The Paradox of Climate Science

Written by Ross McLeod

Australian climate analyst explains why climate science may have got the effect of greenhouse gases back to front. Below we examine how the real “heat trapping” gases are Nitrogen (N2) and Oxygen (O2) and the Earth actually relies on “greenhouse gases” to keep cool!

Ross McLeod writes:

The Earth’s atmosphere is composed of a mix of gases and research reveals the following information. From http://en.wikipedia.org/wiki/Atmosphere_of_Earth I quote the following:

These figures are for a “dry” atmosphere which does not include Water Vapour (H2O) at ~0.25% by mass over full atmosphere, locally 0.001%–5% by volume.

Climate science says that Nitrogen (N2) and Oxygen (O2) (i.e. – 99% of the atmosphere ) are not “ greenhouse gases”. “Greenhouse gases” absorb infra-red radiation and the principal “greenhouse gases” in the atmosphere are Water Vapour(H2O) and Carbon dioxide (CO2).

The theory says:

1. The Sun heats the Earth and this is a “constant” varying little over time; and,

2. The only mechanism whereby the Earth in the vacuum of space can lose energy – which equates to “cooling” – is by radiating infra-red radiation to space.

3. The power of this radiation has been measured by satellites and is of the value of about 239 Watts per square metre (W/m 2 ).

4. 239 Watts per square metre (W/m 2 ) is the radiation emitted by an object at a temperature of about minus 18 degrees C.

This is of course the source of the famous quote:  http://www.giss.nasa.gov/research/briefs/ma_01/

“Without naturally occurring greenhouse gases, Earth’s average temperature would be near 0°F (or -18°C) instead of the much warmer 59°F (15°C).”

Thus the theory says that “greenhouse gases” absorb most of the radiation emitted by the heated surfaces of the Earth. This causes the Earth’s atmosphere to “trap” heat and therefore adding more “greenhouse gases” – read Carbon dioxide (CO2) – will result in even higher temperatures.

To support this hypothesis the IPCC has drawn information from various scientists and regularly compiles this information in reports. One oft quoted diagram is this updated Energy Budget radiation diagram.

What this shows is average radiation energy flows between Earth’s surfaces, atmosphere and space. We need not concern ourselves with any argument about these figures and what they mean – this has already been widely canvassed.

Let us simply accept this represents a significant summary of the “settled science” of the “greenhouse effect” and the warming caused by these “greenhouse gases”.

What I want to focus on is the “Outgoing Long wave Radiation” shown in the diagram as approximately 239 W/m2 .

So let me summarise the situation:

1. The Sun continuously warms the Earth’s surfaces and atmosphere; and,

2. The Earth radiates 239 W/m2 continuously to space; and,

3. This radiation is the only method of removing energy from the Earth’s atmosphere and land surfaces to space; and,

4. Of this 239 W/m2 a mere 40 W/m2 is emitted directly from the surface; and,

5. 199 W/m2 is absorbed by greenhouse gases – Water Vapour (H2O) and Carbon dioxide (CO2) and then emitted to space by these gases.

Thus 199/239 or ~83% of the Earth’s ability to lose energy is because “greenhouse gases” radiate infra-red energy and a mere ~17% is directly emitted from the surface.

The other part of the story is the claim that because Nitrogen (N2) and Oxygen (O2) are not “greenhouse gases” they do not absorb any significant amount of the radiation emitted by the Earth’s surface and accordingly the do not radiate any significant amount of infra-red radiation to space.

There is certainly evidence that Nitrogen (N2) and Oxygen (O2) are essentially transparent to infrared radiation.

Now for the “Paradox”!

We all know the temperature of the atmosphere changes over time. The air can be hot or cold depending on the season, time of day, clear skies etc. It is unrealistic to claim that the temperature of Nitrogen (N2) and Oxygen (O2) does not change – they constitute 99% of the substance we call air after all.

Obviously they absorb energy and increase in temperature by contact with warm surfaces – climate science says they are not “greenhouse gases” so do not absorb radiation but they are subject to heating and cooling.

We all know that hot air rises. So the really significant question is:

1. If 99% of the atmosphere obviously increases in temperature by absorbing energy from the Earth’s surfaces – but not by radiation; and,

2. This 99% cannot radiate this energy to space as claimed by climate science,

Then how does the 99% of the atmosphere ever lose energy and thus cool down?

Convection involves rapid movement of energy but because convection is limited to the atmosphere it does not allow for energy loss to space – it merely moves energy around and transforms it into different forms – it may cool the surface but it simply transfers the energy to the atmosphere, not to space.

Conduction depends on thermal contact with other matter and the space surrounding the Earth is essentially a vacuum. The Earth cannot lose energy to space by conduction.

So how does the 99% of the atmosphere – Nitrogen (N2) and Oxygen (O2) – ever lose energy and thus cool down?

We all know it does – air temperature can be over 40 degrees C in some locations in summer and below freezing in winter.

Well, climate science clearly shows that the only mechanism available to Nitrogen (N2) and Oxygen (O2) to lose energy and thus cool down is by transferring their energy to the “greenhouse gases” which heat up and in turn radiate it to space.

The energy Budget diagram clearly shows ~83% of the radiation to space is from “greenhouse gases”.

It is accepted science that there are three mechanisms of thermal energy transfer – conduction, convection and radiation. 99% of the atmosphere – Nitrogen (N2) and Oxygen (O2) – apparently doesn’t radiate infra-red radiation – they aren’t “greenhouse gases” after all.

Convection simply moves energy around and transforms it into other forms.

So 99% of the atmosphere can only cool down by conduction of their energy to “greenhouse gases” by collisions between gas molecules and the “greenhouse gases” alone radiate the energy to space.

But the chance of Nitrogen (N2) and Oxygen (O2) colliding with a “greenhouse gas” are of the order of about three in one hundred for Water Vapour and less than four in ten thousand for Carbon dioxide (CO2).

If this is all true, and climate science says it is, the real “heat trapping” gases are Nitrogen (N2) and Oxygen (O2) and the Earth relies on “greenhouse gases” to keep cool!

So can someone please explain to me how increasing the concentration of the major coolants in the atmosphere and thus increasing the likelihood that Nitrogen (N2) and Oxygen (O2) will be able to transfer their energy to space will result in more “heat trapping” causing global warming? Logic clearly says the opposite is more likely. And that is the Paradox of Climate Science!

PS: http://www.livescience.com/19700-hottest-place-earth.html tells us:

“Satellites, on the other hand, can get a reading on these hard-to-reach, harsh places because they can scan every piece of the Earth’s surface.”

“The single highest land skin temperature recorded in any year of the study was found in the Lut Desert in 2005 and measured a stunning 159.3 F (70.7 C).”

How can that happen if the Sun supplies only 161 W/m 2 as indicated in the Energy Budget and “back radiation” supplies 333W/m 2 for a total of 494 W/m 2 ? 494 W/m 2 is equivalent to about 32.5 degrees C – where did the other 37.5 degrees C come from?

 

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Comments (46)

  • Avatar

    Dense

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    Ed Bo

    Ed Bo

    April 5, 2017 at 2:18 am | #

    “But NO energy is being transferred in this situation. The chair needs no energy source — batteries, electrical outlet, or anything else — to maintain this force on my body. By your logic, the chair would need such an energy source to hold me up.”

    “I have no idea what you are trying to say”

    I am trying to say that gravity, which is a downward acting force of radiative character, requires energy supply equal to its documented amount of force acting on mass above the surface. The amount of energy required to act on mass with a force of 9.8m/s is equal to 9.8W/s when acting in a point at the center of mass. If the source is earth and the force is of radiative character, it would need a power emitted from the surface which is 4 times the force acting on a square meter above the surface according to the inverse square law. If that force acts in a point at the center of the mass of that square meter, the least amount of energy required for that force is f=sqrt(f/4)=9.8W/s, and the power emitted from the source has to be at least f^2*4=384W/m^2.

    If you dissect the numbers, which part is wrong? I´m not saying I´m right, just that it adds up and I see no problem with it. I´m not making claims about pink unicorns vomiting backradiation into a pot of gold at the end of a brown rainbow here, or waving photon-blankets in “all directions” making dry ice hotter than the sun. I am trying to not claim anything strange at all, and that seems to solve some problems. I try to avoid assumptions about gravity and radiation and just set the units to be the same all over, and there is nothing strange about converting Joule/second to Watts/second/m^2 since J/s is Nm/s, when tracking it backwards to the source.

    The background is that I realized that the only energy/force we observe in the universe, on earth and in space, is thermal radiation at different intensities, and action of gravity on stars and our solar system. Everything we see can be traced to an effect of thermal radiation and gravity acting on mass.

    If the only thing we observe both here on earth and in space, is thermal energy and the force of gravity, would it be illogical if they were connected? I think it would be the only reasonable conclusion to make based on the information we have. And when it is only a matter of converting units and using the same principles we use for thermal radiation to see that they are equal in power, I think it´s hard to argue against it. Do you know any better explanation of gravity?

    Dark matter and such hocus pocus is as illogical as photon-blankets, and just say that we have to use “ghost”-stuff to explain it. That makes me look elsewhere in proven physics instead.

    Reply

    • Avatar

      Ed Bo

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      Dense: Now you are just talking complete nonsense!

      You say: “gravity, which is a downward acting force of radiative character, requires energy supply equal to its documented amount of force acting on mass above the surface.”

      WTF is a “force of a radiative character”? This force, of whatever character, requires NO energy supply.

      You continue: ” The amount of energy required to act on mass with a force of 9.8m/s is equal to 9.8W/s…” 9.8m/s is a velocity, not a force. 9.8W/s is not a unit of energy, or even power — it is the second derivative with respect to time of energy.

      I could go on, but there is no point. You have gone off the rails before you even have gotten started…

      Reply

      • Avatar

        Dense

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        “WTF is a “force of a radiative character”? This force, of whatever character, requires NO energy supply.”

        Try thermal radiation. Work is force acting on mass causing displacement. Work acts through thermal energy on matter by causing excitation like vibrations and particle speed. That means that radiation is a force doing work. Radiative character, like in radiation and radiating. Which part of it do you have a hard time with? Gravity is predicted to be a radiative force. Thermal radiation is a radiative force and it fulfills the criteria for work in how it can be measured in mass.

        Gravity decline by the inverse square law, it is the source causing energy of mass to increase when it moves in direction of the surface. That is a definition of work being done on mass, the energy of mass increase when gravity act on it and cause a displacement.

        When mass is stationary on the surface, where you say no energy is needed and no forces are acting, forces balance the net transfer so that no work is done. That does not mean they don´t exist or that they are zero.

        https://en.wikipedia.org/wiki/Work_(physics)

        “Another example is a book on a table. If external forces are applied to the book so that it slides on the table, then the force exerted by the table constrains the book from moving downwards. The force exerted by the table supports the book and is perpendicular to its movement which means that this constraint force does not perform work.”

