Ode to Misinterpretations of the Second Law

October 21st, 2014 by Roy W. Spencer, Ph. D.

Inspired by a couple comments from my solar eclipse post.

He said an object that was cold
Could not make something warm still warmer
So he donned his coat, went out the door
To prove the truth of former.

“See?” he said, “the sky is cold”
“and so it cannot warm”
Then back inside he merrily went,
Removing the cold coat he’d worn.

-Burma-Shave


211 Responses to “Ode to Misinterpretations of the Second Law”

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  1. jimc says:

    Now that’s downright disrespectful.

  2. I supposed I could have called it “Ode to the Greenhouse Effect”

  3. jimc says:

    Ode to the greenhouse effect:
    Those little atoms, the heat they reflect
    And cause a negative temperature gradient.
    You see, its all radiant.

    No. Never mind.

  4. Dave says:

    Make something warm or keep something warm…..there is a difference.

    • Scott says:

      It depends on how heavy your coat is. I could put a coat on you that would make you die of heat stroke in the dead of winter!

      • Geoff wood says:

        Hi Scott, Dave doesn’t mean that. Place your warming coat about an object that doesn’t produce heat like a human body does, and rotate it upon a pedestal in front of a tungsten lamp as the only heat source.

        Does the coat then heat the object?

  5. Kelvin Vaughan says:

    Is there a formula for the relationship between sky, ground and air temperature?

    • jimc says:

      One simple/gross rule of thumb I got out of my meteorology seminar was a negative 10 deg C per mile for the troposphere starting at +19 deg C at surface to -62 deg C at the tropopause (7.5 mi up and 80% of the atmospheric mass heated from below by the earths IR). For the upper 60% of the stratosphere, it slopes from -45 deg C to 0 deg C (from 30 mi to 50 mi heated from above by the suns UV). Other places, not so simple. The cloudless sky (read with my IR thermometer) reads about 30 to 35 deg cooler than the surface (perhaps an average for the troposphere?). Coincidently (or not), that is about the temperature the earth radiates to space as.

      • jimc says:

        Oops. Sorry, those stratosphere numbers are in km (about 18 to 30 mi up).

        • Geoff wood says:

          Look up ‘derivation of the adiabatic lapse rate’

          • jimc says:

            What I have is that the dry adiabatic lapse rate is -16 deg C per mile (not enough to overturn in the tropospheres -10 deg C). The moist adiabatic lapse rate can become high enough (e.g. -8 deg C per mile) locally to do it (billowing clouds, a thunderstorm, or something even more dramatic).

          • jimc says:

            The troposphere is inherently vertically stable. Local instabilities are worked out of it by the conversion of water vapor to water or ice in clouds, rain, and storms.

          • Geoff wood says:

            Jim, look up the derivation mate. It is not radiative. The lapse is thermodynamic. A consequence of gravitational containment.

          • jimc says:

            Except for special forced situations (e.g. turbulence, wind moving air up and over a mountain, or air nearing saturation), there is no vertical airflow rapid enough to fit the definition of adiabatic. Instead the air moves so slowly vertically that it reaches an equilibrium temperature dominated/controlled by other factors. Thats where radiation comes in. People use things like lapse rates theoretically to test if the air is vertically stable. On the grand scale, it is. On that scale, there is no adiabatic process going on.

          • Geoff wood says:

            jimc. So take tropical air at around 35degC, with a moisture content of around 30g per kg of air. By adiabatic I am requiring that no energy enters or leaves. ie the energy of the air at the surface is the energetic equivalent at equilibrium of the air aloft. If we define the upper tropopause at -63degC then dry air adiabatic would give an upper tropospheric height of -35+63=93K/9.98, or 9.32km

            However add the energy from 30g of water, present in the humid air, cooled to ice point and we add around 90,000J per kg to the energy content of the air at the surface.
            For 1kg of air this equates to ~9km of potential energy so raises the equilibrium height by 9km to 18.32km without any energy added or subtracted.

            The energy to do this exists within the surface air energy content.

            Therefore the approximation of the troposphere to an adiabatic which by definition is conservation of energy requires little else to explain it. It does not require the dominating diabatic processes (radiative) you speak of when the bulk of the troposphere exists in an ‘equal total energy’ state.

          • jimc says:

            Geoff, I have to beg off. Being a novice, I cant keep up without a lot of reading and reflection.
            I do, however, have an objection to your example. Some of the numbers I have for vapor capacity of air are:
            -20 deg C 0.75 g/kg
            -10 deg C 2 g/kg
            0 deg C 4 g/kg
            10 deg C 8 g/kg
            20 deg C 15 g/kg
            30 deg C 28 g/kg
            40 deg C 50 g/kg
            I.E. roughly half for every decrease of 10 deg C. In other words, your kg of air 35 deg C and 30 g/kg of moisture is at saturation. If you hoist that kg of air to the tropopause with a normal temperate of -63 deg C, it could only hold about .027 g/kg. Something non-adiabatic has to happen (heat has to depart embedded in water or ice). Admittedly, something lifting as much air as say a hurricane could temporarily raise the temperature of the upper troposphere above normal; but how much, for how long, and what percentage of the earth is effected by a hurricane at any given instant of time?
            Water complicates things immensely. Thats why I tried to avoid it in my postulation. And it is a transient thing (it gets dumped). Can you give an example I can cope with? 🙂

          • Geoff wood says:

            Jim, the lapse rates you quoted are based upon the thermodynamic principles that govern any gravitationally bound atmosphere. The common derivation of the lapse is from the 1st law of thermodynamics. However a simpler derivation is available through kinetic theory in a gravitational field. The existence of a thermal gradient running downwards counters the undeniable gravitational potential energy gradient running upwards as a mechanism of containment.

            If you throw an object aloft you can see the effects of gravity. Potential energy at the zenith and kinetic energy at low altitudes. Molecules have mass and therefore feel gravity. At these pressures they spend most of their time between collisions where gravity modifies the mean free path. In a gas 1/2mv^2 is 1/2kT where v is the mean thermal velocity.

            Returning to the example given. Firstly part of the calculation was clumsy on my part and should have read 35+63=98K (surface to upper troposphere) difference. Running the dry adiabatic lapse of 9.8K/km gives an upper tropospheric altitude of 98/9.8 or 10km for energy equivalence from temperature alone. This is around 10km less than the equatorial height of the tropopause start. However, the lapse is about the available energy storage states that can and do modify the thermal profile. Evaporation of water and incorporation of water vapour within the surface air greatly reduces the surface temperatures through latent heat capacity. This energy once part of convection is the major heat source of the atmosphere. The 30g per kg of water as you have noticed available within tropical air cannot be contained as circulation cools (by doing work against gravity) the cell to saturation, releasing 2257J per gram of water condensing into every kg of air as it ascends. This then powers the ascent further. The total of water vapour contained is sufficient to describe the upper tropospheric height if all the latent heat and heat capacity of the water is accounted for. In reality the profile is stepped regularly by near isothermal heights as water condensation and fusion controls and reduces the total lapse to the wet profile of around 5.5K/km from 9.8K/km for dry air.

            Hope that provides some clarity.

          • jimc says:

            Geoff, again I have to beg off the water thing, so Ill stick to the first two of your paragraphs. First, just a nit, I think the KE of a gas particle is x*k*T where x varies between 0 and 1.5 depending on which particle velocity you use (most probable, average, or rms).
            Next, it seems you are saying (or implying) that gravity affects the velocity of the gas particles (i.e. the ensembles temperature). My counter is that if I had a weightless, rigid, insulated box containing a fixed quantity of air at some pressure and temperature at infinity (no gravity) and lowered it to the surface of our planet (1 g of gravity), the change in PE was absorbed externally by the crane that lowered the box (or by the buckling of the earth and box in the crash if dropped). Like height and velocity, PE and KE have to be relative to something. In this case for the particles in the box, it is the box. The net PE of the particles in the box did not change. If they shifted within the box, then the increase in one part was offset by a decrease in another part for no net change. Likewise, if you want to talk about particle ballistics inside the box, a particle that loses PE to KE recovers the PE again on the downward path for no net change in PE. And no change in KE (temperature) since this process is adiabatic.
            The application or change in gravity by itself does not modify temperature within the box (or anywhere else). Something else has to occur (e.g. the box gets smaller).

          • jimc says:

            Oops. x varies between 1 and 1.5.

          • Geoff wood says:

            Jim don’t bother with the ‘nit’, there is no ‘think’, 1/2mv^2=1/2kT per degree of freedom, where v is the mean thermal velocity.
            Next I ‘am saying’ not implying, that gravity affects the velocity of gas particles in a gravitational field. Why would you think there might be exceptions? At these pressures molecules spend most of their time between collisions where their trajectory ‘is’ modified by gravity.

            On what level so you think gravity switches off?

            Gravity is the containment. All self gravitating bodies are hot in the middle whilst supported on the surface by even a piddling external flux.

            Forget the box and the crane. Thermodynamic work is done by or from ‘spontaneous’ change of volume. PV is energy, and for the surface layer, for instance, heated by the Sun, volume increases at constant pressure. Pressure is the result of the weight of atmosphere above and that doesn’t change as the atmosphere at the bottom is heated. The box and the crane immediately rules out reality. Firstly buoyancy or lack of does not require a crane. It is a spontaneous process. Secondly Cp describes a thermodynamic atmosphere not restrained by a box. Cv describes conditions in your box and Cp-Cv=thermodynamic work (R).

          • Massimo PORZIO says:

            Hi Geoff,
            for what it’s worth, when I came here this night my intention was to write more or less what you wrote above.
            I fully agree with you.

            What I can’t realize is how do the ones who support the isothermal atmosphere explain the outer layer of the atmosphere?
            If the temperature at the TOA was the same of the ground zero level, then the molecules of the outer layer should have converted all the vertical KE received from the layers below into an horizontal KE, this because they can’t rise no more.
            How could a vertical movement being converted into an horizontal one without gravity?

            Have a nice day.

            Massimo

          • jimc says:

            Geoff and Massimo, I applied gravity to the particles without changing the pressure (the box was rigid). I separated the change in gravity from the accompanying change in pressure you mentioned. The change in gravity alone does not affect the temperature of the box contents; you need to change pressure as well. There is already physics for that (ideal gas law, etc.).

          • Massimo PORZIO says:

            Hi jimc,
            you missed to state that your box must be an imaginary perfect thermal insulator (which doesn’t exist) otherwise when your box has dropped to the ground it is no longer in thermal equilibrium until the inner gas molecules regain the external gas molecules KE.
            With your imaginary box you made an insulated system where gravity works anyways but its influence on temperature profile is so little that you can’t measure it. The molecules inside your box get so much KE that they collide against the top and the bottom of the box without having the space to show any thermal gradient for our current thermal resolution measurement.
            It’s the very same Maxwell issue.

            Have a nice day.

            Massimo

          • jimc says:

            Massimo, Geoff said that gravity affects the velocity of gas particles [temperature] in a gravitation field. The idealized experiment shows that the application of gravity alone does not change the temperature of the gas. Thats precisely my point. It has to be accompanied by something else such as an increase in pressure to change temperature. Things like the ideal gas law take care of that.

          • Massimo PORZIO says:

            Hi jimc,

            Ok, now I get your point, and I agree.

            Have a nice day.

            Massimo

          • jimc says:

            Massimo, thanks for enduring my long winded explanation. I was trying to plug all the leaks.

          • jimc says:

            Uh … loopholes.

          • Geoff wood says:

            Jim, careful with the wording.

            “Geoff said, “that gravity affects the velocity of gas particles (temperature) in a gravitational field””

            Gravity does affect the velocity of gas particles in a gravitational field!!!!!

            Gravity is a force and a force produces an acceleration and therefore a change in the vector quantity that is velocity in the direction of the force upon all matter (and energy).

            It produces a density, pressure and thermal gradient.

            Lower the box via the crane onto a neutron star.

            When all the molecules are twitching on the lower surface then you will have to acknowledge that at some time earlier a density, pressure and thermal gradient started to invade even the weird world of ‘the box’.

          • Geoff wood says:

            Jim. If you require a reference point then it’s the ‘effective mean radiative height’. Not the outside of your ‘box’

          • Geoff wood says:

            Jim, from kinetic theory we can derive the macroscopic gas laws. You seem to treat them as separate entities.

          • jimc says:

            Sorry Geoff, youre getting too incoherent for me to even form a response. I demonstrated two ways, macro- (first law and an adiabatic process) and mirco- (particle PE/KE). Peace.

          • Massimo PORZIO says:

            Hi jimc, hi Geoff,
            jimc, I thought a little more about your box examples and indeed it demonstrates that Geoff is perfectly right.
            In fact, when your rigid box (even if perfectly thermal insulated) touch the ground, it discharge its KE which is the result of the gravity converted PE of the box but also the one of the inner gas molecules. If it was empty (it had vacuum inside) then it weighted less, so it released less KE to the ground.
            IMHO this demonstrates that even if you constrain the gas molecules in a box and move the box into a gravitational field, the molecules convert their PE to KE too.

            Have a great day.

            Massimo

          • jimc says:

            Massimo, the individual particles are moving relative to the earth during the decent to earth. PE like height has to be relative to something. I separated the height of each particle into two parts the height relative to the horizontal center plane of the box and the height of that plane above the earth. The particles dont know whats going on outside the box and the crane doesnt know whats going on inside the box. All the particles are aware of is that there is an increasing gravitation pull toward the bottom of the box. They are positioned and move relative to the box and make only short temporary excursions away from the center plane some above and some below for no net PE change within the box. The crane removes the PE of the box (and the ensemble of particles in the box) EXTERNALLY, not effecting the adiabatic-ness of the box. There is no energy passing through the walls of the box via the ether or any other spooky means. The net PE change within the box is zero as the box descends. Therefore, the net KE (temperature) change is zero (the box and particles are an adiabatic system). If you dont believe me, imagine the box fixed in space and the earth moving toward the box instead (with the crane holding the box in place). Easier to visualize? Force may be passing though the walls of the box, but work is not.

          • Massimo PORZIO says:

            Hi jimc,
            uhmmm…
            Maybe be I’m misunderstanding something, because you wrote:
            “The crane removes the PE of the box (and the ensemble of particles in the box) EXTERNALLY, not effecting the adiabatic-ness of the box. There is no energy passing through the walls of the box via the ether or any other spooky means.”
            In the first part of the sentence you asserts that the crane removes both the PE of the box and those of the particles inside, but in the second part you state that no energy passes through the box wall.
            Ok, I know that PE is just potential energy, so it is not temperature for gasses, I fully agree with that, but the only reason you don’t have it converted to KE is because you constrained the gas in the wall of the box.
            While this doesn’t exclude that free falling gas molecules into a gravitation field can convert PE to KE, it confirms that the crane increased its work during the descent for the particles too, not for the box itself only. IMHO this imply that when the molecules fall free they have to “put” their PE somewhere, don’t you?

            Have a nice day.

            Massimo

          • jimc says:

            The crane got hot,
            The gas did not. 🙂

          • jimc says:

            Massimo, sorry, I wasn’t being smart. I posted that last as you were posting yours. I’ll have to think a minute about yours.

          • jimc says:

            Massimo, yes, a free falling particle (relative to earth) would gain speed and convert PE (relative to earth) to KE. But if I tie a string to it and lower it slowly, it gains no KE and I absorbed the PE. I upped the particle to a box of particles so I could measure temperature (and create an adiabatic system). Have I destroyed my assertion that the application of gravity alone does not heat the box contents? I dont think so. Your opinion?

          • Massimo PORZIO says:

            Hi jimc,
            sorry but, I’m Italian and English is not my native language.
            So sometimes I miss some important details, especially when the post are long.
            It seems that we are agreeing on this argument indeed. 🙂

            Have a nice day.

            Massimo

          • jimc says:

            Massimo, Ah, a long distance discussion (Im in the US, I think Geoff is in Australia). Your English I light years above my Italian.

          • Massimo PORZIO says:

            Yes, this is a “a long distance discussion”, here we say it’s a “mirror of these times”.
            Not sure it has a meaning in English, but here is the same as “not wonder these are normal things supposed to happen these days”.

            Have a great day.

            Massimo

          • Geoff wood says:

            Jimc. With respect. You have said,

            “Sorry Geoff, youre getting too incoherent for me to even form a response. I demonstrated two ways, macro- (first law and an adiabatic process) and mirco- (particle PE/KE). Peace.”

            What have you demonstrated? The second sentence is not complete.

            Later, you repeat,

            “PE like height has to be relative to something”.

            In my ‘incoherent’ answer was,

            “Jim. If you require a reference point then its the effective mean radiative height. Not the outside of your box”

            Then you say in explanation of this weird box experiment,

            “Have I destroyed my assertion that the application of gravity alone does not heat the box contents?”

            At no point Jim have I implied in any way that gravity added to the energy of the atmosphere. At no point have I implied that the atmosphere was heated by gravity. Gravity provides gradients in density, pressure and temperature at equilibrium with the supporting flux.

            Gravity provides the initial heating of a collapsing object like a star, or Neptune, or Saturn or Jupiter. Once collapsed the thermal gradient is supported by equilibrium over the last optical depth by the external flux. No heat is added by gravity once the collapse is supported externally. Without a supporting flux the gravitational potential would be converted to thermal energy through collapse and radiated to space through interaction. However, the position of the mean effective radiative height is in many objects in space in a position of a temperature at a massive potential energy. The temperature equivalence of this to the core, with zero potential energy yields the core temperature through the respective heat capacities.

