Gravito-thermal discussion thread

by Judith Curry

There has been much discussion of this topic on open thread, which is getting unwieldily.  Here is a new thread to continue this discussion.

957 responses to “Gravito-thermal discussion thread

  1. The Sun is our solar system’s supreme creative and sustaining natural force. It bathes us with warmth and light in ways we still do not fully understand. In its gravitational harmony with the Earth and other planets, it irradiates us with an awesome spectrum amidst a complex play of cycles. The Sun alone has the power to determine whether we live and prosper in that warmth or descend into an ice age of almost 100,000 years of lethal cold. ~John L. Casey, Dir., Space and Science Research Center

  2. Pierre-Normand

    Here are some essential references for this topic. If someone wishes to read only one, read the first. And for toying with a molecular dynamics simulator, follow the last two links.

    Coombes and Laue, ‘A Paradox Concerning the Temperature Distribution of a Gas in a Gravitational Field,’ Am. J. of Physics, 1985.
    http://tallbloke.files.wordpress.com/2012/01/coombes-laue.pdf

    Pekka Pirilä, ‘Kinetic gas theory for gas in gravitational field’
    http://pirila.fi/energy/kuvat/barometric_derivation.pdf

    Berberan-Santos, Bodunov, Pogliani, ‘On the Barometric Formula’, Am. J. of Physics, 1997.
    http://web.ist.utl.pt/ist12219/data/43.pdf

    http://scienceofdoom.com/2010/08/16/convection-venus-thought-experiments-and-tall-rooms-full-of-gas/#comment-81966

    http://physics.weber.edu/schroeder/md/InteractiveMD.html

    • That really is a generalized discussion considering most of the CO2 in the Earth’s atmosphere is around our ankles.

    • Pierre-Normand, I read the Coombes and Laue paper you linked to. I didn’t think it presented a very strong argument, mainly for two reasons:

      1. It claims to “restrict analysis to a one-dimensional model in which molecules can only move in the vertical direction”. This scenario would constrain all molecules to move along the same path, a path defined by a vertically oriented line. The molecules would not be able to pass by one another. At any given height there could only be at most one molecule. The molecules would exchange kinetic energy, and momentum, up and down the path and both would decrease as a function of height due to gravity. This result would be contrary to the paper’s conclusion.

      2. The analysis seems to ignore collisions (eg in the appendix), which would also affect the paper’s conclusion.

      As another point, I am not a big fan of the proof because it assumes a fixed M-B distribution over a finite vertical distance. I consider that starting with a premise such as this is begging the question somewhat. The M-B distribution is a theoretically derived distribution, derived from assuming an isothermal condition. It therefore becomes uncertain to me whether the proof is possibly starting with the assumption of a constant temperature over a narrow vertical band (dz) and then extrapolating it across the full height being considered.

      • Pierre-Normand

        Hi Willb,

        I had postponed discussion of this topic until there would have been an open thread mainly because of you. And then Judith created this special thread when it became unwieldy in the previous one. I’m glad you showed up!

        I think you are overstating the scope and significance of the simplifying assumptions in Coombe and Laue. The aren’t considering a 1 dimensional gas, but rather a three dimensional ideal gas (in a 3D box) while simply abstracting from the two horizontal components of the speeds (or momenta). They are demonstrating the vertical speed distributions corresponding to the isothermal case to be stationary, contrary to naive expectation. This result also is achieved much more directly (but less illuminatingly) from applying the Bolzmann distribution law to the full 3-dimensional case (as shown in section D of the “On the Barometric Formula” paper, and a couple other references that I provided earlier.

        Now let me make two more specific points in response to your objections. For an ideal gas in thermodynamic equilibrium, the rate of collisions isn’t relevant. The random collisions simply enable the gas to achieve equilibrium in the first place. After this has been achieved, the collisions can be ‘turned off’, as it were, and all the relevant statistical configurations (speed and density distributions at various locations) will be maintained. See further some of the comments at the end of Pekka’s note, regarding molecular collisions.

        Secondly, you are right that the vertical component of kinetic energy isn’t conserved in individual 3D collisions. However the three components of the momenta *are* conserved separately. The Coombes and Laue treatment is based on a consideration about two volumes in phase space that are shown to be equal. (This is an interesting special case of Liouville’s theorem). This result is true, independently, for any of the three 2D planar projections of the whole box (as does the Liouville theorem, more generally) because phase space is a space of positions and momenta, and both positions and momenta do factorize along the three spatial dimensions. Of course, after it has been shown that the momenta distributions are equal for all heights, and also for the two horizontal dimensions, then it follows trivially that kinetic energy distributions are equal too.

      • Pierre-Normand

        Rather more accurately, I should say that the Coombes and Laue 1D model readily generalizes to 3 dimensions because it is equivalent to a full 3D model where the two horizontal dimensions simply are abstracted (or summed over).

      • Pierre-Normand

        Willb wrote:

        “As another point, I am not a big fan of the proof because it assumes a fixed M-B distribution over a finite vertical distance. I consider that starting with a premise such as this is begging the question somewhat.”

        It is not question begging against their intended target, and this target is the claim that speed distributions *must* vary vertically because molecules collectively lose speed while rising against the gravitational field.

        “The M-B distribution is a theoretically derived distribution, derived from assuming an isothermal condition.”

        I think it rather derives from the assumption that entropy is maximized at equilibrium. It applies to the canonical ensemble of microstates of a mechanical system that is free to thermally exchange energy with an external thermal bath, and it is true that if the external thermal bath itself isn’t isothermal (with multiple points of contact at different temperatures) then it becomes unclear how the formalism applies to the system.

        However, proponents of a gravito-thermal effect take it to be realized in a fully isolated system and the finding by Velasco et al. — that the result for the microcanonical case is equivalent (at the thermodynamic limit) to the result for the canonical case (within a thermal bath) — is instructive. It would be queer that a temperature gradient spontaneously develops in an open system (with some mechanism that ensures no heat flows at the boundary though continuously adjusting the external gradient in the bath to match the gradient that develops in the system) and then slowly vanish again after the system is isolated.

      • Pierre-Normand

        “…that ensures no heat flows at the boundary [through] continuously adjusting…”
        Sorry.

      • Pierre-Normand

        I wrote: “Secondly, you are right that the vertical component of kinetic energy isn’t conserved in individual 3D collisions.”

        I am referring to a fact that you had highlighted in our earlier discussion and that I had overlooked at first.

      • Pierre-Normand, from reviewing the last few posts at CE I see you have attracted quite a large fan base :)

        With respect to the Coombes and Laue analysis, if their paper is not actually based on a one-dimensional scenario then I think they should state their specific assumptions explicitly. The claim that the paper is going to “restrict analysis to a one-dimensional model in which molecules can only move in the vertical direction” is otherwise somewhat misleading.

        I don’t see a problem with their abstracting the vertical dimension from the two horizontal dimensions to conduct an analysis. However, their analysis result is radically different from that obtained in analyzing a one-dimensional model. I think they need to explain why their abstraction is valid in the face of this difference.

        Concerning the M-B distribution, you say “I think it rather derives from the assumption that entropy is maximized at equilibrium.” You say potato and I say Solanum tuberosum. I still think assuming a fixed M-B distribution over a non-zero, finite vertical width is begging the question somewhat if your subsequent analysis results in an isothermal condition across the entire height.

        Concerning your point about “turning off the collisions”, I expect you could get away with this only if you “froze time” while deriving statistics. Moving molecules across a distance ‘dz’ to derive statistics, as Coombes and Laue do, would not be allowed.

      • Pierre-Normand

        Willb,

        For some reason I have a really hard time locating our previous short discussion on this topic. I’d like to review your original argument again before commenting further.

      • Pierre-Normand

        OK, at long last I found your earlier argument. I guess I was looking too far back:

        https://judithcurry.com/2014/10/25/week-in-review-30/#comment-641674

      • Pierre-Normand

        Willb,

        Let me run through some steps of your earlier argument.

        “3. In the horizontal direction, the mean atom velocity is zero ==> therefore all kinetic energy is thermal energy.”

        If the initial state at t_0 is isothermal, and the molecules have the same Maxwell speed distribution everywhere, then this is true also in the vertical dimension at t_0.

        “4. In the vertical direction, the mean atom velocity is shifted by gravity in the downward direction (i.e away from zero) ==> therefore some portion of the vertical KE is no longer thermal energy.”

        Coombes and Laue demonstrate that this can’t possibly result from the effects of gravity alone. Notice that if it were true for the entire molecular population at some higher level, at any time t > t_0, then the speed distribution would not be isotropic anymore and there would be a net downward motion of the gas at that level. (Since, by the same argument, molecules that fall to this level from higher up all have a *larger* average KE than they had at t_0, and hence also a preferentially downward motion). The density profile would not possibly be stationary.

        “5. At height H + Δh, some of the non-thermal vertical KE is ‘lost’ to PE. To re-balance horizontal and vertical KE and maintain equilibrium under equipartition, a portion of horizontal KE must be converted to vertical KE.”

        So, now, this doesn’t follow anymore since it depends on (4). Though particles that are able to rise from the lower level to some z level higher up lose some kinetic energy, the average kinetic energy of all the molecules that will have been able to end up at this level z at some time t+dt will have the very same isotropic speed distribution that they had earlier at any level (as demonstrated by Coombes and Laue).

      • Pierre-Normand,

        Just to remind us both, my previous argument was based on a simulation in which balls were bouncing in 2D space. Now, as to your comments on my argument:

        Regarding my step 3, I agree with your comment. I also think that the molecules, at any given height, will have the same M-B speed distribution in all directions.

        Regarding my steps 4 and 5, I disagree with your comments. Even though gravity shifts the mean vertical velocity in the downward direction, there is no overall, average downward motion of the molecules at any level. This is because, under equilibrium and at each height level, horizontal KE is instantaneously transferred to vertical KE through collision. The speed distribution remains isotropic through collision. However as height increases, the continual removal of horizontal KE to maintain this balance results in a non-zero lapse rate.

        IMHO the Coombes and Laue analysis has a hole in it because it ignores collisions.

      • Pierre-Normand

        Willb,

        Did you read Pekka’s note (which I listed in my reference list near the top of this thread? He deals with collisions. See also his recent comment in this thread:

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653204

        Consider the case of a diffuse gas where the mean free path is very much larger than the size of the box. The molecules would still collide once in a while. The spatial density and speed density functions of an ideal gas will be the same at equilibrium regardless of the collision rates (or mean free path). That’s because the partition functions of the phase space are unchanged by those assumption. It’s just the size of a volume in phase space, and the number of cells (defined by the relevant partition function) that it covers, that determines the probability for the system to realize a state that is located within that volume. It would just take a longer time for a gas to equilibrate in such conditions (provided only the initial condition with lower entropy nevertheless yields an ergodic trajectory).

        If you feel Pekka’s rather simpler explanations don’t fully address your issue about collisions, then feel free to tell me why.

      • Pierre-Normand,

        It will take me some time to fully absorb Pekka’s note. In the meantime I’d like to comment on the comment you linked to. First I will paraphrase the experiment description so you can see if I understand it correctly:

        1) There are three main components to his experiment: a horizontal planar surface, a gas above the surface and a gravity field pulling the gas towards the surface.

        2) The gas molecules are in a state of thermal equilibrium with an M-B speed distribution.

        3) The gas is extremely rarefied such that collisions can now be ignored when analyzing the motion of the molecules in the gas.

        4) Pekka claims that, even though collisions can be ignored, it can be shown that the gas molecules have the same speed distribution at all heights and in all directions, including along the axis normal to the surface.

        Is this a correct interpretation of the experiment? I have a feeling I’ve missed something because I don’t understand how Pekka arrives at his claim. For example, I get this paradox:

        Consider the subset of molecules at the surface, i.e. at height h = 0, at time t = 0.
        a) The equilibrium density profile establishes that there are more molecules at h = 0 than at any other height h.
        b) The equilibrium vertical speed distribution of the molecules is identical for all values of h, with a velocity for the mode (the maximum value of the distribution) of |V|_mv.
        c) From a) and b), there are more molecules at h = 0 with |V|_mv than at any other height h.
        d) At time t = T, the molecules that were at h = 0 travelling at vertical speed |V|_mv are now at h = H1 = ½gT² with vertical speed |V|_mv-gT.
        e) There are now more molecules at H1 with vertical speed |V|_mv-gT than with vertical speed |V|_mv. This contradicts b) above and the claim that the speed distribution is the same for all heights and for all directions.

      • … h = H1 = |V|_mvT-½gT² …

      • –Pierre-Normand, I read the Coombes and Laue paper you linked to. I didn’t think it presented a very strong argument, mainly for two reasons:

        1. It claims to “restrict analysis to a one-dimensional model in which molecules can only move in the vertical direction”. This scenario would constrain all molecules to move along the same path, a path defined by a vertically oriented line. The molecules would not be able to pass by one another. At any given height there could only be at most one molecule. The molecules would exchange kinetic energy, and momentum, up and down the path and both would decrease as a function of height due to gravity. This result would be contrary to the paper’s conclusion.–

        I would say in troposphere, that faster or slower molecules are not going up or down [or sideways or whatever direction]. And that faster or slower molecule is only “one” of the faster or slower for fraction of nanosecond. A molecule of air can travel through out that atmosphere- but faster one or slower one can not.
        Air packets are not the same molecules moving up [or whatever direction] a air packet are more like a memory of energy being transferred. It’s kinetic energy momentum which goes up [or whatever direction] not a molecules or pack of molecules moving, but the result of the past movement which moving.
        Once get into more rarefied air, then a faster molecules might be able to travel some distance for more than a fraction of nanosecond.

        My problem with article is:idea that kinetic energy of molecules decease
        with elevation. It’s only that density is reduced with elevation- or million molecules at 500 meters don’t have less kinetic energy as a million molecules at 5000 meter- assuming dry air/ideal gases are only involved. So 5000 meter air per number rather than volume have same kinetic energy- plus they have PE which is not expressed as temperature. It is only expressed if air falls- atmosphere as whole contracts. the 5000 meter air colder because there are less of them per a volume..

      • Pierre-Normand

        Gbaikie wrote: “My problem with article is:idea that kinetic energy of molecules decease with elevation. It’s only that density is reduced with elevation […]”

        That’s indeed what Coombes and Laue conclude. The isothermal profile (same kinetic energy distribution at all elevations) together with the barometric vertical density gradient constitutes a stable configuration.

      • David Springer

        Absolutely amazing that Coombs and Laue only got two citations since 1985mover 30 years for a paper that “proves” a controversy about gravity acting at the quantum scale that’s never been proven since proposed since 19th century golden age of physics. We still don’ t to THIS day have a theory of quantum gravity. Amazing that such definitive abstract “proof” can be asserted about an imaginary physical system that doesn’t actually occur in nature.

  3. Pierre-Normand

    David Springer wrote: “What you are overlooking is that the Boltzmann energy distribution will ensure that as molecules bump together at the same level there will be a continuous supply (but diminishing in frequency) supply of them with the energy required.”

    The Boltzmann energy distribution law as applied to an ideal monatomic gas column under gravity also yields an isothermal profile (with the Maxwell speed distribution for T at all heights) while the vertical density distribution conforms to the barometric formula. This is demonstrated at p.210 of Reif, Fundamentals of Statistical and Thermal Physics.

    https://judithcurry.com/2014/10/21/ethics-of-communicating-scientific-uncertainty/#comment-643688

    It is also demonstrated in section D of this paper:

    Berberan-Santos, Bodunov, Pogliani, ‘On the Barometric Formula’, Am. J. of Physics, 1997.

    http://web.ist.utl.pt/ist12219/data/43.pdf

    • Pierre-Normand

      Willb,

      I think you are overlooking that fact that the molecules that have the most likely speed |V|_mv at height H0 (h = 0) are going to sweep a layer of thickness dz at height H1 at a different speed than those from the same layer (which departed from H0 at another time) that are sweeping the same layer with the terminal speed |V|_mv. Hence they will not contribute equally to the speed distribution at H1 (since they don’t spend the same amount of time at this level of thickness dz) and this consideration ought to resolves your paradox.

      • Pierre-Normand

        Consider also the case where the level H1 is such that m*g*H1 > (1/2)*m*(|V|_mv)^2. That is, molecules that start up with speed |V|_mv at H0 don’t even have enough (vertical) kinetic energy to reach the level H1 before falling back down. Then, a fortiori, *those* molecules aren’t going to be more numerous than those that have speed |V|_mv at H1. There will be zero of them.

      • Pierre-Normand,

        Your first comment doesn’t make sense and your second comment is irrelevant to my argument that a paradox exists. I suspect you are probably too pre-occupied at the moment with JimD and perhaps we should continue this discussion at another time.

      • Pierre-Normand

        Willb,

        You are saying that constancy of speed distribution at all height together with a density gradient yields a paradox. That’s because, if I understand you correctly, you are saying that molecules at the peak of the speed distribution (where the probability density is largest) that reach a higher level H1 (where they will be slower) will be more numerous than those that have this very same initial speed at that level, and this contradicts the assumption that the speed distribution is the same. But you didn’t correctly account for the ratio of the number of molecules with those two speeds at the two levels (that is, the molecules that have speeds [v1, v1+dv], and [v2, v2+dv]) at levels H0 and H1. Here, I am using v1 for your maximum value ‘|V|_mv’ at H0, the level at the bottom of the box, and v2 is v1 – gt.

        My point is that if you want to compare the number of molecules that have speed v1 and v2 at height H1, in order to see if the ratio is the same as the ratio between molecules that have those two speeds (within the same dv intervals) at height H0 at any given time, then you must check how many of them there are at that level at some given time. However, in order to backtrack molecular populations in time, or track them forward in the future, you must define your populations precisely in phase space. You must not just delimit the initial or final speed ranges (e.g. [v1, v1+dv]) but also the initial and final volumes where they are located at some definite time (e.g. [H0, H0+dz] and [H1, H1+dz]). That’s because, molecules that start up with an initial velocity range in some small volume can end up (as it is the case here) in a larger volume at a later time, and if this change in volume isn’t identical for two different initial speed ranges, then this will incorrectly skew the ratio between them. Coombes and Laue do this correctly because they define volumes in phase space that jointly define velocity and ranges and location ranges, and map them exactly with the equations of motion.

        The explanation from shifting of exponential functions is simplest. When the Maxwell distributions of speeds (Gaussian) are the same, then so are the kinetic energy distributions. The latter are (negative) exponential functions of EK (Boltzmann energy distributions). The effect of gravity on rising molecules merely shifts the exponential function (probability distribution) along the KE axis. This causes a drop off of the molecules that don’t have enough KE to reach the new level, and accounts for the barometric density formula, but the new shifted exponential function that represents the new (normalized) distribution is invariant under this transformation.

      • Pierre-Normand,

        “You are saying that constancy of speed distribution at all height together with a density gradient yields a paradox.”
        Ans) Yes, that is correct.

        “That’s because, if I understand you correctly, you are saying that molecules at the peak of the speed distribution (where the probability density is largest) that reach a higher level H1 (where they will be slower) will be more numerous than those that have this very same initial speed at that level, and this contradicts the assumption that the speed distribution is the same.”
        Ans) Yes, that is correct. All of the molecules that leave H0 with a vertical speed component v1 will arrive at H1 at precisely the same time with a vertical speed component v2. The only thing I might change in your wording is to remove the word “initial” to clarify my meaning, since the distribution is constant over time.

        “But you didn’t correctly account for the ratio of the number of molecules with those two speeds at the two levels (that is, the molecules that have speeds [v1, v1+dv], and [v2, v2+dv]) at levels H0 and H1. Here, I am using v1 for your maximum value ‘|V|_mv’ at H0, the level at the bottom of the box, and v2 is v1 – gt.”
        Ans) I didn’t calculate the ratio explicitly. I showed that, at height H1 and at time T, the number of molecules with vertical speed component v2 is greater than the number of molecules with vertical speed component v1. This follows from my statement a) and violates my statement b). This is the paradox.

      • Willb,
        The mode of the vertical velocity is zero at every altitude. Thus molecules at the mode are at just at the point, where they switch from rising to falling.

    • Pierre-Normand

      …Coming to think of it, indeed, in three dimensions the Maxwell speed distribution peaks at some finite speed, but in one dimension, the distribution is Gaussian around zero. So, the mode is zero. This rather short circuits the formulation of your paradox. I should have thought of that first.

      http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/maxdev.html#c5

      • The velocity mode is 0, not the mode of the speed, which is the magnitude of the velocity.

      • Pierre-Normand

        “The velocity mode is 0, not the mode of the speed, which is the magnitude of the velocity.”

        There is a rather unfortunate tendency in kinetic theory to speak of “speed” to designate velocity (vector) rather than speed (magnitude). This use, however, matches the common usage in French (vitesse; vecteur, et vélocité; module). But here, it’s inconsequential, I think. When the distribution of the one-dimensional velocity (vector) along z is Gaussian, then the mode of speed (magnitude of velocity) also will be zero!

      • If the speed mode were 0 then no molecules would be moving at all.

      • Pierre-Normand

        “This use, however, matches the common usage in French (vitesse; vecteur, et vélocité; module).”

        I just checked, and it seems that I am wrong about the French usage!

      • Pierre-Normand

        “If the speed mode were 0 then no molecules would be moving at all.”

        Why? It’s just the peak of the probability distribution. For the molecules to be moving at v, where dv > v > 0 is more probable than them moving at any other speed +- dv/2.

      • I am assuming the speed follows a Maxwell-Boltzmann distribution.

      • Pierre-Normand

        Also, the reason why the Maxwell speed distribution has a mode larger than 0 for particles that are free to move in three dimensions in spite of the fact that the mode of each one-dimensional component of the velocity distribution being 0 is because of the properties of the volume integral, as explained in the hyperphysics item referenced earlier. Though it is most likely for the speed projection along any one direction to be around zero, it’s much less likely for it to be around zero along three orthogonal directions all at once.

      • Pierre-Normand

        Yes, that’s correct. The two expression are used interchangeably. “In physics, particularly statistical mechanics, the Maxwell–Boltzmann distribution or Maxwell speed distribution describes particle speeds in idealized gases where the particles move freely inside a stationary container” — Wikipedia

        Still, in one single dimension, the projection of the isotropic velocity vector distribution that satisfies this scalar speed distribution is a Gaussian distribution rather confusingly called the one dimensional Maxwell ‘speed’ distribution.

      • Pierre-Normand, thank you. I see now where I was going wrong.

      • Pierre-Normand

        Willb, you are welcome. And thanks to you for the probing questions.

      • P-N, “…Coming to think of it, indeed, in three dimensions the Maxwell speed distribution peaks at some finite speed, but in one dimension, the distribution is Gaussian around zero. So, the mode is zero. This rather short circuits the formulation of your paradox. I should have thought of that first.”

        I believe that is why a spherical shape has greater entropy since it more closely matches the shape of the speed distribution. So if you keep adding pressure to your box it would tend to want to become a sphere or a cylinder rounded at the ends.

  4. Pierre-Normand

    Captdallas wrote: “No, thermodynamic temperature isn’t entropy it includes entropy, 1/T=dS/dE. Ideal models eliminate entropy so processes are 100% reversible.”

    The models don’t eliminate entropy. They maximize it. All defenders of the gravito-thermal effect have taken it to apply to an ideal gas in thermodynamic equilibrium. It’s not my assumptions that the system is under equilibrium or that the effect applies to an ideal gas. It’s their’s. They don’t argue that it’s a real effect that stems from deviations from ideality. There is one person in the world (that I am aware of) who claims that the gravito-thermal effect only is exhibited in a real atmosphere that doesn’t satisfy those assumptions, and that’s you. But your theory isn’t well worked out yet. Indeed, everyone of your posts so far seems to have express a completely different ad hoc ‘non-ideal’ theory.

    • Pierre-Normand,

      What’s, in your opinion, the correct explanation for the vortex tube effect?

      http://en.wikipedia.org/wiki/Vortex_tube#Method_of_operation

      • Pierre-Normand

        Edim,

        I don’t know what the correct explanation is, but since the device requires the continuous injection of compressed air this seems to be a process that is far from equilibrium and hence has little bearing on the alleged gravito-thermal effect.

    • P-N, “But your theory isn’t well worked out yet. Indeed, everyone of your posts so far seems to have express a completely different ad hoc ‘non-ideal’ theory.”

      Indeed they do. As I mentioned before this thought experiment is a bit of a waste of time since the parameters can be changed. Call it “Thought-Ball” The real atmosphere is an open system the thought experiment is a closed system that can include any number of assumptions.

      My “theory” is and always has been that it is an un-physical problem, a waste of time. There is no such critter as an atmosphere in any kind of equilibrium. Energy is always flowing through an atmosphere making it a “conditional” steady state. You can use “local” thermodynamic equilibrium or hydrostatic equilibrium with proper respect, but they are both steady states with no definite source or sink when you just pick a parcel. You can’t just tap into an energy flow and declare you have usable source of energy because there is a temperature differential.

      Now when you say your model is maximizing entropy don’t you mean it is maximizing conductive heat transfer within a volume? Since there is no source or sink, just a volume, you are assuming conduction/convection always wins no matter how tall the z dimension is or how strong the force of gravity is. So there is never a chance of there being a temperature difference between the top and bottom because all of the Ideal relationships derived are truly “ideal”. I believe you have used the term “exactly” a number of times. To me that is just the opposite extreme of “declaring” you have found 2lot violation. In all likelihood there can be a small, completely unusable temperature differential due to gravity in a tall enough box under just the right cherry picked conditions.

      • Pierre-Normand

        CD: “Now when you say your model is maximizing entropy don’t you mean it is maximizing conductive heat transfer within a volume?”

        No. A system in thermodynamic equilibrium is, a fortiori, in thermal equilibrium. There is no heat flow anywhere in the system.

      • David Springer

        Yup. It’s a classic of example of arguing about how many angels can dance on the head of a pin.

      • P-N, “No. A system in thermodynamic equilibrium is, a fortiori, in thermal equilibrium. There is no heat flow anywhere in the system.”

        Again that is solving the problem by decree. The problem started with a thought experiment challenging Boltzmann’s H-theorem.

        http://bayes.wustl.edu/etj/articles/gibbs.vs.boltzmann.pdf

        That is a paper by E. T. Jaynes comparing Gibbs and Boltzmann entropies. Once you add gravity there will be some kintetic energy converted to potential energy which would create a density and entropy gradient. that should lead to a very small “theoretical” temperature difference. I have no idea if it would be measurable and I am pretty confident it would never be usable, but if the molecules are above 0K they will be transferring energy between themselves and a gravity field, Each one of those microscopic transfers would need to be 100% reversible for there to be no fluctuations of temperature within the volume.

        That lead I believe to the Fluctuation theorem and is a part of the Arrow of Time problem. Pretty obviously, the effect is very small so defining an “equilibrium” using the Boltzmann distribution and H-theorem would eliminate the effect since that is why the effect was proposed to begin with. So far in this conversation we have circular reasoning and angels dancing on pin heads.

      • Pierre-Normand

        CD wrote: “Again that is solving the problem by decree. The problem started with a thought experiment challenging Boltzmann’s H-theorem.”

        No, it doesn’t prejudge the relevant issue. Proponents of a gravito-thermal effect characterize this effect as a temperature gradient that *subsists* in a gas column under gravity at equilibrium. There is no vertical heat flow in that case either. If there would remain a net vertical heat flow then the vertical temperature profile would evolve and the system would not be in equilibrium yet.

  5. The ideal gas model explains the temperature profile of the troposphere, if you can specify a temperature/altitude point on the lapse rate curve. To do that, you have to consider the radiation balance for each layer of the atmosphere. This gives you an altitude where the Stefan–Boltzmann equation is satisfied (the equivalent emissions height). Now, you can plot the equilibrium temperature profile of the troposphere and begin to argue about the multitude of other things that affect it (the actual non-ideal, non-equilibrium state).

    The problem is that it is not an either/or proposition. Both sets of equations need to be considered, the layer-by-layer lapse rate, and the layer-by-layer, line-by-line radiation balance. Not an easy problem.

  6. The question of what’s the thermodynamic equilibrium temperature profile has perhaps been discussed sufficiently. What’s more important for the real Earth system, are the mechanisms that determine and maintain the circulation and the actual environmental lapse rate.

    Circulation can be driven only by an atmospheric heat engine that requires the hot end of low latitude surface and the cold end in the upper off-tropical troposphere. Solar radiation and GHE are essential ingredients of this layout.

    Isaac Held wrote two weeks ago a related post, where he lists some constraining factors:

    1) Rising air that has reached saturation follows the moist adiabat, i.e. cools at low altitudes less rapidly than the typical environmental lapse rate of 6.5 C/km.
    2) Total precipitation is constrained by the rate of heating of the oceans by solar SW.
    3) Subsiding air has an absolute moisture controlled by the minimum temperature that the parcel of air has met (the rest has condensed and been removed as precipitation).

    Rising and subsiding air get mixed. How that takes place must be an essential factor in explaining, why lapse rates seem to be more uniform at different latitudes (excepting polar winter) than one might expect from simplistic arguments that propose a significantly lower lapse rate in tropics and a higher one at mid latitudes, where subsiding air is common. These questions are related to the case of the “tropical hot spot” as well.

    • Pierre-Normand

      Pekka wrote: “The question of what’s the thermodynamic equilibrium temperature profile has perhaps been discussed sufficiently. What’s more important for the real Earth system, are the mechanisms that determine and maintain the circulation and the actual environmental lapse rate.”

      But that’s completely off topic! Nevertheless, I agree.

      Thanks for the pointer to Held’s post. Let me also refer again to this recent discussion thread on Nick Stokes’s blog:

      http://moyhu.blogspot.ca/2014/10/calculating-environmental-lapse-rate.html

      Pekka’s insightful posts within it are worth a read.

  7. Pierre-Normand

    CD wrote: “You battle that with statements like, (P-N)’The following paper does exactly that and finds out that the two effects indeed cancel out exactly when the vertical profile initially is isothermal and the density profile is barometric (as a consequence of the ‘drop off’ effect). Hence, this is a stationary state.’

    CD: “Amazing, the two effect “exactly” cancel out if there is a lapse rate.”

    You misunderstand. Isothermal means *no* lapse rate — i.e. the temperature is uniform. There is a *density* gradient. That’s not a lapse rate.

    You really ought to read at least the first few paragraphs of the paper if you want to talk about the topic knowledgeably. It’s a very simple paper aimed at college students and that targets heads on you own “basic premise of the gravito-thermal effect as [you] understand it”.

    http://tallbloke.files.wordpress.com/2012/01/coombes-laue.pdf

    • P-N, Apologies, that should have been no lapse rate. The main point was the “exactly” cancels. Since the average velocity in the +z only direction decreases with altitude. For a layer near the top, the average velocity would still be determined by the Kinetic temperature and as long as there is elastic collision, the M-B distribution would not change since it is not dependent on the direction of the velocities. The average -z velocity of the layer would be slightly higher than average. Given enough time and pressure the system will try to reach an isothermal equilibrium, but the velocity distribution in z is skewed toward -z. The density is skewed toward -z. So the heat capacity is skewed toward -z. To say it “exactly” cancels is a stretch in my opinion.

      • Pierre-Normand

        CD, read the paper. It’s a rigorous mathematical proof, and so it isn’t a “stretch” that the two effect that I have mentioned regarding speed distributions cancel exactly. Also, in all your replies to this argument, you only consider *one* of the two effects discussed in the paper (the asymmetrical variation in speed of individual molecules — both moving up and down) and display no awareness of the second one (the changes in the molecular populations due to some molecules not having enough energy to rise from a lower level to some higher level). How can you doubt that the two effects cancel out when you haven’t so much as acknowledged the second effect?

      • P-N, I have said a number of times that if the system reaches equilibrium, which can take a long time depending on how an experiment is set up, there is no usable work to be found inside but there could be a small measurable difference in temperature because energy is always flowing in the volume. There is no form of energy transfer that is 100% efficient so “exactly” cancelling would imply that there is no such thing as entropy. You can build as ideal of a container as possible to reduce entropy, but it can never be eliminated. so the problem will eventually boil down to does 1/ KEave=dS/dE. Because of that I would say “for all practical purposes, the system will be isothermal”, but I would not say it exactly will or exactly how long it will take to reach a “true” equilibrium.

        When I mention a number of things that could impact results, you respond by saying “my theory” isn’t well derived, when all I am doing is pointing out that “your model” requires perfection. Since I am pretty confident that perfection doesn’t exist I am equally confident entropy does exist meaning with enough perseverance some dolt can discover some higher order “effect” that might actually be “real” and large enough to measure in an un-physically large, “assumed” to be in perfect equilibrium, volume.

        I believe when the problem was originally posed, Boltzmann had derived,

        and that, ” In 1876, Loschmidt pointed out that if there is a motion of a system from time t0 to time t1 to time t2 that leads to a steady decrease of H (increase of entropy) with time, then there is another allowed state of motion of the system at t1, found by reversing all the velocities, in which H must increase. ”

        A jumbo sized paradox. So I would never use the term “exactly”.

  8. Although I have found Pierre-Normand’s comments to be the most focused, he and Capt. Dallas may have failed to identify the distinction between the kinetic and thermodynamic definitions of temperature.

    As I believe Pierre-Normand recognized, gravity does indeed impose a non-zero (but immeasurably small) kinetic-energy gradient in a thermally isolated gas column at equilibrium. So there’s a non-zero kinetic-theory temperature gradient. But, by definition of equilibrium, no net heat flow occurs, so under the thermodynamic definition of temperature there’s no temperature difference: the temperature gradient is zero.

    As this layman sees it, much of the confusion in these discussions arises from scientists’ failure to recognize that the several definitions of temperature, although experimentally indistinguishable, have slight differences theoretically: you can have a kinetic-theory lapse rate that differs from the thermodynamic one.

    • Pierre-Normand

      Joe Born,

      The alleged gravito-thermal effect that has been discussed on this site, but for a short intermission, isn’t the tiny effect on speed distributions discussed by Velasco et al. (in connection with the microcanonical ensemble). It’s rather the alleged effect that corresponds to the second alternative of the apparent paradox discussed by Coombes and Laue. This alleged effect is dozens of orders of magnitude larger in macroscopic systems than is the effect discussed by Velasco et al. The controversy over *this* effect (C&L) doesn’t have anything to do with any controversy over the definition of temperature. I just wanted to keep those two issues separate.

      • To the extent that “the alleged effect that corresponds to the second alternative of the apparent paradox discussed by Coombes and Laue” implies a measurable lapse rate, it is indeed different from the effect that Velasco et al. discuss. I haven’t yet been able to explain to myself, though, why the effect that Velasco et al. describe does not ultimately result from molecules’ losing kinetic energy between collisions as they rise.

      • Pierre-Normand

        Joe Born,

        The effect that Velasco et al. describe certainly does result from the slowdown of molecules as they rise. I didn’t see anyone deny this on the SoD thread. It was denied that there is a sensible definition of temperature such that we can claim a temperature gradient. I largely agree with that. But your own position isn’t completely indefensible either. If the perfectly isolated column for which the microcanonical ensemble formalism applies were suddenly put into thermal contact with a small thermal bath that has a well defined temperature T, either at the top or bottom of the column, there would be a higher probability that the bath placed at the bottom would be warmed up a little after just a few molecular collisions.

        This is most obvious in the degenerate single molecule case. The single molecule that was bouncing up and down in the isolated column would hit the bottom bath with a higher speed than it would have hit the higher bath (if it would reach it at all!). So, the resulting collision would *likely* result in giving up more energy to it (since that depends also on the speed of the molecule from the bath that it will hit).

        This probabilistic effect from one single molecule is just about the largest effect that would be exhibited in such a scenario, whatever the number of molecules in the fully isolated system was. This is why the gradient in average kinetic energy of the molecules is so tiny.

      • Pierre-Normand: “. . . . This is why the gradient in average kinetic energy of the molecules is so tiny.”

        I overlooked that comment previously. Excellent explanation.

        Not inconsistent with that, and possibly untrue because I haven’t thought it through completely, is my guess that the ensemble of microstates of the system consisting only of the single erstwhile-isolated particle is the same when the (finite) heat bath is coupled to it at a high altitude as when the heat bath is coupled a low altitude.

        Neither here nor there, I suppose, but thinking about it gave me a nice respite from dealing with estate issues in which I’m currently embroiled.

  9. Willis Eschenbach

    While it is rare to be able to prove anything in science, I believe I’ve proven that not only gravity, but no other such process can warm the surface of a planet with a non-radiative atmosphere (i.e. no greenhouse gases) to a temperature warmer than the theoretical Stefan-Boltzmann blackbody temperature. The proof is here</u).

    Don't like my proof? There is also a lovely proof by Dr. Robert Brown of Duke University here.

    Here’s the short version of my proof. Imagine a rocky planet with no atmosphere in outer space with no nearby stars. It is in radiative equilibrium with the ~ 3 W/m2 of the cosmic background radiation. This means that the surface of the planet is radiating 3 W/m2, and receiving 3 W/m2 of radiation from the cosmos. As a result, its temperature is unchanging.

    Now, add a non-radiative atmosphere to the rocky planet, one with no greenhouse gases. IF the crackpot gravity theories were correct, the surface of the planet would warm, and stay warmer than it was with no atmosphere.

    But at that point, the surface of the planet would constantly be radiating more energy than it is receiving … which is an obvious physical impossibility.

    Short answer? Gravity alone cannot warm the surface of a rocky planet.

    Best regards,

    w.

    • Pierre-Normand

      This is a perfectly valid argument Willis. But it will appear as a no-brainer to any atmospheric physicist or climate scientist.

    • Wow, is your transparent atmosphere made of atoms and molecules and held by gravity? If so I argue that this atmosphere will have its own lapse rate even if there were no suns and no cosmic background radiation. The energy that creates this lapse rate is completely unrelated to SB and radiation. There is no violation of conservation.

      • Willis Eschenbach

        gymnosperm, IF your situation warms the surface, then the surface will be radiating more energy than it is receiving. This is a violation of conservation of energy.

        To argue otherwise, you’d have to identify where the energy to warm the surface is coming from. Surely you don’t think that a lapse rate creates energy from nothingness?

        w.

      • I don’t know if it warms the surface. Doesn’t really happen on earth as the surface is almost always warmer.

        Think about your rocky planet alone. It has its own “lapse rate” in the rock as pressure increases towards its core. If some of this mass converted to thermal energy happened to be conducted to the surface and radiated away would you argue it was a violation of conservation? Mass and energy transmography. Energy = Mass x that weird square light thing.

      • Pierre-Normand

        Gymnosperm,

        It doesn’t matter if there is or isn’t a lapse rate. After a steady state is achieved, there will not be any conductive flow from the surface to the atmosphere since the atmosphere is unable to cool to space and doesn’t gain any energy from the Sun. So, the only energy flows will be from the Sun to the surface and from the surface to space. The latter only is a function of surface temperature. So the atmosphere is irrelevant to the steady state surface temperature when the atmosphere is not radiatively active .

      • Pierre, we took the suns and all external energy sources away. I argue that there will be a purely adiabatic lapse rate due to the mass of this transparent atmosphere just as there is non diabolical lapse rate in the rock of the planet itself.

        Let’s be trite and say the lapse rate in this atmosphere is from 1K at the surface to 0K at the top. You get radiation from this 1K that may warm the rock depending on the rock’s temperature or just zoom off to space because this is not a resonating atmosphere.

        I do not believe any of this violates conservation.

      • The atmosphere – not active in the IR bands by assumption – would be warmed by conduction and convection. The energy would accumulate until emissions again balanced incoming energy. You would need to know the makeup of the ‘atmosphere’ to have any idea of the atmospheric physics which is where it all becomes inuterably ridiculous.

        What you need to do is divest yourself of the notion that strange little thought experiments explain anything interesting at all.

      • Willis, a sideways digression, If the planet is big enough , Say 6 times the size of the sun what would happen??
        Basically it would become a sun right, as it collapses under the effect of the enormous gravity. This would then lead to heating up of the compressed atoms and a sun would be born. All powered by the effect of gravity causing heat.
        This overcomes your objections about the planet being a source of energy when as you say, small enough like the earth or a meteor There is not enough gravitational force to stoke up the engines.
        Tell me where I am going wrong.

      • “Basically it would become a sun right, as it collapses under the effect of the enormous gravity.” “All powered by the effect of gravity causing heat.” The height from which it collapses is potential energy. If a surface packet of the your planet were to fall to the center of mass which was small, when it hit it, its speed would be cashed in, probably as heat. The heat potential was already there. We just realized it. If there is water behind hydro power generating dam, gravity allows potential storage. A timing difference again. Where does the energy reside? Perhaps in the distance between the center of gravity and the water. Maybe it’s an acceleration potential? The potential comes from separating two things that attract each other. When they move closer together it’s a transfer, the creation of this potential came when they got further apart in the first place. Something banked the energy.

    • Willis,

      In your thought experiment, what sort of density curve of the non-radiative gases would you expect from the top of the atmosphere to the bottom? How are you measuring the “temperature”. Are you assuming an even density of the gas across the entire height of this atmosphere? This assumption would seem to indicate a certain conclusion as to the temperature at the surface of this planet versus the top of the atmosphere.

      • Pierre-Normand

        R. Gates,

        The temperature and density profiles of a non-radiating atmosphere would not have any incidence on the surface temperature. Since such an atmosphere would not radiate heat to space, then the conductive flux at the surface would tend to average to zero over diurnal/seasonal/etc. cycles at (average) steady state. Hence the surface would only warm enough to radiate to space with the same power received from the Sun. Whatever the atmospheric temperature profile would be (as measured by thermometers!) the temperature of the lowest layer would closely match the surface temperature.

    • Thanks Willis,

      Interesting and clear. Good enough for me.

      And like nearly all the down-in-the-weeds climate science is irrelevant for input to policy analysis.

    • Willis, is there any matter that does not absorb, at all, in the Far-IR and microwave region?

      • Willis Eschenbach

        Hey, Doc, good question. Yes, any mono-atomic gas doesn’t either absorb or emit thermal IR. The usual example is argon. Gases can only absorb and emit IR by flexing, stretching, twisting, or otherwise disturbing the bonds between the molecules that make up the gas … but being mono-atomic, argon has no such bonds.

        Argon can absorb and radiate at other frequencies, absorbing radiation by kicking an electron up to a higher orbit and radiating it when the electron drops down again. But thermal radiation at the temperatures found on earth doesn’t have enough energy to kick an electron out of orbit.

        w.

      • Pierre-Normand

        “Gases can only absorb and emit IR by flexing, stretching, twisting, or otherwise disturbing the bonds between the molecules that make up the gas …”

        …and rotating! (for molecules that have permanent electric dipoles, such as CO2 and H2O)

      • and collisional perturbations. Noble gases are not ideal.

    • Willis, where does the energy come from to “float” your non-radiative atmosphere?!?!

      • Willis Eschenbach

        kuhnkat | December 1, 2014 at 9:24 pm

        Willis, where does the energy come from to “float” your non-radiative atmosphere?!?!

        Thanks, kuhnkat, but I fear I have no Idea what you mean. What does “float” signify regarding atmospheres?

        w.

      • Pierre-Normand

        Kuhnkat, it is in thermal contact with the surface.

    • David Springer

      I agree with P-N it’s obvious that gravity doesn’t add energy to the atmosphere. Basic knowledge of star formation provides that understanding, Willis. An isolated cloud of gas in interstellar space collapse in on itself under the force of gravity eventually bringing pressure and temperature to fusion ignition point. Gravity concentrates mass- energy that’s already there. In an isolated cloud it would be violation of conservation for total energy to increase.

    • Willis, please be careful about talking in absolutes (“No other such process”). Imagine a source of a neutron flux (the Sun qualifies) which causes a certain (probably small, granted) amount of activation / isotopic shift resulting in radioactive heating or even fission. And therefore … heating.

    • Colin Wernham

      “add a non-radiative atmosphere to the rocky planet”
      Be a bit difficult adding an atmosphere to a rocky planet at such a low temperature…

      • David Springer

        Bingo. The proposed PM2K by Eschenbach-Brown stop running when the height of the columns is reduced below the contact points for the hot and cold sides of the heat engine or thermopile.

    • Dear Mr. Eschenbach,

      Thank you for the clear example. I would go even further.

      In this specific case, radiative equilibrium with the cosmic background being equal in all directions, there would never be a temperature gradient anywhere in or on this planet, between any part of the atmosphere, or on the surface, right down to the core, no matter what the composition of the rocks or atmosphere.

      A temperature gradient never comes into existence spontaneously. Period.

      • David Springer

        Brownian motion and Boltzman distribution means any atmosphere will have an abudance of molecules both above and below the average temperature. Just for sake of argument say helium remains a gas at 3K. There will be many atoms both above and below that temperature. Gravity simply sorts them into non-Boltzmann distribution. However there is not an energy gradient established as the average mechanical energy is the same at all layers in the atmosphere with potential energy substituting for kinetic in one-to-one correspondence with altitude.

        I cannot stress strongly enough that this configuration, mechanical energy the same at all levels in the atmosphere, is the highest entropy state of the atmosphere and hence the *only* state which doesn’t violate 2LoT.

      • Pierre-Normand

        DS, Individual atoms don’t have a temperature.

      • Pierre-Normand

        DS, Also, gravity doesn’t sort gas molecules (of identical masses) into different layers with different speed distributions depending on height. This naive expectation is disproved by the Coombes and Laue paper cited earlier. You never acknowledged their argument.

        Secondly, the Boltzmann distribution law for energy isn’t countered by the effect of gravity. This law takes the gravitational potential energy of the molecules into account and merely results in a vertical density gradient together with uniform Maxwell speed distributions (for an ideal gas) at all height. See section D in the paper “On the Barometric Formula” referenced earlier.

        Gravity does tend to sort molecules into independent barometric density distributions in accordance with their masses, but this effect doesn’t change the isothermal profile either.

      • Hi,
        Now that I’ve poked my nose into this ants nest, I’ll take my statement a bit further.

        I agree that Willis Eschenbach gives a good example of a rocky planet alone in outer space.

        However, Willis qualifies his argument with the conditions that the planet is dry rock, that the atmosphere was non-radiative (no greenhouse gases) and that there was only cosmic background radiation of 3 W/m2.

        Here I disagree As long as the incident radiation over the entire sky is homogeneous, these qualifiers are quite unnecessary.

        Imagine our Earth, as is, in utter loneliness in some space, with a homogeneous sky radiating eternally at 22C. Sound comfortable?

        Given enough time, the planet would come to equilibrium. What would this be like?

        There would eventually be no temperature differential, anywhere

        Not between surface and atmosphere
        Not between light and shade
        Not between the core of the planet and the top of the atmosphere

        There would be

        No convection
        No wind
        No clouds
        No lapse rate

        Homogeneous humidity everywhere

        Oceans smooth as glass

        No non-random mechanical processes whatsoever

        You could not burn your finger with a magnifying glass

        Solar power would not work because it could never drive anything
        (it would be like trying to drive a turbine in a circular pipe full of water)

        No plants, since they couldn’t transpire

        No wild animal

        All that would remain would be brownian motion, which is the only thing that would prevent the atmosphere from stratifying according to molecular weight

        You could live there, but only as long as you have some sort of energy stored somewhere. (chemical, nuclear) You could then be your own personal temperature anomaly, until your stockpile ran out. I suppose that you could use gravitation to transport rocks from the tops of mountains to the bottom of the sea until everything was flat.

        Anything that happens on our Earth is only because the sun shines on one spot at a time, and the spot keeps moving. A temperature differential is a very fine thing indeed!

      • Pierre-Normand

        Ken W,

        I agree with almost everything you wrote. It’s a nice twist on Willis’s thought experiment thought it makes a rather different point since in this case the reason why there is no surface warming from a greenhouse effect, or any other cause, is because there is no Sun — no source of low entropy (shortwave) radiation; and everything has achieved thermal equilibrium with the background radiation of space.

        “All that would remain would be brownian motion, which is the only thing that would prevent the atmosphere from stratifying according to molecular weight”

        There would still be some stratification since Brownian particles and molecules alike have vertical density distributions according to mass at thermodynamic equilibrium.

        See figure 4 (and associated discussion) as well as section C in the following paper:

        Berberan-Santos, Bodunov, Pogliani, ‘On the Barometric Formula’, Am. J. of Physics, 1997.
        http://web.ist.utl.pt/ist12219/data/43.pdf

      • Pierre-Normand

        …In our out of equilibrium troposphere it isn’t molecular diffusion but rather convective mixing that prevents the aforementioned partial stratification of molecules according to molecular mass.

      • Pierre-Normand

        The stratification can also be seen with this online 2D simulator (not the same one that I linked to earlier). For best visualization, reduce the number of balls to 50, increase both gravity and temperature to the max, and chose three different masses, with uniform sizes in order not to introduce a confounding factor. (Blue balls are lightest, and red ones are heaviest.)

      • Pierre-Normand,
        thank you for the link, also interesting history,
        and you are correct, I hadn’t considered quite to infinity.
        That’s even worse than I thought!

      • KenW, “Now that I’ve poked my nose into this ants nest, I’ll take my statement a bit further.” Then “I hadn’t considered it quite to infinity.”

        The ant’s nest is the nature of the problem. If you assume that a certain state is “equilibrium” then it turns out that isn’t really equilibrium, it appears that work was done in the process of moving toward a “real” equilibrium. Since equilibrium is a concept, not a reality, how you frame the experiment determines the results.

        If you assume that all gases will be equally well mixed in the volume and the temperature is perfectly isothermal, gravity would tend to separate molecules by weight so that the bottom is more well mixed than the top. Your experiment failed. Now you have to change your concept of equilibrium.

        The problem was originally posed over the definition of entropy. Boltzmann’s H-theorem defined a maximum entropy using ideal perfectly hard round balls without real consideration of gravity, so his entropy and therefore equilibrium are only accurate for near ideal cases. The less “ideal” an experiment you design the greater the violation of the 2 law can appear. So if you use the kinetic definition of temperature, you can find a temperature gradient. There is no “real” difference in energy between altitudes other that gravity which is external to the system. when you use an ideal column of air to perform your calculations on.

        If you pose a spherical planet, now the “most” of force of gravity is internal to the system. So if you pose a rock core instead of an all gas planet you will find a different equilibrium IF you use Boltzmann’s H-theorem, which does not consider the interaction of the individual molecules.

        That was all the problem was intended to do, make ya think. So no matter what “proof” someone comes up with, you can continue to pick nits until some “real” truth appears to fit every conceivable case.

        Right now Gibbs Entropy appears to be a better “truth”. There would be some density gradient depending on the mixture in the volume and the total force of gravity “felt” at each point in the volume. Because of the 2nd law, no experiment would have runaway heating or cooling if the system is in a real equilibrium state which should take an infinite amount of time to be realized. In the real world there is always some dissipation until you reach some point where gravity ultimately wins and starts sucking everything back together. At that point physics will have a brand new set of observations to describe.

        So you can continue to defend your concepts of entropy and equilibrium or bite the bullet and consider non-equilibrium thermodynamics which allows various concepts of entropy. On the whole, physics is close enough for government work if you aren’t into tiny nits.

      • Captian Sir,
        my point was that Willis went out of his way to provide his model planet with a non-radiative atmosphere and his sky with a very low temperature, I felt this wasn’t necessary because his sky is uniform. If his planet exists long enough, even the rocks will stratify.
        Good evening to you, Ken

    • David Springer

      There won’t be an atmosphere on your rocky planet, Willis. Even helium is a liquid at 3K.

      • Willis Eschenbach

        Yeah, you’re right, David … so make it a rocky planet with a natural nuclear reaction at the core that keeps the surface at 15°C. Then add an atmosphere without any GHGs. The principle is identical, as is the result—gravity cannot heat the surface with a non-GHG atmosphere, because then the surface would be radiating more heat than it gets from the core. This, of course, violates conservation of energy.

        w.

      • Conduction would heat the atmosphere with energy unable to be dissipated to space radiatively. Is that 2 silly little thought experiments in a row from Willis?

        Along with agreeing with FOMBS about not getting stellar temperatures in centrifuges. Yeah – you don’t.

      • David Springer

        Thanks. The surface gas would be 15C but does not follow there isn’t a lapse rate so upper atmosphere is less.

      • David Springer

        Willis you seem to be under the mistaken impression that gravito-thermal effect makes the surface warmer than it would be without gravity. That is not true. The mechanical energy is 100% kinetic at the surface and the temperature the same as if there were no gravitation confinement.