        As you can see in the wiki-article, there is twice the amount of force present. When they are undbalanced they do work but otherwise they are equal.

        https://en.wikipedia.org/wiki/Force

        “Third law
        Main article: Newton’s third law

        Newton’s Third Law is a result of applying symmetry to situations where forces can be attributed to the presence of different objects. The third law means that all forces are interactions between different bodies,[16][Note 3] and thus that there is no such thing as a unidirectional force or a force that acts on only one body. Whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction.”

        You need to provide something a little sharper than:

        ” You have gone off the rails before you even have gotten started…”

        You are contradicting Newton and gravity with precision when you write:

        “This force requires NO energy supply.”

        That is a very strange statement. I have never heard anyone say that a force doesn´t need energy supply. You should not use your personal lack of understanding for what you observe, as a statement about the universe in general. You shouldn´t trust your own eyes, just because you can´t see the energy doesn´t mean that you have support for saying it doesn´t need energy.

        All forces need energy. It would be beneficial for you to support your statements with some sources. Start right now.

        Reply

        • Avatar

          Ed Bo

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          Dense: I’m afraid your handle fits you all too well! .From the Urban Dictionary: “if dense is used as figurative (of a person), then dense means that it’s difficult to explain anything to that person because they can’t make sense of complex ideas (because their head is too “dense” or “thick” to get anything through).”

          You say: “Work is force acting on mass causing displacement.” OK. But we are discussing the case when the force causes no displacement of the mass, like my body and chair example, your book and table example, or the atmosphere and surface issue we started on. No displacement, no work. It’s a very simple deduction from your own argument, but you can’t get there.

          You say: “Work acts through thermal energy on matter by causing excitation like vibrations and particle speed.” No! Work is an organized transfer of energy (non-entropy-increasing) –you are describing the randomized nature of heat transfer (entropy-increasing). Here you have contradicted your own argument about work causing displacement!

          Electromagnetic radiation is not a force, and gravity does not radiate. This is basic physics.

          You say: “When mass is stationary on the surface, where you say no energy is needed and no forces are acting, forces balance the net transfer so that no work is done. That does not mean they don´t exist or that they are zero.”

          Again, you don’t even understand the argument. I have explicitly stated that forces are acting — my weight on the chair, and the chair’s opposite supporting force on my body. But what you cannot understand no matter how many times it is explained to you is that because there is no displacement, there is no energy transfer (work) in the static case.

          You say: “I have never heard anyone say that a force doesn´t need energy supply.” I have asked you repeatedly, where is the energy supply for my chair that supplies the force to hold my body up, or the energy supply for your table to hold your book up? (Hint: There is none!)

          (For the chair to be able to actively lift me, it would need an energy source, because then its force would be displacing in the direction of the force. Can you understand the distinction? I am losing hope.)

          Reply

          • Avatar

            Dense

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            “You say: “Work is force acting on mass causing displacement.” OK. But we are discussing the case when the force causes no displacement of the mass, like my body and chair example, your book and table example, or the atmosphere and surface issue we started on. No displacement, no work. It’s a very simple deduction from your own argument, but you can’t get there.”

            So molecules and atoms doesn´t move when heated? Ok.

            And you ignore Newton?

            “Whenever a first body exerts a force F on a second body, the second body exerts a force −F on the first body. F and −F are equal in magnitude and opposite in direction.”

            What is it in there you are having trouble with?

            “You say: “Work acts through thermal energy on matter by causing excitation like vibrations and particle speed.” No! Work is an organized transfer of energy (non-entropy-increasing) –you are describing the randomized nature of heat transfer (entropy-increasing). Here you have contradicted your own argument about work causing displacement!”

            Every photon absorbed cause increased vibrations and collisions . They move around more from increased internal energy. Heat is the collision and transfer of energy, or emission and absorption of photons. Both cause a local relative decrease in entropy, relative to before absorption or collisional transfer. Try flying a balloon sometime, then you can feel the force of heat lifting you up by doing work on the surroundings.

            A molecule that absorbs a photon decrease in entropy when the energy cause it to move around more. That is energy doing work and displacing mass.

            ” No! Work is an organized transfer of energy (non-entropy-increasing) –you are describing the randomized nature of heat transfer (entropy-increasing). Here you have contradicted your own argument about work causing displacement!”

            Yes. It is heat transfer. Are you saying gravity is organized when an object fall towards the surface? Organized by who?

            Heat is entropy increasing when transferred from higher density to lower. But in the receiving end entropy is decreasing on the molecular level. A molecule receiving energy and get excited, clearly decrease in entropy.

            “Electromagnetic radiation is not a force, and gravity does not radiate. This is basic physics.”

            http://math.ucr.edu/home/baez/physics/Relativity/GR/grav_radiation.html

            “You can accelerate any body so as to produce such radiation, but due to the feeble strength of gravity, it is entirely undetectable except when produced by intense astrophysical sources such as supernovae, collisions of black holes,”

            https://en.wikipedia.org/wiki/Gravitational_wave

            “In 1905 Henri Poincaré first proposed gravitational waves (ondes gravifiques) emanating from a body and propagating at the speed of light as being required by the Lorentz transformations.[2] Predicted in 1916[3][4] by Albert Einstein on the basis of his theory of general relativity,[5][6] gravitational waves transport energy as gravitational radiation”

            You should start to check in with reality before saying dumb stuff. You are really just making shit up as you go along. Both radiation and energy can be seen in those quotes.

            “Again, you don’t even understand the argument. I have explicitly stated that forces are acting — my weight on the chair, and the chair’s opposite supporting force on my body. But what you cannot understand no matter how many times it is explained to you is that because there is no displacement, there is no energy transfer (work) in the static case.”

            You are really not the smartest guy. Your body use energy constantly. One example is keeping your blood pressure up by a beating heart. That pressure makes sure that your fat ass prevents your bones from cutting through your flesh. Each beat transfer energy into the chair and also heat it.

            The forces are kept up by a steady flow of energy, being equal in size, no work is done. Until you drag your ass out of the chair.

            “You say: “I have never heard anyone say that a force doesn´t need energy supply.” I have asked you repeatedly, where is the energy supply for my chair that supplies the force to hold my body up, or the energy supply for your table to hold your book up? (Hint: There is none!)”

            Drop the temperature enough and the chair will break when its brittle enough. That happens when the energy supply for the force of the chair gets so low that your body acts with a larger force downwards. If nothing else, heat is the force upholding mass. Heat expands, you know. Expansion needs energy to do work on its surroundings.

            So, you have contradicted Einstein, Poincare and Newton. Whos next?

          • Avatar

            Ed Bo

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            Dense:

            You continue to demonstrate that you don’t understand the most basic concepts of thermodynamics or physics.

            I point out that if there is no displacement, there is no work. Physics 101. Your response is “So molecules and atoms doesn’t move when heated?”

            In the first chapter of any introductory thermodynamics text, the authors will make a clear distinction between the “internal energy” of an object (from the random movement of molecules within the object resulting in no overall motion of the object) and the “kinetic energy” of the object (from the “organized” motion of the overall object). I was talking about the second case, and you responded with the first case. I guess you slept through that lecture.

            A standard beginning physics problem would be: How much energy is required to lift a 10kg mass by a height of 5 meters in earth’s gravity (9.8 m/s^2)?

            A similar question would be: How much energy is required to hold a 10 kg mass at a given height in earth’s gravity?

            I have already reminded you that I was the one who introduced Newton’s Third Law of equal and opposite forces to this discussion long ago with my example of my weight on my chair. Why do you continue to claim that I “ignore Newton”???

            Next, you claim that absorption of energy causing increased internal energy causes a “decrease in entropy”. 100% WRONG!!! Absorption of heat causes an INCREASE in entropy. This is part of the most basic definition of entropy: DeltaS = DeltaQ/T, where DeltaS is the change in entropy of the object, DeltaQ is the heat in (positive) or out (negative), and T is the absolute temperature. Did you sleep through that lecture as well?

            You ask: “Are you saying gravity is organized when an object fall toward the surface?”

            Yes!! It is organized in the thermodynamic sense that the motion it creates is all in the directly downward direction, increasing the kinetic energy and not the internal energy of the object.

            I don’t know what you are trying to argue with those quotes on gravitational waves, which say that these waves are “entirely undetectable except when produced by intense astrophysical sources such as supernovae, collisions of black holes…” We are talking about books on tables and atmosphere on the earth’s surface, so you are citing what your own source says is “entirely undetectable”.

            And I ask you once again, if my chair needs to expend energy to provide the (equal and opposite) force to support my weight, where is the source of that energy? I can’t find a battery pack in my chair, nor an electric plug. It does not get colder over time (which it would have to do to use its thermal energy). So where???

      • Avatar

        Dense

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        “9.8m/s is a velocity, not a force. 9.8W/s is not a unit of energy, or even power — it is the second derivative with respect to time of energy.”

        It is the derivative, yes. Like heat is the derivative of internal energy, the rate of change in a body. Gravity is the rate of change in a bodys energy as well.

        They are both forces that do work and they both need supply of energy. Do you think that the surface doesn´t heed an energy supply when the air has the same temperature and there is no heat transfer at the location?

        If you don´t have any arguments better than: W is not convertible to m/s, it doesn´t really make you right. And you better call Porsche that their engines doesn´t work. Because Watt can´t be transformed to acceleration in m/s.

        I really don´t know what to say.

        Reply

        • Avatar

          Ed Bo

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          Dense:

          When you talk about “a force of 9.8 m/s”, your statement is literally meaningless.

          When you talk about an “amount of energy … equal to 9.8W/s”, your statement is literally meaningless.

          Kids are taught very early in a technical education to use “dimensional analysis” to check their work. If the units are wrong, their analysis CANNOT BE CORRECT!

          You seem to have missed this lesson, along with the most basic physics. I have explained to you what the proper units are elsewhere in the thread — you don’t have a chance of making a coherent argument until you get these right. You would immediately flunk out of an introductory physics class with your present state of knowledge.

          You keep arguing that an energy supply is required to produce a mechanical force, even if that force causes no displacement, but nothing you cite backs that up at all!

          Reply

          • Avatar

            Dense

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            When you talk about “a force of 9.8 m/s”, your statement is literally meaningless.

            “When you talk about an “amount of energy … equal to 9.8W/s”, your statement is literally meaningless.”

            Ok, you can have any religion you want,

            “Kids are taught very early in a technical education to use “dimensional analysis” to check their work. If the units are wrong, their analysis CANNOT BE CORRECT!”