            I am describing equilibrium equivalence. Not heat added by gravity.

            Is that incoherent?

            Potential energy reducing as kinetic energy increases through to the core?

            Gravity produces within any atmosphere a thermal gradient. No energy added.

            The same thing happens within a box given sufficient gravity. But you are off on tangent.

          • Geoff wood says:

            Massimo, I thank you for your comments and appreciate that language issues can be a problem. Having read many of your replies on many threads without commenting I an aware that your responses are sensible, honest and measured. We may not agree over every detail but please don’t hesitate to question. I can assure you that I have taken on board details of your opinions on many occasions.

            Regards,

            Geoff

          • Massimo PORZIO says:

            Hi Geoff,
            yes, I know that sometimes I wrote silly things, this is because I’m not a scientist at all, I’m just an electronic engineer which does some thoughts about climate issues which I feel a little “hard to believe” (such as the isothermal GHGs free atmosphere).
            Anyways I appreciate the ones who discuss without fall in squabble.
            When people “write aloud” sometimes they hide some their misunderstandings of the issue indeed.

            Thank you for write here.

            Massimo

      • MikeB says:

        You cant use an IR thermometer to measure the temperature of the sky. They are specifically designed to operate in the atmospheric window and thus not see the air. The IR thermometer must also assume an emissivity for the surface it is looking at, usually 1. What constituent of the air do you think has an emissivity of 1?

        • Kelvin Vaughan says:

          Can you use one for measuring the temperature of clouds?

          • jimc says:

            I have. As Mike pointedout, though, there is an interpretation problem. Is the emissivity of a collection of water droplets 1? Judging from some measurements I made of liquid water heated on a stove with a contact thermometer and the IR thermometer, I think liquid water at IR must be close to 1.
            Specifically, I did measure a cloud and got a reading of about 10 deg C colder than surface. At the standard lapse rate, that would put the cloud about 1 mi up. You got a way of measuring the height of clouds? 🙂

        • jimc says:

          Mike. I agree, I did not measure the temperature of an object. Someone asked a question, and I gave the best answer I had. I did see IR radiation though something far more than the coldness of space (which I guess is 4 Kelvin). The question is, what did I see? I think it must have something to do with GHGs which are warm and warm the adjacent (transparent) air molecules. But you raise an interesting point: Why can I measure an object embedded in air that is not the same temperature as the object, is the air such a poor radiator? Apparently so (If its not a good absorber, its not a good emitter). But GHGs can absorb and can therefore emit (same wavelength?).
          I suspect that if I could see with IR eyes, I would have seen a milky white soup hiding anything on the other side. Maybe I can think of the atmosphere as being the equivalent of a radiant sheet at the temperature I measured. Is that the green house blanket?

          • MikeB says:

            Your IR thermometer probably operates at wavelengths around 10 to 12 microns. The instruction leaflet may explain its exact limits.

            Greenhouse gases (GHGs) can only emit at wavelengths that they absorb. The common GHGs do not emit at all in this wavelength band, neither do oxygen and nitrogen, although CFCs and aerosols such as Nitric Acid will absorb and radiate slightly in this band.

            The emissivity of liquid water is about 0.99 over the 10-12 micron region.

            If your eyes operated in the 10-12 micron band then they would see pretty much as you do in the visible region (around 0.5 microns). You have probably seen images from IR cameras on police helicopters.

            On the other hand, if your eyes were to operate on a wavelength of 15 microns all you would see is a thick fog. Your visibility would be about one metre. This is the CO2 absorption band.

          • Geoff wood says:

            Interesting though that Ian Stewart, on TV used a candle at 1700K and massive amounts of CO2 to obscure the flame when he could have then apparently chosen 15um and 390ppm to illustrate this point.

          • jimc says:

            Mike, as you suspected, my thermometer works from 8 to 14 um with a 95% calibration. It seems you are saying that Kevin can measure the temperature of (the underside of) clouds.
            The question of what did I measure remains though.

    • wayne says:

      Kevin, testing, if Roy’s site allows some simple latex like:
      $latex \displaystyle half=\dfrac{2}{4} $
      I can give you some equations you might find useful. Unfortunately without latex they would look a mess.

  6. BBould says:

    Wow a Burma-Shave reference! Popular road signs back in the day.

  7. Chuck L says:

    Dr. Roy, Did you see Abraham’s arrogant & pompous missive in the Grauniad claiming to refute your and Dr. Braswell’s simple climate model?

  8. Mike Flynn says:

    First, the coat wearer’s temperature rose not at all, if he was in good health. Within a fairly wide range, core temperature remains constant, From the searing heat of the Libyan desert, to the somewhat chilly Tierra Deo Fuego, coats of one sort or another are usually desired.

    There is a desert Berber saying “If I had known it was going to be this hot, I would have worn a thicker robe.” So coats can apparently cool, as well.

    Second, all joking aside, clothe the nearest recently deceased corpse you can find in your finest, warmest, coat, and tell me how well it is warming the corpse. An ordinary poorly calibrated and imprecise anal thermometer should suffice. My prediction is that the corpse will continue cooling to ambient, rather than warm up ie, increase its temperature.

    And so with the Earth. It’s cooling, and will continue to do so for some considerable time.

    Live well and prosper,

    Mike Flynn.

    • Pierre-Normand says:

      Mike Flynn wrote: “And so with the Earth. Its cooling, and will continue to do so for some considerable time.”

      It’s also being warmed by the Sun and is going to be warmed by it for a long time. While the Sun warms it, increasing CO2 concentration reduces the amount of cooling. So it’s like adding insulation to your house while the central heating system produces a steady heat output.

      That’s the trouble with your dead man analogy. The dead man doesn’t have a warming source anymore (metabolism). The Earth is alive so long as the Sun is. If the Sun were to suddenly shut down, then you would have a point.

  9. Physicist.↓ says:


    Roy and others:

    The Clausius (“hot to cold”) statement of the Second Law is really just a corollary which has certain pre-requisites, namely that it only applies if other forms of internal energy remain constant. We know this because thermodynamic equilibrium (which the Second Law says will evolve) must take into account entropy and thus all forms of energy.

    In particular, the mean gravitational potential energy of any small region in which a process takes place must be homogeneous if we are only considering mean molecular kinetic energy which determines temperature. In other words, for conduction, diffusion, advection and convection you can only claim that thermal energy transfers from hot to cold in a horizontal plane.

    In a vertical plane any change in gravitational potential energy must be offset by an equivalent change in kinetic energy (temperature) if there is a state of thermodynamic equilibrium.

    Unless you understand this Roy, you will never understand the thermodynamics which explains why it’s hotter than Earth at the base of the nominal troposphere of Uranus, even though it’s nearly 30 times further from the Sun than we are.

    The Second Law of Thermodynamics can be applied to explain why a planet’s troposphere has a density gradient. This has nothing to do with upward convection from a surface heated by solar radiation. There is no such surface on Uranus, but the density gradient is still there. When you have such a density gradient caused by gravity acting on molecules between collisions, then you must also have a temperature gradient. Each is the state of thermodynamic equilibrium and, until you understand why, you do not understand thermodynamics. I suggest you heed what Josef Loschmidt said in the 19th century, because he was right.

    When you realise that the sloping thermal profile acts like a level surface of a lake and spreads new thermal energy (rain) in all accessible directions, then, and only then, will you understand how downward convection is what supplies the missing energy that James Hansen though he could explain as being due to back radiation. But it isn’t, and thus the whole greenhouse conjecture is false.

    That’s physics, Roy!

    • Tim Folkerts says:

      Hey “Physicist”

      Wouldn’t your argument also apply to other forms of potential energy?

      In particular, the mean gravitational potential energy chemical energy of any small region in which a process takes place must be homogeneous if we are only considering mean molecular kinetic energy which determines temperature. In other words, for conduction, diffusion, advection and convection you can only claim that thermal energy transfers from hot to cold in a horizontal plane the same chemical compound.

      So if higher altitudes with more gravitational PE are “naturally cooler” than lower altitudes after thermal equilibrium has been reached, then logically a tank full of gasoline and oxygen with more chemical energy should be “naturally cooler” than an adjacent tank of H2O & CO2 that has lower potential energy once they are allowed to come to thermal equilibrium. Or a compressed spring with more potential energy should be cooler than an adjacent uncompressed spring.

      (Not to mention that I could easily build a perpetual motion machine out of your “different temperature but in thermal equilibrium” situation.)

      • Physicist.↓ says:

        The IPCC models and mine postulate and work within an “ideal” atmosphere, just as do the Ideal Gas Laws which climatologists resort to (unnecessarily) to calculate the “lapse rate” in a roundabout clumsy way in which pressure is introduced only to be then cancelled out.

        Weather conditions (and that includes release of latent heat when it rains) and any chemical reactions are ignored in what I am describing. As explained in my book, we only need to consider what does change in “ideal” calm conditions, namely gravitational potential energy and kinetic energy.

        Now, you talk about thermal equilibrium. I don’t and neither does the Second Law which says the system tends towards thermodynamic equilibrium which is a more comprehensive set of circumstances.

        Neither you, Robert Brown nor I can build a perpetual motion machine. What you don’t understand is that, if you combine two systems, the new system just moves to its own state of thermodynamic equilibrium (with an intermediate temperature gradient) and because it is such an equilibrium state, there are no unbalanced energy potentials and thus no further net energy transfers, let alone any cyclic motion.

        Your inane comments just demonstrate to me your complete lack of understanding of thermodynamic equilibrium.

        Who’s next?

      • Physicist.↓ says:


        To Roy and silent readers:

        Tim Folkerts had a similar argument with “BigWaveDave” over two and a half years ago here. Tim never learns and never thinks – he just calls upon (and demands) “authority” all the time, even though any pass level graduate in physics ought to understand the thermodynamics involved.

        BigWaveDave was writing about the Loschmidt gravitationally induced temperature gradient, which stems from the process described in statements of the Second Law of Thermodynamics (that is indeed so misunderstood) and which gradient wipes out any need for a radiative greenhouse effect. I quote …

        BigWaveDave March 1, 2012 at 4:19 pm

        Tim Folkerts:

        You asked

        “what qualifications do you have to judge a disagreement between PhD physicists on issues of fundamental thermodynamics?”

        I have been earning a living as an engineer specializing in cutting edge technology for very large scale thermal energy transfer processes and power systems for close to 40 years. My credentials include BS, JD and PE, and I have four patents.

        As for my qualifications to engage in argument with PhDs, I have many times been part of and have led teams with PhD team mates. I was also married to a PhD for 20 years.

        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 GHGs. And, because this is based on long established fundamental principles that were apparently forgotten or never learned by many PhDs, it is not something that can be left as an acceptable disagreement.

        • Tim Folkerts says:

          Not that it will help you understand, but the perpetual motion machine is easy with your version of physics. This is jsut for the edification of other readers.

          * Take two insulated columns.
          * Fill them with different gases that have different specific heats.
          * Thermally connect them at the bottom (perhaps by removing the insulation from the bottoms of the columns and putting the bottom of both columns into a water bath).
          * Since the temperature gradient depends on the specific heat, the tops of the two columns will be at different temperatures.
          * Since the tops are at different temperatures, I can connect a heat engine between the warm reservoir at the top of the one column and the cool reservoir at the top of the other column.

          The heat engine could be a thermopile, or a Stirling engine, or any other engine you can imagine. I can now run the heat engine as long as a temperature difference exists between the tops of the two columns. Which you say will be forever since that is your proposed equilibrium condition for the tops of the two columns.

          • Massimo PORZIO says:

            No Tim,

            1) It will be “forever” just until you heat the bottom. It’s not a perpetual machine indeed, when the source of energy at the bottom ends the machine stops. And more, if you apply your very same argumentation to the current atmosphere, you should get a perpetual machine too. Think a little: just connect your heat engine between to altitudes and you get it.

            2) Constrain two gases columns in a specific space is not the same system as a free atmosphere kept together by a gravitational field only.

            Have a great day.

            Massimo

          • Physicist.↓ says:

            Tim, if you think you have invented something that would violate the Second Law of Thermodynamics then forget it. Entropy cannot decrease and perpetual motion is impossible. I am quite aware of Robert Brown’s weak attempt, which is why I named him. I also explained that a new state of thermodynamic equilibrium would evolve with a temperature gradient in between. Your problem is that you just don’t understand what thermodynamic equilibrium is, nor the explanation of the gravitationally induced temperature gradient by the brilliant 19th century physicist Josef Loschmidt who was first to estimate the size of air molecules fairly accurately. If you throw a stone in the air, does its kinetic energy convert to gravitational potential energy – yes or no? I wonder if you even know why the question is relevant. When I teach my students physics I try to help them understand first by asking questions.

            How do you think the gravitationally induced temperature gradient is maintained over 350Km of the Uranus troposphere?

          • Physicist.↓ says:


            Water vapour cools, Tim temperature data proves it.

            But the IPCC wants you to be gullible enough to believe that a mean of less than 2% water vapour raises the surface temperature about 30 degrees. So 4% in a rain forest raises it 60 degrees does it? But 1% in a desert raises it 15 degrees does it?

            I guess you are gullible enough, are you? Yes or no, Tim.

          • Tim Folkerts says:

            “If you throw a stone in the air, does its kinetic energy convert to gravitational potential energy yes or no? I wonder if you even know why the question is relevant.”

            The real question is “If you throw an air molecule in the air, does its kinetic energy convert to gravitational potential energy yes or no?” how about if you drop an air molecule — does it fall to the ground, converting potential energy to kinetic energy?
            I wonder if you even know why these questions are more relevant than your question about a stone.

            “if you think you have invented something that would violate the Second Law of Thermodynamics then forget it.”
            That’s the thing — *you* have invented something that would violate the Second Law of Thermodynamics! I just pointed out how you did it. 🙂

            PS Water vapor has nothing to do with this discussion. Please don’t bring in red herrings.

          • Tim Folkerts says:

            Massimo says “when the source of energy at the bottom ends the machine stops.”

            That’s the thing — there is NO SOURCE OF ENERGY NEEDED! The whole gradient is set up spontaneously!

            Suppose I have two tall columns of gas that are each insulated except for a small area at the bottom where they are thermally connected. Suppose I could somehow get them isothermal (with various heaters/coolers).

            Now I disconnect the heaters/coolers, so the whole thing is thermally isolated from the universe. If a lapse rate naturally occurs, then the tops of the two columns will eventually become different temperatures. It might take an hour; it might take a week; it might take a decade. But once it exists, I can use that temperature difference to run a heat energy for a while. This cools the warm top and warms the cool top. This makes the lapse rate “too big” on the warm side and “too small” on the cool side. When I get tired of this (ie when the temperature difference gets rather small), I disconnect the heat engine. The lapse rate reforms, bringing the tops back to their “natural” temperature”. So I reconnect the heat engine. And repeat. The “naturally created” temperature difference would keep reforming and would keep being source of a temperature difference to run a heat engine.

          • Massimo PORZIO says:

            Hi Tim,
            maybe I missed your point.
            But if you remove the heat from the below once the energy is all dissipated, both the two columns of gas collapse to the ground and there is no longer differential of temperature at all on the top of the columns.
            IMHO it is not a perpetual motion.

            By the way, if you attach two thermocouples in series at two different places of a thermal engine head, one far from the ignition chamber and one close to it, and you turn on the motor you get a little energy on the thermocouples when the engine is running, but when you turn off the engine, slowly the energy vanish away when the whole engine head stabilize at the surrounding temperature. IMHO the theoretical two columns work almost the same.

            Also, IMHO the gravitational thermal lapse rate exist not only because of gravity but also because there is a continuous energy flow in the atmosphere.
            No matter where the energy escapes to space, because the whole atmosphere is continuously exchanging energy with the surface, until the surface is at a temperature greater than 0 K.

            About the buoyancy of gas molecules in air, IMHO your example is not pertinent because in that case there is just an exchange of energy with the surrounding gas molecules (both KE & PE). If you drop an air molecule it immediately gains the average KE of the surrounding molecules for that altitude and it shares its PE with the same surrounding molecules because of gravity.
            When Maxwell did the experiment to evaluate the impact of gravity on the temperature, he missed the point that no matter whether the ceiling surface was thermal conductive or not, because that ceiling returned the molecules back to the ground.
            For example, put a ball on a loudspeaker driven by a pulsed voltage so that the ball can jump 10 cm above the loudspeaker and measure the vertical speed profile, you obviously get a parabolic function. Now put an obstacle, say at 1 cm above the ball, do you believe you still get a parabolic profile?
            And no matters whether the obstacle was elastic or inelastic.

            Have a great day.

            Massimo

        • Ball4 says:

          Tim F. 2:44pm – “So I reconnect the heat engine. And repeat.”

          You will have to decrease the entropy of the water to continue any energy transfer or otherwise force the system entropy to decrease. If not, when you reconnect nothing happens, the entropy will still be maxed out. Remember your system is isolated from the universe and has achieved max. entropy at disconnection.

          There is no perpetual motion in the column. Entropy increases to max. whether or not a temperature(z) gradient exists (that just depends on initial conditions).

