably well to the tropospheric part of the Standard Atmosphere.” So I suggest the closed container is a better model since we are interested in the lapse rate and the solution to the experiment. And this solution is recent not classical; it proceeds with the complicated integration by invoking 1st & 2nd law and ideal gas law not just assuming T constant to make life easy. This general solution 2b resulting in non-isothermal profile is way more elegant & attractive than 2a.
Trick: “Cold on top and warm on bottom will be transient on the way to real non-ideal LTE.”
br1 responds: “So you don’t believe in a temperature gradient due to gravity, or how am I to interpret this?”
Interpret it as you wish. “As time goes by…”. I do believe in temperature gradient due to gravity b/c the 2b math – consistent with all the thermo laws – shows it.
I also interpret it to mean the experimental initial conditions will 1st change to the non-isothermal solution before the non-perfect but GOOD insulation succumbs and allows heat exchange with the basement as it must, after which the profile will change to something else. Call it what you will.
Q. Daniels 2:57am: “The device has two thermocouples, each of which are in equilibrium with the interior. The Zeroth Law says they should be in equilibrium with each other. The evidence seems to indicate otherwise.”
The evidence AND theory seem otherwise b/c you are invoking the 0th law with the three bodies incorrectly (inconsistent w/1st & 2nd, ideal gas laws) when the ideal gas column is enclosed & perfectly or GOODLY insulated in the presence of gravity field.
In that case, thermocouple 1 is in equilibrium with gas column interior at h1; thermocouple 2 is in equilibrium with gas column interior at h2. 0th law says nothing about thermocouple 1 and thermocouple 2 being in equilibrium with each other even though we have 1 gas column.
Amazing! The 1st and 2nd law coupled with ideal gas law come in to tell us thermocouple 1 and thermocouple 2 are not at same T(h) for an enclosed, insulated column in gravity field case – non-isothermal solution case.
Open that column at top and bottom to the atmosphere and those same laws tell us thermocouple 1 and 2 are at same T as the interior of that open gas column – the isothermal solution case.
David Socrates is right – this is addictive, just like expensive scotch. I need to find some other kool aid.
Q. Daniels:
“I’d love to see that. How can I help?”
I’ve since written another one which swaps energy, definitely conserves momentum, and gives no temperature gradient. So while it’s fun, it still needs a bit of sorting. Not sure how you can help – what is the best way to share the actual code?
“Posted here for your delectation:
I looked at it. It’s very easy to make a subtle assumption.
ncd = nab (8)
This is only valid for collisionless gases. ”
Glad somebody has read it! Well spotted, but this is as stated in the ‘Notes on assumption’ at the end.
I was afraid the presentation had got lost in the discussion about whether 1=1=1 and whether that can be generalised to whether 1=1=1=1 or not.
Also note that Walton makes the point that if the gas is under equilibrium then you can take collisions into account by saying that the average number knocked into a set will equal the average number knocked out of a set, so that doesn’t change the conclusions. In my case I’m happy to say there are no collisions at all, because the fundamental question is how does *gravity* affect the distribution, not about how collisions affect it.
Trick:
“All of this classic work, in whatever formulation, runs up against an integral that contains T(z) or T(h) or here T(w). Every one of them & I mean EVERY, passes this step to easily by invoking equilibrium to make T constant (isothermal) ”
Mine doesn’t!
I’m sure you’ll find this version convincing, I can feel it! 🙂
Tim F says:
Graeff has produced some evidence. The set-up needs to be independently examined for any stray energy inputs; any stray voltages; …. Then it needs to be independently confirmed, hopeful with a very different set-up and very different instrumentation to confirm the same gradient appears.
Excellent, an institutional physicist has got the message. 🙂
I completely agree Tim F. So can we now expect you to assist Graeff and the Talkshop in lobbying an accredited lab to perform the replication and test alternative methodologies?
Graeff as done some impressive work, but it is only a start. It has not reached the point where thermodynamic textbooks need to be completely re-written.
Agreed. However, until his work has been put to the test by an accredited lab, there is a question mark around their validity. Experimentum summas judex, as Einstein was fond of saying – Experiment is the final arbiter.
Trick and br1: I’m endlessly impressed by the civility time and effort you are putting into this discussion. Thank you both.
Trick:
“So I suggest the closed container is a better model since we are interested in the lapse rate and the solution to the experiment. And this solution is recent not classical; it proceeds with the complicated integration by invoking 1st & 2nd law and ideal gas law not just assuming T constant to make life easy. This general solution 2b resulting in non-isothermal profile is way more elegant & attractive than 2a. ”
The constraints are not about closed and open containers, they are about constrained H and constrained L!
To see this (as my previous arguments seem to have had no effect…) let’s take Verkley2a (again). Let the system do its thing and come to an isothermal distribution.
Now imagine a thermos flask with the top open so the gas freely fills the flask, held stationary in the isothermal column, and then close the top on the flask. The thermos flask has perfect insulation.
Big question – what happens to the thermal profile of the gas in the flask?
If you answer that a thermal gradient develops, which as Verkley states in table 1 has a lower entropy than the isothermal situation that preceded it, then how does this gradient get established?
br1 9:44am: “Mine doesn’t.”
Uh-huh, well…
br1 continues: “I’m sure you’ll find this version convincing….”
I am convinced your eqn. [1] for phi1 shows a term T. This is the thermal wall temperature Tw in the paper you reference let’s say Tw could be 288K as an initial condition.
Your eqn. [4] then integrates phi1 wrt dv holding Tw constant. Ok, we have a horizontal thermal wall with a constant temperature Tw1 at the ground plane h=0 which just might be 288K.
You then introduce us to “plane 2 at height h” being hit by a gas molecule with reduced KE under gravity g.
br1 then tells us: “We do not know…the temperature at height 2…” (i.e. Tw2) which reduces with h because you tell us the plane 2 gets hit by a molecule “its KE will be reduced” from its reduced velocity meaning mathematically Tw2 must be some function of v I’ll denote Tw2(v). (You drop the w but that’s immaterial.)
br1 integrates an equation [10] for phi2 wrt dv containing the unknown Tw2(v) dv easily by just assuming Tw2(v) is a constant that – hey, what do you know just happens to = Tw1 (= 288K here).
Therefore br1 announces your theoretical work derives the enclosed, insulated isothermal gas column T plane 1 (T=288k here) = T plane 2 (= 288K here) as the result and, oh man, even gives the barometric eqn! And furthermore your computer simulation of the enclosed, insulated ideal gas column verifies the isothermal result!
Hey wait….isn’t that EXACTLY what Walton does and the classic thermo grandmasters before him (which he references)?
I stand gobsmacked and unconvinced.
To reduce my gobsmackedness, explain just how you integrate to get Tw2(v) dv? How does the integration & simulation satisfy thermo law 1 and 2? You didn’t mention those thermo laws or even need the ideal gas law all of which B&A 1998 text invoke to do the integration of Tw2(v) dv.
Hmmmm…. might br1 be invoking the magic equilibrium demand again?
br1 12:46pm – “… imagine a thermos flask…”
Fill it with scotch and I’m there.
br1 continues: “…thermos flask has perfect insulation..”
Oh wait , nuts, we are back in theory land again. Forget the scotch, fill flask with famous isothermal gas at entropy = 4.3783, close the top. Wait for LTE.
br1 asks: “what happens to the thermal profile of the gas in the flask? ”
I answer: A non-isothermal, isentropic profile develops at LTE.
br1: ” If you answer that a thermal gradient develops , Verkley states in table 1 has a lower entropy than the isothermal situation that preceded it, then how does this gradient get established?”
Nature establishes the thermal gradient. The entropy cannot decrease even in theory. However thermo theory IS ok with constant entropy therefore isentropic is ok. Here the entropy of nature’s newly non-isothermal air in the flask increases to max. isentropic point above 4.3783 and remains, if perfectly insulated, constant. My that’s a big flask.
Trick:
“To reduce my gobsmackedness, explain just how you integrate to get Tw2(v) dv?”
That’s the beauty of my derivation – I *don’t* integrate Tw2(v)!!!
The temperature at level 2 is never mentioned, and also note that the integral in Eqn(9) is of phi2 (no mention of temperature dependence, so be careful of notation).
There are no assumptions on the form of phi2(v), what it’s temperature is, or what distribution of velocities it has. Hence I circumvent (as far as I can see) all your objections to such assumptions.
“How does the integration & simulation satisfy thermo law 1 and 2? You didn’t mention those thermo laws or even need the ideal gas law all of which B&A 1998 text invoke to do the integration of Tw2(v) dv.”
Because I don’t integrate phi2(v) dv, I feel my derivation is perhaps even a little superior to other approaches. B&A eat your heart out!
The trick is in Eqn(8) and Eqn(9). Because only particles with velocity va>v1min are counted at ground level, then Eqn(8) holds. Because phi2 is a probability density function, it’s integral has to be particle number, ncd, hence Eqn(9) holds. No dependencies are assumed.
Isothermal follows, QED.
I wrote:
“The trick is in …”
If you wonder about how to get from Eqn(10) to Eqn(11), just look at Eqn(4) and Eqn(5), the answer is done already (maths is maths).
Alternatively, if you choose not to appreciate that Eqn(10) has been solved already, just *differentiate* both sides of it. This removes the integral and leaves you with a bare phi2(v). You don’t actually need to assume anything about phi2.
br1 2:53pm – “I *don’t* integrate Tw2(v)!!!”
I know. I can see that. It is my point. Yes…you DO NEED to do that properly to find the integral of phi2. B&A do and it takes about 12 steps all told, not your 1. This is freshman calculus. Maybe soph. calculus when you run up against proving the 1st deriv. of entropy set = 0 is really a max.
You’ll need to make a trip over to the math dept. then and verify whether you or B&A are correct. What the math dept. will tell you is to perform the integral of Tw2(v) dv from v1 to v2 you need to show them how the function for Tw2(v) naturally varies in that interval from h=0 to h = h + dh. Exactly. They will really want to know.
Tell them v (hence Tw2(v)) will vary as h varies according to the 1st & 2nd thermo law, ideal gas law & gravity. Maybe they can work THAT for you, but they will probably send you to the thermo dept. Then when you come back from there, the math dept. will delight in: changing variables, integrating the function constrained to show const. of energy, ideal gas law and then finding 1st deriv. of enthalpy (set = 0) and then proving that 0 IS the max. point.
It is really, really NOT trivial as you write. Once you get B&A the light will click on. I remember my light clicking on from my HS algebra teacher.
You will even value your trip if you just say Tw2(v) = 288K constant and see what they say. The math dept. TA will laugh and say that’s easy. See what I mean. Too easy.
Maths is maths. Non-isothermal follows for the enclosed, insulated column. QED.
Trick:
“Yes…you DO NEED to do that properly to find the integral of phi2.”
No I don’t!
It seems that this is a proof unlike any you have seen before. Please read carefully, and with a fresh outlook.
The logic goes as follows:
1, phi2 is an unknown probability density function describing particle number over velocity (by definition).
2, The integral of *ANY* such PDF between *ANY* limits is the particle number within those limits (by definition. Because PDF’s are normalised to unity, one has to multiply by N, the total particle number in the system, but that is just a constant prefactor).
3, I know the particle number within any two limits.
4, Hence I know phi2.
Maybe that is so unexpected that you don’t believe it, yet it is certainly correct.
I reckon that upgrades the elegance of my solution a few notches. Doubt I’ll get the Fields Medal for it though 😦
I don’t make any comparison to B&A, this is a stand alone proof.
Tallbloke says: “Agreed. However, until his work has been put to the test by an accredited lab, there is a question mark around their validity.”
There is ALWAYS a “question mark around the validity” of EVERY theory in science. Scientists are aware that any long-held belief may be wrong. But not all “question marks” are equal. We have an extremely robust theory (actually two — thermodynamics and statistical mechanics) that predict an isothermal profile for an isolated gas in a gravitational field. These theories have been validated by wide ranging experiments for more than 100 years, with “gas in a cylinder” being just one interconnected piece. When the theory has broken down, then new theories have successfully been found (eg Fermi-Dirac or Bose-Einsten distributions rather than simply Maxwell-Boltzmann distributions).
If there was something wrong with something as fundamental as either the 0th or 2nd Laws, it would have been damn near impossible to hide it for this long. As such, searching for violations in a familiar setting (insulated gas containers) seems unlikely to be fruitful. Similarly, amateur interpretations of fundamental laws are unlikely to uncover ideas that have not already thought through thoroughly by experts.
Consequently, it will be a hard sell to convince any experimentalists OR theorists to take on a project they deem to be almost certainly doomed to failure. If you and your cohort want to invest your own time and effort, then that is your prerogative. I doubt you will convince a panel at any university or national funding source to pursue this topic.
[Reply] Enough of the negative waves! You were the one who said Graeff’s results were interesting and needed replication, remember? As for ‘hiding it this long’, it’s only recently that measuring equipment has become sensitive enough to detect a 0.07C gradient. Stuff your negativity, on with the science. Elitist comments about amateurs don’t cut it with qualified technicians and engineers who do the real legwork around experimental design for the scientists.
br1 – “I know the particle number within any two limits”
No you do not. The PV = nRT. eqn. knows it; But you do not know Tw2(v) as you wrote. Thus you cannot know n at any h of plane 2. Your integration fails to tell you with 1st & 2nd law abiding formulas..
Hint: You are not getting any closer to the right theory AFAIK..Try it.
br1 wrote:
In my case I’m happy to say there are no collisions at all, because the fundamental question is how does *gravity* affect the distribution, not about how collisions affect it.