            Stress and pressure is in units N/m^2. They seem related to gravity, don´t they?

            Once again physics hit you in the head.

            https://en.wikipedia.org/wiki/Stress_(mechanics)

            Read it, you will find lots of things in there that fit gravity, force and the energy needed for it.

            I would say that you:

            “CANNOT BE CORRECT!”

            “You seem to have missed this lesson, along with the most basic physics. I have explained to you what the proper units are elsewhere in the thread — you don’t have a chance of making a coherent argument until you get these right. You would immediately flunk out of an introductory physics class with your present state of knowledge”

            Yeah, I guess you would know about flunking.

            You find a problem with units for pressure and stress for gravity?

            “You keep arguing that an energy supply is required to produce a mechanical force, even if that force causes no displacement, but nothing you cite backs that up at all!”

            Try reading the sources I give you throughout my responses. I still haven´t seen you provide a single one.

  • Avatar

    Dense

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    “No! Absolutely not! The chair is NOT doing work continuously on my body, even though it is exerting a force continuously on my body. There is NO energy transfer because there is no motion induced by the force. (Where would the chair get this energy from???) It is vital that you understand the distinction, or everything you analyze afterwards will be 100% wrong.”

    Work is done when there is a displacement from where the force is acting. The force can act without displacement and you still need energy for the force. Push your hand on the table and the table acts with equal force in opposite direction.

    ” If a body is in equilibrium, there is zero net force by definition (balanced forces may be present nevertheless). In contrast, the second law states that if there is an unbalanced force acting on an object it will result in the object’s momentum changing over time”

    https://en.wikipedia.org/wiki/Force

    “I am totally serious (and I mean this constructively) that you are totally wasting your time in any analysis you do if you do not understand these very, very basic points. If you want to continue in this, you should first take a course or two in Newtonian mechanics (maybe try Khan Academy) and really understand these distinctions.”

    Ok, do you have a source that invalidate the quote from wiki? In that case I need a course, yes.

    Reply

    • Avatar

      Ed Bo

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      Dense: You are simply demonstrating that you have no idea what you are talking about.

      You say: “Work is done when there is a displacement from where the force is acting.”

      So far so good…

      You continue: “The force can act without displacement and you still need energy for the force. Push your hand on the table and the table acts with equal force in opposite direction.”

      NO!!! You do NOT need energy for the force! Where would your table (or the chair in my example) get this energy? Does your table have a battery or electric plug?

      Your Wikipedia post is simply a statement of Newton’s 2nd Law of Motion. It never mentions, or even implies, anything about energy. You REALLY need to take those courses!

      Reply

      • Avatar

        Dense

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        Ed Bo

        April 6, 2017 at 4:33 am | #

        “You REALLY need to take those courses!”

        You should stop before you make a complete fool of yourself.

        Which part of this sentence is hard for you to understand?

        “F and −F are equal in magnitude and opposite in direction.”

        https://en.wikipedia.org/wiki/Force

        Reply

        • Avatar

          Ed Bo

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          Dense:

          “Which part of this sentence is hard for you to understand? ‘“F and −F are equal in magnitude and opposite in direction.'”

          Three days ago I wrote (April 4, 2017 at 2:16 am):

          “As I sit here typing this, I am exerting a downward 784-Newton force on my chair due to my mass (80 kg) in earth’s gravitational field (9.8 m/s^2). By Newton’s 3rd Law, the chair is exerting a 784-N upward force on me in our static situation with no motion.”

          You don’t even understand the arguments at hand! You are the one making a complete fool of yourself.

          My argument is that when these forces cause no displacement, there is no energy transfer due to them. Nothing you have cited or argued contradicts this.

          Reply

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    nickreality65

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    All these bogus thermodynamic gymnastics (cold to hot flow, perpetual looping, energy from nowhere), misapplications of S-B equations (Q=σ(T_TSI^4-T_surface^4) / 4 = σT_effective^4, non-participating media, relative areas), the absurd multi-concentric atmospheric shells acting as thermal diodes or gated transistors are desperate attempts to justify two notions:

    1) That at 396 W/m^2 the surface/ground/earth loses heat, i.e. cools so rapidly (it doesn’t) that icy Armageddon would result if not for the 333 W/m^2 RGHE cold to hot perpetual loop downwelling from the sky and warming the ground (it’s colder and can’t),

    2) That the earth is 33C warmer with an atmosphere than without,

    two notions that are either 1) patently false and/or 2) contradicted by physical evidence.

    Reply

    • Avatar

      Dense

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      “All these bogus thermodynamic gymnastics (cold to hot flow, perpetual looping, energy from nowhere), misapplications of S-B equations (Q=σ(T_TSI^4-T_surface^4) / 4 = σT_effective^4, non-participating media, relative areas), the absurd multi-concentric atmospheric shells acting as thermal diodes or gated transistors are desperate attempts to justify two notions”

      I think you are too agitated to think clearly. “icy armageddon”, how would that happen? Solar radiation is balanced to the red hot inside of earth. The earth surface is in equilibrium with a red hot and even hotter inside, and the sun.

      The shells don´t act as anything, that is only temperatures represented in a state where those energy densities are necessary at a minimum to keep the steady state. The whole point was to avoid the questions around absorption since we are observing inside mass, not a surface. Emission happen according to temperature and heat transfer as well. When radiation enter at the speed of light at the boundary and exit at the same speed, I argue that the surface must obey that condition as well, and all other mean values. The surface is in a confirmed equilibrium, since the addition of 90mW is really small. Non-participating media? Ok, I didn´t say that. What I did was modify the blackbody to what I observe in reality, and it did fit very good. Whether it is true or not I leave to others, but when I see how bitter and agressive you are, I doubt you accept that Work is a true physical term even if it hits you in the head in the shape of a rock.

      If the surface is in equilibrium and only has an addition of 90mW from the internal state, it is not strange if media appears non participating, in this -entirely-fictive-model- of mean values that is an unreal representation of reality. But the sun and the surface is pretty constant in temperature, using them for heat transfer should be fine. And the inverse square law, do I need to validate that?

      I think one main point of it was that the instantaneous emission combined with transfer was, if anything, showing how the surface absolutely not can be 33 degrees hotter with an atmosphere than without.

      Cool it, I´m not forcing anything on anyone.

      Reply

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    nickreality65

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    The only reason RGHE theory exists is as an explanation for the notion that the earth is 33C warmer with an atmosphere than without as explained in the ACS climate change tool kit. This notion is total rubbish.

    The 33C is determined by subtracting 255 K, the alleged surface temperature without an atmosphere, from 288K the alleged “average” surface temperature.

    The 255K is actually the SB BB equivalent temperature of 240 W/m^2, the radiative balance at the ToA, i.e. atmosphere.

    The surface average of 288K is pulled out of the WMO’s butt and has no physical meaning.

    So what would the earth be like without an atmosphere?

    The average solar constant is 1,368 W/m^2 with an S-B BB temperature of 390 K or 17 C higher than the boiling point of water under sea level atmospheric pressure, which would no longer exist. The oceans would boil away removing the tons of pressure that keep the molten core in place. The molten core would rupture flooding the surface with dark magma changing both emissivity and albedo. With no atmosphere a steady rain of meteorites would pulverize the surface to dust same as the moon. The earth would be much like the moon with a similar albedo (0.12) and large swings in surface temperature from lit to dark sides. No clouds, no vegetation, no snow, no ice a completely different albedo, certainly not the current 30%. No molecules means no convection, conduction, latent energy and surface absorption/radiation would be anybody’s guess.

    Whatever the conditions of the earth might be without an atmosphere, it is most certainly NOT 240 W/m^2 and 255K.

    BTW the S-B IDEAL BB radiation equation between two bodies applies only in a vacuum, boundary conditions including thermal equilibrium, thermal transparency and a non-participating medium. For an object to radiate 100% of its energy per S-B there can be no conduction or convection, i.e. no molecules, a “non-participating” medium or aka vacuum. That’s why a laboratory demonstration of the S-B constant is performed in a vacuum chamber. The upwelling LWIR S-B calculation (no physical reality) of 15 C, 288 K, 390 W/m^2 only applies/works in vacuum.

    Applying the IDEAL S-B equation to a surface temperature of 15C/288K/390W/m^2 (Trenberth Figure 10 16C/289K/396W/m^2) where there is significant participating media in the form of latent heat, conduction and convection processes is 100% WRONG!!!!!!!!

    I have several IR thermometers and have used the IR technology in industry. Beyond 30 meters you have zero idea what you are actually measuring, certainly not a “surface.”

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      Dense

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      nickreality65:

      “The only reason RGHE theory exists is as an explanation for the notion that the earth is 33C warmer with an atmosphere than without as explained in the ACS climate change tool kit. This notion is total rubbish.

      The 33C is determined by subtracting 255 K, the alleged surface temperature without an atmosphere, from 288K the alleged “average” surface temperature.
      The 255K is actually the SB BB equivalent temperature of 240 W/m^2, the radiative balance at the ToA, i.e. atmosphere.”

      The “surface” temperature is the blackbodys temperature, 255K. A blackbody has an infinitely thin surface that is perfectly black, it has a perfectly even temperature through its body, it absorbs all radiation that falls on the surface and it emits depending on its temperature only. The surface has to be located where absorption happens, it is the same altitude where the incident radiation is measured. And the blackbody absorbs from a photon gas in a state of equillibrium, from all directions. That is the ideal case, an ideal model can be used as reference and give the information needed when analyzing the problem. So lets look at the differences between earth and the perfect blackbody.

      The first obvious difference is that earth does not absorb from a cloud of photon gas, it absorbs on one side, and emits from both sides twice the area of absorption.

      On earth the equivalent of the blackbody surface would be where the heat source fluxdensity is absorbed at the TOA, it is 1361W/m^2, in the greenhouse wherever the altitude is that absorbs 240Wx4W/m^2 since it is the inverse square law that gives us the effective temperature after albedo. The albedo is the missing 33C, if you don´t use albedo there is no need for a greenhouse effect.

      “The surface average of 288K is pulled out of the WMO’s butt and has no physical meaning.”

      I´m not that sure about that, I came to another conclusion. When comparing earth to the ideal, I see that earth has no perfectly black surface where the incident radiation is measured, at the tropopause. If we measure the solar constant at the tropopause, that is the surface of the ideal blackbody, but the first obvious difference is that absorption happens in a volume below the blackbody surface, absorbed in depth. The absorption then does not happen in a spherical surface, it happens in a spherical volume. Then it is absorbed again in the solid body of earth, which means we have another obvious difference to the ideal blackbody. It is absorbed in a second spherical volume.