        Understand the question before answering please.

    • Would the theoretical planet with an atmosphere now have an ERL which would be seen to be radiating 3 watts out. This would be at some height above the surface which would have a higher temperature than the ERL unless the theoretical planet has no lapse rate.

  10. That Mr. Eschenbach’s and Dr. Brown’s proofs are invalid is explained here: http://wattsupwiththat.com/2014/08/18/monday-mirthiness-spot-the-troll/#comment-1711550.

    • Willis Eschenbach

      Joe, I’ve just given a very simple proof above. Perhaps you can explain where it is wrong.

      Thanks,

      w.

      • Mea culpa. In light of the Coombes-Laue discussions and Dr. Brown’s post which was directed to the zero-lapse-rate proof, I didn’t read further. It is the proof of a zero equilibrium lapse rate that I object to.

        No, I see nothing wrong with the proof that a transparent gas blanket won’t warm the planet.

      • Pierre-Normand

        Yes, Willis’s reference to Brown’s proof was misleading. I hadn’t paid attention to that link. (Unless Brown also has a proof similar to this one?)

      • David Springer

        I can and did explain a few comments above this one.

    • Pierre-Normand

      Joe Born, that’s not the same proof. What Willis talks about here isn’t the lapse rate, but rather the surface temperature, whatever the lapse rate may be.

    • I have argued with Joe Born on his claim at SoD. That’s a point about the properties of perfectly isolated systems of finite number of molecules. The effect disappears, when the system is allowed to interact with the rest of the world. Any measurement performed on the system introduces so much interaction that nothing can be observed even in principle. It’s also impossible to create such an isolation, because that requires exactly zero energy transfer with the walls that enclose the system. We are discussing here effects that are changed by a single molecular collision with external world.

      • “The effect disappears, when the system is allowed to interact with the rest of the world. Any measurement performed on the system introduces so much interaction that nothing can be observed even in principle.”

        I’ve never contended that the effect would be great enough to measure. But it will take more than Dr, Pirilä’s mere assertion for me to accept that the (admittedly immeasurable) effect that according to Velasco et al. does prevail will disappear suddenly when some interaction, however slight, is permitted. While it is true that the absolute isolation on which the microcanonical ensemble is based is not encountered in real life, neither is it true that the infinite-heat-capacity heat bath on which the canonical ensemble is based encountered in real life. Therefore, the exactly zero kinetic-energy gradient isn’t either. In any event we’re talking about ideal cases, not what we see in the world around us.

        Consider two identical initially isolating containers of the same-molecular-weight ideal monatomic gas. The containers extend infinitely high in a uniform gravitational field. One of them has twice as much gas as the other. Velasco et al. says that the first should have a lower, but non-zero, kinetic-energy gradient than the second.

        Now couple the two containers thermally, say at two different altitudes, without allowing the amounts of gas in either container to change. Will the interaction–which, notice, now permits a variation in each gas column’s total energy and therefore makes its possible microstates no longer constitute a microcanonical ensemble–cause either gas column’s kinetic-energy gradient suddenly to drop to zero?

        Dr, Pirilä never seems to give me a straight answer to this. I say the answer is no, and I say that Roman, White, and Velasco demonstrate that it’s no. I don’t see how Dr, Pirilä’s quote above is correct if Roman et al. are right.

      • Pierre-Normand

        Joe Born wrote: “Now couple the two containers thermally, say at two different altitudes, without allowing the amounts of gas in either container to change. Will the interaction–which, notice, now permits a variation in each gas column’s total energy and therefore makes its possible microstates no longer constitute a microcanonical ensemble–cause either gas column’s kinetic-energy gradient suddenly to drop to zero?”

        The (grand) canonical, or microcanonical ensembles represent the sets of microstates that are accessible to the non-isolated (in equilibrium with a thermal bath), or fully isolated systems, respectively. The statistical properties of those sets are relevant to the probabilities of finding the system, at any given time, to occupy some subset of those accessible microstates. Temperature not only is a macrostate, is also is an ensemble property rather than a microstate represented by a small cell in phase-space. So, yes, when the two columns are put in thermal contact and aren’t isolated anymore, then whatever ensemble property you may wish to call ‘temperature’ immediately changes since the set of accessible microstates to both systems immediately changes.

      • Pierre-Normand: “So, yes, when the two columns are put in thermal contact and aren’t isolated anymore, then whatever ensemble property you may wish to call ‘temperature’ immediately changes since the set of accessible microstates to both systems immediately changes.”

        While I’m not sure, I think I agree with this, but it depends on which definition of temperature you use. Perhaps muddied the water by gratuitously using “suddenly.” But the point I was intending to make was instead that the mean-molecular-kinetic-energy gradient–whether or not it would change “suddenly”–would not fall all the way to zero.

        Of course, even in the context of ideal monatomic gases there are definitions of temperature that differ from mean molecular kinetic energy. But in Dr. Brown’s proof, in which he assumes a non-zero equilibrium lapse rate–and therefore different temperatures at different altitudes–that’s the only definition that is consistent with his initial assumption. Under the thermodynamic definition, equilibrium by definition means no lapse rate, and, as you say, the statistical-mechanics definition makes temperature an ensemble property that it is conceptually difficult to assign to different values for different altitudes of the single molecule set that the macrostate characterizes.

  11. “the surface of the planet would warm, and stay warmer than it was with no atmosphere.”

    It clearly does.

    “But at that point, the surface of the planet would constantly be radiating more energy than it is receiving … which is an obvious physical impossibility.”

    At that point the surface radiates 255K to space but the other 33K is locked into convective adiabatic overturning whereby conductive ascent takes heat from the surface exactly as fast as convective descent returns it to the surface so that the surface cannot radiate to space at more than 255K

    • Pierre-Normand

      Stephen Wilde,

      The emission power from a surface only is a function of its temperature and emissivity. You can’t reduce it through allowing the surface to conduct heat away (or though latent heat release). If it’s allowed to conduct heat away, then it will just lose heat at an even higher rate. If this weren’t the case, one would think it would have been observed in the laboratory already.

      • The emission power at the surface remains at 288K (for Earth). It is just that it can only radiate to space at 255K because the surface is exchanging energy with the atmosphere at the same time.

        During the very first convective cycle the surface does indeed lose energy at a faster rate than 255K because as well as trying to radiate to space at 255K it loses energy into the developing convective cycle.

        Thus, during that first convective cycle surface temperature temporarily dipped below 255K

        When the first convective cycle completed then radiation to space recovered back to 255K but the other 33K remained locked into the continuing convective overturning.

        And there it will remain as long as mass, gravity and insolation remain constant.

      • Pierre-Normand, December 1, 2014 at 12:23 pm:

        “The emission power from a surface only is a function of its temperature and emissivity. You can’t reduce it through allowing the surface to conduct heat away (or though latent heat release). If it’s allowed to conduct heat away, then it will just lose heat at an even higher rate. If this weren’t the case, one would think it would have been observed in the laboratory already.”

        It should be a pretty obvious fact, almost to the point of being intuitive, that a body of water struck with the first round of a solar SW flux and thus heated to depth by it, will not be able to also shed this particular amount of energy within a similar period of time by radiating (LWIR) back out from its surface. Why? Because the energy coming in as SW is absorbed across the three dimensions of the volume of water, but the energy going back out as LW is only emitted from the two-dimensional surface.

        The SW flux is after all not instantly able to warm the body of water up to the point where its surface temperature correlates to a LW emission flux equal to the originally absorbed SW one.

        In a hypothetical, purely radiative setting, this would pose no problem. The system would simply move towards a steady state (dynamic equilibrium), by incoming (SW) energy gradually accumulating inside the volume, until the point where the surface temperature did finally reach exactly that level. At this point, the incoming and outgoing radiative fluxes would match and there would be no more warming.

        (This is how normal, real-world objects warm, after all, by energy gradually accumulating inside them (internal energy, U, growing), fast at first, then progressively slower up to the point of balance between heat (Q) IN and heat OUT. Then the increase in U, and in T, stops.)

        On our planet, however, there’s an atmosphere on top of the solar-heated body of water, and the ocean releases most of its absorbed energy back out into this atmosphere by way of evaporation to stay in balance with its incoming from the Sun.

        This means that on real Earth, the water surface will reach its steady state temperature long before radiative balance has been accomplished, i.e. as much energy is released from the surface as taken up by the volume of water per unit of time even at much lower temperatures than the ideal Stefan-Boltzmann blackbody situation would seem to demand. Courtesy of the release of latent heat through evaporation.

        So yes, it would seem that putting an atmosphere on top of a water surface would enable it to reach its heat IN/OUT balance point at a lower steady state temperature than without such an atmosphere. This, however, is not because the evaporative loss comes IN ADDITION TO the radiative loss. It is because it comes INSTEAD OF a large part of the previous radiative loss.

        You see, emissive power from a surface is only a function of its temperature and emissivity alone if that surface happens to be a pure/ideal emitter, that is, its entire heat loss comes via radiation, meaning it’s isolated in a vacuum at 0 K or simply very much hotter than its surroundings. There is nothing in the Stefan-Boltzmann equation suggesting otherwise. The S-B law applies only to purely radiative situations. Just like Planck’s and Kirchhoff’s laws do. The idea that it applies no matter what has no empirical basis and it ignores simple energy budgetting principles.

      • “putting an atmosphere on top of a water surface would enable it to reach its heat IN/OUT balance point at a lower steady state temperature than without such an atmosphere.”

        The heavier an atmosphere the warmer the water has to become to supply the necessary latent heat of vaporisation to fuel evaporation.
        The amount of energy required to effect the phase change is pressure related. If there were zero atmospheric pressure the oceans would evaporate instantly, freeze and fall to the ground as a solid.

      • Stephen,

        Indeed. But first we have to get Pierre-Normand and his kind to understand that radiation is not some divine entity that operates completely independent from all other physical processes in this universe, like for instance other heat loss mechanisms for a warm object (radiation is, after all, simply one of several heat loss mechanisms).

      • Pierre-Normand

        “Indeed. But first we have to get Pierre-Normand and his kind to understand that radiation is not some divine entity that operates completely independent from all other physical processes in this universe […]”

        This claim is to vague for me to either endorse or disown. My claim simply is that a solid surface emits the same gross radiant power, as a function of its temperature only (and emissivity), and this power is unaffected by the rate of conduction with a gas it is in thermal contact with. Why would the mere existence of the nearby gas molecules changes the rate of emission of the surface?

        The rate of conduction just relates to the probabilities that some molecules at the surface will exchange various amounts of energy in random collisions with gas molecules. If the gas is warmer than the surface, then individual collisions are more likely to transfer energy from the gas to the surface, and conversely. In between the collisions, the molecules of the surface emit photons with some probabilities dictated by temperature only. Before some molecules at the surface will emit a photon, they don’t care at all how likely they are to either lose or gain some energy in the future from incoming gas molecules. This is basically why the rates of the two processes, conduction and radiation, are independent. (Though the interactions of the gas with the surface may have a small effect of the surface emissivity, but that’s something else.)

      • Pierre-Normand

        “may have a small effect [on]… “

      • Pierre-Normand,

        So you of course chose to not read my main comment, just two stops up from the one you replied to.

        I’ll quote from it:

        “You see, emissive power from a surface is only a function of its temperature and emissivity alone if that surface happens to be a pure/ideal emitter, that is, its entire heat loss comes via radiation, meaning it’s isolated in a vacuum at 0 K or simply very much hotter than its surroundings. There is nothing in the Stefan-Boltzmann equation suggesting otherwise. The S-B law applies only to purely radiative situations. Just like Planck’s and Kirchhoff’s laws do. The idea that it applies no matter what has no empirical basis and it ignores simple energy budgetting principles.”

        You repeat:
        “My claim simply is that a solid surface emits the same gross radiant power, as a function of its temperature only (and emissivity), and this power is unaffected by the rate of conduction with a gas it is in thermal contact with.”

        No, Pierre-Normand. This is just the flawed idea that has been allowed for too long to disseminate among the rGHE believers of this world. It only constitutes a fact in a purely radiative setting. All these laws are originally and fundamentally made for (and thus apply to) a radiation only scenario.

        Beyond that world, you need to start accounting for you energy gains and losses. You can’t just add whatever suits you. You need to have the energy available. This is the domain of THERMODYNAMICS.

        Niels Bohr once said, and pay close attention to this, Pierre-Normand, all the old guys knew this (Planck too), they had no illusions about being able to actually observe directly what they described:

        “There is no quantum world. There is only an abstract quantum physical description. It is wrong to think that the task of physics is to find out how nature is. Physics concerns what we can say about nature …”

        We only ever detect the HEAT moving between two objects in heat transfer (that is, at different temperatures), that’s the actual TRANSFER of energy across the radiation field between them. The whole principle of a two-stream photon exchange between the two objects is just that … a principle, a model, a theoretically based assumption of how things might work. Such a two-stream photon exchange in a heat transfer has NEVER experimentally been observed. It CANNOT physically be observed, you cannot detect two separate fluxes moving in opposite directions inside one continuous, integrated radiation field. All that nature reveals to us is the heat, the actual transfer of energy through that field. Which CAN be detected. Because it is actually able to affect the two bodies directly, something we can measure, a physical change, in their internal energy (U) and thus their temperature (T).

        “Why would the mere existence of the nearby gas molecules changes the rate of emission of the surface?”

        Enter thermodynamics. You need to account for your energy gains and losses, Pierre-Normand. Not just assume that your radiative law remains untouched and independent no matter what happens. When the surrounding conditions change, physical relationships change too. Other laws and processes come into play.

        An object just seeks (dynamic) equilibrium with its surroundings. Radiation is no more than a method of attaining such balance. It is a heat loss mechanism. Pure and simple. It is there to balance the heat gain. This is thermodynamics at its most fundamental level.

        If a no-atmo BB planet in space after equilibration with its heat source gets 2 parts IN and emits 2 parts OUT, all as radiative heat, then if we surround it with an atmosphere and keep the heat input from the source constant, then after new equilibration, the heat IN to the surface will still be 2 parts and the heat OUT from the surface will still be 2 parts, only now there is 1 part escaping through conduction > convection and 1 part escaping through radiation. The surface at any one time only absorbs 2 parts of heat from the heat source, meaning that it always has 2 parts only of heat to shed. If 1 of these parts go out through conduction > convection, there is only 1 part left to radiate. And this is regardless of the new steady state surface temperature.

        The mean global surface of the Earth gains 165 W/m^2 worth of radiative HEAT from the Sun, its only source of energy (except geothermal). This flux is balanced by the total heat OUT, which is made up of 53 W/m^2 of radiative heat and 112 W/m^2 of conductive/evaporative heat. 165 W/m^2 IN, 165 W/m^2 OUT. Still, the surface temperature is 289K.

        The postulated 398 W/m^2 radiative flux postulated to leave the surface is a mere mathematically derived temperature POTENTIAL. Based directly on the surface temperature (289K). It has never been separately detected. It cannot physically be separately detected. Only the HEAT flux can ever be detected. The 398 W/m^2 would be the radiative HEAT flux from a BB surface radiating into a vacuum at 0 K. It is NOT a real flux. It is a temperature potential, a temperature potential facing some air temperature potential, generating in between them a spontaneous radiative HEAT flux from warmer to cooler, up from sfc, towards atm. If the air warms without the surface doing the same, the difference in temperature potentials is reduced and the heat flux from the surface up is consequently reduced as well.

        P/A = es(T_sfc^4 – T_atm^4)

        P/A is the only actual radiative flux involved here, the heat transfer. The righthand side of the equation simply shows the temperature POTENTIALS facing each other. These potentials would constitute real radiative heat fluxes only if the two bodies were isolated from each other and surrounded by perfect vacuums at 0 K.

        Too many people look at this formula thinking that they see two physically opposing thermodynamic fluxes (transfers) of energy. That concept is only mathematically derived. In reality, in the real physical world, there is but one flux, one transfer of energy, and in a spontaneous heat transfer situation, it always and only moves from hot to cold.

        This is what Bohr was pointing to. The only thing we KNOW is the heat and the temperatures. The quantum explanation behind is pure speculation, simplistic models of reality in an attempt to make some sense of it.

      • Pierre-Normand

        “So you of course chose to not read my main comment, just two stops up from the one you replied to.”

        I can’t always reply to all the posts that are addressed to me.

      • Sure. No one can force you not to cherry-pick your responses. It just speaks volumes when you do …

      • Steven Mosher

        Gentleman here is Kristian at his best.

        ” I took for granted that it was empirically shown (out there in the real earth system) how more CO2 in the atmosphere will and does in fact help to warm the surface, and that the reason we are able to even live on this planet at all is because of an ‘Atmospheric Radiative GreenHouse Effect’ (rGHE) where so-called ‘GreenHouse Gases (GHGs)’ – like CO2 – make it so that some of the energy that leaves the surface of the earth never manages to escape the system as a whole to space, but is rather ‘recycled’ internally between atmosphere and surface, creating ‘extra’ atmospherically induced warming of the surface on top of the original solar warming.”

        When he starts from a mistaken view of the greenhouse effect, as Kristian does, there is no point talking to him.

  12. The same parcel of kinetic energy at the surface cannot both radiate to space and hold the atmosphere up against gravity simultaneously.

    If it radiates to space it cannot replenish the PE reservoir in the atmosphere and if it is replenishing that reservoir it cannot be radiating to space.

    • Pierre-Normand

      Stephen Wilde,

      Sure. Your first sentence is true. Which is why a surface that is allowed both to conduct heat away and radiate to space will cool faster than a surface that only is allowed to cool radiatively. Still, the radiating power only is a function of temperature. It not an either/or proposition. When it does both, then it cools faster.

      • “When it does both, then it cools faster”

        Not once the first convective cycle has completed it doesn’t because the energy going into conduction on the ascent is from then on being returned on the descent for a net zero thermal effect.

        In that scenario you get back to the original rate of surface cooling to space (255K) but in addidtion you have that separate net zero energy exchange in the adiabatic convective loop which holds up the atmosphere against gravity for as long as the sun keeps shining.

      • A non-radiating atmosphere would not be able to lose heat to space, which means it would eventually come into thermal equilibrium with the surface – at which point the surface would no longer lose net heat to it by conduction.

      • phatboy,

        The atmosphere can never come into equilibrium even without GHGs.

        In the vertical plane an atmosphere is always in thermodynamic DISequilibrium (with or without GHGs) due to work done against gravity in holding the mass of the atmosphere off the surface converting KE to PE to produce an inevitable lapse rate even in a GHG free atmosphere.

        For a GHG free atmosphere all radiation to space would indeed have to be from the surface but the surface for Earth would still need to be at 288K to get 255K out to space due to conduction and convection enabling the mass of the atmosphere to obstruct the free flow of radiation to space.

        The ERL would still be at the height where the temperature is 255K i.e. some distance off the surface and not at the surface.

        You can only have 255K at the surface if there is no atmosphere at all.

        Mass not radiative capability creates the greenhouse effect.

  13. There is also a lot of confusion about the connection between the greenhouse effect and the lapse rate. You cannot have one without the other.

    Gravity merely sets the scale of the adiabatic lapse rate in a planetary atmosphere. It does not ‘generate’ a lapse rate through barometric pressure or some other magic. Heat needs to flow through the atmosphere to maintain convection. That is the role of greenhouse gases. Radiative transfer from the surface causes a temperature gradient steepening towards radiative equilibrium. Convection (and evaporation) are then ‘powered’ to limit the temperature gradient to the Dry/Moist adiabatic lapse rate.

    Now imagine that you could somehow switch off gravity while keeping an atmosphere of pure CO2 held to the surface by some membrane or other. There would be no greenhouse effect at all! The temperature of the CO2 atmosphere would thermalize to Tsurf and an equal flux of radiation would be emitted to space from the top of the atmosphere as was absorbed at the bottom.

    http://clivebest.com/blog/?p=6305

    • Pierre-Normand

      Clivebest, Fully agreed.

    • Willis Eschenbach

      clivebest | December 1, 2014 at 12:24 pm | Reply

      There is also a lot of confusion about the connection between the greenhouse effect and the lapse rate. You cannot have one without the other.

      Actually, they have nothing to do with each other. You don’t even need an atmosphere at all to have a greenhouse effect. See my posts The Steel Greenhouse and People Living In Glass Planets for an explanation.

      Regards to all,

      w.

      • Willis,
        In a sense your examples have something equivalent to lapse rate in form of a radiating layer at a lower temperature than the surface.

        That’s part of the point of radiative GHE: A layer of lower temperature absorbs IR coming from below and emits less upwards, because it has the lower temperature.

        The second part of the argument explains, why the temperature of the outer layer is lower than that of the surface.

        My above sentences can be applied both to your examples and to the atmosphere.

      • Willis,

        Both your examples generate a ‘lapse rate’ although with just 2 data points.

        The fully transparent IR shell corresponds to the case of an atmosphere without greenhouse gases – say 100% N2 or argon.

    • Clivebest,

      “Now imagine that you could somehow switch off gravity while keeping an atmosphere of pure CO2 held to the surface by some membrane or other. There would be no greenhouse effect at all!”

      There is a heat flux from the surface passing through the atmosphere until full dissipation. Heat flux mean gradient. And thus a higher surface temperatrue than at your upper atmosphere. Gravity is not necessary to have a greenhouse effect.

      • Phi,

        There is an equal and opposite radiative flux throughout a 100% CO2 atmosphere without any gravity. There can be no temperature gradient because there can be no lapse rate. Convection is meaningless. Exactly the same number of 15 micron photons are absorbed at the bottom of the atmosphere as will be emitted from the top of the atmosphere to space.

        I think this fact should be more widely known.

        CO2, per se, cannot warm anything at all. Lab experiments purporting to explain global warming by placing some beaker of CO2 in front of a furnace have absolutely nothing to do with the greenhouse effect on earth and are mostly junk.

      • Clivebest,

        “There is an equal and opposite radiative flux throughout a 100% CO2 atmosphere without any gravity.”

        Why is that? There is only one IR flux, it goes from the warm source (the surface) to the cold source (space).

        “There can be no temperature gradient because there can be no lapse rate”

        The temperature gradient is related to heat flux, not to gravity.

        “Convection is meaningless.”

        There is no convection and thus the heating effect of CO2 is even more important.

        “Exactly the same number of 15 micron photons are absorbed at the bottom of the atmosphere as will be emitted from the top of the atmosphere to space.”

        1. You must not limit to the center frequency.
        2. The photon model is not suitable for thermodynamic reasoning.
        3. The 15 micron frequency is practically saturated in your example, there is virtually no thermal effect on that wavelength throughout the thickness of the atmosphere. By cons, 15 microns emits very efficiently into space at the top.

      • In the hypothetical case of CO2 held to a surface without and gravity, I see the atmosphere as being an integral part of the surface. There is no pressure gradient and acts like a perfect gas. It has to be isothermal in the same way as sand on a beach is isothermal.

      • Clivebest,

        The sand will be isothermal only if there is no heat flow therethrough. Your atmosphere without gravity is cooled at the top by IR emission and heated from below by conduction and radiation. Thus, there is a thermal flow from bottom to top and thus a temperature gradient.

      • It is also heated from below by radiation (absorbed 15 micron photons). There can be no lapse rate because the CO2 gas must reach thermal equilibrium in all 3 spatial directions.

      • Clive,
        Without gravity the atmosphere would act essentially as a layer of solid insulator on top of a heated surface and cooled from the upper surface. There’s no reason to assume that it would be isothermal. The density differences between the warmer parts and the cooler parts would not induce convection in absence of gravity.

      • Pekka,

        yes but it’s not a solid. The molecules are moving at an average speed of over 500 m/s and will rapidly thermalize to a constant temperature throughout the bulk of the atmosphere. Radiation will escape to space from a skin layer at the top exactly balancing the absorbed skin radiation at the bottom while 90% of surface radiation passes straight through the atmosphere to space.

        I admit that the top skin layer will not be in perfect thermal equilibrium so there could be a very small greenhouse effect.

      • Clive,
        Both solid insulators and gas conduct heat, but heat conduction is a not an efficient way of heat transfer in gas. We know that it can be ignored relative to both convection and radiative heat transfer in most applications. In your example no convection is present, because nothing will initiate convection in absence of gravity, but the relationship between radiative heat transfer and conduction remains unchanged, i.e. conduction can be ignored, when gases like CO2 are present.

      • Pierre-Normand

        “The molecules are moving at an average speed of over 500 m/s and will rapidly thermalize to a constant temperature throughout the bulk of the atmosphere.”

        The relation between heat transfer in a gas (or liquid) and molecular speed distribution is quite indirect. If there is no convection, and no radiative heat transfer, then there remains only conduction. And the heat conductivity of air is extremely low, as you must know, contrary to naive expectation from high average molecular speed.

      • Pierre-Normand

        …[thermal] conductivity…

  14. Build a sphere of 9 billion miles in diameter with its temperature set at 4K, incorporating a gravity neutralizer, that gets rid of all the gravity within the sphere. Add 2×10^30 Kg of 4He, at 4K, and allow the gas to come to thermal equilibrium within the large flask. When at equilibrium and when the gas is equally dense throughout the flask switch the gravity neutralizer off.
    The gas will become attracted to the center of the container and form a very hot body. The body will radiate the same amount of energy as the inward radiation from the vessels walls, however, the vessel has a much greater surface area than the solid/liquid/gas spherical body in the center.

    • Pierre-Normand

      DocMartyn,

      But the internal surface of the isolating vessel ought to have the same temperature as the spherical body, at equilibrium, and so, if both are black-bodies, we have a paradox, right?

      • Pierre-Normand

        I think the solution of the paradox (if I understood your puzzle correctly) simply is that, though the internal surface of the vessel emits with the same power (per unit surface), most of the radiation that it emits misses the core and falls back unto other internal areas of the vessel. If it emits as a Lambertian radiator, as a black-body should, then everything will appear at the same brightness at any point withing the vessel, either from some vantage point on the internal surface of the vessel, or from the surface of the spherical body.

      • No Paradox.
        The matter inside the sphere is in equilibrium when gravity is switched on and when switched off.
        In one case we have a dispersed gas and in the other we have a hot, dense mass.
        In one case there is a thermal gradient and in the other there isn’t.
        You can do P=TV when gravity is switched off and get the same numbers, but not when switched on.

      • Pierre-Normand

        DocMartyn, then I don’t understand the point of the thought experiment. After gravity is switched on, there is a collapse and gravitational potential energy is converted to internal energy during the collapse. Things heat up withing the collapsing body. The initial equilibrium is lost for a while until the inside surface of the container will warm up to the same temperature. (I had imagined it as a black-body with an external adiabatic coating.)

      • P-N. After a few million yours you will have a hot body in the center in dynamic equilibrium with the flask.
        The system will have extreme temperature and pressure gradients, when gravity is switched on.

  15. And then there is also Miskolczy’s theory which states that the effect of increasing co2 on the radiative imbalance is counteracted by a decrease in water vapour so that there is no net radiative warming effect.

  16. A radiatively inert atmosphere must still have a reducing temperature with height even without GHGs simply because the work done against gravity during convective uplift converts KE to PE and PE is not heat and cannot radiate.

    Then all one needs to initiate convective overturning is uneven surface heating which places parcels of air of different densities next to one another in the horizontal plane. Uneven surface heating is inevitable for a rotating sphere.

    So there is never going to be an isothermal atmosphere even without GHGs.

    The examples that purport to show that an isothermal atmosphere will develop all rely on a thermal equilibrium being attainable. In reality it can never be attained because of that conversion of KE to PE with height.

    In the vertical plane the atmosphere around a planet will never be in equilibrium so all those examples are invalid.

    There will always be convective overturning and a lapse rate.

    There will always need to be an additional kinetic energy store at the surface to sustain the ongoing convective overturning and that requires a surface temperature enhancement above S-B.

    • “A radiatively inert atmosphere must still have a reducing temperature with height even without GHGs simply because the work done against gravity during convective uplift converts KE to PE and PE is not heat and cannot radiate.”
      ___
      Yep.

  17. Quite relevant to this discussion, the formation of early atmosphered around planets, where the gravity of the planet determines the ultimate size of the atmosphere. Interesting charts on this site related to atmospheric pressure/temperature curves that many should find quite enlightening:

    http://astro.berkeley.edu/~ormel/science.html

  18. David L. Hagen

    Temperature gradients due to Gravity
    For the impact of gravity on conduction (WITHOUT convection OR Solar radiation etc.) see the experiments and theory developed by Roderich W. Graeff.

    Viewing The Controversy Loschmidt –Boltzmann/ Maxwell Through Macroscopic Measurements Of The Temperature Gradients In Vertical Columns Of Water Roderich W. Graeff
    Version: Descript 372_dec6 December 9, 2007
    At FirstGravityMachine

    The Production of Electricity out of a Heat Bath
    RODERICH W. GRAEFF , AIP Conf. Proc. 1411, 193 (2011); http://dx.doi.org/10.1063/1.3665239

    Abstract. In order to clarify the dispute between Loschmidt and Boltzmann/Maxwell concerning the existence of a temperature gradient in insulated vertical columns of gas, liquid or solids, macroscopic measurements of the temperature distribution in air, water and solids were performed. A negative temperature gradient, cold at the top and warm at the bottom, is found in insulated vertical tubes, while the outside environment has a reverse gradient. This is explainable by the influence of gravity. It allows the production of electricity out of a heat bath.

    • Graeff is a rare case as he thinks that his experiment proves that a perpetum mobile of second kind can be used to produce work. That’s contrary to most of those, who argue for a heat gradient in equilibrium, and think that such a situation is consistent with the Second Law and that it doesn’t make perpetum mobiles of the second kind possible.

      Both Graeff and those others are wrong. Graeff’s experiments are far too crude to tell anything about the real phenomena. His results are surely only empirical errors.

      • David L. Hagen

        Pekka
        Graeff’s “engine” is insignificant power wise. However I encourage you to review Graeff’s experimental data and show where they are off. I see how they could be improved 10x or more, but the base case with no circulation looks interesting.

      • The most important reason that makes the experiment extremely difficult is that the mechanism that brings the system to the isothermal state according to standard physics is conduction. Conduction in gas is a really weak process, but the experiment tells nothing unless all other heat fluxes combined are guaranteed to be weaker than internal conduction in the gas volume. Very weak heating or cooling of the walls of the enclosure are likely to lead to larger heat fluxes and to convective flows that produce the lapse rate inside the volume.

        For an experiment to have any value it must be shown by reliable measurements and calculations that no external influence affects the outcome. I have looked at the papers of Graeff and found that he does not present such analysis. I have also observed that he has not done the experiment with such care that there were a change that he ends up with valid measurements.

        This is a deceptive experiment. Most errors in the execution of the experiment lead towards the same wrong result. Only extremely careful work has any change of avoiding on error of that type. He is looking at an issue, where the standard physical understanding predicts that he finds a “confirmation” of his “theory” even when he is wrong, because he is unlikely able to avoid conductive fluxes inside his apparatus.

  19. A fan of *MORE* discourse

    FOMD’s Self-Evaluation Test of
    Thermodynamic and Heat-Transport Intuition

    Your spouse gives you a vacuum-flask thermos in which boiling-hot tea loses temperature at an initial rate of 20 C per hour.

    As an experiment, you dismantle the thermos, and pack the vacuum-space with thin aluminized mylar, then reassemble the thermos (finishing by pumping a fresh vacuum).

    Question  The modified thermos will show:

    (1) much faster heat-loss

    (2) about the same heat-loss

    (3) much slower heat loss

    Answer the question with both a closed-form mathematical analysis and references to the engineering literature; include a quantitive description of the effects of aluminizing the mylar, and of the significance (or not) of the thickness of the mylar. How good does the vacuum need to be, in order to reach the point of diminishing returns?

    Self-assessment  People who firmly believe that they understand the basic principles of thermodynamics and energy transport, and yet can’t answer the test question (or answer it incorrectly), and/or are entirely ignorant of the engineering literature, are displaying Dunning-Kruger cognition!

    \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • A fan of *MORE* discourse

      Follow-on “thermos” self-evaluation questions

      (2)Blackened mylar:  does it insulate better than aluminized mylar? Does it insulate better than no mylar all?

      (3)Infrared-absorbing, optically transparent plastic film:  does it insulate at all?

      Note  By design, option (3) has similar radiation and heat-transport properties to a non-advective CO2/H20/CH4-laden atmosphere!

      Conclusion  Yes Virginia, there is a CO2-driven greenhouse effect!

      \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • WAG, just for fun, from a not-so-confident idiot: if “packing” implies thermal contact then it will lose heat faster due to conduction.

    • Why use a vacuum?
      Why not use CO2 to reflect the IR back into the thermos?

      • FOMBS was armwaving about the symplectic manifold of the Hamiltonian – somehow in connection to a Wikipedia cartoon of a box of ideal gas molecules. A fantastic and improbable juxtaposition with the symplectic manifold seeming to have only a symbolic significance. A talisman of cargo cult science.

        For my sins – I introduced the Fronsdal paper as an example of a Hamiltonian of the atmosphere – that – intriguingly – suggested a gravito-thermal effect.

      • Rob, “For my sins – I introduced the Fronsdal paper as an example of a Hamiltonian of the atmosphere – that – intriguingly – suggested a gravito-thermal effect.”

        Yep, it is all your fault.

      • A fan of *MORE* discourse

        Rob Ellison perceives a “symplectic manifold seeming to have only a symbolic significance”

        Confident ignorance by Rob Ellison, well-regarded dynamics textbook by the celebrated mathematician Vladimir Arnold!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • JC SNIP

        FOMBS use of the symplectic manifold is symbolic. As is the vague arm waving toward arbitrary bits of science and maths that he has little understanding of. A symplectic manifold simply says that the Hamiltonian is differentiable. But there is no sense in which FOMBS uses it in for any concrete purpose – merely arm waving about a ‘well-regarded dynamics textbook by the celebrated mathematician Vladimir Arnold!’ whose only purpose is a symbolic one of demonstrating a ‘confident ignorance’ of deniers. FOMBS cargo cult science – in which the symbol supplants the reality.

      • It’s not my fault – btw – and my banter – which a guy would understand was banter and not take it seriously – got snipped. It is getting way too politically correct around here.

      • Banter is fine, hitting them with the facts repetitively is better.
        JC SNIP suggests the banter was disappearing and beyond a guy’s simple understanding.
        Personally I think use of the symplectic manifold is better in a car magazine blog, because I don’t understand it while the two of you do.
        Is there a simpler expression?

      • The banter was aimed at Capt. Dallas – and wouldn’t have been a problem.

        The Hamiltonian is an extension of Newtonian dynamics using energy instead of force. The symplectic manifold is of interest as the solution space of the Hamiltonian.

        http://arxiv.org/pdf/physics/0004029v1.pdf

      • FOMBS use of the jargon is a talisman intended to show how clever he is and how dumb deniers are. Does he even understand the words let alone the math? Over the time he has been here making this same symbolic use of scientific and mathematical talisman – there is little to show that he does. It seems all sham and bluster – and it seems far from an isolated case.

        A rational person would wonder why.

      • A fan of *MORE* discourse

        Rob Ellison proclaims [wrongly] “[mathematical] jargon is a talisman”

        A more productive lesson, Rob Ellison, is that climate-change skeptics who seek to remake the foundations of thermodynamics are well-advised to heed the sage advice of Landau and Lifshitz:

        STATISTICAL PHYSICS: Preface

        Statistical physics and thermodynamics together form a unit.

        All the concepts and quantities of thermodynamics follow most naturally, simply, and rigorously from the concepts of statistical physics.

        Although the general statements of thermodynamics can be formulated non-statistically, their application to specific cases always requires the use of statistical physics.

        We have tried in this book to give a systematic account of statistical physics and thermodynamics together.

        Conclusion  Young researchers (especially) are well-advised to study thermodynamics the modern way … via the language of geometric dynamics … with particular care that every macroscopic thermodynamical assertion is solidly and explicitly grounded in statistical mechanics and (microscopic) Hamiltonian flows.

        The scientific rewards of this program are *AWESOME* !!!

        Best wishes are extended to you Rob Ellison, and to all Climate Etc readers, for continued enjoyment and success in appreciating thermodynamics and statistical mechanics and geometric dynamics the modern way: as one unified subject!

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  20. Is it true that this is like talking about balance sheets and income statements? Starting with Eschenbach’s example a 3 watts incoming planet. Adding any kind of gas at most delays radiation back to space. Any GHG effect is simply borrowing. It’s having a policy of holding all bills for two weeks before paying them so that we have short term cash or heat. Assume the gas has some vertical gas circulation. That’s not income. It is a transfer like from a money market to a checking account. In the end I think the books must balance. Beginning stuff plus changes equals ending stuff.

    • Yes, if you add GHGs they allow radiation to space from within the atmosphere which is earlier expenditure than if it were returned to the surface in convective descent before being radiated out to space.

      Then the energy having been lost to space (spent) the convective descent returns less to the surface than was taken up in the ascent so there is net deficit (cooling).

      But the convective overturning having been weakened the atmosphere contracts a little which increases density at the surface and that increased density picks up more energy from the surface (earned) and that balances the expenditure from GHGs.

      But since it is all about mass in the first place you would never notice the variation caused by GHGs. It would disappear in the rounding.

  21. ‘It is important to ask to what extent the observed polytropic relations are to be attributed to the intrinsic properties of the gas or to radiation. Although the question is somewhat academic, since it does not directly affect the main applications, it is natural to ask: what are the natural configurations of an isolated atmosphere, one that is not exposed to radiation? Our understanding of atmospheres will not be complete without an answer to this question.

    The statement that any two thermodynamic systems, each in a state of equilibrium with a well defined temperature, and in thermal equilibrium with each other, must have the same temperature is a central tenet of thermodynamics. A natural generalization is that the temperature, in an extended, but closed system in a state of equilibrium, that must be uniform, and there is a wide spread opinion that this remains true in the presence of gravitational fields. This is important for the understanding of terrestrial and stellar atmospheres, where the gravitational forces create a non-uniform density distribution.’
    http://www.mdpi.com/1099-4300/16/3/1515

    So this is the question being investigated by Christian Frønsdal with his ‘action principle’. It is a very old question with very little in the way of convincing exposition either way. In a sense – it is merely academic. It may or not underlie radiation and convection processes in the real atmosphere – which are likely far more significant. Indeed as the original protagonists – Loschmidt and Boltzmann – concluded long ago.

    I propose two corollaries:

    1. That silly little thought experiments resolve nothing of any significance.

    2. That it is hugely unnecessary to take a position on each and every point of contention. Science advances by observation and experiment – see proposition 1.

    • As we discussed with P-N in the previous thread his approach lacks exactly those mechanisms that lead to the correct isothermal solution. I cannot be absolutely certain that we found the correct reason for his strange results, but the explanation seems to be as follows.

      His hydrodynamical equations (Euler’s equations) do not include viscosity. He doesn’t even mention the word in his paper. He does discuss thermal conduction at one point in the introductory discussion, but he does not include that in his theory, as far as I can see. Thus his theory lacks both dissipative mechanisms that are always present in real gases. Without dissipation he can maintain convection indefinitely long without any input of energy. His equations allow the non-equilibrium states persist for ever, and he cannot describe the approach to equilibrium. Therefore he gets wrong results.

      Furthermore he proposes the centrifuge experiment using erroneous arguments that he just assumes without presenting any support for them beyond a few words that just state the wrong argument.

      • It’s a very long time since I used the Action Principle for anything. Thus I had to refresh my knowledge on that. A suitable source seems to be here

        http://www.scholarpedia.org/article/Principle_of_least_action

        One thing that we can learn from this presentation is that the Action Principle cannot be used for dissipative processes (with some exceptions). For that reason the principle is not at all suitable for the study of the thermodynamic equilibrium. It can describe only isentropic processes, but the approach to equilibrium is fundamentally not isentropic, but dependent on the rise of entropy towards its maximal value.

      • The alternatives suffer from a circularity of argument that is unconvincing at best. There is very little in the way of mathematical exposition or physical theory. There is little to distinguish it from symbolic logic.

        There is certainly no notions of viscosity, convection or conduction in the alternatives – merely handwaving at fundamental principles. I rather fear that Pekka’s quibbles fall into the same trap. We have some maths that suggests that a gravito-thermal gradient exists in an isolated system under gravity – and Pekka arm waves about conduction and convection countering the effect to get the ‘true’ result of a uniform temperature. It is so far from proven as to be laughably unconvincing.

        Here’s the experimental apparatus as visualised by Clive Best I presume.

        It relies only on the centrifuge simulating a large enough gravity to enable easy measurement of a temperature gradient. Seems simple enough.

      • A fan of *MORE* discourse

        Rob Ellison wonders about a “centrifuge simulating a large enough gravity to enable easy measurement of a temperature gradient.”

        These babies pull 70,000 “g’s” (and more!).

        After a run, when you take out the fluid-filled tubes, their tops aren’t frozen and theirs bottoms aren’t hot … the tubes are the same temperature from top-to-bottom … precisely as orthodox thermodynamics predicts.

        It’s a pleasure to increase your knowledge, Rob Ellison!

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      • I would assume the bottom of the test tubes doesn’t heat up because the centrifuge spins them around like an air-cooled WW-I rotary engine on steroids. They will remain exactly at the ambient air temperature.

      • ‘This experiment would be of moderate cost and well within the means of a university physics department. My personal opinion is that initially adiabatic compression will create a temperature gradient which then rapidly conducts through collisions to a single uniform temperature. Despite this, such an experiment is still worthwhile because it would resolve this dispute once and for all. Perhaps it has even already been done !’ http://clivebest.com/blog/?p=4101

        The effect – if it exists – is consistent with conduction. For me this is just more arm waving around fundamental principles. Again – hardly conclusive but it is just framed as an opinion.

        The predictions – consistent with Lochsmidt’s original calcualtions – assuming a gravito-thermal effect with a quite ill defined physical mechanism – is here. Although the right hand axis seems a bit odd.

        I presume that once the device is switched off
        the gas quickly returns to it’s original state – hence the need for thermocouples to measure temp under the simulated gravity.

        It is a pleasure to demonstrate what an utterly superficial bore FOMBS is.

      • fan of *MORE* discourse | December 1, 2014 at 6:07 pm |

        Rob Ellison wonders about a “centrifuge simulating a large enough gravity to enable easy measurement of a temperature gradient.”
        After a run, when you take out the fluid-filled tubes, their tops aren’t frozen and theirs bottoms aren’t hot.
        Really??
        Perhaps it is because they cool the whole darn system?

        “The continuous cooling starts with each rotor standstill of the Centrifuge 5418 R in order to maintain the preselected nominal temperature of the rotor chamber not only during centrifugation”.
        Details on frictional forces and heating in centrifuge tubes are very hard to find with a google search. Perhaps someone or Fan himself could quote the relevant science and articles.

      • Rob,

        I have explained in two comments reasons that make all experiments of this nature extremely difficult

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-651710

        https://judithcurry.com/2014/11/27/open-thread-thanksgiving-edition/#comment-651583

        In the first I discuss Graeff’s experiment, in the second the proposal of Frønsdal, which is essentially the same as that discussed by Clive Best.

        Finding a temperature difference in those experiments proves nothing without a lot of additional evidence to verify that the result is not due to an imperfection in the experimental setup, because most errors in the setup would lead to a spurious positive answer by maintaining circulation. A very weak circulation is enough to dominate over conduction, and to make the experiment faulty.

        A related effect applies to atmospheres. An extremely weak GHE is enough to maintain some circulation and a lapse rate that might actually be closer to the adiabatic lapse rate than the present one. Under such conditions the Earth surface would be as cold as without atmosphere, and the whole troposphere much colder than the present one. Thus the strength of the GHE and the difference between the surface temperature and the tropopause temperature are not strongly linked (although there’s limited causal dependence). This observation is background to my first comment in this thread

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-651632

      • Pierre-Normand

        If this centrifuge experiment give positive results (that aren’t errors) then it will provide the basis for building a perpertual motion machine of the second kind. Indeed, if the positive result comes from measurements with a thermocouple, then it will already be such a perpetual motion machine.

      • A fan of *MORE* discourse

        angech requests “Perhaps Fan himself could quote the relevant science and articles.”

        It is a pleasure to assist Climate Etc knowledge seekers!

        The relevant science  Under the Gravito-Thermal hypothesis, a fluid body (whether gas or liquid) of particles of mass m, in an acceleration field g, of height h, will come to an equilibrium having a top-to-bottom temperature difference \delta T given by the simple expression

           \delta T = m g h / k_B,

        where k_B is the Boltzmann constant.

        Checkpoint  No other expression for \delta T, depending upon the variables {m, g, h}, is dimensionally consistent.

        Now load three test tubes, each of height two centimeters, respectively with benzene, xenon gas, and mercury, into a high-end Beckman-Coulter Optima™ MAX-XP desktop ultracentrifuge, with the all-titanium MLA-130 Rotor. This bad boy pulls one million “g’s” at a rotational speed of 130,000 rpm.

        Is this spinning rotor comparably dangerous to a stick of dynamite?

        You bet these rotors are dangerous!

        Predictions of gravito-thermal theory

           \displaystyle\delta T = \begin{cases} 1876\ \text{K (benzene)}\\ 3158\ \text{K (xenon)}\\ 4825\ \text{K (mercury)} \end{cases}

        These temperature differences are large enough to carbonize the benzene, heat the xenon vapor to incandescence, vaporize the mercury, and melt-through the titanium.

        Observation  In ordinary laboratory usage, none of these disasters happens.

        Experimental conclusion  Ordinary laboratory experience with desktop ultracentrifuges falsifies the Gravito-Thermal hypothesis.

        Thank you for your well-considered question, angech!

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      • Thanks Fan.
        But.
        They do have cooling mechanisms to prevent it heating up, yes?
        And they do have equations for this heat which you supplied. Can you explain why the equations do not match your version of reality.
        Can you state categorically that there is no extra heat in the distal test tube.
        And, seeing the whole tube is spinning at these high rates and hence under the same G, would one really expect gigantic amounts of heat in such a small separation distance. We are talking 30 mm, not 30 Kilometers.

      • We have arm waving at perpetual motion, experimental error without an experimental design and a litany of mad FOMBS assertions that are typically not worth reading.

        There would be horrendously more energy being put in than could be extracted, there is a time to critique experimental design and that would seem to be after the design is complete and really we are not discussing the creation of energy by the mere fact of rotation. One can’t create horrendous temperatures in a centrifuge out of nothing. On a scale of silliness – the last ‘bad boy’ is a million.

      • Pierre-Normand

        FOMD wrote: “Checkpoint No other expression for \delta T, depending upon the variables {m, g, h}, is dimensionally consistent.”

        This consideration yields proportionality at best, not equality. Also, the dry adiabatic lapse rate depends on c_p rather than molecular mass, so there are other possibilities that are dimensionally consistent.

      • A fan of *MORE* discourse

        angech wonders “Thanks Fan. But. They do have cooling mechanisms to prevent it heating up, yes?”

        No. Unltracentrifuge rotors spin hour after hour with no rotor-cooling whatsoever.

        Were it not for friction in the bearings, the rotor would spin forever with no heating.

        No rotor-cooling is needed for this simple reason: there is no Gravito-Thermal effect.

        It is a pleasure to assist your understanding, angech!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • The alone have Iranians have 20,000 centrifuges running ! The Lochsmidt experiment test that I originally proposed has therefore been performed many times, and no-one reports any temperature gradients. I am convinced it does not work!

        The Ranque-Hilsch vortex tube separates a pressurized gas into hot and cold components. However the explanation is likely mechanical and energy is being consumed in the process!

      • A fan of *MORE* discourse | December 2, 2014 at 5:59 am |
        ” angech wonders “Thanks Fan. But. They do have cooling mechanisms to prevent it heating up, yes?”
        No. Unltracentrifuge rotors spin hour after hour with no rotor-cooling whatsoever.”

        This is absolutely untrue as any simple Google search would show. Why do you post untrue statements?

        One of many quotes,
        “Refrigerated centrifuges offer the added benefit of cooling to protect from sample degradation caused by heat generated by the action of spinning.”

        You also said
        “Were it not for friction in the bearings, the rotor would spin forever with no heating.”

        But there is friction in the bearings which the rotors spin on.

        No rotor-cooling is needed for this simple reason: there is no Gravito-Thermal effect.

        The rotors are cooled. but not because of the the GT effect which is a separate question.

        It is a pleasure to assist your understanding, angech!
        Rob is right, you really get a kick out of deliberate misrepresentation. But unlike him I do encourage you to keep going as efforts with this sort of input only fool people a little while until they check, then you lose all support. Hm, are you really a skeptic using negative psychology?
        No.

      • A fan of *MORE* discourse

        Summary  Gravito-thermal theory straightforwardly predicts temperature differentials of several thousand degrees in ultracentrifuge tubes.

        Question  Why don’t the ultra-high temperatures predicted by gravito-thermal theory utterly destroy any and all ultracentrifuge samples, and/or the titanium rotors that hold the sample-tubes?

        The world wonders!

        Oh wait, the world doesn’t wonder … because there is no gravito-thermal effect!

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      • Pierre-Normand

        FOMD: “Question Why don’t the ultra-high temperatures predicted by gravito-thermal theory utterly destroy any and all ultracentrifuge samples, and/or the titanium rotors that hold the sample-tubes?”

        (Note: I don’t believe the effect exists, myself)

        That could be because the hypothesized gradient is an equilibrium state. As it tends to be established in the gas sample, though internal diffusion, any external conduction with the solid parts of the device tends to destroy it at the very same time. One would need to put the sample in a good insulator, and, as already noted, inhibit any convection that would have the same effect as the hypothesized gravito-thermal effect. However, I now grant you that the effect is larger than I thought. (1,000,000 g, holy c**p !!)

      • Concentric rotating cylinders have been considered in the following papers.

        Howard Brenner, Steady-state heat conduction in a gas undergoing rigid-body rotation. Comparison of Navier–Stokes–Fourier and bivelocity paradigms, International Journal of Engineering Science, 70 (2013) 29–45.

        a b s t r a c t
        This paper proposes a seemingly unequivocal experimental and/or molecular dynamics simulation test of the viability of the compressible Navier–Stokes–Fourier (NSF) equations for gases in the continuum region, namely for near-zero Knudsen (Kn) numbers. While experimental gas kineticists have long known of the inadequacy of the NSF equations for rarefied gases (i.e., noncontinua), for which Kn is no longer small, it is nevertheless believed by fluid mechanicians that the NSF equations remain valid for gaseous continua. The author is, however, unaware of the existence of any unequivocal experimental (or simulation) data to support this view. Indeed, based upon recent work by the author and others on the subject of bivelocity hydrodynamics [Brenner, H. (2013). Proposal of a critical test of the Navier–Stokes–Fourier paradigm for compressible fluid continua. Physical Review E 87, 013014]; [Brenner, H., Dongari, N., & Reese, J. M. (2013). A molecular dynamics test of the Navier–Stokes–Fourier paradigm for compressible gaseous continua. arXiv:1301.1716 [physics.flu-dyn]], ample reasons exist for believing that the NSF equations may not, in fact, be valid for compressible continua. Given the fundamental role played by the NSF equations in both theoretical and applied physics, it would obviously be well if the assumption of the viability of the compressible NSF equations was put to a variety of experimental tests, particularly if the interpretations of those tests were seemingly unequivocal as a consequence of the simplicity of their interpretation and ease of execution. This paper adds one such test to that cited above. It involves contemplating a gaseous continuum confined in the annular space between two co-axial circular cylinders rotating steadily at the same angular velocity while, at the same time, both cylinder walls are maintained at a common temperature, say Tc. In such circumstances the NSF equations, assumed to govern the resulting rigid-body rotation of the fluid, trivially predict the gas’s temperature to be uniform at the value Tc throughout the entire annular body of gas between the cylinders. The proposed test consists of measuring the temperature distribution so as to establish if, in fact, this predicted temperature uniformity is actually observed in practice — and, if not, of establishing whether the observed temperature distribution varies with such experimentally- controllable parameters as the cylinders’ angular velocity or the gas’s mean pressure. Was the temperature distribution found to be nonuniform the compressible NSF equations for continua would have to be abandoned as physically unsound. Anticipating that outcome we have, for the prescribed experimental arrangement and protocol, solved the corresponding bivelocity equations in order to establish if this model accords better with the test data. For, in contrast with NSF behavior, the bivelocity temperature distribution is predicted to be nonuniform.