I don’t think the questions are independent. See my comments about n=1.
Also note that Walton makes the point that if the gas is under equilibrium then you can take collisions into account by saying that the average number knocked into a set will equal the average number knocked out of a set, so that doesn’t change the conclusions.
Once you start talking about collisions, the degrees of freedom are no longer fully independent. The probability of a collision depends on the total velocity, not just the 1D velocity.
To visualize this, consider two particles with the same vertical velocity, one of which is moving straight up, the other moving arbitrarily fast in horizontal plane. They do not have the same probability of collision per vertical unit.
TB wrote:
Elitist comments about amateurs don’t cut it with qualified technicians and engineers who do the real legwork around experimental design for the scientists.
Q. Daniels says: http://sotak.info/sci.jpg
Thanks for the laugh. 🙂
And of course, science would grind to a halt without grad students, professors, or technicians. All are vital and contribute important work. Over the years I have had the privilege of working with some great people each category.
br1, I noticed you are still hanging on to the distributions and I’d be very careful here. As you know I programmed (modeled) an exact duplicate of your simulation except instead of using monte carlo “tosses” and grouping the number of particles within a dh level, I calculated the exact z velocity as each particle crossed a level boundary. You got some noise in you results, mine was numerically exact so there is never any noise and my two plots at any given level plot exactly on top of each other.
But I’ve gone further. Since there are only a few equations used it was easy to trace “why” this was so at any level, any temperature, any dimensions and here is what I found. Both you and I are using the same type of distribution equations, yours phi1, mine 3d M-B, but each has ‘T’ terms and if you never vary the temperature in the simulations, of course, the distributions will always be exactly the same.
But if you then integrate into the distributions a varying ‘T’ potential temperature at each altitude level, now the distributions no longer match and it instead matches closely to the dry adiabatic lapse rate we see within the atmosphere.
I’d just be careful exactly what ‘T’ you are using at each vertical level for if ‘T” never changes in the simulation then you have made the results isothermal. I found I was the one forcing mine to be isothermal. It was circular logic.
Tim Folkerts says, June 7, 2012 at 5:34 pm: If there was something wrong with something as fundamental as either the 0th or 2nd Laws, it would have been damn near impossible to hide it for this long. As such, searching for violations in a familiar setting (insulated gas containers) seems unlikely to be fruitful. Similarly, amateur interpretations of fundamental laws are unlikely to uncover ideas that have not already thought through thoroughly by experts.
I agree with you Tim.
But aren’t you and many others here missing an important point. Graeff has produced an apparently inexplicable result. It may be due to experimental error but it may be correct. The only way to find out is to repeat the essence of the experiment. So it is far too early (and a waste of time) to talk about violations of the Laws of Thermodynamics.
The only thing that needs verifying (or refuting) is that a very carefully insulated vertically mounted column of water, heavily doped with glass powder to minimise convection, settles down to a negative temperature gradient of around 0.05K/m.
This doesn’t require expensive laboratory equipment or crazy schemes with centrifuges or any other wacky solutions. As far as I can see (unless Graeff has a very cogent, but so far unstated, explanation) it doesn’t even require his complicated multiple insulation cylinders – just a single insulation jacket. It doesn’t even require two tubes side by side. And it certainly doesn’t need the complexity of multiple thermocouples and thermistors. All it needs is just one thermocouple with one of its junctions positioned at the top of the single water+glass column and with its other junction positioned at the bottom. It doesn’t even need digitised voltage capture – just a sensitive enough DVM that can resolve down to a few microvolts and which the experimenter can go and look at every day to see whether or not the system has settled down to the predicted 0.05K/m steady state.
Personally, having read his paper and looked at his graphs, I wouldn’t be at all surprised if Graeff is essentially right. If so, and if the essence of his experiment can be successfully replicated with air as well as water, it would simply confirm the viewpoint of those such as Nikolov & Zeller, Huffman and many others now who contend that the temperature profile of a column of air in a gravitational field is solely a function of the correponding pressure profile induced by the weight of the air in that column and is not a function of greenhouse gases.
The point is that neither Graeff, nor N&Z, nor Huffman, nor anybody else should then be compelled to explain whether or nor the experimental finding (if proven true) violates any particular Law of Physics. That would be completely beside the point because the greenhouse gas theory would be busted, surely our overarching objective?
I have two suggestions for alternate apparatus.
1) Use SF6 or some other heavy gas in place of air.
2) Use water as a measurement device, particularly in combination with a heavy gas. To do this, place a pool of water at the bottom of the cylinder, and a wet (damp) sponge in a catch basin near the top. If there is a real temperature difference, water should accumulate at the sponge and basin. That’s because the temperature difference due to the majority gas is substantially larger than the partial pressure difference vs vapor pressure for the minority gas (water vapor).
If #2 isn’t clear, I can take another cut at writing it up, or maybe someone else can try translating that into English.
Trick:
“No you do not”
Could you please put down the glass of Scotch and read the presentation again. Eqn(9) is undeniable.
[Moderation note] easy, tiger.
David Socrates 8:23pm: “Graeff has produced an apparently inexplicable result.”
No, the experiment apparent result is remarkably explicable. The thermo theory is fully explained in Verkley 2b, B&A et. al.
“…it would simply confirm… the temperature profile of a column of air in a gravitational field is solely a function of the corresponding pressure profile induced by the weight of the air in that column and is not a function of greenhouse gases.”
No it would not, GHGs are still able to function. The real standard atmosphere profile is different than theory. Still lotsa’ work to ‘splain that..
Tim Folkerts 5:34pm: “We have an extremely robust theory (actually two — thermodynamics and statistical mechanics) that predict an isothermal profile for an isolated gas in a gravitational field.”
Yes, and the right constraints are shown in Verkley 2a after all these years (no lapse in this work out so not as interesting). The non-isothermal profile 2b theory is also extremely robust, shows a remarkably close lapse profile and Graeff’s results seem promising to prove it by experiment. The real standard atmosphere profile of interest up thru about P=225 hPa lies right between these two ideal solutions.
So other physics in action is still fair game.
Wayne:
“I’d just be careful exactly what ‘T’ you are using at each vertical level”
Eh? Now I’ve no idea what you have done.
I don’t use any T at any level except ground level. There is absolutely no way I can set the temperature at any other altitude, or program any dependence of T into the simulation. The whole point is that T vs altitude is an output. Anything else breaks Newton’s law of F=-mg which is the only effect acting on the particle.
If you can set T to drop off with altitude, then your simulation has no predictive power, so as you say will be circular logic. It seems my sim is fundamentally different.
Q. Daniels
“I don’t think the questions are independent. ”
In toy models they can be completely independent. However, the closer one gets to ‘reality’, the more effects one has to include, so of course eventually one would include collisions.
The other part of the post deals with *equilibrium*, so I’m not sure how your comment applies.
br1 10:15pm – Glass is now a flask. You are a bad influence but it is ok mods.
Looks like [9] only contains Tw1, constant 288K at h=0 here. So might be ok notwithstanding the no g term in phi.
Though, I would much prefer you start asking something like “Sayyy, in B&A ed. 1 p.186 where they change variables to do the integration with enthalpy constrained constant, why does…”
br1 wrote:
In toy models they can be completely independent. However, the closer one gets to ‘reality’, the more effects one has to include, so of course eventually one would include collisions.
I think whether or not you properly include collisions changes the answer.
Trick: June 7, 2012 at 1:46 pm
“br1 asks: “what happens to the thermal profile of the gas in the flask? ”
I answer: A non-isothermal, isentropic profile develops at LTE.”
We didn’t finish this discussion, let’s have another go. I’ve now read and derived every equation in Verkley, so am getting a finer grip on the paper.
“The entropy cannot decrease even in theory”
Yet you have just decreased it by saying an isentropic profile develops from an isothermal one.
Note Verkley, Introduction:
“one considers an ideal gas in a gravitational field and seeks the state of maximum entropy under the constraints of 1) a constant mass and 2) a constant energy (internal plus potential). The answer—the profile will be isothermal—was rigorously proven”
and in the Conclusion:
“Of course, the actual atmosphere is subject to processes like convective mixing. They prevent the atmosphere from ever coming close to thermodynamic equilibrium, that is, the ultimate state of maximal entropy. In this sense, these processes lower the maximum value that the entropy is allowed to attain.”
Yet you might say that Verkley 2b also maximises entropy, and yet a gradient is found. But notice Table 1 again. For the isothermal case, L=2.3326. The *constraint* in 2b is to set L=2.3150 and find the T profile when entropy is maximised. This reduced value of L also gives a reduced value of S. S is not as high as it was previously because the system in 2b is not in thermodynamic equilibrium.
You also seem to think that the difference between 2a and 2b is that 2a is for an open column while 2b is for a closed one. Yet this is not true, as Verkley points out in the introduction:
“Constraint 2 manifests itself as the requirement that the vertically integrated (absolute) temperature be constant. As we will show below, the same requirement is found if one relaxes 2 by allowing neighboring layers to do work on the layer under consideration; constraint 2 is then to be replaced by 2′, a constant enthalpy. As a result, here too the outcome is that of an isothermal profile.”
Please note that Constraint 2 works for a closed container! And the result is isothermal. Constraint 2′ which works for an open column and is discussed in section 2a also gives an isothermal answer.
I don’t know what B&A are up to, but the Constraint 3 (given by setting Eqn(13) equal to the measured value) will also work perfectly well in either a closed or open container. The T profile is not about containers! Indeed, Verkley says that Constraint 3 has not been properly justified by anyone, and I sympathise – the only justification I can give it is that the atmosphere is ‘fully mixed’ by convection, almost like it is boiling. Any system that is not ‘fully mixed’ and actively jumbled around, will not be subject to the constraint that L must have any particular value. In particular, a gas in an insulated sealed jar with no heating will NOT be fully jumbled around and fixed L will not be a constraint. In contrast, such an insulated gas will have a fixed mass and a fixed internal energy, hence Constraint 1 and Constraint 2 apply, and the profile is expected to be isothermal, even under gravity.
The explanation to the Graeff experiment is not to be found in Verkley, unless Graeff’s setup is not what we think it is.
br1 4:52pm – “Yet you have just decreased (entropy)…”
NB: My attention span much easier to maintain w/your switching back to Verkley/B&A.
Oh man, geez! At least the inefficiencies of the internet blog mode of discussions always seem to increase entropy. Still, no entropies were harmed in the making of my post. Here’s what Trick 6/7 1:46pm wrote exact clipped:
“…fill flask with famous isothermal gas at entropy = 4.3783…, close the top. Wait for LTE….the flask increases to max. isentropic point above 4.3783…”
No typo’s I see. No entropies harmed; entropy actually increased.
br1 example started with an open ended air isothermal 2a air flask at LTE max. entropy of 4.3783 and closed the cap. It is now an enclosed, perfectly insulated 2b ideal gas column isothermal & thus below the max. entropy point now enabled. Verkley 2b shows nature moves T profile from isothermal constant to non-isothermal, new higher entropy. AS I WROTE. N’est ce pas?
Now, before going further on the rich targets provided, I would like to see br1 write back solving this example below (putting the new skills developed from deriving Verkley eqn.s to the test):
Open the same flask cap again! All else same. What happens to flask T profile & current entropy above 4.3783?
Trick:
” nature moves T profile from isothermal constant to non-isothermal, new higher entropy.”
‘Scuse me butting in for a minute. As I understand it, entropy relates to energy not heat. So although the heat distribution is non-isothermal, the energy distribution is isentropic?
If so, how is entropy affected?
TB 7:09pm – “Scuse me butting in… how is entropy affected?”
No problem, butt in more. Good point. I just received a copy of your book rec. today, you’ll need to get a copy of my rec. now – B&A 1998 text as amended to get your exact answer.
I will offer an answer but those B&A guys TELL you in no uncertain terms right from 1st principles, believe me, I’ve been there.
All MY thermo stuff always starts with step 1 “Put a control volume (cv) around…” Not doing this just stops many getting it right like in a recipe where the chef says Step 1: “Take a clean plate…”. Either the thermo prof. or chef cracks my knuckles until I comply.
Step 1: Put a control volume (cv) around any column (parcel?) of air in earth’s atmosphere from Temp. = 288K to 225K and pressure 1000 hPa and 300 hPa.
Step 2: Define the innards of cv has constant mass, we define no mass allowed across cv.
Step 3: Define the cv to be perfect insulation, constant volume, the famous adiabatic assumption, no energy in or out.
Step 4: Let gas captured in cv come to LTE (no more heat flow but w/vigorous mixing), I have no supercomputer
Before LTE, that air will have some Temp. profile & it will have a random entropy below the max. possible but not the min. either. If cv happens to have captured the standard atmosphere, you can see this profile in Fig. 2 of Verkley but cv more than likely captured a different T profile not at max. entropy point.
Constant enthalpy (Law 1), PV=nRT law, and entropy always increases (Law 2) will energize nature to move the entropy to the max. (can be proven SAME as employing Hamiltonian to minimize energy for a mechanical system.).
Verkley 2b/B&A math show that max. point entropy is non-isothermal profile… barring typo’s T varies!!! & varies WITH/pressure altitude …..pretty cool. (Sorry long, I took typing in HS.)
Return to step 1 again: Redefine cv to be open to atmosphere only at the top & bottom.
Your turn.
Hint: 1) You’ll see gotta’ get an accountant working with green eye shade across cv for enthalpy counting now 2) Check Verkley 2a.
