Experimental test of Loschmidt effect

Does  a lapse rate exist in an isolated gas under a gravitational field, or is it only a bulk convective effect? This essentially was the dispute between Lochsmidt, Maxwell and Boltzman at the end of the 19th century. Maxwell and Boltzman argued that such an isolated atmosphere would reach an isothermal temperature.  Nearly every physicist since then would agree with Maxwell since a persistent temperature gradient appears to violate the second law of thermodynamics. A modern analysis of this problem has recently been given by Christian Fronsal, a theoretical physicist at the University of California [1]. He proposes using a centrifuge to resolve this issue one way or the other. Renewed interest  has also been triggered by a seemingly positive temperature gradient observed in an isolated water column under gravity [2]. Such an  experimental test using a gas centrifuge is shown in figure 1, and described  below.

Fig 1: Experiment to decide whether Loschmidt could have been right.

Gas centrifuges such as those used for Uranium isotope separation can spin at well over 1000 revs/sec simulating huge gravitational fields. For a gas chamber with radius R and an  inner bore of radius ‘a’ spinning at an angular velocity \Omega, the effective gravitatonal acceleration at a distance z inwards from the rim is g =\Omega^2(R-z)

For a gas with specific heat Cp the lapse rate is

\frac{dT}{dz} = -\frac{\Omega^2(R-z)}{C_{p}}

T(z) = -\frac{\Omega^2}{C_{p}} \int_0^z (R-z)\mathrm{d}z = -\frac{\Omega^2}{C_{p}}[_R^zRz -\frac{z^2}{2}+C]

putting  z=0, -\frac{\Omega^2}{C_{p}}C = T_{0}

T(z) = T_{0}+\frac{\Omega^2}{C_{p}}(\frac{z^2}{2}-Rz)

Assuming a centrifuge with an outer radius of 10cm and an inner radius of 1 cm spinning at 100 rev/sec, the predicted lapse rate temperature gradient  is shown in figure 2. It is assumed that the centrifuge is surrounded by a heat bath held at 15C and that the inner bore is a vacuum. Using air  in the centrifuge the temperature at the inner radius is then predicted to be 2 degreec C. lower than the outer radius.  Such a large effect would be easy to measure with thermocouples. Now substituting CO2 instead of air increases slightly the effect by a further 0.3C, while replacing it with argon produces the largest effect with a drop of 3.7 degrees C. Lochsmidt originally calculated a lapse rate  g/Cv  instead of g/Cp. This corresponds to zero work being done (PdV) ( see  understanding the lapse rate ). The results assume in figure 2 assume g/Cp.

Fig 2: Predicted lapse rates for a Loschmidt effect in a gas cyclotron.

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 !

References

1.  Heat and Gravitation. I. The Action Principle, Christian Fronsal, Univ. Califormia, 2011.

2.  Graeff, R.W.  see http://firstgravitymachine.com/descript_B372_V_5.pdf   2007.

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52 Responses to Experimental test of Loschmidt effect

  1. I just left a comment on another of your posts saying that setting up the laps rate in a laboratory is impractical, and here you are with a ingenious idea for how to do it. I don’t have such a centrifuge in my laboratory, so I can’t try it. The cost would be hundreds of thousands of dollars, I’m guessing.

    Now, suppose you had a centrifuge with perfect bearings, so this thing could keep spinning around forever. And suppose we arranged to remove heat from the gas so that at time zero it was all at the same temperature. Is there someone out there that claims this gas would spontaneously develop a temperature gradient?

    If so, we could use this gradient to run a heat engine until we cooled the gas down almost to condensation, then feed heat in from a Big Reservoir to warm it up, and repeat, thus converting the heat in our Big Reservoir into energy with 100% efficiency, which violates the Second Law of Thermodynamics (see here for drawing).

    The gas in the frictionless centrifuge will settle down to a uniform temperature by conduction, and stay that way.

    • Clive Best says:

      Yes I agree that the gas will quickly thermalize to a uniform temperature. However, for a tiny instance when the cyclotron starts a lapse rate will be formed which conduction instantly restores to thermodynamic equilibrium. There are people out there who believe that a lapse rate in an isolated gas is maintained because at first sight it appears that gravity reduces entropy. see for example http://firstgravitymachine.com and http://ruby.fgcu.edu/courses/twimberley/EnviroPhilo/FunctionOfMass.pdf

      All known planets maintain lapse rates but they are not in thermodynamic equilibrium. Energy is continuously flowing through the “active” atmosphere and induces convection. I don’t think energy has to flow from the surface upwards. It can also flow from the atmosphere downwards as on Venus.

      • Can you enlighten me about energy flowing downwards through the atmosphere of Venus?

        • Clive Best says:

          90% of incident solar energy is absorbed in the upper atmosphere on Venus – mostly in the thick H2SO4 clouds. Only 17 watts/m2 reaches the surface, which is the equivalent of one light bulb heating the surface of a small room. Differential heating in the atmosphere causes strong winds mixing the atmosphere acting like a heat pump. At the equator low pressure pumps air upwards and gas spirals down at the poles. The poles are very hot and under high pressure. A lapse rate is maintained with Teff=260K essentially 60 km above the surface.

          • I read that a few times and I still can’t understand what you are describing. I can’t conceive of any process that would allow convection to transport heat downwards. If that were possible, the oceans would be warm at the bottom. But they are not. At the same time, the 17 W/m2 reaching the surface of Venus is perfectly adequate to heat the surface until convection starts, thus setting up the lapse rate and transporting heat to the altitude where it can be radiated into space.

  2. Clive Best says:

    Let’s imagine Venus without any solar radiation at all, but with 17 watts/m2 of internal geothermal heating from radioactive decay. Would the surface temperature still reach 500 deg.C ? I don’t think so – I think you need the extra 150 watts deposited in the atmosphere.

    Then we also have to explain why both the polar regions and the night-time side of the planet are both just as hot. Here there is zero surface heating. The polar regions are a swirling vortex of air and clouds – like a hurricane or a tornado. The atmosphere is being churned up by strong heat induced winds resulting in convection.

    Yes the surface reaches enormous temperatures and then you can describe it somehow as convective cooling of the surface towards a lapse rate. However I don’t think the traditional (Earth like) greenhouse effect trapping heat radiated in the atmosphere radiated just from the surface can apply to Venus. The atmosphere is actively involved in distributing 90% of the heat absorbed by it around the planet. So for example air driven downwards at the poles compressed under gravity thereby heats the surface.

    I remain open to be convinced otherwise !

    • “Let’s imagine Venus without any solar radiation at all, but with 17 watts/m2 of internal geothermal heating from radioactive decay. Would the surface temperature still reach 500 deg.C ?”

      Yes, almost.

      It does not matter whether the heat comes from within the planet or from the Sun. The surface will warm up until convection carries those 17 W/m2 up to the top of the atmosphere, where it can be radiated into space. Now, you say “without any solar radiation”, so the top of the atmosphere, where the SO2 clouds end, will be at around 130 K so as to radiate 17 W/m2. The temperature at the surface will be around 650 K, or 371 C. I’m guessing there that the total lapse from land surface to cloud tops will be the same, even though the cloud-top temperature is 130 K less than you assert in one of your comments above.

      So, without the sun, and 17 W/m2 geothermal heating, the surface of Venus would still be very hot indeed.

