*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.

br1 said:

“Could be a good chance for Stephen to do a bit of grokking too”

I’m not sure which point I need to grok.

I’ve agreed with br1 and Almaev that if one could prevent convection (actually ANY movement up and down) then in theory one should get an isothermal column despite a pressure difference at each end.

However I don’t think that such complete suppression of movement is possible at a molecular level in a gas. For example, even the movement of the container through the gravitational field as the Earth spins would produce forces within the column tending to induce movement.

If one could achieve an immobile and thus isothermal setup with pressure discrepancies at each end the kinetic energy would be the same throughout but the molecules at the lower pressure end would additionally have more potential energy.

Trick rightly points out that that would offend the gas laws and of course it would because such immobility is characteristic of solids and not gases.

As soon as molecules at the isothermal temperature start to move towards the low pressure end they will lose kinetic energy (cool) but gain potential energy and in the process force molecules at the other end to move in the opposite direction and gain kinetic energy (warm) at the expense of potential energy.

That would then create a non isothermal temperature gradient from high pressure to low pressure with highest temperature where pressure is highest.

So unless Graeff can somehow get his gases to behave in an immobile fashion as if the gas were a solid then he is never going to be able to create an isothermal outcome.

As far as I can see the introduction of movement within the gravitational field as the critical factor rather than the presence of gravity itself resolves the problem between br1 and Trick.

If a scenario could be created whereby there is no movement then br1 would be right. If movement is inevitable then Trick is right.

Trick:

“energy conservation constraint in the layer.”

no, there is no ‘energy conservation constraint in the layer’ in Verkley 2b!

Verkley himself refers to this in phrases in section 2b such as

“Even when on the scale of the turbulence the heating rate J would be zero, this now needs not to be the case. Indeed, (14) is expected to have a nonvanishing right-hand side”

So heating of the layer under consideration is explicitly allowed.

One can also see this simply by noting that H is unconstrained in Verkley 2b! H is allowed to vary to any value consistent with the constraint that L has the measured value of L. So again, conservation of energy is very much not assumed.

Verkley points out it is this that B&A make a mess of:

“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”

So if you strictly follow B&A you get an isothermal profile!

Maybe this has caused confusion.

“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,”

no, he says it is lower than the isothermal profile, hence the isentropic profile is not in thermal equilibrium. Yet again I quote Akmaev p192:

“TL(p) cannot be in thermodynamic equilibrium” where TL(p) is the isentropic profile (=Verkley 2b). Check it out, Akmaev says this comes straight from his own Eqn(2) which is Verkley Eqn(13), it really can’t get any clearer.

“Using a refrigerator just moves the air columns to say the North Pole surface T < 288K avg. Doesn’t change anything."

The example of the refrigerator compartment is to point out a very clear example where S is pinned to a lower value of S than you would get if you didn't have "dynamical dissipative processes" (Akmaev p195) which can take up the entropy slack. The analogy goes as:

1, fridge compartment at room temperature vs fridge compartment at cold temperature

2, isothermal profile vs DALR profile

In each case one needs a power supply to reduce S, which becomes pinned at a value lower than what it would be in thermodynamic equilibrium. Once you turn off the power supply, the system reverts to 'true' equilibrium.

Stephen Wilde:

“I’m not sure which point I need to grok.”

don’t worry, I was just play around with the word ‘grok’, which is one of the better contributions Trick has made to this discussion :)

“If a scenario could be created whereby there is no movement then br1 would be right. If movement is inevitable then Trick is right.”

actually, we are both caught short in any case – Graeff’s experiment shows ten times the DALR!

Verkley Fig. 1: “The column is assumed to exchange no net heat with its surroundings…”. Verkley continues: “…the assumption of no net heat exchange, which is the essence of keeping H fixed…”. B&A figure for V2b: “adiabatic.. column boundary assumed to allow no heat in or out…”.

br1 9:56am – “no, there is no ‘energy conservation constraint in the layer’ in Verkley 2b!”

Emden 1926: ‘‘Periodisch wiederkehrende Irrtu¨mer’’. I agree with Emden (and Exner).

And building on Eddington in Lucy’s 2nd post, Trick: “…if your theory is found to be against the 1st law of thermodynamics I can give you no hope.”

br1 needs to better explain not being a law abiding poster.

Trick:

“br1 needs to better explain not being a law abiding poster.”

will do:

Note that your first two Verkley quotes lead up to Eqn(6), where he says:

“If in the variational process no net heat is to be transferred to or from the column, the enthalpy

Eqn(6) is to remain constant.”

which he uses in section 2a, Eqn(9).

But he doesn’t use this in 2b, instead he *only* uses Eqn(13) and Eqn(15). In these equations, constant H is not mentioned. That H is allowed to ‘float’ can be seen from Table 1 when comparing isothermal and isentropic profiles.