  10. Physicist.↓ says:


    Regarding convection, everyone needs to understand that “heat transfer by bulk fluid flow” is not a bulk flow caused by an external mechanical energy supply. All convection is driven only by a higher level of mean kinetic energy in some region where it is hotter than the state of thermodynamic equilibrium (with its associated temperature gradient) would normally be. So when direct solar radiation strikes an asphalt surface (emissivity 0.92) it can raise the temperature thereof because its intensity may be, say, 450W/m^2 which would support a temperature of about 305K and this causes air molecules at the surface interface to be hotter than what they were in the cool of the pre-dawn hours of the morning – ie near the supporting temperature. So upward diffusion and advection (that is, convection) occur. But warm air does not rise in parcels, and warmed air may even fall – convection is just a net movement of molecules during their normal free path motion between collisions. Extra kinetic energy in the surface “pushes” more molecules away so that “sheets” of warmed air appear to rise, but in fact it is mostly movement of the warmer temperature that gives this effect. A light fan will turn slowly, but the heat transfer can be faster than the apparent movement of the air which is very slow. But extra thermal energy absorbed from solar radiation in the upper troposphere of a planet can drive convection downwards.

  11. Robert B says:

    Just a slight dig at the choice of a coat for an analogy.

    To measure the insulation effect of a coat you use the body temperature and the outside temperature, not the lining.

    To calculate what it would be, you figure out the drop in rate of heat transfer to the outside and not the increase in rate of heat transfer to the body because it became hotter. The better insulation makes the outside colder.

    Most of the GHE (ab. of LWIR) is between the surface and warming upper troposphere (supposedly)

    People are having a problem with the description of how it works and how to calculate the effect and not that there can’t be transfer of energy in both directions if one object is colder (transfer of heat always being the net transfer of energy).

  12. Robert B says:

    I would also like to add that the equilibrium between lower and upper troposphere only needs the molecules to move individually to be established (very slowly). The GHE should be the physics behind a warmer upper troposphere and a colder stratosphere, not a warmer surface and a warmer upper troposphere.

  13. Physicist.↓ says:

    The Ranque Hilsch vortex tube is the best “experimental test” of the fact that (in the process described in the Second Law of Thermodynamics) molecular kinetic energy is redistributed in a force field, forming a temperature gradient.

    Because the temperature gradient in a planet’s troposphere is the state of thermodynamic equilibrium which the Second Law of Thermodynamics says will evolve, the planets supported surface temperature is autonomously warmer than its mean radiating temperature, so warm in fact on Earth that we need radiating gases (mostly water vapour) to reduce the gradient and thus cool the surface from a mean of about 300K to about 288K, this being confirmed by empirical evidence (as in the study in my book) which confirms with statistical significance that water vapour cools rather than warms, all these facts thus debunking the greenhouse conjecture.

    • Tim Folkerts says:

      “The Ranque Hilsch vortex tube is the best experimental test ”

      A vortex tube is a HIGHLY non-equilibrium condition. It does not in the LEAST confirm nor refute anything about behavior in thermodynamic equilibrium. If you want an experimental test of thermodynamic equilibrium, then you need an experiment where things are in thermodynamic equilibrium.

      • Massimo PORZIO says:

        Hi Tim,
        why do you state that it’s not in thermal equilibrium?
        Once the input pressure and the temperature is stable, its outputs should return constant temperatures and pressures too.
        Am I missing something?

        Have a nice day.

        Massimo

  14. Tim Folkerts says:

    Massimo, you are missing the distinction between “equilibrium” and “steady-state”.

    “A collection of matter may be entirely isolated from its surroundings. Then if left undisturbed for an indefinitely long time, classical thermodynamics postulates that it reaches a state in which no changes occur within it, and there are no flows within it. This is a thermodynamic state of internal equilibrium.” — Wikipedia

    Equilibrium can have no flows, no temperature variations, no changes of any sort.

    PS Regarding your point earlier about perpetual motion machines, you have a valid point that that is probably not the best way to critique this whole concept of “temperate differences at thermal equilibrium in columns of gas”. There are simpler, more direct ways.
    1) This violates the 0th Law of thermodynamics, which basically destroys the concept of “temperature”.
    2) We can run a heat engine from a *single* heat reservoir. As you suggested, the engine would run down as the thermal reservoir at the bottom cooled, but there is no heat going to any other cool heat reservoir.

    • Massimo PORZIO says:

      Hi Tim,
      Ok, I missed that you were writing about internal equilibrium, but that’s what happens to the atmosphere too, or not?
      Also the atmosphere is considered into a “steady state” when we write about the thermal gradient.

      About 1):
      Why do you suggest that the two columns are into an internal thermal equilibrium?
      They are in what you just call “steady state” condition, since they are “feed” by the bottom temperature and they returns that KE to the bottom via gravity. There are continuous flows of energy there, those flows have just a zero net exchange when the bottom temperature is constant.
      IMHO 0th law is not pertinent here.

      About 2):
      the cool heat reservoir is the outer space which exchange energy by radiation directly from the ground.
      If you connect an heat engine between the tops of the columns, you could just move the exiting point of the outgoing energy to the outer space.

      • Tim Folkerts says:

        We can specify any conditions we want. If were talking about “typical” conditions in the atmosphere, with energy arriving at the bottom (sunlight) and leaving from the top (IR to space), then that is a steady-state condition and a temperature gradient is to be expected.

        The problem is that many people want to say that the true *equilibrium* condition in a tall column of gas is also a temperature gradient (and would be even without energy in & out). This is a very different claim. It is this claim that I reject — and that is rejected by the basic laws of thermodynamics. To get such a gradient, you would have to re-write thermodynamics starting from the most fundamental levels.

        *************************************

        “the cool heat reservoir is the outer space which exchange energy by radiation directly from the ground.”

        No, the heat engine I postulated had the hot side at the top of one air column, and the cold side at the top of the other air column (if such a thing actually existed). Both columns are connected together at the bottom. So heat would be flowing up from the bottom heat reservoir in one column, and flowing back down (at a lower rate) from the other air column into the heat reservoir, with the difference being the work done by the engine.

        • Massimo PORZIO says:

          Hi Tim.
          Ok, I get your point about the 0th law. It seems that we were writing about different situations. Sorry.

          About the heat engine, what I can imagine is that the it’s work changes the lapse rates of the two columns, but I don’t understand why should this not possible?
          Also in my heat engine head example, if you connect 2 points at different temperature with a good thermal conductor, you get the same effect; the hotter became a little colder and the colder became a little hotter changing the temperature gradients of the two heat paths which extend from the ignition chamber and the two points.

          Anyways, excuse me if I appear a little silly about this, but my thermodynamic studies go back to the time of school, so much time ago.

          Have a nice day.

          Massimo

        • Massimo PORZIO says:

          Tim,
          sorry, maybe I misunderstood your “with energy arriving at the bottom (sunlight) and leaving from the top (IR to space)”

          If you are arguing that if the IR energy escapes from the ground only there is no lapse rate, I don’t agree.

          Gas molecules are not different from any other body in a gravitational field.

          Try to put two balls of the same weight into a vertical glass pipe having the bore just a little greater than the balls, so you can see them inside to go up and down, then apply an upward energy impulse to the lower ball.
          Do you believe the upper ball runs faster or slower than the lower one, or the same?

          Have a nice day.

          Massimo

  15. Gordon Robertson says:

    Roy…you and other 2nd law deniers are hung up on the myth that infrared energy is heat. The 2nd law is based on heat, not IR.

    IR is a product of atomic structure, it is electromagnetic energy. EM has an electric field and a magnetic field, a specific frequency, and it contains no heat.

    Heat is not a product of atomic structure, it is applicable to various forms of energy, it has no associated fields, and it has no associated frequency.

    Heat and IR have nothing in common physically.

    Heat is essentially a description of the kinetic energy of atoms en masse. If the atoms of a gas in a container have a certain average kinetic energy and the atoms of gas in a nearby container have a lower kinetic energy, they have different levels of heat. Temperature is a means of measuring relative levels of KE.

    If the containers are brought close together, and the temperature difference is significant, heat will be transferred from the warmer container to the cooler container.

    What does that mean? It means some of the kinetic energy in the warmer container will be transferred to atoms in the cooler container.

    How is it transferred? By infrared radiation. IR can flow both ways between containers but only the more intense IR in the warmer container can raise the kinetic energy level in the cooler container. The reverse process is not possible since the lower intensity IR from the cooler container cannot raise the KE level in the warmer container.

    In solid bodies, it makes no sense that atoms with a lower KE can transfer energy to more energetic atoms in a warmer area of the solid. If you heat one end of a steel bar with a torch, the atoms become excited as their KE increases. That increased KE is transferred to the cooler end of the bar.

    That’s how physics works, Roy. In the solid bar, heat is transferred atom to atom. In the atmosphere, IR transfers energy from a warmer region to a cooler region by transferring KE. The reverse process is not doable unless you supply some kind of external energy and mechanism to extract heat from the cooler region and transfer it to the warmer region.

    • Tim Folkerts says:

      Gordan saays “If the atoms of a gas in a container have a certain average kinetic energy and the atoms of gas in a nearby container have a lower kinetic energy, they have different levels of heat.”

      Sorry, but your description of “heat” is at odds with thermodynamics. You are describing “U” (the internal energy) not “Q” (the heat).

      “In solid bodies, it makes no sense that atoms with a lower KE can transfer energy to more energetic atoms in a warmer area of the solid.”

      But because of the distribution of velocities within an object, some atoms it colder object have higher energy than some atoms in warmer objects, and thus can and do occasionally transfer energy from the cooler to the warmer object. This is not against any laws of thermodynamics.

      • Massimo PORZIO says:

        Tim:”But because of the distribution of velocities within an object, some atoms it colder object have higher energy than some atoms in warmer objects, and thus can and do occasionally transfer energy from the cooler to the warmer object. This is not against any laws of thermodynamics.”
        You are right, but the average goes from the warmer to the colder, and the energy centroid of the warmer body never move to higher levels of energy.

        Have a great day.

        Massimo

        • Tim Folkerts says:

          Certainly the average energy transfer is from warmer to cooler, but that doesn’t stop some collisions or some photons from transferring energy the other way. This can “slow the cooling” of an object, thus allowing some other actual energy input (eg sunlight) to elevate the temperature more than without the “back-collisions” and “back-radiation”.

          • Massimo PORZIO says:

            I fully agree.
            The increase of temperature, it’s just a question of back-radiation and (IMHO in lower troposphere especially a question of) gravity induced back-collisions.

            Have a great day,

            Massimo

          • Gordon Robertson says:

            Tim… “…the average energy transfer is from warmer to cooler..”

            You are confusing infrared energy with heat. The 2nd law is about heat, not IR, and the 1st law is about the conservation of all energy, IR in particular when radiation is involved.

            There is no such thing as average energy transfer in terms of heat. There is also no such thing as summing energies (IR) to validate the 2nd law.

            Also, photons do not exist as real particles. Photons are theorized particles of EM that have momentum but no mass. They were developed to particalize EM but Einstein stated long ago that no one knows if EM is a wave or a particle.

            Once again, from Clausius, heat can only be transferred from a warmer body to a cooler body under ordinary means. That means to transfer heat from a cooler body to a warmer body, as in a refrigerator, you must supply electrical power to drive a compressor.

            There is no such mechanism in the atmosphere hence heat transfer in the atmosphere is governed by the 2nd law. That heat cannot be transferred from cooler GHGs in the atmosphere to a warmer surface.

            This is compounded when it is understood that the surface is warming the GHGs. I would think incoming IR from the Sun would warm GHGs as well but AGW gurus insist that all GHG warming is initiated from surface flux.

            That’s what the IPCC preaches. So answer me this. If the surface flux is warming the GHGs, how the heck can you back-radiate energy to raise surface temperature? If it worked, you’d have a perpetual motion machine.

            That’s why Clausius developed the 2nd law. The 1st law as it stood back then would allow perpetual motion in certain instances. Carnot thought initially that a heat engine had no losses but Clausius straightened that out.

            Please don’t talk to me about slowing down the escape of heat due to opacity of GHGs. Physicist/meteorologist Craig Bohren, who as written a book on atmospheric radiation, summed that theory up nicely. He said it is a metaphor at best, and at worst, plain silly.

            Try to map radiative flux from the surface to GHGs is nigh impossible. There is no straight-forward one-to-one association between radiation and GHGs. It involves complex Feyman diagrams, and at that, it can’t be done.

          • Ball4 says:

            Gordon 10:58am: “Please dont talk to me about slowing down the escape of heat due to opacity of GHGs.”

            It is ok to talk about atm. opacity according to Bohren 2006 text. Bohren actually writes p. 34: “…the notion that the atmosphere traps radiation is at best a bad metaphor, at worst downright silly.”

            Just above that: “As the (atmosphere) emissivity increases, so does the radiative equilibrium temperature…” Pg. 33 shows the calculations. By formula Bohren shows increasing atm. opacity increases atm. emissivity therefore increases radiation from the atm. toward surface.

      • Gordon Robertson says:

        @Tim “Sorry, but your description of heat is at odds with thermodynamics. You are describing U (the internal energy) not Q (the heat)”.

        refer to: http://www.humanthermodynamics.com/Clausius.html

        Tim…internal energy is part of heat, it depends on whether you are taking a macro view or a micro view. I am taking the bit about internal energy straight from Clausius. He talked about the disgregation of atoms within the context of heat and he was referring to how atoms increase their mean paths as their internal energy is increased.

        After all, how can you talk about internal energy and kinetic energy of atoms as being separate? And how can you separate internal work done in atoms from the external work?

        Q is applied to heat in thermodynamics when a macro view is taken. That’s because the internal energy does not matter when describing energy transfer externally as long as it is cyclical in nature. However, there is a part of thermodynamics (statistical mechanics) in which heat is regarded as internal energy.

        In the day of Clausius, I don’t think statistical mechanics had been developed. He knew enough about atoms, however, to understand the value of internal energy in a substance.

        From his Mechanical Theory of Heat, he claimed:

        In calculation, however, it amounts to the same thing if we assume an alteration in the quantity of heat equivalent to each of the two kinds of work. Let Q be the quantity of heat which must be imparted to a body during its passage, in a given manner, for one condition to another, and heat
        withdrawn from the body being counted as an imparted negative quantity of heat. Then Q may be divided into three parts, of which the first is employed
        in increasing the heat actually existing in the body, the second in producing the interior, and the third in producing the exterior work. What was before
        stated of the second part also applies to the first it is independent of the path pursued in the passage of the body from one state to another: hence both parts together may be represented by one function U, which we know to be
        completely determined by the initial and final states of the body. The third part, however, the equivalent of exterior work, can like this work itself,
        only be determined when the precise manner in which the changes of conditions took place is known. If W be the quantity of exterior work, and A the equivalent of heat for the unit of work, the value of the third part will be
        A*W, and the first fundamental theorem will be expressed by the equation:

        Q = U + A*W

        It’s important to note that in the macro model of heat, work and heat are equivalent. The motion of atoms in their mean paths in a substance can be considered as work.

        He states later:

        “..in my former paper, I wished to avoid everything that was hypothetical, I entirely excluded the interior work, which I was able to do by confining myself to the consideration of cyclical process that is to say, operations in which the modifications which the body undergoes are so arranged that the body finally returns to its original condition”.

        He says at another point:

        “Mechanical work may be transformed into heat, and conversely heat into work, the magnitude of the one being always proportional to that of the other”.

        That applies to internal processes as well, according to Clausius. The work done by atoms moving in their mean free paths can be converted to heat. If the KE increases, hence the motion, the work increases and the heat.

        Then the bonus round:

        “Heat can never pass from a colder to a warmer body without some other change, connected therewith, occurring at the same time”.

        If I am reading him correctly, a warmer sources requires a source of heat which a colder source does not have relative to a warmer source. In the case of the surface/atmosphere, the surface is the source of heat for atmospheric GHGs but the GHGs cannot draw heat from the surface then return more heat than what was drawn.

        In fact, given the dearth of ACO2 in the atmosphere, and the confined frequency space of its IR, there’s no way it can have an effect on the surface and it’s massive IR flux.

      • Gordon Robertson says:

        @Tim “But because of the distribution of velocities within an object, some atoms it colder object have higher energy than some atoms in warmer objects…”

        Within an object, do you not visualize the velocity of atoms as being fixed and confined to a path around the atom’s mean position? In gases, what you say might be true if the temperatures of the warmer and cooler body’s are close.

        However, velocity is a component of kinetic energy, which is internal energy. That motion is also part of the definition of work, and since work and heat are equivalent, kinetic energy must be directly related to heat.

        In gases, solids, and liquids, the velocities of individual atoms can be calculated. Therefore, the average velocity is taken and combined with the mass of the atoms (presuming we are talking about a pure substance with atoms of one element) the kinetic energy can be estimated.

        Therefore, individual velocities should not be an issue. The average kinetic energy can be verified by taking the pressure of a gas exerted on the walls of the container. I don’t have the formula at hand, but I am sure KE is related to that pressure.

  16. Tim Folkerts says:

    Gordon says: “Within an object, do you not visualize the velocity of atoms as being fixed and confined to a path around the atoms mean position?”
    No, I don’t. The atom has some mean position, but the speed/amplitude of vibration/KE all vary randomly about some average value.

    “refer to [some web page]”
    When looking for fundamental science information, I tend not to refer to a random, one page, non-peer-reviewed, summary of 150 year old science at a .com address with no indication of who created the webpage.

    “Timinternal energy is part of heat …”
    No. It isn’t. Not in any modern understanding of thermodynamics. Read any physics or engineering thermo text written in the last 50 years.

    “… but the GHGs cannot draw heat from the surface then return more heat than what was drawn.”
    Good thing that no one (who understand the basic physics) makes such a claim. Typical numbers are 350 W/m^2 average from surface to GHGs (and clouds) and 324 W/m^2 average from GHGs (and clouds) back to the surface. 324 is less than 350, not greater than 350.