      • Clive Best says:

        You are assuming the lapse rate as a given fact. Supposing instead the heat is simply radiated away from the surface through the IR opaque atmosphere. The atmosphere will absorb all surface IR and then acts like a blackbody. We know that 17 watts/m2 is emitted by the atmosphere to space by energy balance, so the atmosphere must emit an equal 17 watts/m2 down to the surface. So the surface receives a net 34 watts/m2 (17+17). Surface temperature should then be 156K, and the Atmosphere temperature = 132K

        i.e. Ts^4 = 2*Ta^4

        We can only get very high surface temperatures via the “magic” lapse rate, which itself assumes strong convection. I still find the lapse rate somewhat mysterious. At exactly the DALR no energy is needed for air to move adiabatically. It is stable against convection.

  3. I know what you mean about the lapse rate being confusing, and you make a good point about the back-radiation of the atmosphere. You’ll find a treatment of this subject for the Earth’s atmosphere here. If the bottom of the opaque atmosphere is at the same temperature as the land surface, the radiation between them will be equal, and the land surface will be unable to dissipate its 17 W/m2 through radiation. Only through the existence of a temperature gradient will radiative transfer be able to carry heat up through the atmosphere, and in our simulations we always find that this gradient was greater than the lapse rate, so that convection would always occur to carry the bulk of the heat.

    I’ll try to prove that statement. Consider a layer of the atmosphere of thickness b that is just thick enough to be opaque. It is at temperature T, and altitude z, losing a net Q by radiation. Let a be Stefan’s constant. We require:

    Q = aT^4 + aT^4 – a(T+dT/dz*b+)^4 – a(T-dT/dz*b)^4
    = -6a(T*dT/dz*b)^2

    From which we get after re-arranging and integrating, for altitudes z=0 and z=h we get:

    Th^2 = T0^2 – 2z*sqrt(Q/6/a)/b

    Inserting T0=600, Q=17W/m2, h=20km, a=5.7^-8, and b=1km we see that there is no Th low enough to support the transfer. Indeed, the more opaque the atmosphere, the more resistant it is to radiative heat transfer. But convection will start when Th is around 150K, according to dT/dz = -zg/Cp (see here).

    Thus I claim that the land surface will heat up until it warms the atmosphere resting upon it, and this atmosphere will rise, to be replaced by cool atmosphere, and slowly the lapse rate will be set up, until convection occurs easily. We see this happening in a simulation of the Earth’s atmosphere here and subsequent posts.

  4. Oops: -6a(T*dT/dz*b)^2 should be -12a(T*dT/dz*b)^2 and subsequent calculations must change accordingly.

  5. Clive Best says:

    OK, I am 90% convinced now that you are basically right. All planets can only lose heat through IR radiation to space thereby balancing incoming solar radiation. For a planet without an atmosphere say Mercury or the Moon the mean surface temperature is simply Teff = (S*(1-a)/4*sigma)^0,25. For a planet with a (partially) opaque atmosphere the surface heat must be “transported” from the surface up to a variable level where the density decreases enough so that radiation can freely escape to space. This level varies continuously with day/night, summer/winter, latitude, cloudiness, humidity and albedo. There are two ways heat gets transported upwards 1) Radiative Transfer 2) Convection. On Earth there is a third way – latent heat release of water vapor. Venus has an opaque atmosphere up to 60 km and convection is a far more important energy flow than radiative transfer. On Earth net radiative transfer is 40 watts/m2 through the IR window and 23 watts/m2 by radiative transfer. Convection and Latent Heat transfer combined are about 140 watts/m2 on Earth. Therefore on Earth convection carries more than twice the energy as radiative transfer up to the tropopause. Note I think “back radiation” is a bit of a red herring since it is net energy flow that matters.

    For your proof – I am still not understanding your formula. I get energy balance for the level

    In from below: a(T+dt/dz*b)T^4 and in from above a(T-dT/dz*b+)^4

    Out from surface: (up+down) = 2aT^4

    So : Q = aT^4 + aT^4 – a(T+dT/dz*b+)^4 – a(T-dT/dz*b)^4 is correct

    Binomial expansion doesn’t get me -12a(T*dT/dz*b)^2

    thanks

  6. I have not checked your specific values (like 60 km tropopause for Venus), but I agree with your description of the phenomena. So: the lapse rate is pretty neat, right? And it sounds like you enjoyed working out the expansion for radiative transfer as much as I did.

  7. Rog Tallbloke says:

    Hi both,
    I’m having trouble following the equations, but I have a question.

    Is the radiation the cause of the lapse rate or a symptom of it?
    I’m asking this in relation to Clive’s comment that the back radiation is a red herring because it’s the net radiation that counts.

    On Venus, which turns very slowly, a lot of solar energy is transferred to the night time side where loss of energy to space is more efficient due to a higher differential in temperature between atmosphere and space. This sets up strong advective winds, which then cause lots of vertical heat transfer due to the resulting turbulence.

    Don’t you have to account for that advection in your calculations?

    • Clive Best says:

      I think that the lapse rate is “caused” by convection. The adiabatic lapse rate is exactly the rate at which convection stops. At the DALR there is a perfect balance between internal kinetic energy (temperature) and gravitational potential energy. Surface heating increases the gradient during the day and induces convection restore the DALR temperature gradient. Heat transferred upwards by convection or latent heat radiates to space at frequency dependent levels where the atmosphere becomes transparent for IR photons of that frequency. A planet can only loose heat through radiation to space. I view radiative transfer as the multiple scattering of photons diffusing energy upwards emitted by the surface through the atmosphere. On Earth it is more efficient to transfer energy by convection and latent heat than radiative transfer. Heat radiates out from the tropopause as well as directly from the surface (IR window).

      I am sure that advection plays an important role on Venus where there are violent winds. The atmosphere also absorbs 90% of solar energy and this heat must be spread around the planet and some transported down to the surface particularly at the poles. I have read that the poles are like a giant tornado. However, the basic net heat loss must still be from hot to cold via convection and the lapse rate. The tropopause is 90 miles above the surface so there is a huge lapse rate.

  8. tallbloke says:

    I understood that the lapse rate is actually about the same as on Earth, taking into account their different distances from the Sun. So there is a huge temperature gradient due to the much higher tropopause, but only around the same lapse rate per km.

    I agree with everything else you aid here though.

    • Clive Best says:

      Sorry – yes you are right. The lapse rate is always g/Cp. I really just meant the same as you – that there is a huge net temperature gradient from the surface to the tropopause.

      Incidentally I discovered that Dyson vacuum cleaners use “cyclone centrifuges” generating gravitational forces up to 150000g to to fling dust from the air. One of those scaled up would work to set up a lapse rate for atmospheric physics experiments. Cost would be a few hundred quid !

    • Doug Cotton says:

      Clive, Roger and others

      The “lapse rate” does not require upward convection from a surface heated by the Sun’s radiation. There is no such thing happening at the base of the nominal Uranus troposphere where it’s hotter than Earth and the thermal gradient in the troposphere is about 95% of the -g/Cp value. These gradients evolve spontaneously at the molecular level, as I have been saying since mid-2012, even before Teofilo’s book was published in October 2012, though I have just learnt of it.