Furthermore in 2c he says:

“Keeping M and H fixed leads to a uniform absolute temperature (isothermal profile); keeping M and L fixed leads to a uniform potential temperature (isentropic profile).”

It’s the very fact that 2b doesn’t fix H that allows you to get a different profile.

Switching over to Akmaev, apart from all the quotes I’ve quoted numerous times, here is another one (Akmaev, p294, bottom right):

“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well, as suggested by Verkley and Gerkema (2004).”

It took me a few goes to grok that one, but it is important.

So one of the highlights of Akmaev’s paper, where he actually goes about pinpointing the physical processes which can reduce column entropy (by increasing it elsewhere via dissipation terms), does not conserve column enthalpy. Verkley noted this in his discussion on heating in section 2b.

Of course, there *is* conservation of energy, but it is just not conservation of energy *within the layer*. That B&A screwed up here is noted by both Verkley and Akmaev who say that B&A would end up with an isothermal profile if they followed their own instructions.

“actually, we are both caught short in any case – Graeff’s experiment shows ten times the DALR!”

Probably because the effect of the circulation (that he has been unable to eliminate) inside the container is proportionately ten times greater than the effect of the atmospheric circulation in the real world outside the container.

I don’t think he will ever be able to eliminate such circulation within any container.

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.

==============

Gravity is an energy source. There is nothing that violates thermodynamics if nature finds a way to make use of this energy source.

Molecules move. Under the influence of gravity their paths are modified from straight lines to curved lines. This curvature reflects a conversion between potential and kinetic energy.

When the sun evaporates water, and this later falls as rain and eventually powers a turbine, how is this any different? You are seeing the conversion between kinetic and potential energy by the sun, which we have learned to harness.

Gravity is energy. Harnessing gravity is not a violation of thermodynamics.

Ferd 2:51pm – “Gavity is energy.”

Strictly speaking gravity is a force field F = mg. When gravity is able to move mass a distance d then energy = F * d becomes part of the total energy in any control volume.

br1 1:01pm – “But (Verkley) doesn’t use (constant H) in 2b… “

Yes, Verkley does use constant H in 2b control volume. He must, it is the law. Even though no mention can you find. Some things (esp. in a condensed paper like this) are just assumed – the informed, critical reader might have to dig thru assumptions or cites, in this case the diggin’ leads to B&A cite.

br1 continues: “…constant H is not mentioned…”

You can’t see it explicitly in the condensed Verkley writing cited from B&A but in B&A which you can’t see, they do use constant H in 2b. See step 4 for 2b derivation of non-isothermal closed insulated column above 6/18 12:56pm: “…realize energy conservation (Law 1) imposes a constraint that total dry static energy is constant in the layer (within adiabatic control volume). “

br1 continues: “It’s the very fact that 2b doesn’t fix H that allows you to get a different profile.”

No. Maybe I’ve said a few times in different ways, but grok it please: the control volumes of 2a and 2b are defined different. So Verkley gets different H in each. But in each control volume, 2a & 2b, H must be conserved – it is the law. H (enthalpy) is the total gas energy in the control volume, energy cannot be harmed or created, energy H must be fixed. Account for energy.

Let me say it again, energy H is fixed in 2a and 2b control volume. It is the law. But H can be different if the control volume is different and I have shown br1 why above. (Hint: find the eqn. for enthalpy H.)

br1 continues: “Of course, there *is* conservation of energy, but it is just not conservation of energy *within the layer*.”

Let me say it a 3rd time, energy H is fixed or conserved or constant call it what you will & yes, even within the layer. Any control volume you name, I’ll respond to say total energy is conserved in there, just gotta’ do accounting right. Sometimes that’s tough, dang tough.

“That B&A screwed up here is noted by both Verkley and Akmaev who say that B&A would end up with an isothermal profile if they followed their own instructions.”

No. B&A conserves energy in the control volume. Always. B&A follow the law(s). Perfectly.

Trick:

“the informed, critical reader”

mmm hmmm…

“Let me say it a 3rd time, energy H is fixed or conserved or constant call it what you will”

do I detect a touch of exasperation?

“B&A conserves energy in the control volume. Always.”

there is a teensy weensy problem with this. This itsy bitsy problem has been pointed out before, but seeing as it hasn’t gone away, let’s look at it again.

Here is what Akmaev has to say about B&A (Akmaev p190):

“Bohren and Albrecht (1998), apparently anticipating that P cannot be conserved exactly under condition (2)…”

ooh, what’s this? B&A can see that enthalpy cannot be conserved during the L optimisation process?

Let’s continue the quotation:

“…explicitly assume that the conservation of L approximately implies the conservation of P”

what’s this? B&A are conserving enthalpy? But really they shouldn’t?