        Howard Brenner, Nishanth Dongari, Jason M. Reese A molecular dynamics test of the Navier-Stokes-Fourier paradigm for compressible gaseous continua

        Knudsen’s pioneering experimental and theoretical work performed more than a century ago pointed to the fact that the Navier-Stokes-Fourier (NSF) paradigm is inapplicable to compressible gases at Knudsen numbers (Kn) beyond the continuum range, namely to noncontinua. However, in the case of continua, wherein Kn approaches zero asymptotically, it is nevertheless (implicitly) assumed in the literature that the compressible NSF equations remain applicable. Surprisingly, this belief appears never to have been critically tested; rather, most tests of the viability of the NSF equations for continuum flows have, to date, effectively been limited to incompressible fluids, namely liquids. Given that bivelocity hydrodynamic theory has recently raised fundamental questions about the validity of the NSF equations for compressible continuum gas flows, we deemed it worthwhile to test the validity of the NSF paradigm for the case of continua. Although our proposed NSF test does not, itself, depend upon the correctness of the bivelocity model that spawned the test, the latter provided motivation. This Letter furnishes molecular dynamics (MD) simulation evidence showing, contrary to current opinion, that the NSF equations are not, in fact, applicable to compressible gaseous continua, nor, presumably, either to compressible liquids. Importantly, this conclusion regarding NSF’s inapplicability to continua, is shown to hold independently of the viability of the no-slip boundary condition applied to fluid continua, thus separating the issue of the correctness of the NSF differential equations from that of the tangential velocity boundary condition to be imposed upon these equations when seeking their solution. Finally, the MD data are shown to be functionally consistent with the bivelocity model that spawned the present study.

    • David Springer

      +1

  22. Just out of interest I used equation 107 from page 70 of G&Ts paper

    http://arxiv.org/pdf/0707.1161v4.pdf

    This was to contrast the radiative loss from a bottom face of a cubic metre of dry air with the radiative loss from the top face of a cubic metre of water.
    This is the most common interface on our planet covering almost 70% of the surface area.
    For both the calculated temperature drop is a modest one unit from 300K to 299K
    The loss of internal energy would be typical for night time conditions

    For air answer is 2 milliseconds
    For water it is 2.54 hours

    We don’t even need to invoke the second law to note that radiative warming of the surface by the atmosphere is pretty far fetched.

    • Yep. Yet when I mention that there is really two greenhouses to consider, the oceans and the atmosphere all you hear is crickets.

      • David Springer

        Two greenhouses to consider… agreed. Greenhouse gases are fluids with the defining properties of being transparent to shortwave and opaque to longwave. Water is a fluid possessing both definitive properties just as water vapor is a fluid with both properties.

  23. For those that seem to enjoy here the mental gymnastics of the gravito-thermo paradox as laid bare both logically and mathematically by Pierre-Normandes link to Combes-Lau, I highly recommend Thinking Physics is Gedanken Physics. No math (which after all is just a precise way to describe for calculation purposes the Gedanken). Willis’ posted ‘proof’ above of the gravito-thermo paradox is a classic Gedanken. As were Einstein’s elevators and trains for relativity.
    Book covers all introductory physics subfields: mechanics, vibrations (mechanics with complications), optics (light), EM, fluids, heat (a fluid problem with complications), and even relativity (mostly special). Really hones critical thinking skills useful for climate science and this blog.
    Author is Lewis Epstein. My version is the third edition ISBN 0-935218-08-4.

    Dedication paraphrased from the inside cover (subbing math for algebra):
    “Mathematics is a wonderful invention. It allows fools to do physics without understanding.”
    Another of those ‘sipping’ books (as I hope some will find Blowing Smoke) to be savored one chapter at a time.

    And, for those who cannot live without the math, Feynman’s 1962-63 Cal Tech Lectures on Physics provide all the requisite rigorous math for every problem in Thinking Physics (and then some), just in a different order. RF always emphasized the physical intuition first–“if you cannot explain it simply, then you do not really understand it yourself”. His treatment of fluid dynamic boundaries (aka ‘the religious epiphany in Volume 2 chapter 41 verse 6’) is a classic (having to do with with the mysteries hidden inside the Navier-Stokes equations and the use of renormalization groups, just as he also did with his quantum electrodynamics).
    Hope some the more technically inclined on this thread find this reference suggestion both useful and enjoyable. And lest you think thinking about physics has lttle to do with climate, see ‘Hot and Sticky’, in my edition the problem beginning page 238.
    Regards all.

    • Pierre-Normand

      “And, for those who cannot live without the math, Feynman’s 1962-63 Cal Tech Lectures on Physics provide all the requisite rigorous math for every problem in Thinking Physics (and then some), just in a different order.”

      They also include their fair share of enlightening thought experiments. And they now are available online. I had been delighted as an undergraduate student (not in his classroom!) by his discussion of the “ratchet and pawl” perpetual motion machine of the second kind, and how Brownian motion prevents the machine to work as intended, and thereby rescues the second law of thermodynamics.

      http://www.feynmanlectures.caltech.edu/I_46.html

  24. Yes, I think we are in an equilibrium state at present, unlike the twentieth century when we had two rises in global average temperature. We are in equilibrium now because the global average temperature has been almost constant since 1997. Why is that? The 1940 singularity provides the answer. Before 1940 the global average temperature had been rising at 0.15C/decade, after 1940 global average temperature actually fell despite extra CO2 concentration, until 1970. That 30 year gap was enough to start overcoming the transport lag in the oceans, so the oceans started to respond to the1910- 1940 ramp in temperature. But why did it stop in 1997? Carbon in CO2 can have up to six neutrons/carbon atom. Neutrons are heavy particles and can readily vibrate, absorbing energy from the earth’s IR at 14.99 microns, when the vibration reaches 100% of its absorption capability, it can absorb no more heat, so heat escapes to space and the earth cools. That is what I believe happened in 1040. But heat was still being absorbed by N2, O2 and H2o and that provides the equilibrium of today.

  25. Darn.
    There is friction of the water on the seabed, of the waves crashing against each other and the shore, of raindrops through air and gravity waves upwelling the earth constantly by deforming said earth and sea.
    Where there is a rotating planet air and water particles that move away from the mid line gain or lose potential energy due to the Coriolis force. This in turn generates the jet streams and currents and possibly even shapes the land very slowly.
    It is enough over time to wear out steel railway wheels in trains traveling North /South. We had a German machine in Darwin to cut the wheels back into shape.
    There is heat in the center of the Earth put down to Nuclear heating. Why? Because Gravity ensures enough energy input to force out these reactions.
    The Sea is compressed at 5000 meters and way away from any sun but does not freeze. The earth crust heat is very hot at 5000 meters depth but still not enough to warm that top comparative 10 microns of the Earth’s surface and radiate any heat to space Probably 1 meter in temperate zones in Winter, referred to as the Frost line for cement fillings.
    The mystery of the way heat, and geothermal heat interacts on this planet and the fact that no one here can convince 97% of people here, some of them very good brains, leaves me baffled.
    Rob or Fan or Dallas is there a text cited on this please.
    Or so help me I’ll go read Science of Doom.

  26. The problem for “the gravito-thermo paradox as laid bare both logically and mathematically by Pierre-Normandes link to Combes-Lau, ”

    is that the question assumes that a thermal equilibrium is attainable.

    Since work against gravity changes KE to PE and work with gravity chanmges PE to KE the vertical structure of an atmosphere is in permanent thermal disequilibrium so there can be no paradox and the whole issue is a straw man.

    • Pierre-Normand

      Stephen, the reason why the atmosphere is out of equilibrium is because it is heated by the warm Sun (mainly via the heating of the surface below) and it is radiating back to the cool space. Gravity alone can’t prevent a gas (either a planetary atmosphere or some gas in a box) from attaining thermodynamic equilibrium. Your argument about the effect from gravity on the balance of kinetic and potential energy of individual molecules is refuted in the papers that I have linked to at the beginning of this thread.

      https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-651619

      I also summarized the flaw in the PE+KE balance argument here:

      https://judithcurry.com/2014/11/27/open-thread-thanksgiving-edition/#comment-651519

      • Combes-Lau uses an assumed statistical relationship to achieve an assumed outcome. This sort of hand waving at so called fundamental principles is convincing only for those already convinced. Little math – little data – and little minds. A dangerous conflation.

        As I said – there seems little need or reason to take a position on something so fundamentally peripheral to the real issues – and something with so little real science – as in experimentation – involved. I guess it makes for a perfect playground for pedants.

      • Pierre-Normand

        Rob Ellison: “Combes-Lau uses an assumed statistical relationship to achieve an assumed outcome. ”

        What they did was to show why a seemingly valid argument is invalid. This invalid argument has been endorsed quite explicitly by Captdallas (who called it the “basic premise of the gravito-thermal effect”), David Springer, Stephen Wilde, and Doug C. You yourself seem to have been impressed by it in earlier threads.

        When the logical flaw in the argument is disclosed, then the standard result from kinetic theory doesn’t seem paradoxical anymore. This standard result, as Coombes and Laue note, can be obtained directly from the Boltzmann energy distribution law, but this isn’t going to impress college students since it is too abstract and it doesn’t highlight in an intuitive way what really is occurring among the different molecular populations at various heights. That’s the pedagogical virtue of their contribution.

        They aren’t proving anything controversial. The standard result is also proven in standard statistical mechanics textbooks (I referenced two of them) and a few other physics papers (I referenced three of them, not including Pekka’s useful note). It also is exhibited in the newer 2D molecular dynamics simulator that I linked to. (This also rebuts your earlier claim that molecules bouncing around elastically in a box don’t tend to exhibit a vertical density gradient at equilibrium.)

      • ““The time integral of the kinetic energy of any atom will be equal to the time integral of the kinetic energy of any other atom. This truism is simply and solely all that the Boltzmann–Maxwell doctrine asserts for a vertical column of a homogeneous monatomic gas.” Kelvin – See more at: http://www.mdpi.com/1099-4300/16/3/1515/htm#sthash.hzabWZj7.dpuf

        I have been consistent all along – hand waving at so called fundamental principles – as indeed Boltzman and Maxwell did – inspires little confidence. P-N infinitely less so.

      • I have of course stopped reading P-N – simply endless repetition of the same points. So few of them – so trivial – so little actual distillation and synthesis of science. Much verbiage and some few references waved around as talisman.

        But I caught the tail end of the last comment. 2-D bouncing balls with a free surface is not remotely the box of gas at the classical limit. It’s pretty. It ‘rebuts’ nothing at all.

      • Pierre-Normand

        Rob Ellison wrote: “But I caught the tail end of the last comment. 2-D bouncing balls with a free surface is not remotely the box of gas at the classical limit. It’s pretty. It ‘rebuts’ nothing at all.”

        In what relevant respect isn’t the 2D simulation relevant to an ideal gas at equilibrium in a box? What might be the cause of the vertical density gradient in the simulation such that it would not occur for an real monatomic gas? What is missing in the simulation?

      • Pierre-Normand

        I think your rebuttal of the gravito thermal effect is flawed because you are seeing kinetic energy simply as forward or sideways motion in a single plane.

        Kinetic energy also arises when gas molecules are forced together under pressure so that their vibratory movements are confined within a smaller space and become faster so as to release kinetic energy in the form of heat.

        It is that type of kinetic energy that changes back and forth between KE and PE not simple movement up, down or sideways.

        What happens is that the sun is constantly pumping the atmosphere up from the surface in regions of ascending air and gravity is constantly compressing the atmosphere in regions of descending air so that the heat of compression is being constantly replenished once the first convective cycle has been completed.

        That heat of compression is then locked into the adiabatic process in the form of PE and cannot escape to space as radiation. The process requires a reserve of kinetic energy at the surface to continue holding the atmosphere off the ground. That is our 33K greenhouse effect.

        That is a completely different scenario to simple one off gravitational compression which allows the heat to dissipate once the pumping and contraction stops.

      • Pierre-Normand

        “Kinetic energy also arises when gas molecules are forced together under pressure so that their vibratory movements are confined within a smaller space and become faster so as to release kinetic energy in the form of heat.”

        You are talking about adiabatic compression of a descending parcel of air, right? The source of the increase in internal energy (including kinetic energy), in that case, is the work from the surrounding on the boundary of the compressed air parcel. It isn’t gravitational potential energy.

      • I for one am very impressed both by Pierre-Normand’s persistence and his understanding of the issue. This is particularly so since I have observed on the blogosphere that most scientists blow smoke on the issue, making incorrect or irrelevant arguments that force us laymen to figure it out for ourselves.

        The only place I think he may have gone wrong is his (https://judithcurry.com/2014/11/27/open-thread-thanksgiving-edition/#comment-651592) of the purported refutation of an equilibrium lapse rate by way of the perpetual-motion argument. Although I can’t say I know exactly what the result of connecting a thermocouple would be, I believe that such an approach suffers from the same logical flaw that Dr. Brown’s purported proof had. As I mentioned above, I had dealt with that flaw here: http://wattsupwiththat.com/2014/08/18/monday-mirthiness-spot-the-troll/#comment-1711550.

        But a shorter, different way to look at it is this: it seems unlikely that connecting a thermocouple to an otherwise-isolated ideal-gas column will magically convert its ensemble of possible microstates into a canonical ensemble, i.e. into the ensemble that would result from being bathed in an infinite-heat-capacity heat bath.

        Be that as it may, I take my hat off to Pierre-Normand’s stamina–and ability to stick to the issue.

      • Pierre-Normand

        Joe Born,

        Thank you for the kind words.

        Notice that in the post that you linked to, where I was replying to David Springer, the effect at issue was the ordinarily discussed gravito-thermal effect which is very many orders of magnitudes larger than the effect discussed by Velasco et al. This alleged effect doesn’t rely on the system being fully isolated from its surrounding. In that case, the consequence from using a thermocouple is that energy can be extracted from the column while this energy can be replenished spontaneously from a thermal bath that is is in contact with it (anywhere along the length of the column). This is in violation of the second law and it constitutes a perpetual motion machine of the second kind.

      • David Springer

        @Joe Born

        I refuted Brown’s PM2K years ago on WUWT. In a gravity confined gas column the volume is variable with temperature. As you cool the hot side and warm the cool side the column heights diminish and grow respectively until the difference in lapse rate is nullified. The PM2K grinds to a halt well before you’ve extracted enough useful work to reduce both gas columns to solid ices at the bottom.

      • Pierre-Normand

        David Springer, the second law of thermodynamics doesn’t merely imply that you can’t extract enough energy from a unique thermal bath and thereby perform enough macroscopic work to usefully reduce your energy utility bills. It states that you can’t extract *any* energy in that fashion.

      • David Springer

        Your understanding of perpetuum mobiles of the second kind is wrong.

        Suggest you read up on 2LoT PM’s here:

        http://lectureonline.cl.msu.edu/~mmp/kap12/cd333.htm

        PM2K’s are not prohibited by thermodynamics. They are prohibited by engineering reality that ideal things like frictionless bearings, perfect vacuums, isolated systems, ideal gases, and so forth are physical impossibilities. No device known can turn heat into work with 100% efficiency thus all PM2Ks stop running at some point.

        In the world of engineering we simplify as much as practical but no more than that. For instance if we are timing an air race we don’t make relativistic corrections for differences in local gravity and inertial reference frames at different race courses. However if we are meausuring nanosecond differences in precision clock rates in orbiting satellites vs. the clock in the ground receiver in GPS applications we must correct for relativistic effects or live with errors large enough to spell the difference between hitting a schoolhouse with a GPS guided missile instead of the munitions factory across the street from it. In space exploration we don’t try to compute intractable effects of multi-body orbital mechanics we instead make mid-course corrections based on empirical measurement of location and velocity. For a friction example we don’t worry about the effect of air-drag on some wheeled vehicles like earth movers but we must do it for Formula 1 cars to be competitive. It’s all about context.

      • Pierre-Normand

        David Springer: “PM2K’s are not prohibited by thermodynamics.”

        Yes, perpetual motion machines of the *second* kind are prohibited because they violate the *second* law of thermodynamics. (While perpetual motion machines of the *first* kind violate the *first* law: conservation of energy.) That’s because they operate in such a manner that heat extracted from a thermal bath is converted into macroscopic work with no compensating heating of a colder thermal bath (as is the case with ordinary thermal engines) and, as a result, the total entropy of the Universe is reduced.

        If work is extracted from a column of gas that exhibits a spontaneous gravito-thermal gradient, then the column can be warmed back through putting it in contact with a large thermal bath. Then the initial temperature profile will be spontaneously restored after a while and some more work can be extracted. Through iterating this process, one is steadily converting heat from the large thermal bath into usable macroscopic work (and hence reducing its entropy) without generating nearly enough waste heat to satisfy the second law.

      • If a gravito-thermal gradient exists – we may presume that it doesn’t contravene the first or second law.

      • Pierre-Normand

        Rob Ellison: “If a gravito-thermal gradient exists – we may presume that it doesn’t contravene the first or second law.”

        But you do agree with me that perpetual motion machines of the second kind contravene the second law. Don’t you? That was my issue with Singer.

    • A fan of *MORE* discourse

      Rob Ellison spews “[logic-free personal abuse redacted]”

      Your penchant for personal abuse is appreciated, Rob Ellison!

      Especially to assist your understanding, a simplified (yet vivid!) mathematical analysis/laboratory experiment has now been posted on this thread.

      It is a continuing pleasure to assist your understanding of basic thermodynamical and statistical mechanical principles, Rob Ellison!

      \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • Yes I noted. I just replied.

        We have arm waving at perpetual motion, experimental error without an experimental design and a litany of mad FOMBS assertions that are typically not worth reading.

        There would be horrendously more energy being put in than could be extracted, there is a time to critique experimental design and that would seem to be after the design is complete and really we are not discussing the creation of energy by the mere fact of rotation. One can’t create horrendous temperatures in a centrifuge out of nothing. On a scale of silliness – the last ‘bad boy’ is a million.

        And as I have said before – the whining by such as FOMBS about insults and abuse is simply sanctimonious hypocrisy.

        Beside I was being helpful – as I have tried to be many times with P-N. I have stopped reading and this is why. A few few hundred comments saying the same things – and I might pass over fewer.

        FOMBS I nearly always pass over – unless he directs his comment at me. Then I want to be as abrasive as possible simply to discourage the behaviour. Negative reinforcement. The level of FOMBS silliness is just too great to want to have much to do with it.

      • A few hundred less…

      • A fan of *MORE* discourse

        Rob Ellison states policy “FOMD I nearly always pass over – unless he directs his comment at me.”

        FOMD notes  Ultracentrifuge rotors spin hour after hour with NO ROTOR-COOLING WHATSOEVER.

        Were it not for friction in the bearings, the rotor would spin forever with no heating.

        No rotor-cooling is needed for this simple reason: THERE IS NO GRAVITO-THERMAL EFFECT.

        Please consider the ultracentrifuge test-case to be cheerfully directed toward gravito-thermal enthusiasts in general!

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      • The rotors spin without creating energy? What a shame – we would otherwise be able to power the world using FOMBS comments.

      • Pierre-Normand

        FOMD wrote: “Were it not for friction in the bearings, the rotor would spin forever with no heating.

        No rotor-cooling is needed for this simple reason: THERE IS NO GRAVITO-THERMAL EFFECT.”

        I don’t get this argument. There ought to be some friction in the bearings and hence some warming. If no rotor-cooling mechanism is supplied that must be because the warming isn’t large enough to need it. If there were a gravito-thermal effect, and if it would affect the rotor as a whole, then it would carry heat away from the rotor (since the simulated gravitational field would be directed outside, radially). So, the complete absence of any heating near the axis would rather *favor* the existence of a gravito-thermal effect!

        I rather think, for reasons Pekka gave, that the alleged effect is too small to be detected with this setup.

  27. Pierre-Normand said:

    “You are talking about adiabatic compression of a descending parcel of air, right? The source of the increase in internal energy (including kinetic energy), in that case, is the work from the surrounding on the boundary of the compressed air parcel. It isn’t gravitational potential energy.”

    If the process is adiabatic then the work done is against gravity (ascent) or with gravity (descent). By definition.

    Any work done on surrounding molecules is diabatic. By definition.

    How is the PE derived from or lost to the adiabatic process not gravitational potential energy ?

    Does it matter anyway?

    Whether the PE locked into the overturning cycle is gravitational or not it still requires a store of KE at the surface to sustain continued convective overturning once the first convective cycle has completed and that KE at the surface cannot be radiated to space or the atmosphere will fall to the ground.

    Hence 288K instead of 255K at the surface.

    This is not a simple matter of one off gravitational compression. It is a continuing process that needs fuel in the form of KE at the surface to sustain it. That is where the radiative theory is flawed. It makes no provision for such an ongoing mechanical process or the surface energy needed to keep it running.

    • Pierre-Normand

      “If the process is adiabatic then the work done is against gravity (ascent) or with gravity (descent). By definition.

      Any work done on surrounding molecules is diabatic. By definition.”

      That’s incorrect. A process is adiabatic if it doesn’t involve flows of heat or matter. One typical example is the second stage of the Carnot cycle. The gas in the cylinder expands and thereby performs work on the receding piston. Though the gas loses internal energy (and cools) there is no flow of heat through the piston. Adiabatic compression and decompression of descending and raising air parcels in the atmosphere are likewise adiabatic processes that involve work on (and by) the surrounding. The amount of heat gained or lost in those processes is completely unrelated to the gain or loss of gravitational potential of the air parcels.

      See “Stage 2” in the following link:
      http://www.massengineers.com/Documents/carnot_cycle.htm

      • “Stage 2: In the second stage, the heat source is removed; the piston continues to move downward and the gas is still expanding while cooling (lowering in temperature). It is presented by the graphic from point B to point C. This stage is called a adiabatic expansion (Energy stays)”

        Energy stays during convective uplift of gases. It is just converted in form from KE to PE
        Work done with or against gravity does not involve a flow of heat or matter.
        That is why it is adiabatic.

        Once a parcel of air detaches from the surface during uplift all further upward movement is adiabatic because all work is done against gravity and not against surrounding molecules.

        You think that the word ‘surroundings’ means the adjoining molecules but in fact it means the gravitational field in relation to adiabatic uplift within an atmosphere.

        The reason for the lack of interaction with adjoining molecules is that the pressure on the parcel reduces at the same rate as the pressure in the surroundings so the initial temperature differential that caused uplift to begin is maintained throughout the period of adiabatic uplift. If work were being done against adjoining molecules then that would not be the case. Instead the initial temperature differential would quickly decline as heat flowed out of the parcel to the adjoining molecules.

        “The amount of heat gained or lost in those processes is completely unrelated to the gain or loss of gravitational potential of the air parcels”

        Using work against gravity to drive molecules upward and further apart (thereby cooling) within a gravitational field always results in more gravitational potential energy whether the movement be relative to the surface below or relative to the adjoining molecules.

      • Pierre-Normand

        Wildeco2014 (Stephen Wilde?) wrote:

        “Energy stays during convective uplift of gases. It is just converted in form from KE to PE Work done with or against gravity does not involve a flow of heat or matter. That is why it is adiabatic.”

        Work done on the surrounding through expansion also is an adiabatic process. You are incorrectly mixing up two different processes.

        (1) When an air parcel rises in the atmosphere then it gains potential energy. Work therefore must be performed by the surrounding air in order to lift the parcel against gravity. This work is performed by the buoyant force that results from the vertical barometric pressure gradient around the air parcel. This produces a net force that performs the required work. This works is equal to the gain in potential energy of the rising parcel. But where does the energy come from for performing this lifting work? As the parcel rises, it displaces some air above and other parcels of air replace the evacuated space. The net motion of those other displaced air parcels is downward. Those descending air parcels thus are losing gravitational potential energy at the same rate that the rising parcel gains some. That’s the first process. It conserves gravitational potential energy between *all* the involved air parcels and it has no incidence on molecular kinetic energy. It has no direct effect at all on the internal energy or temperature of the rising and descending air parcels.

        (2) The second process, quite independent from the first one just described, is the the adiabatic expansion (or compression) of the rising (or descending) air parcels. This process is governed by variations in ambient pressure around the moving parcels. This is the process that account for the variation in internal energy though work:

        dW = P(h)dV, integrated over the vertical displacement of the parcel.

        It is quite unrelated to variation in potential gravitational energy of the parcels.

        Though mixing up conceptually those two different processes you are not accounting correctly for the energy exchanges. The adiabatically expanding air parcel *must* lose internal energy while it performs work through expanding within its pressurized surrounding. The parcel can’t use up this same energy to lift itself up against the force of gravity. That wouldn’t make sense anyway unless this kinetic energy would result from a net initial upward motion of the bulk of the molecules. But the velocities are random and mostly isotropic. The gain in potential gravitational energy already is accounted for by the external buoyant force.

  28. A fan of *MORE* discourse

    Pierre-Normand claims [absurdly] “The alleged gravito-thermal effect is too small to be detected [in million-g ultracentrifuges]

    LoL … in what sense are wrongly-predicted temperature differentials several thousand degrees “too small to be detected”?

    Riddle  What spins faster than an over-stressed ultracentrifuge rotor?

    Answer  Gravito-thermal enthusiasts confronted by physical evidence!

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    • Pierre-Normand

      FOMD: “LoL … in what sense are wrongly-predicted temperature differentials several thousand degrees “too small to be detected”?”

      Yes, it’s true 1,000,000 g is quite a large gravitational acceleration. If the gavito-thermal gradient is about as large as the dry adiabatic lapse rate (about 10°K/km), as some proponents allege, then over a radial distance of 1cm, the equilibrium gradient would produce a 100°K differential. That’s indeed larger than I thought. As Pekka notes, though, just a little amount of convection would also produce it, so this would still be a confounding factor that needs to be eliminated.

      • A fan of *MORE* discourse

        Yes, and filling the centrifuge-tubes with heavier fluid-molecules like organic solvents (which is routine) or filling the tubes with even-heavier xenon gas or mercury (which is not so common, but is perfectly feasible) increases the gravito-thermal effect in proportion to the increased atomic mass … to several thousand degrees.

        Yikes!

        Conclusion  Naive gravito-thermal theory has perished … at the hands of everyday experiments.

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      • When the centrifuge is turned on, there will be compression at one end and expansion at the other. In addition some work is converted to heat. Thus the effect of heating one end of tubes and cooling of the other end takes place anyway initially. No “gravito-thermal effect” is needed for that.

        Whether a “gravito-thermal effect” would enhance the initial heating of the outer end is impossible to tell, because no theory exists that describes the effect, there are only incomplete proposals that cannot answer questions of this type. The existing complete enough theory tells that the temperature difference disappears gradually, if the gas samples are really isolated well enough from external heat sources and sinks.

      • A fan of *MORE* discourse

        Pekka Pirilä imagines “When the centrifuge is turned on, there will be compression at one end and expansion at the other […] Thus the effect of heating one end of tubes and cooling of the other end takes place anyway initially.”

        Compressive/decompressive effects are transient, and moreover are small for nearly-incompressible liquids (such as organic solvents and mercury in the worked example).

        Whereas gravito-thermal theory predicts a temperature-differential of several thousand degrees that is sustained throughout centrifuge-runs that routinely last for hours and even days.

        And yet  even the most delicate ultracentrifuged proteins show no signs of being hard-boiled by gravito-thermal heating.

        Conclusion  gravito-thermal theory is dead as a dodo!

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      • Pierre-Normand

        FOMD: “Conclusion gravito-thermal theory is dead as a dodo!”

        Ever since the Dodo went extinct, the Dodos who are dead have been outnumbered by the Dodos who will never be born.

      • What theory?

        I have seen only vague and highly incomplete hypotheses that very likely cannot be expanded to a full theory.

        But what could be the source for the energy that results in those high temperatures, if a theory could be formulated. When the transient phase is over, no more energy enters the system.

        Dodos have once existed in contrast to theories that include the “Gravito-Thermal effect”.

      • Pekka said, “What theory?”

        Funny how this thought experiment keeps getting promoted. I believe it has led to a few theorems though. The Fluctuation Theorem comes to mind.

      • It came to my mind that there are centrifuges that operate in the regime, where diffusion is significant relative to convective and turbulent mixing. All gas centrifuges used for isotope separation are such devices. Thus measurements of temperature distributions in such gas centrifuges should provide the evidence.

      • Pekka, “All gas centrifuges used for isotope separation are such devices. Thus measurements of temperature distributions in such gas centrifuges should provide the evidence.”

        I can’t find any. From what I have seen most applications involve heating or cooling to improve separation. What effect there is is so small it would be difficult to measure.

    • I have a question for Fan & Pierre-Normand.
      How do you explain the Atmospheric Temperatures of the Solar Systems Gas Giants?
      They are much further from the Sun than Mars and therefore get even less radiation from the Sun and yet their Atmospheric Temperatures are 100s of times higher than Mars and Earth.
      It is accepted science that it is due to the Massive Gravity of those worlds.

      • Pierre-Normand

        I’m not sure about that A C Osborn. I’ve heard that a small rate of gravitational collapse accounts for some of the excess energy radiated compared with the solar input but that may not fully answer your question. I don’t know anything about the atmosphere of other planets.

  29. This is interesting:

    http://www.teachastronomy.com/astropedia/article/Thermal-Radiation-from-Gas-Giant-Planets

    It seems that there are a number of possible reasons why a planet might emit more than it receives from the sun but in the case of Uranus they say:

    “Interestingly, Uranus doesn’t appear to have significant excess heat”

    but given the very high gas temperatures they must simply mean that it does not emit much excess heat to space i.e. it is in approximate thermal balance with the sun.

    Obviously the ‘surface’ (however defined) is hot enough to emit at 83,000 times more than it receives from the sun but the heat is not actually getting out in the form of thermal emissions.

    That fits with the proposition that the excess heat within the compressed atmosphere is a result of adiabatic recycling of energy being conducted and convected up and down within the gases in just the same way that I propose the Earth radiates to space at 255K from a surface at 288K but recycles the other 33K adiabatically.

    That is different from one off gravitational compression and I think that is where the AGW proponents go wrong.

    The adiabatic process involves continuous pumping up by the sun in regions of ascent simultaneously with continuous compression by gravity in regions of descent and so there is constant replenishment of the surface kinetic energy needed to sustain the process and thereby keep the weight of the atmosphere suspended off the surface.

    Adiabatic decompression and compression is quite different from simple gravitational compression and given that Uranus has very hot gases but is nonetheless in approximate thermal balance with solar radiation it is a good example to use.

  30. To correct my previous post the temperature of Uranus is not high but the temperature is in excess of that expected from solar input so the basic proposition holds:

    “Weather on Uranus functions much as it does on other gas giants. Like Jupiter and Saturn, the planet has bands of zones and belts that orbit parallel to the equator, which is warmer than the poles. The warm temperatures that drive the planet’s weather come from the interior of the planet, rather than from the sun. The significant distance to Uranus from the sun may play a role in why the planet’s interior heat overpowers the faint light from the star.”

    from here:

    http://www.space.com/18707-uranus-temperature.html

    • The temperature of Uranus is very high, just not at the outer surface of it’s Atmosphere, as you would expect.

    • Pierre-Normand

      The Wikipedia article about Uranus states that it’s internal heat is unexpectedly low and it isn’t well understood why that is so.

      http://en.wikipedia.org/wiki/Uranus#Internal_heat

      • Yeah, right, “The outermost layer of the Uranian atmosphere is the thermosphere and corona, which has a uniform temperature around 800 to 850 K”
        Note that is the outer most layer.

      • Pierre-Normand

        No surprise there. The thermosphere of a planet is heated up by Solar UV. UV radiation is very hot. Further the volumetric heat capacity of the thermosphere is extremely low since it is so sparse. Being largely ionized also gives is a large cross-section to capture UV photons.

      • You do realise how far away from the Sun the Gas Giants are?
        The closest is 5 times as far as the earth.

      • Pierre-Normand

        Sure, A C Osborn. This might explain why the thermosphere of Uranus is so very much colder than the thermosphere of the Earth.

  31. More detail about Uranus:

    “The troposphere is the lowest and densest part of the atmosphere and is characterized by a decrease in temperature with altitude. The temperature falls from about 320 K at the base of the nominal troposphere at −300 km to 53 K at 50 km. The temperatures in the coldest upper region of the troposphere (the tropopause) actually vary in the range between 49 and 57 K depending on planetary latitude. The tropopause region is responsible for the vast majority of the planet’s thermal far infrared emissions, thus determining its effective temperature of 59.1 ± 0.3 K.
    The troposphere is believed to possess a highly complex cloud structure; water clouds are hypothesised to lie in the pressure range of 50 to 100 bar (5 to 10 MPa), ammonium hydrosulfide clouds in the range of 20 to 40 bar (2 to 4 MPa), ammonia or hydrogen sulfide clouds at between 3 and 10 bar (0.3 to 1 MPa) and finally directly detected thin methane clouds at 1 to 2 bar (0.1 to 0.2 MPa). The troposphere is a very dynamic part of the atmosphere, exhibiting strong winds, bright clouds and seasonal changes”

    The troposphere seems to exhibit similar convective phenomena to ours and is as warm as 320K as against Earth’s 288K despite the weakness of insolation so I think that gravito-thermal activity is present in that region just as it in our troposphere.

    Lower and denser levels from the base of the troposphere become colder which suggests to me something similar to the structure of our oceans which cool with depth.

    Overall it seems a suitable analogue given that the troposphere temperatures cannot be accounted for by insolation or heat coming up from below.

    The convective region does seem to require its own store of KE to maintain constant overturning just like on Earth.

  32. From Wiki “The outermost layer of the Uranian atmosphere is the thermosphere and corona, which has a uniform temperature around 800 to 850 K”

  33. Pierre,

    I’ve come across the buoyant force argument before but think it flawed in that buoyancy requires no force against surrounding molecules when the movement is into a region of lower pressure. In that situation the work done is all against gravity and so is purely adiabatic.

    The thing is that once a parcel of air heats up and becomes less dense relative to its surroundings then it develops buoyancy within itself and rises spontaneously.

    It also applies in reverse. A descending parcel of air does no work against surrounding molecules when descending into a region of greater pressure because the parcel is being compressed at the same rate as its surroundings.

    The concept of a buoyancy force external to the parcel has confused a lot of people so you are not alone.

    Obviously no process is purely adiabatic so there is some small amount of work done against surrounding molecules but that is then a diabatic process whereby some of the energy in the adiabatic process is able to ‘leak’ out to the surroundings.

    I think it is you who is conflating the diabatic and adiabatic processes.

    • Pierre-Normand

      Stephen Wilde,

      You must have misread me. I was precisely arguing that the work preformed by the buoyant force is done against gravity and the process involves no change in internal energy. Both the processes that I have described are adiabatic, but only the second one (compression or expansion) yields a change in internal energy of the moving air parcel — and hence a change in average molecular kinetic energy.

      An air parcel can’t lift itself up against gravity. That’s nonsense. Likewise, you can’t lift yourself up through pulling on your own bootstraps. Of course, for the buoyant force to begin working against the force of gravity (when the parcel is initially at rest), it must overcome this force. For this to occur, the density of the air parcel must be reduced compared with its surrounding. This can occur through latent heat release (through water vapor precipitation), or radiative heating, etc. In that case, the air parcel begins to rise.

      Since the buoyant force is the result of the net external forces from the surrounding air (equal to the pressure of the surrounding air all around the parcel integrated directionally over the whole surface) applied on the parcel, and this force acts over the vertical displacement of the parcel, it performs work. This work is supplied from outside the parcel, and the energy for performing it mainly comes from the lost gravitational potential energy of the air that must descend to fill up the volume evacuated by the rising parcel.

  34. the density of the air parcel must be reduced compared with its surrounding. This can occur through latent heat release (through water vapor precipitation), or radiative heating, etc. In that case, the air parcel begins to rise.”

    I don’t see why that cannot be interpreted as the parcel rising spontaneously from its own internal buoyancy with no need for pushing from outside.

    “This work is supplied from outside the parcel, and the energy for performing it mainly comes from the lost gravitational potential energy of the air that must descend to fill up the volume evacuated by the rising parcel”

    I think that illustrates where the idea of a buoyancy force acting on the parcel from outside goes wrong.

    It is a matter of getting cause and effect in the right order.

    For you to be right the descending air has to make the first move in order to push the lighter air up by way of exerting an external buoyancy force.

    However, in reality the first move is by way of the warmed lighter air starting to move away from the surface due to its own internal buoyancy as imparted to it by surface warmth.

    There is a test as to which interpretation must be correct and that is the development of lower pressure beneath ascending warm air as seen in low pressure cells.

    Thus the parcel rises first, pressure below it reduces and air is pulled in from the sides which is the opposite of your scenario.

    • Pierre-Normand

      SW wrote: “I don’t see why that cannot be interpreted as the parcel rising spontaneously from its own internal buoyancy with no need for pushing from outside.”

      Because that violates Newton’s second law of motion. For the parcel to begin to rise as a whole, you need a net upward force applied on it from outside. Gravity is one component of that force, while buoyancy is another. The internal forces are irrelevant to the acceleration of the parcel as a whole.

    • Pierre-Normand

      SW: “For you to be right the descending air has to make the first move in order to push the lighter air up by way of exerting an external buoyancy force.”

      That’s not true. Suppose two identical buckets of water are suspended from two ends of a string that goes over a pulley. If you remove some water from the first bucket, then it will start to rise while the heavier bucket starts to descend. The descending bucked didn’t need to “make the first move”. This first bucket made the first most just through becoming less heavy.

      • Gases are different to liquids and solids because of their low density and internal mobility.

        The more kinetic energy they carry the higher they will drift against gravity without any additional external force being applied.

        That is why we need the Gas Laws.

        You are focusing on the behaviour of liquids and solids which do operate under Newtonian physics because they are so much heavier and denser with more powerful inter molecular bonds than gases so that they do need an external push.

        As for your bucket analogy you are removing mass which is an entirely different situation to reducing the density of the same mass. If you kept the same mass in the bucket but reduced its density it would still have the same weight and neither would move. The difference is that the water in the buckets is not subject to the Gas Laws but to Newtonian physics as you point out.

      • Pierre-Normand

        Stephen, the gas laws apply to gases in addition to the laws of mechanics. They aren’t replacement for them. A system can’t violate Newton’s second law of motion just because is has gaseous or liquid parts.

      • Pierre-Normand

        GW: “If you kept the same mass in the bucket but reduced its density it would still have the same weight and neither would move. The difference is that the water in the buckets is not subject to the Gas Laws but to Newtonian physics as you point out.”

        My argument simply refutes your “has-to-make-the-first-move” causal argument. In the bucket case, the net forces on the buckets are the sum total of the force of gravity and the (opposite) tension in the supporting string. In the air parcel case, its the sum total of the weight of the air and the (external) buoyant force on the parcel. The only difference is that the change in net force results from the buoyant force being modified by an initial change in volume (rather than a change in weight). This difference is not relevant to the flaw in your argument from causality. The expanding parcel of air is free to “make the first move” (e.g. though initiation of a condensation process). It will still be caused to rise within the denser fluid as a result of its spontaneous expansion.

    • Pierre-Normand

      SW: “There is a test as to which interpretation must be correct and that is the development of lower pressure beneath ascending warm air as seen in low pressure cells.

      Thus the parcel rises first, pressure below it reduces and air is pulled in from the sides which is the opposite of your scenario.”

      The pressure below the rising parcel may become lower at this location than it was before the parcel began to rise (just from Bernouilli’s principle). This doesn’t show that the pressure is lower below the parcel than it is *above* the parcel. If this were the case, and the less dense rising cell would not decelerate, then Newton’s second law of motion would be violated.

      • In adiabatic uplift the less dense rising cell does not decelerate.
        Gases are ruled by the Gas Laws and not Newtonian physics.

      • Pierre-Normand

        SW wrote: “In adiabatic uplift the less dense rising cell does not decelerate.
        Gases are ruled by the Gas Laws and not Newtonian physics.”

        The laws that govern the motion of macroscopic air parcels also include laws of thermodynamics. They are consistent with the laws of mechanics and can’t conflict with them. At any moment in time, the acceleration of an air parcel relates to the sum total of the external forces that are acting on it in accordance with F = ma (where ‘F’ and ‘a’ are directed vectors).

      • Pierre-Normand

        Also, for sure the uplifted air parcel doesn’t decelerate at first. That’s just *because* the external buoyant force applied on the less dense parcel of air is larger than the weight of the parcel. (Though viscous forces also ought to be taken into account and, eventually, will equilibrate the two other forces such that the rising air parcel will reach a terminal velocity even if the buoyant force is constant, just like sky divers do.)

  35. “only the second one (compression or expansion) yields a change in internal energy of the moving air parcel — and hence a change in average molecular kinetic energy.”

    There is no change in internal energy from gravitational compression or decompression. There is only a transformation of energy between KE and PE.

    Decompression converts KE to PE for a cooling effect.

    Compression converts PE to KE for a warming effect.

    Total internal energy (KE+PE) remains the same.

    As per the Gas Laws.

    • Pierre-Normand

      “There is no change in internal energy from gravitational compression or decompression. There is only a transformation of energy between KE and PE.”

      I said nothing about “gravitaitonal decompression” whatever that is. When the parcel expands adiabatically within its pressurized environment, then it performs work on its surrounding dW = PdV, and the law of conservation of energy ensures that the change in internal energy dU = -dW. This is also explained at the molecular level by the fact that molecules that bounce back from receding molecules on the periphery of the expanding mass of air bounce back with a lower velocity on average. The process has nothing to do with gravity.

      • If it does work on surrounding molecules such that there is a change in total internal energy (KE+PE) then that is diabatic and not adiabatic.

        To be adiabatic it must do work against gravity.

      • Pierre-Normand

        SW: “To be adiabatic it must do work against gravity.”

        You are making up definitions. The second stage of the Carnot cycle is a paradigmatic example of an adiabatic expansion process that performs work on the surrounding (the piston, in that case). A Carnot engine works just the same in the absence of gravity.

      • Pierre-Normand,

        The sooner you withdraw from this line of discussion, the better. Stephen Wilde has been arguing his nonsense physics about the adiabatic process against a whole truckload of pretty hardcore sceptics on Tallbloke’s blog for a long time. He’s been shown a gazillion times exactly what the adiabatic process IS and is NOT, and still he simply refuses to listen, to read, completely and utterly tangled up in his own warped ideas of how the world supposedly functions …

        Just a warming.

      • Hehe, warNing, not warming.

      • David Springer

        Pierre-Normand | December 2, 2014 at 11:31 am |

        “A Carnot engine works just the same in the absence of gravity.”

        Who cares? Convective cells in the earth’s atmosphere won’t work in the absence of gravity. Duh.

      • David Springer

        Pierre-Normand | December 2, 2014 at 11:31 am |

        “A Carnot engine works just the same in the absence of gravity.”

        Carnot defines “work” as “weight lifted through a height”.

        What does something weigh in the absence of gravity?

      • Steven Mosher

        “The sooner you withdraw from this line of discussion, the better. Stephen Wilde has been arguing his nonsense physics about the adiabatic process against a whole truckload of pretty hardcore sceptics on Tallbloke’s blog for a long time. ”

        he’s a lawyer. Note you will never see a formula or quantitative statement from him that can be tested.

      • Pierre-Normand

        David Spinger,

        This is completely beyond the point. SW is denying that adiabatic expansion of a rising parcel of air performs work of the surrounding. He rather believes the loss of internal energy of the gas to be converted into gravitational potential energy of the rising gas. That’s a crude misunderstanding of the work of adiabatic expansion (and it also is quantitatively wrong). Also, work isn’t defined as work against gravity. The definition of a Carnot cycle leave it completely open what the work perfomed by the adiabatically expanding gas on the walls of the expanding container (or piston) is converted into on the other side of the walls (or piston).

      • Pierre-Normand

        Kristian,

        I am not arguing with people on discussion fora because I have a high expectation that they will change their minds anytime soon, if ever. Those are not private discussions.

      • Yes, I have the same approach. I’m just saying that this particular discussion with this particular person is particularly pointless. We ALL know what the adiabatic process is and is not. There is basically nothing to discuss. SW, however, adamantly asserts we’re ALL fundamentally wrong, even Carnot and Clausius themselves.

      • Pierre-Normand

        “We ALL know what the adiabatic process is and is not.”

        “We” seems to include very few people on this blog.

  36. Here is a brain teaser for you.

    Consider a small heat source suspended high above the earth’s surface with a much higher temperature than the surrounding air. For example we could simply use a hot air balloon made of high emissivity material and carrying a heavy load.

    Now take an IR spectrometer and measure the spectra emitted by the balloon from a fixed distance but at 3 different positions.

    1. At the same heigh as the balloon
    2. Directly above the balloon
    3. Directly below the balloon

    How the spectra differ if at all ?

    • Pierre-Normand

      Clivebest, is the gravitational red-(blue)-shift of the radiation relevant?

      • The interference with the earth’s IR radiation will complicate things. However the main point is that any difference between the up and down spectra would prove that there is a net radiative transfer upwards. All measurements made in any direction at the same height should give the same result.

      • Clivebest,
        This is a calibration issue. Simply take a pyrgeometer, the temperature of the cell is part of the measurement. If the cell is at ambient temperature you will have three different flux measured, three different corrections and three identical results.

    • Pierre-Normand

      Is pressure broadening of GHGs absorption lines relevant?

    • There is no trick question here.

      I would expect to see a BB spectrum at the balloon temperature, with pressure broadened absorption lines dependent on the viewing position.

      From below you would see the largest absorption and from above you should see less absorption. However, in the horizontal you should see a perfect BB spectrum with no absorption at all !

      This experiment should all be done at night of course and maybe also over a desert. Otherwise the background IR from the earth’s surface and atmosphere would wash everything out.

      • Pierre-Normand

        Clivebest “From below you would see the largest absorption and from above you should see less absorption. However, in the horizontal you should see a perfect BB spectrum with no absorption at all !”

        Why is that? I was also expecting more broadening from below, just because the pressure of the air is larger. However, horizontally, the pressure of the air is intermediate in between the (average) pressures along the lines of sight from above and below. Why wouldn’t the GHG molecules absorb any IR from the warm black-body in the horizontal direction?

      • Pierre-Normand

        Clivebest,

        OK, I think I see what you were driving at; bit there may have been an unwanted assumption in your setup. You said that the black-body is much *warmer* than the surrounding air! In that case I would expect abortion to occur in all three cases (though with variable amounts).

        If, on the other hand, the black-body were at the *same* temperature as ambient air at that level, and there is a lapse rate, then one would see absorption in the GHG bands above, more emission in those bands below, and a pure unmodified BB spectrum horizontally.

  37. It is not the pressure differential between the top and bottom of a hot air balloon that causes it to rise because that is offset by the gravitational field for a net zero effect.

    It is the lower density of the air within the balloon that causes it to rise spontaneously.

    • Pierre-Normand

      Stephen Wilde, this is a false dichotomy. The balloon will begin to rise if and only if the net force is up. This net force is the sum of the weight of the balloon (down) and of the buoyant force (up). The pressure differential around the balloon determines the strength of the buoyant force. In conditions of hydrostatic equilibrium within the surrounding, it is equal to the weight of the displaced fluid. The expansion of the balloon increases the buoyant force because it then displaces more air around it. It doesn’t change the weight of the balloon. But just changing the external buoyant force thereby changes the net external force.

      • I found an aeronautics site which explained that the effect of the pressure differential between the top and the bottom of a balloon is negated by the pull of gravity on the balloon for a zero net effect.

        That applies to solids such as the material that comprises the fabric of the balloon.

        Gases are different so if you have a lower density of gas at a given temperature as compared to the surroundings then it will rise spontaneously towards a region of lower pressure.

        The ballooon will rise only when the reduced density within it swings the balance between the pressure differential from top to bottom and the gravitation pull downwards.

        The point is that:

        Solids to not rise upward when heated. An external buoyancy force is needed.

        Liquidfs do not rise up when heated. An external buoyancy force is needed.

        Gases DO rise up when heated.An external buoyancy force is NOT needed.

        Gases do that without any external buoyancy force pushing them up simply because warmer, lighter gases are inherently more buoyant than colder, heavier gases and will alter their position along the lapse rate slope spontaneously if they are not at the correct height for their temperatutre and density.

        I can sense that you are unwilling to consider the possibility that it is so.

      • Pierre-Normand

        Stephen Wilde,

        SW: “I found an aeronautics site which explained that the effect of the pressure differential between the top and the bottom of a balloon is negated by the pull of gravity on the balloon for a zero net effect.”

        As I said earlier, under conditions of hydrostatic equilibrium, the buoyant force on a body (or air/liquid parcel) is equal and opposite to the weight of the displaced fluid. So, of course, when the density of the balloon is the same as the density of the ambient air, the two forces will be equal and cancel exactly.

        When the density of the gas in the balloon is less than the density of the surrounding air, the the buoyant force is exactly the same (since it still is equal to the weight of the displaced outside air) but the weight of the gas is reduces, so the two forces don’t exactly cancel anymore.

        “Gases DO rise up when heated.An external buoyancy force is NOT needed.”

        The gas rises when heated because when it thermally expand the buoyant force increases as a result. That’s because the gas thereby displaces more surrounding fluid. The weight of the gas remains the same, and now the weight doesn’t cancel the buoyant force anymore. So, the net external force is up. Newton’s second law isn’t falsified.

    • Forget the buoyancy. Just assume there is a black body held at a fixed temperature like 350K suspended at say 4000 m.

      • Pierre-Normand

        Clivebest, I am just trying to address SW’s simple misconception that rising parcels of air exchange internal energy for gravitational potential energy.

      • Pierre-Normand

        Clivebest, sorry. I hadn’t noticed your brain teaser.

  38. Pierre said

    “The gas rises when heated because when it thermally expand the buoyant force increases as a result. That’s because the gas thereby displaces more surrounding fluid. The weight of the gas remains the same, and now the weight doesn’t cancel the buoyant force anymore. So, the net external force is up. Newton’s second law isn’t falsified.”

    If the expanded gas displaces more surrounding fluid then the buoyant force is coming from the expanded gas and NOT from the surroundings.

    Note that it is the expanded gas that was heated and not the surroundings that were cooled.

    My point stands but I have long since forgotten why you thought it relevant anyway.

    Didn’t I point out previously that such a semantic point made no difference to the existence or otherwise of the gravito thermal effect which requires a store of KE at the surface to sustain the continuing convective cycle without being radiated to space ?

    In the case of Earth 33K above S-B

    • Pierre-Normand

      SW wrote: “If the expanded gas displaces more surrounding fluid then the buoyant force is coming from the expanded gas and NOT from the surroundings.”

      This is like saying that when you sit on a table, the force that holds you up “is coming from” your making contact with the table and not from the table. Maybe that’s true in some sort of causal sense. From the point of view of Newtonian mechanics, it’s still an external force that holds you up against gravity. Likewise it’s still the pressure differential within the surrounding air that pushes the expanding air parcel up against gravity.

      “Note that it is the expanded gas that was heated and not the surroundings that were cooled.”

      Yes, in a sense, the expansion of the gas created the conditions whereby the net external force on the gas is directed up. It’s still the external force that does the mechanical work.

      “My point stands but I have long since forgotten why you thought it relevant anyway.”

      You also forgot the point of your own argument. Your point was to deny that the work that is responsible for the increase in gravitational potential of the rising air parcel comes from the surrounding (and ultimately, as I argued, from the gravitational potential energy lost by other parcels of air that are descending while the expanding parcel rises in their midst). That’s because you ultimately were arguing that the gain in potential gravitational energy of the rising parcel rather results from the conversion of (some of) the kinetic energy of the molecules of the air parcel. If this energy rather comes from descending parcels of air around it, your point doesn’t stand.

      You were confusing two different processes: one that is responsible for the increase in gravitational potential energy (buoyancy), and a different process (adiabatic expansion) that is responsible for the decrease in internal energy. Since those processes have different energy sources, they need no be in balance with one another. There is no process of conversion of molecular kinetic energy into gravitational potential energy.

  39. “the work that is responsible for the increase in gravitational potential of the rising air parcel comes from the surrounding (and ultimately, as I argued, from the gravitational potential energy lost by other parcels of air that are descending while the expanding parcel rises in their midst). ”

    The surrounding is the gravitational field and not adjacent molecules.

    The loss of GPE in the descending air becomes KE since the descending air warms.

    The loss of GPE in the descending air does NOT become GPE in the ascending air.

    At all times the total energy content of both parcels of air tremains constant and no energy passes between them.

    Each parcel works against gravity, one against gravity as it rises and the other with gravity as it descends.

    I think you are totally confused and wasting my time.