NB: where you say “..although the heat distribution is non-isothermal, the energy distribution is isentropic..”
Crack your knuckles for me. You didn’t start with cv. And you mean temperature; no more heat flow at max. isentropic point.
Trick:
“br1 example started with an open ended air isothermal 2a air flask at LTE max. entropy of 4.3783 and closed the cap. It is now an enclosed, perfectly insulated 2b ideal gas column isothermal & thus below the max. entropy point now enabled. Verkley 2b shows nature moves T profile from isothermal constant to non-isothermal, new higher entropy. AS I WROTE. N’est ce pas?”
Tres non!
There you go totally messing up about open column is 2a and closed column is 2b. That is plain wrong.
I’ll quote some Verkley for you, here are the four constraints:
Constraint 1: Fixed amount of M.
Constraint 2: Fixed amount of E.
Constraint 2′: Fixed amount of H.
Constraint 3: Fixed amount of L.
He says in the Introduction that for a gas under gravity with fixed M and fixed E, it will become isothermal. This has been known for over 150 years now, and is hardly worth comment at this stage (apart from that some people still make a mess of it!). For example, take the statement: “A gas in a closed insulated stationary container has a fixed amount of M and a fixed amount of E” – can you please say whether you agree with this statement or not?
Next Verkley generalises Constraint 2 to give Constraint 2′, and this is worth some comment so he writes section 2a. To have a fixed amount of E is a special case of having a fixed amount of H, it is one where there is no work done at the boundaries. But he makes the point that even if there is work done at the boundaries, so long as H has to remain constant, then the gas will still be isothermal.
So the example with the flask goes from a case of
Open = Fixed M , Fixed H (in column as a whole) = isothermal
to
Closed = Fixed M, Fixed E = isothermal.
Opening and closing the lid will make no difference.
So what of Constraint 3? Well one has to understand what L is, and what a constrained amount of it means. That’s complicated and I think that requires another post.
Before that, what do you say to the above?
br1 wrote:
He says in the Introduction that for a gas under gravity with fixed M and fixed E, it will become isothermal. This has been known for over 150 years now, and is hardly worth comment at this stage (apart from that some people still make a mess of it!).
Graeff’s empircal data seems to contradict this. Given contradictory empirical data, it seems reasonable to me to suggest that there might be errors in the theory.
Under sufficient weight of empirical data, the theory bends or breaks.
br1 9:37pm: “Tres non! There you go totally messing up about open column is 2a and closed column is 2b. That is plain wrong…what do you say?”
Very good French. Tres non though! You just won’t find the closed container constraint in Verkley paper anywhere or specific in 2b as much as you search b/c it simply isn’t there. Very unfortunate they do not make that clear. As I think back now that may have been a big driving point for me to go get B&A text they cite & once I found that picture of the 2b closed column & I compared to Verkley 2a Fig. 1, the light snapped right on. Getting this right is very important in discussions.
In B&A discussion, they CLEARLY teach you Verkley 2a, the long standing classic isothermal exact general integrated solution (by dropping the equilibrium demand!) is for an open ended mid-air gas column as does Verkley with Fig. 1. And Walton even implies that – though gives no picture in your link. Partly I think it is reason Exner was so exasperated. (Just like me right now.)
B&A 1998 goes one step further and discusses why they show closed container figure that Verkley 2b 2004 had available and could have used. Maybe Verkley had a space constraint & it was available elsewhere, so they didn’t, dunno.
Instead of using up TB bandwidth debating this point, just go to some trouble get B&A see for yourself the closed cartoon figure B&A draws for case 2b & their reasoning for it. I would send some pix to TB to post but I no longer have B&A, would take a week+ to get it again – maybe TB will do so.
I have tried to prove to you consider open & closed by theoretical reason enthalpy is different for you, posting my work & see! look! H is different in table 1. You can trace the only reason enthalpy can be different is the term where 2a is open to environment and 2b is not open to environment. Tres oui! Vive la différence!
Q. Daniels says, June 8, 2012 at 11:30 pm:
br1 wrote:
He says in the Introduction that for a gas under gravity with fixed M and fixed E, it will become isothermal. This has been known for over 150 years now, and is hardly worth comment at this stage (apart from that some people still make a mess of it!).
Graeff’s empirical data seems to contradict this. Given contradictory empirical data, it seems reasonable to me to suggest that there might be errors in the theory.
Under sufficient weight of empirical data, the theory bends or breaks.
Well said.
What is it with these theoreticians that they cannot see that good data always trumps good theory? It is the reason that the scientific process has been so successful for the last three hundred years. Yet many educated, intelligent people just don’t get it.
In any case, as I have said earlier in this blog trail, even if Graff’s experiment is confirmed this does not necessarily mean that the Laws of Thermodynamics are thereby proven wrong.
The logical consequence of this position? Concentrate on replicating the Graeff experiment and, in the meanwhile, stop theorising.
And armchair theorists who think the experiment is not worth replicating should, in my view, simply shut up.
Q. Daniels and David Socrates:
“What is it with these theoreticians that they cannot see that good data always trumps good theory? It is the reason that the scientific process has been so successful for the last three hundred years. Yet many educated, intelligent people just don’t get it.”
Maybe it slipped your notice, but not only have I pledged support for Lucy’s replication, I have also attempted my own replication.
In the meantime, the theorising won’t slow Lucy down! If you are interested in thermodynamics, you might even learn something. And if a sound explanation could be found, that would be VERY big news!
Trick:
You don’t seem to be trying to apply Verkley’s constraints to a closed container.
Let’s see what a closed (insulated, isolated, etc) container has:
It definitely has fixed M (tick for constraint 1).
It definitely has fixed E (tick for constraint 2).
It definitely has fixed H (tick for constraint 2′).
You can’t actually deny any of these, surely?
Then the big question comes up about constraint 3, fixed L. What does that mean, and how does it apply to a closed insulated container?
I could answer, but instead of taking my word for it (which seems highly unlikely at this stage), I’ll let Akmaev say it. He wrote a paper
(hope that link works, title is ‘On the energetics of maximum-entropy temperature profiles’) where he quotes Verkley and B&A loads, discusses their results, and develops them a bit more. This paper is definitely worth reading!
It may take a while to digest, but he gives the same constraints (Eqn(1)=Verkley 2a, and Eqn(2)=Verkley 2b), which he pursues for the rest of the paper and gets to p192
‘the isotherm corresponds to the state of maximum entropy’
and
‘This also means that the dry-adiabatic profile TL(p) maximizing entropy under condition (2) always corresponds to a state with a lower column entropy SL than the maximum SP attainable with the same total energy: TL(p) cannot be in thermodynamic equilibrium.’
In other words, you can’t go from an isothermal profile to a lapse profile without energy input.
He states this explicitly in the abstract:
‘It can be rigorously shown, for example, that maximization of entropy under the conservation of potential enthalpy in an initially stable stratification always requires additional energy, and so cannot be accomplished in an isolated layer. Conversely, a temperature profile satisfying the second condition always possesses less entropy than an isothermal state with the same total energy, and so cannot be in thermodynamic equilibrium.’
The application of these results is that if one closes the lid on a flask which has an isothermal profile inside, a lapse profile can’t be produced unless energy is added. As the flask is insulated and isolated, then that can’t happen.
At least, not according to Verkley and Akmaev.
Help mods – the forum is eating links again! Try this?
http://onlinelibrary.wiley.com/store/10.1002/qj.209/asset/209_ftp.pdf?v=1&t=h38tkx7m&s=3764a2e08c27a79e1a9448f8c9ed6d224cf75ed7
grr, if that link doesn’t work just google
akmaev On the energetics of maximum-entropy temperature profiles
It is freely downloadable.
br1 – Ok. I see the new term “conservation of potential enthalpy”. This would be like conserving ideal gas energy Law 1 but .i.e. conserving the ability to potentially open the flask.? Not a law. If that is conserved, then where is it hiding when the flask (cv proxy) is closed? My 1st guess would be the ability to move the flask higher when it is closed doing work outside the insulated system. Could be a control volume issue making the difference. Have internet, will google.
Trick:
“I see the new term “conservation of potential enthalpy”.”
A new term but an old idea. This is Verkley Eqn(13), the whole discussion of Verkley 2b.
Apparently L is called ‘potential enthalpy’. I had never heard of this before (hey, I *am* learning from this exchange, thanks!) and there is precious little of use that I could find. Ball 1956 might have started it, but I can’t access that paper. That such a thing might be conserved requires considerable justification – Verkley repeatedly says that it is not fully justified but he thinks there might he some way. Akmaev also says it is a delicate subject which goes beyond classical thermodynamics, and both Verkley and Akmaev say that B&A made a mess of it!!!
br1 – “…Verkley and Akmaev say that B&A made a mess of it!!!”
Geez, just like the wonderful world of blogging while waiting for Lucy. Godot Act 2 “In the meantime let us converse calmly. For we are incapable of keeping silent.”
“In other words, you can’t go from an isothermal profile to a lapse profile without energy input.”
Density differentials with a gravitational field at one end would do it but to get such differentials in the first place one needs the capacity for increasing expansion with height as in a planetary atmosphere so that you get a circulation. A closed container is not a suitable analogy.
You also need the medium to become ‘lumpy’ so that some parcels are warmer and some cooler than others so that they can change position relative to one another within a gravitational field. That is provided by a rough planetary surface and / or rotation of the planet and /or a three dimensional shape so that insolation varies from place to place.
Once you have such a circulation involving decreasing pressure with height you get constant adiabatic decompression and compression as air rises and then falls.
You then get highest temperature and density nearest the source of gravitation and there is your lapse rate.
But what is doing the ‘work’ ?
I would say that work is done when a lower density parcel of air moves upward against the force of gravity but I’ve heard it said that since work takes up energy that would require gravity to slowly dissipate but it doesn’t.
Therefore I suggest that any gravity ‘used up’ in the work that produces the rising air is then ‘given back’ when the air descends again so that gravity never gets depleted.
So it is correct that you do need energy to create a lapse rate but the cost of that energy to the gravitational field is zero due to the constant recycling.
Energy is available, it is being used, but it is never lost because the atmosphere gives back what it utilises in a fully sustainable process.
br1 6:25pm – Akmaev paper is a very tough read. Verkley/B&A much more clear. Spinach time, you can skip this Akmaev and still live an enjoyable law abiding thermo life.
Examples of paper’s obtuseness: changed universal symbol of H for enthalpy to P, changes “non-isothermal” profile term to “dry, adiabatic” profile, writes “clearly” a lot and “well known” instead of showing the informed, critical reader what he means. No cartoons of control volume/constraints or example graphs or numbers.
Refers to Verkley eqn. 13 as the potential enthalpy introducing a new verbiage.
My initial impression is the paper completely 100% supports Verkley/B&A, better explains w/exact math, e.g. adds an improved exact math foundation for the isentropic case 2b to better model the “turbulence” (Verkley term) or “vigorous mixing” (my term).
Akmaev conclusion: “Formally, maximization of entropy in a closed system is always a conditional extremal problem. Applied to the atmosphere, the classical concept of thermodynamic equilibrium as the state of maximum entropy results in different temperature profiles depending on the constraints imposed.” Just like Verkley conclusion.
Then continues “This has been a source of confusion, especially in view of possible influence by the planetary gravity field.”
Trick and Exner both in unison erupt & stand in applause: “Bravo. Bravo. I’ll say”.
When obtuseness happens I figure the writer wants to hide something from the reader & is trying hard to do so. But not in this case. I searched and found your clip: “TL(p) cannot be in thermodynamic equilibrium.” My version of paper found is: Q. J. R. Meteorol. Soc. 134: 187–197 (2008).
In this statement IV p. 191-2, Akmaev is discussing if the two differently constrained columns have the same enthalpy then tells us one of the columns then cannot be at thermal equilibrium. This is quite obvious but he’s being rigorous. Check Table 1 in Verkley: the isothermal and non-isothermal solutions at thermal equilibrium DO have different column enthalpy H. Akmaev pointing out had they been the same, one of the two would be at thermal non-equilibrium, again this is obvious but he is covering all bases.
br1 continues clipping: “stable stratification always requires additional energy cannot be accomplished in an isolated layer”.
This is a tough one – the turbulence issue. The body of the paper seems to just be putting this on a solid math basis. Remember there is a gravity force across the cv for both columns so as the gas particles move they have energy INSIDE the cv. This is the energy he means & cannot be limited to isolated layer so turbulent or vigorous mixing happens across the small layers but NOT across extended ones, he writhes a little over that. I like my Chebyshev raw, how about you?
Akmaev also writes confirming support of Verkley/B&A: “It is not mere coincidence that the thermosphere is generally much closer to the isothermal state than any other atmospheric layer.”
Check Verkley Fig. 2 for the confirm: standard atmosphere changes from non-isothermal profile (i.e. dry adiabatic) to isothermal profile ~220K above 200 hPa due to molecular level effects & et. al. high altitude physics.
Recommend just need Verkley/B&A to get thru class, like the Prof. telling you not to worry about Akmaev in the last section in the chapter, won’t be on the test.
Nonobtusively yours…
everyone, 🙂 🙂
I love the chatter though I have to ‘fess I understand little of it. But it has reminded me to send off for Bohren and Albrecht which Graeff showed me and looked like a cracking good read. br1 “Verkley… say that B&A made a mess of it” – no they don’t quite. The problem is that BOTH B&A AND Verkley et al are talking from theory, whereas Graeff is talking from experiments.