And there’s more:

“…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…”

B&A’s position is not defensible? What about that law we’ve been hearing about?

But what difference does it make? Read on:

“…because the maximum-entropy temperature profiles corresponding to the two constraints differ drastically (e.g. Verkley and Gerkema, 2004, their figure 2).”

so you get a completely different profile if you actually follow B&A than what they are claiming?

Well, let’s see what profile we would get if we followed B&A properly. I notice Verkley was quoted, what is his view on the subject?

Let’s see – Verkley discusses B&A at the top of p932:

“had no approximation been made, one would have found an isothermal instead of an isentropic profile”

Sounds clear to me. OK, so let’s grok that all together – B&A enforce conservation of enthalpy, but this is not correct for the problem they are trying to address and leads to the wrong solution.

But surely nobody could claim enthalpy is not conserved, what about the law?

Well, Akmaev has this to say (Akmaev, bottom of p194):

“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well”

seems like you blipped over this quotation in your last post. Akmaev has identified a process that doesn’t conserve enthalpy. But there’s more if one finishes that sentence:

“as suggested by Verkley and Gerkema (2004).”

in other words, if one reads Verkley properly, he also says this.

Maybe a trip to Specsavers might come in useful?

Watching Trick and br1 go around and around, I’m not sure either has the math completely right, and maybe neither.

Can either of you reconcile your understandings with Graeff’s results?

br1 11:10am – Trick blipped over: “Akmaev has identified a process that doesn’t conserve enthalpy….if one reads Verkley properly, he also says this.”

Is br1 really prepared to blip down the path purporting the Akmaev paper has found the 1st law of thermodynamics has been overturned with some identified process AND with support from Verkley paper?

If so, please go into more detail and better ‘splain the Akmaev identified process along with Verkley support precisely. Show your work and where I fall off the tracks. Until you do, I’ve already shown how both papers along w/B&A correctly support & invoke the 1st law of thermodynamics.

Q. Daniels 12:35pm: “Can either of you reconcile your understandings with Graeff’s results?”

If Graeff’s 1/2L inner dewar in B74 is really filled with 1/2L of powder as Lucy writes, then no. The ideal gas law is not applicable as PV=nRT is not the eqn. of state anymore.

Trick wrote:

If Graeff’s 1/2L inner dewar in B74 is really filled with 1/2L of powder as Lucy writes, then no. The ideal gas law is not applicable as PV=nRT is not the eqn. of state anymore.I don’t understand this statement.

Within the beads, V and n are constant. Reading the Data Sheet, the beats account for 55 to 68% the total volume. The material of the beads accounts for roughly 5% of the volume, with the balance being air within the beads.

The space between the beads also has a constant, though irregularly shaped V. If the tubes are fully sealed, then n is constant as well. This volume also accounts for 32%-45% of the total volume.

Trick:

“If so, please go into more detail and better ‘splain the Akmaev identified process along with Verkley support precisely. ”

It’s not that 1lot is violated, it is simply a matter that the column can no longer be considered an isolated system. Energy passing into and out of the column changes the column enthalpy. At a smaller scale, energy passing into and out of a particular layer is absolutely necessary if one is to redistribute potential temperature in an optimised way. This is said so many times across Verkley and Akmaev that I won’t quote it all (and I’ve probably already done so), but here are two quotes to note:

Verkley 2b, p934,

“Even when on the scale of the turbulence the heating rate J would be zero, this now needs not to be the case. Indeed, (14) is expected to have a nonvanishing right-hand side”

Akmaev says a lot more about heating, but here is one, (Akmaev p188):

“It may be shown, for example, that for any initially stable stratification, maximization of entropy under condition (2) always requires additional energy, and so cannot occur spontaneously in an isolated layer.”

Note that ‘initially stable’ includes isothermal. ‘Additional energy’ clearly refers to energy input from outside the layer in cosideration, so this will not conserve the enthalpy of the layer. An ‘isolated layer’ is one that can have no energy transfer across it’s boundaries – such a layer clearly won’t do.

Another important one can be found on Akmaev p193

“Physically, turbulence in a stable stratification (or forced convection) cannot be sustained without a continuous energy supply”

So to get to the isentropic state, one needs a power supply. For the Earth’s atmosphere, this ultimately means the Sun’s input (note Akmaev’s discussion of radiation on p192), though at a local level it may manifest as turbulence due to, for example, “the reservoir of large-scale kinetic energy, for example by a background wind-shear instability.” (Akmaev p193). Such a process can dump energy into a layer, changing it’s enthalpy and sending it to the isentropic lapse rate. It is clear though that such a situation can hardly be described as ‘thermodynamic equilibrium’!

Grok or non-grok?