      The gravito-thermal effect can be verified by using Kinetic Theory in conjunction with the modern entropy statement of the Second Law of Thermodynamics, because the state of maximum entropy that evolves must have no unbalanced energy potentials, and so must be isentropic. This is explained in my new book “Why it’s not carbon dioxide after all” available on Amazon late April 2014.

      It is not the energy in the oceans which controls climate just because there’s far more energy there than in the atmosphere.

      Valid physics tells us it’s the other way around. It is the atmosphere (all the troposphere in particular) that autonomously comes into radiative balance with incident solar radiation, because the whole Earth+atmosphere system is what acts similar to a blackbody – not the surface, which is mostly transparent wherever there’s water.

      The thermal gradient (aka lapse rate) evolves spontaneously due to gravity acting at the molecular level, and so the whole thermal profile in the troposphere is pre-determined.

      Now, it doesn’t matter that the atmosphere holds far less thermal energy than the ocean. All that matters is what happens when molecules at the interface of the air and water collide. That “evens out” the temperatures and it is (eventually) thermal energy absorbed in the atmosphere that “creeps” up the (sloping) thermal plane and into the ocean. Of course the Sun adds some energy to the oceans, but its radiation passes almost entirely through the first 1cm of the surface and so its radiation is not determining the surface temperature – the troposphere is doing that by non-radiative diffusion and conduction.

      Similarly, the Sun is not affecting the Venus surface temperature much with its direct radiation that is barely 20W/m^2, but that surface is over 730K. So exactly the same happens enabling energy to get into the Venus surface (by diffusion and downward convection) and this non-radiative process causes its temperature to rise during its 4-month-long daytime.

      Planetary atmospheric, surface and even sub-surface temperatures are not controlled primarily by so-called greenhouse radiative forcing. That is why it’s not carbon dioxide after all.

      • Clive Best says:

        On Earth the excess heat from the tropics moves towards the poles through convective cells so there is mixing. Likewise latent heat reduces the lapse rate towards the moist lapse rate especially in the tropics. So heat energy is transported upwards to the top of the troposphere. I also think that without greenhouse gases energy would not be able to radiate to space from the top of the atmosphere so this would lower the the tropopause nearly to ground level. On Venus there are also strong convective movements and the poles end up being very hot.

        However the logic of gravitational potential energy balancing kinetic energy is attractive. Would there still be a lapse rate in an atmosphere or pure argon ?
        or would conduction eventually equalize temperatures ? I think the centrifuge experiment could settle this.

        • Doug Cotton says:

          If you want a more clear cut example, the Uranus troposphere is the best in our Solar System for demonstrating the gravito-thermal effect.

          Uranus is noted for not having any convincing evidence of any significant net energy loss, so it is not just cooling off or generating internal energy due to collapsing, probably because it has a solid core about 55% the mass of Earth and at about 5,000K.

          It could easily have cooled right down in billions of years, but for the Loschmidt gravito-thermal effect which enables gravity to “trap” thermal energy under the pre-determined thermal profile. For the Uranus troposphere that gradient is about 95% of the -g/Cp value over the 350Km high nominal troposphere. There is no surface at the base thereof and no significant incident solar radiation, yet it’s 320K.

          Note that it is not actually latent heat release which reduces the gradient, but inter-molecular radiation which increases with the percentage of water vapour and CO2. Uranus only has a little methane in the lower troposphere, so the gradient is reduced less. CO2 reduces it on Venus somewhat more, but less than a third as happens on Earth.

          There is a very comprehensive coverage of all this in my book “Why it’s not carbon dioxide after all” due out on Amazon and Barnes & Noble by late April 2014.

        • Doug Cotton says:

          Clive: Where you say “or would conduction eventually equalize temperatures” you seem to miss the point that diffusion is conduction. Physicists usually use the word “conduction” for solids and “diffusion” where a gas is involved. See second paragraph here.

          Furthermore, “convection” can be “diffusion”, “advection” or both.

          Conduction in a vertical insulated tall gold bar does not lead to isothermal conditions. You could probably measure the temperature difference in gold in about 2 to 3 metres because it is around 60 to 70C/Km.

          The Loschmidt effect occurs in all solids, liquids and gases, as he claimed.

    • Doug Cotton says:

      Roderich Graeff presented the derivation below about ten years ago, but he mistakenly multiplied by degrees of freedom. It shows why the lapse rate has nothing to do with pressure, density or expanding gas. This is from my book ….

      Let us consider a thought experiment in which a region of a non-radiating gas of mass M all happens to move downwards by a small height difference, H in a “closed system” where g is the acceleration due to gravity.  The loss in PE will thus be the product M.g.H because a force Mg moves the gas a distance H.  But there will be a corresponding gain in KE and that will be equal to the energy required to warm the gas by a small temperature difference, T.  This energy can be calculated using the specific heat Cp and this calculation yields the product M.Cp.T. Bearing in mind that there was a PE loss and a KE gain, we thus have …

      M.Cp.T = – M.g.H 

      T/H = -g/Cp

       But T/H is the temperature gradient, which is thus the quotient -g/Cp.

      This result is well known, as is the fact that the atmospheres of all planets exhibit a similar temperature gradient that can be calculated from the gravitational force on that planet and the mean specific heat of the gases in its atmosphere.

  9. tallbloke says:

    oops – said, not aid

  10. Teofilo Echeverria says:

    Lots of your doubs about this subject will be cleared out if you read the “Theory of Equilibrium in Temperature Gradient” from my book “Waiting for Sunrise Till Dusk”

  11. Doug Cotton says:

    This is discussed in great detail (and with empirical evidence) in my book “Why it’s not carbon dioxide after all” which will be available through Amazon and Barnes & Noble late April 2014.

  12. Doug Cotton says:

    Teofilo (who posted above) and I each came to the same conclusion independently in our publications around October/November 2012 and we each postulated heat diffusion downwards in a planet’s troposphere restoring thermodynamic equilibrium. This is fact, I am sure, and is backed up with mountains of evidence from throughout the Solar System.

  13. Doug Cotton says:

    Yes the environmental lapse rate is the rate when convection stops, but the gradient forms at the molecular level and does not need advection. This is blatantly obvious in the troposphere of Uranus as I explain in my book.

    What actually happens is that the direction of convection reverses because a true temperature inversion is when the top temperature is just a little warmer than that which the environmental lapse rate would indicate it ought to be. The reasons for the environmental lapse rate being about 30% less than the -g/Cp value for Earth are explained in my book, though for Uranus it is only about 5% less, which I have also explained.

  14. Doug Cotton says:

    In the 2004 paper (by Verkley et al) “On Maximum Entropy Profiles” here, they assert that …

    “convective turbulent motions are now taken into account, albeit implicitly. Their role is to mix the potential temperature field, to strive to homogenize it.”

    This is not necessary, as there is no reasonable evidence of such convective turbulence existing on some other planets, notably Uranus. Instead it is the actual movement of molecules between collisions which provides the random mixing they claim is requiring advection. (They are not even precise in their terminology, because “convection” can include diffusion.)

    They deduce in fact two conclusions using different constraints. However the constraint that leads to their deduction of isothermal conditions is not appropriate. It involves assuming constant enthalpy and this implies that there is a compensating increase in mean molecular total energy that is offset by the reduction in density at higher altitudes. This means that the molecules would be retaining equal kinetic energy, whilst gaining gravitational potential energy, that being offset by the reduction in total numbers so that total enthalpy remains constant. There is no justification for this assumption and the constraint is not a reality.