    • David Springer

      “the work that is responsible for the increase in gravitational potential of the rising air parcel comes from the surrounding (and ultimately, as I argued, from the gravitational potential energy lost by other parcels of air that are descending while the expanding parcel rises in their midst). ”

      Totally confused and wasting myeveryone’s time.

      Spooky potential energy exchanges at a distance. Now I’ve heard everything.

      The standard model is missing yet another particle now… a particle that carries the force of gravity and a particle that carries the force of gravitational potential energy. ROFL

      • Pierre-Normand

        “Spooky potential energy exchanges at a distance. Now I’ve heard everything.”

        That’s not spooky at all. If you forcefully sink a cork near the bottom of a bucket of water (pushing it down with a thin metal rod, say), this forces some of the water to rise in the bucket and thereby gain gravitational potential energy. Even after the cork is fully immersed, as it moves down, the average height of the water mass increases. If the cork is then allowed to rise again from the buoyant force alone, then the average height of the total mass of water will progressively diminish as the cork rises (even before it is begins to emerge at the surface). So the water will progressively lose gravitational potential energy at the exact same rate as the work that it perform through lifting the cork (some of which is converted in gravitational potential energy of the cork, some in kinetic energy, and some converted to heat through viscous dissipation). This potential gravitational energy loss of the surrounding fluid is the source of the work performed by the buoyant force.

  40. David Springer

    “This is like saying that when you sit on a table, the force that holds you up “is coming from” your making contact with the table and not from the table.”

    O…M…G that is one of the least informed things I’ve heard here in a while.

    There is no “force” holding me up when I sit on a table. The muscles in my legs provided the force to lift my a$$ up to the table height. Energy isn’t expended to sit there once elevated and potential energy provided by my legs isn’t bleeding off.

    And where I come from we sit on chairs not tables. And we don’t have to fuel our chairs so they have enough force to hold us off the ground indefinitely. ROFLMAO

    • Pierre-Normand

      David Springer, you are jumping on all my comment while ignoring what I am responding to. SW was claiming that the surrounding air can’t perform work on a buoyant air parcel just because the existence of force is caused by the expansion of the parcel. This is a non sequitur and my table sitting example merely illustrates why it is.

  41. A fan of *MORE* discourse

    It appears that Climate Etc don’t uniformly appreciate these engineering facts:

    • ultracentrifuge rotors spin in a vacuum, and

    • the rotors themselves are not cooled in any way, and

    • the drive-motors are cooled by an ordinary heatsink (no different from the heatsinks of an desktop computer).

    Needless to say, the thousand-degree temperatures that the gravito-thermal effect (if it existed!) would generate in million-g untracentrifuges would swiftly destroy biological samples.

    Observation  This doesn’t happen.

    Conclusion  There just plain ain’t no such thing as a “gravito-thermal” effect!

    \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • Fan do you have a site for anyone that claimed the effect was huge?

    • AFOMD,

      Surprise! I agree with you.

      When using ultracentrifuges for uranium enrichment, external heat is required to obtain the required convection. Zippe centrifuges have been the subject of intense research, and there is a distinct lack of gravito-thermal effect in the peer reviewed literature, noted by users, or claimed by manufacturers.

      Like the CO2 greenhouse effect, it purely doesn’t exist. Wishing will not make it so.

      Live well and prosper,

      Mike Flynn.

    • Willis Eschenbach

      Dang, me and FOMD agree!! Stars collide, the sun rises in the west, this doesn’t happen often …

      Nice example, FOMD, well done.

      w.

    • FOND,

      Thank you. You are clearly such a wise man.

      What a pity you don’t contribute without all the silliness. I would not have seen the comment without Willis’s comment pointing to it.

    • Yeah – heating to Sun like temperatures doesn’t happen everyday in centrifuges.

      Seriously – where is this energy meant to come from? Ambient temperatures? FOMBS as usual is not worth a first glance.

  42. Note that during the very first convective cycle, whilst KE (equivalent to 33K) is being converted to PE and no PE is yet being returned to the near surface, the surface is losing energy by radiation to space AND to the first convective cycle via conduction.so the surface temperature will drop below 255K for a while.

    When the first convective cycle completes and adiabatically warmed air is being returned to the surface at the same rate as surface warmth is being taken up adiabatically then the surface temperature rises to 288K because:

    i) Incoming solar energy continues to arrive at a rate commensurate with a surface temperature of 255K AND

    ii) Descending adiabatically warmed air is returning to the surface at a rate commensurate with surface warming of 33K AND

    iii) The next convective cycle is simultaneously removing KE from the surface at a rate commensurate with surface warming of 33K thereby locking it away in PE which is not heat and does not radiate.

    The net outcome is that the surface temperature of 288K is provided by 255K from the sun plus 33K from descending air but the surface cannot use that 33K for radiation to space because it is recycled into further convective uplift. Nevertheless the surface temperature will still be enhanced by 33K because the energy required by the ongoing convective cycle is still held at the surface.

    The result is a surface temperature of 288K, 255K escaping via radiation to space and 33K continuing to support the weight of the atmosphere.

    Some AGW proponents say that adiabatically descending air that has been warmed cannot heat the surface.

    It doesn’t have to warm the surface directly. All it has to do is provide a less steep lapse rate which inhibits convection so that the surface warms from solar irradiation more than it otherwise would have done.

    We see that all the time within atmospheric high pressure cells containing descending air and often they create an inversion which blocks convection altogether.

    The irony is that reducing or blocking convection is what the glass in greenhouses does.

    That descending adiabatically warmed air acts just like the glass on a greenhouse roof by reducing convection so it is mass acting via the adiabatic convective overturning cycle that is and always was the true greenhouse effect.

    The term ‘Greenhouse Effect’ must have been originally coined by a meteorologist maybe 100 years or more ago who realised that adiabatically warmed air by virtue of being transparent and inhibiting convection would act exactly like a greenhouse roof.

    The radiative chaps never learned that so they often say that the term is misleading.

    The term might be misleading in terms of their attempt to use it in connection with their imagined radiative only scenario but it is spot on in terms of long established meteorology.

    The thing is that meteorology was a very arcane subject in the 20th century and earlier.

    Hardly anyone knew anything about it and a lot of the ways the laws of physics play out within an actual atmosphere are counterintuitive and hard to envisage such as the thorny concept of adiabatic processes.Everyone I discuss it with is oblivious to its true nature.

    In the 1980s or thereabouts a bunch of astrophysicists thought they could take over climate science with no knowledge of basic meteorolgy and they have spread nothing but confusion in their wake.

    Made them rich and famous though.

    The Greenhouse Effect is perfectly described as a process that occurs as a result of a transparent layer of warm air that inhibits convection so as to allow the sun to warm the surface to above the S-B figure of 255K.

    On average for the Earth as a whole the surface temperature enhancement for the mass induced ghreenhouse effect is 33K leaving no room for any contribution from radiative gases.

    Believe it or not. Your choice :)

  43. ‘Needless to say, the thousand-degree temperatures that the gravito-thermal effect (if it existed!) would generate in million-g untracentrifuges would swiftly destroy biological samples.’

    Temperatures created presumably by energy from quantum vacuum zero-point energy spontaneously emerging in this space/time as the rotor spins?

    But it is potential energy of the atmosphere that I wanted to discuss briefly. The atmosphere has no potential energy as a whole – it is sitting in a gravity well and has no potential to flow anywhere by and large. Within the atmosphere there are multiple processes that change the height of individual molecules as some rise and some fall – but this doesn’t change the net zero of the PE of the atmosphere as a whole. The table sits on the floor – it has zero PE assuming it doesn’t collapse.

    The processes for air movement are turbulence and convection – these change the bulk properties of specific regions of the atmosphere and these processes – extending down to high frequency micro-eddies – change the velocities of individual molecules as energies ebb and flow. .

    The point is that simple Newtonian consideration of potential and kinetic energy at the molecular level do no begin to approach the complexities of the real atmosphere.

    • “The atmosphere has no potential energy as a whole – it is sitting in a gravity well and has no potential to flow anywhere by and large.”

      It has potential energy relative both to the centre of gravity of the Earth and that of the Universe.

      “Within the atmosphere there are multiple processes that change the height of individual molecules as some rise and some fall – but this doesn’t change the net zero of the PE of the atmosphere as a whole. ”

      No it doesn’t but rising molecules gain more PE at the expense of KE and falling molecules gain more KE at the expense of PE. It may be a net zero process but it means that there is constant work being done and so to maintain the circulation a KE store (33K for Earth) needs to be present at the surface over and above the KE needed to radiate to space at 255K.

  44. Interestingly enough, the Moon experiences higher surface temperatures than the Earth, even though it lacks significant atmosphere, and has less gravity than the Earth.

    So it would appear that an atmosphere combined with higher gravity reduces the maximum surface temperature, devout belief to the contrary notwithstanding.

    Gravito-thermal effect? Excellent for promoters of free energy, perpetual motion, or purveyors of tinfoil hats. Merely a source of mild amusement for the person of average intelligence, it would appear.

    Live well and prosper,

    Mike Flynn.

    • NASA gives the Bond albedo of the moon as 11% – and all energy is radiated from the surface. No convective because no water or atmosphere of course. This energy balance balance is what determines the temperature of the moon’s surface.

      It is not of course relevant to the Fronsdal paper – which Mike understands perfectly he says. Perhaps you can fill us in on where he goes wrong Mike?

      • Rob Ellison,

        I can’t figure what you are trying to say. Are you disputing the facts as I understand them?

        I cannot understand the relevance of your first sentence, or the second, or the third.

        I am not sure where I said I understood Fronsdal’s paper perfectly. you might care to provide a copy of my words, if you have them handy.

        Are you really interested in understanding where Fronsdal’s reasoning is at fault? I will be happy to help, but first could you let me know why Fronsdal has been unable to obtain the use of an ultra centrifuge to perform his proposed experiment. He has been endeavouring to get someone to perform the experiment for at least 8 years that I am aware of. I suspect that like experimental verification of the greenhouse effect, attempts to show the existence of the gravity-thermal effect in fact show there is none.

        I will defer to your superior knowledge, if you can back it up with some evidence.

        Live well and prosper,

        Mike Flynn.

      • Oh Mike – stop prevaricating. The situation of an atmosphere under gravity is obviously different to that of no atmosphere under a lesser gravity.

        There are several other differences between the Earth and the Moon that seem quite obvious – but that I won’t labour.

        I of course have no clue as to why an expensive piece of lab equipment modified for the specific needs of the experiment is not readily available – I will bow to your infinite wisdom here.

        The first mention of the centrifuge experiment I am aware of was a 2010 arxiv.org paper. Perhaps you help me by linking this much earlier reference?

        But you did say you understood the 2014 paper – complex and advanced math and physics that I struggle so hard with. So where did he go wrong? Don’t be shy – we all want to know.

      • Oh – and as to what you don’t remember claiming. I won’t bother with the thankless task of searching through comments in far flung post – well a couple of weeks at least.

        I don’t know – perhaps we can ask P-N if he can refresh your memory. P-N was interested – seeing as you understand it so well – to have it explained to him as well.

      • Rob Ellison,

        You obviously wouldn’t be able to understand my critique of Fronsdal’s paper, if you cannot understand the original. I don’t believe your demand is sincere, but I’m willing to change my mind.

        Nice try. A fact or two would be appreciated.

        I can’t help it if you have difficulty understanding reasonably straight forward concepts. Maybe your Warmist fervour exceeds your ability – what do you think?

        Both the gravito-thermal effect and the CO2 greenhouse effect remain firmly ensconced on the same shelf as phlogiston, caloric, and the luminiferous aether, until demonstrated otherwise.

        By the way, it might serve you better to cut and paste what I wrote, rather than baselessly assert what you wish I had written. This will help other readers to distinguish fact from fiction, I would hope.

        Live well and prosper,

        Mike Flynn.

      • I am sorry that you found the paper difficult. I did not. I disagree with some of the writer’s speculations. Indeed, he indicated he would be much more satisfied if experimental verification was possible. Alas, to date, no.

        Live well and prosper,

        Mike Flynn.

        Pierre-Normand | October 21, 2014 at 12:35 am |

        It’s been a while since I’ve studied thermo and GR at an undergraduate level and we covered nary anything at all of the interaction between those two fields in the GR course. So, if I get round to reading this paper, and encounter some problems, I possibly will rely on you for some help.

        Emphasis mine. Happy to refresh you memory from all of a month or so ago.

        By all means elucidate on these simple ideas. Which ‘speculations’ did you disagree with? In which specific aspect of the Lagrangian or the Hamiltonian did the error leading to the suggestion of a gravito-thermal effect enter the derivation? Please be assured that I have have reasonably level math skill and will attend with due diligence.

        I await your words of wisdom with bated breath.

      • … reasonably (high) level…

    • Rob Ellison,

      I am happy to indulge you – slightly, at least.

      You wrote –

      “It is not of course relevant to the Fronsdal paper – which Mike understands perfectly he says. Perhaps you can fill us in on where he goes wrong Mike?”

      When I queried your assertion, you responded –

      “Oh – and as to what you don’t remember claiming. I won’t bother with the thankless task of searching through comments in far flung post – well a couple of weeks at least.”

      Apparently you had a change of heart, and decided to bother after all – and discovered that I had written –

      “I am sorry that you found the paper difficult. I did not. I disagree with some of the writer’s speculations. Indeed, he indicated he would be much more satisfied if experimental verification was possible. Alas, to date, no.”

      Of course, as usual, I have been provided with no reason to change my opinion. I point out that I never claimed perfect understanding, as you claimed.

      I still disagree with some of Fronsdal’s speculations. Given the epithets and vituperative aspersions which you have tossed my way in the past, I assess your present desire that I provide you with knowledge which you lack, as somewhat insincere – I trust you understand why.

      You have previously stated, as I recollect, that you ascribe precisely no value to anything I say. So be it. If you wish to ignore me, feel free. I promise I won’t burst into tears!

      Live well and prosper,

      Mike Flynn.

      • Well – you seemed to have had a lapse of memory and and insisted that I provide the quote or be found lacking in veracity. That’s 10 minutes I won’t get back – never mind. At least it will in your words help the onlooker to distinguish fact from fiction.

        But I am quite disappointed that you – with your deep understanding of atmospheric physics – ‘CO2 has never warmed anything’ – and planetary energy dynamics – ‘the planet has been cooling for billions of years’ – do not give us the benefit of your insights into the Lagrangian and the Hamiltonian for the atmosphere under gravity – and not under gravity – in the Fronsdal paper. As P-N ans I at least am struggling with it and you find it so simple.

        I am deeply wounded that you would think me insincere – and don’t answer my detailed questions. Believe me – I don’t ascribe your memory lapse to bad faith but to you having many other things of far greater import on your mind – despite being prompted not once but twice. But be assured that it was for your benefit that I spent the 10 minutes looking for your comment – because I just wanted to clear this up for you as it clearly didn’t ring a bell and P-N hadn’t turned up to confirm. It was clearly not fictionalised on my part – which not having a memory of such a recent claim seemed obviously to you to be a possibility. I hope this corrects the record and sets your mind at rest. .

        You have not supplied me with the much earlier reference either. I suppose that’s out of the question too?

      • And please – far from ignoring you I will continue to benefit from your profound scholarship..

    • An atmosphere with gravity lowers the highs and raises the lows but overall raises the average.

      The sun constantly pumps up the atmosphere against gravity (surface low pressure cells) and gravity constantly pulls it down again (surface high pressure cells).

      The ongoing convective overturning needs a KE source at the surface over and above S-B to sustain it.

      Basic meteorology, read some.

  45. Can I ask something really simple.
    We take an inert gas like helium.

    We take a all wavelength transparent container filled with a 1 atmosphere of He at 285K, with the container walls at the same temperature.
    What is the emission, W/m2, through the walls of our completely transparent container?

    If we lower the pressure to 0.1, 0.01 and 0.001 atmospheres does the emission rate remain constant at the same temperature?

    • DocMartyn,

      I’m not sure where you are heading with this, but the mass of He will reduce, as the pressure drops, assuming the container retains the same surface area.

      As the mass drops, the concept of temperature as generally understood becomes somewhat less than useful.

      Eventually, you could have a situation where the container encloses one atom of He. Assuming this atom has a temperature of 285K, the pressure will be very low, and the absorption and emission of the relevant photon will result in little average emission through the walls, although the concept of average emission is as useful as that of average anything in this case.

      I extrapolate this situation to infer that as the pressure drops, the emissive power will drop in proportion to the mass of the enclosed matter, all else being held constant – which is probably unlikely!

      I’m curious – what was your intent?

      Live well and prosper,

      Mike Flynn.

      • Perhaps Professor Flynn has had another memory lapse.

        ‘Emission spectra are produced when atoms of a dilute gas are `excited’ — in effect, heated — by an electrical current, ultraviolet radiation, or some other source of energy. Excited atoms have electrons in high orbits, and these emit photons with specific wavelengths when they jump back down to lower orbits.’ http://www.ifa.hawaii.edu/~barnes/ASTR110L_F05/spectralab.html

        Dilute gases – unlike liquids, solids and compressed gases – do not emit a continuous Planck spectrum. An emissions spectra for helium is shown at the site above. Helium absorbs and emits at quite high energies – well above 285K. At lower temps – energy is transferred at the relative snails pace of particle velocities and collisions and the energy flux across Doc’s transparent walls depends on the temperature of the gas and the particle density. I suspect that the walls would emit in IR and the whole package cool down toot sweet.

        We have to be a bit forbearing with Professor Flynn – he’s forgotten more than we ever knew.

      • Rob Ellison,

        A helium balloon in a room containing air at 285K will also stabilise at a temperature of 285K. No cooler, no warmer. The balloon will be 285K. The helium will be at 285K. The string holding the balloon will be 285K.

        If the air in the room is cooled to 273K, the balloon and its contents will cool to that temperature, unless you use magical Warmist helium which does not radiate light – EMR – and therefore does not fall in temperature.

        Conversely, magical Warmist helium will not absorb EMR – light – and will not rise in temperature, for if it did, without being able to cool, it will keep increasing its temperature without pause or obvious limit. This is obviously nonsensical.

        A gas cylinder of helium at a pressure of several hundred bars can be both warmed and cooled. The contents, not surprisingly, are likewise warmed or cooled, until they are the same temperature as the container. Helium at 1K, 100K, or 1000K emits radiation, the intensity of which is proportional to the fourth power of the absolute temperature.

        However, if you choose to believe otherwise, that is your choice. If you believe that photons move at different speeds dependent upon their wavelength, I beg to disagree.

        As I have said before, there seems to be some confusion relating to heat, energy, EMR, temperature, and related matters. All matter can absorb energy. All matter can radiate energy. The laws of thermodynamics – in all their disparate framings – do seem to endure.

        Live well and prosper,

        Mike Flynn.

      • As I am sure Professor Mike realises – in the atmosphere of the Earth nitrogen and oxygen are transparent to infrared radiation.

        This doesn’t mean of course that the atmosphere doesn’t warm and cool – just that it occurs through molecular collisions. It is greenhouse gases that interact with photons at these frequencies.

    • Pierre-Normand

      That’s a good question. Helium doesn’t have any spectral line in the infrared region. Its least energetic line is at 667.815 nm. That’s well into the red visible spectrum. So, for gaseous helium to emit any photon at all at room temperature and 1atm would possibly require some sort of transient bound state from especially energetic collisions that may occur very seldom. This would be even less common at lower pressures. The emissions power must be extremely small (I would think).

      • Pierre-Normand

        “…would possibly require some sort of transient bound state…”
        Maybe ‘transient interaction’ would sound less like an oxymoron.

      • Pierre-Normand,

        With respect, DocMartyn stated that the He had a temperature of 285K. It is therefore emitting photons with the requisite energy appropriate to that temperature. Which is to say, unless my understanding is defective, exactly the same energy contained by photons emanating from any other matter at the same temperature.

        A sample of any matter at 285K is just that. It doesn’t matter whether at that temperature said matter is gaseous, liquid, or solid. It is at 285K, and is indistinguishable by temperature from other matter at the same temperature. Helium breathed into the lungs and exhaled is the same temperature as the other exhaled gases.

        Helium/oxygen filled dive cylinders will cool down if hotter than the surroundings, and heat up if cooler. The gas mixture is identical in temperature to cylinders of compressed air, or empty cylinders, if allowed to stand for a bit. The spectral absorption properties of the gas are irrelevant to the temperature.

        Correct me if I’m wrong, but any gas at room temperature must be emitting photons. It is only at absolute zero (which is a practical impossibility if you think about it), that emission ceases. Hence, no temperature – otherwise known as absolute zero.

        There is no such animal as a non radiative gas. All this talk of short wave, long wave, infrared, upwelling, downwelling, serves only to confuse the issue. Things heat up. Things cool down. The laws of thermodynamics don’t distinguish between oxygen, hydrogen, or Bombay Gin – at least at the macro level.

        What do you think? Are my facts in error?

        Live well and prosper,

        Mike Flynn.

      • Pierre-Normand

        Mike Flynn: “With respect, DocMartyn stated that the He had a temperature of 285K. It is therefore emitting photons with the requisite energy appropriate to that temperature. Which is to say, unless my understanding is defective, exactly the same energy contained by photons emanating from any other matter at the same temperature.”

        285°K just is 10°C. That’s not very warm. Gases at room temperature (or even colder) don’t glow in the visible spectrum. If a gas doesn’t have any bands in the infrared spectrum, then it will not emit more radiation in the visible light spectrum to compensate for this incapacity. It will emit exactly as much energy in its visible spectral bands as a black-body at the same temperature does at those very same frequencies. At a temperature of 10°C, this means it will emit almost nothing. Look up the Plank radiation curve (Plank’s law) as a function of temperature. The radiation power of a gas within its spectral bands is contained within the envelope of that curve.

        http://en.wikipedia.org/wiki/Planck%27s_law

      • Pierre-Normand

        P-N wrote: “It will emit exactly as much energy in its visible spectral bands as a black-body at the same temperature does at those very same frequencies.”

        That was incorrect. Replace “exactly as much” by “no more”. Only if the gas volume were very large would it approach the emission power of a solid black-body in those bands.

      • Mike Flynn I agree re the temp
        in your example of 1 molecule it could not have a temperature of 285 K in any physical sense. It is in a vacuum and the concept of temperature becomes meaningless.

        Pierre Normand
        The gas is at 285 K , It has to emit energy if it is cooling down or staying at the same temperature due to incoming energy. It will emit that energy at some frequency that that degree of energy can provide. If not visible light, red light or some infra red light then infra red at a lower frequency. It has no choice.

        Doc Martin, The spheres could also be of different sizes to account for the different pressures. To be at the same temperature the low pressure large sphere would have to have extremely fast moving particles. The surface areas for emissivity would be quite different.
        Despite all this I guess Helium is helium because it has an unchanging signature and hence an unchanging rate of emission until one reaches very cold conditions.
        Personally I expect the larger surface area (lowest pressure or thinner gas) would have to emit more heat as less of the emitted heat is reabsorbed by the gas in the container as there are less atoms in the way of the outgoing radiation but this effect would be minuscule.

      • Pierre-Normand

        angesh wrote: “If not visible light, red light or some infra red light then infra red at a lower frequency. It has no choice.”

        Real helium gas in the laboratory still chooses not to emit any radiation at all in the infrared spectrum (or any lower frequency). Contrary to your expectation, helium atoms may be endowed with free will.

      • Pierre-Normand,

        Real helium gas can absorb energy. It warms. If it radiates energy, it cools. If you have helium gas in a laboratory which radiates energy not at all, by definition it is at absolute zero. If it is above absolute zero, it is radiating energy at a wavelength dependent on temperature.

        Neither helium nor any other gas refuses to emit photons if it is above absolute zero, in a laboratory or otherwise. Whether you agree or not, Nature doesn’t care. If something has a temperature, it is radiating energy.

        As a side note, many people still view heat related phenomena in classical terms, which can lead to incorrect conclusions. Conduction is an example. As you know, I’m a CO2 warming non believer. The Warmists have to have non-radiating gases, perfectly opaque gases, an initially cold Earth, surfaces which radiate energy without a concomitant drop in temperature, and other similar strange things.

        Luckily for me, Nature seems to be similarly inclined – at least at present.

        Live well and prosper,

        Mike Flynn.

      • Pierre-Normand

        Mike Flynn: “Real helium gas can absorb energy. It warms. If it radiates energy, it cools. If you have helium gas in a laboratory which radiates energy not at all, by definition it is at absolute zero. If it is above absolute zero, it is radiating energy at a wavelength dependent on temperature.”

        Yes, it radiates energy at some definite frequencies. There just aren’t any such frequencies in the infrared spectrum in the case of gaseous helium. It doesn’t have any spectral line in that region. That’s an empirical fact that confirms quantum theoretical models of the helium atom. It can emit in the visible spectrum where it has some spectral bands. But those photons are much more energetic than infrared photons and hence require inter-molecular collisions that are more energetic than is typical for a helium gas at room temperature. Hence is emits such photons very seldom. It radiatively cools, but extremely slowly.

  46. I may be missing something here, but wouldn’t a transparent container containing a transparent gas simply transmit whatever radiation entered from outside the container? The whole assembly will still be in local thermal equilibrium with its surroundings and a thermometer inside the container will read the same as one in the room outside of the container. Otherwise, you could have a perpetuum mobile.

    • Pierre-Normand

      Pochas, yes, it would transmit it. It would transmit it equally well whatever its temperature is. If radiation is the only kind of thermal interaction, then if the initial temperatures are different, it will possibly take centuries for thermal equilibrium to be attained between the helium in the container and its surrounding (maybe isolated though some vacuum) through the exchange of the odd photon energetic enough to interact with the gaseous helium.

      • Nitrogen cylinders are a frequent item of commerce, and I am unaware that they behave any differently than compressed air as they equilibrate to room temperature. At temperatures below 1000 F convection dominates and radiation is relatively unimportant. Radiatively “transparent” gasses convect like any other gas. The thing is, that thermodynamic temperature is related to the kinetic energy of the gas molecules, and not to radiation flux. In order for radiation to translate to temperature, a photon must be absorbed by an atom which kicks an electron to a higher orbital, then that electron decays to its ground state, and its energy can be released either by emitting another photon(s) of lower energy, or as kinetic (heat) energy and that second process results in an increase in temperature. Please forgive the lecture, but if we are to grasp anything at all about thermodynamics, we must understand these things.

      • As I’m sure Professor Mike meant to say – despite being being transparent the walls can be heated by particle collisions and will emit in IR even into a surrounding vacuum.

      • Pierre-Normand

        Pochas,

        I was addressing DocMartyn’s question, which is specifically about radiation. Conduction and convection is something else. Nobody denies that it would be hard to prevent a compressed helium gas to conduct heat to the container. That’s besides the point. The question is about the amount of radiation from the helium gas at 385°C.

  47. Speaking of greenhouses.

    They warm up inside because the glass lets solar radiation in and then prevents convection.

    Consider the atmosphere, in which air cools adiabatically as it rises and then warms adiabatically as it descends.

    That adiabatically warmed descending air becomes largely transparent because it causes clouds to dissipate which is why surface high pressure cells are usually cloud free.

    That adiabatically warmed descending air reduces the lapse rate slope and often reverses it in so called inversions and that inhibits or even blocks convection.

    Convection being reduced or prevented, the solar heated surface is then able to rise 33K above S-B on Earth.

    The term ‘Greenhouse Effect’ is a perfectly accurate description of the surface warming effect from descending adiabatically warmed air.

    That must have been the original thought behind the first use of the term and it applies to up and down transport of mass within the atmosphere and not radiative fluxes.

    The term ‘Greenhouse Effect’ has been hijacked by the radiative physics crowd and applied to an entirely different scenario. Enhanced radiation at the surface increases convection rather than reducing it so the term is wholly inappropriate for a radiative situation..

    Hence all the confusion.

    • Stephen Wilde reminds me of someone who pretends to speak French.
      To others who know no French, he might even sound convincing.
      However those who speak French know its all gobbledygook.
      They would advise Stephen to go and take some French lessons,but alas he will not
      Its good that Stephen is interested in thermodynamics but well wishers would advise Stephen to read a good textbook on the subject.

  48. A fan of *MORE* discourse

    Mike Flynn appreciates [first rightly, then wrongly] “Both the gravito-thermal effect and the CO2 greenhouse effect remain firmly ensconced on the same shelf as phlogiston”

    LoL  Mike Flynn, it’s mighty good that folks like you and Willis Eschenbach and even Peter Lang have all come to appreciate that ultracentrifuge evidence renders the gravito-thermal hypothesis dead as a dodo.

    However, it’s not clear that Climate Etc folks (as yet) appreciate that, by the similar arguments of FOMD’s self-evaluation test of
    thermodynamic and heat-transport intuition
     …

    …  NASA’s hugely successful multilayer superinsulation technologies show plainly that the fundamental mechanism of CO2-amplified greenhouse physics is entirely correct, in that:.

    • each increment of NASA-superinsulation increases the ability of a thermos-bottle to retain heat, and similarly

    • each increment in CO2 concentration increases the ability of the earth to retain heat.

    Of course, phenomena like lapse rates, aerosols, and circulation considerably enliven planetary physics relative to thermos-bottle physics … but the basic physical mechanism of the greenhouse effect still works.

    Conclusion  Good on `yah, NASA engineers and thermodynamicists … for showing the world plainly that the fundamental heat-transport mechanism that drives AGW ain’t rocket-science.

    Oh wait … yes it is!

    \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • I suspect that star like temperatures don’t evolve in FOMBS centrifuge for good reason – the lack of zero point quantum energy perhaps.

      But the essential atmospheric physics of the greenhouse effect is interaction with outgoing IR and subsequent emission in specific frequencies in random directions. The scattering effect can be observed from space using narrow aperture observing devices. The difference between 1970 and 1997 is shown here as an equivalent black body temperature.

      Although we have it on impeccable authority – Professor Mike – that this is incapable of warming the surface through increased downward radiation – and that the Earth is in fact cooling from the core out. This beats heck out of zero point centrifuge energy – so suck eggs FOMBS.

      • Pekka, “I agree with Gibbs and Jaynes.”

        I am glad, sorry I missed your other comments

      • Rob Ellison,

        Any warmer body will emit radiation which if absorbed by a cooler body will result in both bodies being in thermal equilibrium, eventually, in the absence of changing external energy inputs.

        If you are referring to the nonsensical CO2 warming concept, may I respectfully point out that energy emanating from the surface results in a lowering of surface temperature. Even if one hundred percent of the outgoing energy is absorbed by the atmosphere, and radiated back to the surface, the best that can happen is that the surface temperature does not fall.

        Now the only thing that will return one hundred percent of the outgoing radiation is a perfect insulator, which by definition will allow no energy from the Sun to reach the surface. Of course, Warmists continue to ignore physics, and espouse the nonsense that arises from their ignorance. The fact that the CO2 warming effect, like the gravito-thermal effect, cannot be shown to exist, deters the Warmists not one jot.

        In relation to the Earth having cooled since its creation in a molten state, you seem to dispute this being fact. I must admit to the assumption that the Earth’s surface was originally molten, and observe that currently it is not, and draw the inference that it has therefore cooled.

        You may assume that the Earth was created at absolute zero or thereabouts, and due to the magical and mysterious effects of CO2, in combination with the distant Sun, heat has been absorbed to the point where the interior of the Earth is now molten, and the surface will continue to warm until it too, is molten. AFOMD appears to subscribe to this view, as does another commenter who proposes the mechanism of heat creep to overcome the inherent defects of the initial cold Earth theory.

        I suppose I should ask if you believe that the Earth was created with a colder surface than is now the case, and if you believe that the core, if hotter than space surrounding the Earth, is either losing no energy at all, or accruing energy and heating up. I presume you can adduce some logic to support your contention, and if so, I would be most grateful to be apprised of same.

        I suggest, once again, that you quote me directly, rather than ascribing to me words which you think or wish I uttered.

        Live well and prosper,

        Mike Flynn.

      • Dear Professor Mike

        Any warmer body will emit radiation which if absorbed by a cooler body will result in both bodies being in thermal equilibrium, eventually, in the absence of changing external energy inputs.

        Far be it for me to correct such a distinguished scholar of quantum mechanics. But dilute gases may not be a body in the sense of a fluid or a solid – and may not emit in a continuous spectrum.

        If you are referring to the nonsensical CO2 warming concept, may I respectfully point out that energy emanating from the surface results in a lowering of surface temperature. Even if one hundred percent of the outgoing energy is absorbed by the atmosphere, and radiated back to the surface, the best that can happen is that the surface temperature does not fall.

        Nonsensical? Right – got it. Added energy at the surface does not result in a temperature rise.

        Now the only thing that will return one hundred percent of the outgoing radiation is a perfect insulator, which by definition will allow no energy from the Sun to reach the surface. Of course, Warmists continue to ignore physics, and espouse the nonsense that arises from their ignorance. The fact that the CO2 warming effect, like the gravito-thermal effect, cannot be shown to exist, deters the Warmists not one jot.

        Right – the increase in IR interaction with gases in the atmosphere between 1970 and 1997 is not observational proof in the atmosphere of laboratory results.

        Git it – sorry.

        In relation to the Earth having cooled since its creation in a molten state, you seem to dispute this being fact. I must admit to the assumption that the Earth’s surface was originally molten, and observe that currently it is not, and draw the inference that it has therefore cooled.

        You may assume that the Earth was created at absolute zero or thereabouts, and due to the magical and mysterious effects of CO2, in combination with the distant Sun, heat has been absorbed to the point where the interior of the Earth is now molten, and the surface will continue to warm until it too, is molten. AFOMD appears to subscribe to this view, as does another commenter who proposes the mechanism of heat creep to overcome the inherent defects of the initial cold Earth theory.

        Obviously I had assumed heat transfers from the core and mantle at 0.05W/m2 was some 3000 times less than the 161W/m2 hitting the surface from the Sun – and even the changes in the energy flux – incoming and outgoing – were orders of magnitude greater then the core and mantle flux.

        Sorry. Where would I be without Professor Mike.

        I suppose I should ask if you believe that the Earth was created with a colder surface than is now the case, and if you believe that the core, if hotter than space surrounding the Earth, is either losing no energy at all, or accruing energy and heating up. I presume you can adduce some logic to support your contention, and if so, I would be most grateful to be apprised of same.

        You have obviously forgotten again that we have discussed this – and I have gone into the various processes that create core and mantle heat.

        I had thought that the surface warms and cools due to the transient difference between incoming and outgoing energy at a nominal top of atmosphere where all energy flux is radiative. This is obviously a fairly popular. I apologise for falling for it.

        I suggest, once again, that you quote me directly, rather than ascribing to me words which you think or wish I uttered.

        You may suggest again – but I did quote your comment of a month ago that I referenced twice – and which you had forgotten – despite being reluctant to search through the back issues for something that was clear in my memory. You then complained that I went searching through the back issues. As I said – you have many more important things on your mind than a throwaway comment that you were sorry that I was finding the paper difficult but that you understood it perfectly. I am disconcerted – what I claimed you said – was what you had said.

        Equally my shortform versions – CO2 has never warmed anything by surrounding it – the Earth is cooling from the core out – seem quite accurate depictions of your positions on the topics.

        Having memorized each and every one of your uniquely valuable contributions – I humbly suggest that this may be another – quite understandable – memory lapse and not bad faith at all.

      • .Rob Ellison,

        I am pleased that you are in the process of abandoning the tenets of of the false prophets of Warmism, and moving towards the lights of logic and reason, not to mention science.

        Now that you have accepted that the surface of the Earth has, indeed, cooled, and continues its remorseless and relentless slow cooling as evidenced by geophysical measurements by real scientists, you are well along the road to reality.

        In relation to your attempt to claim that dilute gas cannot be treated the same as any other matter, I suggest you measure the temperature of the gaseous contents of a mercury vapour lamp as instanced in the common fluorescent lamp, when in the off, or unexcited, state.

        Alternatively, radio valves of some types contain very dilute gas, or failing that, gas discharge tubes using noble gases such as neon, can be examined. The contents, after stabilising at 20C, will be found to be at 20C. If the surrounding temperature is reduced to 10C, the tube and its contents will be found to fall to 10C. You will notice that the dilute gas manages to achieve this as does all other matter, by emitting radiation in line with its temperature. No glowing, no emission spectra – it cools by the same mechanism as tube within which it is contained.

        You may not like it, you may choose to avoid the fact by performing a lateral arabesque and then flying off at a tangent while whirling like a Dervish, but Nature cares even less than do I!

        You mention falling for the TOA energy balance scam. No need to apologise, smarter people than either of us have been sucked into this one!

        You have rather accurately summarised the facts of the Earth cooling, and the inability of CO2 to warm anything at all by increasing its energy content, merely by surrounding it. Hold fast to these truths!

        I am pleased to be able to assist in your attempts to throw off the Warmist shackles. Many have escaped the cult, and subsequently resumed a normal life. Although some have found difficulty dealing with the ostracism and perverted attempts to reinculcate Warmist dogma, most feel a sense of relief at once more being able to exercise free will.

        I sincerely hope you succeed in your endeavours. Remember I am always there to assist, if you need my help.

        Live well and prosper,

        Mike Flynn.

      • Professor Mike

        I am pleased that you are in the process of abandoning the tenets of of the false prophets of Warmism, and moving towards the lights of logic and reason, not to mention science.

        I am quite confident that there has never been any logic, reason or science to match it.

        Now that you have accepted that the surface of the Earth has, indeed, cooled, and continues its remorseless and relentless slow cooling as evidenced by geophysical measurements by real scientists, you are well along the road to reality

        I still find it quite difficult to get past data showing the Earth and oceans being warmed from the top down for several hundred metres. No doubt purely invented by false prophets. .

        In relation to your attempt to claim that dilute gas cannot be treated the same as any other matter, I suggest you measure the temperature of the gaseous contents of a mercury vapour lamp as instanced in the common fluorescent lamp, when in the off, or unexcited, state.

        We should probably consider luminesence, incandesence and the energies available in a maercury vapour lamp – not just a gas at 285K.

        A heated gas in a tube – when turned of – will thermalise the walls through elastic collision and the walls will emit IR to cool the whole – even without emitting radiation directly from the gas. As I said before. However – I look forward to your instructions on this and other matters.

        Alternatively, radio valves of some types contain very dilute gas, or failing that, gas discharge tubes using noble gases such as neon, can be examined. The contents, after stabilising at 20C, will be found to be at 20C. If the surrounding temperature is reduced to 10C, the tube and its contents will be found to fall to 10C. You will notice that the dilute gas manages to achieve this as does all other matter, by emitting radiation in line with its temperature. No glowing, no emission spectra – it cools by the same mechanism as tube within which it is contained.

        It will cool off toot sweet – bot traditional quantum theory says gases absorb and emit emit in specific frequencies. This is no doubt another false prophecy. Thanks for clearing that up. Funny – we used to call them vacuum tubes and they got much hotter than that. I stand corrected.

        You may not like it, you may choose to avoid the fact by performing a lateral arabesque and then flying off at a tangent while whirling like a Dervish, but Nature cares even less than do I!

        You mention falling for the TOA energy balance scam. No need to apologise, smarter people than either of us have been sucked into this one!

        You have rather accurately summarised the facts of the Earth cooling, and the inability of CO2 to warm anything at all by increasing its energy content, merely by surrounding it. Hold fast to these truths!

        An arabeque involves standing on one leg while turning out the other leg and extending it behind – performed en point, demi pointe or a flat foot. But it is not clear to me what the mixed metaphor is intended to convey.

        Hardly – what I have done is accept on your authority that increased energy at the surface does not result in a warming surface.

    • A fan of *MORE* discourse

      Climate Etc’s many students of thermodynamical climate-science are well-advised to heed the sage advice of the great Lev Landau:

      Preface to Statistical Physics

      Statistical physics and thermodynamics together form a unit.

      All the concepts and quantities of thermodynamics follow most naturally, simply, and rigorously from the concepts of statistical physics.

      Although the general statements of thermodynamics can be formulated non-statistically, their application to specific cases always requires the use of statistical physics.

      Conclusion  Nowadays, young researchers (especially) are well-advised to study thermodynamics the modern way … in the language of geometric dynamics … with particular care that every macroscopic thermodynamical assertion is solidly and explicitly grounded in statistical mechanics and (microscopic) Hamiltonian flows.

      This unified program afford WONDERFUL levels of thermodynamical understanding!

      Best wishes are extended to all Climate Etc readers, for continued enjoyment and success in appreciating thermodynamics and statistical mechanics and geometric dynamics in the modern style … as one unified subject!

      \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • Actually, the Loschmidt Paradox aka gravito-thermal effect is a bit of a statistical anomaly. The Maxwell-Boltzmann distribution assumes perfect chaos believe it or not, which is not really correct. Seems even chaos is not allowed to be perfect.

        So while your centrifugal star equivalent heat generation is amusing, Gibbs entropy which allows for a more reasonable number of degrees of freedom and more likely phase space is the more accurate of the Gibbs versus Boltzmann entropies, at least according to Jaynes. But then entropy is an Anthropomorphic concept, not a property of a physical system which would make “equilibrium” a useful concept, but a property of an experiment, thought or otherwise, not a physical reality.

        There is no second law violation using the Gibbs concept, but there can appear to be a second law violation using the Boltzmann concept. Since you are an absolute physics whiz I would have expected you resolve this little paradox quickly.

        http://bayes.wustl.edu/etj/articles/gibbs.vs.boltzmann.pdf

      • CD,

        A couple of my comments starting from this one

        https://judithcurry.com/2014/11/27/open-thread-thanksgiving-edition/#comment-651077

        seem to be directly related to the difference of Gibbs and Boltzmann H functions. I agree with Gibbs and Jaynes.

      • missed it by this much, Sorry I missed your other comments.

      • The other comments are a little further in the same thread, but what they add is related to the comments they respond to.

        I tend to switch immediately to QM when I consider issues like interaction between molecules. Many concepts of statistical mechanics seem easier for me that way, but that cannot be true to everybody. Having lectured several courses of QM before lecturing thermodynamics for the first time must have something to do with my preference.

        This may also be an issue that’s more obvious in Quantum Statistical Mechanics.

    • A fan of *MORE* discourse

      There are many paths to appreciating thermodynamics, statistical mechanics, and microscale (quantum) dynamics. Here is one such path … it’s optimized for fun learning AND practical applications.

      —————
      FOMD’s Thermodynamical Reading List

      Knuth’s “A=B” essay  for overall philosophy (e.g. “Science is what we understand well enough to explain to a computer. Art is everything else we do.”)

      Zia et al on Legendre transform  to make sure your mathematical foundations are tuned-up.

      Frenkel and Smith on classical simulation algorithms  to get your feet wet, Knuth style.

      Onsager on entropy  to define entropy the modern way … as a thermodynamic potential, whose hessian is specified by local fluctuations, from which all other potentials can be derived (per Zia et al.).

      Nielsen and Chuang on quantum dynamics  to appreciate how quantum measurement back-action enters into the Third Law.

      Murcko’s video on quantum simulation agorithms  to advance to the super-hip frontier of modern quantum simulation research.
      ————–

      Happy learning to all Climate Etc thermodynamicists!


      @inbook{Knuth:1996it, Author = {Donald E.
      Knuth}, Publisher = {A.~K.~Peters, Ltd.},
      Title = {$A=B$}, Year = 1996}

      @article{Zia:2009lr, Author = {Zia, R. K.
      P. and Redish, Edward F. and McKay, Susan
      R.}, Journal = {American Journal of
      Physics}, Number = {7}, Pages = {614-622},
      Title = {Making sense of the {L}egendre
      transform}, Volume = {77}, Year = {2009}}

      @book{Frenkel:1996ly, Address = {San
      Diego}, Author = {Frenkel, Daan and Smit,
      Berend}, Publisher = {Academic Press},
      Title = {Understanding {M}olecular
      {S}imulation: from {A}lgorithms to
      {A}pplications}, Year = {1996}}

      @article{Onsager:1953fk, Author = {Onsager,
      L. and Machlup, S.}, Journal = {Physical
      Review}, Pages = {1505 - 1512}, Title =
      {Fluctuations and irreversible processes},
      Volume = {91}, Year = {1953}}

      @book{Nielsen:00, Address = {Cambridge},
      Author = {Michael A. Nielsen and Isaac L.
      Chuang}, Publisher = {Cambridge Univ{.}\
      Press}, Title = {Quantum {C}omputation and
      {Q}uantum {I}nformation}, Year = 2000}

      @inproceedings{Murcko:2014aa, Author =
      {Mark Murcko}, Booktitle = {Advances in
      Drug Discovery and Development}, Month =
      {24 September}, Organization = {Chemical
      \&\ Engineering News (Virtual Symposium)},
      Title = {Accelerating Drug Discovery:
      The Accurate Prediction of Potency"},
      Year = {2014}}

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      • A fan of *MORE* discourse

        Hmmm … let’s try this link to

        Mark Murcko’s Accelerating Drug Discovery: The Accurate Prediction of Potency

        Note  For reasons unknown, these simulation quantum algorithms work better than we expect them to.

        Welcome to cutting-edge quantum simulation-science!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • Physics is of interest but usually in climate is the uncountable in Einsteins bon mot. Mistaking arm waving at general principles – as I have said – for fundamental physical enlightenment is unlikely to be of much value in a large, complex and dynamic system. Of far greater relevance than quantum mechanics to the macro systems state is that third great idea in 20th century physics – after relativity and quantum mechanics.

        I tend to read quite broadly in the Earth sciences – here’s a recent article fully referenced for instance – and fill in a picture from a wide variety of credible internet sources – including on physics.

        http://watertechbyrie.com/2014/06/23/the-unstable-math-of-michael-ghils-climate-sensitivity/

        Does FOMBS have a big picture – or is this merely vaguely waving at the barely relevant? Don’t bother answering – it’s a rhetorical question.

    • AFOMD,

      Your NASA thermos-bottle contains no CO2 as insulation. It is known as a vacuum flask for good reason. NASA has more sense than to introduce CO2 into the vacuum of the thermos-bottle.

      You may be unaware that Thermos is a trade name. I assume that NASA would prefer that you use the generic terms vacuum flask or Dewar flask to describe the insulated container.

      In any case, NASA has not managed to produce any insulators with greater R values than previously known. Aerogels are vastly inferior to vacuum insulated panels, although more practical in many cases. In any case, not even NASA would be so foolish as to use CO2 in an aerogel – although they have some other strange things, and employed some odd people. If you are aware of any available insulator with better R values than a properly constructed Dewar flask, I would be most interested to hear from you.

      NASA scientists and engineers have managed to overlook or ignore some basic principles which have resulted in avoidable deaths. This is rocket science in the real world, unfortunately.

      It might behove you to use NASA as a reference more appropriately, if you wish to avoid being seen as desperately clutching at straws whilst attempting to remain erect in the absence of any legs on which to stand.

      Phlogiston, CO2 warming effect, gravito-thermal – all shattered dreams, I fear.

      Live well and prosper,

      Mike Flynn.

      • A fan of *MORE* discourse

        Mike Flynn randomly blithers “[about aerogels?]”

        Three simple points:

        Multi-layer insulation (MLI) ain’t aerogels (this video shows MLI being laid-down)

        • Yes, MLI does supply far better insulation (typically 0.1 mW/m-K) than any solid insulation.

        • In climate-science, CO2 plays the role of the heat-trapping mylar film (not the vacuum between the film).

        It is a pleasure to help further increase your understanding of climate-science, Mike Flynn!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • AFOMD,

        MLI needs high vacuum between the layers presently – around 1×10^-4 Torr. Even at these pressures, perlite powder meets or exceeds MLI under almost all circumstances. No super insulation here. A vacuum is hardly a solid, so your comparison is specious.

        Mylar film no more traps heat than CO2 does. If you can use Mylar film to trap some heat, maybe you can sell it to Siberians or Minnesotans during the winter. I’m only joking, of course. That’s probably a waste, because I suspect you are being serious. Ah, the sublime certainty of the Warmist faith, eh, AFOMD?

        If you can produce any evidence that the study of averages derived from past weather events has any greater claim to being called a science than phrenology, please do so.

        Strident self acclamations from purported Nobel Laureates such as Mike Mann, or the strangely bizarre Death Trains Hansen, are not evidence of anything except possibly shared delusional psychosis. You may not agree, but that’s life, I guess.

        Live well and prosper,

        Mike Flynn.

      • A fan of *MORE* discourse

        Mike Flynn claims [wrongly] “perlite powder meets or exceeds MLI under almost all circumstances”

        Uhhhh … mebbe yah overlooked the difference between “mW” and “W”? `Cuz NASA’s MLI insulation leaks heat at 0.1&nbsplmW/m-K = 0.0001 W/m-K.

        Perlite don’t compete!

        Conclude  It’s mighty good that NASA’s engineers are neither careless nor ignorant nor arrogant nor abusive, eh Climate Etc readers?

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      • AFOMD,

        I apologise. I inadvertently followed the Mannian procedure of reading the profile inverted. Test results indicate that up to 1 Torr, 60 layer MLI outperforms perlite K108. Past that point, perlite is better.

        So no, I didn’t overlook the difference between mW and W. Sorry.

        Your test results may differ.

        As to the arrogance, ignorance, carelessness, or abusiveness of individual NASA employees, I make no particular comment. I suggest that perusing the testimony of various employees contained in the Rogers Commission report will provide the opportunity for readers to form their own opinion.

        Richard Feynman, I believe, made the following comment –

        “What is the cause of management’s fantastic faith in the machinery? .. It would appear that, for whatever purpose, be it for internal or external consumption, the management of NASA exaggerates the reliability of its product, to the point of fantasy.”

        And people died as a result. So, was arrogance, ignorance, or carelessness involved? Or just bad luck? Repeated bad luck, perhaps?

        In any case, wrapping CO2 around the Earth still cannot provide additional energy resulting in increased temperature. Neither the fantasies of NASA, AFOMD, or Rob Ellison can change the laws of physics. Good luck with trying!

        Live well and prosper,

        Mike Flynn.

    • A fan of *MORE* discourse, December 3, 2014 at 11:13 am:

      “… NASA’s hugely successful multilayer superinsulation technologies show plainly that the fundamental mechanism of CO2-amplified greenhouse physics is entirely correct, in that:.

      • each increment of NASA-superinsulation increases the ability of a thermos-bottle to retain heat, and similarly

      • each increment in CO2 concentration increases the ability of the earth to retain heat.”

      Which goes to show that AFOMD doesn’t understand the difference between Earth’s surface/atmosphere system and a blanket.

      # Each increment of ANY insulating layer of ANY kind will reduce the ability of a body to shed its energy as heat to its surroundings. Any person who wears clothes and/or who prefers to sleep under a blanket knows this intuitively. The point of contention here is not with your first bullet point. It’s with the second one:

      # Each incremental increase in atmospheric CO2 concentration will NOT reduce the ability of the body beneath the atmosphere, the solid surface, to shed its energy as heat.

      First of all, notice how AFOMD in his first point above talk about the thermos bottle, the container inside the insulating layer, while in the second point, he moves his argument out to the outside of the insulating layer itself.

      We’re talking about the solid/liquid surface here. The surface is the thermos bottle. The ToA is the outside of the insulating layer, the atmosphere.

      Does more CO2 in the atmosphere manage to reduce the ability of the solid/liquid surface of the Earth to shed its energy as heat? Of course not. That ability is completely and utterly ruled by convection/evaporation. The only way more atmospheric CO2 could ever reduce general convective power would be by forcing the lapse rate (the temperature gradient away from the solar-heated surface), globally and on a permanent basis, to become less steep. It cannot possibly do this. It works the opposite way.

      The main difference between the atmosphere and any other piece of insulation is that there is no solid, immovable top surface facing the bottom (solar-)heated one. If there’s a top layer acting as a rigid lid and which is warmed by the outgoing heat, then the temperature gradient down to the heated surface at the bottom will naturally reduce, and you get warming. This cannot and will not happen in the atmosphere.

      The environmental lapse rate (ELR) is constrained by the tight interaction between radiation and convection to fluctuate around the average (gravity-based) adiabatic lapse rate (ALR), so that the mean global ELR finds a so-called ‘radiative-convective equilibrium’, equalling the globally averaged ALR (which is on the continuum between the dry ALR (DALR, ~10 K/km) and the saturated ALR (SALR, ~5 K/km), at 6.5 K/km).