Q. Daniels:

“Can either of you reconcile your understandings with Graeff’s results?”

I certainly can’t.

But I also can’t find a flaw in Graeff’s experiment. I would love for the experimental results to stand, so I keep looking for possible ways to explain them. One reason I keep up the detailed discussion with Trick is that I want to learn all the little details about temperature gradients, and I have certainly learnt a lot since I came to this forum. For that: thanks!

br1 11:09am: “It’s not that 1lot is violated, it is simply a matter that the column can no longer be considered an isolated system.”

NO! Akmaev is tough to read but he is not telling us here what you write. The column in 2b CAN continue to be an isolated system non-isothermal, isentropic ideally. See the 13 steps above. They are cool with all laws so Akmaev is ok with that (he says in the conclusion).

br1 continues: “Akmaev says a lot more about heating, but here is one, (Akmaev p188): “It may be shown, for example, that for any initially stable stratification, maximization of entropy under condition (2) always requires additional energy, and so cannot occur spontaneously in an isolated layer.””

Watch the pea under Akmaev’s thimble

verycarefully. You are not doing so. Here Akmaev is writing about any initially stable column including one such as V2b (or one in the monotonic atmosphere at large) BEFORE reaching max. entropy point – see: “maximization of entropy under condition (2) always requires additional energy and so cannot occur spontaneously in an isolated layer.”This is true – if additional energy cannot occur spontaneously in one isolated layer then another isolated layer has to give up energy. All authors agree on this point for 2b entropy rising up to max. entropy point at LTE. You even seem to explain this but I fear your context “outside the layer” means you think this is across the control volume – it is not crossing the cv. All this is happening

insidethe cv.This (inside cv) is confirmed in the very NEXT Akmaev sentence which you didn’t clip (blipped by it I guess): “Conversely, energy has to be released by any initially unstable layer to maximize entropy and satisfy condition (2).” All inside cv.

br1: “So to get to the isentropic state, one needs a power supply.”

Yikes. See my fears are confirmed. No, here you are wrong; there is no power supply needed in Verkley 2b or in any step 1-13 above, or any need for one in theory as Akmaev JUST WROTE.

Then Akmaev extends all this to say generally: “…with the same column enthalpy, a temperature profile maximizing entropy under condition (2) possesses less entropy than any monotonic temperature profile that is closer to the isotherm (i.e. has a smaller lapse rate), and so cannot be in equilibrium.”

Here Akmaev is telling us if a column of real atmospheric air happens to attain non-isothermal, at the adiabatic max. entropy, this max. entropy is lower than what can be attained when let loose in the monotonic profile (real atm.) & therefore it cannot be in LTE anymore and so moves to higher entropy towards the “isotherm”. This is not surprising, Verkley Fig. 2 shows us that IS the case.

You are getting closer to the “Aha!” grok moment that when you encounter something puzzling in Akmaev – you can stop and eventually parse an understanding. If stop too long, expect fullback Chebyshev to run you right over though, ha.

Trick:

Just to let you know I’ll be travelling for the next four days, so won’t be able to correct all your errors for the next while. Apart from the string of errors in the first half of your post there was some comfort in your second last paragraph where you seem to say an isothermal profile has higher entropy than an adiabatic one at the same enthalpy.

I’ll have to come back to the rest of your post on Tuesday, but I can’t believe you said some of those things! Please try and think of the consequences before I have to point them out to you.

Back in a few days.

right, sorry for the delay, where were we… Ah yes:

Trick:

“NO! Akmaev is tough to read but he is not telling us here what you write.”

Oh yes he is! (this could take a while).

“Akmaev is writing about any initially stable column including one such as V2b (or one in the monotonic atmosphere at large) BEFORE reaching max. entropy”

yes, I know that, glad we agree.

“This is true – if additional energy cannot occur spontaneously in one isolated layer then another isolated layer has to give up energy.”

This is nonsense – isolated layers cannot exchange energy – by the *definition* of isolated. However, I may let you away with this one, as I will presume you know better and just mis-spoke. But worse, you sound like you are simply asserting Akmaev Eqn(25) which I hope we all agree is rubbish, as Akmaev is at pains to point out.

“All this is happening inside the cv.”

not only. Think about what happens for a whole column – if the column has a lapse rate less than the DALR and is therefore stable, how can the whole column get to the DALR without input from outside?

“This (inside cv) is confirmed in the very NEXT Akmaev sentence which you didn’t clip (blipped by it I guess): “Conversely, energy has to be released by any initially unstable layer to maximize entropy and satisfy condition (2).” ”

umm, this refers to a low entropy unstable system that is available to give up energy – did you ever consider how power supplies are defined in thermodynamics??? If you have a power supply in the column then fine, but the situation of an initially stable column is not like this, the power supply has to come from outside the cv. Otherwise one is just asserting Akmaev Eqn(25) which breaks 2LoT. His whole point in discussing Eqn(25) is to *identify the power supply*.