      For absorption of solar radiation in the earth system compared to the ideal blackbody, the difference in structure of the bodys surface is that earth absorbs radiation in depth, and the structure consists of two surfaces of concentric spheres of semitransparent character at even temperature throughout the two enclosed volumes.

      Now, I´m not saying that this is the correct model of earth, just that the difference between the ideal case of the blackbody and the real planet earth, is what I described above, among lots of other things. But since the blackbody is only surface and volume of a sphere, we can compare to how it would change with two concentric volumes to see if we are on the right track.
      This is not really a model of absorption as much as a snapshot of the distrubution of energy that is necessary to account for observed mean values.

      If a sphere with a concentric shell is irradiated with an intensity of 1361W/m^2 on one side, the energy density of absorbed solar energy in the inner sphere, if all radiation is absorbed, would be:

      TSI ⁄ (4/3)^2=2σT_surface^4

      Since insolation is only on half the surface area an emission is from the whole surface, the surface temperature of the inner surface would be:

      1/2*TSI ⁄ (4/3)^2=σT_surface^4 = 382.7W/m^2 or 286.6 Kelvin

      Again, I am not saying this is correct, but just correcting for obvious differences we get surprisingly close to the mean temperature. Remember that this is a snapshot where everything has to be in balance in a single second, and now we can see how much heat is transferred from irradiation to 1 m^2 of surface area in that second. The amount of heat transferred from the irradiation to 1m^2 of the inner surface in one second would be:

      Q=σ(T_TSI^4-T_surface^4) / 4 = σT_effective^4

      The inverse square law gives 1/4 of the amount transferred to the surface is radiated towards the surroundings of the sphere and gives the effective temperature of 256.2 Kelvin. Now we can find the amount of heat transferred to the atmosphere from the surface:

      Q=σ(T_surface^4 – T_effective ) = σT_tropopause^4 = 138.2W/m^2 = 222 K

      Then we have:

      Q=σ(T_TSI^4 – T_effective^4=1116
      TSI ⁄ (4/3)^2 = 2σT_surface^4
      and
      σT_surface^4 = σT_effective ^4 + σT_tropopause^4

      =TSI ⁄ (4/3)^2 = 2σT_surface^4
      =σT_surface^4 + σT_effective ^4 + σT_tropopause^4

      If we use heat transfer to find the rate of transfer to the atmosphere and the surface in total:

      Now when we have the surface temperature, we can use heat transfer to see the amount transferred to the surface

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        Dense

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        Wasn´t quite finished and accidently hit the post button.

        What I was going to continue with was, the rate of transfer to the entire system between insolation and effective temperature:

        Q=σT_TSI^4 – T_effective^4 = 1116W/m^2

        This is very close to the measured total irradiance at zenith on a clear day with both direct and diffuse solar radiation according to wikipedia:

        https://en.wikipedia.org/wiki/Sunlight

        The amount of direct solar radiation at zenith under a clear sky, 1050W/m^2, is close to:

        TSI ⁄ (4/3) = 1020.75W/m^2

        In my calculation that is the flux density at the bottom of the outer shell towards the surface.

        Like I mentioned above, I don´t say that it is a correct model, it is just modification of the obvious parameters and then heat transfer. A ball with a shell that absorbs on one side while emitting from the whole surface, shows a distribution of energy that is very close to what we see on earth.

        If we want to see what a surrounding stratosphere would emit, we can just use the real blackbody surface emission:

        TSI ⁄ 4 = 340W/m^2 = 278 Kelvin, which is a pretty close fit to the highest temperature of the stratosphere.

        I think this model is so close to reality that its hard to ignore. I think it is correct, it explains all different parts and their connections to each other, and I can not see any reason to why this structure and distribution would not apply. The distribution of incoming energy can be a coincidential fit, but heat transfer confirms that it is correct, and my opinion is that the mean values at the outer and inner boundary has to obey the radiative transfer equation since the energy coming in and going out of the system is radiation at light speed from real or fictive surfaces. There is more to the model if one starts to pick it apart and I find that all different parts can be connected and is reversible, which must be a positive thing. I think earth has a more or or less fixed energy density as long as temperature stays the same at the sun and in the internal volume of earth. If surface temperature increase, according to heat transfer, it will absorb less since there will be less transferred from insolation. When co2 increase, and absorption increase which is confirmed by observed imbalance, it means that the temperature of the atmosphere drops and the surface increase the rate of transfer to the air. That might explain why surface emission is a bit higher than the empty shells. So in a way, co2 might increase the temperature at the surface, but not in a sensitive way where it might spin out of control. If it is an increase of the rate that heat transfers, it is a small increase at the surface that is giving all it got to keep the temperature up where it is. That would mean that the sensitivity is in the other way, temperature will drop at the surface as soon as the temperature increase in the atmosphere. But the troposphere doesn´t seem to increase in temperature, it is cooled by the extra tiny amounts of potent heat absorber we have put there, tiny amounts of dry ice.

        Why does so many people think that adding a potent heat absorber to a system with limited and constant energy supply that is already present in the system, would increase the temperature?
        More heat absorbing molecules means less energy per molecule, and that can never add up to higher temperature. It does increase the rate of heat transfer to the larger number of heat absorbers though, but I don´t think that is anything like a greenhouse model.

        I don´t know, but I use only basic concepts and mean values, and it fits very good. I think occhams razor would apply, it is the simplest explanation and it doesn´t break the first and second law of thermodynamics, which the greenhouse does. The problem of the greenhouse is that they use a sun at a quarter of the intensity. It is not a model of reality. To me it seems like it is a matter of energy density since I think it is necessary to balance the system in a single second. It is in a steady state and balance is at maximum from mill ions of years of insolation, it would be strange if it wasn´t.

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          nickreality65

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          Those concentric spheres (besides being a stupid, simplistic, unrealistic and useless model) are at best concentric colanders, 99.96% hole and 0.04% reflective stuff. They CAN NOT simultaneously absorb and reflect as that violates conservation of energy.

          The Sun’s photosphere expended to the average earth orbit has a power flux of 1,368 W/m^2. This is spread evenly over the ToA and a sphere of r has 4 times the area as a disc of r means 1,368 W/m^2 / 4 = 342 W/m^2. (Another stupid, simplistic, unrealistic and useless model.)

          S-B BB can be used on the surface of the sun. S-B BB CAN NOT be used on the surface of the earth because the atmospheric molecules constitute participating media and S-B must be corrected with significant emissivity.

          “The “surface” temperature is the blackbodys temperature, 255K.”

          This surface is a S-B BB calculation at the 240 W/m^2 radiative balance at 100 km, NOT the actual earth surface.

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            Dense

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            “Those concentric spheres (besides being a stupid, simplistic, unrealistic and useless model) are at best concentric colanders, 99.96% hole and 0.04% reflective stuff. They CAN NOT simultaneously absorb and reflect as that violates conservation of energy.”

            Wow, that was rude. I haven´t seen you deliver anything of value at all, you seem mostly confused about it all.

            What du you think the blackbody is?

            It is only a volume and a surface, and as I wrote, I only correct for the obvious differences and see if it gets close. It did, in every single part of the system.

            I don´t think you should tell the universe what the stefan-boltzmann law should be used for. That you can´t handle it is not defining the universe. Do you think physics should follow your words?

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            Dense

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            “They CAN NOT simultaneously absorb and reflect as that violates conservation of energy.”

            I did not say that, or that is not what I mean, if I accidently wrote it like that.

            My point is, Every part of the system emits and absorbs at the same time. The values i calculate represent the distribution of energy in a single moment, I balance everything in a second. So it is an equation of state for earth, the flux densities represent the temperature that is the minimum amount necessary for earth keeping a steady state in a single moment. And emitted power IS and instantaneous value, the immediate effect.

            What do you mean by:

            “They CAN NOT simultaneously absorb and reflect as that violates conservation of energy”

            Happens all day and night at earth. You are aware of that this is supposed to be a fictive model of mean values, reality will of course fluctuate and absorption/emission/transmission/reflection will happen in varying amounts at varying locations. But the balance would follow this mean.

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          Dense

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          Sorry about the messy part in the end of the first post, copy pasting and accidental post before correction.

          Anyhow, don´t you think that this:

          TSI ⁄ (4/3)^2 = 2σT_surface^4

          can be written as ampitude of a standing wave seen from the node:

          Amp=TSI ⁄ (4/3)^2 + 2σT_surface^4

          With the more detailed outgoing distrubution of the force:

          σT_surface^4 + σT_effective ^4 + σT_tropopause^4

          Wouldn´t that be logical for a point in space emitting according to its temperature in a radiation field of electromagnetic waves at the strength of the solar constant in that point? A total force of:

          Amp=TSI ⁄ (4/3)^2 + σT_surface^4 +
          σT_effective ^4 + σT_tropopause^4

          I´m not sure, but wouldn´t that be a 4-way manifold at a point in space?

          I like to look at TSI as c and 2σT_surface^4 as c, where ingoing and outgoing radiation at a single moment would be c^2, and in combination with mass they become together a total mass-energy-density of E=mc^2. But that is just to dream away to a place where I don´t have any questions any more. Oh, well, one more question, gravity.

          What if gravity should be treated as a thermal force? I just read the other day about how molecules act in spectroscopy using a flame. Apparently they are drawn into the flame an stay there, attracted to it. That gave me a little confidence to show this:

          If gravity acts on an atmosphere that stays floating above the surface, but doesn´t move, there is an equal force working upwards. The only observed forces all are of thermal origin, and they are all in units of m^2/s. Since gravity is equal to Watt and it has the unit of m/s, if it acts on a stationary object pushed in the other direction with equal force, and gravity is considered to have radiative character, and we use a surface area of 1m^2 for gravity to act on, the necessary force radiated from a surface would be radiated to a surrounding sphere at an intensity of:

          g=m/s
          g^2=m^2/s
          f =sqrt(x/4) = g

          If gravity is 9.8/m/s then the force emitted by the source would be:

          4g^2=384.15W/m^2

          If gravity is a thermal force of radiative character, radiated from a surface to where we measure it in the atmosphere at 9.8m/s average, the surface would emit the energy needed for that force in a second at the same intensity as earth emits thermal energy, to perform work on a square meter above the surface. Right?

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            Dense

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            Forgot the detail that gravity does have a thermal effect, from the work done every second expressed as pressure at the surface. Many people seem to forget that gravity is a force that acts by the second, that means it uses equal amounts of energy per second, and that has to be accounted for. Since all energy we observe is thermal in the beginning and at the end, it makes sense to account for gravity in the same way we do thermal energy, per square meter in watts, as a radiated force. Emission of thermal energy heats the atmosphere and acts as a force outwards, and gravity counteracts with the same amount. observed in pressure that is 1 atm.