I’ve just about done the next piece on the experimental work – but see I need to do another whole piece after that, on the theory and the way forward. I think it is important to build carefully. I am sticking to experiments first, theory after that, and only the minimum theory needed to embrace the experimental results in terms of already-accepted science.
Good on you, Lucy.
The world waits with bated breath (or at least I do!)
Trick:
“My initial impression is the paper completely 100% supports Verkley/B&A, better explains w/exact math, e.g. adds an improved exact math foundation for the isentropic case 2b to better model the “turbulence” (Verkley term) or “vigorous mixing” (my term).”
I agree! I even feel we are getting a little closer to a common understanding.
“TL(p) cannot be in thermodynamic equilibrium.”
Bravo! Encore!
“Akmaev is discussing if the two differently constrained columns have the same enthalpy then tells us one of the columns then cannot be at thermal equilibrium. This is quite obvious but he’s being rigorous. ”
Glad you find it obvious, but beware of the immediate consequence.
“Check Table 1 in Verkley: the isothermal and non-isothermal solutions at thermal equilibrium DO have different column enthalpy H. Akmaev pointing out had they been the same, one of the two would be at thermal non-equilibrium, again this is obvious but he is covering all bases.”
No, he is saying that the non-isothermal solution is NOT in thermal equilibrium! You can *always* find a column with the same H but has a higher entropy than the one with constrained L.
This is the meaning of
“This also means that the dry-adiabatic profile TL(p) maximizing entropy under condition (2) always
corresponds to a state with a lower column entropy SL than the maximum SP attainable with the same total energy: TL(p) cannot be in thermodynamic equilibrium.”
Maybe the confusion lies in the idea that because Verkley constraint 3 (=Akmaev condition 2) is at maximum entropy, it is in thermodynamic equilibrium. But Akmaev is pointing out that this is NEVER the case – the constraint itself has imposed that the entropy you get (while subject to that constraint) is always lower than the entropy you would get if the column was isothermal. The first sentence on p192 is
“In other words, the further a monotonic temperature profile is from the isotherm, the further it is from the thermodynamic equilibrium.”
This applies even when entropy is maximised under the constraint of fixed L (Verkley 2b, etc).
Equivalently, it is saying that the vigorous mixing/convection required to produce the constraint of Verkley 2b in the first place requires an energy source.
For a pot of soup, the energy required to produce convection is clear – the cooker. The soup transports heat around, cooling at the upper surface. For an atmosphere it is also clear – the sun heats the ground which heats the lower air. The atmosphere is basically simmering like a pot of soup, and likewise is cooling at the top (though more by radiation than evaporation in this case). When one turns off the cooker, ones sees the convection die down and stop, and the heat transport eventually stops. It is the lack of heat transport which defines *thermodynamic equilibrium*.
While the soup is simmering, it’s entropy is constrained to be less than what it would be in thermodynamic equilibrium, but this is only physically achievable by heat transport. So the soup is doing its best to maximise S, but it can’t go all the way to isothermal because there is a heat flow which mixes up the fluid and constrains L. The simmering can last for a very long time, as long as the heat source is applied, so in that sense it is steady state, but it is not *thermodynamic equilibrium*. According to thermodynamics, the heat source can’t last forever (even the sun will burn out), at which time the constraint L will be removed, entropy will increase so that everything will return to isothermal, and things will return to *equilibrium* (which implies no heat transport).
Verkley 2b describes things like Hadley cells and simmering soup, which can last a heck of a long time, but these are not thermodynamic equilibria as they necessarily imply heat transport.
I like Akmaev’s paper, because he teases out the heat transport issues much more explicitly.
Lucy:
Glad you are proceeding and enjoying the discussion. As I wrote to you, I would be delighted to help in any replication and I hope it works. I have nothing but goodwill.
“br1 “Verkley… say that B&A made a mess of it” – no they don’t quite.”
Well here are the quotes I was thinking of:
Verkley p932:
“Bohren and Albrecht arrive at their constraint 3 by starting with a constraint similar to 2′, and then
modify it in an approximate way, which in fact amounts to replacing 2′ by 3. This way of obtaining 3 can be criticized on the grounds that, had no approximation been made, one would have found an isothermal instead of an isentropic profile, which in itself shows that the approximation is problematic.”
Akmaev p190:
“Bohren and Albrecht (1998), apparently anticipating that P cannot be conserved exactly under condition (2), explicitly assume that the conservation of L approximately implies the conservation of P, as the ratio T/θ = Pi(p) does not change much for sufficiently close p1 and p2. However, this assumption is hardly defensible for any extended layer, because the maximum-entropy temperature profiles corresponding to the two constraints differ drastically.”
However, both Verkley and Akmaev proceed with the B&A constraint, so it is not like they flat out disagree with it, but they do say that B&A’s *justification* is incorrect. Verkley tries to justify it a bit better but admits he doesn’t really succeed, whereas Akmaev seems to make a better go of it.
Like others, I’m looking forward to the next write-up!
and two more snippets from Verkley to back up my last two posts:
Verkley 2b:
“For this particular form of [b]heating[/b] it makes sense to assume that the [b]integrated potential temperature[/b] θ remains constant as its main effect is to redistribute u… This is what the variational problem posed by Bohren and Albrecht (1998)—maximize S for fixed M and L—is [b]intended[/b] to express, even though a proper foundation of the latter constraint would [b]require further scrutiny[/b].”
Verkley Conclusion:
“Here, we have taken, following Ball (1956) and Bohren and Albrecht (1998), constancy of the integrated potential temperature as a single additional constraint 3, but this choice is of course [b]open for debate[/b]. In our view, this particular constraint still [b]lacks a solid physical basis[/b]”
Br1 12:26pm – “No, he is saying that the non-isothermal solution is NOT in thermal equilibrium!”
Then how? does br1 reconcile eliminating 1of2 Verkley/B&A profiles not being LTE to a single one that is LTE when Akmaev writes uses the plural in conclusion: “…concept of thermodynamic equilibrium state of maximum entropy results in different temperature profiles depending on the constraints imposed.”
Trick:
“Then how? does br1 reconcile..”
by reading the rest of the conclusion! Akmaev is bringing up your very point in order to clarify it.
Here is a fuller quote, with a bit of emphasis:
“the *classical* concept of thermodynamic equilibrium as the state of maximum entropy results in different temperature profiles depending on the constraints imposed. *This has been a source of confusion*”
then to get rid of the confusion he writes:
“Gibbs’s isothermal solution is the maximum-entropy state in a thermally-isolated (energy-conserving) atmosphere.”
quickly followed by
“On the other hand, *dynamical dissipative* processes dominating in the lower atmosphere ‘feel’ the presence of gravity, and may produce entropy subject to other conditions, such as conservation of the column potential enthalpy. The maximum-entropy solution under this condition is the dry-adiabatic profile.”
Note that he says the dry adiabatic profile is the *maximum-entropy* solution, not the *thermodynamic equilibrium* solution.
Note also he said this is due to “dynamical dissipative” processes, which effectively means heat transport, and so is definitely not a thermodynamic equilibrium state.
A little further on in the conclusions:
“Generally, a temperature profile that is monotonic (as a function of pressure or height) corresponds to a state of lower entropy than an isotherm with the same total potential energy (column enthalpy). Moreover, the further a monotonic profile deviates from the isotherm, the further it deviates from the thermodynamic equilibrium.”
so again he says that a lapse profile is not in thermodynamic equilibrium. It all looks reconciled to me, what do you reckon?
br1 – “what do you reckon?”
In the balance of the conclusion, Akmaev is telling us about a 3rd non-equilibrium monotonic profile which is the standard atmosphere profile (or any other transient real atmosphere T profile p2 to p1 at any given moment) that is dominated by lower entropy & heat transport therefore can’t ever be at thermal LTE.
His conclusion is none other than Verkley Fig. 2 showing the two ideal solutions and a 3rd real atmosphere solution somewhere in between containing effects of GHGs, aerosols, et. al.
A couple cartoons showing 2 ideal constraints, a Table and a graph in Akmaev would have been helpful. Instead, gotta’ see Verkley/B&A for that non-academic stuff.
Trick:
“In the balance of the conclusion, Akmaev is telling us about a 3rd non-equilibrium monotonic profile which is…”
not really. He mentions the solutions as follows:
“Gibbs’s isothermal solution is the maximum-entropy state in a thermally-isolated (energy-conserving) atmosphere” = Verkley 2a.
“*dynamical dissipative processes* dominating in the lower atmosphere ‘feel’ the presence of gravity, and may produce entropy subject to other conditions, such as *conservation of the column potential enthalpy*. The maximum-entropy solution under this condition is the dry-adiabatic profile.” = Verkley 2b and B&A. Note his reference here to conservation of potential enthalpy, Verkley Eqn(13), of which he comments earlier in the paper “The maximum-entropy solution under condition (2) is a dry-adiabatic profile (Bohren and Albrecht, 1998; Verkley and Gerkema, 2004).”. It is clear he is saying this is due to “dynamical dissipative processes” and that this is Verkley 2b, not 2c.
Next in the conclusions he writes:
“Comparative energetics and thermodynamics of *these and other* temperature profiles in a dry static atmosphere have been examined.” = may possibly include Verkley 2c, but Akmaev never explicitly reproduces the simultaneous constraints of Verkley 2c, he looks at stratified temperature profiles instead.
But no matter what the profile is, he says:
“some *exact* statements are available for a *wide class* of relevant temperature distributions. For example, in a stable stratification, the adiabatic profile cannot be reached via an entropy-generating process without an *additional energy input*”
and that
“the further a monotonic profile deviates from the *isotherm*, the further it deviates from the thermodynamic equilibrium.”
which clearly applies to both Verkley 2b and Verkley 2c. If you want to bring Verkley Fig. 2 in, then please do it here.
There’s no escape in saying that Verkley 2c requires energy input but Verkley 2b doesn’t. Please note that Verkley 2c is a maximum entropy solution just as much as 2b is!
This Akmaev paper is so obtuse, gives me a headache discussing it. I stand by my statements so far.
br1 – “note that Verkley 2c is a maximum entropy….”the further a monotonic profile deviates from the *isotherm*, the further it deviates from the thermodynamic equilibrium.” which clearly applies to…Verkley 2b.”
I observe the Akmaev paper is causing you some confusion – understandable. Verkley 2c makes no claim about max. entropy. 2c is just an attempt to model the standard atmosphere which is Akmaev’s monotonic.
No that clearly doesn’t apply to 2b. Geez. To translate the above Akmaev-ese into Verkley-ese correctly: “..the further 2c profile deviates from 2a the further it deviates from thermodynamic equilibrium.” Obviously but obtuse in Akmaev-ese.
[…] Comments Trick on Lucy Skywalker: Graeff’s…br1 on Lucy Skywalker: Graeff’s…Gerry on Frequency combs to join the hu…Trick on […]
Trick:
“Verkley 2c makes no claim about max. entropy.”
yikes! Have you read Verkley2c recently? Time to brush off the cobwebs.
Here is what he says in 2c, p935:
“The problem of maximizing S, now for fixed M, H, and L, leads to (Eqn(19))”
Clearly he is about to maximise S.
Compare Eqn(19) to Eqn(15) and Eqn(9) – they are formally identical. So when Eqns(15) and Eqns(9) give maximum entropy solutions, so will Eqn(19). All the solutions of Vekley have maximised S.
Going back to 2b, to understand it a bit better, note that after introducing Eqn(13) he says that B&A’s condition requires some comment, and the *first thing* he does is talk about heating. Indeed he says that if you don’t consider heating, dtheta/dt, the S maximisation procedure is meaningless!!!
Then he gets to Eqn(14), which is about heating, he says it will have a “nonvanishing right-hand side”, which means that the heating term J will be a non-zero value. One can consider whether the heating applies to the whole column, and whether the heat is redistributed around inside the column under “convective turbulent mixing”. He talks about heating some more with “For this particular form of heating it makes sense to…”.
Now compare to Akmaev who says that you can’t get to a lapse profile without heating. So this agrees absolutely 100% with Verkley.
Clearly with all this heating going on, one can’t say that the system in Verkley 2b is in thermodynamic *equilibrium*. But even in the face of all the above, you can still find the maximum S consistent with all that heating!
br1 8:26am – Look and ye shall find NO maximization of eqn. 19 in 2c. This is getting comical. Eqn. 19 is combining eqn. 9 from 2a and eqn. 15 from 2b.
“The problem of maximizing S…” refers back to that maximizing problem being considered in 2a and 2b! I kid you not, no spiders at work. The result of that earlier maximizing is combined in 2c eqn. 21. Which Verkley then plays around with to do some curve fitting to model standard or any 3rd 2c atmosphere profile NOT in LTE between 2a iso-LTE and 2b non-iso-LTE (translates to Akmaev’s 3rd “monotonic” profile). The range between the 2 LTE “endpoints” shown in Verkley Fig. 2.
As far as your discussion of heating, think about the vigorous mixing process going on in 2b for reasonable earth initial T&P profiles. What the authors struggle with (B&A uses “approximate”, Akmaev “Chebyshev”, Verkley “turbulent”) is the microscopic heat flow going on as in 2b nature forcing the column to climb up the entropy curve to the max. S at LTE. (Same as mechanical system settling down to min. energy w/g field.) This all happens w/in each cv with which Exner & Trick are ok.
The BIG difference in assuming idealism is that 2b air particles are ideally colliding w/o making entropy – as in the real world there really are elastic collisions that produce entropy climbing the hill and thus microscopic heat flows – this can’t be happening in 2b so the authors all struggle with how that ideal process makes entropy. Akmaev going to the extent of bringing Chebyshev into the fray for a precise and entirely 2b consistent answer.