“Here Akmaev is telling us if a column of real atmospheric air happens to attain non-isothermal, at the adiabatic max. entropy, this max. entropy is lower than what can be attained when let loose in the monotonic profile (real atm.) & therefore it cannot be in LTE anymore and so moves to higher entropy towards the “isotherm”. ”

YES! Something we can agree on.

Which is also why a DALR column needs a power supply not only to reach DALR but also to *maintain* the lapse rate. It is that supply which prevents it from being ‘let loose’ from its non-equilibrium state. Note that when the column is at a steady-state DALR, the active power supply doesn’t actually heat up the column or increase enthalpy any more, but it is needed to replenish the dissipated energy of the ‘dynamical dissipative processes’ which by their nature are dissipative. The clue is in the phrase ‘dynamical dissipative processes’. Once the dissipation equals the supply then the column steady-states at the DALR and the net column enthalpy no longer changes. This is the whole discussion of Akmaev after Eqn(25), but it does not mean that because the enthalpy no longer changes that the system can be closed and still maintain the DALR! Maybe you could share what ‘dissipation’ means to you?

br1 – “This is nonsense – isolated layers cannot exchange energy – by the *definition* of isolated….you just mis-spoke.”

No. Here the term “isolated” means only separate; isolated does not mean insulated. Of course separate layers w/in an insulated column cv CAN exchange energy.

Trick: “All this is happening inside the cv.”

br1 – “not only”

Yes, only. Again, Verkley 2b control volume is perfectly insulated, enclosed & shown precisely in the B&A cartoon figure.

br1 – “..the power supply has to come from outside the cv.”

Nope. Read the line right above. The cv is perfectly insulated; no power supply allowed from outside cv in V2b. Akmaev is just telling us the power supply comes from inside cv, isolated layers exchanging energy (heat i.e. your power supply) INSIDE the cv up to max. entropy point of LTE.

br1 – asks: “Maybe you could share what ‘dissipation’ means to you?”

br1 answers correctly (& I have nothing to add except the column enthalpy is not net of anything): “Once the dissipation equals the supply then the column steady-states at the DALR and the net column enthalpy no longer changes.”

br1 –“umm, this refers to a low entropy unstable system that is available to give up energy – did you ever consider how power supplies are defined in thermodynamics??? If you have a power supply in the column then fine…”

Power supplies are important in electronics, aerospace, et. al. So YES, I have considered them in thermo also. Here though, br1 is getting closer to the real crux of the matter.

All authors of interest here discuss the problem of the V2b column being at lower than max. entropy & initially unstable then climbing to the max. entropy point. How does that happen? Well, the energy to do so comes from the isolated layers exchanging energy w/in cv to dissipate any excess energy (heat) imbalance over the stable DALR. When that V2b 13 step process ends, a zero & unstable DALR is voila converted at LTE to non-isothermal, isentropic (max. point) stable column in perpetuity (ideally anyway) & heat no longer flows. Any random perturbation is damped out quite quickly since there is no power supply across the cv. How that happens exactly is proved by Akmaev mathematically. Quite formally, actually.

Folks that get conduction in solids confused with enthalpy/entropy in ideal gases cannot see this without additional study.

[…] UnderwayErl Happ: Climate changes – oh so naturallyMDGNN: Limits on the Co2 Greenhouse EffectLucy Skywalker: Graeff's Second Law SeminarEarth motion, LoD and SORCE, mysteryLucy Skywalker: Graeffs experiments and the second law of […]

Trick:

“Here the term ìisolatedî means only separate”

oh dear, redefining the meaning of thermodynamics terms – please don’t.

” Verkley 2b control volume is perfectly insulated, enclosed & shown precisely in the B&A cartoon figure.”

no, Verkley 2b says that heating is allowed, Akmaev confirms it, and both say that B&A screwed up.

“Akmaev is just telling us the power supply comes from inside cv, isolated layers exchanging energy (heat i.e. your power supply) INSIDE the cv”

Not when you apply the statement “According to statement I, a stable temperature profile cannot be rearranged to maximize entropy subject to condition (2) with no additional energy input” to a whole column. If you have a *static* stable *column* and want it to have an isentropic temperature profile of the same L, you have to add energy to it to increase H – if you only swap energy between layers inside the column (conserving column H) then you reduce its entropy, thus violating 2LoT (Akmaev Eqn(25)).

“Power supplies are important in electronics, aerospace, et. al. So YES, I have considered them in thermo also.”

then I’d expect you to recognise them a bit better. Shame you don’t seem to.