    Furthermore, they introduce “constancy of the integrated potential temperature as a single additional constraint” and then they admit “but this choice is of course open for debate.” Well, of course it is open for debate because there is no logic supporting it. What they are doing is trying to find a reason for the wet lapse rate being less than the dry one. They know that isentropic conditions lead to the dry rate (-g/Cp) but what they don’t realise is what I have explained in my book about the temperature levelling effect of inter-molecular radiation.

    As I have said all along, the empirical evidence that water vapour cools rather than warms supports the fact that the gravito-thermal effect produces the dry gradient which is then reduced in magnitude by the inter-molecular radiation, not primarily the release of latent heat.

    All in all, this is a very wishy-washy paper. Whilst their computations are OK, they do not engage in any detailed discussion or reasoning as to what would be the correct constraints. It would have been appropriate to start by considering a sealed perfectly insulated cylinder of ideal non-radiating gas. If they had done this there would have been no ambiguity about the constraints or any need to discuss advection. This it the approach I have taken in my papers and the book. Once we accept that the gravito-thermal gradient evolves spontaneously at the molecular level without any need for advection, then it is not hard to extend the concept to a troposphere which has a propensity to approach such a thermal gradient, modified by inter-molecular radiation.

  15. Doug Cotton says:

    The second law of thermodynamics states that the entropy of an isolated system never decreases, because isolated systems always evolve toward thermodynamic equilibrium—the state with the maximum possible entropy.

    An isothermal profile in a gravitational field is not isentropic, for the simple reason that, firstly you are assuming all molecules have the same kinetic energy, but secondly, we know the ones at the top have more gravitational potential energy.

    So, consider the following thought experiment, starting with …

    Molecules at top: More PE + equal KE

    Molecules at bottom: Less PE + equal KE

    In such a situation you have an unbalanced energy potential because the molecules at the top have more energy than those at the bottom. Hence you do not have the state of maximum entropy, because work can be done.

    Let’s consider an extremely simple case of two molecules (A & B) in an upper layer and two (C & D) in a lower layer. We will assume KE = 20 initially and give PE values such that the difference in PE is 4 units …

    At top: A (PE=14 + KE=20) B (PE=14 + KE=20)

    At bottom: C (PE=10 + KE=20) D (PE=10 + KE=20)

    Now suppose A collides with C. In free flight it loses 4 units of PE and gains 4 units of KE. When it collides with C it has 24 units of KE which is then shared with C so they both have 22 units of KE.

    Now suppose D collides with B. In free flight it loses 4 units of KE and gains 4 units of PE. When it collides with B it has 16 units of KE which is then shared with B so they both have 18 units of KE.

    So we now have

    At top: B (PE=14 + KE=18) D (PE=14 + KE=18)

    At bottom: A (PE=10 + KE=22) C (PE=10 + KE=22)

    So we have a temperature gradient because mean KE at top is now 18 and mean KE at bottom is now 22, a difference of 4.

    Note also that now we have a state of maximum entropy and no unbalanced energy potentials. You can keep on imagining collisions, but they will all maintain KE=18 at top and KE=22 at bottom. Voila! We have thermodynamic equilibrium.

    But, now suppose the top ones absorb new solar energy (at the top of the Venus atmosphere) and they now have KE=20. They are still cooler than the bottom ones, so what will happen now that the previous equilibrium has been disturbed?

    Consider two more collisions like the first.

    We start with

    At top: B (PE=14 + KE=20) D (PE=14 + KE=20)

    At bottom: A (PE=10 + KE=22) C (PE=10 + KE=22)

    If B collides with A it has 24 units of KE just before the collision, but then after sharing they each have 23 units. Similarly, if C collides with D they each end up with 19 units of KE. So, now we have a new equilibrium:

    At top: C (PE=14 + KE=19) D (PE=14 + KE=19)

    At bottom: A (PE=10 + KE=23) C (PE=10 + KE=23)

    Note that the original gradient (with a difference of 4 in KE) has been re-established as expected, and some thermal energy has transferred from a cooler region (KE=20) to a warmer region that was KE=22 and is now KE=23. The additional 2 units of KE added at the top are now shared as an extra 1 unit on each level, with no energy gain or loss.

    That represents the process of downward diffusion of KE to warmer regions which I call “heat creep” as it is a slow process that happens in which thermal energy “creeps” slowly up the sloping thermal profile. It happens in all tropospheres, explaining how energy gets into the Venus surface, and explaining how the Earth’s troposphere “supports” surface temperatures and slows cooling at night.

    See also ..

    http://hockeyschtick.blogspot.com.au/2014/03/why-ideal-gas-law-gravity-atmospheric.html#comment-form

  16. Doug Cotton says:

    There never will be evidence of carbon dioxide or water vapour warming, because each cools. On Venus, carbon dioxide leads to considerably lower supported surface temperatures than would be the case with an atmosphere of less-radiating gases like hydrogen and helium.

    Any skeptic who still thinks carbon dioxide is having any effect at all is a warmist in my view. There can be only one truth. Those denying the truth that these gases actually cool, as is blatantly obvious in temperature data (as far as water vapour is concerned) is a denier of the truth.

    The truth that nearly all are denying, lukes or warmists, lies in a whole new paradigm based on the now-proven gravito-thermal effect.

    This retired physics educator puts it succinctly in summarising it all after reading the text of my book

    Essential reading for an understanding of the basic physical processes which control planetary temperatures. Doug Cotton shows how simple thermodynamic physics implies that the gravitational field of a planet will establish a thermal gradient in its atmosphere. The thermal gradient, a basic property of a planet, can be used to determine the temperatures of its atmosphere, surface and sub-surface regions. The interesting concept of “heat creep” applied to diagrams of the thermal gradient is used to explain the effect of solar radiation on the temperature of a planet. The thermal gradient shows that the observed temperatures of the Earth are determined by natural processes and not by back radiation warming from greenhouse gases. Evidence is presented to show that greenhouse gases cool the Earth and do not warm it.

    John Turner B.Sc.;Dip.Ed.;M.Ed.(Hons);Grad.Dip.Ed.Studies (retired physics educator)

  17. Doug Cotton says:

    Below is a comment I have just posted on Lucia’s Blackboard in response to a common thought experiment attempting to disprove the existence of the gravito-thermal effect that is obvious in all planetary tropospheres.

    The “argument has been put to me several times and is obviously yet another attempt among climatologists to rubbish what is of course a very threatening postulate, because it smashes the greenhouse.

    The argument … does not display a correct comprehension of Kinetic Theory, or indeed the manner in which molecules move and collide.

    If a perfectly isentropic state were to evolve then all molecules in any given horizontal plane would have equal kinetic energy, and of course equal potential energy, just as after the first two collisions in the 4 molecule thought experiment above.

    Now, the direction in which a molecule “takes off” in its next free path motion just after a collision is random – rather like what happens with snooker balls.

    So two molecules with equal KE set out in different directions after the collision, but there is no requirement that they must have more KE to go upwards. They don’t travel far anyway. It’s not as if any one molecule goes up a matter of several cm before colliding with another, for example. In fact, they nearly all travel in a direction that is not straight up or down.