      The ‘radiative-convective equilibrium’ works like this:

      The solar input heats the surface, which then in turn heats the lowermost layers of the troposphere, while the radiatively active gases (the so-called ‘GHGs’) cool the air masses to space through radiation from aloft in the troposphere. These more or less continuous radiative processes work towards STEEPENING the ELR above the ALR. This incites convection, which task it is to bring the original solar (radiative heat) energy from the heating end (low towards the surface) up to the cooling end (high towards the tropopause), LOWERING the ELR back down to (or below) the ALR in the process. At this point it stops. And the cycle can start again.

      Radiation STEEPENS the ELR, convection brings it back DOWN.

      CO2 is not convection.

  49. ‘Atmospheric and oceanic forcings are strongest at global equilibrium scales of 107 m and seasons to millennia. Fluid mixing and dissipation occur at microscales of 10^−3m and 10^−3s, and cloud particulate transformations happen at 10^−6m or smaller. Observed intrinsic variability is spectrally broad band across all intermediate scales. A full representation for all dynamical degrees of freedom in different quantities and scales is uncomputable even with optimistically foreseeable computer technology. No fundamentally reliable reduction of the size of the AOS dynamical system (i.e., a statistical mechanics analogous to the transition between molecular kinetics and fluid dynamics) is yet envisioned.

    This reality in nature and computers has given rise to two pervasive AOS practices: (i) AOS solution fields are nonsmooth near the space–time discretization scales (i.e., the “resolution” of the model) imposed on the known governing principles expressed mostly as partial differential equations. (ii) AOS models contain essential parameterizations for unresolved or highly simplified processes whose specifications are not at a fundamental level of known governing principles.

    Both practices are coping strategies to span as large a subrange as feasible for the uncomputably broad scale in the most fundamentally grounded fluid dynamical problem.’ James C. McWilliams – Louis B. Slichter Professor of Earth Sciences – UCLA Institute of Geophysics and Planetary Physics and Department of Atmospheric and Oceanic Sciences

    There are statistical mechanical principles to do with pressure, temperature, energy states, etc. But we have a a large, dynamic and complex system that is not reducible by any equivalent statistical method and is more generally represented by set of hydrodynamic equations in which physical properties – mass and momentum – are conserved across grid boundaries. More or less smoothly.

    There is a hierarchy of models – http://weather.unl.edu/RCM/IDB_Mexico/PDF/modelprelim.pdf – in which a Hamiltonian is one approach to low order models – http://www.nonlin-processes-geophys.net/13/125/2006/npg-13-125-2006.pdf

    Each of these approaches has intrinsic limitations. There are emerging methods that may be of more fundamental interest at the systems level.

    ‘Considering index networks rather than raw three-dimensional climate fields is a relatively novel approach, with advantages of increased dynamical interpretability, increased signal-to-noise ratio, and enhanced statistical significance, albeit at the expense of phenomenological completeness. Climate indices represent distinct subsets of dynamical processes. One could consider these indices – the nodes of our network – to be climate oscillators, each node, by itself, an intrinsic, self-sustaining system. When coupled with other self-sustaining oscillators of the network, the collective choreography of interlinked nodes generates a hemispherically spanning, propagating teleconnection signal – our “stadium wave” – an atmospheric and lagged oceanic teleconnection sequence that communicates an Atlantic-born climate signal of multidecadal warming and cooling (superimposed upon longer-time-scale temperature trends) across the Northern Hemisphere. Significantly, a warm North Atlantic generates a decadal-scale lagged cooling hemispheric response; a cool Atlantic generates a warming one.’ Marcia Wyatt

    The white dots cascading down the screen at this site – btw – seem to be real and not a visual hallucination. Let me know if they aren’t and I will make an appointment with a neurologist.

    Although the warming and cooling signal is global originating at the both poles with a see-saw of atmospheric pressure driving ocean circulation.

    Multi-decadal variability in the Pacific is defined as the Interdecadal Pacific Oscillation (e.g. Folland et al,2002, Meinke et al, 2005, Parker et al, 2007, Power et al, 1999) – a proliferation of oscillations it seems. The latest Pacific Ocean climate shift in 1998/2001 is linked to increased flow in the north (Di Lorenzo et al, 2008) and the south (Roemmich et al, 2007, Qiu, Bo et al 2006) Pacific Ocean gyres. Roemmich et al (2007) suggest that mid-latitude gyres in all of the oceans are influenced by decadal variability in the Southern and Northern Annular Modes (SAM and NAM respectively) as wind driven currents in baroclinic oceans (Sverdrup, 1947).

    The systems approach to climate as suggested by complexity theory reveals a fundamentally different view of climate variability that cannot be approached by the reductionist methods of the Hamiltonian or any other technique.

    ‘Climate is ultimately complex. Complexity begs for reductionism. With reductionism, a puzzle is studied by way of its pieces. While this approach illuminates the climate system’s components, climate’s full picture remains elusive. Understanding the pieces does not ensure understanding the collection of pieces.’ Marcia Wyatt

  50. In the discussion above there is reference to planets in space with no nearby stars and just the cosmic background radiation. The planet is either atmosphere-less, has a non-GHG atmosphere or has a GHG atmosphere. It also has either a water surface or a rock surface.

    What about an atmosphere-less planet (called Vulcan) that is both solid (rock) and translucent at the surface with an optical depth of 50 metres. It is a grey body with an emissivity of say 0.85. What is the surface temperature? Is it higher than the cosmic background temperature?

    The planet Vulcan is not transparent and it is not opaque. It is made of a substance intermediate between black obsidian and transparent silica glass.

    Now put the planet into our solar system at the orbital position of the Earth. What is the new surface temperature of Vulcan? Presumably it is not the standard SB temperature of 255 k. It must be significantly higher.

    If the surface temperature is substantially greater than 255 k then this must reduce the approximate warming that can be attributed to the greenhouse effect on Earth. This is because the Earth has an ocean that is an analogue for the translucent obsidian. The Earth’s translucent ocean lifts the average surface temperature regardless of the presence or absence of greenhouse gases in the atmosphere, so long as its surface does not freeze over.

    An argument is made in a previous comment that with a non-GHG atmosphere the Earth’s oceans would evaporate, condense, and fall to the surface as ice, thereby freezing all water at the surface. This does not apply to Vulcan. Nor does it actually apply to the real Earth as the real Earth always has an atmosphere. In other words the argument is a red-herring.

    So on Earth how much of the observed greenhouse effect is caused by the the translucent nature of about 70% of its surface, given that most of the ocean surface is not frozen, and how much is caused by GHG in the atmosphere?

    • Rob R. “So on Earth how much of the observed greenhouse effect is caused by the the translucent nature of about 70% of its surface, given that most of the ocean surface is not frozen, and how much is caused by GHG in the atmosphere?”

      In addition to the basic black body estimate of TSI/4 adjusted for albedo there is a sub surface temperature estimate of TSI/pi. TSI/pi gives an estimate of the sub surface assuming there is perfect insulation. To do it right, you would integrate over the entire liquid ocean surface considering the solar azimuth, elevation, seasonal variation, optical depth etc. to determine the actual input energy. This is what you would consider designing a solar gradient pond. Then you would have to consider all the heat loss. The heat loss estimates would include solar energy absorbed in all the atmospheric layers plus energy transferred from the sub surfaces, each layer of the atmosphere/ocean system. That is what the climate models try to do, but the ocean part is pretty difficult. Since the two “greenhouses” ocean subsurface and atmosphere are very tightly coupled, I don’t think you can completely separate the two. A rough estimate though is around 17% of the ocean part is not properly accounted for which is part of the reason the models tend to track on the hotter side of reality in addition to the cloud issues. That 17% is roughly what Guy Callendar listed as “other” in his paper on the GHE.

      BTW, I am not a physicist, I just stayed at a Holiday Inn Express once.

  51. A fan of *MORE* discourse

    Mike Flynn is curious “If you are aware of any available insulator with better R values than a properly constructed Dewar flask, I would be most interested to hear from you.”

    It is a pleasure to oblige your curiosity Mike Flynn!

    Spacecraft engineers never orbit plain vacuum insulated dewars, because multi-layer insulation of cryogenic tanks reduces heat-leaks by further factors of (typically) hundreds to thousands.

    Just as CO2 similarly increases the heat-trapping capabilities of the Earth’s atmosphere.

    It is an ongoing pleasure to improve your aerospace, climate-science, and thermodynamical understanding, Mike Flynn!

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    • A fan of *MORE* discourse

      Mike Flynn jokes “Mylar film no more traps heat than CO2 does. If you can use Mylar film to trap some heat, maybe you can sell it to Siberians or Minnesotans during the winter. I’m only joking, of course.”

      Lol … Mike, yer joke conveys profound thermodynamic insight!

      Stay cozy, Climate Etc readers!

      \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • AFOMD,

      So you cannot provide an R value for your MLI. This is understandable, as in a vacuum, each pair of highly reflective surfaces acts in the same manner as a single Dewar flask. Nesting Dewar flasks in similar fashion provides far greater insulation, but as NASA discovered, MLI suffers from serious defects in some aspects, which is why NASA used other forms of insulation in preference to, or in conjunction with MLI, depending on circumstances.

      The matter is, in any case, moot. NASA seems to be using Russian rocket motors, or simply buys Russian tickets on Russian rockets. Invoking the name of NASA, third rate pseudo scientists, minority religious leaders, or a collection of assorted representatives of some 5% of the world’s population, doesn’t support the pseudo-science of Warmism terribly well.

      Once again, you make the groundless assertion that the endless reexamination of averages derived from past weather events qualifies as a science. It doesn’t. Not according to me, anyway. I certainly wouldn’t have given Michael Mann a Nobel prize for any sort of science, but then neither did the Nobel Committee, so maybe I’m in good company.

      Live well and prosper,

      Mike Flynn.

    • AFOMD,

      Wrapping a corpse in aluminised Mylar film provides no warmth, nor does CO2.

      Wrapping a live human body in aluminised Mylar film, provides no warmth. It does, of course, retard the rate at which the body exchanges radiant energy with the external environment. However, a properly designed four seasons down filled sleeping bag will provide even better comfort. You won’t survive long sleeping in sub-zero temperatures clad only in a space blanket, NASA notwithstanding.

      Surrounding a live human body with CO2 will achieve not much except bemused laughter from the onlookers, and agonised wonderment from the object of such lunacy. Maybe you might choose to volunteer to enter a blast freezer filled with CO2, clothed only in your dignity and suitable breathing equipment, or maybe not.

      I would certainly pay a small fee to see your faith in the warming ability of CO2 evaporate in the face of freezing reality. When may we look forward to your demonstration?

      Live well and prosper,

      Mike Flynn.

  52. Too much fun!

    Rob Ellison refuses to accept that it was quite a long time before the oceans formed, and some time after that before the average water temperatures dropped below 50C. And so it goes. The Earth is warming – except that it isn’t. The surface radiates away energy – which causes it to warm.

    The list goes on. I leave it to you, gentle reader.

    Live well and prosper,

    Mike Flynn.

    • Dear Professor Mike,

      You know I believe religiously everything that you say. Including that the Earth was a pretty uncomfortable place to be during the early planet forming days.

      But the question that presents itself – is what has it done for you lately? The core and the mantle continue to lose energy at a rate of 0.05W/m2 or so – and to gain energy at the surface at a rate of some 3000 times greater – warming the oceans and the ground to hundreds of metres depth at least. Even more at the place high above the planet where all energy flux is radiative sometimes called top of atmosphere for short – although i know you don’t like the term.

      The problem is that this not a steady 3000+ plus times energy gain – but varies over time from snowball Earth to blue-green planet – repeatedly. So that the planet surface does appear to warm and cool and then warm again and cool again and so on. But I take it that you mean that the surface isn’t actually warming at the moment?

      ‘It is hypothesized that persistent and consistent trends among several climate modes act to ‘kick’ the climate state, altering the pattern and magnitude of air-sea interaction between the atmosphere and the underlying ocean. Figure 1 (middle) shows that these climate mode trend phases indeed behaved anomalously three times during the 20th century, immediately following the synchronization events of the 1910s, 1940s, and 1970s. This combination of the synchronization of these dynamical modes in the climate, followed immediately afterward by significant increase in the fraction of strong trends (coupling) without exception marked shifts in the 20th century climate state. These shifts were accompanied by breaks in the global mean temperature trend with respect to time, presumably associated with either discontinuities in the global radiative budget due to the global reorganization of clouds and water vapor or dramatic changes in the uptake of heat by the deep ocean. Similar behavior has been found in coupled ocean/atmosphere models, indicating such behavior may be a hallmark of terrestrial-like climate systems [Tsonis et al., 2007].’ http://onlinelibrary.wiley.com/doi/10.1029/2008GL037022/full

      These synchronization events suggest that the Earth system is deterministically chaotic – and that the changes in the trajectory of surface temperature – around 1910, the mid 1940’s, the late 1970’s and 1998/2001 are quite natural changes in the system state involving abrupt changes in ocean and atmosphere circulation.

      Clouds did seem to change about the right time.

      ‘Earthshine changes in albedo shown in blue, ISCCP-FD shown in black and CERES in red. A climatologically significant change before CERES followed by a long period of insignificant change.’

      But if anything the rate of ocean warming seemed to decline in Argo – although with great uncertainty.

      As you have explained – this is a false prophecy and the gentle and cliched reader may cheerfully ignore the data on long term – and recent – climate change because it is clear from first principles that the planet surface is cooling at a steady rate. What would we do without Professor Mike? .

  53. A fan of *MORE* discourse

    The Summary Judgment of Climate Etc Discourse

    • The reality of the Gravito-Thermal Effect  is denied by macroscopic thermodynamic theory, denied by microscopic dynamical theory, denied by observation of nature, and denied by laboratory experiment.

    • The reality of the Greenhouse Effect  is affirmed by macroscopic thermodynamic theory, affirmed by microscopic dynamical theory, affirmed by observation of nature, and affirmed by laboratory experiment.

    That’s why the comments from Climate Etc’s denialists have distilled down to irrelevant quibbling, angry ranting, impotent frothing, and pointless abuse … eh Climate Etc readers?

    \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • Fan, it’s rare, so I’ll say it. I agree with you.
      Have a good on ya day.

    • ‘The equilibrium state of any ideal gas with a finite adiabatic index is essentially polytropic. Here, it is important to make precise what we mean by equilibrium. We use the term in the context of a mathematical model; it refers to a solution of the equations of motion with the property that all flows vanish and all the fields are time independent. Polytropic models of earthly and stellar atmospheres are very widely used, and the stationary configurations of such atmospheres are equilibria in this sense, although the physical configurations that they are meant to represent are not states of true thermodynamic equilibrium. An issue that we wish to understand is the precise role that is played by radiation. We should hope to develop an understanding of what would happen if the intensity of radiation were continuously reduced to zero. The possibility that the limit might turn out to be other than isothermal is not easy to accept, for it goes against one of the basic tenets of thermodynamics: Clausius’ statement of the second law (Section 3.7). The question is not entirely academic, but it has no direct bearing on the validity of our approach, for we apply it to the standard, polytropic atmospheres, just as has been done since the pioneering work of Lane (1870) [7].

      It is important to incorporate the isothermal atmosphere into a dynamical framework; that is, to develop a system of equations of motion that predict the uniformity of the temperature at equilibrium. We suggest that the standard theory of polytropic atmospheres should incorporate a parameter or variable to represent the intensity of radiation and that would allow the effect of radiation to be reduced parametrically to zero, resulting in an isothermal equilibrium in the limit. In the event that an experiment should validate the isothermal atmosphere, the need to construct such a theory would become urgent.

      There seems to be a dearth of experimental data. We study an ideal gas in a centrifuge and invoke the equivalence principle to relate this situation to atmospheres. Experiments are proposed. (Section 3.8).’

      – See more at: http://www.mdpi.com/1099-4300/16/3/1515/htm#sthash.NZC7a5DJ.dpuf

      As predicted – some hundreds of comments later – usually repeating the same arguments over and over – but failing to restrain scientific curiosity and the scientific method in a straitjacket of absurd and misguided gedankenexperiment

      FOMBS lack of stellar like temperatures in a centrifuge is an especially extreme example of ideas gone feral. But then he has ulterior motives and science in this seems little more than cargo-cult science. Icons of a powerful and jealous God who will raise the believers on high. There is no solid basis in down in the dirt science – merely smoke and mirrors. Is this not obvious to everyone?

      That’s why the comments from FOMBS has ‘distilled down to irrelevant quibbling, angry ranting, impotent frothing, and pointless abuse … eh Climate Etc readers?’ Is his purpose, practice and lack of substance not abundantly evident? Oh well.

      • Pierre-Normand

        In section 4.2 of his paper Frønsdal writes: “Suppose we start with a vertical column that, for one reason or another, is isentropic. At a certain moment, we turn off the incoming radiation and isolate the gas column from its environment. Assume that the column eventually becomes isothermal and that the dissipation of the temperature gradient is a slow process, during which the gas passes through a sequence of adiabatic equilibrium configurations. The question is this: what are those intermediary configurations, and what is the adiabatic dynamics around the ultimate equilibrium? Recent work by Fronsdal and Pathak (2011) [39] has attempted to answer this question, but this work may raise more questions than it set out to answer.”

        I am unsure what an “adiabatic equilibrium configuration” means in this context. Maybe you have some suggestion. Frønsdal assumes the initial state to be “isentropic”. When the vertical temperature profile exactly matches the adiabatic lapse rate, then the gas not only is stable under convection (as it also is when the temperature gradient is smaller) but it also has a stationary temperature profile under convection. Adiabatic compression or expansion are isentropic processes. So it makes sense to say, as Frønsdal does, that the initial state (that exemplifies the adiabatic lapse rate) is isentropic since it is stationary under convection, both in respects of temperature, pressure and entropy. Though, this is only true in the case of an ideal gas, neglecting the effects of viscous dissipation that occur in all real gases, as Pekka notes.

        Even more troubles arise when we consider the other configurations in the sequence. When the temperature profile doesn’t match the adiabatic lapse rate anymore, then it isn’t stationary under convection. So, what does it mean to say of such configurations that they are in “adiabatic equilibrium”? And the initial configuration itself only was stationary under convection, as we noted, under the assumption that there is no viscous dissipation.

        Though the terminal isothermal profile *is* convectively stable and maximizes entropy, it isn’t stationary under convection. However, no convection occurs. Convective motions have been fully dissipated when the gas achieved that state, and it is stable. So, that’s not a problem. It’s the only true thermodynamic equilibrium state, so far as I can see.

      • Pierre-Normand

        I wrote rather misleadingly: “Convective motions have been fully dissipated when the gas achieved that state, and it is stable.”

        Or it may be better to say that dissipation prevents convection to occur, or be maintained (without external energy input), as soon as the temperature profile reaches or drops below the adiabatic lapse rate, because of viscous dissipation. From then on, the evolution towards the isothermal profile occurs very slowly as a result of diffusion (conduction) only. It is convectively stable over that whole sequence, and thermodynamically stable at the end.

    • David Springer

      Wrong.

      • Pierre-Normand

        David, your idea that maximization of entropy entails, for a column of gas under gravity, that the average mechanical energy of the molecules be the same at all heights isn’t just simplistic and wrong, it is quite absurd. You clearly didn’t think it through.

        If the temperature at the bottom of a tall column of air under gravitational acceleration g, at equilibrium, is 300°K, then the average kinetic energy of the molecules is about (3/2)kT, which is 6.21*10^-21J. This means that nitrogen molecules (mass = 4.65*10^-26 kg) that have this average kinetic energy, have speed

        V = sqrt(2*KEavg/m) = 517m/s.

        So, a N2 molecule that starts up at the bottom of the column with an upward trajectory, and the average kinetic energy, will rise to the height of h = 1/2g*t^2/2 (assuming g uniform for simplicity), where t = (517m/s)/g =

        h = 6.8km.

        This is the height at which a nitrogen molecule with the average kinetic energy at ground level has converted all this energy into gravitational potential energy. Now, according to you, it’s impossible for any nitrogen molecule (or any heavier molecule) to rise past this height h, at equilibrium, since else the average mechanical energy for the molecules at this higher level would be larger than the average at lower heights (which you deem to be constant). *All* the molecules at ground level that have a kinetic energy larger than average would somehow be prevented from rising any higher than the height h at which the molecules of that kind with the average kinetic energy are potentially capable of rising. By what magic would they be so prevented?

  54. ‘The equilibrium state of any ideal gas with a finite adiabatic index is essentially polytropic. Here, it is important to make precise what we mean by equilibrium. We use the term in the context of a mathematical model; it refers to a solution of the equations of motion with the property that all flows vanish and all the fields are time independent. Polytropic models of earthly and stellar atmospheres are very widely used, and the stationary configurations of such atmospheres are equilibria in this sense, although the physical configurations that they are meant to represent are not states of true thermodynamic equilibrium. An issue that we wish to understand is the precise role that is played by radiation. We should hope to develop an understanding of what would happen if the intensity of radiation were continuously reduced to zero. The possibility that the limit might turn out to be other than isothermal is not easy to accept, for it goes against one of the basic tenets of thermodynamics: Clausius’ statement of the second law (Section 3.7). The question is not entirely academic, but it has no direct bearing on the validity of our approach, for we apply it to the standard, polytropic atmospheres, just as has been done since the pioneering work of Lane (1870) [7].’

    All I can do is quote the literature.

    • Pierre-Normand

      Rob Ellison (only) wrote: “All I can do is quote the literature.”

      Agreed.

      • This was actually meant for somewhere else – but it obviously doesn’t stop smarmy little … feeling they can leave one line piles of utter goop behind. Something that is standard practice for CE it seems with nothing in the way of effective sanctions.

        And although I do quote actual and relevant science a lot – as opposed to arm waving in the general direction of a scientific talisman – I am not known for not commenting at length in an erudite, informed, educated and urbane if somewhat and sometime abrasive style of someone not given to suffering fools willingly – and occasionally waxing humourous or lyrical – on a broad range of scientific, social, technological and economic issues. Waxing humorous or lyrical seems a sure fire way to generate snide and spiteful snark and is something else I have stopped indulging in here.

        My position from the start on this paper was that here was an interesting but difficult paper that reached – via Lagrangian evolution of a Hamiltonian based on a Gibbs free energy action principle – some challenging conclusions.

        That a conflation of twits futilely pontificating on many and varied less than convincing thought experiments doesn’t challenge my initial belief that I didn’t know enough to draw an informed opinion on the paper. My considered opinion after 100’s of comments is that there little light but at least in once case amusingly stellar like heat – and that discourse on advanced atmospheric physics topics at CE is energetic but naive, quite superficial and very frequently mind bogglingly misguided.

      • Pierre-Normand

        “[…] but it obviously doesn’t stop smarmy little … feeling they can leave one line piles of utter goop behind […]”
        And you constantly complain about my being excessively verbose…

      • Pierre-Normand,

        It took me a while to understand. Very clever. I often deride Warmist for one or two word dismissive declarations. Sorry.

        I dips me lid to you, sirrah! I can only add – touché.

        Live well and prosper,

        Mike Flynn.

      • So not commenting is problematic as is commenting? This is a rather obvious ploy. I suppose if I shut up and went way – everything would be fine. My words however have nothing in common with P-N’s. I have actual word-smithing skills. I can be terse – I can be florid – I can layer meaning within meanings and sometimes amuse myself by saying something while implying the direct opposite in nested irony. As I did above. Or rollick in a Grand Guignol cascade of linguistic calamities. Right over his head of course. Every word has a place and a meaning in the ordinary English usage.

        Have these people never read any of the great stylists of world literature? Fireworks exploding across a mindscape – diamond bright images illuminating to the depths of the soul. Or even within the utterly astonishing American tradition? Have your schools failed your population? Looking only at P-N and Maxy – grave fears would be held for the health of American culture. I cut my teeth on Henry Miller – a truly great American stylist.

        “In the days to come, when it will seem as if I were entombed, when the very firmament threatens to come crashing down upon my head, I shall be forced to abandon everything except what these spirits implanted in me. I shall be crushed, debased, humiliated. I shall be frustrated in every fiber of my being. I shall even take to howling like a dog. But I shall not be utterly lost! Eventually a day is to dawn when, glancing over my own life as though it were a story or history, I can detect in it a form, a pattern, a meaning. From then on the word defeat becomes meaningless. It will be impossible ever to relapse.

        For on that day I become and I remain one with my creation.

        On another day, in a foreign land, there will appear before me a young man who, unaware of the change which has come over me, will dub me “The Happy Rock.” That is the moniker I shall tender when the great Cosmocrator demands-” Who art thou?”

        Yes, beyond a doubt, I shall answer “The Happy Rock!”

        And, if it be asked-“Didst thou enjoy thy stay on earth?”- I shall reply: “My life was one long rosy crucifixion.”

        We may not aspire to poetry – but some fluidity and facility is the seasoning to the meat of reason and meaning.

        There are 2 problems with P-N’s style. First the convoluted and impenetrable syntax consisting of physics like terms strung together in a way he fondly imagines is meaningful . It is not. The usual stricture applies. If you cant describe it simply you don’t understand it – and if you don’t understand it you can’t explain it at all. Visualisation process is the key to understanding both poetry and physics.

        ‘Visualization in some form or other is a vital part of my thinking…half-assed kind of vague, mixed with symbols. It is very difficult to explain, because it is not clear. My atom, for example, when I think of an electron, I see an atom and I see a vector and a y written somewhere, sort of, or mixed with it somehow, and an amplitude all mixed up with x’s … it is very visual … a mixture of a mathematical expression wrapped into and around, in a vague way, around the object. So I see all the time visual things associated with what I am trying to do.’ Feynmann

        Kip Thorne commented on Stephen Hawing.

        ‘As Stephen gradually lost the use of his hands, he had to start developing geometrical arguments that he could do pictorially in his head. He developed a powerful set of tools, you may develop other tools, and the new tools are amenable to different kinds of problems than the old tools. And if you are the only master in the world of these new tools, that means there are certain kinds of problems you can solve and nobody else can.’

        Einstein said that the ‘words or the language, as they are written or spoken, do not seem to play any role in my mechanism of thought. The psychical entities which seem to serve as elements in thought are certain signs and more or less clear images which can be “voluntarily” reproduced and combined. There is, of course, a certain connection between those elements and relevant logical concepts. It is also clear that the desire to arrive finally at logically connected concepts is the emotional basis of this rather vague play with the above-mentioned elements. But taken from a psychological viewpoint, this combinatory play seems to be the essential feature in productive thought–before there is any connection with logical construction in words or other kinds of signs which can be communicated to others.’

        The 2nd problem is simply reproducing the same arguments hundreds of times in a post. A little excessive. This latter combines with the former in what I call extreme verbiage. Sadly – a mess of pottage sans any literary seasoning or precise yet simple explication of whatever it is he is on about.

        Oddly – my trying to help him out here is not being greeted in the spirit it is offered.

      • Pierre-Normand

        Einstein also said:

        “In the interest of clearness, it appeared to me inevitable that I should repeat myself frequently, without paying the slightest attention to the elegance of the presentation. I adhered scrupulously to the precept of that brilliant theoretical physicist, L. Boltzmann, according to whom matters of elegance ought to be left to the tailor and to the cobbler.”

      • ‘Mathematics, rightly viewed, possesses not only truth, but supreme beauty — a beauty cold and austere, like that of sculpture, without appeal to any part of our weaker nature, without the gorgeous trappings of painting or music, yet sublimely pure, and capable of a stern perfection such as only the greatest art can show. The true spirit of delight, the exaltation, the sense of being more than Man, which is the touchstone of the highest excellence, is to be found in mathematics as surely as poetry.’ Bertrand Russell

        The theory of relativity has a sparing and simple elegance that unfolds in the mind for decades.

        Is P-N comparing his prolix prose to that of Einstein? No contest there.

      • Albert Einstein — ‘If you can’t explain it to a six year old, you don’t understand it yourself.’

      • A fan of *MORE* discourse

        Rob Ellison quotes Albert E.  “If you can’t explain it to a six year old, you don’t understand it yourself.”

        Following Let’s follow Donald Knuth’s advice

        “Science is what we understand well enough to explain to a computer. Art is everything else we do. […] Science advances whenever an Art becomes a Science. And the state of the Art advances too, because people always leap into new territory once they have understood more about the old.

        It was Knuth who inspired FOMD to take one further step … explaining gas-column physics to a symbolic programming language!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • ‘Everything that can be counted does not necessarily count; everything that counts cannot necessarily be counted.’ Albert Einstein

      ‘Sensitive dependence and structural instability are humbling twin properties for chaotic dynamical systems, indicating limits about which kinds of questions are theoretically answerable. They echo other famous limitations on scientist’s expectations, namely the undecidability of some propositions within axiomatic mathematical systems (Gödel’s theorem) and the uncomputability of some algorithms due to excessive size of the calculation.’ James McWilliams

      ‘In sum, a strategy must recognise what is possible. In climate research and modelling, we should recognise that we are dealing with a coupled non-linear chaotic system, and therefore that the long-term prediction of future climate states is not possible.’ IPCC

      FOMBS has explained everything mathematically? One of the first things we learn in engineering is the sanity check. I take it we have gotten past not getting stellar like temperatures in a centrifuge then?

  55. A fan of *MORE* discourse

    Per requests:

    ————————-

    Assignment  Demonstrate that the maximum entropy principle, for a vertical gas-column in a gravitational field, implies that the temperature of the gas-column is uniform, such that there is *NO* gravito-thermal effect. Show all work explicitly.

    Showing the work  Here we derive temperature-uniformity solely from the assumption of maximum entropy, (using the symbolic programming language Mathematica to avoid sign-errors).

    To anticipate, here is the output of the code (which will follow in a separate comment).


       Implications of the Sackur-Tetrode entropy function
       ***************************************************
       ideal gas law: T -> (2*U)/(3*k*N)
       ideal gas law: P -> (2*U)/(3*V)
       Verified: P*V == N*k*T
       ideal-gas law: \[Epsilon] -> (3*P)/2
       ideal-gas law: \[Rho] -> (m*P)/(k*T)
       maximum-entropy gas-column: T -> T0
       maximum-entropy gas-column: P -> P0/E^((g*m*z)/(k*T0))
       *******************************************************
       Verified: the maximum-entropy equilibrium is isothermal
                 (i.e., there is no gravito-thermal effect)

    Result  The maximum-entropy state of an ideal gas-column in a gravitational field has uniform temperature, such that there is *NO* gravito-thermal effect.

    \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

    • A fan of *MORE* discourse

      Assignment  Demonstrate that the maximum entropy principle, for a vertical gas-column in a gravitational field, implies that the temperature of the gas-column is uniform, such that there is *NO* gravito-thermal effect. Show all work explicitly.

      Appendix: Working Code

      Just paste-and-run in Mathematica (hopefully).


      (* *********************************** *)
      "\<\
      Implications of the Sackur-Tetrode entropy function
      ***************************************************\>"//
        Print;
      (* begin with the Wikipedia definition 
         of the Sackur-Tetrode entropy *) 
      U::usage = "kinetic energy of the gas (units J/m^3)";
      V::usage = "volume of the gas (units m^3)";
      N::usage = "number of particles in the gas";
      h::usage = "the Plank constant";
      m::usage = "the mass of a gas particle";
      k::usage = "Boltzmann constant";

      (* define intrinsic thermodynamical variables *)
      z::usage = "vertical gas-column spatial coordinate";
      g::usage = "gravitational acceleration (units m/s^2)";
      \[Rho]::usage = "mass-density of the gas: \[Rho] \[Congruent] N*m/V";
      \[Epsilon]::usage = "kinetic energy-density of the gas: \[Epsilon] \[Congruent] U/V";
      \[ScriptCapitalE]::usage = "total energy-density of the gas: \[ScriptCapitalE] = \[Epsilon]+\[Rho]*g*z";

      (* **************************** *)
      (* specify the entropy function *)
      (* **************************** *)

      Sfun$UVN::usage = "Sackur-Tetrode entropy function";
      Sfun$UVN[U_,V_,N_] = k*N*Log[(V/N)(U/N)^(3/2)] + 
          (3/2)*k*N*(5/3 + Log[(4 Pi m)/(3 h^2)]);

      (* ************************ *)
      (* derive the ideal gas law *)
      (* ************************ *)

      T$UVN::usage = "temperature of the gas";
      P$UVN::usage = "pressure of the gas";

      T$UVN = {1/(T) == D[Sfun$UVN[U,V,N],U]}//
        Solve[#,T]&//Flatten//
          ReplaceAll[T,#]&;
      Print["ideal gas law: T -> ",T$UVN//InputForm];

      P$UVN = D[Sfun$UVN[U-P*\[Delta]V,V+\[Delta]V,N],\[Delta]V]//
        Limit[#,\[Delta]V->0]&//Solve[#==0,P]&//Flatten//
          ReplaceAll[P,#]&;
      Print["ideal gas law: P -> ",P$UVN//InputForm];

      (* verify that ideal gas law is satisfied *)

      P*V == N*k*T//
        ReplaceAll[#,{T->T$UVN,P->P$UVN}]&//
      If[#//TrueQ,
          Print["Verified: P*V == N*k*T"];
          ,
          Print["Invalidation: P*V != N*k*T, yikes!"]; 
          Throw["Invalidation"];
      ]&;

      (* derive rules for change of variables *)
      toPVrules = {T == T$UVN,P == P$UVN}//
        ReplaceAll[#,{U->\[Epsilon]*V,N->\[Rho]*V/m}]&//
          Solve[#,{\[Epsilon],\[Rho]}]&//Flatten;
      Map[Print["ideal-gas law: ",#//InputForm]&,toPVrules];

      (* ******************************* *)
      (* find gas-column maximum entropy *)
      (* ******************************* *)

      (* compute the equilibrium equations *)
      \[Lambda]1::usage = "Lagrance multiplier for energy-conservation";
      \[Lambda]2::usage = "Lagrance multiplier for mass-conservation";
      equilibriumEquations = (Sfun$UVN[U,V,N]/V + \[Lambda]1*\[ScriptCapitalE] + \[Lambda]2*\[Rho])//
        ReplaceRepeated[#,{
            U->V*(\[ScriptCapitalE]-\[Rho]*g*z),
            N->V*\[Rho]/m
        }]&//{D[#,\[ScriptCapitalE]],D[#,\[Rho]]}&//
          ReplaceRepeated[#,{\[ScriptCapitalE]->\[Epsilon]+\[Rho]*g*z}~Join~toPVrules]&//
            Simplify//PowerExpand//Simplify//
              Map[#==0&,#]&;

      (* it is simpler to verify the equilibrium solution 
         as an ansatz than to construct it ab initio (although
         the ab initio construction is entirely feasible) *)

      equilibriumSolutionAnsatz = {
        T -> T0,  (* NOTE: *no* gravito-thermal effect *)
        P -> P0 Exp[-g*m*z/(k T0)]
      };

      equilibriumSolutionAnsatz//
        Map[Print["maximum-entropy gas-column: ",#//InputForm]&,#]&;

      equilibriumEquations//
        ReplaceAll[#,equilibriumSolutionAnsatz]&//
          PowerExpand//Solve[#,{\[Lambda]1,\[Lambda]2}]&//Flatten//
            FreeQ[#,z]&//
      If[#//TrueQ,
          Print["\<\
      *******************************************************
      Verified: the maximum-entropy equilibrium is isothermal
                (i.e., there is no gravito-thermal effect)\>"];
          ,
          Print["Invalidation: isothermal disequilibrium, yikes!"]; 
          Throw["Error"];
      ]&;

      \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • This is an example of why we lawyers think science is too important to be left to scientists. Scientists too often think they’ve proved a conclusion when in fact they’ve assumed the conclusion to begin with. There are circles in which this is known as begging the question.

      • A fan of *MORE* discourse

        Joe Born opines  “We lawyers think science is too important to be left to scientists.”

        Renowned legal scholars like Roberto Mangabeira Unger certainly lend weight to your opinion!

        Good on `yah, Joe Born, for helping to inspire Climate Etc readers to inquire more deeply and learn more broadly!

        \scriptstyle\rule[2.25ex]{0.01pt}{0.01pt}\,\boldsymbol{\overset{\scriptstyle\circ\wedge\circ}{\smile}\,\heartsuit\,{\displaystyle\text{\bfseries!!!}}\,\heartsuit\,\overset{\scriptstyle\circ\wedge\circ}{\smile}}\ \rule[-0.25ex]{0.01pt}{0.01pt}

      • Steven Mosher

        Jesus Joe.

        You lawyers dont even understand that science isnt about proof.
        you dont prove anything in science.

        A) you explain.
        B) you predict.

        There are some areas of science where you cannot predict, or cases where you can predict, but testing is practically not feasible. In these areas science is limited to explanation. explanation is “making sense” of observables with the most parsimonious set of rules.

        Proof belongs to math and logic and geometry. Science is explanation and prediction. You dont prove you are right. you show that your explanation accounts for observables and that it makes predictions that are in principle testable, even though you may not be able to explicitly test them.

      • No amount of experimentation can ever prove me right; a single experiment can prove me wrong. – Albert Einstein

        FOMBS treats science and maths as a talisman. It doesn’t seem at all practical.

        ‘1. An object marked with magic signs and believed to confer on its bearer supernatural powers or protection.
        2. Something that apparently has magic power.’

        Pretty much the antithesis of science.

      • Steven Mosher, “You lawyers don’t even understand that science isn’t about proof. You don’t prove anything in science.”

        That is so true. I think Joe though was commenting more on how Fan presents his science than how science is supposed to be.

        Fan is assuming that the maximum entropy principle as it applies to ideal monoatomic gases accurately demonstrates that an isolated atmosphere under the influence of gravity will be “perfectly” isothermal. The way Fan presents his “demonstration” has more of a “proof” feel to it than a more scientific “feeling” argument that includes limits of his model and potential sources of uncertainty etc. etc.. Sackur-Tetrode entropy is generally recognized a being less than ideal at lower temperatures, ie lower molecular velocities, where real gases tend to behave less like ideal monoatomic gases, though it is close enough in the majority of cases.

        For Fans case though, covering the Earth with his Mylar MLI could produce an equilibrium temperature in the low thousands of degrees which would pretty much ensure an isothermal equilibrium. Unfortunately, that might inspire some of the Sky Dragon types to believe Fans Demonstration of *no* gravito-thermal effect is actually a demonstration “of” the gravito-thermal effect. I think it has something to do with the assumption that there is Zero potential energy at the lowest level of the atmosphere due to gravity when the mass of the entire volume actually creates the gravitational potential. Kind of a chicken or egg thing as I understand it.

        Joan Baez, the physicist not the singer, believes he has a solution but I don’t think he has published it.

      • This does not imply anything about temperatures in the column. These two simple equations – Sackur–Tetrode and the ideal gas equation are statistical expressions of the system state. We get a number for entropy – which is maximum for the equilibrium state and for pressure-volume-temperature relationships. So knowing pressure and volume we can determine temperature of the gas – for instance – and use it to calculate entropy for the equilibrium state.

        To imagine that the solution to these equations – if that is in fact what happens – lacking assumptions for pressure and volume – or a coherent approach to the logic of progression in these simple equations – is a mathematical proof of anything at all is a complete sham or utter insanity. Take your pick.

      • AFOMD,

        As you probably know, I don’t follow your links.

        However, I can see a representation of a sign which might lead one to believe that Buddhism is a religion. I might disagree, but would you mind expressing your personal definition of religion?

        Just curious. I have no intention of arguing – it is of course personal to you. I don’t follow your links, I’m interested in your personal opinion.

        Feel free to ignore me if you think it’s a bit personal. I understand.

        Live well and prosper,

        Mike Flynn.

      • A fan of *MORE* discourse

        Mike Flynn  asks [respectfully] “Would you mind expressing your personal definition of religion?”

        A respectful and sincere question surely deserves a respectful and sincere reply.

        Two words: continuing revelation.

        Caveat  Everyday experience, and history too, remind us that God and the Devil alike reside “in the details” … the details everyday life … and that the line between good and evil “runs down the middle of every human heart.”

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      • Pierre-Normand

        Most everyone who responded to FOMD’s Mathematica calculation seem to overlook that it is a valuable contribution since it purports to prove (I haven’t checked) that for an ideal gas under gravity, the state that maximized entropy is isothermal. Since this result flatly contradicts the expectations of many advocates of a gravito-thermal gradient (such as David Springer, Doug C and probably also Christian Frønsdal) also regarding ideal gases, it is a valuable contribution to the debate. A valuable response to it would be to verify the soundness of the proof (under the assumption that the gas is ideal) or pointing out any error.

      • Pierre-Normand

        Rob Ellison: “This does not imply anything about temperatures in the column.”

        Yes, looking more closely at the code — for how little I understand — it seems to me Rob might be right. I don’t see how FOMD allows for various vertical temperature profiles (with T an explicit function of z) and select for the profile that has the maximum entropy. This may need more work.

      • A fan of *MORE* discourse

        Technical Note  Physically, an ideal gas column is specified physically by the temperature of the column and the pressure at its base; mathematically the gas column is specified by the thermodynamic potentials \lambda_1 and \lambda_12, which enter in the Mathematica code as Lagrange multipliers.

        In thermodynamical lingo, \lambda_1 is the “inverse temperature” of the gas-column; \lambda_2 is the “chemical potential” of the gas-column; and *BOTH* are spatially uniform.

        In engineering practice, temperature and pressure are the preferred variables for the pragmatic reason that they are easy to measure, even though they are *NOT* as mathematically natural as the potentials \lambda_1 and \lambda_2.

        Thorough understanding of thermodynamics requires familiarity with *BOTH* sets of variables.

        ————

        Pierre-Normand  notices [correctly] “Most everyone who responded to FOMD’s Mathematica calculation seem to overlook that it […] purports to prove […] that for an ideal gas under gravity, the state that maximized entropy is isothermal.

        Pierre-Normand, this is precisely correct!

        However, the result is *not* limited to ideal gases only, or to the Sackur-Tetrode entropy function in particular.

        Advanced exercise  By abstracting and extending the methods of the Sackur-Tetrode example, show that for any convex entropy function whatsoever, for any number of conserved quantities whatsoever, subject to any spatially-varying fields whatsoever, in any (possibly non-euclidean) geometry whatsoever, in any number of spatial dimensions whatsoever: at the maximum-entropy equilibrium *ALL* of the thermodynamic potentials are spatially uniform.

        Happy thermodynamical computing, Climate Etc readers!

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      • P-N, there are two groups that any “solution” needs to address. One group, Sky Dragons and fans of perpetual motion, are literally looking at the limits of Maxwell-Boltzmann statistics and the ideal gas laws thinking that the G-T effect is “significant” enough to patent and another group that would like to fine tune “equilibrium” to include things like relativistic effects. It would probably be better to address the more sane of the two instead of rehashing “Ideal” gases and “Ideal” definitions of equilibrium.

        The second group could be onto something by the way, at least in my opinion.

      • Pierre-Normand: “A valuable response to it would be to verify the soundness of the proof (under the assumption that the gas is ideal) or pointing out any error.”

        My sell-by date is too imminent to spend enough time slogging through the Mathematica code and learn the physics arcana, so I’ll have to decline that invitation.

        I did look “Sackur-Tetrode” up here: http://en.wikipedia.org/wiki/Gibbs_paradox, though, and at least at first blush its derivation appears to assume the absence of gravity: the phase space is said to be the surface of a hypercylinder, not a hypercone.

        As I said above in a previous comment, that smells a lot like begging the question. Again, though, I have not waded through the whole thing.

      • Joe Born,

        A hypercone would be a better choice but I think the “Idealists” would still defend their concept to the ultimate heat death.

      • Pierre-Normand

        Joe Born wrote: “I did look “Sackur-Tetrode” up here: […], and at least at first blush its derivation appears to assume the absence of gravity: the phase space is said to be the surface of a hypercylinder, not a hypercone. As I said above in a previous comment, that smells a lot like begging the question.”

        Good point. There is a discussion of and ideal gas in a gravitational field in the following link where they adapt the S-T formula to account for gravity, though the isothermal profile is assumed (which as they seem to imply goes hand in hand with the assumption that diffusive equilibrium is achieved and that therefore the chemical potential doesn’t vary with altitude — see also the last paragraph of Pekka’s note) and the density gradient is derived.

        http://www.physics.usu.edu/torre/3700_spring_2013/lectures/04.pdf

        Let’s see of FOMD can run his code with the adapted S-T formula or show why the result ought not to change. Meanwhile, have a look at the more complete discussion that I linked to earlier:

        http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%282004%29061%3C0931%3AOMEP%3E2.0.CO%3B2

        There seems not to be any question begging in this treatment of our problem,

      • Pierre-Normand: “Meanwhile, have a look at the more complete discussion that I linked to earlier”

        Thanks for the link, but it’s something I had looked at before and found unsatisfying. Possibly it was just that I was leery of the what assumptions may go into the concept of potential temperature. (I’m a layman–e.g, before I looked it up yesterday I would have thought Sackur Tetrode was some kind of vacuum tube–and it takes me some time to think these things through.)

        My interest in this issue is not so much whether at equilibrium a gas column is isothermal but rather how much we layman should accept when we hear that “scientists say” such and such. Even in their particular areas of expertise, and where there actually is a known solution, the discussion above shows scientists are likely to be unaware of the tacit assumptions they are making. Particularly when humanity’s welfare depends on it and there is no known solution, we laymen should challenge their opinions.

        I also think the Roman et al. and Velasco et al. papers are the ones that make the fewest assumptions (except for ignoring quantum mechanics). They don’t speak explicitly in terms of maximizing entropy, but they do compute the probability density of a particle’s having a particular altitude and speed, and assigning distribution in accordance with that density should, I think, maximize entropy. So far, therefore, I think Roman et al. and Velasco et al. together provide the best treatment. And, as you say, they find isothermality to within the limit of thermodynamics’ applicability.

      • Pierre-Normand

        Joe Born wrote: “So far, therefore, I think Roman et al. and Velasco et al. together provide the best treatment. And, as you say, they find isothermality to within the limit of thermodynamics’ applicability.”

        Indeed, and their mathematical treatment at the level of partition functions of mechanical states simply vindicates what can be found independently from entropy consideration in classical thermodynamics. This corroborate the sound statistical mechanical foundation of classical thermodynamics and explains why Verkley and Gerkema find results consistent with those from kinetic theory. That’s no accident.

    • Pierre-Normand

      CD wrote: “Joan Baez, the physicist not the singer, believes he has a solution but I don’t think he has published it.”

      That would be John Baez

      • Pierre-Normand

        FOMD: “By abstracting and extending the methods of the Sackur-Tetrode example, show that […] at the maximum-entropy equilibrium *ALL* of the thermodynamic potentials are spatially uniform.”

        Yes, that would be valuable as an *supplementation* (not just an extension) of your result. If I understand correctly, it would amount to a demonstration of the fact that lambda_1, the inverse temperature (and hence also T), is spatially uniform. However, as your previous demonstration stood, this was simply assumed, right?

      • A fan of *MORE* discourse

        No. In the example, the unique entropy-maximizing state has uniform temperature. It’s true that we guessed this as an ansatz … which is not the same thing as assuming it … for the simple reason that we might have discovered that the uniform-temperature ansatz failed!

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      • Pierre-Normand

        FOMD: “No. In the example, the unique entropy-maximizing state has uniform temperature. It’s true that we guessed this as an ansatz … which is not the same thing as assuming it … for the simple reason that we might have discovered that the uniform-temperature ansatz failed!”

        OK, I think I am getting it now. So, T is a function of z, after all, and you are making use of a variational principle to check that the the isothermal solution yields an extermum for entropy, correct? That would indeed be valid and a special case of your general statement about uniformity thermodynamic potentials being a consequence of entropy maximization.

      • Pierre-Normand

        “extermum (sic)” …extremum…

      • A fan of *MORE* discourse

        Pierre-Normand gets it “T is a function of z, after all, and you are making use of a variational principle to check that the the isothermal solution yields an extremum for entropy, correct?”

        That is exactly correct!

        Further reading  Here is an excellent (and recent) discussion of the Legendre transforms that go back-and-forth between conserved densities and their thermodynamic potentials.


        @article{Zia:2009lr, Author = {Zia, R. K.
        P. and Redish, Edward F. and McKay, Susan
        R.}, Journal = {American Journal of
        Physics}, Number = {7}, Pages = {614-622},
        Title = {Making sense of the {L}egendre
        transform}, Volume = {77}, Year = {2009}}

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      • Pierre-Normand

        Very nice reference, FOMD. Thanks. It’s available at arxiv:

        http://arxiv.org/pdf/0806.1147.pdf

        The American Journal of Physics (published by the American Association of Physics Teachers) is simply awesome. As an undergraduate student, 20 years ago, I would spend whole afternoons at the library perusing old issues to find fun puzzles about QM and the theory of relativity. I also spent way too much time doing this shortly before my exams.

      • So we have morphed from a Lagrance (sic) multiplier to the Legendre Transform?

        ‘The Legendre Transform (LT) is a commonly used
        mathematical tool in upper division and graduate physics
        courses, especially in Classical Mechanics (CM) [1], Statistical Mechanics (SM) and thermodynamics. [2]
        Most physics majors are exposed to the LT first in CM,
        where it provides the connection between the Lagrangian
        L(q) and the Hamiltonian H(p), and then in SM where
        it creates relations between the internal energy, E and
        the various thermodynamic potentials, e.g., S, G, H
        and A.’

        Let me know when you get past hand waving at 100 year old equations.

      • Pierre-Normand

        There is no “morphing” of anything into something else. The method of Lagrange multipliers was used to find extrema of equations subject to equality constraints (such as follow from conservation energy, and of mass). Discussion of the Legendre transform just was relevant to the definition of the dimensionless thermodynamic potentials. (Read a bit more of the paper). They’re two independent topics, though both are relevant to explaining FOMD’s calculation.

      • Pierre-Normand

        P-N: “They’re two independent topics, though both are relevant to explaining FOMD’s calculation.”…
        Though the latter isn’t necessary to validate the result, it explains how thermodynamics potentials relate to conserved quantities. See specially the two subsection entitled: “The route of physics: interpretation of the
        equilibrium condition” and “How does the Legendre transform enter into
        thermodynamics?”

      • You may believe anything you like P-N. I am certainly not falling for the saga of using a Lagrange multiplier to determine equilibrium entropy in Sackur-Tetrode. Utter BS. Nor am I inclined to wonder what a Legendre transform means for the potentials in this utterly mad riff on Sackur-Tetrode and the ideal gas equation.

        Let’s keep it at basics. What can we possibly know from these 2 simple equations? Anything more – and there has been a great deal more said with much fanfare and little to no substance – and it is all BS.

      • Pierre-Normand

        “Nor am I inclined to wonder what a Legendre transform means […]”

        Yes, I know.

      • The thermodynamic potential is explicit in the formula. All the rest is crazy talk. You have to have something with a modicum of relevance – not simply arm wave at a paper. Or simply arm wave if you find it interesting – but not claim relevance in some context in which it is simply bafflegab.

      • Pierre-Normand

        Rob Ellison wrote: “The thermodynamic potential is explicit in the formula.”

        Indeed it is. The Langrange multiplier method just is used to check that the isothermal profile, together with the vertical barometric density profile, jointly realize an extremal entropy state (and it’s obviously not a minimum). Any continuous deviation from this profile, either in density or temperature, would therefore reduce entropy. It is therefore an equilibrium state.

        You can now forget about the Legendre transform paper if you aren’t interested to learn more about its applications to classical thermodynamics and statistical mechanics. FOMD’s calculation doesn’t depend on it.

      • Pierre-Normand

        For people who unlike Rob aren’t completely incurious about the direct relevance of Legendre transforms to entropy maximization (and the dual principle of energy minimization), for simplifying calculations with the use of thermodynamic potentials, there is also this neat online presentation:

        http://casey.brown.edu/research/crp/Edu/Documents/00_Chem201/6_thermodyn_pot/6-thermodyn_pot-frames.htm

      • Sorry we have the Sackur-Tetrope expression.

        http://www.wolframalpha.com/input/?i=Sackur-Tetrode+equation

        And the ideal gas equatuion.

        http://www.ajdesigner.com/idealgas/

        Do you really expect that the rest of the apocryphal bafflegab holds any meaning at all? Such mendacious and empty verbiage from the acolyte of extreme verbiage. I always hope for a higher standard of personal integrity – but am never bothered when it doesn’t eventuate.