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            Ed Bo

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            Dense:

            Mechanical work is force acting over a distance (the integral of force over distance). The force must cause movement for thermodynamic work to be performed. A static force, such of that of the atmosphere on the surface due to the pull of gravity does NO work and transfers NO energy — it has no thermal effect.

            (This is very different from a dynamic compression, where the gravitational force does act over a distance and cause motion. This happened in the earth’s formation … 4.5 billion years ago!)

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            Dense

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            Ed Bo April 2, 2017 at 9:47 pm | #

            Dense:

            “Mechanical work is force acting over a distance (the integral of force over distance). The force must cause movement for thermodynamic work to be performed. A static force, such of that of the atmosphere on the surface due to the pull of gravity does NO work and transfers NO energy — it has no thermal effect.”

            Thank you for your comment. I think the atmosphere moves quite alot, don´t you? The reason I said it lays still is to visualize that there is an expanding force outwards and an attracting force inwards that cancel out.

            By your reasoning, if I strap myself to two cars that accelerate in opposite direction, I will not experience a force if they act with an exactly equal force at the center and the forces cancel out with both cars stationary burning rubber. Right?

            The accelerating force acts every second and has units of Watt, you counteract it constantly when staying upright with your muscles. If there were no force, why do you fall to the ground if your muscles would be instantly paralyzed? And why do astronauts suffer from muscle loss in weightlessness? It seems like our bodies are built on mechanical stress as a frame, and we all know that the product of a life form is nothing else than thermal energy at the end.

            When an object falls to the surface, the kinetic energy is released as heat on impact. Gravity, whatever it is, reclaims the energy you use in work when you raise an object and drop it, as thermal energy. Right?

            “(This is very different from a dynamic compression, where the gravitational force does act over a distance and cause motion. This happened in the earth’s formation … 4.5 billion years ago!)”

            But dynamic compression and the connected heat release happens internally on earth today as well. The atmosphere circulate by compression and thermal expansion all the time. And gravity causes the surface pressure, and if it didn´t, the gas would drop in temperature without it (among other things). The temperature of the surface air has the essential component of pressure as part of the conditions. Pressure answers for keeping the density of matter in the atmosphere at the surface constant which is fundamental for the energy density and temperature. Without gravity it would be cold in the air, even if radiative heat would heat surfaces.

            I´m not saying I´m right, but I cant see how gravity should be excluded from temperature at the surface. And if you don´t agree, then we still has to explain where the energy comes from, that can accelerate mass with 9.8m/s, increasing its kinetic energy with 9.8 Watts every second all the way to the surface. Sure, if we put it at altitude, we added the energy doing work, but the mass has no increase of energy in itself, it is all energy that is added in the fall with speed. And it would apply for the surprising event of a beaver materializing at 5000km altitude, from an alternative universe, and it didn´t get there from added work. Sure, unrealistic scenario, but I think you can come up with something else yourself, where an object comes into the gravitational field without any work. On impact heat is added to earth. If gravity cause heat release when acting on surroundings, I would think that gravity itself has a thermal origin. It seems logic to me, I don´t say anyone else should believe it, just that it can be calculated to equal surface emission, and heat is the effect of gravity when it has acted on mass from starting point to end point at the surface.
            Personally I think that is much less BS than the mystical version where it fails to explain observations and resort to some force that is everywhere, can´t be explained, and governs the universe. That is too religious for me. When gravity is taught to be a force that is explained with “Dark Matter”, another even more mystical force that we can see even less, but some dude assures us is there just because don´t have the capacity to figure it out without it, then I have already left the room and opened a beer. I believe in what I see and the forces we know and the theories that stands over time. If they add upp to explain what we know with some certainty, then I´m happy.

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            Ed Bo

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            Dense:

            Thank you for your response.

            As I sit here typing this, I am exerting a downward 784-Newton force on my chair due to my mass (80 kg) in earth’s gravitational field (9.8 m/s^2). By Newton’s 3rd Law, the chair is exerting a 784-N upward force on me in our static situation with no motion.

            But NO energy is being transferred in this situation. The chair needs no energy source — batteries, electrical outlet, or anything else — to maintain this force on my body. By your logic, the chair would need such an energy source to hold me up.

            (Note carefully that I am NOT claiming that there is no force here. There most certainly is. If the 784 N is more than the chair can support, it will break. The 784 N of force a strong-enough chair uses to hold me up can make my backside sore!)

            Look at the units carefully: 1 Newton = 1 kg*m/s^2 (as with my mass multiplied by earth’s gravitational acceleration). This is a unit of force, NOT of energy or power.

            Now, look at the units of energy. 1 Joule = 1 kg*m*m/s^2 (or 1 kg*m^2/s^2). This is the units of force times distance. The force must cause motion for there to be a work transfer, as work is a form of energy transfer. This is one of the first topics covered in any introductory physics course, and it is vital that you understand it at a very deep level.

            (Similarly the fundamental unit of power is the Watt. 1 W = 1 J/s = 1 kg*m^2/s^3. This can be expressed as 1 W = 1 N * 1 m/s. Mechanical power transfer is force times velocity.)

            If I drop a 1 kg ball of putty from a 4.9 meter height over the surface, it reaches a velocity of 9.8 m/s by the time it hits the surface 1 second later (ignoring the minimal air friction). It has converted its potential energy of height into (organized) kinetic energy of motion.

            This kinetic energy of m*v^2/2 is 1 kg * 9.8^2 m^2/s^2 / 2 = 48 Joules.

            As it “splats” on the floor in an inelastic collision, this organized 48 J of kinetic energy of motion is converted to the “disorganized” kinetic energy of the “internal energy” of thermal energy. But this is a one-time conversion — it is NOT an ongoing transfer.

            Now, there can be downdrafts in the atmosphere, but they must be exactly compensated for by updrafts, because the atmosphere as a whole is neither rising nor falling. So for the atmosphere as a whole, there is NO energy transfer with the surface by virtue of the atmosphere’s pressure/weight.

            When hydroelectric power is generated from the potential energy of water, the water ends up lower than where it started. It has fallen. The atmosphere has already fallen, so there is no energy to be transferred from the potential energy of the atmosphere as a whole.

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            Dense

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            Hello again, Ed Bo

            April 4, 2017 at 2:16 am | #

            “Look at the units carefully: 1 Newton = 1 kg*m/s^2 (as with my mass multiplied by earth’s gravitational acceleration). This is a unit of force, NOT of energy or power.”

            Well, I make no difference between them, since they can be converted to each other, and you need the equivalent energy input to have that force do work. And in the end it is all thermal.

            “Now, look at the units of energy. 1 Joule = 1 kg*m*m/s^2 (or 1 kg*m^2/s^2). This is the units of force times distance. The force must cause motion for there to be a work transfer, as work is a form of energy transfer. This is one of the first topics covered in any introductory physics course, and it is vital that you understand it at a very deep level.”

            I am aware of this, but nevertheless, 1 Watt is equal to 1 newton-meter per second, and Watt is just Watt per second. In gravity we use it for center of mass, not a surface. But I don´t think we can say gravity doesn´t act on surfaces, and if we want to know the relationship between radiation and gravity we need to have them acting in the same dimensions.

            “As it “splats” on the floor in an inelastic collision, this organized 48 J of kinetic energy of motion is converted to the “disorganized” kinetic energy of the “internal energy” of thermal energy. But this is a one-time conversion — it is NOT an ongoing transfer.”

            And if you dropped a cube falling flat on one side, it is easy to see that the force of gravity that turns into thermal energy would be in m^2. Right?

            You are right, that particular case is not one of ongoing transfer, but the case of you sitting in a chair or the ball laying on the floor is. If gravity didn´t do work continously on your body, and if it was as you say, no forces were acting, you would float away into space, at least if you didn´t grab on to anything but then you would replace the work that gravity did.

            Same goes for our planet, it has a high speed and to preserve energy it would have to go straight forward into space at constant speed. But it turns constantly in its orbit. Some people seem to think of it like centripetal force which we observe on the surface, but in that case it is downward force balanced to the rotation which acts in two horizontal directions. Earth has to have an equal force outwards from its orbit, as inwards, but also forwards, to find balance. If there were no forces acting on it, it would be at rest travelling straight ahead with constant speed.

            My view of it is that only straight forward at constant speed is the case where no forces act on earth. If there was only outwards and inwards forces cancelling out, it would wind up like a chord around the sun closing in fast.

            “Now, there can be downdrafts in the atmosphere, but they must be exactly compensated for by updrafts, because the atmosphere as a whole is neither rising nor falling. So for the atmosphere as a whole, there is NO energy transfer with the surface by virtue of the atmosphere’s pressure/weight.”

            That was kind of my point, emission is a clear force acting on mass making it rise, without it the atmosphere would not move at all. My view of it is that gravity balances that energy that moves mass around the atmosphere, by acting in the opposite direction pulling mass down. The updrafts are clearly caused by energy, but without an equal force pulling it down again it would just drift of into space. It cannot sink by itself, there is no reason for it not just continuing outwards. But gravity *pulls* it down.

            The outwards energy is in W/m^2/s, and gravity is Nm/s, or W/s, that is why I did that. And I am aware of the units, its just that it doesn´t matter since gravity is not accounted for in m^2. And I don´t think we can say that gravity doesn´t act on surfaces.

            Thanks for taking your time and keeping a good tone, not like the blanket people.

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            Ed Bo

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            Dense:

            In physics, force, energy, work, and power have very specific and distinct meanings. I am afraid you are conflating them completely.

            You say: “You are right, that particular case is not one of ongoing transfer, but the case of you sitting in a chair or the ball laying on the floor is. If gravity didn´t do work continously on your body, and if it was as you say, no forces were acting, you would float away into space.”

            No! Absolutely not! The chair is NOT doing work continuously on my body, even though it is exerting a force continuously on my body. There is NO energy transfer because there is no motion induced by the force. (Where would the chair get this energy from???) It is vital that you understand the distinction, or everything you analyze afterwards will be 100% wrong.

            I am totally serious (and I mean this constructively) that you are totally wasting your time in any analysis you do if you do not understand these very, very basic points. If you want to continue in this, you should first take a course or two in Newtonian mechanics (maybe try Khan Academy) and really understand these distinctions.

            A lot of your other claims are literally meaningless — I have no idea what you are even trying to say when you claim: “And if you dropped a cube falling flat on one side, it is easy to see that the force of gravity that turns into thermal energy would be in m^2. Right?”

            or: “The outwards energy is in W/m^2/s, and gravity is Nm/s, or W/s, that is why I did that. And I am aware of the units, its just that it doesn´t matter since gravity is not accounted for in m^2.”

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    nickreality65

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    Upwelling of 396 W/m^2 double counts the energy, like entering your paycheck twice. One or the other, but NOT both.