    At thermodynamic equilibrium (as you can see in the 4 molecule experiment) when any molecule has an upward component in its direction, it loses KE that is exactly the amount of energy represented by the difference in gravitational potential energy between the height of the molecule it last collided with and that of the next molecule. With the thermal gradient in place, the next molecule it strikes will have KE that is less than the one it last struck, and its own KE will have been reduced to exactly the same KE that the next molecule already has.

    So, at thermodynamic equilibrium all collisions involve molecules which had identical KE before the collision, and so they exit the collision process still having the same KE which is the mean KE for all molecules in the horizontal plane where the collision occurred.

    Now, for a small height difference, H in a “closed system” where g is the acceleration due to gravity, the loss in PE for a small ensemble of mass M moving downwards will thus be the product M.g.H because a force Mg moves the gas a distance H. But there will be a corresponding gain in KE and that will be equal to the energy required to warm the gas by a small temperature difference, T. This energy can be calculated using the specific heat Cp and this calculation yields the product M.Cp.T. Bearing in mind that there was a PE loss and a KE gain, we thus have …

    M.Cp.T = – M.g.H

    T/H = -g/Cp

    But T/H is the temperature gradient, which is thus the quotient -g/Cp. This is the so-called “dry adiabatic lapse rate” and we don’t need to bring pressure or density into the calculation.

    • Clive Best says:

      This argument is correct at the molecular level. At the bulk level the normal explanation is based on convection and movement of heat upwards by the work done adiabatically -PdV You need to get in touch with Stephen Wilde. He argues that the atmospheric mass of the atmosphere alone determines surface temperature.

      • Doug Cotton says:

        Sorry Clive, but I disagree wildly with Wilde who has an incorrect concept which does not allow him to explain correctly how the additional thermal energy gets from the colder atmosphere into the Venus surface in order to raise its temperature 5 degrees during the course of its 4-month-long day.

        Nor can upwards advection be the reason for the thermal gradient in the Uranus nominal troposphere which has a thermal gradient very close to the -g/Cp value, because there’s no source of internal or direct solar energy there, and no surface – but it’s hotter than Earth.

        I have explained all this in my book “Why it’s not carbon dioxide after all” being printed at the moment.

        There is no reason why my four molecule model cannot be extrapolated by inductive reasoning to the whole troposphere. You can think of it representing two horizontal layers of molecules where the layers are separated by a distance equal to the mean free path between collisions.

        The evidence in my study of real world data (to appear in the book in April) also confirms the gravito-thermal gradient, and that makes it inevitable that water vapour cools. .

        Neal J. King (from SkS) tried to argue with me on Lucia’s Blackboard here.and a copy of my reply is in this comment just above.

  18. Doug Cotton says:

    Following on from my comment #126658 on Lucia’s Blackboard, it appears that SkS team member Neal J. King made a huge error in assuming any molecules would run out of kinetic energy when they are moving upwards between collisions. In my four molecule thought experiment #126576 we are talking about a distance averaging the mean free path of air molecules between collisions. That’s about 68 nanometres! Even in a whole kilometre air molecules only lose about 3% to 5% of their kinetic energy because that’s how much the temperature drops.

    So may I suggest that Neal J. King goes back to his Skeptical Science Team to work up a better “answer” to the trillion dollar question (which many will be asking when my book comes out) what’s wrong with the Loschmidt gravito-thermal effect theory, which eliminates any need for explaining things with GH radiative forcing?

  19. Doug Cotton says:

    Just remember there’s going to be a genuine $5,000 reward for the first to come up with proof I’m wrong and proof IPCC are right about water vapour – see last paragraph..

    In a horizontal plane you can observe diffusion of kinetic energy in your home. Just run a heater on one side of a room, turn it off or even remove it quickly from the room, and you will temporarily have measureably warmer air on one side of the room. Molecules then keep on colliding and as they do, kinetic energy is shared. Statistical mechanics tells us that temperature (that is, mean kinetic energy per molecule) will even out across the room assuming it’s well insulated.

    Suppose now that the room has double glazed windows and it’s cooler outside. Which is more effective at insulating the room?

    (a) A window with dry air or even argon
    (b) A window with moist air – say 4% water vapour or water gas
    (c) A window full of carbon dioxide only, like the Venus atmosphere?

    The answer is the dry air or argon, as is well known in the construction industry. Why? Because radiating “pollutants” like water gas and carbon dioxide send the energy across the gap (and up through the troposphere) with inter-molecular radiation. Such radiation only ever transfers thermal energy from warmer to cooler regions. Otherwise what happens is as described in “Radiated Energy and the Second Law of Thermodynamics.”

    Why then does the thermal gradient reduce in magnitude because of the inter-molecular radiation between carbon dioxide molecules in the Venus atmosphere, or between a few methane molecules in the Uranus troposphere or between water vapour molecules in Earth’s troposphere and Earth’s outer 9Km of its crust?

    All these thermal gradients (aka lapse rates) are less steep than they would have been in dry air or (nearly) non-radiating gases. Gravity would have induced a steeper -g/Cp gradient.

    The thermal gradient in the Uranus troposphere does not level out (despite no solar radiation or any surface) because to do so would violate the Second Law of Thermodynamics. It seems most of you don’t understand why, but the reason is that entropy would decrease. If somehow a state were to evolve with more gravitational potential energy per molecule at the top, but no compensating reduction in kinetic energy per molecule (ie temperature) then there would be unbalanced energy potentials at the top, so work could be done and thus entropy would not have been at a maximum. The four molecule experiment demonstrates this and how it happens at the molecular level.

    The vortex tube demonstrates it, and kinetic energy is re-distributed such that the inner tube gets far colder than the air that was pumped in. So you can’t blame friction for heating the outer tube. Nor does pressure alter temperature, because pressure is proportional to the product of temperature and density: temperature is an independent variable and only varies when mean kinetic energy per molecule varies.

    Finally, none of you can explain how the Venus surface actually rises in temperature from 732K to 737K during its four-month-long day, unless you start by understanding that the thermal gradient is the state of thermodynamic equilibrium. Then you need to understand the mechanism of “heat creep” explained in the second part of the four molecule experiment.

    There will be a $5,000 reward for the first to prove me wrong with conditions explained in public advertisements and on all of a dozen or so of my websites. To win the award you will also have to show empirical evidence of the IPCC postulate that the sensitivity to water vapour is of the order of 10 degrees of warming for every 1% increase in the Earth’s troposphere.

  20. Doug Cotton says:

    Clive the centrifugal experiment has already been done,

    Read about the Vortex tube and note that it generates a huge force of about 10^7g which, in the 1cm radius (10^-5Km), using Cp=1 gives a temperature difference of 10^7/10^5 = 100 degrees. The observed results are of this order of magnitude, which means the Loschmidt gravito-thermal effect seems the most probable explanation.

    • Doug Cotton says:

      Actually I should correct those calculations. 10^7g is about 10^8 because g=9.8. Now I know that then gives 1,000 degrees and the quoted range is 250 degrees, but the factor of 4 in the difference is not unacceptable because of all the variables involved, especially the fact the g force is much less in the central tube. So we would expect that the mean g force might indeed be between 10^7 and 10^8.

    • clivebest says:

      Doug,

      That is remarkable ! They even give the most likely explanation as the gravito-thermal effect -g/Cp. I can’t think of another explanation !

      I also wonder why the vortex tube effect is not more widely known about.