    • Gordon Robertson says:

      @Tim “When looking for fundamental science information, I tend not to refer to a random, one page, non-peer-reviewed, summary of 150 year old science…”

      I guess then, we’d better throw out all of Newton’s science.

      “Timinternal energy is part of heat
      No. It isnt”.

      Tim…Clausius said it is and when it comes to thermodynamics, for you to state flatly that he is wrong, smacks of arrogance, especially when you base your opinion on the age of the science and whether or not it has been peer reviewed.

      The guy wrote the 2nd law of thermodynamics and introduced the scientific world to entropy. He coined the term entropy.

      Clausius is of the order of several times more brilliant than you will ever think of being. And he is the same order more intelligent than anyone who argues that heat can be transferred from a colder body to a warmer body.

      • Tim Folkerts says:

        “I guess then, wed better throw out all of Newtons science.”
        Newton’s initial writings have been studied, refined, peer reviewed, and accepted — VASTLY different from the unknown webpage you referenced. The day he published his work, Philosophi Naturalis Principia Mathematica would not have counted as a good reference either! Today, Newton’s works are part of the foundation of mechanics, while “humanthermodynamics.com” is just an obscure webpage.

        Nothing I said remotely suggests that Newton’s works should be thrown out.

        ” for you to state flatly that he is wrong, smacks of arrogance “
        1) For you to state flatly that science has not progressed one iota in 150 years smacks of arrogance! Ideas — even those from great scientists like Clausius or Einstein — are honed and improved upon by subsequent generations of scientists.
        2) Even Clausius recognized in his original paper that Q & U were two distinct quantities, rather than Q being a part of U as you claim. Since you are disagreeing with Clausius here, it seems that you are the one stating flatly that he is wrong! 🙂

    • JohnKl says:

      Hi Tim Folkerts,

      You claimed:

      “Good thing that no one (who understand the basic physics) makes such a claim. Typical numbers are 350 W/m^2 average from surface to GHGs (and clouds) and 324 W/m^2 average from GHGs (and clouds) back to the surface. 324 is less than 350, not greater than 350.”

      Hmmh! Really Tim?!!! If GHG’s and clouds supposedly absorb all the radiation emitted from the earth’s surface (or even all the IR radiation emitted from the earth’s surface) how do you explain the ability of thermal imaging satellites to routinely map the earth’s surface from space using many bands of IR radiation?!!! You do realize that CO2 only absorbs a small band of infrared energy, don’t you?

      If facts impress you, this website may help you.

      http://www-users.math.umn.edu/~mcgehee/Seminars/ClimateChange/presentations/2013-1Spring/20130212ThermalIRandCarbonDioxideintheAtmosphere.pdf

      Have a great day!

      • Tim Folkerts says:

        “If GHGs and clouds supposedly absorb all the radiation emitted from the earths surface … ”
        Where did you get this idea? The earth emits ~ 390 W/m^2 on average. About 350W/m^2 on average gets absorbed by clouds and GHG, and about 40 W/m^2 oln average escapes straight to space for those satellites to observed. (Much more than 40 W/m^2 escapes when it is clear; ~ 0 W/m^2 escapes when it is cloudy).

        All of this is exactly in agreement with the reference you gave (and your discussion about thermal imaging and narrow absorption bands), so I can’t figure out what your objection is.

        • JohnKl says:

          Hi Tim Folkerts,

          My apology for misreading your statement to suggest the surface only emits 350W/m^2. Thank you for the clarification.

          Have a great day!

  17. Physicist.↓ says:

    People (well climatologists anyway) don’t realise that the Second Law is about all forms of energy, not just the kinetic energy in molecules which gives them their temperature. The Second Law says thermodynamic equilibrium (not just thermal equilibrium) will evolve, and that includes mechanical equilibrium.

    So, when a density gradient forms in a gravitational field that is an example of the Second Law in operation, and when a lake levels out again after rain falls on a small section of it, that also is the Second Law operating as entropy approaches the maximum level that is accessible within the constraints of the system.

    I believe Josef Loschmidt (a brilliant physicist from the 19th century) realised this about the Second Law, and he deduced correctly that we need to consider the gravitational potential energy in molecules when determining the maximum entropy state, that is the state of thermodynamic equilibrium. This leads to the inevitable conclusion that we only eliminate unbalanced energy potentials (maximising entropy) when the sum of molecular kinetic energy and gravitational potential energy is homogeneous. This means that there is always a propensity to form a temperature gradient in a gravitational field. However, radiating molecules do radiate to each other and have a temperature levelling effect that means that the overall state of thermodynamic equilibrium (taking this radiation into account) has a less steep gradient. In water it is almost eliminated, but not in Earth’s outer crust for example.

    Now, the important point is that, when thermodynamic equilibrium is attained (being the same as what climatologists like to call hydrostatic equilibrium, even though they don’t really understand why it is equilibrium at all) then all net non-radiative heat transfer stops. If we add thermal energy at the bottom (for example, the Sun warms the Earth’s surface after dawn) then non-radiative heat flow starts again with net upward transfers. But if instead, we add new thermal energy at the top, as at dawn on Venus and Uranus, then new energy transfer starts again with a net downward direction, this being what I call “heat creep” up the temperature gradient. And this is what really supplies the necessary thermal energy to supplement the Sun’s energy entering Earth’s surface, not back radiation which can’t transfer thermal energy from a colder troposphere to a warmer surface.

    • Tim Folkerts says:

      But your temperature gradient means (compared to an isothermal atmosphere with the same total energy) that we have warmer gas at the bottom = expansion at the bottom = lifting the atmosphere as a whole = more potential energy.

      And inevitably objects (when the option is available) will fall, losing potential energy and converting it to KE and then eventually to thermal energy. Since we can lower the PE by changing to an isothermal profile, you need to at least *consider* this as a lower energy/higher stability/higher entropy state. (And of course, this is what everyone has decided and what the laws of thermodynamics predict)

      • Physicist.↓ says:

        No Tim, the Second Law process doesn’t work that way. The density gradient and the temperature gradient form simultaneously and each is the one-and-the-same state of thermodynamic equilibrium. There can only be one state of maximum entropy and that’s it. Hence no further expansion or compression takes place, for if it did then entropy would decrease.

        As I said, people don’t understand thermodynamics, yourself included, as I can detect from experience in helping students learn physics for about five decades. You my friend are not a good learner, because I’ve explained all this in my book and numerous comments on several climate blogs. And BigWavDave told you about the gravitationally induced temperature gradient in the final comment on the Robert Brown thread about the Loschmidt effect (in which Brown botched it) on WUWT over two and a half years ago. And still you don’t learn what physicists have known since the 19th century.

        This is not child’s play for school kids. Temperature and density are the independent variables, not pressure. I repeat, when the sum (at every altitude) of mean molecular kinetic energy and mean molecular gravitational potential energy is homogeneous then we have thermodynamic equilibrium with no unbalanced energy potentials. You have to think at the molecular level and use Kinetic Theory which, after all, was what was used to derive the Ideal Gas Law in the first place.

      • Physicist.↓ says:

        Maybe this will help you understand, Tim. When we have thermodynamic equilibrium we have a temperature gradient in which, if we consider two horizontal planes of molecules separated by the mean free path, then the difference in mean KE equals the difference in gravitational PE. When a molecule with mean PE and KE at the top level moves downwards between the layers, by the time it collides with one on the lower level its KE has increased and its PE decreased so as to match one with mean KE and PE at the lower level. Hence when they collide they have the same KE and so the combined KE after the (assumed elastic) collision is the same as before, causing no further warming or cooling. The opposite happens for molecules moving upwards between the two planes. Hence we have thermodynamic equilibrium. And this also demonstrates why the density gradient is not altered once that same state of thermodynamic equilibrium evolves, simply because there’s no further redistribution of KE during collisions.

        • Tim Folkerts says:

          I know you won’t try to learn, but this is how it goes …

          But what you miss is the number of molecules that actually go up and down. In particular, consider the molecules in a lower layer. Some of them head upward to collide with molecules higher up. Some. Not all. Specifically, molecules in the lower layer with low energy never make it up to the higher layer because they simply don’t have enough upward velocity to make it up that high. Only a self-selected set of higher-than-average-kinetic-energy molecules get up to the higher layer (hence the decreasing density). By the time they make it up, they have lost some energy. The fascinating thing is that the extra energy those self-selected high-energy molecules had is exactly the same as the energy they lose going upward. So the few that make it to the next level have the same average energy as the larger set in the lower layer. Which means they have the same temperature.

          It is really not worth continuing. You have your interpretation of thermodynamics and I have mine. At some point you have to ask yourself why only you and one guy from 150 years ago come to your conclusion. Are you really that much smarter after tutoring all those freshmen than the myraid professors that actually teach the course to the freshmen? Smarter than the slew who teach the grad students and write the textbooks?

          • Massimo PORZIO says:

            Hi Tim,
            I’m not sure I understand you here:
            “So the few that make it to the next level have the same average energy as the larger set in the lower layer. Which means they have the same temperature.”
            AFIK, they have the same “heat content” not temperature.
            Isn’t the temperature in gases the KE only?
            That is, as yourself stated:
            “Only a self-selected set of higher-than-average-kinetic-energy molecules get up to the higher layer (hence the decreasing density). By the time they make it up, they have lost some energy. The fascinating thing is that the extra energy those self-selected high-energy molecules had is exactly the same as the energy they lose going upward.”
            It seems to me that you are confirming a temperature gradient because of gravity.

            Again, I miss something?

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            Maybe I’ve been not clear above.
            I meant: all the molecules above have higher KE, they are lesser than the below ones, that is their “heat potential” is lower than the below ones, but thanks to their higher KE (temperature) the average available energy is the same of the below ones (that is in steady state).

            Please, I know that “heat potential” is a silly definition, but read “availability to exchange heat to the surrounding”

            Have a great day.

            Massimo

          • Massimo PORZIO says:

            Uhmmm,
            I did a little confusion (I was writing during the coffee break), I would write:
            all the molecules above had higher KE, they are lesser than the below ones, that is their heat potential is lower than the below ones, but thanks to their lower KE (temperature) + their higher PE, the average shared energy is the same of the below ones (that is in steady state).
            In few words, the energy flow carried by the upward molecules is the very same of the downward ones.

        • DougCo t t o n says:

          This statement by Tim Folkerts is absolute garbage copied from elsewhere as I know, having read and responded to the same several times in recent years.

          ” Specifically, molecules in the lower layer with low energy never make it up to the higher layer because they simply dont have enough upward velocity to make it up that high.”

          Typical molecular velocities between collisions are in the vicinity of 1,800 Km/hour. That’s quite enough kinetic energy to take them to the mesosphere and beyond if they could somehow avoid collisions.

          See this model. Then read this comment I wrote on Jeff Condon’s “The Air Vent” …

          Sadly the model doesnt appear to give data for the mean KE in, say, the top quarter compared with the bottom quarter. If it did we would have evidence of the temperature gradient. Of course the density gradient is obvious, and we can deduce that the temperature gradient evolves at the same time. What you can see, however, if you click DATA and introduce gravity (sliding the bar) then, each time you PAUSE, the sum of KE and PE remains close enough to the same, regardless of the mean KE. This is the condition of homogeneous (PE+KE) that I have been talking about and which directly implies there is a temperature gradient when this state of thermodynamic equilibrium evolves.

  18. Tim Folkerts says:

    I’ll try one more time, Massimo.

    Imagine some “layer” of gas @ temperature T. The average KE is ⟨KE_original⟩ =(3/2 kT), with individual KE’s varying from 0 to values much larger than (3/2 kT).

    Now imagine these molecules moving up to collide with molecules in a layer a height h higher up. A naive conclusion would be that each molecule lost some PE = -mgh so that the new average KE would be reduced to ⟨KE_higher⟩ = ⟨KE_original⟩ – mgh.

    However, that ⟨KE_original⟩ included some molecules with KE = 0. It is mathematically & physically impossible for these molecules to reach the higher level — they would have to have negative KE! So we have to exclude many of the molecules from consideration when we look at collisions in the higher layer.

    Since we have removed the lowest KE molecules from consideration, the average KE of the remaining molecules is *higher* than (3/2 kT) by some amount ⟨KE_original⟩+ΔKE . But as they rise, they *lose* PE, so the new correct average is
    ⟨KE_higher⟩ = ⟨KE_original⟩ + ΔKE – mgh

    Depending on the specific values of ΔKE & mgh, the new ⟨KE_higher⟩ could be larger or smaller than ⟨KE_original⟩ (ie the temperature could be higher or lower than before). It would take much more rigorous calculations to figure it all out. But it is at least highly plausible that the two factors are the same size and thus cancel out, leaving the temperature constant.

    PS. When “Physicist” shows the calculations (or even a link to such calculations)for ΔKE that show it is less than mgh , then his idea would be worth a more detailed look. Until then, his verbal conjectures are just that — unsupported conjecture.

    • Massimo PORZIO says:

      Hi Tim,
      Not sure what you wrote:
      “Since we have removed the lowest KE molecules from consideration, the average KE of the remaining molecules is *higher* than (3/2 kT) by some amount ⟨KE_original⟩+ΔKE . But as they rise, they *lose* PE, so the new correct average is”

      Are you arguing that the average temperature above the reference layer is greater just because there are less molecules?
      If it is the case, I would tell you that it is impossible, because if you analyze the issue at the very reference level at which some molecule have 0 KE, they are the only ones which can go up the very above layer, so the average KE is the same except the -PE of the delta H. So it’s mathematically & physically sure that the above layer has less KE than the reference one.
      This can be repeated for each layer, imposing a gravitational lapse rate.
      Have a great day
      Massimo

      • Massimo PORZIO says:

        Again:
        “When Physicist shows the calculations (or even a link to such calculations)for ΔKE that show it is less than mgh , then his idea would be worth a more detailed look. Until then, his verbal conjectures are just that unsupported conjecture.”

        No, it is supported by the analysis of the issue at the infinitesimal delta H, that I proposed above indeed.

        Have a nice day.

        Massimo

      • Tim Folkerts says:

        I was trying to use a common notation with angle brackets (chevrons) to indicate an average (http://en.wikipedia.org/wiki/Bracket). Apparently the brackets don’t work consistently.

        • DougCo t t o n says:

          You just have no clue as to what thermodynamic equilibrium is, Tim my boy.

          If the molecules at the top had the same mean KE as those at the bottom, but of course have higher gravitational PE at the top, then you have unbalanced energy potentials, and so entropy can and will still increase until the state of thermodynamic equilibrium is achieved. Then, and only then, is the mean (PE+KE) homogeneous at all levels, and then of course you have the gravitationally induced temperature gradient which is the only way we can explain what happens in the Uranus troposphere, now isn’t it. I’ve tried to teach you all this many times.

          But, as I’ve said many times, you’re a beggar for punishment Timmy.

  19. Tim Folkerts says:

    This is WAY more detail than I intended, but the calculations only took 10 min and they illustrate the point perfectly.

    Suppose I have 10,000 particles with an average KE of E. The energies follow an exponential distribution (common int thermo).

    956 particles have energies between 0 and 0.1 E
    856 particles have energies between 0.1 E and 0.2 E
    782 particles have energies between 0.2 E and 0.3 E

    6 particles have energies between 5.0 E and 5.1 E

    1 particles have energies between 6.8 E and 6.9 E
    0 particles have energies above 6.9 E

    The average energy is 0.985 E. (It should be 1.000 E, but we were rounding off a bit and only have 10,000 particles)

    ********************************

    Now, let those particles fly upward so that they have lost 0.1 E of PE. We cannot say the average energy is now 0.985 – 0.1 = 0.885 E assuming all the particles lost 0.1 E of energy. 956 f those particles never made it that high because they simply didn;t have the energy to get there.

    Now we have
    956 particles never made it and don’t get counted.
    856 particles have energies between 0 E and 0.1 E
    782 particles have energies between 0.1 E and 0.2 E

    6 particles have energies between 4.9 E and 5.0 E

    1 particles have energies between 6.7 E and 6.8 E
    0 particles have energies above 6.8 E

    When you find the average, it is now 0.984 E. The average dropped by 0.001 E, not by 0.1 E (and that can be attributed to the limitations of the small # of particles). The average energy is essentially unchanged at 1.0 E, even though individual particles lost 0.1 E of energy.

  20. Massimo PORZIO says:

    Hi Tim,
    you still need to convince me that your method is applicable to reality, just one question:
    you started at an average energy of 0.985 E.
    By your computations, what it happens when your last molecule (the one which is between 6.8 E and 6.9 E) reaches the its top?
    Please compute that average, we are talking about physic effects, so it must be possible to do.
    How could be the last top layer of your atmospheric model at the very same temperature of the ground if it is populated by only one molecule?
    It seems to me that you are hypothesizing that the most energetic molecule stops to fly upward when it reaches 0.985 E (or 1 E without rounding).
    Why should it stop there?

    Have a nice day.

    Massimo

    • Tim Folkerts says:

      “By your computations, what it happens when your last molecule (the one which is between 6.8 E and 6.9 E) reaches the its top?”

      *****************************
      FIRST — A CORRECTION
      This comment made me spot a small error. When I calculated how many should be in each ‘bin’, I truncated, rather than rounded to the nearest integer.

      This correction means that 1 *or more* particles would be between 6.8 and 6.9 on average. The bin at 7.5-7.6 turns out to be the bin that averages at least 0.5 particles (which is then rounded up to 1). (This also means fewer particles in the low energy bins to maintain 10,000 particles total.)