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    nickreality65

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    So here’s my similar thermos/heat transfer based version.

    References:

    Trenberth et al 2011jcli24 Figure 10

    This popular balance graphic and assorted variations are based on a power flux, W/m^2. A W is not energy, but energy over time, i.e. 3.4 Btu/eng h or 3.6 kJ/SI h. The 342 W/m^2 ISR is determined by spreading the average discular 1,368 W/m^2 solar irradiance/constant over the spherical ToA surface area. (1,368/4 =342) There is no consideration of the elliptical orbit (perihelion = 1,415 W/m^2 to aphelion = 1,323 W/m^2) or day or night or seasons or tropospheric thickness or energy diffusion due to oblique incidence, etc. This popular balance models the earth as a ball suspended in a hot fluid with heat/energy/power entering evenly over the entire ToA spherical surface. This is not even close to how the real earth energy balance works. Everybody uses it. Everybody should know better.

    An example of a real heat balance based on Btu/h is as follows. Basically (Incoming Solar Radiation spread over the earth’s cross sectional area, Btu/h) = (U*A*dT et. al. leaving the lit side perpendicular to the spherical surface ToA, Btu/h) + (U*A*dT et. al. leaving the dark side perpendicular to spherical surface area ToA, Btu/h) The atmosphere is just a simple HVAC/heat flow/balance/insulation problem.

    http://earthobservatory.nasa.gov/IOTD/view.php?id=7373

    “Technically, there is no absolute dividing line between the Earth’s atmosphere and space, but for scientists studying the balance of incoming and outgoing energy on the Earth, it is conceptually useful to think of the altitude at about 100 kilometers above the Earth as the “top of the atmosphere.”

    The top of the atmosphere is the bottom line of Earth’s energy budget, the Grand Central Station of radiation. It is the place where solar energy (mostly visible light) enters the Earth system and where both reflected light and invisible, thermal radiation from the Sun-warmed Earth exit. The balance between incoming and outgoing energy at the top of the atmosphere determines the Earth’s average temperature. The ability of greenhouses gases to change the balance by reducing how much thermal energy exits is what global warming is all about.”

    ToA is 100 km or 62 miles. It is 68 miles between Denver and Colorado Springs. That’s not just thin, that’s ludicrous thin. 99% of the atmospheric mass is below 32 km. Above 32 km there are very few molecules. Without molecules, energy, heat, cold, hot concepts get a tad iffy.

    The GHE/GHG loop as shown on Trenberth Figure 10 is made up of three main components: upwelling of 396 W/m^2 which has two sub parts: 63 W/m^2 LWIR and 333 W/m^2 and downwelling of 333 W/m^2.

    The 396 W/m^2 is calculated by inserting 16 C or 279K in the S-B BB equation, a calculation that does not actually exist in the real world. The result is 55 W/m^2 of power flux more than ISR entering ToA, an obvious violation of conservation of energy, i.e. created out of nothing. That should have been a warning.

    ISR of 341 W/m^2 enter ToA, 102 W/m^2 are reflected by the albedo, leaving a net 239 W/m^2 entering ToA. 78 W/m^2 are absorbed by the atmosphere leaving 161 W/m^2 for the surface. To maintain the overall energy balance and a steady temperature (not really a requirement) 160 W/m^2 rises from the surface (0.9 residual in ground) as 17 W/m^2 convection, 80 W/m^2 latent and 63 W/m^2 LWIR (S-B BB 183 K, -90 C or emissivity = .16) = 160 W/m^2. All of the graphic’s power fluxes are now present and accounted for. The remaining and perpetual looping 333 W/m^2 are the spontaneous creation of an inappropriate application of the S-B BB equation violating conservation of energy.

    But let’s press on.

    The 333 W/m^2 upwelling/downwelling constitutes a 100% efficient perpetual energy loop violating thermodynamics. There is no net energy left at the surface to warm the earth and there is no net energy left in the troposphere to impact radiative balance at ToA.

    The 333 W/m^2, 97% of ISR, upwells into the troposphere where it is allegedly absorbed/trapped/blocked by a miniscule 0.04% of the atmosphere. That’s a significant heat load for such a tiny share of atmospheric molecules (aren’t any above 32 km) and they should all be hotter than two dollar pistols.

    Except they aren’t.

    The troposphere is cold, -40 C at 30,000 ft, 9 km, < -60 C at ToA. Depending on how one models the troposphere, an evenly distributed average or weighted by layers from surface to ToA, the S-B BB equation for the tropospheric temperatures ranges from 150 to 250 W/m^2, a considerable, 45% to 75% of, less than 333. Radiation is a surface phenomenon. There is no “surface.”

    (99% of the atmosphere is below 32 km where molecular energy moves by convection/conduction/latent/radiation & where ideal S-B does not apply. Above 32 km the low molecular density does not allow for convection/conduction/latent and energy moves by S-B ideal radiation et. al.)

    But wait!

    The GHGs reradiate in all directions not just back to the surface. Say a statistical 33% makes it back to the surface that means 50 to 80 W/m^2. An even longer way away from the 333, 15% to 24% of.

    But wait!

    Because the troposphere is not ideal the S-B equation must consider emissivity. Nasif Nahle suggests CO2 emissivity could be around 0.1 or 5 to 8 W/m^2 re-radiated back to the surface. Light years from 333, 1.5% to 2.4% of.

    But wait!

    All of the above really doesn’t even matter since there is no net connection or influence between the 333 W/m^2 thermodynamically impossible loop and the radiative balance at 100 km ToA. Just erase this loop from the graphic and nothing else about the balance changes.

    BTW 7 of the 8 reanalyzed (i.e. water board the data until it gives up the “right” answer) data sets/models show more power flux leaving OLR than entering ASR ToA or atmospheric cooling. Obviously those seven data sets/models have it completely wrong because there can’t possibly be any flaw in the GHE theory.

    The GHE greenhouse analogy/theory not only does not apply to the atmosphere, it doesn’t even apply to warming a real greenhouse. (“The Discovery of Global Warming” Spencer Weart) In a real greenhouse the physical barrier of walls, glass, plastic trap the convective heat, not some kind of handwavium glassy, transparent, multi-layer, radiative thermal diode.

    The surface of the earth is warm for the same reason a heated house is warm in the winter: Q = U * A * dT, the energy flow/heat resisting blanket of the insulated walls. Same for the atmospheric blanket. A blanket works by Q = U * A * dT, not S-B BB. The composite thermal conductivity of that paper-thin atmosphere, conduction, convection, latent, LWIR, resists the flow of energy, i.e. heat, from surface to ToA and to make that energy flow (heat) requires a temperature differential, 213 K ToA and 288 K surface = 75 C. The atmosphere is just a basic HVAC system boundary analysis.

    Open for rebuttal. If you can explain how this upwelling/downwelling/”back” radiation actually really works be certain to copy Jennifer Marohasy as she has posted a challenge for such an explanation.

    Nick Schroeder, BSME, PE

    Additional observatons:
    http://writerbeat.com/articles/15582-To-be-33C-or-not-to-be-33C
    http://writerbeat.com/articles/15855-Venus-amp-RGHE-amp-UA-Delta-T

    Reply

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    Quokka

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    Of course O2 and N2 hold (not ‘trap’) most of the heat in the atmosphere. That’s how greenhouse gases heat the atmosphere: by absorbing energy, then colliding with other molecules and passing the energy on to them. And, of course, the reverse is true; greenhouse gases also absorb energy from other gases then radiate it away again. When you add greenhouse gases to the atmosphere, the equilibrium state that is reached is hotter then it was without them. By your logic, because black surfaces radiate better than shiny ones, painting your car black will allow it to cool down better. In reality, your car gets hotter, for the same reason that the atmosphere gets hotter when you put molecules in it that are superior absorbers and emitters of radiation.

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    Ed Bo

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    Rosco — You ask: “Then how does the 99% of the atmosphere ever lose energy and thus cool down?”

    That’s easy! If there are radiatively active gases in the mix (the other 1%), they transfer energy to those gases through collisions, permitting those gases to radiate energy to space. That you acknowledge.

    But with or without radiatively active gases, if the atmosphere at the surface has a higher temperature than the surface, it transfers heat to the surface through conduction, and the surface can radiate that energy away.

    This happens frequently on earth at night, particularly on still, clear nights with low (absolute) humidity that permit the surface to radiate a lot of power past the lower atmosphere. (This is why you can get frost on nights when the surface air temperature always stays above freezing.)

    This is what creates temperature inversions. The surface is radiating power very effectively past the lower atmosphere, so it has a lower temperature than the air just above it. This causes the lowest layers of air to transfer heat to the surface, and higher layers of air to the lower layers, setting up the inversion which is likely to last until the sun starts to warm the surface.

    This inversion lasts for months over Antarctica in the long polar winter night when it gets no sunlight at all.

    I emphasize that this mechanism would be present in an atmosphere without ANY radiatively active gases at all. A transparent atmosphere would absorb energy from the surface when it is cooler than the surface (daytime), and transfer energy to the surface when it is hotter than the surface (nighttime).

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      Carl Brehmer

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      I am aware that this explanation of nighttime, near-surface, atmospheric temperature inversions is very common, so common, in fact, that it is being taught at the Penn State College of Earth and Mineral Science. It suffers though because it is based on a demonstrably false assumption—an assumption that your post repeats.
      “if the atmosphere at the surface has a higher temperature than the surface, it transfers heat to the surface through conduction, and the surface can radiate that energy away.” You go on to assert that “This happens frequently on earth at night, particularly on still, clear nights with low (absolute) humidity . . .”

      I tested this hypothesis using three thermometers and found it to be false. Here is the experiment:
      1) I chose an open, flat patch of ground unobstructed by trees or structures so as to measure the thermal relationship between the ground and the atmosphere without interference.
      2) I laid one thermometer directly on the ground to measure the temperature of the dirt
      3) I positioned a second thermometer about 10cm above the ground
      4) I positioned a third thermometer about 1.5 meters above the ground
      5) I then recorded these three temperatures every 20 minutes 24 hours a day for two months (the thermometers where attached to electronic data loggers so I didn’t have to be physically present at each reading)

      The nighttime results:
      a) The temperature of the ground was virtually always warmer than either of the air temperatures, which contradicts your assertion that at night the ground is frequently cooler than the air above it.
      b) The air above the ground virtually always cooled faster at night than did the ground below it, i.e., as the night progressed the temperature differential between the cooler air and the warmer ground increased and this temperature differential was widest on cloudless night.
      c) At night the temperature of the air 10cm off of the ground was always cooler than the air 1.5 meters off of the ground, which was always cooler than the ground beneath both of them.
      d) The temperature difference between the air 10cm off of the ground and the air 1.5 meters off of the ground was on average about 2 °C. This accounts for the Hoar Frost to which you referred in your post, i.e., the the appearance of frost on grass and low growing plants at night even when the measured air temperature is above freezing. On close inspection when Hoar Frost appears at the same time that the air temperature at 1.5 meters is above freezing you will note that the ground itself is not frozen. The only thing frozen will be those structures that are surrounded by the cold air just up off of the ground—things like grass and low growing plants, etc.