      Clive

  21. Doug Cotton says:

    Neal J. King, a Skeptical Science team member, is still tied in knots with his fictitious fissics fantasy./b>

    I wrote to him on Lucia’s blog …

    Don’t forget you are still trying to calculate the thermal gradient at the state of maximum entropy which the Second Law of Thermodynamics says evolves spontaneously. Remember? I did it quite easily with mathematical induction from the four molecule experiment. You’re wandering off track and now getting side-tracked about the vortex – just read Wikipedia about that.

    And please don’t misquote me, Neal, as when you wrote “His claim is that the isothermal assumption is self-contradictory … “ because you know that gets my back up. What I did say was that an isothermal state would not be a state of maximum entropy because it still has unbalanced energy potentials. You can see from the four molecule experiment that there is still a potential transfer of kinetic energy across the boundary between the two layers 68 nanometres apart.

    Now you know that we can’t have kinetic energy transfers across a boundary if there were thermal equilibrium, don’t you, Neal? And to have thermodynamic equilibrium we must have thermal equilibrium, mechanical equilibrium, radiative equilibrium and chemical equilibrium. The inter-molecular radiation reduces the -g/Cp gradient by about a third on Earth when we have the radiative equilibrium as well as the thermal, mechanical and chemical equilibrium.

    I know it’s a little involved, but that’s why we really shouldn’t try to teach it to young school kids or politicians, now should we, or get it all mixed up at SkS?

    When you realise you are off the track with all those radiation calculations about insolation that passes straight through the ocean surface, and back radiation that just raises electrons through energy states, rather than adding kinetic energy, then I assume you will tell them all on the SkS team just how wrong has been the carbon dioxide hoax, now won’t you?

  22. Doug Cotton says:

    To pinpoint the main problems that are often apparent in climatology false, fictitious, fanciful fissics …

    (1) Temperature is the independent variable. Temperature is only changed if there is a physical addition or removal of kinetic energy which causes the mean kinetic energy per molecule to vary.

    (2) Altering density does not necessarily alter the mean kinetic energy per molecule, and nor does altering pressure.

    (3) The Ideal Gas Law tells us that pressure is proportional to the product of both temperature and density.

    (4) Gravity affects density, obviously, and (via the gravito-thermal effect) it also affects temperature by a different process.

    (5) In a planet’s troposphere, pressure is merely the end result of the gravitational processes which increase both temperature and density independently. It is not pressure that maintains the lapse rate and the higher temperatures at lower altitudes. It is gravity affecting density and temperature gradients.

    (6) There are still some people who seem to think that reducing the enthalpy of a system necessarily reduces the temperature. If all you do is remove some of the gas without altering the mean kinetic energy of the remaining molecules, you do not alter temperature. Likewise if you increase the density, temperature does not necessarily have to increase. In fact, temperature decreases if you pump in much colder gas.

    So, you do not have to involve pressure or density in your calculations of the thermal gradient which occurs at thermodynamic equilibrium. You just equate KE gain with PE loss, using the fact that KE gain is the energy required to raise the temperature of mass M by dT and so, using specific heat Cp we get M.Cp.T. Gravitational potential energy is of course M.g.dH where dH is the height differential. So …

    M.g.dH = -M.Cp.dT

    Thermal gradient: dT/dH = -g/Cp

    So remember, gravity controls the temperature gradient (aka lapse rate) in any planet’s troposphere, and energy is “trapped” for the life of the planet under the pre-determined sloping thermal plane. This provides a supporting temperature for every layer of the troposphere and the surface itself. Yes the Sun makes Earth’s surface a bit hotter during the day, but the supporting temperature stops the surface cooling anywhere near as much as it would otherwise do at night, and this helps the Sun to warm it to a higher temperature the next day, because it has limited hours in which to do so. Climate has nothing to do with radiative balance, radiative forcing, greenhouse gases or back radiation.

  23.  Doug Cotton  says:

    It gets better Clive. SteveF on Lucia’s Blackboard now says the radius of the Vortex tube is only 6mm not 1cm. So I did some revised calculations as I wrote this reply …

    We expect the thermal gradient to be able to be calculated from -g/Cp just as it can be in Earth’s troposphere and all other planetary tropospheres, but reduced by up to about a third in some cases by inter-molecular radiation.

    SteveF tells us the radius is 6mm. Wikipedia tells us that the temperature gradient is most likely caused by the gravito-thermal effect and the difference in temperature is 250 degrees.

    The calculations tell us the gradient is -g/Cp where Cp is specific heat (not heat capacity) but the force is about 10^6 x 9.8 according to SteveF. Hence the gradient is about 9.8 x 10^6 / 1.0 which is about 10^7K/Km. Applying that over 6mm we get about 600 degrees. Possibly the actual distance between the mid points of the two cylinders is only 4mm, which would give about 400 degrees, and reducing by a third gives about 270 degrees.

    Given that the 10^6 has only one significant figure, and the 4mm is a guess, I claim that 270 is close enough to 250 considering the magnitude of the figures involved.

  24. Doug Cotton says:

    Because the Second Law has to (and does) apply to thermal energy apparently transferred by radiation, back radiation from a cooler atmosphere does not penetrate water surfaces, even though such surfaces are almost completely transparent to the infra-red radiation that makes up about 48% of the incident solar spectrum. If back radiation were to penetrate and warm the water beneath the surface, this would be a violation of the Second Law, despite what climatologists teach climatologists about so-called net effects.

    Radiation one way is a completed independent process, and cannot be combined with any other process in order to derive a “net” result. You cannot have entropy decreasing in any such natural (spontaneous) process. You cannot justify a decrease in entropy just because entropy may increase more in some subsequent independent process.

    I was probably the first in the world to publish in March 2012 a comprehensive explanation as to why the apparent transfer of thermal energy by radiation is in fact a one-way process, with the amount being transferred corresponding to the area between the Planck curves.

    The rest of the radiation is common to both the Planck function for the warmer source and the Planck function for the cooler target. It is this radiation which undergoes what I called “resonant scattering” but others were starting to call “pseudo scattering.”

    This process involves photons raising electrons between energy states, but then the electro-magnetic energy (that became electron energy, but not kinetic energy) is immediately re-emitted as part of the target’s Planck function, because the target can indeed radiate that frequency and intensity. Hence the target uses less of its own molecular kinetic (thermal) energy and so its radiative cooling rate is slowed. However, non-radiative cooling is not affected and can indeed accelerate to compensate.

    Anyway, this is how and why the Second Law of Thermodynamics applies to radiation. So back radiation never transfers thermal energy to a warmer surface. Thus it cannot raise the maximum temperature to which the Sun’s radiation can heat a surface.

    In any event, it doesn’t work that way, and you need to understand the whole new paradigm relating to temperatures that are supported by the tropospheric thermal profile, the gradient for which is formed by the gravito-thermal effect.

  25. Doug Cotton says:

    To all readers: this is important.

    Skeptical Science team member Neal J King writes on Lucia’s Blackboard, referring to thermodynamic equilibrium: “a transfer of energy ?E between two sub-components, j = 1 and j = 2, will change neither E_total nor, to 1st order, S_total”

    Yes, and that is exactly what happens when there is a thermal gradient such that the difference in mean kinetic energy per molecule (temperature) exactly matches the negative of the difference in mean gravitational potential energy per molecule.