“All authors of interest here discuss the problem of the V2b column being at lower than max. entropy & initially unstable ”

no that is not the case. Akmaev spends more time dealing with how a *stable* profile can reach isentropic, as that is where the problems lie. He says “the isothermal stratification is also statically stable in a gravity field”, so a *static* isothermal column can’t reach isentropic by itself. So one needs “a continuous energy supply from the reservoir of large-scale kinetic energy, for example by a background wind-shear instability”. Maybe you wish to include this power supply inside your cv, but what if it is not in the cv? Winds aren’t always present. And of course wind-shear instabilities don’t come for free! They themselves don’t arise without power supplies (the sun being the most obvious). If for example, one has a small open thermos flask in a closed room on a windless day and you close the lid, do you think wind-shear instabilities will spontaneously arise inside???

Finally, if one considers what happens if one has used a power supply to get from an isothermal column to a ‘steady-state’ isentropic column – we find that “TL(p) cannot be in thermodynamic equilibrium”. So the isentropic profile will *not* last ‘in perpetuity’ as you say, but can only be sustained with continued application of a power supply to continually replenish the dissipated energy of the wind turbulence, maintaining steady-state enthalpy that way. Once the power supply runs out, the system reverts to isothermal, “maximization of entropy in a thermally-isolated atmosphere yields an isothermal temperature distribution”.

This is now so obvious to me that I feel I’ve nothing more to learn here. While it’s actually been a pleasure discussing this with you for the last few weeks, maybe it’s time to move on.

br1 wrote:

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 functionV_phi1=sqrt((-2*kB*T/m)*log(1-rand))

where ‘rand’ is an evenly weighted random number selected between 0 and 1.

I finally got around to checking this. I get a different ICDF for phi1, ending in:

log(1-sqrt(rand))

It lines up better graphically, but I don’t trust it. Do you have a reference I could use for checking my math?

Also of note, whether I use (1-rand) or (1-sqrst(rand)) I get a very interesting artifact. I’m plotting the entire distribution, and then cutting out the lower half of the population, and subtracting out the minimum KE (sqrt(v^2-vmin^2)). It’s decidedly not the same temperature.

I did not expect that result.

Q. Daniels:

Quick reply:

The PDF is

phi1=abs(v).*(m/(k*T)).*exp(-m*v.^2/(2*k*T))

This can be integrated to give the CDF via

http://integrals.wolfram.com/index.jsp?expr=x*%28m%2F%28k*T%29%29*exp%28-mx%5E2%2F%282*k*T%29%29&random=false

Note that the integral goes from 0 to v, as the initial distribution is only defined over positive v (due to it representing a solid wall where the particles leave the wall in the positive direction).

This gives

CDF=1-exp(-m*v.^2/(2*k*T))

Solving for v goes as

exp(-m*v.^2/(2*k*T))=1-CDF

-m*v.^2/(2*k*T)=log(1-CDF)

v^2=(-2*k*T/m)*log(1-CDF)

v=sqrt(-2*k*T/m)*log(1-CDF)

Replace CDF with a random number and hey presto.

About cutting off the lower velocities – this will surely change the temperature, but I’m not sure what you are doing with that – can you elaborate a little?

ah, the problem is in your last step. The last formula should be:

v=sqrt((-2*k*T/m)*log(1-CDF))

The distribution of sqrt(log(1-CDF)) != log(1-CDF).

My solution wasn’t correct, either.

I’m working in Excel, using the CDF form. It’s an easy way to visualize the entire population.

What I was doing was selecting a delta-V such that half the population dropped out, and subtracting out that much energy from each line of the remaining population. Using the wrong distribution produced a positive temperature gradient.

Using the distribution above, I get the exact same curve with a reduced population. No change in temperature.

The assumption I used in doing that was that the population was not interacting. That is, all particles which left z=0 with sufficient velocity reached z(1/2) with energy E(1/2)=E0-mgz(1/2).

br1, drop me a line to q_daniels_t-2l

at

yahoo com

Q. Daniels:

“ah, the problem is in your last step. The last formula should be:

v=sqrt((-2*k*T/m)*log(1-CDF))”

Yes, apologies for the copy-paste error. That is the formula I used.

“Using the distribution above, I get the exact same curve with a reduced population. No change in temperature.”

Yes, that’s what it does alright. So if one starts with this distribution under gravity, one gets an isothermal temperature profile. I’ve tried including collisions and intermolecular forces but still get an isothermal profile.

One thing I’m toying with is NOT starting with this velocity distribution. After all, it is like an ideal blackbody emission, all very well in theory but real objects are not perfect blackbodies. Haven’t got a temperature profile so far, but haven’t exhausted the possibilities yet. We can continue via email if you like.

br1 wrote:

One thing I’m toying with is NOT starting with this velocity distribution.I’ve been considering this. In particular, the beads interiors are Canonical Ensemble systems. The interiors of the beads appear to be about 1 micron in diameter.