      • David Springer

        None of the recent links in this thread mentions gravity confinement thus there is no analysis applicable to gravito-thermal effect. One must first accept the fact that potential energy and kinetic energy are freely interchangeable in gravitational confinement. The sum of PE and KE is called mechanical energy. ME is of highly practical concern to engineers who build things in gravity wells i.e. engineers living on the planet earth. Like duh. From there we merely consider what is the state of maximum entropy in a non-convecting atmosphere. It cannot possibly be isothermal IMO because PE at altitude will give a mole of atmosphere more usable energy than a mole of atmosphere at no altitude if both moles have the same kinetic temperature.

        No one has ever disproven this since Loschmidt described the gravito-thermal effect 150 years ago in the golden age of physics. No amount of blog banter from undistinguished physicists who can’t even rise to having their opinion on the matter placed in the peer reviewed literature. It’s a comedy of inferiors thinking they don’t need experimental physics because they have the world perfectly modeled inside their pointy little heads like much of the rest of warmist dingbats whose failure to predict what happens in the real world is legendary. Hiding the decline in temperature is a knee jerk reaction among that crowd whether it comes to paleo-temperatures from tree rings, meausured temperatures in the southern hemisphere, the Little Ice Age, the missing tropospheric hot spot, and now a lapse rate in a non-convecting atmosphere that is demanded by maximum entropy considerations in free mechanical energy.

    • A fan of *MORE* discourse

      Technical Note  Physically an ideal gas column is naturally specified by the temperature T_0 of the column and the pressure P_0 at its base; mathematically the gas column is naturally specified by the thermodynamic potentials \lambda_1 and \lambda_2 (which enter in the Mathematica code as Lagrange multipliers).

      In thermodynamical lingo, \lambda_1 is the “inverse temperature” of the gas-column; \lambda_2 is the “chemical potential” of the gas-column; and *BOTH* are spatially uniform.

      In engineering practice, temperature and pressure are much-preferred variables — even though they are *NOT* as mathematically natural as the potentials \lambda_1 and \lambda_2 — for the pragmatic reason that they are easy to measure.

      Pierre-Normand  notices [correctly] “Most everyone who responded to FOMD’s Mathematica calculation seem to overlook that it […] purports to prove […] that for an ideal gas under gravity, the state that maximized entropy is isothermal.

      Pierre-Normand, this is precisely correct!

      However, the result is *not* limited to ideal gases only, or to the Sackur-Tetrode entropy function in particular.

      Advanced exercise  By abstracting and extending the methods of the Sackur-Tetrode example, show that for any convex entropy function whatsoever, for any number of conserved quantities whatsoever, subject to any spatially-varying fields whatsoever, in any (possibly non-euclidean) geometry whatsoever, in any number of spatial dimensions whatsoever: at the maximum-entropy equilibrium *ALL* of the thermodynamic potentials are spatially uniform.

      Happy thermodynamical computing, Climate Etc readers!

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      • Pierre-Normand

        FOMD: “In thermodynamical lingo, \lambda_1 is the “inverse temperature” of the gas-column; \lambda_2 is the “chemical potential” of the gas-column; and *BOTH* are spatially uniform.”

        This may most likely be the case, but just assuming it seems to be the question begging against someone who would insist that lambda_1 might not be uniform if there is a gravito-thermal effect.

      • A fan of *MORE* discourse

        What the calculation demonstrates is a special case of this general theorem:

        Theorem  Given any state with non-uniform thermodynamic potentials, there is a state having (1) the same globally conserved quantities, and (2) spatially uniform thermodynamical potentials, and (3) higher entropy.

        Indeed, from both a physical and mathematical point of view, thermodynamical potentials are defined in terms of the entropy function, precisely so as to ensure that this theorem holds!

        For the common case that energy is the sole conserved quantity, this potential-uniformity theorem is equivalent to the Zeroth Law of Thermodynamics

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      • Really? The usual sanity check is to cite literature along these lines. But of course there is none. Nothing remotely bears any resemblance to this utterly mishmash of physics madness in support of an ‘antatz’.

        I have never seen such horrifying triple plus blogospheric unscience. Is there a taint of lunacy here – an immensely over the top Bose-Einstein meltdown – or is merely a cynical duping of the gullible. I can’t imagine any rational purpose for the latter.

    • Pierre-Normand

      CD wrote: “P-N, there are two groups that any “solution” needs to address. One group, Sky Dragons and fans of perpetual motion, are literally looking at the limits of Maxwell-Boltzmann statistics and the ideal gas laws thinking that the G-T effect is “significant” enough to patent and another group that would like to fine tune “equilibrium” to include things like relativistic effects. It would probably be better to address the more sane of the two instead of rehashing “Ideal” gases and “Ideal” definitions of equilibrium.”

      Some people I choose not to argue with at all. When I choose to argue with someone, I try refrain from pigeonholing him/her. You yourself seemed to endorse “the basic premise of the gravito-thermal effect as [you] understand it”. This ‘basic premise’ actually is an *argument* that can be applied with no modification to an ideal gas at equilibrium in a box. So long as you don’t acknowledge the flaw in this basic argument, you are pigeonholing yourself into the same box as the Sky Dragon Slayers, and David Springer; at least on that issue. It’s difficult to move forward to examining more complex cases, or more subtle arguments, when this simple misunderstanding isn’t acknowledged and corrected.

      • P-N, Interesting. So it is like your “ideal” against the world. Sounds like a religious thing. Since my position is that no ideal model is ever truly ideal, I am a heathen.

        btw, in the newest paper you linked comparing isothermal to isentropic the author finds a value of alpha equal to 1.6868. There is a whole new group of “deniers” out their that would likely find the ratio to be this

        That particular group is into Self-Organizing Criticality. They may be considered the Zen group that believes there is no true equilibrium only preferred states of relative balance in nature :)

      • David Springer

        Pierre, you don’t know enough to distinguish my position from the dragon slayers. I would be most happy though if you didn’t waste my time by responding to me in the future as you’re too uninformed to engage in any productive dialog with me. You’re a warmist troll and you might better find a blog with more of your kind on it to engage in mutual back patting and congratulatory activities that is your only source of self-esteem.

      • Pierre-Normand

        “P-N, Interesting. So it is like your “ideal” against the world. Sounds like a religious thing.”

        CD, for the Nth time, most of your fellow gravito-thermalists — including yourself! — practice the same religion since your “basic premise” applies to ideal gases. If the premise is correct, then there ought to be a temperature gradient in an *ideal* gas column *at equilibrium* because it requires no extra assumption regarding deviations from ideality or equilibrium. Else the premise is flawed and must be abandoned. You can’t have you cake (endorse the ‘basic premise’) and eat it too (claim that the premise is consistent with the absence of a temperature gradient for an ideal gas under gravity at equilibrium).

      • Pierre-Normand

        DS: “Pierre, you don’t know enough to distinguish my position from the dragon slayers.”

        It’s not my goal to distinguish or compare you with them. You misread my post. I said to CD that he can’t dissociate himself from the slayers on the issue of the G-T effect when he (and yourself) endorses basically the same “basic argument” about constancy of mechanical energy as a function of height, and is seemingly unaware (as you are) of the flaw in this argument. If you don’t want to lose time with me, and are unwilling to respond to some criticisms or your arguments… don’t respond.

      • P-N, “Else the premise is flawed and must be abandoned.”

        Spoken like a true authoritarian :) Ideal Gas Theory is a valiant attempt to force nature to be a linear as humanly possible. T is related to E by 5.67e-8*T^4 +/- a touch. So as T linearly decreases E, exponential decreases. When you pick a range where the relationship between T and E can be assumed linear or more accurately more predictable, your model that you are assuming the universe obeys is more accurate.

        When you get down in the ballpark of 184K degrees, about 65 Wm-2 with CO2 in the mix, there are interesting coincidences. When you get to around 273K with water there are interesting coincidences. So assuming a fixed enthalpy and a predictable relationship between T and E gets more complicated.

        Now if you assume away those issues, you would predict the temperature change between the Earth’s Stratopause and the Turbopause to be one thing, but when you include those coincidences you get a lapse rate from 273K to 184K over approximately 50 kilometers. Which is about 1.78K/km

        You can pick another planet with another mixture of gases and find another lapse rate in some portion of its atmosphere that behaves “oddly”. In my experience with dealing with water vapor, assuming a “fixed” enthalpy is nutso cuckoo. So I am completely convinced that someone looking hard enough can find a special case.

    • Pierre-Normand

      Joe Born,

      Regarding a topic that interests both of us, I found the discussion in the following paper, regarding the point of making use of different partition functions such as the canonical, grand canonical or microcanonical ensembles (there are more; there even exists a grand microcanonical partition function!), for the study of systems under gravity to be illuminating.

      P. H. Chavanis, ‘Gravitational instability of isothermal and polytropic spheres’
      http://cds.cern.ch/record/566951/files/0207080.pdf

    • Dr Alex Hamilton

      This is easily shown to be wrong for the simple reason that if a vertical column of gas were isothermal in a gravitational field then there would obviously be unbalanced energy potentials because molecules at the top would have more gravitational potential energy whilst still having the same kinetic energy. And if you have unbalanced energy potentials then you do not have maximum entropy.

      • Pierre-Normand

        Dr Alex Hamilton,

        That’s a common misunderstanding. The following post by Pekka addresses your question:

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653688

        See also the following post:

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653483

      • Pierre-Normand

        Consider also that it just doesn’t make sense for a column of a weakly interacting ideal gas at equilibrium to have balanced potentials in the way you conceive. This would mean that the average internal energy of the molecules at the bottom of the gravity well determine a maximum height above which the average kinetic energy would be negative. Any level above this maximum height would be thus be forbidden. This would be the height H where the potential gravitational energy of a molecules m*g*H would be equal to the average internal energy of a molecule at the bottom of the column ((3/2)kT for a monatomic gas). But clearly, since this is an average, half the molecules at the bottom have enough kinetic energy to climb above that height (provided they are suitably oriented).

  56. Pierre-Normand

    I just found another relevant paper:

    Verkley and Gerkema, ‘On Maximum Entropy Profiles’, Journal of Atmospheric Science, 2004

    http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%282004%29061%3C0931%3AOMEP%3E2.0.CO%3B2

    From the abstract: “[…] what vertical temperature profile maximizes the total entropy of the column? Using an elementary variational calculation, it is shown how the result depends on what is kept fixed in the maximization process. 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”

  57. David Springer

    The second response contains the phrase “constant potential temperature”
    which I think is perfectly suited in that it’s a concise and apt summation of my opinion on the equilibrium state of a non-convecting atmosphere. I wish I coined the term myself. The wiki entry http://en.wikipedia.org/wiki/Potential_temperature should also be read and understood. The atmosphere isn’t a box and air isn’t an ideal gas. Insisting that these idealized states are reflective of and predictive of the real non-ideal world can lead to significant mistakes. Climate science in particular, not being amenable to experimental science to confirm postulates, is rife with such mistakes.

    http://www.phys-l.org/archives/2012/1_2012/msg00075.html

    Re: [Phys-l] Temperture profile in a graviational field

    From: John Mallinckrodt
    Date: Tue, 17 Jan 2012 07:21:19 -0800
    Historically you will see this referred to as the Loschmidt Gravito-Thermal
    effect. Loschmidt argued precisely as you have while Maxwell and Boltzmann argued as John Denker has and as I would as well. As far as I can tell there are respectable people who still consider this an unresolved controversy. You can find plenty of material on the web by searching undder some combination of loschmidt boltzmann maxwell gravity gas etc. One example:

    http://bit.ly/ywdmz3

    I think it’s at least pretty clear that the situation is not as simple as it might seem from your (and Loschmidt’s) arguments. The question is not whether
    individual molecules moving to lower altitudes (without collisions, of course)
    will have higher kinetic energy, the question is what does the *distribution*
    of energies look like at different altitudes. It’s not particularly hard to imagine that molecules moving to higher altitudes simply take their place as a lower velocity component of the *same* distribution that simply has a lower overall density.

    John Mallinckrodt
    Cal Poly Pomona

    On Jan 17, 2012, at 6:04 AM, Folkerts, Timothy J wrote:

    > Here is an interesting question that I have been seeing in the context
    > of climate science and the “Greenhouse effect”. (I may have more points
    > to make later, since this topic is interesting and important — for
    > science and for society.)
    >
    >
    > Suppose you have a perfectly insulated column of air. Let’s minimize
    > concerns about IR by making the inner walls of the container highly
    > reflect and making the gas N2 (which emits/absorbs minimal amounts of
    > IR). Suppose the column is a few km tall, with the base at the surface
    > of the earth.
    >
    > 1) What will be the temperature profile? Certainly there is a pressure
    > gradient and a density gradient in the column. I would say there is a
    > temperature gradient as well. On a microscopic scale, between every
    > collision, if the molecule gains altitude it will gain PE and lose KE
    > (ie it will cool). Any molecule moving downward will warm. On a
    > macroscopic level, this can discussed in terms of the “dry adiabatic
    > lapse rate”.
    > http://en.wikipedia.org/wiki/Lapse_rate#Dry_adiabatic_lapse_rate and
    > the “potential temperature”
    > http://en.wikipedia.org/wiki/Potential_temperature
    >
    > In either case, it is clear to me that the equilibrium condition (both
    > in this ideal column and in the real atmosphere) would be a temperature
    > gradient (cooling ~ 10K/km). Do others agree?
    >
    >
    > 2) If this is true, how can this best be squared with the 2nd law of
    > thermodynamics? If the top and bottom of the column were held at a the
    > same temperature, there would be a continuous flow of energy from top to
    > bottom, even though they are the same temperature. Even if the top were
    > slightly cooler than the bottom, there would be a downward flow. This
    > would violate a standard statement of the 2nd law, since we have
    > spontaneous heat from cool to warm.
    >
    > I’ve been trying to think of a good way to explain that this is not
    > indeed a violation. I suspect the best explanation will have to involve
    > the more fundamental statement of the 2nd law — that entropy tends to a
    > maximum. The adiabatic lapse rate leads to an isentropic gas and a
    > constant potential temperature.
    >
    >
    > _______________________________________________
    > Forum for Physics Educators
    > Phys-l@carnot.physics.buffalo.edu
    > https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l

    _______________________________________________
    Forum for Physics Educators
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  58. David Springer

    http://en.wikipedia.org/wiki/Potential_temperature

    “Since parcels with the same potential temperature can be exchanged without work or heating being required, lines of constant potential temperature are natural flow pathways.”

    Implied is that non-constant potential temperature in a planetary atmosphere requires expenditure of energy (work or heating) to maintain. In an isolated non-convecting atmosphere the equilibrium state is constant potential temperature i.e. a constant mechanical energy gradient such that no work can be accomplished across the gradient.

    The resistance of dogmatic (warmist) individuals is indicative of their lack of critical thinking skills and almost total reliance on concepts supported by the ad populum fallacy. Such thinking characterizes a pernicious philosophy of science that stunts progress.

    • Pierre-Normand

      “Implied is that non-constant potential temperature in a planetary atmosphere requires expenditure of energy (work or heating) to maintain.”

      Since the assumption is that the convective process is *adiabatic*, heat conduction (from molecular diffusion) is neglected. Viscous heat dissipation from the relative movement of adjacent convective cells also is neglected. This is why the isentropic temperature profile doesn’t represent a state of thermodynamic equilibrium for a real gas. An external energy source is required to maintain convection.

      “In an isolated non-convecting atmosphere the equilibrium state is constant potential temperature i.e. a constant mechanical energy gradient such that no work can be accomplished across the gradient.”

      On the contrary, the paper argues that the profile is isothermal in non-convecting atmospheres (at equilibrium) and tends to approach an isentropic profile when conviction is large enough such that effects from dissipation and diffusion can be neglected. Actual convection, or the absence thereof, is the criterion that separates the isothermal from the isentropic cases.

      In any case, the isentropic solution exhibits a barometric density profile together with an the adiabatic lapse rate, so it is quite different from the strange ‘iso-mechanical-energetic’ profile that you were defending. The profile that you are defending has a cut-off height where all the mechanical energy is transformed into potential gravitational, and molecules can’t rise any further, rather than the density dropping exponentially (with a temperature dependence of the exponential function).

      • David Springer

        Maximum entropy is attained when mechanical energy in a mole of atmosphere at any given elevation is equal to a mole of mechanical energy at any other elevation. No work can be extracted from that state. Potential and kinetic energy are freely interchangeable so nothing exists to prohibit an isolated non-convecting atmosphere from reaching this maximum entropy state. Your argument simply doesn’t hold water, Pierre. It is disputed by basic statistical mechanics.

      • Pierre-Normand

        “Maximum entropy is attained when mechanical energy in a mole of atmosphere at any given elevation is equal to a mole of mechanical energy at any other elevation. No work can be extracted from that state.”

        (Let me first pick a nit: when a gas loses internal energy through performing adiabatic expansion work on the surrounding, radiating, or conducting heat away, then some of the energy is lost from internal rotational and vibrational states of di- and polyatomic molecules, not just from (translational) kinetic energy. But let us ignore this and imagine that ‘air’ is constituted of monatomic molecules for simplicity, and all the internal energy is therefore kinetic. (Hence Uavg = KEavg = (3/2)kT). I also had overlooked this in a previous commentary.)

        If a mole of air at ground level on Earth has a temperature of 300°K, then, assuming thermodynamic equilibrium as you conceive it, then *all* its molecular kinetic energy is converted into gravitational potential energy at a height of 5.8km. This would entail (again, ignoring internal rotational states of diatomic molecules, etc.) that the kinetic temperature of one mole of air at that height would be 0°K. This is also the height potentially reached by a molecule that has the *average* kinetic energy at ground level and that is initially moving straight up. You view is that there aren’t any molecules that can rise any higher than 5.8km at equilibrium. For if some definite proportion did, then the total mechanical energy of one mole of air at this higher level would be larger than it is at ground level — since it can’t be any lower than the total gravitational potential energy at this level. Right?

  59. David Springer

    Pierre-Normand | December 6, 2014 at 8:34 am |

    ” I said to CD that he can’t dissociate himself from the slayers on the issue of the G-T effect when he (and yourself) endorses basically the same “basic argument” about constancy of mechanical energy as a function of height, and is seemingly unaware (as you are) of the flaw in this argument. If you don’t want to lose time with me, and are unwilling to respond to some criticisms or your arguments… don’t respond.”

    That is not an accurate description of the dragon-slayer argument. Unsurprisingly you aren’t familiar with that either. The dragon-slayer argument is that the bottom of the atmosphere is warmer than it would be in the absence of gravity. That is not my argument nor that of Loschmidt. You bring nothing to this dialog arguing against straw men. Your lack of ability to describe the arguments presented in opposition smacks of religious zealotry rather than informed argument. Please stop wasting our time.

  60. David Springer

    Pierre-Normand | December 6, 2014 at 8:34 am |
    DS: “Pierre, you don’t know enough to distinguish my position from the dragon slayers.”

    It’s not my goal to distinguish or compare you with them.

    —————————————————————————–

    It’s not your goal to understand the arguments in opposition to your own. That is certainly a revealing statement. You’re right and what anyone else thinks simply doesn’t matter and isn’t worth your time. LOL

    • Pierre-Normand

      Obviously, I wan’t to distinguish your *argument* from their’s, just no *you* personally who I want to compare. I thought I had made that clear in the post that first provoked your ire. I said that I try not to pigeon hole people who I choose to respond to. It just so happen that your argument about the alleged invariance of total molar mechanical energy density with height is flawed for the same reason CD’s ‘basic premise’ and the slayers make the same error.

      • P-N, Excuse me, my basic premise is that is that in a real gas the range of potential velocities would produce a range of inter-molecular interactions that cannot be properly considered in an “ideal gas” model. In other words, there are complex enthalpy considerations that would impact entropy estimations. I believe that is in the e^-(H-TS)/kT Gibbs free energy. That would make equilibrium, temperature dependent though it is actually energy dependent with T used as a sometimes limited proxy for energy, i.e. kinematic temperature versus a more robust thermodynamic temperature.

        In a *real* atmosphere the temperature and composition would matter. On the whole, “meh, it is insignificant in a real atmosphere.” is my *position* statement. Now what you *think* my position is could differ, but using circular logic isn’t going to firm up whatever your position is.

        I think Rob has already compared this to the Bose-Einstein fiasco. The object of physics is to describe nature not force nature to bend to your will.

      • David Springer

        Again with the straw men, Pierre. I said nothing about energy density with height. It’s not hard to quote me accurately unless doing so weakens your own argument the maybe it would be hard for someone who is convinced of their own infallibility. I said in an insolated non-convecting atmosphere the equilibrium state is that a mole of air at the bottom has the same mechanical energy as a mole of air at the top. Please stop arguing with straw men. It’s a waste of everyone’s time and mechanical energy.

      • Pierre-Normand

        David Springer: “I said in an insolated non-convecting atmosphere the equilibrium state is that a mole of air at the bottom has the same mechanical energy as a mole of air at the top.”

        That’s exactly what I understood you to have said. A mole has the same mechanical energy at any height. I paraphrased your position thus: “the alleged invariance of total *molar* mechanical energy density with height” (added emphasis). That means the same thing. When parcels of air have the same total mechanical energy per mole at any height, then the molar mechanical energy density is invariant with height. Those are two equivalent ways to say the same thing. And my arguments have been targeting exactly this invariance claims of yours, also seemingly supported by your beliefs that entropy maximization entails it, and that thermodynamic equilibrium entails an isentropic profile.

    • David Springer, December 6, 2014 at 10:44 am:

      “It’s not your [Pierre-Normand’s] goal to understand the arguments in opposition to your own. That is certainly a revealing statement. You’re right and what anyone else thinks simply doesn’t matter and isn’t worth your time. LOL”

      I very much agree with this observation. Very much to the point. Thanks :)

  61. It seems obvious to me that this idea has merit. Imagining the same planet with a non-radiative atmosphere that Willis imagined above, the self-evident truths would seem to me to be:
    1) without an atmosphere; radiation in would equal radiation out, and it would radiate in accordance with its temperature^4.
    2) add the atmosphere, and the gas in contact with the surface would attain that temperature through conduction.
    3) Hot air rises, gravity pulls it down again. New colder gas replaces the warmed gas; convection & lapse rate
    4) Until equilibrium was reached the energy in would be less than energy out. Radiation is still occuring from the surface
    5) Once the atmosphere had reached the temperature of the surface, then equilibrium is restored and enery in = energy out once again.
    6) We are left with a planet with a warmer atmosphere than before.

    If, it seems to me that people are suggesting that non-radiative gasses can’t heat up, can I suggest they do an experiment by filling a pressure cooker with a non radiative gas, closing off the pressure valve and putting it on the stove-top on full, and going out for the evening. Take your time. Have that second bottle. dessert. If they’d be happy doing that – I’d be convinced. :-D

  62. Oops.. 4) should obviously be:
    4) Until equilibrium was reached the energy OUT would be less than energy IN. Radiation is still occuring from the surface

  63. Willis Eschenbach

    David Springer | December 5, 2014 at 5:59 pm |

    Willis you seem to be under the mistaken impression that gravito-thermal effect makes the surface warmer than it would be without gravity. That is not true. The mechanical energy is 100% kinetic at the surface and the temperature the same as if there were no gravitation confinement.

    Understand the question before answering please.

    David, I’ve discussed this question with dozens of people. You’re the first one to make the claim that the “gravito-thermal effect” has nothing to do with thermal …

    If (in your view) the “gravito-thermal effect” doesn’t make the surface warmer than it would be without gravity, just what are you claiming it does do? And why is the word “thermal” in the name if (as you say) the gravity doesn’t affect the temperature?

    Thanks,

    w.

    • David Springer

      I am claiming that a lapse rate would exist in a non-convecting atmosphere where mechanical energy in a mole of air at any given altitude is the same as that in a mole at any other altitude. Potential and kinetic energy are freely interchangeable without any work or heat input required in the exchange. The atmospheric column, forced by the nature of gravity to have more potential energy per unit mass as altitude increases, relaxes into a state of maximum entropy by trading kinetic energy for potential as altitude increases.

      This results in measured temperature declining with altitude and the kinetic energy lost per unit mass is replaced in one-to-one correspondence with gravitational potential energy. The result is a zero mechanical energy gradient from which no work can be extracted which is what 2LoT demands i.e. maximum entropy.
      .

      • Willis Eschenbach

        David, I fear that’s not clear.

        If we take a thermally insulated vertical pipe, with no heat either entering or leaving it, are you saying it will have a temperature gradient from top to bottom?

        If so, you need to read Dr. Brown’s proof that such a situation violates the laws of thermodynamics.

        And if not … what are you saying?

        Thanks for the reply,

        w.

      • David – when I first thought about this, I came to the conclusion that there would be no gravitational effect on temperature. I was considering the case of an 11 mile high column of air, say, 1 square meter in area. The column would be perfectly insulated WRT to heat and radiation.

        Thinking about this some more, I think you may be right. On the molecular level, the temperature of an ideal gas depends on the speed of the molecules. Let’s say for simplicity that we are dealing with a single, monatomic, gaseous element.

        The reason I have changed my mind is this. If a given atom moves straight up, it will lose speed, effectively cooling. If it strikes a second atom, the momentum imparted will be smaller the higher the molecule moves upward. At some point, the atom of gas would stop completely due to gravitational attraction. If it hits another, stationary for the sake of argument, atom at its apogee, it would impart no momentum to the second atom.

        If the atom is moving downward, it will gain speed, i.e. it will carry more energy and since temperature depends on the speed of the molecules, it represents a hotter gas. It will impart more energy to a second atom the further it falls.

        Therefore, if we conceptually divided the column of gas into equal volumes, there would be a temperature gradient while at the same time maintaining thermodynamic equilibrium.

      • Although I do not agree with David Springer and jim2, I believe that Dr. Brown’s attempted refutation suffers from faulty logic; as I explained here: http://wattsupwiththat.com/2014/08/18/monday-mirthiness-spot-the-troll/#comment-1711550, it is based on what I called the “Brown-Eschenbach Law of Lapse-Rate Conservation,” a proposition for which they have advanced no proof.

        Basically, the problem is this. The ensemble of microstates available to the gas molecules inside the pipe is changed by the ability to communicate heat to and from the external wire, so there’s no reason to believe that a kinetic-energy gradient that prevailed in isolation would prevail when the wire is connected. Therefore, the assumption of such a gradient in isolation does not, as Dr. Brown contends, imply that net heat would flow indefinitely.

      • One simple example presented in various places considers highly rarefied gas over a warm surface. The purpose is not to prove anything, but to figure out, what happens in this simple case.

        It’s assumed that the gas is so rare that molecules almost never hit each other. They just bounce from the bottom at variable speed in all possible directions. It’s assumed that the speed that they have after bouncing is distributed with a Gaussian type distribution with the same coefficient multiplying each of the squared components of the velocity in the exponential. Each molecule starts to move up in a parabolic orbit. The horizontal velocity does not change during the flight, but the vertical velocity decreases due to gravity until the molecule reaches it’s maximum altitude and starts to fall.

        One might expect that the average vertical velocity decreases with altitude, because every single molecule has a smaller velocity higher up, but a calculation tells that this is not the case. The molecules that had originally small vertical velocity start to fall very soon, and those with the highest vertical velocity reach highest altitudes. At every altitude some molecules have nearly zero vertical velocity. The maximal possible value is infinite at all altitudes (although very high values are exceptional). Doing the full calculation tells that the velocity distributions have exactly the same shape at every altitude, there are only more particles of each velocity near the surface than high up. This result is naturally true only for the assumed initial velocity distribution.

        As I started, this does not prove much, but this should help in understanding, how it’s possible that the average kinetic energy of an ideal gas is independent of the altitude (not the sum of kinetic energy and potential energy, but the kinetic energy alone).

        The exponential distribution of squared velocity is that of Maxwell-Boltzmann distribution.

        My note that Pierre-Normand has mentioned a few times generalizes this argument to the case, where molecules collide and bounce, but I don’t try to prove that any other distribution would develop towards isothermal, only that a distribution that starts as isothermal Maxwell-Boltzmann distribution remains as such. It does not spontaneously develop any non-zero lapse rate.

      • Willis Eschenbach

        Joe Born | December 6, 2014 at 3:23 pm |

        Although I do not agree with David Springer and jim2, I believe that Dr. Brown’s attempted refutation suffers from faulty logic; as I explained here: http://wattsupwiththat.com/2014/08/18/monday-mirthiness-spot-the-troll/#comment-1711550, it is based on what I called the “Brown-Eschenbach Law of Lapse-Rate Conservation,” a proposition for which they have advanced no proof.

        Joe, when you feel you need to ridicule your adversary by making up a bogus “Law” that neither Dr. Brown nor I have advanced, you do your argument great damage.

        Basically, the problem is this. The ensemble of microstates available to the gas molecules inside the pipe is changed by the ability to communicate heat to and from the external wire, so there’s no reason to believe that a kinetic-energy gradient that prevailed in isolation would prevail when the wire is connected. Therefore, the assumption of such a gradient in isolation does not, as Dr. Brown contends, imply that net heat would flow indefinitely.

        I don’t even understand what this means, that the “ensemble of microstates available” is changed. What is the “ensemble of microstates” available to the gas molecules before and after connecting the wire?

        Finally, Dr. Brown’s proof offers two possibilities. Either heat will flow FOREVER through the silver wire, or it won’t. If it does, as you seem to be claiming, this is a perpetual motion machine … and if it won’t flow it means that the top and bottom of the air in the pipe is at the same temperature.

        Your choice … although from your words, it seems you imply a third state but I’m in mystery as to what that state might be.

        w.

      • Willis Eschenbach

        jim2 | December 6, 2014 at 3:01 pm

        Therefore, if we conceptually divided the column of gas into equal volumes, there would be a temperature gradient while at the same time maintaining thermodynamic equilibrium.

        If that’s the case, then Dr. Brown’s proof applies. Connect an insulated silver wire from the bottom to the top of the column. Since you say there is a temperature difference, heat will flow from the bottom to the top of the column.

        But then, according to you, gravity will re-stratify the column and re-establish the thermal gradient … which means that the heat would flow through the wire forever.

        Obviously, however, this is a perpetual motion machine. If gravity could do that, we could run a heat engine off of the temperature difference forever … sorry, but that’s not possible.

        w.

      • Willis. Heat “flows” in a solid by the atoms, of silver in this case, jostling each other. In a body of silver at a high temperature, the atoms vibrate with a higher amplitude than atoms in a body of silver at a lower temperature.

        Analogous to the gas, the atoms in the solid have to vibrate, as gas atoms move in space, in order to conduct heat.

        In the case of your silver wire, when one considers motion of a silver atom vibrating in along the “z” axis (up and down), when the atom moves up, it will be retarded by gravity. When it oscillates back down, it’s downward velocity will be enhanced. Thus, the effect hypothesized by me for the gas applies also to the silver wire.

      • Willis Eschenbach: “Dr. Brown’s proof offers two possibilities. Either heat will flow FOREVER through the silver wire, or it won’t. If it does, as you seem to be claiming, this is a perpetual motion machine … and if it won’t flow it means that the top and bottom of the air in the pipe is at the same temperature.”

        It won’t.

        Before the silver wire is connected, the gas is at equilibrium: no net heat flows, but there’s a non-zero kinetic-energy gradient. That gradient results from the statistics of the gas’s microstates–i.e., combinations of molecule position and momentum–that are available to the isolated gas. Those statistics follow from the constraint that total molecular energy is fixed.

        Once the silver wire is connected, the statistics change; although in steady state no net heat flows on average between the gas and the wire, random exchanges with the wire cause some minute fluctuations in gas’s the total molecular energy, fluctuations that could not occur when the gas was isolated. As a consequence, the gas’s equilibrium kinetic-energy gradient changes. This change in kinetic-energy gradient requires no perpetual net flow through the wire.

        Now, in a sense we’re talking past each other; in accordance with one definition of temperature, both of those (different) equilibrium kinetic-energy gradients are considered temperature gradients of zero. In accordance with another, they are considered (different) non-zero temperature gradients. Failure to keep track of which definition we used causes confusion in these discussions. By taking the starting point of his proof an equilibrium state with a non-zero lapse rate, however, Dr. Brown implicitly chose the latter definition.

        And, no, I don’t know precisely what situation would prevail when the wire is connected. What I think, though, is that there would be an immeasurably small kinetic-energy gradient in the wire despite its being at equilibrium, i.e., despite its conducting no net heat flow. I recognize that this is at odds with Dr. Brown’s understanding of Fourier’s Law. But such is life.

      • Pierre-Normand

        “Before the silver wire is connected, the gas is at equilibrium: no net heat flows, but there’s a non-zero kinetic-energy gradient. That gradient results from the statistics of the gas’s microstates–i.e., combinations of molecule position and momentum–that are available to the isolated gas.”

        Joe Born, I may be mistaken but I would have thought Willis’s argument wasn’t directed at the effect discussed in the Roman, White, and Velasco papers, but rather at the possibility of a true measurable gravito-thermal gradients that purportedly develop in gas colums at equilibrium even under the assumption that the partition function is canonical or grand canonical. His argument may not apply for the case of the tiny height dependence of the speed distribution that follow from the mechanical system having rigorously constant energy, but that doesn’t seem to be the target.

      • David Springer

        Willis, no heat will be conducted along the wire due to gravity effecting it the same as the gas.

        Gravity is weird. In this case it’s not all that strange. It’s not like we’re talking about time dilation for Pete’s sake. This case is perfectly understandable with Newtonian mechanics. In a gravity field a vibrating atom hits an atom above it with less force than it hits a molecule below it. Thus the conduction of heat is altered with more resistance upward than downward. This results in a kinetic energy gradient. In order to balance the books, satisfying the first and second laws of thermodynamics, the imbalance in kinetic energy is balanced by an equal and opposite gradient in kinetic energy.

        You can’t construct a perpetuum mobile from this effect because the device is subject to the force of gravity which defeats perpetual motion every time.

      • David Springer

        Willis, no heat will be conducted along the wire due to gravity effecting it the same as the gas.

        Gravity is weird. In this case it’s not all that strange. It’s not like we’re talking about time dilation for Pete’s sake. This case is perfectly understandable with Newtonian mechanics. In a gravity field a vibrating atom hits an atom above it with less force than it hits a molecule below it. Thus the conduction of heat is altered with more resistance upward than downward. This results in a kinetic energy gradient. In order to balance the books, satisfying the first and second laws of thermodynamics, the imbalance in kinetic energy is balanced by an equal and opposite gradient in kinetic potential energy.

        You can’t construct a perpetuum mobile from this effect because the device is subject to the force of gravity which defeats perpetual motion every time.

      • David Springer

        jim2

        +1

      • Pierre-Normand

        David Springer wrote: “Willis, no heat will be conducted along the wire due to gravity effecting it the same as the gas.”

        Whether or not the gravito-thermal effect will produce the same temperature gradient on everything (or the same ‘effect’ on electricity — whatever that might be) depends how one accounts for the alleged effect. Roderich W. Graeff believes the gravito-thermal effect to produce different gradients on different gases (and he also believes to have verified this experimetally, by accident). And the way Graeff explains the source of gravito-gradients is precisely the same as your own explanation. He believes the total mechanical energy of some molar amount of gas to be invariant as a function of height. That is, he also believes kinetic energy to be converted into potential energy in such a manner that mechanical energy (PE+KE) is invariant with height in a gas column at equilibrium.

        However, Graeff is aware that such an effect, thus explained, would entail that the temperature gradient of a gas column at equilibrium must be dependent on the ratio of kinetic to internal energy of the molecules. That’s because although rising molecules lose some of the vertical component of their kinetic energy as they rise vertically, their internal vibrational and rotational states are unaffected by gravity (thought the molecular energies thereafter are redistributed though collisions among all their degrees of freedom, including the three translational ones). N2 molecules, for instance, at the same ‘ground temperature’ (T at height h = 0) as argon atoms would lose a larger share of their kinetic energy than would N2 molecules as they rise to the same level because at temperature T, the internal energy of the argon molecule averages (3/2)kT, while the internal energy of the N2 molecule averages (5/2)kT, due to its two additional rotational degrees of freedom (for usual temperature ranges).

        Therefore, if you hold that the gravito-thermal gradient would produce the same temperature profile for any gas (let alone have the ‘same effect’ on any imaginable mechanism of vertical energy transmission) then you have to abandon your idea that molar mechanical energy of gas molecules is the same at all heights at equilibrium.

      • Pierre-Norman: “I may be mistaken but I would have thought Willis’s argument wasn’t directed at the effect discussed in the Roman, White, and Velasco papers, but rather at the possibility of a true measurable gravito-thermal gradients that purportedly develop in gas colums at equilibrium even under the assumption that the partition function is canonical or grand canonical.”

        It appears that my response to this comment got lost in the ether.

        You are undoubtedly right that the purpose of Mr. Eschenbach’s and Dr. Brown’s proofs was to refute the Jelbring hypothesis, which, as I recall, contended that the equilibrium kinetic-energy gradient would be on the order of the adiabatic lapse rate. However, nothing in the proofs was limited to that.

        A little history. Mr. Eschenbach gave a prove in which he contended that assumption of an equilibrium lapse rate would imply that a perpetual-motion machine could be built: a heat engine could be connected between different altitudes of the gas column. That proposition has, as we have observed, a certain ambiguity arising the fact that different definitions of temperature are not precisely equivalent. But my initial objection was the following. The proof starts with equilibrium: no net heat flow is occurring. If the assumption is that the difference in the quantity he calls temperature does not cause heat to flow in the gas, why would it drive a heat engine? Only because he assumes it would.

        People attempt to justify his proof, but they can do it only by straddling different definitions of temperature. The first, which his proof’s initial assumption is based on, does not necessarily require heat flow if there’s a temperature gradient. The second, in accordance with which there’s no temperature gradient if there’s no heat flow, applies to the heat engine.

        I have discussed this stuff at length with Dr. Brown, and he has betrayed no comprehension of the distinctions that you and I have discussed. At one point, for example, he implied that Velasco et al.’s conclusion was wrong: http://wattsupwiththat.com/2014/02/06/satellites-show-no-global-warming-for-17-years-5-months/#comment-1567126. If he were restricting the validity of his proof to Jelbring, he need only have pointed out that the kinetic-energy gradient Velasco et al. find is orders of magnitude lower than what Jelbring claims.

      • Pierre-Normand:

        Just in case this comment gets through, I’ll mention that I’ve attempted a couple of times to respond to your exegesis of Mr. Eschenbach’s position, but my somehow failed to post.

      • Willis Eschenbach: “Joe, when you feel you need to ridicule your adversary by making up a bogus “Law” that neither Dr. Brown nor I have advanced, you do your argument great damage.”

        Well, I must admit that misinterpretation is always a hazard when one tries to characterize Dr. Brown’s work. So I’ll quote Dr. Brown’s words from which I inferred use of the B-E Law:

        “Two different columns of gas with different lapse rates. Place them in good thermal contact at the bottom, so that the bottoms remain at the same temperature. They must therefore be at different temperatures at the top. Run a heat engine between the two reservoirs at the top and it will run forever, because as fast as heat is transferred from one column to another, (warming the top) it warms the bottom of that column by an identical amount, causing heat to be transferred at the bottom to both cool the column back to its original temperature profile and re-warm the bottom of the other column. The heat simply circulates indefinitely, doing work as it does, until the gas in both columns approaches absolute zero in temperature, converting all of their mutual heat content into work.”

      • Joe Born – if the two gasses have different lapse rates, one column will be higher than the other, but the temperature profile will be the same.

      • I withdraw that last remark.

      • Here is an abbreviated version of the attempted reply, previously rejected for some reason, to Pierre-Norman’s speculation about the target of Mr. Eschenbach’s and Dr. Brown’s proofs.

        Pierre-Normand is correct that the target was not the minuscule kinetic-energy gradient that Velasco et al. found but rather the large gradient for which Jelbring argued. But the logic of their proofs did not distinguish between those gradients. Dr. Brown, in fact, opined that Velasco et al. were almost certainly wrong. So the gradient’s small size is irrelevant in this context.

        I would explain that in more detail, but for some reason this site has repeatedly rejected my attempts to post fuller explanations.

  64. FOMD presented a theorem

    Theorem Given any state with non-uniform thermodynamic potentials, there is a state having (1) the same globally conserved quantities, and (2) spatially uniform thermodynamical potentials, and (3) higher entropy.

    This is, indeed, a theorem that can be proven in Classical Thermodynamics, because Classical Thermodynamics is an abstract mathematical theory.

    We have also learned that proving is not part of science. The scientific question that cannot be proven is that Classical Thermodynamics is a good description of real physical systems. This fact can be, and has been, observed empirically. Classical Thermodynamics describes many physical phenomena extremely well, but not perfectly. The description is so good in case of typical gases, that we have really convincing reasons to believe that also the result that thermodynamic equilibrium is isothermal also for gases under gravity.

    Classical Thermodynamics can also be derived from mathematical statistics and rather weak assumptions concerning microphysics. That’s possible starting both from Classical Mechanics and Quantum Mechanics.

    Another issue to consider is the definition of temperature. Is it certain that the same variable describes temperature as we are used to observe it and has an unique relationship with one of those thermodynamic potentials FOMD mentions? Deciding on that requires that we study carefully, what temperature as we are used to observe it means for ideal and real gases in gravitational field. Again we can observe that the hypothesis that the variable is really the same leads to predictions that agree very well with the reality. When we consider interacting gases, kinetic theory must be used as discussed by Gibbs. Boltzmann’s original formulation is fully correct only for non-interacting gases.

    • Steven Mosher

      Thanks pekka

    • We have also learned that proving is not part of science.

      Of course it is.

      The scientific question that cannot be proven is that Classical Thermodynamics is a good description of real physical systems.

      But it can be disproven. That can be done by “proving” some theorem in the “classical” system (whatever it is), then demonstrating that it’s inconsistent with reality.

      When we consider interacting gases, kinetic theory must be used as discussed by Gibbs.

      So far as is currently known. It’s always possible, and the subject of this thread raises that possibility, that the “Second Law of Thermodynamics” is no more true than Newtonian Physics. As usually stated, it may well just be a simplification that works for conditions that are normally observed, but must be modified enhanced for certain circumstances that, so far, have not been experimentally demonstrated.

      Just like Newtonian Physics.

  65. ‘Theorem Given any state with non-uniform thermodynamic potentials, there is a state having (1) the same globally conserved quantities, and (2) spatially uniform thermodynamical potentials, and (3) higher entropy.’

    Recognisable instantly as a pompous restatement of the laws of thermodynamics. What all of this neglects is real effects on molecules in a gravity field. There is a difference in particle velocities under gravity than not. In the normal course of events – in what is normally measured as temperature – this would tend to create a small temperature effect. Small because the gravity effects on velocity are small relative to typical molecular velocities at ambient temps. But the question is – does this affect the vertical distribution of velocities?

    There are real differences in molecular dynamics under gravity than not – and these differences cannot be understand in terms of statistics. It is possible to move from the particle to the average state – but not the reverse. It leads me to discount all arguments based on assuming the same statistics apply to both states. Something technically known as begging the question.

  66. In a thermally isolated non-radiative gas with gravity, the conserved quantity is potential temperature, a function of temperature and pressure, not temperature or internal energy itself. An isothermal atmosphere has a gradient of potential temperature. It can be shown that mixing this to a uniform potential temperature (a dry adiabatic lapse rate) gives a higher entropy than an isothermal state. Furthermore once you have a uniform potential temperature, there is nothing you can do without adding energy to further increase the entropy. The highest entropy state for an isolated gas in gravity is therefore one with a constant potential temperature. Only when you take away gravity (or pressure gradients) does this reduces to an isothermal state. The key is that temperature is not a conserved quantity when there is a pressure gradient. Any vertical mixing would change that temperature. Related to the isothermal state not being sustainable in a bounded gas is that there would also be a downward heat flux in such a state, whether by molecular diffusivity or by any turbulent eddies present.

    • JIm D,
      Neither temperature nor potential temperature are conserved quantities in the language of physics. In nonrelativistic physics energy and mass are conserved, and so is electrical charge, but temperature and pressure are not.

      Potential temperature is a useful variable in the discussion of convective atmosphere that may be stationary, but not in thermodynamic equilibrium. A convective atmosphere can be stationary only, when an external source heats continuously the surface and when heat is lost to the space from high altitudes. Such an atmosphere is very far from thermodynamic equilibrium.

      • Potential temperature is a conserved variable in the language of meteorology. Its only source is diabatic heating, and its equation is derived from the laws of thermodynamics.

    • David Springer

      Jim D

      +1

      • Pierre-Normand

        Jim2, as I stated elsewhere, whether or not two different gas columns at equilibrium will develop two different temperature gradients depend on one’s justification for assuming that there ought to be a gradient at all. Many proponents of a gravito-thermal effect (e.g Frønsdal or Graeff) believe the effect to be dependent on the heat specificity of the gas, just as an adiabatic lapse rate is so dependent, and for basically the same reason.

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653300

      • P-N. To gradient or not to gradient shouldn’t be assumed at all. It should be a conclusion arrived at due to considerations of physics.

      • Pierre-Normand

        That’s what I am saying, jim2. I am saying that it is precisely the “considerations of physcis” that lead some people to conclude that there ought to be a temperature gradient that also lead them (or ought also to lead them) to conclude that the gradient will *not* be the same for gases that don’t have the same specific heat capacity, which relates to the kinetic/internal energy ratio of the gas molecules in the relevant temperature and pressure ranges.

      • There will be a gradient, unless you believe gravity doesn’t act on atoms and molecules. As to the gradient for different gasses, obviously the devil is in the details. But you can’t refute the law of gravity and it’s effect on temperature.

    • Pierre-Normand

      Jim D, what you are saying about the maximum entropy state is contradicted by Verkley and Gerkema who found that the isothermal profile maximizes entropy in a gas column under gravity. Have a look at the abstract and conclusion.

      http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%282004%29061%3C0931%3AOMEP%3E2.0.CO%3B2

      If there are eddies, then for sure the profile isn’t going to remain isothermal. However, when there are no net external energy sources, or energy fluxes across the system, then eddies will tend to dampen out in a real gas that has some non-zero dynamic viscosity as soon as the temperature gradient drops below the adiabatic lapse rate, since in that case the gas is convectively stable and remains it so until — as a result of diffusion — the profile becomes isothermal.

      Why do you say that when the profile is isothermal under gravity molecular diffusivity would produce a downward heat flux? Pekka has shown that such a profile (together with the barometric density profile) is stationary for an ideal gas:

      http://pirila.fi/energy/kuvat/barometric_derivation.pdf

      • Pierre-Normand

        Also of relevance is the recent discussion at Nick Stokes’s Moyhu blog:

        http://moyhu.blogspot.ca/2014/10/calculating-environmental-lapse-rate.html

        The bottom line is that when the temperature gradient is larger than the adiabatic lapse rate, then the atmosphere tends to behaves spontaneously as a thermal engine, and when it is smaller, then it can only behave as a heat pump that requires an external energy source in order for the temperature gradient to increase.

      • The diffusivity acts on the conserved quantity, potential temperature, so the only state that does not have a net heat flux due to diffusivity is an isentropic one. Given enough time, even with no turbulence, diffusivity brings the gas to an isentropic state. It may take a long time, but you can’t stop molecular diffusivity. Yes, an isothermal state is quite stable, but ultimately entropy wins, and the maximum entropy state is demonstrably one of constant potential temperature, a dry adiabatic lapse rate. This does not need convection to achieve, but that can accelerate the process. The paper you pointed to was interesting, but I think they ignored molecular diffusivity in their first example, and that is a critical process.

      • Pierre-Normand

        The Coombes and Laue paper, and Pekka’s note, both prove that the isothermal profile for an ideal gas under gravity is stationary. This means that it is retained even as though the molecules freely diffuse around and collide randomly. This result also can be gotten from kinetic theory. It is proved in Reif, ‘Fundamentals of Statistical and Thermal Physics’, p.210. The paper “On the Barometric Formula” derives the equilibrium state for the Boltzmann distribution law for energy in section D. See my short list of references near the top of this thread. Also look up for Roman, White, and Velasco for a derivation of the isothermal profile from the speed density function in the microcanonical ensemble frameworks, (equivalent to the derivation from the canonical ensemble at the thermodynamic limit, as they note).

        https://tallbloke.files.wordpress.com/2012/01/s-velasco.pdf

        I also have a couple more statistical mechanics texbook references if you wish. I haven’t yet seen, though, any suggestion in the published literature that the isentropic profile maximizes entropy, except maybe from Cristian Frønsdal, but what he may have found possibly just is a stationary condition for an ideal gas where convection occurs and isn’t dissipated (because there is no viscosity). I’ve even found recently a demonstration fof the isothermal speed distribution within star clusters at equilibrium (during stages of galaxy formations where the cluster is bathed in a ‘gas’ of stars such that ‘evaporation’ from stars that reach escape velocity is compensated by new captures.)

        You really ought to start either with Pekka’s note, or the very short paper from Coombes and Laue for the most intuitive explanations.

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-651619

        I’d really like to see any reference that you may have for the demonstration that the isentropic profile — the adiabatic lapse rate — maximizes entropy. That just doesn’t square with what I have seen so far.

      • Jim,
        These are two completely different issues, and you seem to be confusing them in a way that’s not correct in any language.

        The thermodynamic equilibrium has been discussed throughout this thread. It’s the configuration of maximum entropy, it’s isothermal, and it’s approached extremely slowly in a totally isolated gas under gravity (in the thermodynamic limit of very many particles, to satisfy Joe Born), isolated meaning that no heat is fed in at any point and no heat is removed at any other point. Randomly fluctuating heat fluxes that average to zero at every boundary point make the system canonical, while excluding even those corresponds to microcanonical ensemble.

        The other issue is a limiting case of convective atmosphere, where heat enters continuously at low altitude (bottom) and is removed at high altitude (top). This is of interest in meteorology, but this is not a state of maximum entropy. This is not approached over any time frame, but reducing the heat fluxes brings the stationary situation closer to this as long as convection remains strong enough to fully dominate over conduction. Because conduction is so weak, it’s possible to get very close to the limiting case, when the heat flux is very weak, but without any continuous heating of the bottom the situation is not stable, but goes very slowly towards the isothermal one. If the system is stationary, it’s total entropy is constant, but the system is not closed. External heat fluxes bring in negentropy (or free energy), which is dissipated adding some entropy internally in maintaining the circulation strong enough to dominate over conduction, which adds entropy as well. The sum of externally introduced negentropy and internally generated entropy is zero.

      • Pierre-Normand

        Thanks Pekka. This not only diagnoses the likely source of the confusion, but it also clarifies for me the significance of free energy for a convective atmosphere heated from below.

        Wikipedia has a useful section “Correlation between statistical negentropy and Gibbs’ free energy” in the ‘Negentropy’ article.

      • That’s written clearly and concisely enough for me to approach understanding, Pekka. Thanks.
        =========================

      • David Springer

        We all seem to agree that in any real planetary atmosphere, none of which are non-convecting, a constant potential temperature is the equilibrium state i.e. energy in equals energy out. In the earth’s troposphere application of the theory yields the observed dry adiabatic lapse rate.

        Moreover, the preponderance of literature suggests that convection, which is caused by the atmosphere being heated from the bottom, need only be faster than conduction for the system to seek the constant potential temperature state instead of the constant absolute temperature state.

        Therefore, since the earth’s atmosphere is heated from the bottom regardless of whether there are greenhouse gases, we can discard the notion that greenhouse gases are required to establish a lapse rate.

        In point of fact the predominant greenhouse gas, water vapor, reduces the dry adiabatic lapse rate from some 10C per kilometer to the standardized environmental lapse rate of 6.5C per kilometer.

        Thus ends the practical portion of the debate. The greenhouse effect is not needed to produce a lapse rate in an atmosphere heated from the bottom and free to convect and in fact in the earth’s troposphere the predominant greenhouse gas drastically reduces the lapse rate instead of causing it.

        Now to the gravito-thermal effect which is predicated on the assumption that kinetic energy (heat) transfer (a.k.a. conduction) is asymmetrical in the Z axis of a gravity field. Conduction is accomplished by vibrating molecules colliding with other vibrating molecules where a faster molecule imparts some of its excess motion to a slower molecule. In the absence of gravity the collisions are symmetrical in each direction as no force acts to inhibit motion in any particular direction resulting in an isothermal state. However, in a gravity field collision strength is lessened when the faster molecule collides with a molecule above it and is increased with a collision below it. Thus the asymmetrical collision strength should result in an asymmetrical distribution of heat energy accompanied by an equal and opposite gradient in gravitational potential energy.