    You can see this in the second stage of the four molecule experiment: when thermodynamic equilibrium is attained we have homogeneous entropy (which must take PE into account) and every collision involves molecules with equal KE, and so KE for the system does not change, but is different per molecule at different altitudes. Similar happens in diffusion in a horizontal plane – KE of all molecules approaches homogeneity. But in a vertical plane you have to remember that KE changes because PE changes whenever there is a non-zero vertical component in the free path vector between collisions.

    The gravito-thermal effect is blatantly obvious when convection stops in the early pre-dawn hours. It is then that the pre-determined thermal profile has a “supporting temperature” at the base of the troposphere on any planet. That is what explains all the observations on all planets with surfaces, and even planets without surfaces. Temperatures are set based on radiative balance and the gravito-thermal gradient.

    The probability of these thermal gradients being so close to the -g/Cp value on all planets with significant tropospheres just because of some assumed warming by the Sun (whose radiation barely reaches some planetary surfaces) is absolutely infinitesimal. The evidence for the gravito-thermal gradient is blatantly obvious everywhere, as is the theory behind it.

    And as for radiation from carbon dioxide supposedly helping the Sun to attain greater maximum temperatures each day (despite the Second Law) or even just slowing radiative cooling – so what? Oxygen and nitrogen slow non-radiative cooling and outnumber carbon dioxide 2,500:1. Radiation from carbon dioxide (with its limited frequencies) is like a picket fence (with most of its pickets missing) standing up against a torrent of full spectrum radiation from the surface. The mean temperature of carbon dioxide molecules in Earth’s troposphere is far colder than the mean temperature of oxygen and nitrogen molecules colliding at the boundary with surface molecules. Rates of cooling depend on temperature gaps, so think!

    But arguing with lukes and warmists is like playing chess with a pidgeon. No matter how good a player I am, the pigeon knocks over the pieces, craps on the board and struts around looking victorious.

  26. Doug Cotton says:

     
    PROOF of EXISTENCE of the GRAVITO-THERMAL EFFECT

    (1) The second law of thermodynamics states that “the entropy of an isolated system never decreases, because isolated systems always evolve toward thermodynamic equilibrium— a state depending on the maximum entropy.”

    (2) “In thermodynamics, a thermodynamic system is in thermodynamic equilibrium when it is in thermal equilibrium, mechanical equilibrium, radiative equilibrium, and chemical equilibrium. Equilibrium means a state of balance. In a state of thermodynamic equilibrium, there are no net flows of matter or of energy, no phase changes, and no unbalanced potentials (or driving forces), within the system. A system that is in thermodynamic equilibrium experiences no changes when it is isolated from its surroundings.”

    (3) When, in the absence of phase change, chemical reaction or inter-molecular radiation, a gas has reached thermodynamic equilibrium, then there will be no net change in the distribution of energy on a macro scale.

    (4) In such circumstances described in (3) for every molecular movement between collisions, any change in gravitational potential energy must be countered by an opposite change in kinetic energy.

    (5) If (2) applies and noting (4) it follows that when any given molecule is about to collide with another, its own kinetic energy must be equal to that of the target molecule so that no net change occurs in the collision.

    (6) Hence, for any pair of molecules at different heights (or altitudes) the difference in gravitational potential energy must be offset by an equal and opposite difference in kinetic energy, thus maintaining a homogeneous sum (KE+PE) for all molecules.

    (7) Thus, because temperature is a function only of the mean KE per molecule, and because PE varies so must KE vary, causing a thermal gradient throughout the whole system,

  27. Doug Cotton says:

    There can be no correlation found between temperature data and carbon dioxide, because it cools only by about 0.1 degree. You need to come to grips with the new 21st century paradigm shift in climate science based upon the gravito-thermal effect.

    This is how absurd the old 20th century paradigm of greenhouse radiative forcing gets. They claim that you can work out Earth’s surface temperature by adding together the radiative flux from both the Sun and the colder atmosphere, and then bunging this total value into the Stefan-Boltzmann equation and out pops your answer 287K or 288K. Well it might well do if you fiddle the back radiation and then use the emissivity value instead of the absorptivity.

    But there’s absolutely no physics to support the calculations. When you consider that about 70% of the surface is a thin transparent water layer, it is obvious that the solar radiation which mostly (like over 99%) passes through this layer into the thermocline is not what is determining the temperature of that thin surface layer. In fact the mean temperature of the thermocline is obviously less, and the mean temperature of all the ocean water is less again.

    Oh, and the back radiation doesn’t even enter the surface layer – it just raises electrons between quantum energy states momentarily, and then those electrons immediately emit another photon which climatologists think is energy coming from the kinetic energy in the surface molecules, but it’s only electro-magnetic energy from the back radiation being thrown back in their red faces.

  28. Clive Best says:

    On the gravito-thermal effect… If you take an inert gas and release it on the surface of a planet without any atmosphere then it will distribute itself according to a balance between kinetic and potential energy and produce a lapse rate -g/Cp – a true gravito-thermal effect. However, the temperature of the planet will remain constant providing there are no GHGs present. The gas cannot heat the surface. Its kinetic energy is comes from the surface. If the gas contains no GH molecules then the atmosphere cannot radiate to space. Therefore I think the Gravito-thermal effect is a transitional state because eventually the MB distribution will slowly converge to exp(-m(2gh+v^2)/2kT unless heat escaped from the top of the atmosphere.

    I think this is also what happens in the vortex tube. However, in this case the gas escapes the from the tube (from the 1000g surface) immediately to show the temperature difference. It never has time to reach thermal equilibrium.

    The question is what happens if you leave the atmosphere for 1000’s of years or if you close the escape nozzle of the vortex tube ?

  29. Doug Cotton says:

    The Second Law of Thermodynamics never mentions thermal equilibrium or heat transfers from hot to cold. It is all about evolving towards thermodynamic equilibrium which is quite a different thing, involving mechanical equilibrium as well, and thus gravitational potential energy.

    The Ranque-Hilsch vortex tube is indeed constantly evolving towards thermodynamic equilibrium and it generates over 10^6g and hence creates an easily measured gravito-thermal effect. How else could the air in the centre cool to around -50C? There has been no other valid explanation in many years and so, as I have said in Wikipedia, the gravito-thermal effect is the most probable explanation.

    The Maxwell-Boltzmann distribution is derived theoretically for a non-gravitational field. Hence it is not strictly correct when a state of thermodynamic equilibrium (or close to such) exists and you are considering molecules at different altitudes, although it will apply in any horizontal plane.

    It is a red herring to postulate a gas that does not absorb any solar radiation and re-emit it. If such an atmosphere did exist it would still exhibit a thermal gradient but, by the assumption made, it would be just as if it weren’t there at all as far as radiation is concerned. Of course if it got too cold near the top, some would solidify and collapse.

    It is energy from the Sun (mostly absorbed in the atmosphere) which heats the surface of Venus, for example, and actually raises its temperature from about 732K to 737K during the course of its 4-month-long daytime. But the whole temperature profile in the troposphere has to rise 5 degrees also for this surface warming to happen, and then indeed the surface is warmed by conduction from the base of the troposphere.

    All planetary temperatures in tropospheres and even beneath any surface are determined by the gravito-thermal effect, and they have nothing to do with any greenhouse radiative forcing or sensitivity to carbon dioxide.