Trick:
“Look and ye shall find NO maximization of eqn. 19 in 2c. This is getting comical.”
It certainly is! It seems your maths is as funny as your physics!
To understand Eqn(19), please understand Verkley’s earlier comment introducing Eqn(9):
“Using a Lagrange multiplier lambda, we can write the variational problem as (Eqn(9))”
In case you don’t know what Lagrange multipliers are for – here’s Wikipedia (reliable as ever 🙂 ):
http://en.wikipedia.org/wiki/Lagrange_multiplier
“In mathematical optimization, the method of Lagrange multipliers (named after Joseph Louis Lagrange) provides a strategy for finding the local maxima and minima of a function subject to equality constraints.”
Have a read of the rest of that article, it’s good stuff. For example you can see it is conventional that the function that is being maximised does NOT have a multiplier in front of it.
After digesting that, look at Eqn(19) again. Here is how Verkley introduces it:
“The problem of maximizing S, *now* for fixed M, H, and L, leads to (Eqn(19))”
where he puts Lagrange multipliers in front of H and L (but not S) in order to find, as Wikipedia describes it, “the local maxima and minima of a function subject to equality constraints.” Also note the word ‘now’ in his text which you egregiously snipped.
It is absolutely clear that Eqn(19) sets up the problem of finding the maximum S subject to the equality constraints that H should equal the measured H and L should equal the measured L.
Likewise, Eqn(9) sets up the problem of finding maximum S subject to the equality constraint that H should equal the measured H.
Eqn(15) sets up the problem of finding maximum S subject to the equality constraint that L should equal the measured L.
If you can’t agree with those last three statements, then the maths of the paper (as well as the physics) has escaped you. Can we at least agree on what those three equations (Eqns(9, 15, 19)) are doing? Not much point in continuing if we can’t.
br1 7:41pm – I know what the Lagrange multiplier (LM) is, who he was, what it is used for and how to use it. It does not escape me.
I kindly showed you that Verkley uses LM in 2a and 2b for the maximization process (of iso and non-iso) then he copies them to 2c & combines. There is no new maximization w/LM in part c – there can’t be b/c Verkley is showing a 3rd profile soln. not at LTE/max. entropy in between 2a and 2b which in the limit converges to either iso or non-iso LTE/max. S. Or kindly show me a maximization process I missed anywhere after eqn. 21 (a restatement just combining 2a and 2b soln.s).
Good evening gentlemen. I see the in depth conversation is reaching right into the heart of the issue, and that you have traded one mild jibe each this evening.
That is all. Carry on. 🙂
Ahh…yes TB. My exasperation leaked out a bit again. Too, taking time away from your fine book selection, Truesdell’s Tragicomical Thermo. I can’t resist to memorialize 2 of the starting quotes here:
1) “The historic development of thermodynamics has been…particularly susceptible to logical insecurity…and there has been no adequate reexamination of the fundamentals since.” Bridgman (1953).
2) “I select…fields of brilliant success like hydrodynamics, elasticity, and electromagnetism (and) one accursed by misunderstanding, irrelevance, retreat and failure. Thus I write of thermodynamics…” Truesdell p. 3, 1980.
Little did he know how this situation would progress as entropy relentlessly increased over the next quarter century as f(delta t). LOL.
Trick:
“There is no new maximization w/LM in part c”
You keep skipping Eqn(19) for some bizarre reason (I wonder why?). With your knowledge of what Lagrange multipliers look like, can you see any in Eqn(19)? Usually they have symbols such as lambda or mu.
“Or kindly show me a maximization process I missed anywhere after eqn. 21 (a restatement just combining 2a and 2b soln.s)”
There are only four eqns in section 2c, Eqn(21) is the third. It clearly comes from Eqn(20), not only because it is obvious but note that Verkley says “(Eqn(20) so that we should have Eqn(21)”. Pretty clear really. But Eqn(20) comes from Eqn(19), not only because it is obvious but as Verkley says “Eqn(19). That gives … Eqn(20)”. Unmissable once you see it. So Eqn(21) comes from Eqn(19), and is not a trivial copy-paste from earlier. But what is Eqn(19)?
That is the question.
I’ve clearly given my interpretation, what is yours?
It’s so rare to see science discussed in a sustained way at this depth where the protagonists manage to keep talking to each other. It’s a credit to both of you. I just want to see it continue – hence the gentle nudge.
Trick 6/11 12:54pm – “Look and ye shall find NO maximization of eqn. 19 in 2c.”
br1 6/12 7:56am– “You keep skipping Eqn(19) for some bizarre reason…? (I wonder why?)”
I can see the wonderment. Might come from: ““The historic development of thermodynamics has been…particularly susceptible to logical insecurity..”
br1 continues – “But what is Eqn(19)? That is the question.”
Trick 6/11 12:54pm: “Eqn. 19 is combining eqn. 9 from 2a and eqn. 15 from 2b.”
Truesdell 1980: “Thus I write of thermodynamics…a field accursed by…retreat.”
br1 continues: “I’ve clearly given my interpretation, what is yours?”
Retreating to Trick 6/11 8:13pm:
“…Verkley uses (Lagrange multiplier LM) in 2a and 2b for the maximization process (of iso and non-iso) then he copies them to 2c & combines. There is no new maximization w/LM in part c – there can’t be b/c Verkley is showing a 3rd profile soln. not at LTE/max. entropy in between 2a and 2b which in the limit converges to either iso or non-iso LTE/max. S. “
That 3rd Verkley profile in 2c (Akmaev’s monotonic) is the real atmosphere profile, shown as the standard atmosphere in Verkley Fig. 2.
There can be no further maximization in Verkley 2004 2c because here the paper seeks to tie the threads of ideal iso and non-iso max. S LTE profiles together to theorize the real standard atmosphere profile always operates somewhere in between these two ideal solutions below max. S.
The reality is the real atmosphere air column doesn’t ever achieve max. entropy S (except maybe in rare transient coincidence). This is a key point of the paper – placing iso and non-iso idealized profiles as the bookends for reality, the real advancement as the paper says.
Then Akmaev 2008 weighs in handing off the ball to fullback Chebyshev who smashes thru nature’s defense and carries the ball across the goal line w/o ever fumbling so far as I can see. Game over.
NB: Truesdell 1980 says thermodynamics didn’t get to be an adult science until somewhere around the early 1960’s. The caloric theory (heat being a physical entity fluid) not being formally dismissed so far as Truesdell can find before 1973.
Truesdell p.41: “…thermodynamics, from its beginnings, has been a sick science, its sores unprobed by conceptual analysis and uncleansed by logical criticism….caloric theory forbids an ideal gas to have constant specific heats will prove fatal to it.”
Trick:
“The reality is the real atmosphere air column doesn’t ever achieve max. entropy S (except maybe in rare transient coincidence). This is a key point of the paper”
Yippee! A statement we can agree upon!
Unfortunately I think the agreement will be short lived (as I disagree with most of your other statements), but common ground is always nice to have.
Staying on Eqn(19) a bit more, can we further agree that it *does* set up the problem to always find the max entropy achievable *under the constraints* of fixed M, H and L, but that the *maximised* entropy it finds does not in general equal the ‘absolute maximum’ entropy that is possible?
This is a slightly different formulation than your statement, but if we tread carefully we may make progress.
br1: “.. the *maximised* entropy (eqn. 19) finds does not in general equal the ‘absolute maximum’ entropy that is possible?”
The only way I can make sense of this is that some possible physics are left out of eqn. 19 that could lead to even higher ideal S entropy value if considered. Conceptually then, I can see that the ideal max. S computed for iso and non-iso “does not in general equal the ‘absolute maximum’ entropy that is possible”.
Now I point out either a) you choose to add a possible physics present in the standard atmosphere (GHG, aerosols, non-adiabatic walls) or b) you choose to add some possible physics NOT present in the real standard atmosphere. To eventually get to “the ‘absolute maximum’ entropy that is possible”
Remember to re-measure fairly:
a) standard atmosphere profile with only the possible physics you choose and
b) alien possible physics acting on the standard atmosphere profile.
Don’t see how this adds anything, but your turn here in our quiet little talk shop area that TB encourages. Effect is to change H/L curve and unique alpha I suppose.
Trick:
“The only way I can make sense of this is that some possible physics are left out of eqn. 19 that could lead to even higher ideal S entropy value if considered.”
In my mind it is simpler than that. I was agreeing with your statement “the real standard atmosphere profile always operates somewhere in between these two ideal solutions below max. S.”
My view revloves around the following two statements:
1, Atmosphere operates below max S.
2, Eqn(19) determines how to maximise S given constrained M, H and L.
It is the constraints which cap S from rising to a value of S that could be achieved if the constraints were different. That may sound tautological, but I think is an important point to consider in trying to answer the following questions: what are the limits on the constraints, and why do the constraints have the values they do?
So let’s look at the constraints again. We can justify a fixed M because there is only so much mass in the atmosphere. We can’t simply double the value of M in the maths without dealing with conservation of matter in practice. Similarly for H, if we take a column from ground level up to space then there is a fixed energy in that column, and we can’t arbitrarily change that energy without dealing with where that energy is coming from – there is after all conservation of energy to consider.
But what of L? That is where I see all the trouble and squirming in the papers. What does ‘potential enthalpy’ really mean? Why should L have a fixed value? Can I arbitrarily double the amount of L in a column of atmosphere – what law would it break?
The answer to that last question is that there is no *fundamental* law describing conservation of L. It is possible in a totally closed insulated container of gas that the value of L can change without any inputs and without breaking any laws. In this sense it is quite different to changing M or H which would have to be supplied from somewhere. But in the case where there is heavy turbulence, it so happens that L tends to a value that doesn’t depend much on what other parameters are doing. B&A tried to justify this, but according to Verkley and Akmaev they failed, Verkley tried a bit harder but admitted failure even in his own paper, whereas Akmaev may have got to some better justification (I haven’t had time to digest that yet).
The point being that it is the constraint on L which limits how close the entropy can get to its ‘absolute maximum’. The word ‘absolute’ is used here because M and H *have* to be constrained (at least for the atmosphere as a whole), you can’t get around them without breaking the fundamental laws of physics, or supplying heat or matter. But the value of L is much more contingent – how heavy is the turbulence? how strong is the external heating?
In a closed insulated container, if the gas starts off in a turbulent state, that turbulence will die down. The value of L therefore can spontaneously change in an isolated system. In such an isolated system, because L can change it will not be constrained, and so can no longer constrain S. S will therefore rise until it reaches a value constrained by H. This is a ‘hard’ constraint (conservation of energy), so S maxes out at *the* maximum value. Akmaev explicitly shows that the S of a system for a given H but unconstrained L is always greater than the S of a system for a given H and a constrained L. Therefore *any* constrained L system will have lower S than the equivalent isolated system with unconstrained L.
I hope you can agree that this is the case for Verkley 2c.
Lots of things above you might not like – have at it.
br1 2:36pm – “.. gas starts off in a turbulent state, that turbulence will die down..”
Not much free time today, or even thru the weekend, this is the big deal Verkley discusses in 2b & any reader has to come to grips with. B&A approximates. And Akmaev settles exactly with fullback Chebyshev. I come to grips with it & grok/write this as vigorous mixing of the molecules.
Sure, in the beginning there are pockets with more or less turbulence (or vigorous mixing call it what you will) in the closed control volume and then at LTE Akmaev shows us rigorously how the turbulence or vigorous mixing has become smoothly continuous at max. S which is non-isothermal eqn. of state PV=nRT and since it is the max. S point, isentropic. Exactly.
br1, either you or I are a bit confused. I keep reading Akmaev alongside you statement that “2, Eqn(19) determines how to maximise S given constrained M, H and L.” but H (enthalpy) *is* both P and L (see equation 1, 2, and first page and a quarter). The only place H is used is as scale height. It seems L is the enthalpy here, not H in this paper so you using H and L in a statement leaves me scratching my head. Are you sure you have the parameters in Akmaev’s paper correct? I agree with Trick that the way Akmaev wrote this paper has little to desire (it’s not easy to follow due to the phrases and re-definition of past used symbols) but I can follow his flow though.
From what I gather from these papers is that the enthalpy depends greatly on how you define the start point of the ‘system’, but I start it as a very tall, totally evacuated and insulated column with ‘x’ amount of gas (~10 Mg here) at the bottom in liquid or solid state and zero K. This is P & L in Akmaev’s paper. So you track the energy slowly added to the system until the bottom reaches and stabilizes at your target temperature, let’s say 288 K. At this point the column is either isothermal or has a lapse rate with the bottom warmer, hence, natural lapse rate.
The total energy added is ‘L’ (or P). If a lapse rate exists, it would take additional heat to bring the system to a constant 288 K with constant agitation so ‘L’ is at a minimum at equilibrium compared to an isothermal state.
Contrary, if it was found the stabilized column to be isothermal, a later lapse rate would be quite impossible without a constant movement of energy from the top to the bottom, which again is impossible internally or without external transfer of energy across the walls or the top and bottom. So one thing you seem to see is that a lapse rate result has lower enthalpy than an isothermal column.
Both Earth and Venus have mechanisms to decrease this natural DALR, water in Earth’s case, sulfuric acid rain in Venus’s case, some radiation from the surface in both cases, along with the accompanying evaporation and convection near the equators.