        One caveat worth mentioning is we have no theory of quantum gravity and the distances involved in conductive heat transfer are quantum in scale. It’s difficult for me to imagine that gravity doesn’t work over quantum distances the same as it does in classical scales but both gravity and quantum mechanics are weird enough that I won’t discount some non-intuitive effect which would somehow cancel the Newtonian asymmetry in collision strength in the vertical direction of a gravity well.

        Unlike the lapse rate caused by bottom heating of planetary atmospheres, even in the absence of greenhouse gases, the gravito-thermal effect should manifest itself in incompressible, non-convecting solids as well as imcompressible liquids which are free to convect.

      • What Dr. Brown failed to show is that two different gasses in isolated columns would have different lapse rates. These gasses can’t interact other than at the bottom and at the top where they interact via flow of heat.

        Taking helium and hydrogen as examples, why would their lapse rates be different?

      • I think your bouncing balls have helped convect some incompressible knowledge in through the listening holes.
        ===============

      • OK, I think I can refute Dr. Brown’s idea that two different gasses in adjacent columns with heat conduction at bottom and top would create a perpetual motion machine.

        Here is the scenario. Two adjacent columns, 11 miles high, one meter square that share a common silver plate at the bottom. There is no gravity. The initial temperature of the plate is T0. Equal volumes of hydrogen and nitrogen are introduced into the two columns, hydrogen in one and nitrogen in the other. The system is allow to come to thermodynamic equilibrium. The plate and the gasses are now at temperature T1.

        Gravity is turned on. Now, we know that the acceleration due to gravity is independent of mass. We also know that the temperature of a gas is dependent on the average speed of the gas molecules. Therefore, the molecules of each gas will be accelerated in the same manner, to the same speed. So, initially, at the bottom of the column, the nitrogen will be hotter. This is because temperature is proportional to both speed and the mass of the molecule.

        The silver plate will conduct heat until the bottom of both columns are at the same temperature. At this time, the nitrogen molecules will be moving more slowly than the hydrogen ones.

        Then, the same argument applies. Upward moving molecules will be losing momentum to the gravitational field and will effectively cool as they rise. Falling molecules will gain speed and are effectively hotter.

      • After equilibrium in the above scenario, if the nitrogen and hydrogen molecules fell equal distances, the nitrogen would be hotter, but it doesn’t happen that way. The mean free path for hydrogen is about twice that of nitrogen, so the molecules attain or lose the same amount of temperature due to gravitational acceleration over the time to collision.

      • Pierre-Normand

        Jim2, both of your gases would produce the same gravito-thermal gradient, according to some proponents of the gravito-thermal effect, because they both are diatomic molecules and have the same kinetic/internal energy ratio. If you rather use N2, and argon, say, then some people will argue that the argon gas column will develop a steeper temperature gradient at equilibrium because while those monatomic molecules rise they lose kinetic energy to the gravitational potential and this loss represents a larger share of their total internal energy. N2 molecules, unlike argon atoms, have rotational kinetic energy (which stores 2/5th of their total internal enregy), and this energy isn’t directly affected by the rise in the gravitational field, so, through inter-molecular collisions, this energy replenishes some of the vertical (translational) kinetic energy lost from the rise.

      • Pekka Pirilä: “Randomly fluctuating heat fluxes that average to zero at every boundary point make the system canonical.”

        Pardon the following nit-picking, but I think it’s relevant to a previous discussion of ours.

        I believe it is true that fluxes could average to zero and still impose a hard limit on the speed of any particle in the gas column. The thermally non-isolated gas column could, for instance, be immersed in a finite (and otherwise thermally isolated) heat bath so that the column and the bath together would have a finite amount of heat. I don’t think the allowed microstates of the column would in that case constitute a canonical ensemble even though, their total energy not being fixed, they wouldn’t constitute a microcanonical ensemble, either.

        If we assume that the universe is finite, therefore, the mere fact that a gas column is not perfectly isolated does not eliminate the possibility of its exhibiting a non-zero equilibrium molecular-kinetic-energy gradient.

        So as not to confuse a casual lurker, however, I hasten to add that this is all theoretical, how-many-angels-can-dance-on-the-head-of-a-pin stuff. The gradient we’re talking about is in any macroscopic case is too small to be measured, so this non-zero gradient should give no comfort to those who contend that the equilibrium lapse rate would equal the dry adiabatic lapse rate.

  67. I’m surprised that this discussion continues.

    In the spirit of explaining why the gravito-thermal effect is impossible, without including mathematics, here is an attempt.

    My assumptions are –

    1. There is a column of gas of cylindrical or other regular cross section containing sufficient energy to maintain the gas in a gaseous state.

    2. The volume is enclosed by boundaries which isolate the system, allowing no energy into or out of the system.

    3. Under these conditions, the gas will occupy the enclosed volume fully, and will be everywhere the same temperature.

    4. A gravitational field is introduced such that the axis of the cylinder is normal to the field – or end on, if you prefer.

    5. Gravity acts on the gas, causing compression at the bottom of the cylinder, and heat results therefrom.

    Unfortunately, assumption 2 prevents assumption 5. Energy is required to perform the work that compresses the gas, and the system allows no energy to cross its boundary. The laws of thermodynamics implicitly state that energy can neither be created or destroyed. Assumption 5 explicitly states that the force of gravity causes heating, therefore energy has crossed the boundary, in contravention of the requirement that this is not allowed.

    If some smarty pants changes my assumption to allow the force of gravity to transfer energy across the system boundary, you must allow the compressed and warmed gas to transfer energy by radiation across the boundary as well. Its temperature will drop, until the enclosed volume will once again be everywhere the same temperature, albeit with a pressure and density gradient due to the effects of gravity.

    Anybody that wishes to introduce the concept of potential energy emerging due to the introduction of a previously non existent gravity field has created additional energy within an isolated system, which is once again forbidden by the laws of thermodynamics.

    The only way that the gravito-thermal effect can occur, is to adopt one way Warmist physics, which allows gravity to do work across a barrier which does not allow it, but does not allow energy to cross the barrier in the other direction. This allows the creation of a perpetual motion machine of the second kind, precisely as described by Graeff and others.

    Obviously, given my well known and characteristic humility, I welcome constructive criticism, particularly if I have failed to express myself clearly. If you wish to find fault, please quote exactly the words with which you disagree. I may have inadvertently included or excluded something that I should not have. I await the barrage.

    Live well and prosper,

    Mike Flynn.

    • MF – the bottom of the gas warms, but the top of your cylinder would cool – no net energy change.

      Next.

    • Correction, there would be some net energy increase via the gravitational field – unless the container blocked gravity. That’s how the energy gets introduced.

      • Jim2,

        Thanks. If energy is allowed into the system, it must also be allowed out. I have stated what happens in this case. Do you not agree with me?

        Live well and prosper,

        Mike Flynn.

      • Mike – you have posited that energy can’t leave the system. Once you turn on the gravitational field, you have added energy. Once it’s on, it no longer changes, so you once again have established an equilibrium.

        When astronauts or pilots are trained on high G conditions, a centrifuge is used to create a gravity-like acceleration. In order to achieve this feat, the centrifuge is spun by an electric motor. You see, energy is needed to create the acceleration.

        In your case, you are hypothesizing a gravitational field that can be turned on or off. No known way to do that, but once you turn it on, mass begins to move, work is done. Therefore, energy has been added to the system, in violation of your own assumption.

        If you want to exclude all sorts of energy, you must also exclude the gravitational field. But then, your model fails to recreate a column of air in a gravitational field.

      • Jim2,

        What part of what I wrote are you in disagreement with?

        Could I please ask you to quote it?

        For example, you wrote –

        “If you want to exclude all sorts of energy, you must also exclude the gravitational field. But then, your model fails to recreate a column of air in a gravitational field.”

        I was merely attempting to show that there can be no gravito-thermal effect.

        Maybe you overlooked the sentence where I wrote –

        “Its temperature will drop, until the enclosed volume will once again be everywhere the same temperature, albeit with a pressure and density gradient due to the effects of gravity.”

        Has my expression been defective, in that I have not managed to express the effects of gravity on a column of gas assuming known physics? If so, I apologise. Maybe you can suggest an alternative way of expressing my sentence, that will make it clearer to yourself and others.

        Thank you for your comment. Anything else?

        Live well and prosper,

        Mike Flynn.

    • Dear Professor Mike

      So if we don’t allow gravity to influence the mass of the air column – there is no gravity effect at all?

      So perfectly true. Congratulations again on your perspicacity.

      Life is too short for bad coffee
      Robert I Ellison
      Chief Hydrologist

      • Rob Ellison,

        You wrote –

        “So if we don’t allow gravity to influence the mass of the air column – there is no gravity effect at all?”

        Thank you. I obviously assumed that people would realise that an absent force would have no effect, not being there. Corrected here. You are correct, in the absence of gravity, there is no gravitational effect.

        Live well and prosper,

        Mike Flynn.

      • Dear Professor Mike,

        Gravity heats the air at the bottom of the column – but then if you let gravity in – you have to let energy out and it cools again.

        Absolutely brilliant – well done old bean.

        Just another tequila sunrise
        Robert I Ellison
        Chief Hyrologist

      • Rob Ellison,

        You have the better of me, I fear.

        You are allowing gravity to act on a previously isolated system, and add energy, it appears. You then repeat the application of gravity, but it seems that this time it operates differently. Have I got this correct?

        Possibly if you quote the parts of my proposition with which you disagree, and then point out where I have erred, I may understand what it is you are disagreeing with.

        You wrote –

        “Gravity heats the air at the bottom of the column – but then if you let gravity in – you have to let energy out and it cools again.” This appears to be logically unrelated to anything I wrote. If you follow what I wrote, you will see you have created a logical inconsistency, either intentionally or otherwise.

        I obviously have to leave you to your belief in non existent forces and effects.

        Live well and prosper,

        Mike Flynn.

      • Dear Professor Mike

        Gravity acts on the gas, causing compression at the bottom of the cylinder, and heat results therefrom…

        If some smarty pants changes my assumption to allow the force of gravity to transfer energy across the system boundary, you must allow the compressed and warmed gas to transfer energy by radiation across the boundary as well.

        “Gravity heats the air at the bottom of the column – but then if you let gravity in – you have to let energy out and it cools again.”

        I was quite sure that I got the concept right – such a brilliant insight I thought. My humble apologies if I have missed some critical subtlety.

        Everybody wants to go to heaven – but nobody wants to die.
        Robert I Ellison
        Chief Hydrologist

      • Rob Ellison,

        Thank you.

        I will complete your selective quotation for you, as your attention span was obviously too short to allow you to complete the task –

        ” 5. Gravity acts on the gas, causing compression at the bottom of the cylinder, and heat results therefrom.

        Unfortunately, assumption 2 prevents assumption 5.”

        I think the form is called reduction ad adsurdum, but I could be wrong.

        Live well and prosper,

        Mike Flynn.

      • Dear Professor Mike

        No I think I’ve got it. If you don’t let gravity in there is no gravito-thermal effect. But if you let gravity in then you have to let energy out – it’s only uncommon sense.

        Brilliant – this redefines classical mechanics for all time.

        He who laughs last didn’t get it
        Robert I Ellison
        Chief Hydrologisr

    • David Springer

      You allowed no energy to enter the system by definition but then you introduced gravitational potential energy into the system. Gravitational potential energy is not imaginary it’s as real as anything else. Pick up a brick. You added gravitational potential energy to the brick. Release the energy by dropping it on your foot. Use small brick just in case I’m right. :-)

      • God no. Professor Mike excluded gravity from the system to demonstrate that there is no gravito-thermal effect. No gravity – no effect. Simples – get with the program Springer. This is much simpler than FOMBS not finding stellar like temps in a centrifuge on Earth.

      • David Springer,

        You might care to read what I wrote.

        I really didn’t contemplate having to explain that it was an example of reductio ad absurdum, (closely related to the reduction ad absurdum that the autocorrect supplied previously), with which some people are apparently either unfamiliar or uncomfortable.

        I apologise for using a form of logical argument you do not understand.

        Live well and prosper,

        Mike Flynn.

  68. I found a good article by E. T. Jaynes on the Gibbs Paradox which according to Jaynes never was a paradox based on Gibbs text on Heterogeneous Equilibrium which many at the time seemed to have over looked or at least not looked at closely enough.

    http://bayes.wustl.edu/etj/articles/gibbs.paradox.pdf

    I am pretty sure it won’t resolve this debate, but it could add a bit to the discussion.

    • Thank you. I can’t claim to have assimilated it yet, but as a layman I find these statistical-mechanics issues intimidating, and any help I can get is welcome.

    • Joe Born, Well is can take a while, I will likely make another pass or three. If you compare it to the link provided by P-N, http://cds.cern.ch/record/566951/files/0207080.pdf Gibbs free energy includes a function not a constant, that needs to be considered for special cases. So “declaring” an isolated atmosphere is isothermal doesn’t mean the atmosphere is stable. If you specify a stable isolated atmosphere, then how much stability depends on, the micro, canonical, grand canonical, micro-grand canonical ensembles etc. or how much detail you really want to get into. None of the “models” are realistic without radiant heat transfer and consideration of the properties of the real gases involved.

      My favorite example is the assumption that CO2 is non-condensable. Depending on gravity i.e. pressure and temperature, CO2 condenses, become super fluid etc. On Venus you have super fluidity near the surface and condensation in the atmosphere. Since the gravity of Venus is pretty close to the gravity of Earth, there is a 184K +/- a touch or around 65Wm-2 effective radiant energy at the real tops of the atmospheres. I haven’t seen that discussed much, but it looks like the enthalpy of CO2 has a bit of a stabilizing effect on the minimum temperature.

      It is a very interesting problem that so many simplifying assumptions just tends to ruin the fun.

  69. Since a number of people on this thread seem actually to have read Velasco et al. and its predecessor, Roman et al., I’d like to ask whether anyone has understood the calculus.

    Specifically, Roman et al.’s Eq’n 5 uses a calculus identity they got from a paper by Fernandez-Pineda et al., but the latter authors based their result on their Eq’ns 3 and 4, which I’ve been unable to derive.

    I don’t suppose anyone could direct me to a derivation?

  70. Pekka and Pierre-Normand, here is a simple thought experiment. Take an isothermal gas in an insulated box in gravity. In potential temperature this displays a gradient with warmer values above colder ones. Mix this thoroughly. Now the potential temperature becomes uniform (equivalent to the maximum entropy state), and the gas becomes cooler at the top than at the base by the dry adiabatic lapse rate. Leave this gas as long as you want and it will not return to an isothermal state because that would imply a decrease in entropy. Instead of mixing this thoroughly you can leave molecular diffusion to act over a long time and get the same result. In gravity, you can’t unmix potential temperature. This takes the place of temperature or any other mixed intensive property, being a conserved property of a parcel.

    • Pierre-Normand

      Jim D,

      You are simply assuming that the isentropic profile (constant potential temperature with height) maximizes entropy. It is not true that mixing the gas can relevantly be likened to letting diffusion occur in the insulated box for the following reasons.

      The diffusion process that brings the insulated system closer to equilibrium doesn’t require an energy input. It is this process that tends to bring the lapse rate from the dry adiabat towards the isothermal profile (heat conduction works perfectly well against gravity, as does internal radiative heat transport).

      Contrariwise, mixing does require an external energy input. That’s because when the lapse rate matches the adiabat, then it is ‘neutrally stable’, and the energy required to sustain mixing motions is the work performed against viscous dissipation of the (bulk) kinetic energy of moving air parcels. What is worse, when the gradient is lower than the adiabat (closer to isothermal), then the lapse rate is stable, and not just ‘neutrally stable’. External work is required to push a parcel of gas at a lower or higher level such that adiabatic expansion or compression will cause the local profile to depart from the lower surrounding gradient. That’s of course because as a parcel begins to rise and adiabatically expand, it immediately becomes ‘heavier’ than the surrounding gas (e.g. its weight exceeds the opposite buoyant force).

      So, you need an external source of energy to sustain mixing in the gas when it is convectively stable. Any spontaneous vertical motion of an air parcel would make it behave like a harmonic oscillator (where the difference between constant weight and variable buoyancy is the restorative force) dampened by viscosity. To bring the temperature profile towards the adiabat from below (i.e., from closer to the isothermal state) therefore can be likened to powering a heat pump. And, of course, powering a heat pump can result in a state with *lower* entropy, such as the state with constant potential temperature. After you’ve pulled the plug on the heat pump (e.g. stopped actively mixing the gas) thermal dissipation will slowly bring the gas back towards the higer isothermal entropy state — passing through a continuous sequence of convectively stable profiles. Any residual convective motion will rapidly be dampened by viscous dissipation, and thereafter be prevented to spontaneously develop.

      • This is the key difference. You think that an initially isentropic profile will go towards isothermal by dissipation. I disagree. This results in a potential temperature gradient “settling out” which is like unmixing a gas. The isothermal state with the same total energy has a lower entropy. It violates the 2nd law to spontaneously go from isentropic to any other state.

      • Pierre-Normand

        Jim D, if the gas is radiatively active, then, within the box, there is going to be a net vertical radiant heat flux. I think that’s a severe problem for your view. You would also need a (molecular) thermal conductive flux in the opposite direction to compensate for the radiant flux. This is related to another objection currently held in moderation. In short, how do you deal with the case of two adjacent gas columns, N2 and argon, say, that have different adiabatic lapse rates? How will they come into joint thermodynamic equilibrium if you put them in thermal contact only at the top and bottom, say? By what sort of process would a unique gradient be maintained?

      • Pierre-Normand

        “This is the key difference. You think that an initially isentropic profile will go towards isothermal by dissipation. I disagree.”

        Yes, we agree that gases can’t ‘unmix’ spontaneously but overturning motions of a gas that is convectively stable doesn’t amount to molecular mixing. It amounts to operating a heat pump. That was my main point.

        Regarding the assumption about which temperature profile maximizes entropy, I’ve provided a few standard references (including textbooks) that show that the Boltzmann distribution law for energy entails a barometric density distribution *and* a Maxwell speed distribution constant with height (isothermal). It follows that this is the maximum entropy state. I’ve also provided a few references that show that such a density and temperature profile is stationary. It is preserved by molecular diffusion.

        As Pekka noted, the main source of your belief that the constant potential temperature profile maximizes entropy seems to come from consideration of systems constrained *not* to be in thermodynamic equilibrium, since they are warmed from below and enabled to cool from above, and there therefore exists a constant source of negentropy opposing the movement towards equilibrium. The solar source of negentropy — or free energy — powers the heat pump that maintains the lapse rate. That’s not equivalent to molecular mixing. It’s rather equivalent to unmixing (with suitable external work provided).

      • Pierre-Normand, I don’t want to bring radiation into this because that does tend to bring in a radiative equilibrium that favors equalizing the temperature itself. This confuses the issue. Consider our gas is just N2 and O2, which is effectively non-radiative in the thermal IR bands. In principle any stably stratified thermal profile is stable to perturbations and there is nothing special about the isothermal one. The main point is what molecular motions do to this, and these equalize the conserved quantity, potential temperature, given long enough, because that maximizes entropy. We don’t need to introduce convection to get this end state. Another way to see this is that the isothermal state has a vertical gradient of total energy per unit mass, because potential energy per unit mass is higher at the top even if the kinetic energy per unit mass is the same. Molecules at higher levels just have more energy. Molecular diffusion equalizes this energy just through collisions until the gradient no longer exists.

      • Pierre-Normand

        Jim D wrote: “Another way to see this is that the isothermal state has a vertical gradient of total energy per unit mass, because potential energy per unit mass is higher at the top even if the kinetic energy per unit mass is the same. Molecules at higher levels just have more energy. Molecular diffusion equalizes this energy just through collisions until the gradient no longer exists.”

        Even if it were true that the adiabatic lapse rate would maximize entropy (which it isn’t), it is inconsistent with your new proposal that molecules at higher levels can’t have higher total energy (internal + gravitational potential). This requirement is absurd since it would entail that at equilibrium there would by no molecules at all above the height where the gravitational potential energy of one mole of gas equals the total internal energy of one mole of gas at the bottom of the column. In other words, molecules would be unable to rise above the level where the gravitational potential energy is equal to the *average* internal energy that they have at the bottom of the column. Else, their average total molar energy at that height would be larger than it is at the bottom and this would contradict your requirement. However, your different requirement of constant potential temperature with height, together with the barometric density distribution, does not entail such a ceiling limit for the maximum height of molecules, so it’s not as bad.

        It’s better to realize that the Boltzmann distribution of energy, that maximizes entropy for a gas in a gravitational field, does *not* have the implication that the total average energy of the molecules is the same at all height. There is no such requirement of equalization of energy through *space* (though there is one such requirement through phase space, relative to the relevant partition function).

      • Pierre-Normand

        Jim D,

        It seems to me you would benefit greatly from reading just the first few first paragraphs from Coombes and Laue (though this paper is just one and a half page long, excluding the appendix) as it addresses precisely your mistake regarding molecular diffusion and the alleged requirement of random distribution of total molecular mechanical energy through *space* under gravity. You have the opportunity for an ‘Ah ha!’ moment. Right now, your thinking about this particular issue is drifting away from mainstream statistical mechanics straight into Doug C.’s territory.

        Coombes and Laue, ‘A Paradox Concerning the Temperature Distribution of a Gas in a Gravitational Field,’ Am. J. of Physics, 1985.

        http://tallbloke.files.wordpress.com/2012/01/coombes-laue.pdf

      • Pierre-Normand

        “The solar source of negentropy — or free energy — powers the heat pump that maintains the lapse rate.”

        This sentence of mine was rather misleading. The warming from below (after the Sun rises) brings first the lower lowest layers to a gradient above the adiabatic lapse rate and induce convective motion that carry the surface heat upwards. This is analogous to a thermal engine that performs work through bringing back the gradient down to the adiabatic lapse rate. When overturning movements are induced elsewhere (e.g. though abduction of air masses) in areas where the gradient is lower than the adiabatic lapse rate, then in *those* regions, convection functions as a heat pump to steepen the gradient, from below, towards the lapse rate. But the second process (heat pump) is harnessed to the first (thermal engine), which is ultimately driven by the sun (warming from below).

    • Pierre-Normand

      Jim D,

      Let me note also that one familiar objection to the idea of a gas dependent gravito-thermal effects also works against your proposal. The adiabatic lapse rate in a gas column depends on the specific heat capacity of the gas. What would occur, in your view, if two tall adjacent insulated gas containers where put into thermal contact only at the top and bottom, where the first column is filled with argon, and the second one is filled with nitrogen? What would be the equilibrium state of the combined system considering that, on your view, they don’t tend to individually diffuse heat internally such as to produce the same temperature gradient?

      • Pierre-Normand

        Maybe, after this post (beginning with “Jim C, Let me note…”) is released from moderation, someone can venture a guess as to the nature of the moderation trigger.

      • Pierre-Normand

        Correction: “Hence a finite amount of work W1 must have been consumed to effect [a complete half-]rotation of the cylinder.

      • Pierre-Normand

        “I hope Pierre-Normand, Vaughan Pratt, and captd also will resist further postings on this. It is very time-consuming.”

        Jim D, That’s fine with me. There is no point in forcing anyone to pursue a discussion about a theoretical topic of very little practical significance. I may still engage other posters in this thread though. If other people bring up the same arguments that you made, I will only respond to them without mentioning you.

    • JIm D,
      Your thought example is erroneous. A closed undisturbed system of gas in gravitational field has the maximal entropy, when it’s isothermal, not when the potential temperature is uniform. When your basic starting point is wrong, it’s not surprising that you reach the wrong conclusion.

      Therefore it does go very slowly towards isothermal from any other state. (Once more: This is valid in the thermodynamical limit, but 10²³ molecules or even a small fraction of that is enough for that.)

      • It can’t go towards isothermal because that is non-conservation of a conserved quantity, the potential temperature. There is nowhere it can go from a state of uniform potential temperature because any motion, molecular or otherwise leaves the system the same. This is a property of a maximum entropy state.

      • Pierre-Normand

        Jim D, potential temperature only is ‘conserved’ (it is rather more properly said to be *invariant*) under the assumption that temperature gradient already matches the adiabatic lapse rate.

      • Pierre-Normand

        … I’ll grant you that potential temperature can properly be said to a quantity that is conserved by an adiabatic process. But an adiabatic process *precludes* thermal transport through molecular diffusion. So, when molecular diffusion is allowed, you can no longer presume that potential temperature is a conserved quantity within air parcels.

      • Pierre-Normand, no, molecular diffusion is an adiabatic process, and is therefore not a source for potential temperature. It can redistribute potential temperature, but can’t change the integral, so when it is well mixed and constant it has no effect at all.

      • Pierre-Normand

        Jim D, I don’t know what you mean by “molecular diffusion is an adiabatic process”. That doesn’t seem to make sense. And adiabatic process precludes material or heat fluxes through the boundaries of the relevant systems. The sorts of systems at issue here are air parcels. Over the relevant (macroscopic) sizes and time frames, diffusion through the boundaries can be ignored for purpose of tracking temperature variations that result from expansion or compression of rising or falling parcels. When those fluxes can’t be neglected, the processes aren’t adiabatic anymore.

        The very idea if potential temperature, conceived as a conserved quantity, is the temperature of an air parcel after it is adiabatically compressed or expanded to some specified (though arbitrary) reference pressure. If the vertical pressure profile is assumed, this can be a reference height. It is assumed that there is no significant molecular diffusion through the air parcel as it is brought to the reference level or pressure. Once diffusion becomes the dominant process, the concept of potential temperature loses much relevance. There is no reason anymore that it be conserved since the relevant ‘adiabatic’ constraint is relaxed.

      • Pierre-Normand, adiabatic means literally no heat added. Diffusion rearranges but does not add heat. The opposite is diabatic where heat is added, e.g. by radiative fluxes or condensation or surface fluxes. Potential temperature conservation relies on dQ=0 in the thermodynamics.
        In a maximum entropy state you can exchange any two parcels adiabatically moving them to each others positions, and not affect the properties of the whole gas. This does not happen with an isothermal atmosphere in gravity (only without gravity). The parcel that started higher would be warmed as it descended and vice versa. As I mentioned, this is because there is a downward energy gradient, which also has a consequence of a downward energy flux, even if only by molecular collisions, and is therefore not a steady state in a bounded gas.

      • Pierre-Normand

        “As I mentioned, this is because there is a downward energy gradient, which also has a consequence of a downward energy flux, even if only by molecular collisions, and is therefore not a steady state in a bounded gas.”

        There is no such law of physics. If there is a *temperature* gradient when there is a *heat* flux. You are seemingly making an incorrect generalization from something like the theorem of equipartition that ensures equal distribution of energy among independent degrees of freedom of molecules. There is no requirement from statistical mechanics that energy density be uniformly distributed among identical molecules located at different potentials within a conservative force field. I think you haven’t considered the correct partition function in phase space for the energy of the microstates. Another instructive example is section B (Kinetic derivation) of the following paper.

        Berberan-Santos, Bodunov, Pogliani, ‘On the Barometric Formula’, Am. J. of Physics, 1997.

        http://web.ist.utl.pt/ist12219/data/43.pdf

        This thought experiment shows that if you start out with a Maxwell speed distribution (with average kinetic energy (3/2)kT) for molecules constrained to move in a thin layer (a pizza box, say), and then allow the gas to expand upwards through a much higher volume against gravity, the average kinetic energy drops to (1/2)kT (the gas uniformly cools) while the rest of the energy is given up to the gravitational field (the total gravitational potential energy of *all* the molecules), though the distribution of kinetic energy remains uniform with height! Molecules are less likely to be found higher up, and this translates as a density gradient, but the speed distribution is maintained at all height.

      • Pierre-Normand

        “…the average kinetic energy drops to (1/2)kT…”

        I mean, of course, the gas cools from T_1 to T_2 such that the final average kinetic energy is (1/2)kT_1 = (3/2)kT_2, and hence T_2 = (1/3)T_1.

      • Pierre-Normand, in gravity the heat flux is proportional to the potential temperature gradient, not the temperature gradient. A gas with a stable stratification (positive potential temperature gradient) would have a downward heat flux. According to you, the heat flux would be upwards in a slightly stable atmosphere making it go more stable and only stopping when it reaches an isothermal state. No such effect exists. Heat flows downwards in a stable atmosphere and upwards in an unstable one.

      • Pierre-Normand

        “Pierre-Normand, in gravity the heat flux is proportional to the potential temperature gradient, not the temperature gradient.”

        This is a *convective* heat flux. It effect on the temperature gradient works *against* molecular diffusion and it must be driven by the constant injection of Gibbs free energy. When you stop this injection (put the gas in an isolated box, say), then the gradient is free to slowly drop towards the isothermal profile as it rapidly becomes convectively stable. You still are mixing up a process that can only occur in a system far out of equilibrium (in an atmosphere warmed from below and driven towards convective instability) with another process, thermal molecular diffusion (conduction), that occurs also under conditions of convective stability so long as there is a thermal gradient.

      • Jim,
        You have some serious misunderstandings on what an conserved quantity means.

        Approach to thermodynamic equilibrium means that entropy increases, it’s not a conserved quantity, neither is potential temperature a conserved quantity. It’s unchanged in the idealized case of truly adiabatic ascent of a parcel of air, but no real motion of air is truly adiabatic.

        These are just two details related to, what you got wrong. The correct physics has been discussed in many comments of this thread.

      • Pierre-Normand

        Jim D,

        Here is a relevant bit from Pierrehumbert’s textbook:

        “The stable stratification of a layer indicates that convection and other dynamical stirring mechanisms are ineffective or absent in that layer, since otherwise the potential temperature would become well mixed and the temperature profile would become adiabatic. An isothermal layer is stably stratified, *because* its potential temperature increases with height; even a layer like that of Mars’ upper atmosphere, whose temperature decreases gently with height, can be stably stratified.” (My emphasis)

      • Pierre-Normand, yes, stable stratification would be a stable configuration for a long time because molecular-scale collisions alone are a very slow process. There is nothing special about an isothermal stratification compared to one that is even more stable (an inversion). However, molecular diffusion erodes these states and provides a slow downgradient heat flux. The only state it doesn’t erode is the dry adiabatic lapse rate.

      • Jim,
        Where do you pick those false ideas? The only state that’s not eroded by convection is isothermal.

      • Pierre-Normand

        Pekka, typo alert. You mean to say “…the only state that’s not eroded by [diffusion] is isothermal”

      • Pekka, you need to investigate some atmospheric model equations to see what variables they use. A common choice is the potential temperature because its equation is simpler than that for temperature which has a compression term. Potential temperature is also a convenient choice for diffusion because it is conserved under vertical displacement, unlike temperature, so its diffusion equation is simply an function of its own gradients. Yes, entropy is not conserved and would make a lousy model variable because diffusion would be a source, but this is not an issue with potential temperature, the log of which is proportional to entropy itself.

      • Alternatively consider diffusion in an isothermal atmosphere. A downward displaced parcel from above to your reference level would be warm, and an upward displaced one would be cooler than the reference level, so temperature behaves like there is a vertical gradient even when there isn’t one. The consequence of this effective gradient is that there is a downward flux of heat because there is a negative correlation of eddy vertical motion and temperature.

      • Pierre-Normand

        Jim D,

        Aren’t atmospheric models yielding isothermal profiles for optically thin stratospheres? What is it, in your view, that erodes the adiabatic temperature profile that you believe ought to result from molecular diffusion alone? Speed isn’t an issue there. Those isothermal states are steady.

      • Pierre-Normand

        Jim D,

        “Alternatively consider diffusion in an isothermal atmosphere. ”

        After that you proceed to ignore diffusion completely and talk instead about adiabatic processes involving macroscopic air parcels that are somehow displaced against the forces (weight minus buoyancy) that tend to restore them to their original height. You need an external energy source to perform such work when the starting profile is isothermal.

      • I certainly recommend Curry and Webster around page 92 for how vertical diffusion is a function of potential temperature gradients in the atmosphere. See if this link works or Google the book.
        http://books.google.com/books?id=mdFzlFfWbiYC&printsec=frontcover&dq=atmospheric+dynamics+equations+curry+and+webster&hl=en&sa=X&ei=XQKFVKK4CMmNyATkvIKIBw&ved=0CCYQ6AEwAA#v=onepage&q=atmospheric%20dynamics%20equations%20curry%20and%20webster&f=false

      • Pierre-Normand

        Jim D,

        “I certainly recommend Curry and Webster around page 92 for how vertical diffusion is a function of potential temperature gradients in the atmosphere.”

        Thanks. Unfortunately I can’t read pages 90 to 100. Are you sure they aren’t talking about eddy diffusion? You would be somewhat equivocating on the word “diffusion”. Eddy diffusion is a species of adiabatic overturning, and quite unlike molecular diffusion. It still must be driven by an external source of energy for the reasons that I already explained. Plus, you have larger viscous forces to contend with — the smaller the scale — which limits from below the size of the air parcels involved to the 1mm range.

      • Pierre-Normand, yes it is eddy diffusion, but they say by analogy with molecular diffusion. I didn’t look thoroughly to see if they discussed molecular diffusion somewhere else.

      • Pierre-Normand, the stratospheric temperature profile is determined a lot by the solar heating and ozone distribution. Any resemblance to an isothermal profile is coincidental. Radiative heating makes it far from an ideal gas to study for long-term diffusion effects.

      • Pierre-Normand

        Jim D, I wasn’t talking about the terrestrial troposphere, which is optically thick, but rather about optically thin tropospheres as discussed in Pierrehumbert’s texbook. They match the temperature of the convective troposphere at the tropopause, but thereafter the profile remains isothermal *because* of heat diffusion.

      • Pierre-Normand

        Arrgh! Why do I type ‘tropo…’ when I am thinking ‘strato…’. Let me try that again:

        “Jim D, I wasn’t talking about the terrestrial [strato]sphere, which is optically thick, but rather about optically thin [strato]pheres as discussed in Pierrehumbert’s texbook. They match the temperature of the convective troposphere at the tropopause, but thereafter the profile remains isothermal *because* of heat diffusion.”

      • Jim,
        All the discussion of atmospheric physics you are referring to excludes molecular diffusion and conduction from the consideration, because it’s so weak that it can, indeed, be dismissed in typical applications. When you discuss diffusion, you seem to consider collective movement of air that occurs both in convection and in turbulent mixing. In these phenomena individual molecules are never considered, but parcels of air that are large enough to have local thermodynamic properties like pressure, temperature and density. It’s also assumed that diffusion through the boundaries of these parcels can be dismissed without making any significant error.

        What P-N and myself have be discussing is molecular diffusion under conditions, where the external input of free energy is exactly zero (no heating at the bottom), not very small but exactly zero. In that case all convection dies out and all turbulent mixing dies also out, because they need free energy to continue. In terms of entropy, the situation is already so close to maximum, that the addition that turbulence would imply is not possible any more. Under these conditions the laws of thermodynamics that control convection are satisfied in the trivial way that convection goes to zero.

        Molecular diffusion remains as strong as ever at the same temperature, because it’s not controlled by the equations of thermodynamics. It’s not part of the physics described by thermodynamics, but it is part of physics described by statistical mechanics. In presence of temperature gradients molecular diffusion leads to conductivity that can be used as input in equations of thermodynamics.

        When everything else goes to zero, the very weak conduction becomes the dominant form of heat transfer. Conduction is not adiabatic as heat is transferred also through the boundaries of the air parcels. This is the only remaining process, and that leads to the isothermal final outcome.

      • David Springer

        No. Conduction doesn’t lead to an isothermal state in a gravity field. That is because conductivity is asymmetrical in the vertical axis. There is more resistance to conductance against gravity than with it. Or at least that’s the postulate behind the gravito-thermal effect which ostensibly manifests in solids as well as fluids.

        Your mission, should you decide to accept it, is to prove that conductance in a gravity well isn’t asymmetrical in the vertical axis and/or that this doesn’t lead to asymmetry in heat distribution in the vertical axis.

        Good luck.

      • David,
        Conductivity is symmetrical,because there are two factors that cancel exactly at equilibrium:
        1) each particle moves more easily down than up
        2) the density of particles decreases with altitude.

      • pekka, “Conductivity is symmetrical,because there are two factors that cancel exactly at equilibrium:
        1) each particle moves more easily down than up
        2) the density of particles decreases with altitude.”

        I believe that conduction is assumed symmetrical, but depends on the symmetry of the medium.

        http://arxiv.org/abs/1409.5364

        This is why I mentioned in the previous thread that limits like the speed of sound and speed of second sound might need to be considered in specific cases.

      • David Springer

        Trivially wrong, Pekka. The gravito-thermal effect ostensibly manifests in solids as well as fluids.

        I would also argue that the pressure difference between two adjacent vertical layers, each one molecule thick, is miniscule while the force of gravity remains pegged at 10m/s/s regardless of distance.

        Pay attention to CaptDallas who immediately pointed to experimental science in sound propagation through a crystal lattice. Your breadth of knowledge and experience of real world phenomena is deficient.

      • David Springer

        A second point I’ve brought up is that vibration distance is in the quantum realm and we have no quantum theory of gravity. The assumptions Pekka incorporates into infallible conclusions is typical of climate science. The question of gravito-thermal effect can only be answered by experiment. But when experiment is difficult the pseudo-scientist instead relies on mental models and ad populum support for same to reach conclusions that are really no more than untested hypotheses. Thank God engineers don’t build bridges and aircraft based on the kind of science practiced by pontificating psuedo-scientists.

      • CD,
        The discussion I have been participating in does not consider all cases, only the case of thermodynamic equilibrium in gases and approach to that. In this case phenomena as violent as sound are not present at a relevant level.

        If someone thinks that this discussion is not really relevant for the real atmosphere, I agree, as I wrote in my first comment to this thread

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-651632

      • Pekka, “If someone thinks that this discussion is not really relevant for the real atmosphere, I agree, as I wrote in my first comment to this thread.”

        I agree completely, but since we have something like 10^23 states there is a fairly large range of possibilities. So if you increase the particle density to the point there can be zero gravito-thermal potential you have introduced other limits. This is a thought experiment that just keeps on giving. As I mentioned to P-N, if you change the shape of the box to a sphere you are closer to maximum entropy. There should be some nit that can be exploited in any ideal model and some of those nits lead to useful discoveries. Gravito-thermal effect is not likely one of those, but that doesn’t mean it cannot exist in some carefully designed experiment.

      • Pekka and Pierre-Normand, you are saying that in some conditions, such as weakly stable ones, molecular diffusion can be in the opposite direction to eddy diffusion. I don’t think this is the case. There is no scale where you can ignore gravity. Molecules colliding from above in an isothermal gas will have systematically more energy than those from below, leading to an effective gradient, equivalent in every sense to a temperature gradient, when collisions occur and redistribute the energy. The only situation that cancels this is the dry adiabatic lapse rate, where those above are also a little cooler. An isothermal gas has all the molecules moving at the same speed regardless of altitude. By virtue of their potential energy, the higher ones do have more total energy and it is available when they descend to collide.

      • Pierre-Normand

        Jim D,

        Gravity isn’t ignored, it is fully accounted for at the molecular level in the demonstration that the isothermal state, together with the barometric density vertical profile, is stationary. This means that in this stationary state the Maxwell distribution of speeds is the same at every level. (Check again Pekka’s short note, and the Coombes and Laue short paper). Since the velocity distribution is isotropic, the molecules that come from above have the exact same average kinetic energy than the molecules that come from below. Collisions don’t break this isotropic distribution. This may be counter-intuitive, but you have to let the calculation control your intuition before you are going to decide that mainstream statistical mechanics and thermal physics are wrong.

        Maybe the most intuitive way to approach the calculation starts with the vertical component of the Boltzmann kinetic energy distribution rather than the vertical velocity distribution. This energy distribution is a negative exponential A*exp(-m*v^2/2kT), where A is a normalization constant and m is the molecular mass (let us assume identical molecules for simplicity). As you rise to higher levels, the molecular density is reduced in accordance with the barometric formula since the slower molecules fall back down and drop from the higher level populations. The molecules that have enough energy to rise from a lower level to another level dz higher, are uniformly down-shifted in kinetic energy by the value m*g*dz. This means that the kinetic energy distribution is uniformly shifted to the left. Only the positive values for kinetic energy have a physical meaning. Shifting an exponential probability distribution leaves it invariant after it is normalized (the area under the curve in the upper right quadrant set to 1). So the vertical component of the kinetic energy distribution is stationary provided only the drop-off rate is matched by the initial vertical density profile, and this drop off rate corresponds precisely to the barometric density formula (as shown in many references that I provided). This concludes the proof that such an isothermal profile is stationary.

      • Pierre-Normand, it is you that is defying thermodynamics. Spontaneously transitioning from an isentropic state to an isothermal one decreases entropy. This is completely counter to the well recognized 2nd law. As far as I knew there was no exception, but you are claiming one. This is a bold claim indeed.
        If the molecules at every level have the same kinetic energy in an isothermal distribution, as we agree, you also have to recognize that those coming down have slightly more energy than those coming up when they reach a level. This is exactly like a temperature gradient would look in terms of what they do to redistribute their energy when they collide, and energy is transferred downwards by this process.

      • David Springer

        Once again, Jim D

        +1

      • Pierre-Normand

        Jim D wrote: “you also have to recognize that those coming down have slightly more energy than those coming up when they reach a level.”

        Jim D, I just provided you with a mathematical proof that this claim is false. Coombes and Laue published the proof why this claim is false in the American Journal of Physics in order to show undergraduate students how intuition can mislead and yield results contrary to what the Boltzmann distribution of energy (for both kinetic and gravitational potential energy) entails. This article has spawned secondary literature that confirms it. The paper is one and a half page long. You really ought to have a look at it.

        Pekka explained to you that isentropic profiles only are stationary under the condition that there is a constant injection of Gibbs free energy in the system that drives convective heat transport (except under idealized circumstances where there is no heat conduction in the classical fluid). When this external heating of the system stops, then convection dies down and the entropy isn’t maximized until heat conduction has brought the gradient down to zero. Pierrehumbert shows that an optically thin atmospheres are isothermal at equilibrium.

      • Pierre-Normand, none of what you cite was looking for a maximum entropy state. By definition an isentropic state is that. There is no way to perturb it adiabatically that leads to higher entropy. An isothermal state with the same energy has a lower entropy. Potential temperature is the generalization of temperature to handle varying pressure in adiabatic processes, as you have with gravity. Take a box with a partition, and different potential temperatures, theta_1 and theta_2. Remove the partition and they will mix to a new value of (theta_1+theta_2)/2. The initial entropy is (log(theta_1)+log(theta_2))/2. The final entropy is log((theta_1+theta_2)/2). This is larger because the mean of any two different numbers is larger than its geometric mean. If these boxes had the same pressure you could substitute T for theta. Imagine the box has a horizontal partition in gravity and that is the case where you may have the same temperature in the two boxes but different theta. The outcome is still the same and the mean theta results from letting the gases mix, even if it takes longer under gravity. The final temperature won’t be the same because it is not a conserved quantity in gravity. You could have two boxes with the same temperature but different pressures and let them mix. You won’t end up with the same temperature.

      • Pierre-Normand

        “Pierre-Normand, none of what you cite was looking for a maximum entropy state.”

        Jim D, I can only refute your objections one at a time. Actually quite a few among my reference look at maximum entropy states ((1) On the Barometric Formula, (2) Roman, White, and Velasco, On a Paradox… (3) Gravitational instability of isothermal and polytropic spheres (4) Venus Atmosphere Profile from a Maximum Entropy Principle. (5) Verkley and Gerkema, On Maximum Entropy Profiles)

        Now you are again running together two fundamentally different processes, convective heat transport and heat conduction. They are processes that usually pull in opposite directions. You are merely asserting that an isentropic profile maximizes entropy (all my references beg to differ). You are ignoring Gibbs free energy and misconstruing the second law. The second law forbids a thermal flow opposite to a *temperature* gradient. The existence of a gravitational field doesn’t negate this formulation of the second law. Potential temperature isn’t temperature. There is no way for a heat flow to move a profile away from an isothermal state. Gravity doesn’t help.

        You propose that molecular diffusion and convective heat transport are basically the same processes operating at different spatial scales. But the steepness of adiabatic lapse rates depend on c_p, the specific heat of the gas. Molecular diffusion doesn’t depend on this, as your arguments show. You argued that if the profile is isothermal then molecules that fall down to some level z move faster on average than molecule that move up to level z. Do you now see why this is false?

        Here is an even simpler demonstration. In a weakly interacting gas column, molecules move along free falling parabolic paths, reaching the maximum height that the vertical component of their initial kinetic energies allows, fall back down, and bounce elastically on the bottom to repeat a similar motion. Each one among those molecules that has enough energy to rise above level z crosses that level twice — once while going up, and once again while falling back down — at the *same* speed. Since this is true for any molecules that crosses the z plane, the speed distributions of the molecules crossing it in must be the same in both directions.

      • JimD,

        You wrote –

        “You could have two boxes with the same temperature but different pressures and let them mix. You won’t end up with the same temperature.”

        I must be misunderstanding your sentence. If you take two dive cylinders of air, one fully charged, and one at say two bar, at the same temperature, open them up, and release them into, say, a flotation bladder that shares the same evironment as the cylinders, that once the cylinders are both at the same pressure, the cylinders will be at different temperatures?

        Or are you referring to an initial temperature drop if high pressure gas is allowed to expand rapidly? Or something else entirely?

        Can you provide more detail? Your assertion seems a bit bald, considering the lack of substantiation.

        Thanks.

        Live well and prosper,

        Mike Flynn.

      • Jim,
        You continue to use a totally false argument

        Spontaneously transitioning from an isentropic state to an isothermal one decreases entropy.

        Transition towards isothermal does not decrease entropy, it increases entropy. That’s true by the most basic formulas of thermodynamics that involve entropy. That’s also part of the proof that a perpetum mobile of the second kind is impossible. All these formulas include temperature, not potential temperature, and that’s not by omission, but because that’s correct even under gravity.

        Molecular diffusion takes place everywhere. It leads towards isothermal also under gravity. Eddy diffusion is a form of collective movement of gas that involves small parcels of gas, not individual molecules. Part of the molecular diffusion and related heat conduction occurs between parcels of gas that are considered in derivations that you seem to be familiar with, but it’s left out of the derivation, because its influence is negligible in most cases. In the phenomena that P-N and I have been discussing everything else is absent or so weak that heat conduction dominates in heat transfer. Therefore our conclusions are totally different from those obtained assuming that this effect is negligible. Those derivation are correct for the applications, where they are normally used, but they cannot tell anything on the mechanisms we have been discussing.

        The Secod law tells that the total entropy of a closed system cannot decrease. That law does not tell that the entropy must increase until it has reached its maximal value. Formulas of thermodynamics that ignore diffusive processes allow for a non-maximal value of entropy to persist for ever, but including the diffusive phenomena of heat conduction and viscosity leads to the additional result that the total entropy of a isolates volume of gas will increase towards the maximum and is strictly constant only at that maximum. The state of the maximal entropy is isothermal,it’s not one of constant potential temperature.

      • Heat conduction is no more than collisions of molecules. These molecules have both kinetic and potential energy and can share both in collisions, just like macroscopic balls can. There is nothing special about the molecular scale where you can ignore their potential energy. It is reasonable to expect the higher molecules to be moving more slowly having given up some of their kinetic energy (temperature) to potential energy to get there. I am sure if you had a computer simulation of elastic balls in gravity, they would reveal this behavior to you. This is why the equilibrium state of a gas in gravity is to be colder with height. The atmosphere naturally goes to states with a lapse rate, not towards isothermal ones, and this has to do with maximizing entropy, which is to equalize conserved quantities such as potential temperature.

      • Pierre-Normand

        Jim D,

        I earlier provided a link to a molecular dynamics simulator, and the relevant discussion at SoD. The simulation indeed makes visually apparent where you are going wrong. The collisions themselves don’t redistribute energy between the potential gravitational and kinetic components. The collisions conserve kinetic energy. *If* the speed distribution of the incoming molecules that are colliding at some level is isotropic, *then* so is the speed distribution of the molecules immediately after the collisions. What you are studiously avoiding considering, then, is the very simple mathematical proof that the combined effect of the asymmetric vertical density gradient and the asymmetric acceleration of gravity cancel each other exactly such that the speed *distributions* of molecular populations remain constant at all heights.

        Molecular dynamics simulator:
        http://physics.weber.edu/schroeder/md/InteractiveMD.html

        Science of Doom discussion of the simulator:
        http://scienceofdoom.com/2010/08/16/convection-venus-thought-experiments-and-tall-rooms-full-of-gas/#comment-81966

        Proof that the combined effects of the density gradient and the gravitational acceleration on the vertical speed distribution cancel out exactly in a gas column at equilibrium:
        http://tallbloke.files.wordpress.com/2012/01/coombes-laue.pdf

      • Pierre-Normand, why are you assuming the speed distribution in colliding molecules is isotropic? Those coming downwards have accelerated since their last collision. This directional asymmetry in the distribution is what brings energy downwards from an initially isothermal state. It manifests as a correlation between vertical motion and kinetic energy, which results in a flux. This, at the molecular level, is the heat flux that is downgradient in potential temperature.

      • Pierre-Normand

        Jim D,

        If you consider all the molecules that contribute to the population at one given level, you must consider that all the molecules that cross a horizontal plane z = h coming down from above also have crossed it from below going up at an earlier time, at the same exact vertical speed (and reversed vertical velocity). So, the contributions of those molecules to the velocity distribution at that level is isotropic. As for the molecules coming from below, the same argument applies. They also are going to fall back down (neglecting the small faction that reaches escape veolocity when the column is open above and the gravity well has a finite depth).

        So, all the molecule that contributes to the density of vertical velocity in the range [v, v+dv] at some level z also contribute equally to the range [-v, -v+dv] at that level, and vice versa.

        This is true for *any* stationary configuration of speed distributions in a vertical column bounded from below (where molecules rebound elastically on the bottom) such that the gas isn’t falling down as a whole and replenished at the top, or some such scenario. Were this not the case, then the density distribution would not be stationary since there would be a material flux across the plane z = h.

        This consideration alone doesn’t prove that the isothermal distribution is stationary to start with. It merely refutes your most recent objection. But I have already proven that it is, and so has Pekka (see his note linked below), and so have Coombes and Laue, for the case where the density distribution is barometric. Such an isothermal + barometric density configuration is stationary. And it is mandated by the Bolzmann distribution law applied to kinetic and potential energy, so it maximizes entropy. Also, if you care to watch the online molecular dynamics simulation that I just linked to, you would see this stationary dynamic state with your own lying eyes! (Check also the SoD post linked above for useful parameter setting suggestions)

        http://pirila.fi/energy/kuvat/barometric_derivation.pdf

      • Pierre-Normand

        “So, all the molecule that contributes to the density of vertical velocity in the range [v, v+dv] at some level z also contribute equally to the range
        [-v, -v – dv] <—(corrected sign error)
        at that level, and vice versa."

      • Pierre-Normand

        Jim,

        Here is another reference that I had provided in another thread:

        “Another special case of the Boltzmann distribution (1.1) is the “barometric”
        distribution, giving the number density ρ(h) (number of molecules per unit
        volume) of an ideal gas of *uniform temperature* T as a function of height h
        in the field of the earth’s gravity. […]” (My emphasis)

        (See p.4)
        http://assets.cambridge.org/97805218/11194/excerpt/9780521811194_excerpt.pdf

        That the temperature is uniform isn’t assumed. It is derived. The Boltzmann distribution of energy for a classical mechanical system maximizes entropy. This result is also derived in Reif, ‘Fundamentals of Statistical and Thermal Physics’, the standard textbook that we used in my undergraduate course in statistical mechanics (p.210).

      • Pierre-Normand, you missed that gravity is asymmetric. When you start isothermal, molecules going down on average will be faster than those going up initially until you settle into a state where the vertical gradient of kinetic energy cancels the acceleration effect. Isothermal is not a sustainable steady state to thermal diffusion in gravity.
        Perhaps you can imagine all the molecules starting with the exact same speed (KE) and distributed uniformly with height. Collisions would be perfectly elastic, so those leave them moving with speed v. Now the only thing changing the speed is gravity. Can constant v be a steady state? No. You end up with faster ones at the bottom and slower ones at the top with the property that KE+PE=const.

      • Jim D
        “Can constant v be a steady state? No. You end up with faster ones at the bottom and slower ones at the top with the property that KE+PE=const.”