At any given instant, the average velocity will not be zero. Over time, it averages to zero, but that’s not the same thing. Instead, the center of mass will be bouncing around the interior with an average energy of 25 meV. At its peak, the vertical component will be converted into potential energy and its temperature reduced by a few millionths of a degree. It will also deliver slightly more collisions to the top wall. Also of note, it spends more time near the peak than bouncing off the bottom.

That’s a really crude approximation of what’s going on.

The behavior of the center of mass (and the system as a whole) only differs from average because of gravity.

I haven’t modeled it out yet. This is just the math I can do in my head. Modeling is up soon.

This is a continuation of the discussion that started on

https://tallbloke.wordpress.com/2012/06/28/graeffs-experiments-and-2lod-replication-and-implications/#comment-28702

copy pasted here in full for convenience:

Trick said:

“br1 4:42pm – “…this is exactly the part that B&A screwed up.”

You haven’t even read B&A yet! Get a copy, you will trace thru & see they have not made a mistake. Give B&A a fair chance. No mistake in Verkley either, you are being run in circles with the sideshow terms & not examining the detail math steps just the summary in V&G. Akmaev confirms B&A and V&G, completely.

Here is where some of your confusion seems to be: Constraint 2 is constant energy = constant enthalpy. Constraint 2’ is also constant energy = constant enthalpy. General by definition in the control volume:

Gas Enthalpy = energy = 1/2mv^2 + mgh + work done on environment (i.e. pV) must be constant by Law 1.

When V&G use the prime on constraint 2 (i.e. 2’ in V2b) they are just distinguishing the third term (pV) which has a non-zero value in V2a because work (pV) is allowed on the environment and in V2b no work is allowed on environment so third term is 0.

Constraint 2 in V2a: control volume (cv) energy = ideal gas enthalpy = 1/2mv^2 + mgh + pV

Constraint 2’ in V2b: control volume energy = ideal gas enthalpy = 1/2mv^2 + mgh + 0

That last 0 in 2’ changes V2a to V2b allowing no work on environment, conserving enthalpy H and arriving at a V2b non-isothermal gradient larger but “remarkably” close to real atm. V2b same enclosed constraint as Graeff dewar B74.

So yes, replace constraint 2 with 2’ in V2b b/c the cv is changed to enclosed so no work on environment (now pV=0) & energy = enthalpy is still conserved constant in V2b, no external heating allowed, cv is adiabatic. Constraint 2 and 2’ conserve energy, constant H must happen in V2a and V2b. By Law 1.

All this is shown in detail in B&A – go thru it! It helps, I was confused with V&G too until I read the B&A derivation details cited by V&G.

br1 continues for V2b: “Note Verkley Eqn(15), you can mathematically see that enthalpy is not conserved.”

Geez. See your confusion? I can see V&G Eqn. 15 DOES conserve enthalpy in 2b as is derived from the 1st law and the 1st law requires conserve energy = conserve ideal gas enthalpy constant as I just showed (again!). Re-read this slow.”

Trick:

“You haven’t even read B&A yet! Get a copy, you will trace thru & see they have not made a mistake”

I’ve traced through Verkley who says:

“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”

What do you make of that?

Akmaev says that B&A setting enthalpy constant in their equivalent of V2b is their mistake. Please re-read slowly:

“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”

More details here:

https://tallbloke.wordpress.com/2012/05/29/lucy-skywalker-graeffs-second-law-seminar/#comment-26892

“Constraint 2 in V2a: control volume (cv) energy = ideal gas enthalpy = 1/2mv^2 + mgh + pV”

whoa, talk about confusion!

Internal plus potential energy is not conserved in V2a, because any work done by the layer will cool it, thus reducing 0.5.m.v^2. External work done is not counted in internal energy. V2a uses Constraint 2′ which has replaced Constraint 2. Constraint 2′ includes work done, Constraint 2 doesn’t. In a closed container, no pV work is done because the boundaries are fixed, hence Constraint 2 applies. What you have written is Constraint 2′.

“Constraint 2’ in V2b: control volume energy = ideal gas enthalpy = 1/2mv^2 + mgh + 0”

Not only is this constraint Constraint 2 (have you mistyped or do you really think this is 2′???), but this constraint is not used in V2b. V2b uses Constraint 3, which has replaced *both* Constraints 2 and 2′. “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 is made even more obvious by considering that V2c would be redundant if Constraint 2′ was already in V2b. Verkley has to *explicitly* write Constraint 2′ into V2c in Eqn(19) – confer with Eqn(15).