If you look at this from an isothermal viewpoint, isothermal is natural and the only reason any planet would have a lapse rate is that there is more energy constantly arriving at the surface and a constant loss of heat at the top, or movement of that energy back down to the surface, radiation here is assumed. In Earth’s case this seems feasible and that is what AGW proponents firmly believe is happening. But if you then turn and look at Venus with an assumed natural isothermal atmosphere, you would need a constant movement of energy from the upper atmosphere to the lower atmosphere, or solar input to the surface, to maintain this huge temperature differential. This would be a continuous cycle and some say this is due to “back-radiation” from all of the co2 in Venus’s atmosphere. But this continuous cycle of energy needed to maintain the ~8.5 K/km lapse rate over 94 equivalent Earth atmosphere thicknesses would be a perpetuum mobile and this where the true violation of the second law of thermodynamics creeps in. You cannot have such a cycle of energy from the atmosphere itself for little solar radiation is absorbed by Venus. In Earth’s atmosphere it is harder to see this same principle but it is there also.
So I keep looking where Akmaev may have made math errors in his derivations, it sure seems it is there somewhere. His explicit constraint equations and surface equation in his development of the Lagrange multiplier sure would have helped. You can assume what he used but it is an assumption.
That’s how I see it so far.
Hi wayne,
No time for a detailed reply at the moment, I’ll be back for your and Trick’s post later.
But a quick clarification, Eqn(19) is from Verkley. Sorry for swapping between papers, but Akmaev is more explicit on a few points (which I’ll return to next time).
P in Akmaev is H in eveybody else’s notation, this is enthalpy.
L is the same symbol across the papers, but this is NOT enthalpy. Akmaev calls it ‘potential enthalpy’, and this is quite a different beastie than enthalpy. It relates to ‘potential temperature’ which is quite a different thing than temperature. Of course there are relations, but examples of the differences are that potential enthalpy is not conserved in an isolated system whereas enthalpy is, and an isothermal column has increasing potential temperature with height.
So the behaviours of these quantities are not always intuitive.
More later.
“But this continuous cycle of energy needed to maintain the ~8.5 K/km lapse rate over 94 equivalent Earth atmosphere thicknesses would be a perpetuum mobile and this where the true violation of the second law of thermodynamics creeps in. You cannot have such a cycle of energy from the atmosphere itself ”
In light of that how would one explain the heating from adiabatic compression as air descends ?
And air MUST descend as much as it rises.
I’m coming round to the view that the atmosphere really doesn’t warm much from solar irradiation but rather from adiabatic compression as air is forced down in one place in response to rising air elsewhere.
Surface warming too would be a minor player because that simply supplies the energy for the initial uplift and the consequent decompression negates it in the rising column.
So we are back to what I suggested in another thread. Surface heating is cancelled out by subsequent decompression and at top of atmosphere solar energy in is cancelled by longwave out.
If surface heating and top of atmosphere heating are both cancelled out then the remaining heat from adiabatic compression is left over as a warming bias and all it is is the energy from the initial compression being constantly renewed and being locked into the system until such time as the sun goes out or the atmosphere dissipates to space.
Note that compression will only generate heat if heat from insolation has energised the gas molecules in the first place by lifting them off the surface. No insolation and gases just lie on the surface in solid form.
It isn’t a perpetuum mobile because rising air working against gravity is exactly balanced by falling air working with gravity and the movement (work) is driven by density differentials and uneven insolation which causes some portions of the atmosphere to be less dense and therefore lighter than other portions.
There is, therefore, no breach of the Laws of Thermodynamics.
“the only reason any planet would have a lapse rate is that there is more energy constantly arriving at the surface and a constant loss of heat at the top.”
That is exactly what does happen in any atmosphere with a circulation.
In the descending air column potential energy is converted to heat as compression increases towards the surface.
Quite simply the compression of the descending air transfers as much energy to the surface as is needed for radiation to space from the surface so as to create a radiative balance at top of atmosphere.
If there were no GHGs at all then ALL the energy that needs to get transferred to space from the surface will be delivered to that surface by adiabatic compression.
You would get a more turbulent circulation in a non GHG world but the lapse rate would be just the same and the top of atmosphere radiative balance will still be maintained.
The presence of GHGs means that the atmosphere needs to be less turbulent since some energy is radiated straight out to space by the GHGs and so does not need to be returned to the surface by adiabatic compression.
The concept of adiabatic compression needs to be incorporated into the planetary energy budget as a means of returning energy to the surface in place of so called downwelling infra red.
There is no such downwelling. What is actually being measured is the IR energy emanating from the air around the sensor which is being generated as a result of the compression process.
You don’t even need to propose any transfer of energy from air to surface. Instead what happens is that the temperature of the compressed air inhibits cooling of the surface until the surface acquires much the same temperature as the air.
However, over water surfaces any temperature variation that results from internal movement of the water will dominate for a time until the water cycle evens things out again back to the temperature dictated by surface pressure.
On Venus the sulphur dioxide cycle does the trick, On Mars the CO2 cycle. On other planets the methane cycle.
On any planet where any material can phase change at ambient temperatures that will help the equilibriation process. In the absence of such materials the planet simply adopts a more vigorous circulation to the same end.
Trick:
“Sure, in the beginning there are pockets with more or less turbulence (or vigorous mixing call it what you will) in the closed control volume and then at LTE Akmaev shows us rigorously how the turbulence or vigorous mixing has become smoothly continuous at max. S which is non-isothermal eqn. of state PV=nRT and since it is the max. S point, isentropic. Exactly.”
But it is not *the* max S point, it is only the max S point under the conditions of heating and energy dissipation, as he discusses in depth from Eqns(25) to (31). Hence the isentropic system (such as in Verkley 2b) is far from equilibrium, as he discusses in depth and states explicitly many times.
I have now re-derived all his equations. I appreciate a bit more the function w(p), and how this gives the whole of Verkley 2a, 2b and 2c and more, including unstable profiles with a lapse rate greater than the DALR. His conclusion across *all* these profiles is “the further a monotonic temperature profile is from the isotherm, the further it is from the thermodynamic equilibrium”, so again Verkley 2b is not in thermodynamic equilibrium.
His whole spiel from Eqn(25) to (31) is to explain how a stable lapse rate (which includes isothermal, Verkley 2a, high entropy) can get to a DALR (=Verkley 2b) which has lower entropy. He notes that “turbulence in a stable stratification (or forced convection) cannot be sustained without a continuous energy supply”. So energy needs to be supplied, and he introduces a *dissipation* term which will increase entropy in order to go from high entropy isothermal to low entropy DALR. The dissipation term takes up the slack in the entropy, but needs to be fed with energy to do so.
That the final isentropic state (=Verkley 2b) is thermodynamically unstable and will revert to an isothermal state once the continuous energy supply is turned off is stated in places such as “the further a monotonic profile deviates from the isotherm (note Verkley Fig. 2), the further it deviates from the thermodynamic equilibrium.” and “TL(p) (=Verkley 2b) cannot be in thermodynamic equilibrium.”
So all in all, Verkley 2b is lower entropy than Verkley 2a, is thermodynamically unstable and will revert to isothermal once the energy supply runs out.
A final Akmaev quotation:
“The classical isothermal solution by Gibbs is the *ultimate* state of *maximum possible* entropy under the conservation of energy … Contrary to what has been suggested in some previous studies, the presence of gravity does not change the outcome”.
“once the continuous energy supply is turned off”
The energy supply must be continuing as long as density differentials caused by uneven surface heating at the bottom of the column continue to give rising parcels of air with consequent decompression and exactly equal falling parcels of air with consequent compression.
In practice, the energy supply in an atmosphere open to space is the sun which causes density differentials in the first place by heating the surface and materials in the atmosphere differently at different heights and at different locations around the planet.
The presence of gravity is important because the same strength of gravitational field varies in its effect on air parcels of different densities, hence rising and falling at different locations.
Furthermore, if there were no gravity all the warmed molecules would drift off into space as soon as they changed from solid to gas.
So, turn off the sun and all the molecules congeal on the surface. Turn off gravity and they all drift into space.
Change anything else and all that happens is an energy redistribution within the atmosphere.
How does that sit with Verkley and Akmaev ?
Stephen Wilde:
“How does that sit with Verkley and Akmaev ?”
Sounds perfect to me.
Now for the argument – what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?
br1 9:47pm – “…what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?”
Reasonable earth gas T profile from surface ~1000 hPa up to ~250 hPa starts off with whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes. Verkley 2b/B&A prove the T profile will become non-isothermal, isentropic at LTE. Akmaev confirms Verkley/B&A and makes available some exact statements for relevant T distributions.
“what is the vertical temperature profile of a gas in a closed insulated container at finite temperature under gravity?”
“whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes.”
Sounds reasonable to me.
Trick:
“Verkley 2b/B&A prove the T profile will become non-isothermal, isentropic at LTE. Akmaev confirms Verkley/B&A and makes available some exact statements for relevant T distributions.”
You are incorrect about Verkley 2b, and Akmaev’s exact statements prove it.
I could repeat the exact statements, but I’ve done that already. You don’t address any of the points about them which flatly contradict your position, so I guess there is no point.
Stephen Wilde:
““whatever monotonic profile initial conditions the weather may be at the time the theoretical insulated container closes.”
Sounds reasonable to me.”
but what does the temperature profile become in the long term?
As it is closed and insulated, then according to Verkley and Akmaev, it will become isothermal no matter what profile it had when the container closed.
According to Trick the profile becomes the DALR no matter what profile it had when the container closed.
According to Graeff’s experiment it becomes about ten times the DALR, but only when convection is suppressed.
Do you have any suggestions as to how Graeff’s experiment could be the case?
For every million molecules of air there is an aerosol. Thermophoresis will induce a net force making the aerosol move to an area of lower temperature. The force of gravity is acting to concentrate aerosols and thermophoresis is inducing them upward to cooler areas where their mass, being greater than air molecules, returns them downward sooner but still retaining some heat from previous encounters with the warmer energetic molecules near the surface.
http://aerosol.ees.ufl.edu/Thermophoresis/section01.html
Don’t suppose aerosols might be causing the results found in these experiments?
br1
I’ll have to confess at trhis point that I don’t really concern myself with what might or might not happen once one closes off the container because at that point it ceases to become an adequate analogue for atmospheric behaviour.
The whole point about a planetary atrmosphere is that it is open to space and readily expands radially with virtually no resistance from external influences.
I know that the point of the Graeff experiments is to show that even in an enclosed container gravity will somehow create a lapse rate but to my mind that is unnecessary in order to account for the effect of gravity on gases under an open sky subjected to uneven warmig from external irradiation.
As I currently see it, one doesn’t need Graeff’s experiments to show any previously unobserved effect of gravity in order to explain the behaviour of planetary atmospheres.
Nonetheless if pressed on the point I would currently say that as soon as one encloses the container one removes the one feature that allows the circulation that could give rise to a lapse rate. That feature is the ability of the vertical column to expand radially with height.
The lapse rate develops because decompression towards the top causes cooling and compression towards the bottom causes warming and one gets neither decompression nor compression unless there can be radial expansion at the top so I would expect that in due course any lapse rate existing at the time of closing would disappear all other things remaining equal.
Reading more of the above somewhat belatedly I see that I’ve not said anything new in relation to Graff’s actual findings which appear to be different from what I and others would have expected.
To summarise for other newcomers to this thread:
i) Graeff’s experiments appear to show a lapse rate type temperature gradient in a closed container in the absence of convection within the container.
ii) That finding is contrary to ‘established’ science which requires convection to set up the lapse rate
iii) Graeff and others are doing all they can to exclude the possibility that the experimental result is flawed due to the influence of factors other than gravity.
iv) If gravity were to have such an effect it would need to be incorporated into current theories about how the actual lapse rate around planets arises.
However, I don’t see why we need Graeff’s experiments to explain observations whether he is right or wrong because the atmosphere is not a closed container. That feature makes all the difference since it allows convection and so far as I am aware all the numbers work well enough as per the gas laws and the standard atmosphere without needing an additional component from gravity at all.
What am I missing here ?
Has anyone considered that the process of getting to an isothermal state from the point where the container is closed might take a very long time and that Graeff might be taking his measurements before the process has been fully concluded ?
Or that external influences however well screened might actually prevent the final isothermal state from ever actually being achieved ?
Indeed, if the stabilisation period is long enough and the scope for disruption large enough even factors such as the rotation of the planet and the pull of the moon could prevent a completely isothermal outcome ever being achievable.
Everything around us is always in motion all the time. That could disrupt the outcome of the experiment whatever the level of screening.
Stephen 11:37am: “…one gets neither decompression nor compression unless there can be radial expansion at the top…lapse rate…disappear…”
Well that’s a new one. This top post has a closed air column pressure gradient caused by gravity – weight from above decreases with increasing height. There is a P(z) gradient.
PV=nRT. V,n,R constant so T(z) varies as P(z), z vertical dimension. No energy in/out. Can’t be isothermal. Move a thermometer up & down at LTE and check. Not sure why br1 can’t see that, gets lost in the context I guess.
Very easy really and 3 informed, critical recent authors prove non-isothermal obeys all the laws issuing tickets to some classic work (& 2 are even easy to read, ha). Yes, no radial expansion continuing work in/out but there you have it, lapse rate does not disappear.
Take away gravity from enclosure somehow, get no pressure gradient, P(z) then constant so T(z) = constant, becomes isothermal. Just that easy.