      With this correction, the average energy starts at 1.00 E and becomes 1.00 E higher up. 🙂
      ******************************

      Now to answer the specific question, the integers were simply the closest whole number for the most likely count of the particles. Due to random statistical fluctuations, sometimes a given ‘bin’ will have more particles and sometimes fewer. In reality, we would expect something more like:

      an average of 952.09 particles between 0-0.1 E

      an average of 6.42 particles between 5.0-5.1 E

      an average of 1.06 particles between 6.8-6.9 E

      an average of 0.527 particles between 7.5-7.6 E
      an average of 0.477 particles between 7.6-7.7 E

      It turns out that the average works out the same. If we go up high enough to lose 6.9 E of PE, then fewer than 1 particle on average will have 0.1E. Slightly fewer will have 0.1-0.2 E. Slightly fewer yet will have 0.2-0.3. Way fewer yet will have 6-6.1 E, but there will be some. Turns out that on average there will be ~ 10 particles that make it that high, and they still have an average energy of 1.00 E

      • Massimo PORZIO says:

        Hi Tim,
        now at home and during my trip I thought a little more to your computations.
        But are you sure that you are using the true thermal distribution?
        IMHO the typical thermal distribution is a Gaussian curve having most of the molecule excited at almost the average temperature and have the molecules with reduced and incremented temperature perfectly distributed on the two sides of the energetic centroid.
        Maybe you had been confused by the fact that at lower altitude there are more molecules than at higher because of the pressure, but this is not the case for computing the average temperature because most of those low altitude molecules have high KE, they just exchange it to the molecules above them and they never fly up.
        As said the curve to compute the average temperature is the Gaussian with the most probabilistic peak of molecule at the middle. That is without PE reduction the average KE will be the same along the whole atmospheric column, but accounting the PE to KE conversion due to gravity you get a lapse rate.

        (Silly me that I didn’t see it before)

        Give a thought about that.

        Now I gonna have a shower.

        Have a nice day.

        Massimo

        • Tim Folkerts says:

          “But are you sure that you are using the true thermal distribution?
          IMHO the typical thermal distribution is a Gaussian curve having most of the molecule excited at almost the average temperature and have the molecules with reduced and incremented temperature perfectly distributed on the two sides of the energetic centroid.”

          I was simplifying for the sake of discussion. Each “degree of freedom” (ie each of the three cartesian directions) will have and exponential distribution for v^2; the combination of the three degrees of freedom leads to the typical Maxwell-Boltzmann distribution (which looks vaguely like a Gaussian distribution).
          http://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution.

          You can either jump into the calculations yourself in gory detail, or accept the results. (Or I suppose you could assume nobody understands physics and reject the results on principle! LOL)

          • Massimo PORZIO says:

            Uhmmm…
            I’m not sure what you mean sorry, maybe I missing some details, as said it’s a long time I studied thermodynamics, but you wrote about Gaussian distribution, and I wrote the same.
            My problem reading your explanation is that you put most of particles at lower KE and few at higher, shouldn’t be the peak of the Gaussian in the middle KE of the molecules?
            Sorry, but I din’t get your point.

            And more, you linked to Maxwell Boltzmann distribution, but that is applicable to gases in thermodynamic equilibrium which is not our case, since we have the bottom which is heating the atmosphere and the gravitational field that returns the molecule to the ground.
            There is a continuous exchange of energy with the bottom. Yes, with a net flux of 0, but it is there.

            “Or I suppose you could assume nobody understands physics and reject the results on principle! LOL”
            No, I never assumed that, I just would have a clear and unequivocal explanation about this physics issue.
            In particular, as said above, why has your distribution an average of 0.477 particles only between 7.6-7.7 E and so much as an average of 952.09 particles between 0-0.1 E?
            I expected to see the mid energy particles at the maximum count, not the ones with lower energy.
            Be patient, but I don’t get it.

            Have a nice day.

            Massimo

          • Tim Folkerts says:

            1) The distribution is only “vaguely Gaussian”. There is a shorter tail on the low energy side (that necessarily stops at KE=0) and a much longer tail on the high energy side.

            2) The original discussion was for a column that was indeed in thermodynamic equilibrium. Even in the real slightly non-equilibrium distribution, the distribution will be pretty close to a M-B distribution.

            Beyond that, we are WAY to far into this topic to hash out the gory details over the internet. I wish you luck, but I have other things to do atm.

          • Massimo PORZIO says:

            Ok Tim,
            maybe much better that I return to my safety grade radio remote controls designs.
            Anyways, one day I would dig into the details. This issue is not clear to me, I still have to imagine those upper particles which received their (not zero) KE from the below but stop their run.

            Good luck to you too.

            Have nice days.

            Massimo

          • jimc says:

            Massimo, a clear head and a sound grasp of basics is best. Sometimes people make things more complicated than they are. Nice to talk with you.

          • Massimo PORZIO says:

            Hi jimc.
            Honestly I’ve no proof who is right or who is wrong.
            I know that mine are just conjectures, but this fact that the PE goes up and down while the average KE is still there till the TOA is really a riddle for me.

            Have a nice day.

            Massimo

          • jimc says:

            Ah, you havent given up. Nothing wrong with not knowing, its pretending you know when you dont thats bad. But being an engineer, you know that. Sorry if I get preachy.

          • DougCo t t o n says:

            Tim Folkerts is talking about an absurdly high proportion of molecules supposedly not having enough energy to rise by the mean free path (a few nanometres) when a molecule with the mean KE has enough energy to reach the mesosphere above the stratosphere. In any event, his molecules at close to absolute zero temperature would simply fall and gain KE which they can then share with molecules at lower altitudes, or gain more KE from them. Furthermore, some of the ones that went up will fall again because, as I explained above, a state with homogeneous mean KE at different altitudes has unbalanced energy potentials because of the additional mean PE. Entropy could increase and so isothermal conditions do not represent the state of thermodynamic equilibrium which the Second Law of Thermodynamics says will evolve.

            “The law that entropy always increases holds, I think, the supreme position among the laws of Nature. If someone points out to you that your pet theory of the universe is in disagreement with Maxwell’s equations then so much the worse for Maxwell’s equations. If it is found to be contradicted by observation well, these experimentalists do bungle things sometimes. But if your theory is found to be against the second law of thermodynamics I can give you no hope; there is nothing for it but to collapse in deepest humiliation.”

            Sir Arthur Stanley Eddington, The Nature of the Physical World (1927)

  21. Massimo PORZIO says:

    Ok, I thought a little more. I apologize if I’m boring you all.
    I’m not 100% sure of this, so maybe I’m wrong, but Tim, IMHO your assumption to average only the molecules which reach the upper layer to get the new KE for that layer seems to me clearly wrong.

    Please try to follow my last conjecture below about it.

    If you do as you proposed, that is you average the KE only for the “surviving” molecules, you don’t get the average KE of any layer (volume of gas) indeed, but only the KE of the space occupied by the molecules, which is a non-sense for a gaseous system. The space between the molecules is very important instead because it plays an important role to establish the probability of collisions.
    In few words, after having removed the delta PE from any single molecule KE for the next layer, you should recompute the Maxwell Boltzmann distribution for these molecules which reach the new layer.
    Doing that, you take account of the space between the molecules and the reduced collision probability of that new layer.
    Your method doesn’t take account of these aspects, instead if you apply the MB distribution, layer by layer, you’ll have the Gaussian-like curve that became even more sharp as the molecules fly up to the next upper layer, and its peak moves to the 0 K position in the graph.
    At the TOA the Gaussian-like curve should become almost a line placed at 0 K.
    This is perfectly representative of molecules that stop their fly up when their KE is fully converted to PE.

    Have great days.

    Massimo

  22. DougCo t t o n says:

    The process described in statements of the Second Law of Thermodynamics (namely that entropy increases to a maximum wherein the state of thermodynamic equilibrium is attained) is precisely what forms the density gradient. I’m sure Tim Folkerts and other alarmists would agree that there is indeed a density gradient in a planet’s troposphere. The density gradient remains stable for one reason only: the molecules at the base have more mean KE and greater density.

    If you rotated a horizontal sealed insulated cylinder of gas to a vertical position, the density gradient forms and at the same time the molecules at the bottom end up with more mean KE simply because more molecules have fallen than have risen, and that’s what also forms the density gradient. A pressure gradient is merely a corollary because pressure is proportional to the product of temperature and density.

    I am posting this comment on four or five other popular climate blogs because I’m sick and tired of the false physics promulgated by this Tim Folkerts alarmist, despite the numerous times I have demonstrated the error in his thermodynamics. He, like many others, should be held to account in a court of law, and some time next year, with funds from companies pooled for class action here in Australia, I hope to organise such and make the criminals who mislead the world pay for their actions. Yes I’m angry because of the waste of money and huge loss of life involved resulting from the biggest hoax in the history of the world.

    • Massimo PORZIO says:

      Hi Doug,
      as always said I’m not a scientist and honestly I know almost nothing about atmospheric physics. I’m just asking Tim to explain me how it’s possible that gas molecules at TOA could keep their ground acquired KE and for this I analyzed the method he used for convincing me. At the moment I have very doubts that the laws of physic have been used the right way for explain this issue. But this doesn’t means that Tim is a bad physicist.
      I know that for the current popular point of view a scientist must never fail, but I believe instead that anybody could fail (me first) and kudos to the one who having found him/his self in failure gets the new info and go ahead with his/her research.
      For example I’m still very worried for the firing of the leading scientist who asked the community about the strange result of the “neutrino faster than light” issue at CERN, because he finally discovered him and his team in failure and he was ready to go ahead if a silly bureaucrat didn’t fire him. But IMHO he is a true scientist, not those who finally “discovered” the “Higgs boson” which except for its mass, it behave completely different than predicted.
      As said, I don’t know if what I wrote above is right or wrong, I’ve not the full knowledge of this field but for the moment what I wrote sounds good to me, or better, it sounds much better than Tim’s method.
      I know you are angry for how the scientific community treat you, and I fully understand you for this. IMHO if they are so sure that you are wrong, they should take just a little of their time to dismantle your theory. They spent so much time playing with their “video games” in last decades, that I really don’t understand why there isn’t one scientist who take a moment to deal with your theory.
      And this is not good.
      Anyways I don’t agree with your opinion to lead scientist in a court of law, because here in Italy we already experienced a similar action for the failed Aquila earthquake prediction. Now the scientists who failed the prediction are at risk of jail confinement (the sentence already passed the 2nd grade judgement). A judge should never enter in the scientific process because his/her consultants are scientists too, and a scientist is a man, could fail.
      Galileo was at risk of imprisonment for his science too, but finally he was right.

      Have a nice weekend.

      Massimo

      • DougCo t t o n says:

        Massimo: No one in the period of over a year since I first explained the “heat creep” process has been able to refute it, or even doubted it after reading my book it seems.

        George Christensen MP from Queensland is initiating a request for a parliamentary enquiry into the temperature data published by the Bureau of Meteorology. My hope is that the matter of failure to carry out due diligence regarding the CO2 issue will also be a matter for a parliamentary enquiry, so they have that opportunity first and I write frequently to PM Abbott, Greg Hunt (Dept of Environment) and the Climate Council. It would only be if such an enquiry were not forthcoming that it may be necessary to take it to the High Court of Australia.

        • Tim Folkerts says:

          Just change the words “Doug”” and “Tim” above, and you pretty much have my reply. 🙂

          • Massimo PORZIO says:

            Hi Doug, hi Tim.
            Ok, it seems really arrived the time to stop this discussion for me.
            I repeat, IMHO science should never be discussed in lawsuits.
            It’s probably because it’s more than 30 years that I fight to live a decent life after that damned car accident where I broken my neck, that it is absolutely unexplainable to me the use of costly and useless lawsuits to discuss science.

            Have both a nice weekend.

            Massimo

          • Doug.C says:

            But governments should be challenged if they waste taxpayers money without paying due diligence in checking the “reasons” for which they are spending that money. That’s what George Christensen is doing and I’ve sent him my book and other information to support his challenge. Meanwhile, I’m waiting on satisfactory response from the Climate Council to my latest letter.

          • DougCo t t o n says:

            Just change the words “Michael Mann” to “Tim Folkerts” in some future case like the recent multi-million dollar law suit that he and/or his supporters forked out for.

          • DougCo t t o n says:

            Massimo doesn’t seem to think that it’s criminal to obtain tax-payer funding for research the “need” for which is based on false science and what has now become a blatant hoax.

          • Massimo PORZIO says:

            Hi Doug,
            “Massimo doesnt seem to think that its criminal to obtain tax-payer funding for research the need for which is based on false science and what has now become a blatant hoax.”

            No, I’m not telling that.
            What I’m arguing is that I don’t believe Tim is dishonest about his GHGs free isothermal atmosphere point of view.
            To be honest, I don’t really know who Tim is. I supposed he is a scientist about physics, just because I always read him about issues related to physics with a good insight on theoretical aspects.
            If you ask me about Mann instead, I think he is a little dishonest man, since the climategate emails IMHO have highlighted a clear propensity of the subject to do anything (even illegal), against anybody who tried to contend the results of his (IMHO) disputable work about tree rings.

            Have a nice day.

            Massimo

          • DougC says:

            Massimo: Well if he’s not dishonest, he is seriously mistaken, as he has been told many times, but because he has pecuniary interests pertaining to climatology he refuses to look into what he has been told by BigWaveDave (in March 2012 mind you) and by myself numerous times. You see, Massimo, people depending on incomes in Climatology Carbonland are not the ones to discuss real physics with, because they have been “assigned” to speak on one side and one side only and, as in any debate, people don’t even speak to their hearts and inner thoughts, just to which ever side of the debate they are asked to speak. And they feign arrogance and superiority, mock and ridicule their opponents etc. It’s all part of their game, but water off a duck’s back to me.

    • Jerry L Krause says:

      Hi Doug:

      I hesitate to comment but you wrote “The density gradient remains stable for one reason only: The molecules at the base have more mean KE and greater density.” I know you know better than this because I am sure I have seen you write about the obvious influence of the earth’s gravity upon the motions of atmospheric molecules and have recognized that the atmosphere is a quite compressible fluid.

      Have a good day,

      Jerry

      • Doug.C says:

        So you should hesitate, for we all know that compression leads to greater density, so your meaningless sarcasm makes no sense anyway.

  23. Doug.C says:

    Come back anyone when you can explain how, when virtually all the solar radiation reaching Uranus is absorbed by a methane layer high up in the atmosphere (at a temperature around 60K) the required thermal energy then gets to the base of the troposphere (where it’s hotter than Earth’s surface) and continues to maintain the temperature there despite upward radiation and upward convection especially on the dark side.

    • Tim Folkerts says:

      All planets are hot inside — from a combination of in-fall of material during formation and continued radioactive decay. Even with *zero* sunlight, larger planets would have a significant thermal gradient from the core up top of the atmosphere.

      Beyond that, in an atmosphere with lots of GHGs and clouds, thermal IR cannot carry energy upward very effectively — it gets absorbed after very short distances. Conduction through gas is miniscule. That leaves convection, but since the thermal gradient is already down near the adiabatic lapse rate (ie the dividing line between convection and no convection), very little convection will take place either.

      Finally, at least a little solar energy penetrates down into the deeper layers of the planet. It may not be much (~ 1 W/m^2), but it is something.

      ********************************

      RECAP:
      1) There is a small energy flow upward — from the hot core and from sunlight absorbed within the atmosphere.
      2) Only a small energy flow is needed to maintain a temperature gradient in an atmosphere where IR is restricted by GHGs and clouds.

    • DougCo t t o n says:

      Tim

      There is no convincing evidence of significant net energy loss due to internal decay on Uranus. Venus cools by 5 degrees at any location on the equator during just 4 months of darkness. But it then warms up again, and you cannot explain the thermodynamics whereby the necessary thermal energy is supplied to the surface to bring about an increase in temperature, because radiation from the atmosphere cannot do that. I have explained it, along with just one other researcher in world literature.

      Planets are not still cooling off from some initial state. They have had plenty of time to do so. The cores of planets, just like the core of our Moon, are kept hot by the Sun. Amazing isn’t it what valid physics tells us. And you can’t prove me wrong.

  24. DougCo t t o n says:

    “The moon has an iron-rich core with a radius of about 205 miles (330 km). The temperature in the core is probably about 2,420 to 2,600 F (1,327 to 1,427 C). The core heats an inner layer of molten mantle, but it’s not hot enough to warm the surface of the moon.” [source]

    Considering how quickly the surface of the Moon cools on the dark side, where does the energy come from to keep the core so hot with no apparent cooling off?

    The process whereby thermal energy is supplied to the core of the Moon requires a sound understanding of the Second Law of Thermodynamics and thermodynamic equilibrium.

    When our friend Tim Folkerts can explain the Moon’s core temperature with valid physics, chances are he’s read my book.

    • Tim Folkerts says:

      “Considering how quickly the surface of the Moon cools on the dark side, where does the energy come from to keep the core so hot with no apparent cooling off?”
      “Keeping something hot” requires no energy. Raiding the temperature requires the addition of energy. Conversely, cooling it off requires the removal of energy.

      So, how quickly does energy leave? As a rough estimate, the thermal conductivity of rock is 1 W/m*K. The core is ~ 1200 K warmer than the surface. The thickness from core to surface is ~ 1,200,000 m. This produces a heat flow on the order of
      (1W/m*K)(1200K)/(1,200,000m) = 0.001 W/m^2.