You also never dealt with Akmaev:

“Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well” when he is describing what “appears to be a characteristic, if not the only possible, explicit representation of a process transporting heat vertically and generating entropy under constraint (2)”

Enthalpy is not conserved in the V2b variational process.

And I thought br1 6/30 10:04pm really meant maybe it was time br1 move on.

Anyway, since br1 will not go look at the detail math in B&A to resolve br1 confusion I find it interesting to try ending this confusion, one small step at a time. When I re-read V&A & Akmaev to do so I appreciate their work even more. Tracing thru V&A won’t end br1 confusion; br1must go get B&A. It is a must to write B&A detail math in this blog avoiding confusion – br1 not doing so gets confused….easily.

br1 11:13pm – “a constraint similar to 2′, and then modify it in an approximate way, ….What do you make of that?…”

Here br1 is confused because cannot see that a constraint similar to 2’ is not 2’ exactly, it is only similar. br1 needs to read B&A math eqn.s for why that is (will find the difference is open and enclosed containers).

Yes a constraint similar to 2’ (what do you know! the similar one is constraint 2) will lead to an isothermal solution not isentropic profile. A constraint exactly like 2’ which I wrote correctly clipped 10:25pm (“have you mistyped?” No, not here at least) will lead to isentropic non-isothermal profile as shown by B&A, V&G and confirmed exactly by Akmaev.

br1 continues: “ 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…What do you make of that?”

Here is constraint 3): a constant integrated potential temperature. Ok, so B&A, V&G and Akmaev avoid the criticism of the approximation by doing the math w/3 showing the

exactsolution of an enclosed adiabatic column with 2’ enthalpy & it is exactly max. S isentropic non-isothermal at LTE. Akmaev adds clarity to the exactness of V2b non-isothermal isentropic and how that works w/o approximation. br1 needs to read the B&A math entirely & not out of context to reduce confusion.br1 continues: “You also never dealt with Akmaev: “Interestingly, it also appears impossible for a physical process described by Equation (31) to simultaneously conserve the column enthalpy (Equation (1)) as well.””

Yes, br1 has a continuing thimble and pea problem as I dealt with before. Here Akmaev develops eqn. 31 to show that V2b entropy can indeed climb up to a max. LTE for V2b and that V2b will not have the same enthalpy as V2a. Just exactly as I wrote & br1 clipped 10:25pm. This answers a question I tried to explain to br1 even earlier. br1 was confused about this and here finds his own answer but cannot see it. Oh well.

br1 11:13pm ends with shocker: “Enthalpy is not conserved in the V2b variational process.”

No shocker exists. Enthalpy 2’ IS conserved in V2b as it must be. Read B&A to see this in the math. It is a good read; br1 will reduce confusion.

Next.

I fear the problem is that it is simply not possible to suppress a circulation developing in a column subjected to a gravitational field and spinning in space around the centre of gravity of a planet.

That is bound to cause differential pressures at different points on the column wall and will force a circulation even if all other factors could be countered.

Once a circulation forms then the theoretically anticipated isothermal profile will be destroyed and I think that Graeff is measuring that effect.

Nonetheless it is true that gravity causes the effect, just not directly. The spinning motion is required to combine with the force of gravity to set up the temperature gradient via the circulatory process.

My first visit to this site,wich is quite interesting,but slightly esoteric,in that,all of the various”effects”pale into insignficance when compared to the major effects.It`s a bit like arguing how many angels can fit onto a pin head.Firstly you have to suspend rational thought and belive in a god.

So lets consider the major factors.

The 2nd Law applies only to a closed system. The Earth is definitely not a closed system,It receives an enormous amount of energy from the sun daily.This is done via EMR(photons),also it can lose heat into space via the same mechanism depending on the well know albido(and other reflective agents). Due to the Earths rotation all heat absorbing components,land,sea, and air are differentialy heated. This naturally will cause turbulance,as the temperatures”try” to equalise.As this is a continuous process there will always be atmospheric turbulance.The Earths rotation also will add to this,thus we have high and low pressue cells rotating in opposite directions.Other atmospheric components,cloud formation,Hadely cells,jet streams etc add to the tubulance.

The oceans cover 2/3 rds of the earths surface,and has the highest heat absorbing capacity,is subject to all of the same differential heating/cooling,so naturally there will be turbulance in the oceans, Also there are tidal effects adding to the turbulance.

There is also precession of the earths axis (Milankovitch cycles,closely associated with ice ages) that means the Earth never returns to the same position relative to the Sun. So that what is now the northern winter, in 130k yrs will be the nortern summer, the seasons (climate) are continuosly changing around the globe.

Finally as a critique to the climate change scientologists, yourr beliefs are a kin to that of a religion.ie it is based on suspension of need of provable scientific evidence.

Regards to all. Terry.

I would like to get some feedback,if you have time.