      Needless to say the cold air just above the warmer ground is not conducting heat to the ground for it to be radiated away through the atmosphere as is so commonly asserted. So, how then do near-the-ground temperature inversions form at night?

      In my view the air is not cooling in place, but rather cooler air is moving into place. As everyone knows who has ridden a motorcycle, the air temperature even at eyelevel is not uniform. Thus at night when the air is not actively being warmed by the sun cool pockets of air sink because of gravity and “pool” on top of the ground. This creates the cold layer of air that is sandwiched between the warmer air above it (which is naturally more buoyant than the cooler air below it) and the even warmer ground below. The entire structure—the ground, the air at 10 cm and the air at 1.5 meters—cools throughout the night as a unit while their thermal relationship remains constant with the ground being the warmest, the air at 10 cm being the coldest and the air at 1.5 meters being cooler than the ground but warmer than the air below it.

      Should you chose to repeat this experiment let me know what you find.

      Reply

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        nickreality65

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        I did the same experiment as a consequence of a discussion with a CSU PhD and observed pretty much the same thing. The popularly observed down welling IR is an illusion.

        Reply

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        Ed Bo

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        Carl:

        Many nights I have seen frost on the ground when the air temperature never got down to 0C. My pool pumps automatically turn on at +4C for freeze protection. What do those system designers know?

        The surface is highly opaque to infrared radiation so it is the temperature of the immediate surface layer that matters. Measurements even a centimeter or two down will be highly misleading.

        Note that I said that you get these temperature inversions “particularly on still, clear nights with low (absolute) humidity”. What conditions did you do your test in?

        If you listen to winter weather reports in hilly or mountainous areas, they will say the overnight temperatures are lowest in the isolated valleys (where there is less wind).

        If you look at vineyards, orchards, or fields where frost damage is a potential problem, you will often see giant fans distributed around these fields. These are turned on when the air is slightly above freezing to mix the warmer air with the colder surface layer and the surface itself. In other words, they use air movement to keep the surface from getting too cold!

        These farmers spend tens of thousands of dollars on these systems, based on a mechanism that you say doesn’t exist. Are they wasting their money?

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          nickreality65

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          Latent heat of evaporation. Dew point. Look at the psychrometric properties of most air. As water evaporates it cools the air below the freezing point. That’s how frost ends up on the windshield. An industrial wet cooling tower will freeze up even thought the dry bulb is above freezing.

          Turning on the orchard fans accelerates convection.

          Drive a motorcycle down the highway. Notice how cold it gets in the valleys. Cold air settles, no mystery, but the ground is still warmer.

          It takes days if not weeks of sub-zero temperatures to freeze the ground and water pipes. If the ground were actually radiating 390 W/m^2 that would not be the case, it would freeze as soon as air temp dropped below 32F. Or if I pitched a big top tent and blocked the down welling IR.

          And at 30m down the ground is a constant temperature, i.e. moving positive heat to the surface.

          As Feynman observed if observations don’t match the theory, the theory loses.

          Reply

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            Ed Bo

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            NIck:

            You say: “As water evaporates it cools the air below the freezing point. That’s how frost ends up on the windshield.”

            Frost occurs when the air is saturated — no evaporation (or sublimation) going on. And the production of frost (desublimation) is the opposite: a transition from gaseous state that “releases” the latent heat.

            You say: “If the ground were actually radiating 390 W/m^2 that [delayed freezing] would not be the case.”

            But the ground IS radiating that amount, as can readily be measured. (At 0C, it’s more like 300 W/m^2.) This is no computer model conjecture, but actual repeatable measurement, spectrum and all! It’s just that ground is a very poor conductor, and there is no convection to speak of.

            You say: “if I pitched a big top tent and blocked the down welling IR…[the ground] would freeze as soon as air temp dropped below 32F.”

            It’s very common on a winter morning with the temperature near freezing to see frost on the part of a rooftop that is directly exposed to the sky, and no frost where there is overhang from an evergreen tree. The tree, like the tent, provides MORE downwelling IR than the upper atmosphere does.

            You say: “Notice how cold it gets in the valleys. Cold air settles.”

            So why is there almost always a negative lapse rate (colder higher) during the day? Why are the lowest altitude valleys — Death Valley, Dead Sea Valley — the warmest places around?

            The amount of stuff you get exactly backwards is staggering! Observations don’t match your theories!

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          Carl Brehmer

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          None of the observations that you present constitute a repetition of the experiment that I outlined in that none of them contain temperature readings of the actual ground. Nor are any of the observations that you present inconsistent with the results of the experiments that I performed. Let’s go through them.

          1) “Many nights I have seen frost on the ground when the air temperature never got down to 0C.”
          The next time that you see “frost on the ground” when the temperature is above zero take a closer look. You will see that the frost is actually on things that are slightly up off of the ground and surrounded by the cold layer of air that has sunk to its lowest point, which is on top of the ground. The frost will be on grass and other plants, wire mess, maybe on the top of rocks protruding out of the ground, etc. The ground itself will not be frozen when the temperature being recorded is above zero °C. The location of thermometers, as I said in my post, that people reference when they say “the temperature is . . .” are invariably some 1.5 meters above the ground, which is above the cold layer of air that has pooled below it on top of the warmer ground.

          2) “My pool pumps automatically turn on at +4C for freeze protection. What do those system designers know?”
          Not knowing anything about the setup of your pool all that I might suggest is that they are being overly cautious because they don’t want you to sue them if any pipes that might be exposed above ground freeze. As you know a little ice on the surface of the pool will cause no problems, but water that freezes within the piping or the pump will crack them open.

          3) “The surface is highly opaque to infrared radiation so it is the temperature of the immediate surface layer that matters. Measurements even a centimeter or two down will be highly misleading.”
          My temperature measurements of the ground were taken by laying a thermocouple right on top of the ground, not buried.

          4) “Note that I said that you get these temperature inversions ‘particularly on still, clear nights with low (absolute) humidity’. What conditions did you do your test in?”
          Throughout the testing the weather varied between completely cloud free nights and completely over cast nights, even rainy nights. The results remained consistent—the ground temperature never dropped below the temperature of the air just above it. The temperature differential, i.e., the amount that the air was cooler than the ground, was highest on cloudless nights.

          5) “If you listen to winter weather reports in hilly or mountainous areas, they will say the overnight temperatures are lowest in the isolated valleys (where there is less wind).”
          Might I suggest that this is not because there is less wind in the valley but rather because air is fluid. As such heavier, more-dense cold air flows down into valleys as does water and pools, as does water, in low lying areas. This phenomenon even has a name. It’s called “downslope cold air drainage” and the pools of cool air are called “cold air lakes”.

          6) “If you look at vineyards, orchards, or fields where frost damage is a potential problem, you will often see giant fans distributed around these fields. These are turned on when the air is slightly above freezing to mix the warmer air with the colder surface layer and the surface itself. In other words, they use air movement to keep the surface from getting too cold!”
          These fans are mixing the warmer air at 1.5 meter with the cooler air lying on top of the ground preventing the pooling of cold air described above. Their intent is not to keep the ground itself from freezing because the temperature of the ground itself will be above freezing as long as the temperature of the air at 1.5 meters is above freezing

          7) “These farmers spend tens of thousands of dollars on these systems, based on a mechanism that you say doesn’t exist. Are they wasting their money?”
          These farmers are not wasting their money because these systems do work to prevent low lying crops from freezing. They are based though not on the mechanism that you say exists, but rather because they prevent cold air from pooling on top of the ground, which causes low lying plants to freeze. These systems have their limitations. They become ineffective at preventing freezing when the air at 1.5 meters drops below freezing.

          Reply

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            Ed Bo

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            Carl:

            Several times each winter, I wake up at dawn to see frost on my rooftop (those sections not shielded from the sky by overhanging trees) and on my cars, with the air temperature several degrees above freezing (and before any morning heating has begun). These are solid surfaces, and it is these surfaces that the air is in conductive thermal contact with.

            You say: “Might I suggest that this is not because there is less wind in the valley but rather because air is fluid. As such heavier, more-dense cold air flows down into valleys as does water and pools, as does water, in low lying areas.”

            You may suggest it, but there are two problems with the idea. First, what makes these valleys “isolated” is that there is nowhere for such a flow to come from. Second, when air flows downward like that, it is subject to (initially) adiabatic heating. If you have any experience in this type of terrain, you know that conditions where the air flows downhill like that are among the hottest of all (Santa Ana winds in California, Chinooks by the Rocky Mountains).

            You say: “These farmers are not wasting their money because these systems do work to prevent low lying crops from freezing. They are based though not on the mechanism that you say exists, but rather because they prevent cold air from pooling on top of the ground, which causes low lying plants to freeze.”

            My claim is exactly that “they prevent cold air from pooling on top of the ground.” They do so by causing it to mix with warmer air higher up. But the only reason that the coldest air is at the lowest levels is that it is in proximity to an even colder surface.

            You say: “These systems have their limitations. They become ineffective at preventing freezing when the air at 1.5 meters drops below freezing.”

            All the systems I’ve seen have the fans about 10 meters above the ground. And these would enhance mixing with warmer air at several times that height.

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    daymo6000

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    Stupid deniers!
    You don’t seem to understand how the atmosphere works.
    Within the Earth’s atmosphere, all electro magnetic radiation must adhere to the local By-Laws of Thermodynamics whereupon a “climate tax” must be “back radiated” to the surface of the Earth in the form of infra red radiation.

    Cheers

    Reply

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    Carl Brehmer

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    QUOTE:
    “Satellites, on the other hand, can get a reading on these hard-to-reach, harsh places because they can scan every piece of the Earth’s surface.”
    “The single highest land skin temperature recorded in any year of the study was found in the Lut Desert in 2005 and measured a stunning 159.3 F (70.7 C).”
    “How can that happen if the Sun supplies only 161 W/m 2 as indicated in the Energy Budget and “back radiation” supplies 333W/m 2 for a total of 494 W/m 2 ? 494 W/m 2 is equivalent to about 32.5 degrees C – where did the other 37.5 degrees C come from?”