      Loosing mW/m^2 of heat from something the size of the moon’s core would be a slow process indeed. Ie the hot core is not going to cool down rapidly. Throw in a little radioactive decay, and it will cool even more slowly.

      Pretty straightforward stuff.

      • Doug.C says:

        You didn’t complete your calculations for a billion years or so. You have no evidence of radioactive decay or net energy loss either for the Moon or Uranus. Sorry, Tim, that’s not the way it works. That’s why you can’t explain the temperature at the base of the troposphere of Uranus yet. There is no greenhouse effect my friend …

        From all the climate research I’ve engaged in for many thousands of hours there is one very simple and cogent argument that smashes the greenhouse conjecture in just a few words …

        They base all their “sensitivity” assumptions upon the false assumption that if Earth’s atmosphere were devoid of all “greenhouse gases” (water vapour, carbon dioxide, methane etc) then its surface temperature would be 255K. To get this figure they use albedo of 0.3 even though, with no water vapour, there would be no clouds shading the surface and reflecting solar radiation back to space. They then assume that the rocky planet without water or vegetation would have emissivity 1.0, but here are some emissivity values for various rocks …

        Basalt: 0.72
        Clay: 0.39
        Shale: 0.69
        Granite: 0.45
        Limestone: 0.95
        Soil: 0.38

        Now IMHO I would say the weighted mean emissivity would be about 0.8 at the most, looking at those figures. So, let’s use Google to find a Stefan Boltzmann on-line calculator and see what we get …

        Their figures: Flux: 0.7(1365/4) = 238.875 W/m^2
        Emissivity 1.00000
        Temperature from SB: 254.8K

        My figures: Flux 1365/4 = 341.25 W/m^2
        Mean emissivity of rocky surface: 0.8
        Temperature from SB: 294.5K

        So, if the mean temperature in the real world is 287.5K we have 7 degrees of cooling, not 33 degrees of warming.

  25. Tim Folkerts says:

    “The cores of planets, just like the core of our Moon, are kept hot by the Sun. Amazing isnt it what valid physics tells us.”
    The core of the earth is hotter than the surface of the sun. You are now violating the 2nd Law!

    The cores are “kept” warm by 1000’s of km of insulating rock and gas above them. Geothermal energy flow is currently less than 1 W/m^2 for earth. It would indeed take on the order of billions of years for the earth to cool at this rate.
    It’s amazing what valid science tells us about heat capacity and thermal conductivity and cooling rates and radioactive decay — http://en.wikipedia.org/wiki/Geothermal_gradient

    “There is no convincing evidence of significant net energy loss due to internal decay on Uranus.”
    Measurements suggest small loses. And small loses are all that are required to maintain a thermal gradient.

    “and you cannot explain the thermodynamics whereby the necessary thermal energy is supplied to the surface to bring about an increase in temperature … “
    No. You cannot understand and accept the thermodynamics. There is a big difference.

    A small amount of sunlight, combined with “venothermal” energy and convection in the atmosphere, would allow for a small amount of warming during the day, followed by a small amount of cooling at night, all the while maintaining quite high surface temperatures.

    Here’s the thing — everyone else’s “valid physics” seems to differ from yours. So rather than vague claims (like the earth’s core would have cooled long ago or Venus’ surface can’t be so warm), show some actual calculations. Then you will get people to start paying attention.

    • JohnKl says:

      Hi Tim Folkerts,

      You claimed: “The cores are kept warm by 1000′s of km of insulating rock and gas above them.” Hmmh! It should be noted that given the Law of Gravity the mass and density of said insulating rocky mass also GENERATES the enormous heat and pressure found within a celestial body’s interior. This very likely includes fissile reactions which themselves generate further heat.

      Thanks and have a great day!

      • JohnKl says:

        Hi Tim Folkerts,

        Just a point of clarification. Previously I stated: “This very likely includes fissile reactions which themselves generate further heat.” In this context I referred to large solid planetary bodies like the Earth and Venus not to smaller bodies like the Moon. Thanks and…

        Have a great day!

      • Doug.C says:

        No JohnKl. Gravity is a force, not a source of energy. There is only apparent energy generation when a gaseous planet is actually collapsing and gravitational potential energy is converted to kinetic (thermal) energy. This is not the case for our Moon or for Uranus which has a small solid core at a temperature of about 5,000K that is maintained by solar energy trapped by gravity and transferred up the temperature profile restoring thermodynamic equilibrium.

    • Doug.C says:

      You Tim Folkerts probably don’t even know what the Second Law says. You imply that you are using the Clausius “hot to cold” corollary which, for non-radiative heat transfer, only applies if all forms of energy other than kinetic energy are kept constant. That means gravitational potential energy is kept constant, which it isn’t in a vertical plain.

      You have not taken into account the error margins in the measurements for Uranus which could even yield net energy gains. That’s why I say there is no convincing evidence of significant net energy loss.

      The surface of Venus cools 5 degrees in 4 months. If it kept cooling at that rate, as it would if the Sun stopped radiating, the temperature gradient between the core and the surface would become far greater and heat flow out of the core would accelerate. Likewise on the Moon it would accelerate every lunar night. Your calculations are thus way out, and all planets could easily cool right down in a few thousand years at the most. The Sun keeps them hot.

      I’m not saying that downward convection is not what supplies the necessary thermal energy to raise the temperature of the Venus surface. It is. That convection originates in regions in the troposphere of Venus where it’s less than about 400K because that’s all the Sun’s radiation can warm. Then the heat transfer is up the temperature gradient and into the 735K surface. Glad you agree. There’s no violation of the Second Law because all this heat flow is doing is restoring thermodynamic equilibrium, as explained over two chapters of my book.

      Not quite everyone else, Tim. Teofilo Echeverria has been talking and writing about this downward convection since the 1950’s and published his explanation in a book about the same time as I did quite independently. Now several physicists realise what we are saying is correct. That’s why water vapour cools, as empirical data confirms.

      And remember Tim, an Earth without greenhouse gases (and thus no albedo) would receive about 341W/m^2 of radiation and, being rocky, would have emissivity no higher than 0.8 and thus a surface temperature at least 7 degrees warmer than the real world has now with all that water vapour cooling us a little.

  26. DougCo t t o n says:

    Roy, whenever a molecule moves with a component of vertical motion its KE alters, keeping the sum (KE+PE)=constant. Roderich Graeff was able to detect very small temperature differences in his sealed insulated cylinders: he did so in nearly every one of over 800 such experiments this century. But the killer is the Ranque Hilsch vortex tube which develops very significant temperature differences in a centrifugal force field. If the gravito-thermal effect did not exist then an Earth paved with asphalt and receiving just 163W/m^2 of direct solar radiation would have a mean temperature around -35C. It is heat creep which supplies the rest of the required energy.

    Roy, you cant do away with the Second Law.

    An isothermal state has unbalanced energy potentials simply because the molecules at the top have more gravitational PE and yet the same KE. If you have unbalanced energy potentials you do not have maximum entropy. If you do not have maximum entropy you do not have thermodynamic equilibrium. But the Second Law says thermodynamic equilibrium (with maximum entropy and no unbalanced energy potentials) will evolve. Thats what happens every calm night in the early pre-dawn hours when convection stops but the temperature gradient remains. If I were wrong then you have no explanation for the heat creep process explained over two chapters in my book. And if there is no heat creep then you have no way of explaining how the necessary energy gets from the upper troposphere of Venus down to the surface and actually raises its temperature. And we do have measured data from Russian probes dropped to the surface of Venus.

    Why would it be Roy, that every planetary troposphere exhibits the gravitationally induced temperature gradient reduced a little (as we would expect) by the inter-molecular radiation that occurs between some greenhouse molecules in its atmosphere? Is it just a coincidence that the core of Uranus is a just the right temperature (5,000K) such that the gases cool to just the right temperature (320K) at the base of its nominal troposphere (thousands of kilometres above the solid core) and continue cooling with just the right temperature gradient so as to get down to just the right radiating temperature (about 60K) at just the right altitude (about a further 350Km up) where there is a methane layer doing nearly all the absorbing and re-radiating of the weak solar radiation? Extend such questions to all planets with significant atmospheres (including gas planets and some satellite moons) in our Solar System and what is the probability that all exhibit a similar temperature gradient? What would happen if they all cooled off in another billion years, rather than being kept at current temperatures with energy from the Sun?

  27. DougCo t t o n says:

    It’s not hard to understand the state of thermodynamic equilibrium. As explained in my book, start by imagining two horizontal planes of molecules with (PE + KE) the same for every molecule. If the planes are separated by the mean free path or less, then a molecule moving down from the top plane gains KE and loses the same amount of PE. The opposite happens for a molecule moving upwards between the planes. The important point is that, when either of these molecules collides with a molecule in the other plane it has gained or lost just the right amount of KE such that it has the same KE as the molecule in the plane it reaches. And because the KE matches in all collisions there is no propensity for further dispersion of KE (conduction, diffusion or convection) and thus we have the state of thermodynamic equilibrium,

    Only in a horizontal plane is there no change in KE during free path motion and so temperature does indeed level out in a horizontal plane, but not in a vertical plane. This was what Loschmidt visualised back in the 19th century and he was right. All attempts to prove him wrong (like Robert Brown’s) don’t demonstrate an understanding of thermodynamic equilibrium and the fact that, if you combine two systems you just get a new state of thermodynamic equilibrium with a new temperature gradient, but certainly no perpetual circulation of energy.

    You all need to come to grips with the process which the Second Law describes, and how it applies to all forms of energy and equilibrium. (Hence it is also the Second Law which tells us why there is a density gradient.) Then you need to understand entropy and the state of thermodynamic equilibrium (which includes mechanical equilibrium) and why that state must have no unbalanced energy potentials.

    Note that the Second Law applies to every independent process or to a sequence of dependent processes. An example of a “net” effect is seen in a siphon where water flows up the short side (the first dependent process) and down the longer side (the second dependent process) and so entropy does increase for the combination of processes or participating systems. But they must be dependent processes or as stated here be “participating systems.” If you cut the siphon at the top you no longer have two participating systems that are inter-dependent.

    Now, the electromagnetic energy in back radiation can only be used for a part of the surface’s own quota of outward electro-magnetic radiation. If its EM energy were converted to thermal (kinetic) energy in the surface then that KE could escape by conduction, evaporation etc and there is no subsequent dependent radiation process. So the first process transferring thermal energy from cold to hot can’t happen, even if those in Climatology Carbonland think it can. All one-way radiation only ever transfers thermal energy from hot to cold because there are no molecules involved and being affected by gravity. Hence the only way the temperature gradient builds up in a high troposphere (as for Uranus and Venus) where no significant direct solar radiation heats a surface (if any) is from the upper colder regions to the warmer regions by non-radiative heat transfer which can “creep” up the temperature plot if thermodynamic equilibrium has been disturbed with new energy at the top, and now needs to be restored by spreading that new energy in all accessible directions away from its source.

  28. Tim Folkerts says:

    “start by imagining two horizontal planes of molecules with (PE + KE) the same for every molecule.”

    Right off the bat, we have a problem. All the molecules in a layer have the same PE, but they will NOT have the same KE if they are in thermal equilibrium (ie they have varying KE’s described by the MB Distribution). The situation you have described is not even thermal equilibrium within a given layer!

    So your hypothesis is basically “assume we have a whole series of layers that are not in thermal equilibrium”. 🙂

    • Massimo PORZIO says:

      Yes Tim,
      the PE for that layer is the average, and the KE is spread around that average, it’s not a thermal equilibrium, a yourself said, it’s a steady state.
      But, I repeat, your Maxwell Boltzmann distribution must be recomputed for any layer, and the vertex of the almost Gaussian distribution is always at the KE which (for that layer) fixes the sum of the integrated KE and the PE for that layer to the constant value of the integrated KE acquired at the ground level.
      At the TOA that MB almost Gaussian curve became a line where the few molecule which have 0 KE are distributed.
      So, IMHO, in the current atmospheric steady state, the gravitational field induces a lapse rate indeed.

      Have a nice day.

      Massimo

      • Massimo PORZIO says:

        By the way Tim, you wrote:
        “So your hypothesis is basically assume we have a whole series of layers that are not in thermal equilibrium. :-)”
        Isn’t it also in case of your in equilibrium supposed isothermal atmosphere?
        Even in that theoretical case you have a MB distribution, that is different KE and different temperatures for the single molecules. Thus following your argument, even there, there is no thermal equilibrium indeed, just a steady state.

        I’m missing something?

        Have a nice day.

        Massimo

      • Tim Folkerts says:

        ” … the KE is spread around that average, its not a thermal equilibrium”
        The spread out KE *is* thermal equilibrium! The collection of fast and slow molecules defines the temperature. (You really can’t even talk about “the temperature” of a single particle. )

        “… fixes the sum of the integrated KE and the PE for that layer to the constant value of the integrated KE acquired at the ground level.”
        No. That is the subtle part that Doug is missing as well. Not all the molecules make it up to the higher level.

        “your Maxwell Boltzmann distribution must be recomputed for any layer”
        Yes, but when you recompute the distribution with only the molecules that have sufficient energy to make it to the higher level (ie molecules with above-average KE to start with), the average KE is still the same! That is not something that we can prove or disprove here, but it is at least *plausible*.

        • Massimo PORZIO says:

          Hi Tim,
          you wrote:
          “The spread out KE *is* thermal equilibrium! The collection of fast and slow molecules defines the temperature. (You really cant even talk about the temperature of a single particle. )”

          Yes, I was just highlight your contradiction, when you write
          about temperature you accept that the KE is averaged, but not when I write about the sum of KEavg+PE. I your case you consider it an “equilibrium”, in my case you consider it a “steady state”.

          “Yes, but when you recompute the distribution with only the molecules that have sufficient energy to make it to the higher level (ie molecules with above-average KE to start with), the average KE is still the same! That is not something that we can prove or disprove here, but it is at least *plausible*.”
          No really so, if it was you miss to account the space between the molecules.
          The temperature is the average KE of the molecules of gas in a given volume. Doing your way, as the layer is upper the reference volume reduces, because you take account only of the volume occupied by the molecules, which is an absolute non-sense for gases.
          A thermometer placed in a layer measures the temperature when it is in thermodynamic steady state, which is imposed by its radiation to the surrounding (a continuous) and the rate of the collision of gas molecules with its probe surface (an heat source not continuous but discretized by that rate). So lesser molecules means a reduced average temperature or KE.

          “Yes, but when you recompute the distribution with only the molecules that have sufficient energy to make it to the higher level (ie molecules with above-average KE to start with), the average KE is still the same! ”
          No, because all the molecule involved in the new MB have reduced KE, and the new MB returns the gaussian-like curve distribution more on the left than for the previous lower layer.

          In few words, you can’t simply average the molecules KE for the upper layer (as you did), because is not that way that they exchange their KE each other.

          Have a nice day

          Massimo

        • Tim Folkerts says:

          “Doing your way, as the layer is upper the reference volume reduces, because you take account only of the volume occupied by the molecules”
          I don’t even know where this comes from. I am talking about all the molecules in a given volume.

          “So lesser molecules means a reduced average temperature or KE.”
          It depends. *Slower* molecules means cooler. The density of molecules and rate of collision will affect the amount of time it takes to come to equilibrium, but a low density gas and a high density gas can have the same average KE and the same temperature and cause a thermometer to come to the same reading.

          “No, because all the molecule involved in the new MB have reduced KE … “
          But as I just pointed out, they all started with enhanced KE. The original average in the lower layer includes some molecules with zero (or close to zero) KE, so they never make it to the higher level (they “fall back” due to gravity before they get there). The particles that we will be looking at in the upper level a short time later started with extra KE. They lost the extra KE on the way up, and these two factors cancel out to give the smaller number of self-selected high-energy particles the same average energy once they get to the higher level. I really don’t know how else to say this.

          The calculations are way up-thread. It works.

          • Massimo PORZIO says:

            Hi Tim,
            “The density of molecules and rate of collision will affect the amount of time it takes to come to equilibrium, but a low density gas and a high density gas can have the same average KE and the same temperature and cause a thermometer to come to the same reading.”
            Please try to imagine the following though experiment.
            Take 2 perfectly thermal insulating boxes of the very same size. Put into the first box the molecules of the layer L1 which as you said have any of them a singular KE.
            Now into the second box put all the same molecule after having reduced their single KE of a known amount and exclude from the box the molecule which KE reached 0.
            At this point we are sure that box 2 has less molecule of box 1 and at the same time it has less KE per molecule than box 1.
            IMHO, this means reduced pressure and reduce average KE in box 2 .
            If you put one thermometer into the 2 boxes, what do you think will be the two measurements?
            Remember that the two boxes are perfectly insulated from the surrounding, so the temperature should stabilize after the thermometer reached the equilibrium.

            Have a nice day.

            Massimo

          • Tim Folkerts says:

            “Put into the first box the molecules of the layer L1 which as you said have any of them a singular KE.”
            No. I said they will have KEs that correspond to the MB distribution. At least that is what I was trying to say.

            Now into the second box put all the same molecule after having reduced their single KE of a known amount and exclude from the box the molecule which KE reached 0….
            At this point we are sure that box 2 has less molecule of box 1 and at the same time it has less KE per molecule than box 1.

            Consider a very simple example. A box contains 3 particles with energy 1 eV, 11 eV, and 21 eV, for an average of 11 eV. Now reduce all of these by 1 eV. We exclude the one particle and are left with two particles of energy 10 eV and 20 eV. The average has INCREASED by 4 eV from 11 eV to 15 eV.