        You are in danger of suggesting that thermal energy (internal energy) can be turned losslessly into mechanical energy.
        This comes close to ignoring the Second Law of Thermodynamics

      • Pierre-Normand

        Jim D,

        I am not ignoring this effect, but you yourself are neglecting some of its consequences on the vertical density distribution and speed distributions. When a collection of N molecules with a one-dimensional vertical Maxwell velocity distribution rises from level z = h to a higher level h+dz (only considering the molecules that are moving up for simplicity) then all the molecules that *can* get to the higher level (N2 < N) lose an amount of kinetic energy m*g*dz. This means that *those* molecules have a lower average speed and lower average kinetic energy. However some molecules (N – N2) had a kinetic energy initially lower than m*g*dz. Those slower molecules drop out from the population that is able to climb to the higher level. This drop off effect accounts for the exponential shape of the barometric density distribution (which is an integral part of the thermodynamic equilibrium state).

        Since the slowest rising molecules among the population at level h don't contribute to the population at level h+dz, this drop off effect skews the speed distribution up. This effect exactly cancels the effect from the deceleration of the molecules that *can* make the travel from h to h+dz. This accounts for the molecular speed distribution being isotropic and stationary at all levels when they have identical Maxwell speed distributions at all heights (isothermal) and the exponential barometric density profile. Coombes and Laue make this calculation, and you can also observe the effect on online molecular dynamics simulators.

      • Pierre-Normand, you are saying that however high you go up in a gravitational field you will always see the same velocity distribution. This is impossible. The speed distribution shifts to slower values, just as it would for the mean ascending molecule. Imagine if you let the molecules go without collision. Each would trace a parabola, and the speed distribution at the top would decrease with height as they lose the vertical velocity component before going back down. Meanwhile at the bottom is where you would see the fastest ones.

      • Pierre-Normand

        Jim D,

        The molecules will indeed all describe parabolas and their vertical velocity of any molecules will drop to zero and reverse at the apex of the parabola. At the bottom of the box, the molecules have a vertical Maxwell velocity distribution that is a function of the Temperature at that level. This is something we can agree. Under those conditions the distribution of kinetic energy will be exponential. The higher you go, the less molecular trajectories (parabolas) reach all the way to that level. The molecular density drops exponentially. The slower molecules progressively drop from the distributions at the higher levels. The average kinetic energy therefore remains constant because although the individual molecules that *can* reach a higher level move slower than they did (or will) at lower levels, the slowest molecules contribute *only* to the distribution at the lower levels.

        “Since the Maxwell distribution depends exponentially on the kinetic energy, a shift in kinetic energy by a constant amount translates into an overall scale factor with no change in distribution of velocities”

        http://tallbloke.files.wordpress.com/2012/01/coombes-laue.pdf

        This is like having a teaching institution with several grades, and only the most successful students move up from one grade to the next while the least successful students drop out. Suppose the examinations get progressively harder at each successive grade. Each student performs a little worse at each successive test (‘lower KE’). Yet the average test results (‘average KE’) remains the same at all grades because only the brightest students move from one grade to the next.

      • Pierre-Normand, why do you think the molecules can rise without on average slowing down? The average molecule will rise and consequently slow down as KE converts to PE. The mean of the distribution will likewise reflect a shift from KE to PE higher up. The density doesn’t matter. It is the mean energy per molecule that gives you the temperature.

      • Pierre-Normand

        Jim D,

        Do you even read my posts? Can you maybe read again the previous one, a little bit slower? I must have acknowledged a dozen times already that *all* the molecules decelerate while going up, and they *all* are losing kinetic energy linearly with height. As you are traveling up, following the center of mass of some representative collection of molecules that are moving up from some lower level, they *all* are decelerating, and *all* are losing kinetic energy. There also always are several molecules at any level or any moment — the slowest ones — that are dropping out from the distribution of the still rising molecules, because they are falling back down, and hence aren’t rising any more. When you eliminate from a population the individuals that score the least at some test, this increases the average score. The two effects cancel out. I’ve proved this result mathematically already but you ignored the proof. You also don’t seem to want to give a single glance at a molecular dynamics simulation that would show you in an instant that there are seemingly as many fast molecules, proportionally, at any height, in spite of the constant downward acceleration of all the individual molecules (and contrary to the claims of your untutored intuition). You are uncharacteristically averting your gaze from the content of my arguments, from online simulations, and from the mainstream literature, all at once. This is a textbook example of a cognitive blind-spot. But I am hopeful that you will recover, eventually.

      • Pierre-Normand, you seem to ave the impression that even if all the molecules slow with height, their mean speed doesn’t. I keep telling you that is impossible. The whole distribution shifts with height. Eventually you get to a height where PE can overcome almost all the KE and the ones left are moving really slowly.

      • Pierre-Normand

        “Pierre-Normand, you seem to ave the impression that even if all the molecules slow with height, their mean speed doesn’t.”

        No. I agree with this sentence. It is true but it involves an equivocation. We ought to be speaking about two different molecular populations. All the molecules that are moving with an upward initial velocity from some level h initially slow down and lose kinetic energy. But not all those molecules reach any given higher level h+dz. So, the molecular population that reaches the higher level can’t be the same population as the initial population that was initially moving up. All the molecules that had an amount of kinetic energy KE < m*g*dz are unable to make the trip. They don't have enough total energy.

        So, the point that you keep missing: The population that reaches the higher level isn't the same population as the one that departed from the lower level with an initial upward velocity. There are very many molecules that don't have enough (vertical) kinetic to make the trip because the mode of the (vertical) Maxwell distribution of velocities (which is a Gaussian distribution) is zero. Being at rest is the peak of the velocity pdf. So, when you are tracking all the molecules that move from h to h+dz, and removing from the initial population the molecules that never are making it to the higher level, the resulting distribution (and average kinetic energy) doesn't change, as calculation shows, because removing from the distribution the slowest molecules skews the distribution *up* and thereby cancels the uniform shift to lower speeds that results from the downward acceleration of gravity

        Coombes and Laue, and Pekka, offer two independent proofs of this fact. In fact, Coombes and Laue offers two proofs. So that makes three proofs. And molecular dynamics simulations seemingly agree with them and not with you. They show the result that you deemed impossible.

      • Pierre-Normand, first of all we are talking about a kinetic energy distribution where the mode is not zero. This is a distribution that can and does shift its peak towards lower values with height because some goes into potential energy. It’s a concept of energy conservation that necessitates this. Individual particles undergo this shift, two particles added together do, ten do, and so on to zillions of particles constituting a gas. These are concepts you should not need to prove.

      • Pierre-Normand

        “No. I agree with this sentence.”

        Sorry, sorry. I meant to say that I agree that the claim is false. I agree with your denial of it. But it involves an equivocation, for the reasons I gave.

      • Coombes and Laue seem to be saying, yes, the KE decreases, but because the shape of the distribution is still MB, the temperature doesn’t. Huh? There are equations that directly relate temperature to the mean kinetic energy. If KE decreases, the temperature decreases.

      • Pierre-Normand

        Jim D, I meant what I wrote. Molecules have both velocity distributions and kinetic energy distributions. KE = m*v^2/2. The two distributions aren’t the same. The former is Maxwell’s distribution of velocities (along the z axis) and the latter is an exponential obtained by the Boltzmann distribution law. At temperature T, the former is

        f(v_z) ∝ exp(-m*v_z^2/2kT),

        while the latter is

        f(KE_z) ∝ exp(-m*v_z^2/2kT) = exp(-KE_z/kT)

        The former is a Gaussian probability distribution of velocities while the latter is a (negative) exponential distribution of energies.

        The mode of the Gaussian is zero, because it is centered at zero.

        My explanation of what happens with velocities stands. The explanation in terms of energies is different but, obviously, consistent with it. When the distribution of energies is shifted to the left by a constant amount, because all the rising particles lose energies while rising to a higher level, the least energetic molecules (with the lowest KE) are shifted to the left of the vertical axis.They are represented as having negative kinetic energies, but this just means that they aren’t part of the distribution anymore. They fell back down before reaching the higher level. f(KE_z = 0) therefore is the mode of the distribution, consistently with the mode of the Gaussian for velocities, since molecules can’t have negative kinetic energies. The most likely kinetic energy is zero.

        All the molecules that still have positive energies after the shift now have the exponential distribution represented on the right of the vertical axis. But since this is a pdf, it must be renormalized to 1. It is a basic property of exponential functions that when shifted and renormalized they are invariant. You still have the same ratio of molecules that are travelling with KE1 and KE2 after a uniform shift in kinetic energy for all the molecules. That is f(KE1)/f(KE2) is invariant for any two values KE1 and KE2. And therefore, the temperature and velocity distributions also are invariant with height. The only change, which we have cancelled through the remormalisation, is a change in the number of molecules that had enough KE to climb all the way from h to h+dz, The drop off rate yields the barometric density formula. It is obtained thus:

        f2(KE_z) = exp((-KE_z – m*g*dz)/kT) = exp(-m*g*dz)*exp(-m*v_z^2/2kT)

        The probability to find a particle with any given kinetic energy decreases exponentially with height (and also decreases exponentially with KE). Since the vertical kinetic energy distribution is the same at all heights, so is the vertical velocity distribution.

      • Pierre-Normand

        Jim D, In case you are wondering how the mode of the speed distribution in three dimensions has a finite value while the mode of the one-dimensional velocity distributions is zero, this is explained here:

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653603

      • Pierre-Normand

        Jim D wrote: “Coombes and Laue seem to be saying, yes, the KE decreases, but because the shape of the distribution is still MB, the temperature doesn’t. Huh?”

        You misread them. The distribution isn’t just *any* Maxwell speed distribution at all heights. It is the *same* Maxwell speed distribution f(v_z) = exp(-mv_z^2/2kT). It’s also the same T at all heights. There is a lower probability to find a molecule at a higher level with a speed in the range [v, v+dv] just because there are fewer molecules at a higher level — because of the barometric density distribution. The speed distribution is the exact same, and so the average kinetic energy also is the same at all levels.

      • Pierre-Normand

        “That is f(KE1)/f(KE2) is invariant for any two values KE1 and KE2.”

        This is very badly expressed, sorry. I mean that for any two values KE1 and KE2 for the kinetic energy, the ratio f(KE1)/f(KE2) is invariant with height. At a higher level there are fewer molecules with either KE1 or KE2 kinetic energies, because the molecular density is reduced, but the ratio is the same. And hence, the average, KEavg, is the same.

      • Jim D,

        In the discussion the sentence appears:

        All molecules are slowing down, but their average velocity does not.

        That’s actually wrong. Half of the molecules are slowing down, while the other half is speeding up (those coming down).

        The correct statement is

        The average velocity of particles is the same at every height.

        The comparison involves different sets of molecules. Those that are slowest at lower altitude are not going much further up and have not come down from much higher, while the fastest may. The average velocity of those molecules that will contribute or have contributed to the average at higher altitude is now higher, but including those that are very slow brings the average to the same value.

        The above is easy to verify by direct calculation for noninteracting molecules and that is what Coombes and Laue, and many others have done.

        You continue to repeat that the result is impossible, but you have clearly never tried to check the calculation, which proves that your argument is wrong.

      • Pierre-Normand

        Pekka wrote: “…The average velocity of those molecules that will contribute or have contributed to the average at higher altitude is now higher, but including those that are very slow brings the average to the same value.”

        That’s a nice, accurate and concise explanation, but I fear that it may confuse some. I am pretty sure you mean to express:

        “The average velocity of those molecules [at lower altitude] that will contribute or have contributed to the average at higher altitude is now higher, but including those that are very slow [and hence can’t contribute to the average at higher altitude] brings the average to the same value.”

      • P-N
        I had difficulties in finding a concise way of getting all parts of the argument together. The main source of difficulty is due to the need to answer to questions that refer to temporal development, when the basic statement

        The average velocity of particles is the same at every height.

        concerns situation at a single moment.

        I expand on the explanation to make it more clear (but less concise).

        At a single moment the particles at different heights are always distinct. Their present vertical velocity tells, however, on their recent history and near future. It tells that those with high positive vertical velocity will soon be at a higher altitude and those with high negative vertical velocity were recently at a higher altitude. Thus these particles belong to the class of particles that contributes also at higher altitudes, while those with near zero vertical velocity are not part of that subset.

        This division into subsets is clear for non-interactive particles allowing for the calculation of Coombes and Laue. When the mean free path is small, all effects are small for the interval between two collisions, but the effects are not zero. Some particles have a positive but too small vertical velocity to make their next collision before they start to fall. That small effect is enough to reduce the denominator (number of particles) in the calculation of the average speed at a slight higher altitude as much as the numerator (sum of speeds of the particles) decreases. When both decrease by the same factor the average speed is unchanged. This is what happens in thermodynamic equilibrium.

      • Pierre-Normand, December 10, 2014 at 11:07 pm:

        “The higher you go, the less molecular trajectories (parabolas) reach all the way to that level. The molecular density drops exponentially. The slower molecules progressively drop from the distributions at the higher levels.”

        This is an important point that people seemingly tend to ‘forget’ about (or try to ‘explain’ away).

      • Pierre-Normand

        Jim D, and other isentropo-equilibrists,

        Back from the coffee shop (Tim Horton’s) I came up with another simple and intuitive demonstration/design that the isentropic profile has a lower entropy than the isothermal profile. No calculation required. No explicit consideration of Gibbs free energy, molecular velocity distributions, etc.

        Imagine two tall and thin cylindrical containers filled with the the same amount of the *same* ideal gas with a short mean free path, and hence a low thermal conductivity. We are using two containers just for purpose of comparison. We are not going to make them interact, so this is quite different from the earlier challenge that Vaughan and I independently proposed.

        The first column initially has an isothermal profile while the second column initially has an isentropic profile (i.e. dry adiabatic lapse rate; constant potential temperature). Both columns are in hydrostatic equilibrium. They both initially have the same total energy TE = U + summation_i(m*g*z_i) (Internal + total gravitational potential energy). They are mounted on pivots attached to their mid sections so that they can be rotated upside down.

        We are going to test the consequences of two assumptions (1) The isothermal profile is the equilibrium state and (2) The isothermal profile is the thermodynamic equilibrium state. We are going to assume that, given a sufficiently long time, a column would evolve towards its equilibrium profile from any other profile through molecular diffusion alone.

        Lemma:

        Rotating the isothermal column by 180° *consumes* some positive amount or work W1, while rotating the isentropic column *produces* some positive amount of work W2.

        Proof of lemma:

        After the isothermal column has been rotated, the barometric pressure profile is reversed (though not exactly). So, the gas parcels that were originally on top are adiabatically compressed and those that were originally at the bottom are adiabatically expanded. This brings the temperature profile closer to the isentropic profile. Therefore the density of the gas parcels at the bottom now has decreased (compared to the earlier situation at the bottom), since the temperature has increased and the weight of the column above is the same, while the density of the highest gas parcels is decreased, since the gas parcels are colder. The overall gas column has been lifted, on average, and it now has more potential gravitational energy. Hence a finite amount of work W1 must have been consumed to effect the full rotation of the cylinder. (We will see shortly that there can’t be any change in internal energy just yet).

        Consider now that while some gas parcels have been adiabatically compressed and others have been adiabatically expanded, all those processes are thermodynamically reversible. The full rotation of the column is an isentropic process, irrespective of the initial temperature/pressure profile, provided only that the time-frame is short enough to neglect diffusion. So, we can rotate again the column back to its original orientation and the initial profile will be rapidly restored. From which we can conclude that a 180° rotation that brings the profile closer to isothermal (further away from isentropic) produces some finite amount of work. This proves immediately the second part of the lemma. (One can start by considering the second cylinder rather than the first, and prove the two parts of the lemma with an analogous reasoning).

        Continued in the next post…

      • Pierre-Normand

        …continuation,

        Consequences of the assumption #1:

        After we have rotated the initially isothermal column, and performed some amount of work W1 to do so, we wait for the gas to come back to equilibrium again through molecular diffusion. Since there will be a release of gravitational potential energy as the gas settles down to the initial temperature and density profiles, the internal energy will increase by the amount dU = W1 (from conservation of energy). We can repeat the whole process n more times and the internal energy will end up being U + (n+1)dU. The entropy will increased as we add mechanical energy to the system (in the form of gravitaional potential energy) and let this energy dissipate withing the system. This device is essentially a heat pump.

        Consequences of the assumption #2:

        After we have rotated the initially isentropic column, and recovered some amount usable work W2 to do so, we wait for the gas to come back to equilibrium again (and restore an isentropic profile) through molecular diffusion. Since there will be an *increase* of gravitational potential energy as the gas settles down to the initial density profile (and the warming gas at the bottom is going to push the rest of the column up), the internal energy will *decrease* by the amount dU = W1 (from conservation of energy). We can repeat the whole process n more times and the internal energy will end up being U – (n+1)dU. The entropy will decrease as we extract mechanical energy to the system (in the form of gravitational potential energy) and let this mechanical energy be restored from internal energy through molecular diffusion withing the system.

        Conclusion, spontaneous evolution from an isothermal profile towards an isentropic profile reduces entropy and enables the conversion of heat into usable macroscopic work.

      • Pierre-Normand

        “We can repeat the whole process n more times and the internal energy will end up being U – (n+1)dU.”

        This is not exact, but may be close enough for the first few iterations of the process. As the gas cools towards absolute zero, the adiabatic profile will change and the amount of work produced will diminish. We can of course warm the cylinder back to the initial internal energy U through recharging it with a large external heat reservoir (such as the Pacific ocean), and therefore extract heat from this reservoir and convert it to work.

      • Pierre-Normand

        Correction: “Hence, a finite amount of work W1 must have been consumed to effect [a complete half-]rotation of the cylinder.”

      • Pierre-Normand, at this moment I don’t have time to study your example, but it really is simpler than that. An isothermal profile has a potential temperature gradient. In conditions of no heating, potential temperature is the conserved quantity (derivable from T*dS=dQ=cp*dT-V*dp, under dQ=0). The maximum entropy state is one of uniform potential temperature, as it is with any conserved variable that is mixed with the fluid. This is all about finding the appropriate conserved variable in a compressible gas under gravity, and it is not temperature. A uniform mixture of a conserved variable is always the maximum entropy state. A closely related variable is the dry static energy, s=cp*T+g*z which can be derived from the thermodynamic relation with the aid of the hydrostatic approximation in gravity. This is also constant in an isentropic state as dT/dz=-g/cp which is the dry adiabatic lapse rate. s explicitly shows you how PE and enthalpy sum to a conserved quantity when there is no heating (dQ=0) and gravity. Enthalpy (H=cp*T) decreases as PE (g*z) increases.

      • Pierre-Normand

        Jim D: “In conditions of no heating, potential temperature is the conserved quantity (derivable from T*dS=dQ=cp*dT-V*dp, under dQ=0).”

        Those statements all are true as applied to the vertical displacement of an air parcel under the condition that the profile already is isentropic. In order to account for the full change in the air column, one must also consider the descending air parcels that fill up space vacated by the rising parcels, and this involves adiabatic compression (and expansion) work, but those processed also are balanced globaly under the condition that the profile (along the travel path of the rising and descending air parcels) already has an isentropic profile.

        What my thought experiment shows plainly, however, is that when one considers not only such invariant changes within the isentropic gas column, as described above, but also the spontaneous diffusion process that you believe would bring an initial profile from isothermal towards isentropic, *then*, mechanical work would be released (in the form of gravitational potential energy) and the gas would lose internal energy as a compensation. This violates the second law and contradicts your claim that the total entropy in the isentropic gas column is larger than the total entropy in the isothermal column (assuming hydrostatic balance, and conservation of total energy, including gravitational potential energy)

        Since you are busy at the moment, here is the gist of my thought experiment.

        If you start with an isolated isothermal gas column and let it relax through internal diffusion towards its equilibrium (or so you allege) isentropic state, then the pressure and density profiles will adjust in a way that raises the center of gravity of the column. If you then pivot the column upside down, accounting for the resulting expansion/compression of all the air parcels which result from the new hydrostatic balance within the inverted column (as explained in more details in my thought experiment) then you will be extracting some usable mechanical work, and the total energy of the column will decrease. This process can be iterated after the column has settled again to an isentropic state through molecular diffusion (as you allege it would), albeit at a lower overall temperature . You are thus able to extract almost all its internal thermal energy as usable mechanical work.

        No such paradox occur when gases are allowed to relax from an isentropic profile towards an isothermal profile. In that case one must *supply* mechanical work to bring the profile closer to isentropic from an initial isothermal state (though pivoting the column upside down), and the internal energy and entropy both increase during the diffusion process that brings the profile back to isothermal (albeit at a higher temperature).

        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-654616

      • Pierre-Normand

        P-N: “From which we can conclude that a 180° rotation that brings the profile closer to isothermal (further away from isentropic) produces some finite amount of work. This proves immediately the second part of the lemma. (One can start by considering the second cylinder rather than the first, and prove the two parts of the lemma with an analogous reasoning).”

        This is not true, actually. The first sentence is correct, but the second part of the lemma isn’t proven. It’s likely false. I now think the mechanical work required to flip the second cylinder (with the initially isentropic profile) upside down is zero. I will have a adjust this thought experiment.

      • Pierre-Normand

        So, thinking more about it, my analysis of the second case — the column with an initially isentropic profile — was flawed. The isentropic profile is invariant under the permutation of any two air parcels (that are allowed to adjust adiabatically to the local pressure). Turning the column upside down, and letting it settle non-dissipatively to a new hydrostatic equilibrium, amounts to permuting air parcels (with equal mass) pairwise around the ‘middle parcel’ (i.e. the parcel at the center of mass of the thin column). So, the column ends up with the exact same density and temperature profile that is had before the 180° rotation. The center of gravity doesn’t change and the total mechanical work is zero. This result ought to please Jim D.

        The flip side, though, is that no amount of pairwise air parcel permutations is going to yield an isothermal profile. There does not exist an adiabatic path between the isothermal and isentropic profiles in any direction. If one of those two states is a thermodynamic equilibrium, then there *is* a diffusive path from the non-equilibrium state to the equilibrium state. (They are both convectively stable, as are all the intermediate states, so the path must be entirely diffusive). This means that even if Jim D were right that the isentropic profile is an equilibrium, it would still not be achievable only through an adiabatic ‘mixing’ of air parcels, however small the parcels are. Diffusion is essentially different from (adiabatic) mixing, since mixing is thermodynamically reversible (the entropy of all the parcels is constant, and therefore so is the total entropy) while diffusion away from a non-equilibrium initial state is irreversible.

        There also was a major flaw in my treatment of the first case that is now apparent. Since the 180° rotation of the isentropic column leaves the resulting profile unchanged, it follow that the 180° rotation of the isothermal profile doesn’t bring it closer to the isentropic profile. It rather overshoots it by a long shot, since, as compared with the initially isentropic column, the bottom parcel is already colder and the top parcel is already warmer even before they are respectively expanded and compressed as a result of the 180° rotation of the column. Hence, the rotation of the isothermal column results into a convectively unstable gradient that is steeper than the adiabatic lapse rate.

      • Pierre-Normand, mixing and diffusion are not at all reversible processes. They conserve quantities but not entropy. You can’t unmix or undiffuse anything unless you gain something like potential energy (e.g. settling out denser fluids). An isentropic gas will not settle out into an isothermal one because that has a lower entropy. An isothermal gas in gravity will eventually diffuse, or can be mixed, into an isentropic state. Without gravity it is fine, but with gravity you have that extra energy gradient that is available to be redistributed to equalize the entropy. An energy gradient leads to an energy flux by even the smallest diffusion, which in the case of an isothermal state is a downward flux.

      • @Jim D: You can’t unmix or undiffuse anything unless you gain something like potential energy (e.g. settling out denser fluids).

        Yes, exactly (though you presumably meant “lose” rather than “gain” since gravity acts to convert PE to KE). But surely that (locally) entropy-decreasing “settling out” process is precisely what’s going on here. The faster molecules in a parcel contribute the most to diffusion. The pressure at the top of the parcel being less than at the bottom, the faster molecules will prefer to exit the parcel through the top. Heat therefore migrates up. But the faster molecules also gain PE as they do so, thereby cooling. The two effects cancel, exactly if Maxwell and Boltzmann are to be believed, with the upshot being an isothermal distribution.

      • Vaughan Pratt,
        “Heat therefore migrates up. But the faster molecules also gain PE as they do so, thereby cooling. The two effects cancel, exactly if Maxwell and Boltzmann are to be believed, with the upshot being an isothermal distribution, ”

        Right, and for the two effects to exactly cancel you would probably need a physical containment system. Well, in addition to the perfectly elastic balls that prefer not to spin very often and have infinitely small mass so the gravitational attraction between the perfectly elastic balls is negligible.

        Other than we are home free :)

        My take on this is if someone says there appears to be a gravito-thermal effect, you let them run with their idea then figure out which “effect” requires the most perfection. To me “equilibrium” is like the Steven Wright joke, “You know how it feels when you’re leaning back on a chair, and you lean too far back, and you almost fall over backwards, but then you catch yourself at the last second? I feel like that all the time…”

      • @cd: for the two effects to exactly cancel you would probably need a physical containment system.

        Agreed. A uniform planar gravitational field, one whose strength does not decrease with altitude, should be a sufficient physical containment system, since escape velocity is infinite . I would expect a rigorous proof of isothermality to depend on uniformity of the field.

        Escape velocity is finite for any body’s gravitational field that falls off quadratically with distance from the center of the body. Your argument that a finite escape velocity militates against exact isothermality is quite convincing (well, it convinces me anyway). I like it because it’s such a simple argument. :)

        However each additional 100 km of altitude only decreases Earth’s gravity by π per cent (a linear approximation that starts to fall off at around 500 km of altitude where gravity has decreased by only 14% rather than 5π = 15.7%). Since only exp(-100/7.6) = 0.000002 of the atmosphere is above 100 km (taking 7.6 km as the scale height of Earth’s atmosphere), the assumption of a uniform gravitational field is not unreasonable for the bulk of Earth’s atmosphere.

        (The π% figure is one of several consequences involving π of defining the meter to be one ten millionth of the distance from the North Pole to the equator via Paris.)

      • Pierre-Normand

        Jim D,

        You misread me. I agree that molecular diffusion in a gas is an irreversible process when it carries the system form lower to higher entropy states. This is indeed the basis for my argument that random molecular diffusion is quite unlike the process of adiabatic ‘mixing’ of arbitrarily small air parcels that are moving about and undergoing adiabatic volume variations, as you conceive ‘mixing’ to be (and requires for going on an external source of energy that you forget to account for — in the case where the initial state isn’t already ‘isentropic’). That’s because although *this* process *does* conserve ‘relative temperature’ in a column of gas that has an adiabatic lapse rate, it can’t reduce the entropy in *any* column of gas, whatever its initial lapse rate, since it is, by stipulation, an adiabatic process. One must allow for heat transfer (diffusion) across the boundaries of air parcels in order to allow the entropy of the gas column to change. ‘Mixing’ as you conceive it conserves entropy by definition. It relies on the ‘air parcel’ idealization for an idealized fluid, which doesn’t allow for any heat conduction across system boundaries and doesn’t allow for any variations in entropy. Real irreversible diffusion acts *against* ‘mixing’ on the vertical temperature profile, and unlike ‘mixing’, can operate in an isolated gas column with no need of a constant external input of Gibbs free energy.

      • Pierre-Normand

        P-N: “(and requires for going on an external source of energy that you forget to account for — in the case where the initial state isn’t already ‘isentropic’ [and the gradient isn’t initially steeper than the adiabat]”

      • Vaughan Pratt, ” the assumption of a uniform gravitational field is not unreasonable for the bulk of Earth’s atmosphere.”

        Agreed, there would be some solar/lunar tidal distractions, if you have that much spare time on your hands. In general, I would give the G-T effect a Snopes rating of possible but not very large. Definitely not something you could count on getting any useful energy from or that would replace all the other physics involved in a real atmosphere.

      • In general, I would give the G-T effect a Snopes rating of possible but not very large.

        I would expect that a gravitational field that decreases with height would create a slightly negative lapse rate, i.e. hotter at higher altitudes (but probably less than a millionth of a degree hotter per km). This is because the rising hotter molecules would lose slightly less KE due to the decrease in gravity at higher altitudes. Also they would be fighting against less pressure from above in order to rise one meter.

      • Vaughan Pratt, “I would expect that a gravitational field that decreases with height would create a slightly negative lapse rate, i.e. hotter at higher altitudes (but probably less than a millionth of a degree hotter per km).”

        With an N2 atmosphere I think there would be a negative lapse rate with closer to 0.5 C per kilometer with the mainly equatorial day air creating the inversion and the surface being closer to the average which would include a diurnal cycle. It is a thought experiment after all.

        With a mix of gases with more degrees of freedom you could find more twists, but an absolutely stable isothermal atmosphere would be my last guess.

      • Pierre-Normand

        Vaughan Pratt wrote: “I would expect that a gravitational field that decreases with height would create a slightly negative lapse rate, i.e. hotter at higher altitudes (but probably less than a millionth of a degree hotter per km). This is because the rising hotter molecules would lose slightly less KE due to the decrease in gravity at higher altitudes. Also they would be fighting against less pressure from above in order to rise one meter.”

        Vertical variations in the strength of the gravitational field ought not cause any deviation from the isothermal equilibrium profile, I don’t think. It would only modify the pressure profile. The terms of the Bolzmann distibution of energies that correspond to the gravitational potential and to the three components of the kinetic energy would still factorize. As for the second point, the molecules don’t care about the rates of collisions, of which there indeed are more in the denser regions. If the incoming molecules before the collisions that occur at any point in space have an isotropic velocity distribution, then the outgoing molecules after the collision at that point also will have an isotropic velocity distributions, and the average kinetic energies of the outgoing molecules will be the same just because of energy conservation for the individual collisions.

        Victor T. Toth makes a more formal argument on the basis of Liouville’s theorem, but the salient point for me is the factorisability of the Bolzmann distribution.

        http://www.vttoth.com/CMS/physics-notes/72-on-the-barometric-formula

      • Pierre-Normand, adiabatic processes don’t necessarily conserve entropy. The isothermal expansion of a gas into a box doesn’t (e.g. by removing a partition), and nor does the mixing of two gases with different potential temperatures. The latter is an analogy of an isothermal state going to an isentropic one. You can conserve potential temperature in adiabatic processes without conserving entropy. The key for conserving entropy is reversibility, which mixing and rapid expansion aren’t.

      • Pierre-Normand

        Jim D wrote: “Pierre-Normand, adiabatic processes don’t necessarily conserve entropy. The isothermal expansion of a gas into a box doesn’t (e.g. by removing a partition), and nor does the mixing of two gases with different potential temperatures.”

        Yes, you are right about the first case — Joule expansion. My focus was on the slow (i.e. slow relative to average molecular velocities) adiabatic volume changes where variations in internal energy match the work performed by the air parcels on the surrounding. The second part of your sentence doesn’t make much sense, though. The potential temperature of a gas — which I mistakenly called “relative temperature” a couple times, I think — is the temperature that it acquires after it is brought to some reference pressure while being allowed to perform work: dW = P(t)dV integrated over the duration of the process. If the process allows for molecular diffusion through the air parcel boundary, or if the rate of expansion is so fast that dW < PdV, then potential temperature isn't conserved. The idea of mixing together two gases with different potential temperatures just doesn't make sense.

        "The latter is an analogy of an isothermal state going to an isentropic one. You can conserve potential temperature in adiabatic processes without conserving entropy. The key for conserving entropy is reversibility, which mixing and rapid expansion aren’t."

        I agree but I think those considerations work against you. For convective heat transport to bring a lapse rate towards the dry adiabat from below, you need a process that can be conceptually decomposed into two steps. First, you need to vertically displace an air parcel up (or down) against a finite hydrostatic force since, as it expands (compress), its temperature drops (increases) faster than the ambient temperature and the buoyant force becomes weaker than its weight. And hence you must perform work. This first part of the process is thermodynamically reversible; it conserves entropy. Second you must allow the air parcel that is now cooler (warmer) than the surrounding to equilibrate temperature with the surrounding through mixing/diffusion. Else, after the external source of work is depleted, the parcel would simply fall back to the level where it came from. This second process is thermodynamically irreversible and it increases entropy. The net result has been to move the temperature profile closer to the dry adiabat, to increase the total energy of the column, and also to increase its entropy (unless it has shed to space as much heat as it has been supplied energy for sustaining the circulation — i.e. the first step — in which case the total entropy decreases).

        In a steady state where the column is allowed to cool (e.g. radiate to space) while an external energy source maintains the circulation required to keep it close to the dry adiabat, the entropy of the column is constant and the rate of cooling equals the rate of energy input. If the column is thereafter isolated, then the energy source required to maintain the first part of the process (adiabatic circulation of air parcels) is shut down. Convection dies off while heat still is allowed to flow conductively against the temperature gradient and brings the column towards the isothermal profile while increasing its entropy.

      • Pierre-Normand

        “…and the buoyant force becomes weaker than its weight”

        I meant: …and the buoyant force becomes weaker (stronger) than its weight, respectively.

      • Pierre-Normand

        P-N: “This second process is thermodynamically irreversible and it increases entropy.”

        …Or maybe not. See the following for a more thorough treatment of the entropy variations:

        http://moyhu.blogspot.ca/2014/10/calculating-environmental-lapse-rate.html

      • @PN: Vertical variations in the strength of the gravitational field ought not cause any deviation from the isothermal equilibrium profile, I don’t think.

        (a) Why would you think that?

        (b) How do you account for the part of the atmosphere that achieves escape velocity? The capn makes a very good argument. No matter what the parameters, the tail of the Maxwell-Boltzmann distribution will extend into that region. And the amount of that tail will depend on the parameters.

        (c) How do you counter my argument that decreasing gravity will move both of the two cancelling effects towards a negative lapse rate (increasing temperature with increasing altitude)?

        For a uniform gravitational field you’re in good and long-established company. With a field decreasing quadratically with distance from the planet’s center I would expect you’re striking out on your own. You’ll need to show us your paddle to persuade us you’re not up the creek without one.

      • I have to agree with Jim D. that potential temperature is conserved in the case of heat transport by convection. After all it’s the motivating property of potential temperature.

        What I don’t understand is why he believes that potential temperature is also conserved by other transports such as conduction or diffusion. So far his only proof seems to be that if it is preserved by convection then it must also be preserved by other transports. That’s not a terribly convincing argument.

      • Escaping particles mean that the state is not in exact thermodynamic equilibrium. Thus it need not be (and probably cannot be) isothermal.

        As long as escaping particles can be ignored, the equilibrium state is isothermal as far as I can judge, but the virial theorem cannot be applied, when the power law is not valid for for the potential at all heights. The power +1 is only approximate and the power -1 is seriously violated by the impenetrable surface of the planet itself. This second point leaves the equilibrium state isothermal, but seems to affect very slightly the average potential energy. A similar correction comes from the difference between the geometry of the spherical planet and the planar case considered. (It’s possible that we find a cancellation of corrections, when a full calculation is done.)

      • I think that there has been some confusion related to different uses of the word isentropic and isothermal in this discussion.

        Both words are used widely for flow processes in gas. In that case a parcel of gas is considered. It’s assumed that the parcel is in local thermal equilibrium, but the whole system is not (otherwise we have not processes, where anything flows or changes). Isothermal processes are not of interest here. Isentropic processes are an idealization. Fully isentropic processes are not possible in real gases as some diffusion is always present when flows occur. In the idealization, where all subprocesses of the full system are isentropic, the total entropy does not change. Thus the system does not approach the equilibrium, but remains at the constant distance from the equilibrium when measured by the difference in total entropy.

        In real gases we have rather a “non-conservation law of entropy” than a conservation law as the entropy of an isolated volume of gas increases always unless the system is in thermodynamic equilibrium that’s reached only asymptotically.

        The concept of an isothermal state is clear, but an isentropic state is not a commonly used concept. Without gravity or other similar potentials isentropic is also isothermal and thus a state of thermodynamic equilibrium. Under gravity an isentropic state is an idealization that’s not fully stable and not thermodynamic equilibrium. It’s stable against flow processes, but it’s not stable against heat conduction, which moves it towards the thermodynamic equilibrium of an isothermal state.

      • Pierre-Normand

        To be clear, in this thread I’ve always used “isothermal profile” to refer to a property of the combined vertical temperature and pressure profiles such that the (slow) vertical displacement of any air parcels would be a reversible *process* that conserves potential temperature as it adjusts to the pressure of the surrounding. So, the therm isentropic really characterizes a process. I agree it is better simply to call this profile the adiabatic lapse rate.

      • Pierre-Normand

        Vaughan Pratt,

        I was specifying equilibrium conditions. As Pekka notes, loss of particles prevents equilibrium from being reached. It’s analogous to evaporation. It cools the TOA. So, what I am saying applies either to a tall gas column enclosed in a box, or a conditions where as many particles come down from infinity as do reach escape velocity.

        The latter isn’t a fantastic idealization either since it is a common property of star cluster formations in young galaxies, or gravitational collapse of gas clouds from initial inhomogeneities in an initially uniform thermal bath of radiation and particles. In both cases, in spite of the initial gravitational ‘heating’ from contraction, the star cluster tends towards an ‘isothermal’ Maxwell distribution of speeds since the loss of stars from the cluster is compensated by the random capture of new stars in the young ‘isothermal’ galaxy. The same is true for the gas cloud after its temperature allows it to balance the gravitational force and it isn’t contracting anymore. The Virial theorem thereafter still applies to the evolving collection of star (or molecules) since the escaping stars have the same speed distribution as the newly captured stars that replace them. It’s akin to enclosing the star cluster into a large spherical container such that the fastest stars, including those that have escape velocity, bounce back with the same speed and kinetic energy.

        Regarding your second objection, I think it is misguided. The two cancelling effects in the case where g is uniform with height are (1) the slowdown of the particles that are climbing the gravitational potential, and (2) the drop off from the molecular populations at any level of the molecules that have the least amount or total mechanical energy (PE+KE). As Coombes and Laue have showed, those two effects cancel out. If the gravitational acceleration decreases with height, then there is less of a slowdown of the rising molecules and also less of a drop off rate. So, the two effects both are smaller and they still cancel out exactly, as Toth demonstrates in the link that I provided (towards the end of the webpage).

      • P-N,
        Do you really mean “isothermal profile” in your latest comment to me. I think, you must have had “isentropic profile” in mind.

      • Indeed, in response to Pekka, isentropic processes are difficult to find, and only exist when the state is isentropic. In an isentropic state, which for the atmosphere would be a dry adiabatic lapse rate, no amount of mixing will change that state. In any other non-isentropic state, mixing will increase the entropy, and this includes the isothermal state as an example. Vaughan Pratt wonders why I think potential temperature is a conserved property under other processes. In a gas, potential temperature is the conserved thermodynamic quantity in the absence of any diabatic effects. It is derived from dQ=T*dS=cp*dT-V*dP. Setting dQ=0 and using the ideal gas law you can find that a conserved quantity is cp*dT/T-R*dP/P and define a quantity theta=T/(P/P0)^R/cp which is potential temperature and where P0 is a reference pressure where theta is chosen equal to T (sea-level pressure). The only condition to conserve this is that dQ=0 (adiabatic processes). Conduction and diffusion are adiabatic processes assuming the gas is isolated and has no heat flux through any boundaries. We also see that dS=cp*d(theta)/theta, so that log(theta) is proportional to entropy.

      • Pierre-Normand

        “I have to agree with Jim D. that potential temperature is conserved in the case of heat transport by convection. After all it’s the motivating property of potential temperature.”

        Yes, I am also agreeing with him on that score.

        “What I don’t understand is why he believes that potential temperature is also conserved by other transports such as conduction or diffusion. So far his only proof seems to be that if it is preserved by convection then it must also be preserved by other transports. That’s not a terribly convincing argument.”

        Exactly. So far as I can see, his main mistake is to assimilate two completely different sorts of ‘mixing’ processes, because both of them can be enhances at once through stirring a pot with a wooden spoon. The processes still are radically different. The first requires (or generates) work and is isentropic. This is the displacement of macroscopic air parcels that are allowed to contract and expand while dU = dW = P(t)dV at all times. The second process is the equilibration of thermal gradients from molecular diffusion, and is spontaneous (doesn’t require work) and irreversible. The first process is reversible. The simultaneous operation of the two processes drives the gradient towards the adiabatic lapse rate, but it requires an energy input (or an initial profile that is steeper than the adiabat and hence unstable). When the external energy source is shut of, the first process dies down and only the second remains. The heat flow bring the column back towards isothermal (regardless of the gravitational acceleration spatial variations, so long as the external force field is stationary and conservative along closed paths).

      • P-N, “To be clear, in this thread I’ve always used “isothermal profile” to refer to a property of the combined vertical temperature and pressure profiles such that the (slow) vertical displacement of any air parcels would be a reversible *process* that conserves potential temperature as it adjusts to the pressure of the surrounding.”

        I was a bit confused on whether you were considering totally isothermal or an isothermal profile. There is not a huge difference between an isothermal profile and an adiabatic profile. The largest changes are when actual specific heat capacity ratios are considered which would be when there are mixed mass particles that could be segregated by gravity, JimD’s case.

        That is when I mentioned for a perfectly isothermal atmosphere you would need real containment, gravity by itself would likely not do the job especially if there is rotational velocity to be considered.

        With an isothermal versus adiabatic profile it is easy to see why some would think the G-T effect might provide usable energy, but then there is an issue with energy density versus a temperature gradient. In other words, potential energy cannot be converted to work without movement and a few bouncing molecules would not produce much energy. You would need the same number of molecule at different kinetic energy to produce a flow and at the top of the column there would be a small percentage of molecules with lower/higher than average KE.

        I could see a very tall useless, perpetual motion toy for a huge skyscraper atrium being built in some place like Dubai or some other place with lots of money to throw away on “art”. I think a serious glass enclosure would be recommended though.

      • Pierre-Normand

        “Do you really mean “isothermal profile” in your latest comment to me. I think, you must have had “isentropic profile” in mind.”

        Yes, absolutely. Thanks Pekka. (I also sometimes say troposphere when I mean stratosphere. Never trust my words.)

      • Jim,

        What is heat conduction?

        It’s transfer of heat from one subsystem to another. It’s transfer of heat from the warmer subsystem to the cooler one. That’s not adiabatic for either of the subsystems. That increases entropy of the combination of these subsystems. You cannot discuss anything of interest in thermodynamics without dividing the overall system to subsystems. Everything that you have discussed involves subsystems.

        You can find statements that thermodynamic equilibrium is isothermal in virtually every textbook that discusses thermodynamics. As an example you can find the following sentences in Pierrehumbert’s book in the chapter 2.3.3 Entropy, reversibility, and potential temperature; the Second Law:

        The Second Law is perhaps more intuitive when restated in the following way: In an energetically closed system, heat flows from a hotter part of the system to a colder part of the system causing the system to evolve toward a state of uniform temperature.

        Further in the chapter he discusses the potential temperature, but nowhere is any hint that the above sentence should be modified or that it were true for the potential temperature. As virtually every physicist, he knows well that the sentence is true for the temperature, not for the potential temperature.

      • Pierre-Normand

        For the discussion of isothermal self gravitating systems at equilibrium such as gas clusters (in some conditions), see the following reference:

        P. H. Chavanis, ‘Gravitational instability of isothermal and polytropic spheres’
        https://judithcurry.com/2014/12/01/gravito-thermal-discussion-thread/#comment-653279

      • There are a couple of ways to address the kinds of concerns Pekka puts forwards.
        First, you end up with a contradiction if you say that an isentropic state will diffuse heat in such a way that it ends up isothermal. The isothermal state with the same total potential temperature, but now a gradient of potential temperature, has a lower entropy, which makes this an impossible process to occur spontaneously.
        Second, mechanistically what is molecular diffusion? It is the transfer of energy by collisions. Starting isothermal, particles from above will have gained kinetic energy from potential energy and those from below will have lost it by the time they collide at some reference level. This systematic correlation between motion and KE means that there is a downward flux of energy through the reference level. Only when the upper layer has cooled sufficiently is this PE gain canceled by the KE gradient. Heat is transferred down a potential temperature gradient, not just by eddies, but also by molecules. There is no reason for these two processes to respond differently to gravity.

      • Jim,

        You continue to claim that constant potential temperature would have a higher entropy than isothermal that has the same total energy. Your statement is wrong. The truth is the opposite as virtually every textbook tells.

      • Pierre-Normand

        “such as [star] clusters…” (Though the discussion generalizes to other self gravitating systems)

      • Pekka, no, a potential temperature gradient with the same total potential temperature has a lower entropy than that same total mixed into a uniform value. You can prove it by using two equal-mass layers as an approximation of a gradient. It follows from entropy going as log(theta).
        0.5*[log(theta1)+log(theta2)] is less than log[0.5*(theta1+theta2)]. It is because the geometric mean of any two numbers is always less than their arithmetic mean.

      • Pierre-Normand

        Jim D wrote: “[…] Second, mechanistically what is molecular diffusion? It is the transfer of energy by collisions. Starting isothermal, particles from above will have gained kinetic energy from potential energy and those from below will have lost it by the time they collide at some reference level.”

        In the stationary isothermal equilibrium state, (which also exhibits a barometric density profile; though the following consideration is valid regardless of this,) at any level z = h, as many molecules cross it going up as do going down. Each molecular trajectory is a parabola that crosses this level with the same identical speed while going up as it does going down. Only the velocity reverses. It follows that collisions don’t carry any heat down since the speed distribution of the molecules coming down is the same as the speed distribution of the molecules going up at any level. The only effect of the acceleration of gravitation is to limit the maximum height achieved by individual molecules and thereby to create an exponential density profile.

      • Jim,

        You argument is based on the assumption that the sum of the potential temperatures does not change in conduction. That’s directly equivalent to assuming that the total entropy does not change in conduction, but neither is true. Both the sum of the potential temperatures and the total entropy increases.

      • Jim

        Consider a situation where the lower parcel is at the temperature T1 and the upper parcel of the same mass at lower temperature T2 < T1. Move some heat from the lower to the upper parcel by conduction. Their temperatures change by equal amounts, because the specific heats are equal. The potential temperature do, however, not change by the same amount, as that of the upper parcel changes more based on the formula of the potential temperature. When you do the full calculation, you see that also the entropy increases in this process of heat conduction.

      • @PN: Regarding your second objection, I think it is misguided. The two cancelling effects in the case where g is uniform with height are (1) the slowdown of the particles that are climbing the gravitational potential, and (2) the drop off from the molecular populations at any level of the molecules that have the least amount or total mechanical energy (PE+KE). As Coombes and Laue have showed, those two effects cancel out. If the gravitational acceleration decreases with height, then there is less of a slowdown of the rising molecules and also less of a drop off rate

        Quite right, my second objection was indeed misguided. I was reasoning that the lower gravity at the top of the parcel would reduce the pressure there, thereby increasing the effect. What I overlooked was that the pressure also decreases at the bottom, and that it is the difference in pressure (equivalently the difference in number density of molecules) that drives the hotter molecules up. That difference decreases with decreasing gravity thereby reducing the pressure on the hotter molecules to rise.

        Apropos of my use of decreasing pressure P instead of decreasing number density n of molecules at higher altitudes, the ideal gas law P = nkT for fixed kT makes P proportional to n, whence it is immaterial whether one couches C&L’s argument in terms of P or n. I prefer P because (a) pressure is an intuitively more obvious driving force than number of molecules, and (b) to the extent that the finite size of molecules relative to their separation makes P larger than nkT it is P that is the driver and not n.

      • @PP: Consider a situation where the lower parcel is at the temperature T1 and the upper parcel of the same mass at lower temperature T2 < T1. … When you do the full calculation, you see that also the entropy increases in this process of heat conduction.

        Actually the “full calculation” is very simple, based on the formula dS = dQ/T for variation dS of entropy with variation of heat content dQ. When a quantity dQ of heat flows from the hotter parcel of temperature T1 to the colder one T2, the entropy dS1 leaving the former is dQ/T1 while the entropy dS2 gained by the latter is dQ/T2. Since T2 dQ/T1, that is, the entropy of the hotter parcel decreases by less than the entropy of the colder parcel increases and so the total entropy of the two parcels increases.

        The environmental lapse rate or thermal gradient of the atmosphere maintains the entropy of the atmosphere at considerably less than the maximum possible were the atmosphere suddenly isolated thermally and allowed to drift adiabatically into equilibrium. With the exception of one detail, the situation is the same as for the thermal gradient developed in a metal rod that is heated at one end and allowed to lose that heat by radiation. If the metal rod is suddenly thermally isolated so as to neither gain nor lose height, it drifts into thermal equilibrium, increasing its entropy as it does so. This makes it clear that the flow of heat through the rod is maintaining its entropy at below its maximum value.

        The same thing happens to the atmosphere. During the day the sun heats the Earth’s surface (land and sea) which then heats the bottom of the atmosphere, and the atmosphere and the surface together radiate that heat to space. This develops a thermal gradient, aka the environmental lapse rate or ELR. The flow of heat through the atmosphere maintains its entropy at below its maximum value. If suddenly isolated thermally, allowing heat to neither enter nor leave, the atmosphere will drift towards its maximum entropy.

        The one detail that makes the atmosphere different from a metal rod is that when the ELR exceeds the adiabatic lapse rate (ALR, appropriately adjusted for moisture content), typically in the afternoon, the atmosphere becomes unstable. Huge hot air balloons with no skin, aka thermals, then break loose and rise, a process that can’t happen with metal rods until they melt. As long as the ELR remains below the ALR the atmosphere will remain stable and (absent winds etc. from some other location) will undergo no convection, merely conduction as with the metal rod along with some absorption and emission of radiation by its GHGs. The ALR sets an upper limit on the amount by which the atmosphere’s entropy can be decreased below its maximum-possible-after-thermal-isolation, much as the melting point of the metal rod does for its decrease in entropy when subjected to a thermal gradient.

      • Damn, foiled by > and < (and only hours after pointing out how to avoid that problem in the megawatts thread). Let me rephrase the first paragraph of my immediately preceding comment, with the text that WordPress deleted in bold.

        Actually the “full calculation” is very simple, based on the formula dS = dQ/T for variation dS of entropy with variation of heat content dQ. When a quantity dQ of heat flows from the hotter parcel of temperature T1 to the colder one T2, the entropy dS1 leaving the former is dQ/T1 while the entropy dS2 gained by the latter is dQ/T2. Since T1 > T2, it follows that dQ/T1 < dQ/T2, that is, the entropy of the hotter parcel decreases by less than the entropy of the colder parcel increases and so the total entropy of the two parcels increases.

      • The arguments by Pekka and Vaughan Pratt are not correct for this system because taking heat from a warm layer and dumping it into a cold layer is not the same as taking the air parcel itself adiabatically from the warm layer to the cold layer. When you start in an isentropic state you find this achieves nothing because the parcel cools as much as the environment in the process. This is an adiabatic adjustment. You just need to evaluate the entropy of the two states, constant theta and constant T (or gradient theta), to see which one has the higher value and is therefore the end-state of maximized entropy. Putting some numbers in for molecular diffusivity of heat for air, we find that this is a slow process, time-scales of about a year to adjust a meter column from isothermal to isentropic, and proportionately larger for deeper layers, so for most intents and purposes an isothermal layer stays isothermal unless you want to wait a long time, or stir it to speed things up.

      • Pierre-Normand, in an isothermal state KE is independent of height, but there is a PE gradient. A downward displaced molecule will have gained some PE relative to its KE at the higher level, and will therefore be faster than the average KE at the level it ends up at, and any collision would tend to accelerate that level. A constant KE in the presence of a PE gradient therefore behaves exactly like a KE gradient in the absence of PE. Downwards displaced particles tend to be faster. This is why there is a thermal flux in the presence of a PE gradient, even in an isothermal (constant KE) profile.

      • Pierre-Normand

        Jim D,

        Your argument about the average kinetic energy of molecules going up or down at some level is invalid because you are losing track of the relevant molecular populations. I offered a very simple proof that it is invalid and you ignored the proof and merely restated the invalid argument. Let me try again to point out the mistaken assumptions that leads you astray.

        You are effectively arguing that an initially isothermal state (at time t) can’t possibly be isothermal at time t+dt because the molecules arriving to some level h at time t+dt will be faster than the molecules arriving at this level from below at the same time. That’s because the former have increased their kinetic energies, and the latter have decreased them, between the times t and t+dt. Therefore, you are arguing, collisions at that level will conduct heat downwards. This argument is invalid.

        The molecular populations that you are considering in your argument only include a fraction of the molecules that arrive at level h at time t+dt. What you mu