    When they drilled the KTB borehole down to 9Km depth in Germany they were surprised at how much water they found underground. This then helps confirm that the gravito-thermal effect is also apparent in solids and liquids. At 9Km depth it was 270C, far hotter than they expected, with a thermal gradient in the outer crust at least 20 times as steep as the mean gradient to the centre of the core. That’s because specific heat increases very significantly with the hotter temperatures in the mantle and core.

    If you plotted just the temperatures between, say, 9Km and 4Km you would find that the near linear plot extrapolates quite well to the actual mean minimum daily temperatures at the surface.

    Why is it so?

  30. Doug Cotton says:

    Clive, unless you understand the physics you will never be convinced of the validity of the gravito-thermal effect. It is a direct corollary of the evolving process which statements of the Second Law of Thermodynamics describe. Hence is cannot be an intermediate state, because it is the state of maximum entropy. You cannot decrease entropy, and that is what you would be doing if you were trying to generate a state with more gravitational potential energy per molecule at the top, and yet the same kinetic energy per molecule as at lower levels.

  31. Doug Cotton says:

    SUMMARY

    In an adiabatic process in a sealed and perfectly insulated vertical cylinder of a solid, liquid or gas a thermal gradient evolves in accord with the process described in statements of the Second Law of Thermodynamics.

    This fact may be used to deduce that such will also occur in calm conditions in a planet’s troposphere if no new energy were being absorbed, such as is close to the case in calm conditions in the early pre-dawn hours, when surface cooling and upward advection almost stops. In such as situation we can observe that there is indeed a thermal gradient, but there is no heat transfer from the lower warmer regions to the cooler regions above, for the simple reason that there is already a state of thermodynamic equilibrium.

    Molecules move in random directions after each collision, and the direction is not significantly dependent upon the kinetic energy in the molecule. So the calculation of the thermal gradient has nothing to do with pressure or density or rising packets of air. There is no such thing as a moving packet of air in adiabatic conditions anyway, because the probability of trillions of molecules all moving in the same direction is absolutely infinitesimal in the absence of wind or forced advection caused by an external energy source like a fan.

    Temperature is the independent variable and only changes if mean molecular kinetic energy changes. Gravity sets up non-zero gradients in density and temperature. Pressure is merely the end result because pressure is proportional to the product of density and temperature.

  32. Doug Cotton says:

    The physics of the gravito-thermal effect gives the right answers.

    Lukes and warmists consider this …

    You cannot explain the gravitationally induced thermal gradient in a vortex tube.

    You cannot explain how the extra energy gets into the Venus surface to raise its temperature with what has to be a net energy input. There cannot be a net energy input brought about by radiation from a colder atmosphere as that obviously would violate the Second Law.

    The Venus atmosphere cannot magnify the incident solar radiation at TOA up to 14,000 to 16,000 watts per square metre that would be needed if radiation were adding energy to the surface to raise its temperature 5 degrees during the Venus day.

    Oxygen and nitrogen molecules in Earth’s troposphere absorb thermal energy by conduction and diffusion processes. They do most of the slowing of surface cooling because there are 2,500 times as many of them as there are carbon dioxide molecules..

    I can explain why surface cooling slows right down and upward convection sometimes stops altogether in calm conditions in the early pre-dawn hours, even though the thermal gradient is still there.

    I can explain why hydrostatic equilibrium is the same as thermodynamic equilibrium, because there can be only one state of maximum entropy.

    Of the incident solar radiation entering Earth’s atmosphere, NASA net energy diagrams showed 19% absorbed on the way in compared with only 15% absorbed on the way back up from the surface. What does that tell you about how the atmosphere gets warmed? It’s like on Venus – more solar energy is absorbed on the way in.

    I can explain why real world data (which I will publish in an Appendix to my book) proves with statistical significance that water vapour cools. The IPCC wants you to believe that it warms by a staggering amount of the order of 10 degrees per 1% of moisture in the atmosphere. That’s simply not what it does, and only the most gullible of people would believe that to be the case.

    I can explain why planets are neither warming or cooling significantly.

    I can explain why the core of our Moon is kept hot by the Sun, as is the case for the cores of all planets and moons.

    I can explain the temperatures in the Uranus troposphere where there is no surface and no significant source of insolation or internal energy.

    I can explain all known and estimated temperature data above and below any surface on any planet or satellite moon.

    You can’t.

  33. Doug Cotton says:

    The greenhouse conjecture would violate the laws of physics. It is totally wrong.

    My study showing water vapour cools is not hard to replicate. To prove me wrong you would have to produce a similar study proving water vapour warms by about 10 degrees for each 1%, as is in effect claimed by GH advocates.

    The Ranque-Hilsch vortex tube provides evidence of the gravito-thermal effect. You would need to provide contrary empirical evidence.

    You would also need to produce a valid (but different) explanation as to how the necessary thermal energy gets into the Venus surface in order to raise its temperature by 5 degrees during its sunlit hours.

    BigWaveDave considers the gravito-thermal effect (seen in the vortex tube) worth your time thinking about …

    “Because the import of the consequence of the radial temperature gradient created by pressurizing a spherical body of gas by gravity, from the inside only, is that it obviates the need for concern over GHG’s. And, because this is based on long established fundamental principles that were apparently forgotten or never learned by many PhD’s, it is not something that can be left as an acceptable disagreement.”

  34. Doug Cotton says:

    You all still follow J.H. and not H.J.

    You follow James Hansen rather than Hans Jelbring, and that’s your problem.

    You think it’s all about radiative heat transfer, when in fact it is mostly about non-radiative heat transfer which obviously is what keeps warm the thin transparent surface layer of Earth’s oceans, the base of the Uranus troposphere, the surface of Venus, the core of our Moon etc etc etc .. throughout the Solar System. It’s all due to the gravito-thermal effect.

    The greenhouse conjecture adds back radiation to solar radiation and uses the total to “explain” the surface temperature. But of course this is wrong. The original NASA net energy diagram showed only about 165W/m^2 entering the surface, but that gave a far too cold temperature in Stefan-Boltzmann (S-B) calculations, so it had to be nearly trebled with back radiation.

    The problem is, no one should be using S-B and be expecting to get the right answer.

    All that S-B calculations can be used for is the mean temperature of the whole Earth-plus-atmosphere system, and it does give about the right value when you deduct about 30% of incident solar radiation due to reflection, but retain about 20% that is absorbed by the atmosphere itself.

    Now, the big problem with all this is that 70% of the real surface is a thin layer of transparent water, let’s say 1cm deep. If you were to use S-B calculations to determine the temperature of that layer, bear in mind that over 99% of incident solar radiation that is not reflected passes right through it. So you should only use 1% or less of the 165W/m^2 of solar radiation and thus get ridiculously low values. Back radiation doesn’t have a hope, because it does not even penetrate a hair’s width into that first 1cm of water, and if all its energy were converted to kinetic (thermal) energy in that hair’s breadth, it sure would be hot and evaporate rather quickly without warming anything else. The fact that it doesn’t, confirms what I wrote in my paper over two years ago about resonant (or pseudo) scattering.

    So you see, you cannot explain Earth’s surface temperature with radiation calculations, for the simple reason that, like the Venus surface, it receives a significant amount of energy by non-radiative processes as is explained in the Amazon book “Why it’s not carbon dioxide after all” by yours truly.

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