But yeah make V variable, allow energy in/out varying rotating big as earth, now need a supercomputer. Not easy.
br1 9:38am – “(Trick’s) incorrect about Verkley 2b, and Akmaev’s exact statements prove it.”
Understanding comes from a few small steps at a time. I’ll stick with it if you want. Clear out the context. Let’s start at the very beginning. Top post defines the analysis control volume. Again, reasonable earth assumptions.
1) Start by computing the total GHG-free air constant mass per unit area of a gas layer in a control volume between any two heights under gravity g.
Any problem with that starting point? I don’t grok latex, would put in the formula if I did. You can ref. B&A 1998 for the formula(s).
2) Add in the hydrostatic equilibrium pressure change with height in the gravity field.
Ok still?
3) Compute the total enthalpy per unit area of the layer realizing the layer possesses potential energy per unit area in earth’s gravity field.
This will look something like H = gas internal energy (or KE+PE) + (gas volume * boundary pressure w/environment). If you are still with me, or even not, I’ll continue.
4) From that, realize energy conservation (Law 1) imposes a constraint that total dry static energy is constant in the layer (within adiabatic control volume).
Following me? Anything incorrect? There are 9 more steps. If you want to see them let me know. They do get more – way more – difficult.
Trick:
In a container with a pressure reduction from bottom to top so that there can be expansion with height:
with circulation – lapse rate.
without circulation – isothermal.
because with no circulation the energy will travel through the material only by conduction which will equalise over time.
Graeff specifies no convection (at equilibrium) which means no circulation which should mean isothermal but his results show otherwise.
How does he ensure that there really is zero circulation ?
Is zero circulation possible in a container on a planet spinning through space ?
Gravity could be causing the non isothermal result not via its vertical pull on the materials in the container but by applying a centrifugal force inducing a circulation.
How could one screen that out from a surface bound experiment ?
“This top post has a closed air column pressure gradient caused by gravity – weight from above decreases with increasing height. There is a P(z) gradient.”
The pressure gradient alone has no necessary thermal consequence. Despite the different densities at top and bottom conduction alone would ensure an isothermal outcome eventually.
To get a lapse rate you have to have constant decompression and compression which requires a circulation between regions of differing pressures.
No energy is added or taken out but the energy that is present from the beginning gets redistributed with more going to the compressed regions to create a lapse rate.
Work is being done by the circulation and that work redistributes the available energy from the isothermal state or possibly prevents an isothermal state ever being achieved.
So if Graeff is getting any sort of lapse rate he has to have a circulation in there somewhere.
Stephen 12:18pm:
Of course you know I will say the real questions are: “..process of getting to a NON-isothermal state…will anything real ever prevent the final NON-isothermal state from ever actually being achieved?”..for the top post setup.
Either way you point out 2 tough questions driving experiments b/c thermo theory tells us precious little about time.
Stephen 12:31pm: “…with no circulation the energy will travel through the material only by conduction which will equalise over time.”
Not physical! This will violate conservation of total gas energy (enthalpy really) in Law 1 so theory says cannot happen. See step 4 above.
Why Graeff seeks to constrain convection escapes me. There is no need theoretically to do so in finding non-isothermal, isentropic point at LTE.
The time to LTE is not known b/c real heat flow overtakes theory at some unknown but experimentally determinable time. Graeff’s curves seem to indicate a short term and long term temperature trend difference; therein is a clue.
Stephen 1:12pm: “The pressure gradient alone has no necessary thermal consequence.”
Yes, there IS a necessary thermal consequence. To rephrase Lucy in 2nd post: “Alas if your theory is found to be against the ideal gas law eqn. of state PV=nRT, then I can give you no hope; there is nothing for it but to collapse.”
br1 9:38am – Continuing on to pinpoint just where br1 thinks Trick is incorrect:
5) From step 4 at 12:56pm, realize and compute the total entropy (S) of the layer over the height of the layer.
Need this eqn. because we want to eventually invoke 2nd law from which there is no escape. But we want to make sure we invoke the proper eqn. of state which doesn’t have height in it so…
6) Transform S integration from height to pressure by invoking hydrostatic eqn. above and ideal gas law eqn, of state: PV=nRT
7) Now just maximize this entropy S to find nature’s required ideal gas temperature profile over the height of the unit area layer from law 2, LTE is achieved at max. S.
8) To do that, confine the math maximization process to reasonable range of pressures found in earth’s atmosphere (roughly 1000mb down to 200mb ~80% of atmosphere) to make no serious errors.
With me still? Lost? Anything incorrect creeping in yet? 5 more steps to go. Increasingly difficult to grok.
Trick, I referred to the pressure gradient ALONE.
Thus referring me to the gas law is a non sequitur.Why do you think a column containing varying pressure cannot become isothermal in the absence of circulation.?
I don’t see how conduction leading to an isothermal state is not physical. Please explain.
As regards your 13 steps I must admit that my technical skills aren’t up to critiquing them but what I would say is that I generally find that if it gets that complex there is likely to be a flaw in there somewhere.
Nature invariably boils down to very simple concepts.
Stephen 2:15pm – “I don’t see how conduction leading to an isothermal state is not physical. Please explain.”
Ideal gas pressure can’t have a gradient ALONE, the eqn. of state controls all things.
The above IS a good question to ask, your expression is intuitive but incorrect. If conduction were able to do this, it would have to violate thermo Law 1. Conduction IS operating normally (as is convection) but neither can violate Law 1. You seem to understand how convection does not violate Law 1 – if something goes up, then something equal and opposite comes down. Ok, that means convection is cool with Law 1.
So what about conduction. Sort of the same thing: think if something heats up by conduction in the column something equal and opposite must cool down since no energy in/out.
This is proven formally & in step 4 where the gas enthalpy is held constant by invoking law 1. If the container is open, then H formula I posted at step 3) has the second term able to vary. However constraints change In this top post to a closed container, the H second term is constant so drops out to zero in the variation. Just left with varying H = gas internal energy (or PE+KE) term with height and hydrostatics tells us then is same as varying with pressure.
Stephen 2:15pm: “I generally find that if it gets that complex there is likely to be a flaw in there somewhere. Nature invariably boils down to very simple concepts.”
Slight mod.: “..there is MORE likely to be a flaw..”
Agreed, like the probe that missed Mars due to a simple unit conversion flaw.
Thermo nature boils down to the 0th, 1st and 2nd law, simple concepts, right? Why then can’t we all come to terms? All 3 simple concepts are invoked correctly in the 13 B&A steps along with the not so simple but really pretty easy eqn. of state ideal gas law b/c after all gas is with what we are dealing.
It is the math that gets hairy, especially in that inscrutable Akmaev paper.
“So what about conduction. Sort of the same thing: think if something heats up by conduction in the column something equal and opposite must cool down since no energy in/out.”
Well yes of course.
So in a sealed container, no convection, no energy added or removed, warmer at one end than the other energy will flow to and fro by conduction until it is all the same temperature.
Wouldn’t that occur whether or not there were a pressure gradient ?
How would it be a breach of the Laws of Thermo ?
Hi Trick,
Glad to take the discussion further. Nice approach to explain things – giving a list of the necessaries.
I got lost at step 7:
“7) Now just maximize this entropy S to find nature’s required ideal gas temperature profile over the height of the unit area layer from law 2, LTE is achieved at max. S.”
First question: what is ‘law 2’? You may say it later, but not clear yet.
Then a note: Akmaev explicitly states that the max. S attainable under certain constraint conditions is not the same as thermal equilibrium. One can have an entropy producing non-equilibrium dissipation term that allows the column S to be pinned at less than that of thermal equilibrium. However, let’s not get into that again at this stage of your presentation, so after dealing with the first question please continue.
“Wouldn’t that occur whether or not there were a pressure gradient ?”
To clarify, the temperature would be isothermal but because of the continuing pressure differential the molecules at the lower pressure would be carrying more potential energy than those at the higher pressure.
Now, that is the position with no circulation as proposed in the top post.
If one then introduces a circulation the molecules at lower pressure move to higher pressure so some of the potential energy converts to heat and the temperature rises as a result of compression.
Meanwhile the molecules at higher pressure move to lower pressure so some of their heat converts to potential energy and the temperature falls due to decompressio.
And then you have the lapse rate with denser molecules warmer than the less dense molecules.
The secret is that the changes in pressure swap potential energy for heat with the sign of the effect being opposite at each end of the pressure gradient so as to produce the lapse rate.
A simple redistribution of the available energy so no perpetuum mobile and no breach of the Laws of Thermodynamics.
Note that the whole process implies that with no circulation the equilibrium state must be or must be capable of becoming isothermal hence my suggestion that Gaeaff’s results are a consequence of a circulation of some sort remaining in his container despite his best efforts to exclude external influences.
Stephen 4:56pm – “So in a sealed container, no convection, no energy added or removed, warmer at one end than the other energy will flow to and fro by conduction until it is all the same temperature. Wouldn’t that occur whether or not there were a pressure gradient ?”
No – only get same temp. when no pressure gradient (i.e.no gravity). There IS convection in the sealed container under gravity since nothing in theory prevents some parcel of air moving up while an equal and opposite parcel moves down as you have earlier pointed out when discussing the real atmosphere. Gravity is a big game changer & only thing allowed to cross the closed control volume.
Why? More specifically with gravity field a closed, constant internal total gas energy (PE+KE) will flow to and fro by conduction, convection, turbulence, vigorous mixing, call it what you will – all physical and eventually given the adiabatic control volume settle down to non-isothermal gradient due to pressure gradient at LTE. Even at LTE, some perturbation might happen randomly or say nonrandom by a big heavy truck rumbling by Graeff’s house late in the night. System settles to non-isothermal again.
Stephen continues: “How would it be a breach of the Laws of Thermo ?”
With gravity & closed, 1st law of thermodynamics is violated when you say constant T. Energy can neither be created nor destroyed – simple, eh? Energy cannot be harmed but if you claim isothermal then energy will be destroyed as the gas particle climbs against gravity (isothermal temperature KE = same, PE goes up with h so non-constant total energy – against the law).
NB: this is the big deal about constraints. If change them to an open container, then get isothermal solution as the total energy is conserved by the air column doing work on the air column above and below (+/-PE = -/+ work on environment) & just the right amount of work to obtain isothermal at LTE. Very cool.
br1 5:31pm – “First question: what is ‘law 2′? You may say it later, but not clear yet.”
A: The 2nd law of thermodynamics, the one that humbles us all. But it is our friend to understand nature. Like friction is our friend, mostly.
br1 continues: “One can have an entropy producing non-equilibrium dissipation term that allows the column S to be pinned at less than that of thermal equilibrium.”
No. Get a ticket from the thermo police or a rap on the knuckles from my thermo prof. S cannot be pinned lower than max. allowed in any control volume, see the very humbling law right above. Not in the closed column of the top post or anywhere. Even in the universe control volume, a whole ‘nother can of worms.
Akmaev is showing how entropy can be produced by the ideal gas to climb up to S max. from a lower S. This is no easy matter, requires REALLY hairy math. You will see it when I reach the end of the B&A 13 steps. It is really step 14, but B&A merely announce it w/o proof. Akmaev formally proves how the entropy needed to get to max. S is produced mathematically.
Ok Trick.
The issue then boils down to whether the presence of gravity inevitably creates a circulation all else remaining equal.
The gravitational effect relies on density differentials arising first so that the gravity can act differentially on separate parcels of the medium thus allowing a circulation to develop.
In the atmosphere the density differentials arise from uneven insolation due to various factors but theoretically in a closed container with no energy added and none removed there should (after a while) be no such thermal unevenness, no significant density variations and so no circulation and the entire contents would become isothermal but the pressure difference induced by gravity at one end would remain.
The only difference would be that molecules at the less dense end would have more potential energy than, but the same sensible temperature as, those at the more dense end.
Thus my suggestion that a circulation is being induced some other way such as via centrifugal force from the spinning of the Earth beneath the experimental chamber.
I don’t see how Graeff could reliably exclude ALL circulation from his experiment and that circulation need be nothing to do with the vertical pull of gravity which for all intents and purposes would be even across the base of the container.Far more likely is a sideways pressure from the medium onto the walls of the container induced by the Earth’s rotation through the gravitational field.
Also possible is an unevenness in the energy distribution from molecule to molecule at the micro level which reacts with the gravitational field to create a circulation whatever Graeff might try to do to avoid it.
I also have in mind the possibility that molecules with different balances of heat and potential energy in a single container might set up a circulation betweeen themselves to destroy the isothermal balance that would otherwise occur and of course one could say that that is a direct effect of gravity as you suggest.
I think this is more a matter of getting the concepts right than doing the maths.
Then check out the concepts with data and the maths must follow.
This issue of constant compression and decompression in the atmosphere and in a closed container has been inadequately explored in the context of the effects of different gases on a planetary atmosphere.
To summarise:
I’m sure that br1 and myself and others are right to suggest that all other things being equal gravity alone does not give rise to a lapse rate.
However there is reason to think that you are right in suggesting that in the real world other things are never equal and the changes in other things inevitably interact with gravity to frustrate the isothermal outcome.
Basically, any circulation however small within the medium will prevent an isothermal outcome and introduce a lapse rate. A circulation is intimately connected to gravitational influence.
Simply proposing a ‘lack of convection’ as Graeff does is not enough. There must be no movement at all for the isothermal outcome to be achieved and I doubt that it is possible to construct a suitable experimental setup anywhere in the known universe.
Stephen 5:33pm: “…because of the continuing pressure differential the molecules at the lower pressure would be carrying more potential energy than those at the higher pressure.”