            At this point I am sure that we have MORE KE per particle for this specific example.

            In other examples, we might end up wit a lower average. In yet other examples (like the one I did way up-thread), the average stays the same. We need to know MUCH more about the specific distribution of energies before we can say whether the average increase, decreases or stays the same. I claim (in agreement with all standard thermodynamics sources) that the average stays the same and that two systems in thermal equilibrium will be the same temperature.

          • Massimo PORZIO says:

            Tim, I also missed to to evidence one point, you wrote “You really cant even talk about the temperature of a single particle.”
            Yes, of course, one system made of a single molecule of gas can’t have any temperature, that’s because to have any KE you must have a mass and a speed of the molecule. Since the speed must be relative to some reference, it’s obvious that a molecule of gas itself has no temperature, but 2 molecules of gas have one. And a system made of one molecule of gas under the effects of a gravitational field has too, because the system is not made by the single molecule alone, if it is under the effect of a gravitational field, must exist a second body which interact with the molecule. So that has a temperature.
            For that, a thought planet at temperature 300 K having an atmosphere made of a single molecule has its atmosphere with a lapse rate determined by the parabolic “jumps” of that molecule.

            Have a nice day.

            Massimo

          • Massimo PORZIO says:

            Tim,
            maybe you missed the point that the MB distribution is exactly a way to compute how the molecule distribute for a specific mass of the column under the gravity applied to it.
            Boltzmann designed it just to compute the distribution starting from macroscopic parameters such as the mass of the column.

            As you go up one layer the mass of the column reduces, so the Gaussian peak moves on the left along the x axis of the graph, and it becomes narrower. At the top the column has little or no mass above that layer, and the Gaussian curve becomes a straight vertical line superimposed to the y axis at 0 K.

            Have a nice day.

            Massimo

          • Tim Folkerts says:

            “For that, a thought planet at temperature 300 K having an atmosphere made of a single molecule has its atmosphere with a lapse rate determined by the parabolic jumps of that molecule.”

            A very good thought experiment! And it supports *my* conclusion perfectly!

            Each time the particle lands against the 300 K surface, it will get some random amount of energy. If you look at enough “jumps” you will find that the single molecule will leave the surface with varying energies as indicated by the MB distribution.

            If you look at the particles at some level like 1 km up, you will discover that the particle rarely makes it that high. None of the low-energy jumps contribute to the average there. Only a small number of high-KE jumps even get to that level to be counted. Of course, for those high-KE jumps the particle has lost a lot of PE on the way up, so when it gets to 1 km they are no longer high KE but only average KE.

            Once again, the calculations way up-thread work perfectly well to describe 10,000 jumps of a single particle over time, rather than 10,000 separate particles at once time.

          • Massimo PORZIO says:

            Hi Tim,
            You wrote:
            “If you look at the particles at some level like 1 km up, you will discover that the particle rarely makes it that high. None of the low-energy jumps contribute to the average there. Only a small number of high-KE jumps even get to that level to be counted. Of course, for those high-KE jumps the particle has lost a lot of PE on the way up, so when it gets to 1 km they are no longer high KE but only average KE.”

            Uhmmm… I’m not sure what you mean, being the MB a Gaussian like curve, I expect to have most of jumps at an average height, but this doesn’t mean that I have at that height any KE for those jumps, in fact it just means that most of the jumps there have 0 KE.
            If you are averaging along time (because you are supposing more jumps are counting for the “average” layers’ temperatures), then you must average the KE of the molecule at any altitude for any jump, otherwise it seems to me that you are averaging only the temperature of the top of the molecule’s run.
            IMHO at any jump the molecule has it’s temperature which starts from the ground value and reduces to zero at the top and then it returns the very same energy to the ground when it reaches the very same temperature that triggered it’s upward run.
            That profile is the very same for any jump, what it changes is the maximum height, which varies with the MB curve profile, but the average temperature of the molecule at any singular altitude must increase with the altitude because for any upward run, it experienced a reduction of KE.

            I apologize, maybe because I’m not a physicist, so maybe I’m missing something very important, but I still haven’t get your point.

            Anyways, now I’ve to go to sleep, it’s 1 AM here in Italy.

            Have a nice day.

            Massimo

  29. Curt says:

    For the commenters here who believe gravitation alone is enough to create and sustain a lapse rate of g/cp, consider the case of the enrichment centrifuges for uranium hexafluoride (UF6).

    These centrifuges typically spin at 100,000 rpm, which is 1667 rev/sec, or about 10,500 rad/sec.

    At a radius of just 0.1 meter, the centripetal/centrifugal force (w^2*r) is over 10 million m/s^2 (one million g’s!).

    UF6 has a cp of 370 J/(kg*K). so the lapse rate g/cp at 0.1m radius would be 10^7/370 = 300,000 K/m, or 300,000,000 K/km!

    Does this lapse rate occur in these centrifuges? I don’t think so!

    • Massimo PORZIO says:

      Hi Curt
      No, it doesn’t happen. Because, except for the initial rotational acceleration where the temperature rises a little at the external border of the centrifuge, it immediately dissipates that power flux to the walls of centrifugal chamber.
      In our atmosphere we have the gravity force and a continuous flux of energy incoming at the gas/ground interface which throw away the gas molecules which cannot dissipate that energy in other manner else by returning it to that very same interface. Doing that, their KE must fall to 0 because of it’s conversion in PE at the top and return to the original KE at the ground, making the lapse rate.
      Without that energy applied at the ground there is no lapse rate of course, it’s not the gravity force itself that build it.

      Have a nice day.

      Massimo

      • Tim Folkerts says:

        “Without that energy applied at the ground there is no lapse rate of course, its not the gravity force itself that build it.”

        YES! The lapse rate is NOT the equilibrium state; it only occurs with continued flows of energy into/out of the system.

        PS. You are now arguing against your own previous statements! Before you yourself tried to claim that gravity itself did indeed cause the lapse rate (all those appeals to parabolic flights of molecules caused solely by gravity changing the MB distribution).

        • Massimo PORZIO says:

          Hi Tim,
          It seems that I’ve been not clear.
          I never intended to relate my argument to equilibrium or steady state, I always wrote about an hypothetical GHGs free atmosphere, which some argue that it must be isothermal.
          Note that I’m not implying with my argument that the GHGs don’t increase the ground temperature, because honestly I don’t know. In fact it is my opinion that the models predicted missing tropical mid troposphere “hot-spot”, could be hidden by the superimposition of the gravitational lapse rate and the GHGs effect.

          Have a nice day.

          Massimo

          • D oug says:

            There’s no warming by 33 degrees for GH gases to do. Read my comment about the errors in the calculation of the 255K figure. The Earth’s surface could very well have been warmer than current temperatures without GH gases if the emissivity of a rocky surface would have been below 0.88 with no water vapour and so no clouds reflecting 30% of solar radiation back to space.

        • Massimo PORZIO says:

          Hi Tim,
          anyways, just to clarify, I don’t remember that I wrote something like “all those appeals to parabolic flights of molecules caused solely by gravity changing the MB distribution”.
          I always implicitly supposed that the gravity itself (being a force) can’t drive the atmospheric molecules, in fact I’m the one who highlighted you that the two gas column thought experiment isn’t a perpetual machine because without the energy from the ground there is no atmosphere at all, since all molecules collapse to the ground.

          Have a nice day.

          Massimo

          • Curt says:

            Massimo:

            For a planet with a “transparent” atmosphere (no absorption either of solar or planetary radiation, the planetary surface would — on average — radiate as much power to deep space as it receives from the sun. It would reach steady state temperature levels necessary to radiate away this much power.

            On such a planet, the only heat transfer mechanism possible for the atmosphere is conduction at the planetary surface (with convection distributing this within the atmosphere). This means that the atmosphere cannot, over the long term, have any net heat transfer to or from the surface. (Over the short term, such as the diurnal cycle, it can exchange to or from the surface, but these must balance out). This further means that the base of the atmosphere will be at the same temperature as the surface in the steady state, even though there is no net heat transfer.

          • Massimo PORZIO says:

            Hi Curt,
            yes I agree, but I’m not sure what you would tell me.
            It’s exactly what I meant in my previous message.
            Once the whole atmosphere is “charged” the net flux at the ground/gas interface is zero.

            Have a nice day.

            Massimo

          • Massimo PORZIO says:

            Hi Curt,
            by the way, if you were arguing that a zero net heat transfer at the ground interface is the same like no fluxes at all (not incoming and not outgoing from the atmosphere), then I don’t agree.

            Those two fluxes (even if the net heat transfer is zero) are responsible for the actual atmosphere build up, in case they never exist all the gas molecules should collapse at the ground.

            Have a nice day.

            Massimo

        • DougCo t t o n says:

          Tim I have proved in my book that the temperature gradient is what the Second Law of Thermodynamics tells us will evolve, namely thermodynamic equilibrium. I see nothing in your assertive claim referring to entropy increasing or balanced or unbalanced energy potentials, so I suggest to silent readers that you haven’t a clue what thermodynamic equilibrium is – you don’t even see the need to talk about it, even though the Second Law does.

          You continue to apply the “hot to cold” corollary of the Second Law which requires that all forms of energy other than kinetic energy are held constant. In that “other forms” include gravitational potential energy, the corollary you depend upon is inapplicable in a vertical plane.

          Read my book if you want the truth, but I doubt that you do because you have a pecuniary interest in maintaining the status quo and plugging away in numerous climate blogs, hell bent in your campaign to promulgate garbage “science” from Climatology Carbonland, regardless of the cost in human poverty and life.

        • D oug says:

          Yes well you Tim Folkerts could start by helping Curt respond to these two comments and then you could try answering my questions about the 255K calculations.

        • D oug says:

          More assertive statements without a word of physics to support such from our Tim Folkerts who’s a beggar for punishment. Try answering my questions about the 255K calculations and the Uranus troposphere.

      • Curt says:

        Hello Massimo,

        I’ve been very busy that last few days, but would like to get back to this. With regard to the centrifuge, you argue that “except for the initial rotational acceleration where the temperature rises a little at the external border of the centrifuge, it immediately dissipates that power flux to the walls of centrifugal chamber.”

        I’m curious about this argument, because if you believe that a gravitational force alone can maintain a lapse rate, there is no “power flux”.

        If you mean that the walls of the chamber will conduct heat from the hotter outside to the colder inside, how can you be sure that it will do so completely. Wouldn’t it depend on the conductivity (both from material properties and geometry) of the walls? What if there were a thermally insulating layer on the inside of the chamber?

        The g/cp lapse rate here is 300K per millimeter! These chambers are at least 100 mm from inside to outside, so that would amount to 30,000K differential. Even if the chamber conductivity could reduce this by 99%, there would still be a 300K difference. I have never seen any discussion of centrifuges that even brings this up as an issue.

        • Massimo PORZIO says:

          Hi Curt,
          I never entered into the details (I’m just an electronic engineer), but some time ago into a physics forum I read a question very similar to your and I just reported what I read there.
          Just like an intuition, I think that since the gas are enclosed into the centrifuge, it behave very differently.
          For example if you take a good insulated room and try to see the temperature of the ceiling and the one of the floor, there is no doubt that you find the ceiling hotter than the floor, this should be due to impossibility of the molecules to do their full parabolic path imposed by the temperature (the air accumulated much KE than the needed to stay at that ceiling “altitude”).
          But maybe I’ve been not clear again, I’m not arguing that the gravity itself maintains a lapse rate, it does it when the heat source is applied at the side which exerts the gravitational force (in this case the ground).

          It’s hard to me to do a parallel from the atmosphere to the centrifuge. In few words without any source of energy (or very little energy) at the ground, I don’t predict any lapse rate, because the molecules are all collapsed to the ground, so they are all sharing the energy by conductivity.
          IMHO, into the centrifuge, you don’t get so much high temperature, because in the long time, the maximum should be fixed by the outside wall of the chamber, the temperature should be initially cooler at the inner wall and hotter at the outer wall because of the small pressure induced by the centrifugal force, but since the small distance between the walls the temperature should be moderated by an effect like the one that I mentioned above for the closed room.

          Have a nice weekend

          Massimo

          • Curt says:

            Massimo:

            Many of the posters here believe that gravity alone is enough to create and maintain a strong temperature gradient in an atmosphere. I thought you were among them. (Even though your English is very good, I know it is not your first language.)

            With my centrifuge example, I thought I would carry that argument to its logical extreme to see where it led.

            This is very different from a temperature gradient created by heating at the bottom and heat loss at the top, as in our real world with radiatively active gases.

          • Massimo PORZIO says:

            Hi Curt,
            “Many of the posters here believe that gravity alone is enough to create and maintain a strong temperature gradient in an atmosphere. I thought you were among them.”
            No I never believed that, but I think that also Doug Cotton don’t believe it. In fact I remember that he somewhere clearly wrote that he knows that gravity is a force and for that, it can’t create the lapse rate without any other source of energy.
            Even the Sun have born because of gravity, but also because the hydrogen particles attracted by their relative gravity had their own KE. Without that KE it never light on.

            In the meantime, I thought a little more to your UF6 question, and I reread your first post above.
            I figured out that you used the lapse rate formula in the wrong way (maybe you did it to highlight me the paradox, but that time I missed your point).
            In fact IMHO, the lapse rate formula g/cp should be used not to establish how much hotter became the outer wall of the centrifugal chamber, but to establish the maximum radial excursion of the molecules under that centrifugal force at the working temperature.
            So, your 300,000,000 K/km at 373 K (I used 100 C because about below 56 C it solidifies) should lead to a tiny 1.24 nm, this means that the whole 373 to 0 K radial excursion should happen in that small space which is well below the size of the molecules themselves. Thus, we discovered that that “atmosphere” in the chamber is collapsed (by the way, it’s probably what the designer wanted to separate the U235 from the U238).
            Probably the post that I read that time was referring to the small amount of heating at the outer wall of the centrifugal chamber experienced during the start-up acceleration, which is simply due to the momentary increasing of the pressure there.
            Anyways, when a gas is constrained to a space by a chamber like that, I believe that the much higher density of the chamber walls compared to the one of the gas itself, plays an important role in averaging the gas temperature.

            You also wrote:
            “This is very different from a temperature gradient created by heating at the bottom and heat loss at the top, as in our real world with radiatively active gases.”
            I believe that it is not important that the loss must be at the top. Also when the loss is performed at the bottom (the very same place of the heating, that is even without any radiatively active gases), the lapse rate should be present. This because the bottom of the column works as a “temperature reference”, the molecules there get their KE and as they raise they convert it to PE. IMHO the Maxwell Boltzmann distribution has noting to do with any “same KE averaged on any layers”. Also the molecules which raise completely vertical after having get the maximum KE of the MB distribution have to stop their upward run, in that place they have 0 KE and all the initial KE converted to PE. The place where they do that is the TOA.
            The MB distribution, just distributed the TOA in a range of layers, not in a single well defined one.

            Of course, as always said, these are just my own conjectures, take them for what they really are… Conjectures 🙂

            Have a nice sunday.

            Massimo

          • Massimo PORZIO says:

            Ehmm… I missed one thing.
            Curt, don’t lie please… I know that my English is not so good!
            I’m almost 50, and at the time of school (so long time ago, sigh), here in Italy the English language was an option alternative to French.
            Even if I have attended some English lessons, they were very basic.
            You can’t imagine how many times I felt silly having misunderstand something.

            Again,
            have a nice sunday.

            Massimo

          • D oug says:

            No Massimo – I didn’t say what you said I wrote.

            Please see my comments (and the support from John, Jerry and numberer) on Roy’s thread for October data.

            Roy, too, should read and digest.

          • D oug says:

            And, Massimo, on the planet Uranus it is the top of the atmosphere which acts “as a reference” because that’s where the solar radiation sets the temperature in the methane layer which is colder than 60K, because that’s all you get when the Sun is 30 times further away than it is from Earth.

            So your guesses are incorrect, Massimo and I’m a little angry at being misquoted.

            You could have saved your time and mine by reading my book, so that you understand what I am in fact saying about the gravitationally induced temperature gradient being the state of thermodynamic equilibrium which the Second Law says evolves as entropy increases towards its maximum. Which part of that statement do you not understand?

          • Curt says:

            Massimo:

            I think you are confusing temperature, which is a potential, like the voltage you work with, or fluid pressure, with heat flow, which is a flux like electrical current, or fluid flow.

            People like Doug definitely believe that gravity alone, with no heat transfer in and out of the column of gas, will create the lapse rate. I think they are simply making a mistake common to first-year physics student, not realizing that the pressure gradient means that the energy per unit volume is constant as you go up, and therefore the isothermal case is the stable steady-state case for an isolated column of gas.

          • D oug says:

            And you are wrong too Curt and you totally and utterly misrepresent what I have written in my book because you obviously haven’t read and understood it or my comments.

          • D oug says:

            You Curt are confusing isothermal states with the state of thermodynamic equilibrium in which mean energy is indeed constant with no unbalanced energy potentials, including mechanical ones. That’s why the Second Law tells us gravity forms a density gradient and a temperature gradient. The pressure gradient is a corollary because pressure is proportional to the product of density and temperature. But I guess your knowledge and understanding of thermodynamics is so primitive that you don’t even understand why the density gradient is the state of thermodynamic equilibrium which the Second Law says will evolve autonomously. Go back and study basic graduate level thermodynamics. Maybe even Wikipedia can help you.