    This final question is oddly out of place in this otherwise logical exposé of the fallacy of “greenhouse warming” by “greenhouse gases”, but just addressing this question consider the following:
    1) Local spot temperatures are not determined by globally averaged insolation.
    2) Since the atmosphere is virtually always cooler than the surface, the surface is not and cannot be “warmed” by “back radiation”. What “back radiation” does is decrease the rate of heat loss from the surface via IR radiation. It has no effect what so ever on the rate of heat lose via conduction, which is enhanced by convection or on the rate of heat loss via transpiration, i.e., evaporation, i.e., latent heat transfer.
    3) The desert of which you speak is located around 32° N, which is under the down-going leg of the Hadley Cell. As air sinks within this down-going leg of the Hadley Cell it warms adiabatically at a rate of ~9.8 °C/km due to the “work” being done on it by its progressively higher pressure surroundings. This, in turn, is what creates the characteristic lapse rate observed when one releases a weather balloon at around 32° N. Just do the math.

    Temperature at the tropopause (about 11 km in altitude) = -60 °C
    Temperature increase due to adiabatic warming of descending air at around 32° N:
    11km x 9.8 °C/km = 107.8 °C
    107.8 °C – 60 °C = 47.8 °C
    Thus adiabatic warming more than accounts for the “extra” 37.5 °C of warming that you are looking for, but beyond that the daytime local insolation in the desert in question is >1300 W/m^2. It is not the 161 W/m^2 that your question implies.

    So, between the high intensity of sunlight in mid-day, in mid-summer when the temperature high in question was recorded and the adiabatic warming of descending air at ~32° N, there is no need to add any other energy source to account for the surface temperatures measured such as “back radiation” and/or a “gravito-thermal effect” as some have attempted to do.

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      Rosco

      |

      I included the Landsat observation simply to show that science does not meaningfully measure temperatures.

      Land surface temperatures are not routinely measured yet ocean surface temperatures are – although the data set for both is too small to make the assertions alarmists make.

      The 70.7°C quoted is the surface temperature – I do not believe the atmosphere heats the desert surface – the reverse is obviously true. As you clearly state the solar radiation easily accounts for the surface temperature.

      I have always had a problem with the assertion that what “back radiation” does is decrease the rate of heat loss from the surface via IR radiation”.

      The experimental evidence from the cavity radiation experiments show that an object emits radiation in proportion to its temperature. The construction of the cavity experiments was to eliminate the ambient radiation effect – any ambient radiation entering the cavity was unlikely to escape, simply becomes part of the internal radiation due to the temperature and hence allowed for the calculations from the observed emission from the cavity.

      If this were not so and ambient radiation had to be considered the solution would not have been possible. There would be more unknowns than equations for solution.

      The “net” form of the SB equation seems problematic to me and algebraic manipulation of it as performed by many is basically wrong. This can easily be verified by performing those same algebraic manipulations on Planck curves and observing that for blackbody radiation the algebra fails totally. If it doesn’t work for blackbody radiation it doesn’t work full stop.

      There is an explicit relationship between the SB equation, Wien’s law and Planck’s law. Only Planck’s fully describes a radiant spectra.

      If the assertion that what ““back radiation” does is decrease the rate of heat loss from the surface via IR radiation” is true then the only possible explanation is the transfer of heat from cold to hot.

      If the foundation of the laws of radiation science are correct then if the emission from the surface is decreased by back radiation there are only 2 possible explanations – the surface temperature decreases or heat transfers from cold to hot.

      I make no claim to know how this stuff works but I do not believe any of the algebraic manipulations performed using the SB equation to “prove” all of the assertions routinely made because I can show that if you apply the same algebra to Planck curves they fail to work at all.

      Reply

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        Carl Brehmer

        |

        Even as you assert, “I have always had a problem with the assertion that what ‘back radiation’ does is decrease the rate of heat loss from the surface via IR radiation,” you contradict this in own article, which says, “This radiation is the only method of removing energy from the Earth’s atmosphere and land surfaces to space; and of this 239 W/m2 a mere 40 W/m2 is emitted directly from the surface.”

        Ergo, what your article calls “back radiation” is reducing the rate at which the surface looses heat via IR radiation from its measured 396 W/m^2 of up-going IR radiation down to a mere 40 W/m^2. That is a significant reduction in the rate of surface heat loss via IR radiation. Note as well that the heat loss via IR radiation is still up-going. “Heat” is not flowing from the cooler atmosphere to the warmer surface.

        In my view the difference between what is called “back radiation” and “surface radiation” in the KT diagram that you included in our article is equivalent to the difference in water pressure at two ends of a pipe or the difference between voltages within an electrical circuit or the difference between the air pressure of two adjacent weather systems.

        1) There is a flow of water from where the pressure is higher to where the pressure is lower. There is not a two way flow of water, with a net flow in one direction.

        2) There is a flow of electricity from where the voltage is higher to where the voltage is lower. There is not a two way flow of electricity, with a net flow in one direction.

        3) There is a flow of air, i.e., wind, from where the pressure is higher to where the pressure is lower. There is not a two way flow of air, with a net flow in one direction.

        4) There is a flow of “heat” from where the IR radiation is higher to where the IR radiation is lower. There is not a two way flow of “heat”, with a net flow in one direction.

        As the difference between the two pressures, voltages, air pressures, IR radiations decrease so does the flow rate of the matter/energy involved. There is never a flow of matter/energy from where the pressures, voltages, air pressures, IR radiations are lower to where they are higher.

        Reply

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          Rosco

          |

          Obviously the values I quoted were from Trenberth et al. I simply use them to illustrate the point that it seems incongruous to assert that having more of the principal radiation to space emitters will result in radiation “trapping”.

          That was the point of the whole article and is supported by all of the satellite data which show a positive anomaly for radiation emitted to space since ~1979.

          As the GHGs are heated by IR radiation I cannot see why the fact that Trenberth et al assert 83% emitted to space comes from GHGs shows any “reducing the rate at which the surface looses heat via IR radiation from its measured 396 W/m^2 of up-going IR radiation down to a mere 40 W/m^2”. The surface does not radiate directly to space and any absorption of its emissions by the atmosphere has zero effect on its radiant emissions.

          The surface still loses the heat via IR – it is just that Trenberth et al claim most is absorbed by the atmosphere. I find that doubtful and not supported by the satellite data anyway.

          For example the image taken from Petty’s book on atmospheric radiation easily found on the web shows the emission observed from space approximates a 290 K emission over a significant part of the monitored spectrum. Given that both water vapour and, especially, CO2 do not emit a continuous spectrum it is inconceivable that emission of the order recorded come from these gases except at higher temperatures and certainly not from higher in the atmosphere.

          I also do not believe one can make the same assertions about algebraic manipulations concerning radiation as one can make over other areas of engineering.

          The only methods whereby the rate of radiant energy loss by the surface can be reduced is by reducing the temperature of the surface or accepting that the radiant energy incident upon the surface from a colder atmosphere transfers heat from cold to hot.

          I used to think that the higher emission from the hot object outweighed the lower emissions incident on it and therefore heating was impossible but slowing the rate was possible but I am not thoroughly convinced. I am not so dumb as to not understand the concept I am not convinced because absorption of lower intensity radiation by the hotter object must equate to transfer of heat from cold to hot if the hot object is prevented from cooling by its higher emissions.

          Besides Pictet demonstrated 2 centuries ago that the radiant emissions from a cold object had only one effect on a thermometer at ambient temperature – the thermometer decreased in temperature despite being in thermal contact with stable ambient air temperatures.

          None of the examples I have seen used to assert that back radiation reduces the rate of “heat” loss take into consideration that none of these examples need to comply with the laws of radiant emission. That includes bank accounts, air or water pressure or electricity.

          In fact, if we accept that Planck provided the only equation that satisfied the emissions of radiant energy observed from the cavity experiments – and he won the Nobel prize for that – and we accept that the other two fundamental laws of radiant emission can be derived from Planck’s law – then any algebraic manipulation using the Stefan-Boltzmann law and black body considerations should produce the same result when that same manipulation is applied to Planck’s law. By that I mean plotting a curve for any temperature, plotting another curve for the original curve added to itself, and plotting a curve for a temperature calculated by sigmaT^4 for the sum of the radiant emission powers DOES NOT produce the same curve.

          I am not disagreeing over obvious fact that all energy flow is from higher to lower but I am saying unequivocally that radiation is not subject to simple algebraic manipulation because Planck’s equation is not amenable to such manipulation.

          Planck’s blackbody equation is continuous over the range of wavelength or all of the other usual variables it is plotted using and therefore it is subject to the laws of calculus.

          Planck’s law can be used to prove that simple algebraic manipulation does not apply as you have asserted. The proof is mathematically sound.

          And why would an equation involving the inverse of the fifth power of wavelength multiplied by the inverse of the function e^(hc/lambda.B.T) simply add up anyway?

          All of the often vitriolic claims I have seen used in insulting manner about how radiation fluxes add up and can be used to calculate temperature fail to give the result when the same algebra is applied to Planck’s law.

          The only one that works is the trivial result that Q(net) = sigmaT1^4 – sigmaT2^4.

          This Q(net) is the area between the 2 Planck curves.

          It is nothing more relevant than a number and cannot be manipulated as if it is a radiant emission – it isn’t and all of the algebra I have seen using this assertion is wrong and I can prove it – I did so in a submission to PSI in 2014 and in the submission recently published.

          So I will respectfully assert that I do not accept there is any evidence in any of what I originally wrote several years ago to support the assertion that the rate of cooling of the surface is reduced by the assertion made by Trenberth et al. The surface is plainly shown emitting 396 W/m2 – obviously they assert that the atmosphere absorbs 356 W/m2 of this.

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            Dense

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            “I used to think that the higher emission from the hot object outweighed the lower emissions incident on it and therefore heating was impossible but slowing the rate was possible but I am not thoroughly convinced. I am not so dumb as to not understand the concept I am not convinced because absorption of lower intensity radiation by the hotter object must equate to transfer of heat from cold to hot if the hot object is prevented from cooling by its higher emissions.”

            How I understand it is that it is about density of populated states at a certain temperature. The hotter mass has all the internal states of the colder mass already populated, and on top of that more higher states populated that the colder body doesn´t have. It would be like a scale of exchange, as when two bodies are at the same temperature there is no transfer of heat, but we know that photons and there energy flow between them, just that there is no net effect. The populated states that are shared by two bodies of different temperatures is a fraction where there is no net transfer. That would be the states populated in the colder body.

            When combining them two, the emission from two bodies and two square meters is combined to one m^2. That would mean dilution of the energy density of the hotter body, and the result would be heat flow into the lower concentration of energy.
            The description at hyperphysics seem to confirm that in simple words:

            http://hyperphysics.phy-astr.gsu.edu/hbase/mod3.html

            That puts my mind at rest.

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