Yes. The light has clicked on. You are lawfully thinking through PV=nRT constrained by control volume unchanging total energy = KE + PE.
Stephen continues: “A simple redistribution of the available energy so no perpetuum mobile and no breach of the Laws of Thermodynamics.”
I am going to nudge your thinking along a little more now that you have puzzled it all pretty much correctly (give or take) thru to this point in 5:33pm post.
I am going to actually suggest what might seem to be some possible heresy: the theory of the closed, perfectly insulated ideal gas DOES allow a perpetuum mobile and no ticket need be issued by the thermo police. The 2nd law, immutable as it may be, is actually ok with that.
Q: Why? Puzzle it out or read below….
.
.
.
.
.
A: 2nd law of thermo allows for constant entropy. It outlaws a decrease in entropy in the control volume LTE. Now granted, this constant entropy only occurs ideally. Graeff’s experiment cannot be ideal, as such Graeff cannot construct a perpetuum mobile, entropy increases in the cylinder unstoppably to LTE. For some unknown time constant, the entropy increase ought to follow thru to non-isothermal but eventually 2nd law says entropy of the Graeff cylinder must settle at LTE with its heat environment b/c no perfect insulation exists. This of course may take a long while. Like this thread…LOL.
“b/c no perfect insulation exists”
Like I said.
And without perfect insulation gravity will always induce a circulation and hey presto a lapse rate.
But that is not a perpetuum mobile in the sense that energy is coming from nowhere.
Instead it is a perpetual recycling of the energy already available within a gravitational field. It is a closed loop with no energy gain or energy loss.
No depletion of gravity either because the work taken up by air rising against gravity is offset by work in the opposite direction when air falls with gravity.
Going one step further, if there is top of atmosphere equilibrium between energy in and energy out then none of the energy in the system and involved in the compression/decompression cycle can ever escape until the sun gets weaker or the atmospheric mass declines.
Once a planet has gained energy in the atmosphere from the initial gravitational compression then it is never lost, merely constantly recycled up and down continually being maintained.
And that is why planets with atmospheres are always warmer than S-B would predict for a planet without an atmosphere.
Stephen 7:08pm: “…the entire contents would become isothermal but the pressure difference induced by gravity at one end would remain.”
You were making good PV=nRT progress but fall off the tracks here. The entire closed, insulated contents would become isentropic and NON-isothermal. Pressure gradient begets temperature gradient at LTE.
Stephen continues to summarize: “I’m sure that br1 and myself and others are right to suggest that all other things being equal gravity alone does not give rise to a lapse rate.”
Think of the enclosed, insulated column no gravity: LTE becomes isentropic, isothermal.
Think of the enclosed, insulated column with gravity: LTE becomes isentropic, NON-isothermal.
Gravity alone DOES give rise to a lapse rate. This has been known in the literature for some time – at least since the mid-1990s. It was unknown classically by the thermo grandmasters though they suspected it. They just couldn’t do the integrations required since integral calculus wasn’t as developed and/or they had not fully mastered it.
Truesdell 1980 points out their handicap really was lacking of the integral & differential math side. He has humorous instances of Fourier running in circles trying to prove stuff page after page and even just announcing he had worked some thermo things out where there is no surviving record of him doing so.
I mean – geez, thermo history & all the famous names just reminds me of blog handles in extremely slow motion. Decades to a whole century of threads to get to non-isothermal solution.
Meanwhile, back at the br1 ranch, where I stopped with step 8): wherein pretty much as usual integral calculus is made easier due to some simplifying assumptions that allow progress to understand nature.
Step 9) Transform variables to make the integral calculation much more simple (math. Dept. instruction will help us here – nothing new in therrno).
10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.
11) To simplify, constrain this maximization process to a linear set of temperature profiles (pretty good for earth’s atmosphere up to 200mb by observing the ICAO standard atmosphere.)
12) The mathematical condition for entropy S to be an extremum is that its derivative vanish, set the derivative of this function to zero.
13) Do the algebra, find the nature of the entropy S extremum IS a maximum i.e. isentropic point & only with non-isothermal temperature profile.
Voilà. QED.
14) Prove that the assumption in step 11) was actually not really needed generally. Akmaev paper shows the work. B&A offer discussion only.
“Think of the enclosed, insulated column with gravity: LTE becomes isentropic, NON-isothermal.”
Without a circulation ?
My position is that it can only become non isothermal with a circulation due to the need to compress and decompress so that density differentials lead to a higher temperature at the end nearest the gravitational field by switching kinetic and potential energy around.
With no circulation as proposed by Graeff for a perfectly insulated container (which I aver is not possible) then you could theoretically have an isothermal outcome. After all, if one prevents exchange of kinetic energy for potential energy nearest the gravitational field then it would become isothermal wouldn’t it ?
Since I don’t think that perfect insulation is possible I favour a non isothermal outcome in the real word as per Graeff’s observations and your contentions.
It may be a semantic issue as to whether the non isothermal outcome is due to gravity alone. I think it is gravity plus a circulation working with and against gravity as a reversible adiabatic process and so if the circulation could somehow be suppressed the outcome would only then be isothermal.
Either way the system is isentropic as you say because there is no change in entropy in a reversible adiabatic process and if circulation is suppressed within a perfectly insulated system then there is no change in entropy there either.
Stephen 8:26pm: Non-isothermal “Without a circulation ?”
Well, that depends on what you really mean. There is nothing in the theory for the closed, insulated container that imposes “circulation” or demands “no circulation”. No thermo law needs “circulation” explicitly. Else explain it better.
By “circulation” my explanation is you mean the ideal gas molecules are free to move & mix naturally from top to bottom w/o constraint – no laws are broken, no enthalpies or energies are harmed. If so, that is certainly ok with ideal gas law, 0th, 1st & 2nd thermo laws.
Graeff’s introduction of the powder in the B74 air experiment is bizarre; reducing convection is stunningly unneeded to find non-classic non-isothermal profile. At LTE all I can suggest happens is that the powder has simply wafted or dropped to the bottom of the cylinder and lies there inert. Like dust on my bookshelf here.
“you mean the ideal gas molecules are free to move & mix naturally from top to bottom w/o constraint”
Correct, with the proviso that such movement would form a pattern (circulation) according to the cause, nature and scale of the disequilibrium between the container contents and the environment outside the container.
Since there can be no perfect insulation there will always be something that will cause such movement in the container thereby preventiing an isothermal outcome.
To achieve an isothermal outcome Graeff needs to eliminate not only convection but also ANY movement of molecules up and down within the container. I don’t think he will ever manage that.
If there is ANY conversion of kinetic energy to potential energy and back again then there will be a lapse rate
The cause isn’t gravity alone but rather movement within a gravitational field and the engine for that is not gravity per se but density differentials that can arise for multiple reasons. Gravity then acts differentially on the density variations to induce movement up and down thereby converting energy back and forth between kinetic and potential in different locations.
The lapse rate is the physical manifestation of the differing proportions of kinetic and potential energy held by individual molecules or groups of molecules in different places.
Trick wrote:
Graeff’s introduction of the powder in the B74 air experiment is bizarre; reducing convection is stunningly unneeded to find non-classic non-isothermal profile.
Graeff wasn’t going for that. He was attempting to isolate the effects of gravity.
At LTE all I can suggest happens is that the powder has simply wafted or dropped to the bottom of the cylinder and lies there inert.
IIUC, he used sufficient volume of powder to fill the tube. This does bring into question the properties of the powder. As noted, they may be filled with air, and even sealed.
One of the models I used some years ago was a vertical column, open at the top, but sufficiently tall that the pressure at the top is negligible.
In the Micro-Canonical Ensemble, it’s pretty easy to calculate the required height for a ‘gravitational seal’. If you assign all of U to one particle as PE, then KE + U = U, or KE = 0. From this, it’s trivial to calculate the height z for which T = 0.
It’s a lot more difficult in the Canonical Ensemble, but I think a similar approach still works. The probability of having particles with extreme PE needs to match up with the probability of the system having the corresponding internal energy.
Thinking about that last phrasing, I might actually be able to crank through the math on that.
Trick:
“10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.”
Constraint of energy conservation is Verkley Constraint 2, which he says gives an isothermal profile. As this doesn’t seem to be what you are after, maybe you could clarify what you mean here?
“No. Get a ticket from the thermo police or a rap on the knuckles from my thermo prof. S cannot be pinned lower than max. allowed in any control volume, see the very humbling law right above. Not in the closed column of the top post or anywhere.”
Sure it can – just think of a refrigeration compartment. This is actually a very good analogy for Verkley 2b and all the Akmaev stuff. Now read Akmaev Eqn(25) and onwards. If one imagines the isentropic profile as analogous to a freezer compartment, one can ask the question – how can one go from a high entropy state (room temperature/isotherm) to a low entropy state (a cold freezer compartment/DALR)? Akmaev even says “the dry-adiabatic profile corresponds to a state of lower entropy than a stable profile with the same column enthalpy” how can you ignore such a statement? If you simply write Eqn(25) (by analogy, ‘the freezer cools down’), then all you get is a breaking of 2lot, which nobody is happy about. So Akmaev points out that one needs a continuous energy supply and a dissipation term, which he calls epsilon_t. This term basically acts (in the freezer analogy) as the plug in the wall, the motor and the compressor/decompressor – it represents the additional machinery at work which can drive the freezer compartment’s entropy lower. After a while, the freezer reaches its ‘equilibrium’ temperature, and this is a steady-state where the entropy of the freezer compartment no longer changes, and the Lagrange multipliers give as a stable solution. By analogy, this corresponds to the isentropic temperature profile, which is the steady-state of the atmosphere in the presence of turbulence and dissipation power supplies. But the freezer compartment is lower temperature than room temperature, so it has lower entropy than what the compartment would have if there wasn’t a motor attached sucking power from the wall socket and driving up the universe entropy at an equal or greater rate. That the freezer compartment is not at thermal equilibrium (despite the fact that the derivative of its S is not changing) is easily noted when one unplugs the power supply or the power supply runs out. In that case it will warm back to room temperature (the profile TL(p) will go back to being an isotherm). Akmaev says as much with “This also means that the dry-adiabatic profile TL(p) maximizing entropy under condition (2) always corresponds to a state with a lower column entropy SL than the maximum SP attainable with the same total energy: TL(p) cannot be in thermodynamic equilibrium.”
I’ve grokked the above, I hope you can grok it too. Could be a good chance for Stephen to do a bit of grokking too.
Stephen 9:38pm – “To achieve an isothermal outcome Graeff needs to eliminate not only convection but also ANY movement of molecules up and down within the container. I don’t think he will ever manage that.”
Yeah, can only manage that in gas by turning off gravity. But my understanding at 1st was Graeff is trying to find the non-isothermal ideal gas lapse rate different from the real atmosphere lapse rate. Of course, I’m second hand – all I know of his work is what others have written.
Q. Daniels 10:07pm – “…(Graeff) used sufficient volume of powder to fill the tube…”
Yes, Lucy was going to double check and in post 2 found: “The innermost Dewar (1) of 1/2 litre was filled with a fine powder in order to eliminate convection currents and radiation between the inner wall surfaces.”
If truly ½ litre of powder is in the ½ litre dewar, then I no longer have any idea what the B74 experiment is trying to accomplish (conduction in powder?). But it is interesting discussing the columns of free ideal air that have no powder – the classic setups.
br1 10:37pm: “Constraint of energy conservation is Verkley Constraint 2, which he says gives an isothermal profile.”
Energy conservation is 1st law of thermo, hence used as a constraint in law abiding Verkley 2a isothermal solution and 2b non-isothermal so to be consistent with thermo law.
What you are trying to grok I think is the gas enthalpy is different between an open air column in 2a and a closed air column in 2b.
2a open: control volume gas enthalpy = internal energy + work on environment = constant1
2b closed: control volume gas enthalpy = internal energy + 0 = constant2
B&A steps are for a closed container so gas enthalpy constant2 = internal energy + 0 = KE + PE + 0 = ½ mv^2 + mgh + 0 for this step:
“10) Obtain the function for which the layer’s entropy S is maximum subject to energy conservation constraint in the layer.”
The rest will take awhile to parse; maybe as long as LTE. Whew.
br1 10:37pm – “Akmaev even says “the dry-adiabatic profile corresponds to a state of lower entropy than a stable profile with the same column enthalpy” how can you ignore such a statement?”
I don’t. Translation from Akmaev-ese: “..the isentropic closed column corresponds to a state of lower entropy than observed standard atmosphere profile with the same column enthalpy”.
Translate into Table 1 Verkley-ese. Isentropic H = 2.0106. Observed H = 2.0032. They are not the same enthalpy because the isentropic is closed container (remember H = KE + PE + 0 = const. = 2.0106) and the observed is open container able to do work on the column environment above and below (H = KE = const. = 2.0032) – see less H.
Now Akmaev says set the SAME enthalpy H. Lower the isentropic enthalpy of 2.0106 down to 2.0032. Akmaev says find S is now lower for the isentropic than observed profile. That’s all, look at S = 4.3769 for isentropic – I suppose it could end up lower than observed S = 4.3764 at same H of 2.0032; I’ll let you do the math.
Ok, then why? Well the observed is higher entropy S than lower S isentropic at same enthalpy H because observed has more entropy S from say galactic rays, GHGs like water vapor clouds & CO2, aerosols, et.al. than ideal isentropic Any problem seeing that (barring typo’s that is)?
Using a refrigerator just moves the air columns to say the North Pole surface T < 288K avg. Doesn’t change anything.