          • D oug says:

            So you explain Curt how the -g/Cp temperature gradient comes about in the 350Km high nominal troposphere of Uranus where the reference temperature is set at under 60K in the methane layer near TOA and the base of the troposphere is 320K. It has nothing to do with the 5,000K small solid core thousands of kilometres further down supposedly cooling off at exactly the right rate, and producing exactly the right temperature gradient to get down to exactly the right 60K temperature all by coincidence at exactly the right altitude thousand of kilometres above near TOA. A few too many coincidences in there I suggest to leave you not cursing, Curt.

          • Curt says:

            Doug:

            I have spent far too much of the only life I will ever have reading your drivel, with its fantastical made-up physics like cutoff frequencies, resonant fogs, heat creep, etc., etc., etc. For a while it amused me in a train-wreck-watching kind of way, but now it is just tiresome.

            You remind me of the confused first-year undergraduates I used to assist who just didn’t have the ability to analyze physical situations in any kind of organized way. With many of them, I ultimately had to convince them to pursue a less demanding course of study.

            There is a reason you have been banned from site after site, alarmist and skeptic alike. You try to inject your ridiculous theories into any discussion, whether remotely relevant or not. People with far more patience than I possess have carefully explained what you are doing wrong, but you simply cannot comprehend it. I am amazed at Roy’s patience in letting you continue to post here.

          • Massimo PORZIO says:

            Hi Doug,
            sorry if I misread you, but you on November 4, 2014 at 6:23 AM
            wrote:
            “No JohnKl. Gravity is a force, not a source of energy. There is only apparent energy generation when a gaseous planet is actually collapsing and gravitational potential energy is converted to kinetic (thermal) energy.”
            I interpreted it as you were saying that there is no way for gravity to increase the temperature directly without any other energy source.

            Have a nice day.

            Massimo

          • Massimo PORZIO says:

            Hi Curt,
            “I think you are confusing temperature, which is a potential, like the voltage you work with, or fluid pressure, with heat flow, which is a flux like electrical current, or fluid flow.”
            I don’t know why you believe that, I know that temperature is the “driver” of the net heat flow, such the voltage is the “driver” of the current flow in a conductor.
            But I was not arguing about that, my problem about isothermal atmosphere is that, in gases, temperature is KE only, and since the mass of the molecule is constant, the molecules must stop their vertical run at the TOA, so their KE there should be 0 for a very little moment.
            You wrote:
            “the pressure gradient means that the energy per unit volume is constant as you go up”
            Maybe it’s a silly question for you, but since the pressure reduces as function of altitude, how could the temperature remain stable to hold PV=nRT?

            Have a nice day.

            Massimo

    • D oug says:

      Centrifugal force in a Ranque Hilsh vortex tube does indeed create an easily measured temperature gradient in the cross-section, and calculations show it to be based on the quotient of the acceleration due to that force and the mean specific heat of the gases. So there! There’s noting like empirical evidence (as in my book) to confirm that the gravitationally-induced temperature gradient is what valid physics says it is, namely the state of thermodynamic equilibrium.

  30. DougCo t t o n says:

    However you look at it, with increasing water vapour the temperature gradient is lower and the thermal profile rotates downwards at the surface end in order to retain radiative equilibrium with the solar radiation, as always happens within 0.5%. That is why real world temperature data proves water vapour cools. Until you produce empirical evidence that it warms Im not interested in you repetitive promulgation of garbage from IPCC Climatology Carbonland which I probably know better than yourself.

    Even if you dont understand the resonance process described in my paper Radiated Energy and the Second Law of Thermodynamics and even if you think that the electro-magnetic energy in a one-way independent passage of radiation from a cooler troposphere to a warmer surface can be converted back to thermal energy and somehow not violate the Second Law with this heat transfer from cooler to warmer, then consider what would happen. The radiation from the cooler source is well known not to penetrate the ocean surface by more than 10 microns, whilst that from the Sun penetrates more than 10 metres. There would thus be a supply of thermal energy (you claim, not me) comparable with that from the Sun but concentrated into less than one-millionth of the depth of water, and obviously boiling it. Suppose the ocean temperature is 290K and this boiling 10 micron surface layer is 373K. Mix it all up and we get (0.000001 x 373) + (0.999999 x 290) = 0.000373 + 289.999710 = 290.000083. So thats a warming of 0.000083 degree. Something to spend billions worrying about?

  31. D oug says:

    An isothermal state does not normally exist in the plane of a force field and we see clear cut evidence of this in a Ranque Hilsch vortex tube which creates a huge centrifugal force and temperature differentials sometimes over 200 degrees.

    There’s nothing quite like empirical evidence to prove our Tim Folkerts wrong.

    But he won’t even try to respond with any argument based on sound physics because, then, unlike myself, he can’t explain the -g/Cp gradient in the Uranus troposphere or the vortex tube.

    • Tim Folkerts says:

      You are persistent — I’ll give you that, Doug.

      A vortex tube is NOT equilibrium — not even close. As such, the fact that it has steep temperature gradients is not evidence for or against what might happen in equilibrium.

      High temperature inside the planets are easy to explain in terms of initial heating from in-falling material during formation and continued radioactive heating. For gas planets, you can add continued contraction and gravitational heating.

      Gradients in atmospheres (pretty much all atmospheres) are nearly -g/Cp because
      1) there is always at least a bit of upward heat flux (either from a tiny bit of sunlight or from ‘geothermal’ heat flows).
      2a) conduction is minimal in gases & radiation is minimal within thick cloudy atmospheres, so even a small heat flux would create a gradient above -g/Cp if those were the only mechanisms.
      2b) if the gradient gets above -g/Cp, convection starts, which can carry heat very effectively, acting as a “governor” that regulates gradients to no more than ~ -g/Cp.

    • D oug says:


      Yawn … more assertive statements from Tim Folkerts copied from Pierrehumbert’s writings mostly.

      The Ranque-Hilsch vortex tube develops a very steep temperature gradient in its cross-section (due to centrifugal force) for exactly the same reason as does a planet’s troposphere. You can even calculate the magnitude of the temperature gradient in the same way.

      All planets could easily have cooled off by now. Venus cools 5 degrees in 4 months; Earth cools 10 degrees during its night. If you switched off the Sun’s radiation the surface would soon be so cold that the crust, mantle and core would also start to cool at a fairly fast rate. Internal energy generation can’t keep up with the current cooling at night, let alone faster cooling after just a few hundred years perhaps.

      It is obvious that the Sun warms planets up again on the sunlit side by as much as they cooled the previous night. There is absolutely no convincing evidence of cooling off or internal energy generation with Uranus. Only collapsing gas giants (without solid cores like Uranus, Venus and Earth etc) are still converting gravitational PE to KE (thermal energy) in the collapsing process.

      No upward convection is necessary for the gradient to evolve for the simple reason that it is the state of thermodynamic equilibrium which the Second Law says will evolve autonomously as entropy increases. As you admit, there is no convection when the gradient is exactly the environmental temperature gradient. That’s what I expect. Even conduction and diffusion will not flatten out the temperature gradient in a billion years.

      And if the gradient gets below the environmental gradient, such as when new solar energy is absorbed each Venus dawn in the top of its troposphere, then conduction, diffusion, advection and convection start with a component of downward motion which is restoring thermodynamic equilibrium.

      —————-

      This summary may help ….

      NASA originally published a net energy diagram without back radiation. But it became obvious that the flux of about 163W/m^2 entering the surface was nowhere near enough to explain the actual surface temperature. The Stefan Boltzmann equation spelled this out, and thats the the first application of the SBL regarding climate. The black body temperature is only 231.55K for 163W/m^2.

      So it was probably James Hansen who dreamed up the back radiation garbage in which such radiative flux is added to the solar flux to explain the 288K temperature that is observed. That black body temperature needs a total of 390W/m^2 which means the atmosphere would be delivering more thermal energy out of its base than entered at its top. But that didnt phase Hansen et al the public wouldnt notice this little problem of assumed energy generation. It must be happening because they thought what else? the SBL “must” be what explains the surface temperature by adding this back radiation.

      So thats the significance of the Stefan Boltzmann equation in the climate hoax, my friends.

      When I am talking about thermodynamic equilibrium in the troposphere this is quite a different matter from radiative equilibrium of the whole planet system with incident solar radiation. Yes that happens and any imbalance (usually no more than 0.5%) is a result of natural warming or cooling, not the cause. (Once again, you could have learnt this in an hour or so reading my book.)

      Now, we dont need Hansens back radiation anyway, and the reason it doesnt do what he thought is explained in the main four sections of my March 2012 paper on radiated energy. The reason we dont need back radiation supplying the extra thermal energy (supplementing the direct solar radiation) is because valid physics can be used to show how and why there will in certain circumstances be net downward thermal energy transfer by non-radiative processes, conduction, diffusion and advection (all being convection in physics) which is restoring thermodynamic equilibrium with its associated environmental temperature gradient. That heat creep process is explained in two chapters of this years book. Its a whole new paradigm in atmospheric physics.

    • D oug says:

      The issue with the vortex tube and a troposphere is that (as the Second Law tells us) entropy never decreases and such systems approach thermodynamic equilibrium. It is this limiting state we are considering. The gradient is forming in the process. It may never get to be the precise calculated amount (due to weather or general motion in the vortex tube) but it is still the limit of what it is approaching that should interest us. My point is that it is not tending towards isothermal conditions. When convection stops, for example, there is not one slightest move due to conduction or diffusion towards isothermal conditions because that would decrease entropy.

      Now you need to consider why the thermal profile in the thousands of kilometres of the gaseous regions on Uranus extrapolates outwards to the exact radiating temperature near TOA where the methane layer (at about 60K) is in radiative equilibrium with the Sun. The fact is that if the planet moved in some elliptical orbit that brought it much closer to the Sun then that 60K might become, say, 160K and all temperatures down to the core would have to increase by 100 degrees or otherwise the temperature gradient would be wrong. Why is it so? Answer in my book.

      • Tim Folkerts says:

        * Of course I haven’t “come to grips” Dave’s explanation — it was wrong then and it is still wrong now.

        * I did follow that other discussion– the top post got the physics right and the intelligent discussion concurred.

        🙂

        • D oug says:

          No Tim Folkerts. Your assertive statements don’t wash yet again.

          Robert Brown got it wrong. Unlike your assertive statements, I will back up what I just said with valid physics….

          The combined system will develop its own state of thermodynamic equilibrium with an appropriate temperature gradient in between. It is impossible for entropy to then decrease from that state of maximum entropy, and so Robert Brown was totally and utterly wrong in assuming there would be a perpetual cyclic flow of energy.

        • D oug says:

          Even Wikipedia could help you, Robert …

          “Systems in mutual thermodynamic equilibrium are simultaneously mutually in thermal, mechanical, chemical, and radiative equilibria. Systems can be in one kind of mutual equilibrium, though not in others. In thermodynamic equilibrium, all kinds of equilibrium hold at once and indefinitely, until disturbed by a thermodynamic operation. In a macroscopic equilibrium, almost or perfectly, exactly balanced microscopic exchanges occur; this is part of the notion of macroscopic equilibrium.

          “An isolated thermodynamic system in its own state of internal thermodynamic equilibrium has a uniform temperature if and only if all forms of energy other than molecular kinetic energy are held constant. For example, molecular gravitational potential energy must be constant.”

          • Tim Folkerts says:

            I gotta hand it to you, Doug — rewriting wikipedia and then quoting it as an authority! Such chutzpah! That is pretty much the ultimate in “assertive statements” being used to justify an answer.

            Fortunately, your efforts there have already been undone. Whoever corrected your changes made me smile: “undid good faith edit that was wrong in physics”.

            **************************************************

            To be a good sport, here is an opportunity for you to demonstrate that your conclusions are not simply “assertive statements”. Show mathematically that a column of air with a temperature gradient has a lower entropy than a similar column of gas with the same total energy but uniform temperature. You can either calculate the total entropy of both columns, or calculate the change as you go from one state to the other.

          • DougC says:

            Firstly, Tim shows his lack of knowledge of physics when he writes “Show mathematically that a column of air with a temperature gradient has a lower entropy than a similar column of gas with the same total energy but uniform temperature.”

            Entropy never decreases in an isolated system Tim. You got it around the wrong way and should have written “higher entropy” not “lower entropy.”

            When we do substitute “higher entropy” that is precisely what I have proved in my book using valid physics, appropriate computations and supporting it all with quite sufficient empirical evidence.

            In fact, I’ve gone further and proved that the environmental temperature gradient is the state with maximum entropy, not just higher entropy. And the very fact that Tim can’t understand why an isothermal state would have unbalanced energy potentials (because of the extra gravitational potential energy at the top) just shows silent readers that he has no understanding of thermodynamic equilibrium and his concept of entropy is limited to the kinetic energy component, even though even Wikipedia would have helped him to understand that thermodynamic equilibrium involves other forms of equilibrium and thus other forms of energy as well.

          • Tim Folkerts says:

            For anyone who is still interested, the answer is provided in this paper:
            http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469%282004%29061%3C0931%3AOMEP%3E2.0.CO%3B2

            The answer — as expected — is that the isothermal atmosphere has the maximum entropy.

            Doug will now simply assert once again that everyone else is wrong.

            Time to leave to to the discerning reader whether Doug’s book is correct or if this paper is correct. (Or Doug could point out which specific equations are wrong in the paper and what they should read instead.)

          • Douyg.C says:

            An isothermal atmosphere has extra gravitational potential energy in higher regions compared with lower regions. If there were no corresponding reduction in molecular kinetic energy then the energy in molecules at the top exceeds that in molecules at the bottom. So some molecules would fall, more than others rise. If I were wrong there would be no density gradient either.

          • Doug.C says:

            The error in the paper you cite Tim Folkerts lies in equation (7) where they are “using the definition of potential temperature” and, since that definition ignores gravitational potential energy, they are in effect begging the question because they are using a definition which in effect assumes isothermal conditions apply – a serious oversight at best, if not a deliberate bluff.

      • Tim Folkerts says:

        Doug says: “It is this limiting state we are considering. “

        But that limiting state is NOT thermodynamic equilibrium! It may be “steady-state”, but that is a rather different beast! Thermodynamic equilibrium can have no flow of mass or energy. Read any source.

        • D oug says:

          The Second Law of Thermodynamics makes it perfectly clear that the limiting state is that with maximum entropy and that is most certainly thermodynamic equilibrium. Yes there’s no net flow of energy in the state of thermodynamic equilibrium because, as I have explained many times, there are no unbalanced energy potentials.

          You really don’t understand thermodynamics, Tim Folkerts, but you’re a beggar for punishment.

          • Tim Folkerts says:

            Once again, you completely miss the point.

            Your vortex tube *DOES* have unbalanced potentials! There is high pressure at one end and low pressure at the other, so that molecules flow in one end and out the other. There is net flow of energy through the vortex tube. Thus by your own words (</i<"theres no net flow of energy in the state of thermodynamic equilibrium"), this is NOT equilibrium. The “limiting state” of the vortex tube running steadily is not approaching equilibrium, and never will.

            The only way to allow the vortex tube to “approach equilibrium” would be to cap the two ends and wait for the air inside to stop moving and heat to stop redistributing. And when this happens, the temperature gradient will also disappear.

            The gradient you are so desperate to use as evidence to support you theory only exists in a non-equilibrium conditions!

          • DougC says:

            Again Tim misses the point. I’m talking about unbalanced energy potentials in the plane of the force field of course that is, the cross section where the temperature varies considerably between the inner and outer regions in the Ranque Hilsch vortex tube. See my note “It works because of the gravito-thermal effect” here …. http://en.wikipedia.org/wiki/Talk:Vortex_tube

        • D oug says:

          You Tim Folkerts are totally bluffed by Pierrehumbert. Even though I have pointed to his huge error (for a planet without water vapour, water, clouds, carbon dioxide, vegetation or any radiating molecules in its transparent atmosphere) in reducing the solar flux by the 30% of solar radiation that is reflected by clouds, when there are no clouds.

          The radiating temperature of such a planet is thus not 255K but about 278K, and out the window goes the IPCC’s “33 degrees of warming” supposedly meant to happen when greenhouse gases are then introduced.

          Well the mean temperature of Earth is not 278+33 = 311K so Pierrehumbert is wrong, and he is just as wrong when he described hydrostatic equilibrium as if it is something different from thermodynamic equilibrium. There can only be one state of maximum entropy and thus only one limiting state which a system approaches in accord with the process described in the Second Law of Thermodynamics – and that is thermodynamic equilibrium. There’s no need for another term, just as there’s no need to use “lapse rate” for “temperature gradient” or to assume air rises in “pockets” or “parcels” which have no physical constraints keeping them together while molecules move in all directions at far greater speeds than that of convection or even most hurricanes.

    • D oug says:

      And Tim, you still haven’t come to grips with the significance of the gravitationally induced temperature gradient which BigWaveDave explained to you way back in March 2012 when he replied to you ..

      “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 GHGs. And, because this is based on long established fundamental principles that were apparently forgotten or never learned by many PhDs, it is not something that can be left as an acceptable disagreement.”

      My book would help you understand the physics, you know. It only takes an hour to read.

      For now, follow the discussion on the October data thread here, as I’m not returning to this old thread.

  32. Atmospheric.Physicist says:

    For further atmospheric physics (pertaining to the Second Law) please see my discussion with Jerry starting here …

    http://www.drroyspencer.com/2014/11/uah-global-temperature-update-for-october-2014-0-37-deg-c/#comment-167464

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