How The IPCC’s Greenhouse Definition Violates Physical Laws

Written by Dr. Antero Ollila

earth sun horizon

The “greenhouse (GH) effect” is the IPCC’s basic concept in global warming. The anthropogenic (man-made) global warming theory is based on the enhanced GH effect caused by GH gases.

There have been many comments on net pages that how carbon dioxide having only 400 ppm concentration can lead to so much warming as proposed by the IPCC.

This story is a piece of simple evidence that it cannot.

The basic information about the energy fluxes affecting the GH effect can be found in Figure 1. The radiation fluxes are practically the same as used in the energy balance presentation of the IPCC.

Figure 1. Energy fluxes contributing to the greenhouse effect in all-sky conditions watts per square meter (Wm-2). Click to enlarge.

The Earth receives net energy of 240 Wm-2 as a difference between the incoming insolation and the reflected shortwave radiation at the top of the atmosphere.

Based on observations, the Earth’s surface absorbs 165 Wm-2 and the missing portion 240-165 = 75 Wm-2 has been absorbed by the atmosphere.

Satellite observations confirm that the Earth radiates 240 Wm-2 longwave radiation into space.

This is also according to physical laws that the Earth must radiate the same amount into space what it receives because otherwise, the Earth would continuously cool down or warm-up.

The magnitude of the GH effect is simple to calculate, but it is a little bit theoretical. It is impossible to estimate accurately what would be the Earth’s surface temperature without the atmosphere and without the GH effect.

It has been generally agreed that theoretical temperature corresponding to the net radiation flux of 240 Wm-2 received from the Sun is the surface temperature without the GH effect.

This temperature can be determined by the radiation law introduced by Max Planck in the year 1901.

According to this law, a black surface having a temperature of -18°C radiates 240 Wm-2.

The average surface temperature is +15°C, and its difference to -18°C is 33°C. This 33°C has been generally accepted to be the magnitude of the GH effect. The GH effect definition is needed to explain what mechanism is resulting in this warming effect.

I think that only a few readers have ever read the GH definition of the IPCC.  This time the IPCC does not refer to any published scientific sources, but it has formulated the definition by itself since the First Assessment Report 1990.

It is easy to think that it is an old definition and it must be correct without any doubts. I show that you should check it because there is a serious flaw.

The definition of the GH effect, according to Assessment Report 5 (2013) is:

“The longwave radiation (LWR, also referred to as infrared radiation) emitted from the Earth’s surface is largely absorbed by certain atmospheric constituents – (greenhouse gases and clouds) – which themselves emit LWR into all directions. The downward directed component of this LWR adds heat to the lower layers of the atmosphere and to the Earth’s surface (greenhouse effect).”

By simple terms, this definition means that GH gases and clouds absorb a certain amount of energy from the infrared radiation emitted by the Earth’s surface and thereafter GH gases and clouds radiate it back to the surface.

So far so good – it is according to radiation and physical laws.

The obvious reason for the GH effect seems to be the downward infrared radiation from the atmosphere to the surface and its magnitude is 345 Wm-2. Therefore, the surface absorbs totally 165 + 345 = 510 Wm-2.

It does not matter whether the surface absorbs shortwave radiation from the Sun or longwave radiation from the atmosphere – the total energy is decisive.

The difference between the radiation to the surface and the net solar radiation is 510 – 240 = 270 Wm-2.

The real GH warming effect is right here: it is 270 Wm-2 because it is the extra energy warming the Earth’s surface in addition to the net solar energy.

The final step is that we must find out what is the mechanism creating this infrared radiation from the atmosphere.

According to the IPCC’s definition, the GH effect is caused by the GH gases and clouds which absorb infrared radiation of 155 Wm-2 emitted by the surface and which they further radiate to the surface.

This same figure has been applied by the research group of Gavin Schmidt calculating the contributions of GH gases and clouds.

As we can see there is a problem – and a very big problem – in the IPCC’s GH effect definition: the absorbed energy of 155 Wm-2 cannot radiate to the surface 345 Wm-2 or even 270 Wm-2.

According to the energy conversation law, energy cannot be created from the void. According to the same law, energy does not disappear, but it can change its form.

From figure 1 it is easy to name the two other energy sources which are needed for causing the GH effect namely latent heating 91 Wm-2 and sensible heating 24 Wm-2, which make 270 Wm-2 with the longwave absorption of 155 Wm-2.

When the solar radiation absorption of 75 Wm-2 by the atmosphere will be added to these three GH effect sources, the sum is 345 Wm2.

Everything matches without the violation of physics.  No energy disappears or appears from the void. Coincidence? Not so.

Here is the point: the IPCC’s definition means that the LW absorption of 155 Wm-2 could create radiation of 270 Wm-2 which is impossible.

This violation of the law of physics is a trick on how the warming effects of GH gases can be practically duplicated as I will show.

Somebody may think that it does not matter if the total driving force of the GH effect is 155 Wm-2 or 270 Wm-2. It makes a big difference.

Or somebody may think that the IPCC may define the GH effect as they like. Yes, they can do so, as anybody can, but in the real world, any correct definition must obey the laws of nature.

The warming impact of carbon dioxide in the GH effect is calculated by dividing the carbon dioxide absorption of 20.1 Wm-2 by the total absorption, which is 155 Wm-2 per IPCC.

Using these numbers, the research group of Schmidt has calculated the contribution percentage of carbon dioxide to be 19% and another research group has calculated 26%.

This latter number is still the most referred figure even though it has been calculated applying a wrong atmospheric composition containing 50% less atmospheric water than the global average atmosphere, thus increasing – practically doubling –  the contributions of other GH gases like carbon dioxide.

By applying the right GH effect number of 270 Wm-2, the contribution of carbon dioxide in the GH effect is only 7.4% corresponding to 2.5°C.

And what is more: this low contribution of carbon dioxide in the GH effect results in the conclusion that in the present atmosphere the warming effect of carbon dioxide model of the IPCC cannot be fitted in the overall carbon dioxide contribution of 2.5°C but it must be reduced by about 70%.

In other words, the climate sensitivity of 0.6°C can be fitted into the overall GH effect but the 1.8°C climate sensitivity of the IPCC cannot be fitted in. The climate sensitivity is the warming caused by the carbon dioxide increase from 280 ppm to 560 ppm.

It turns out that carbon dioxide is not a strong GH gas. A reader must now conclude if this description is scientifically correct meaning that the IPCC science has a massive error in the GH effect definition questioning the scientific basis of the anthropogenic warming theory.

And by the way, Wikipedia uses different numbers in explaining the IPCC’s GH effect covering the breaking of physical laws. Maybe this is the only case when a Wikipedia description is not according to the official science of the IPCC.


Dr. Antero Ollila is an Adj. Ass. Prof. (Emeritus) of Aalto University.


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Comments (98)

  • Avatar

    Richard Wakefield

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    This entire exercise assumes a non-rotating flat disk planet. Downward energy from the atmosphere of 345W/m2 cannot be more than what we get from the sun. That is impossible,. The surface emission of 395W/m2 is more than what comes from the sun. That is impossible.

    The cycle of the atmospheric energy downward contributing to the surface emission is a perpetual motion machine. That is impossible.

    This entire premise is false. IR from the air to the surface cannot warm anything. It is not a heat source. This is akin to saying a slab of dry ice over a hot cup of coffee will increase the coffee’s temperature because the dry ice re-emits the IR from the coffee back to the coffee.

    The sun inputs energy to the planet. Any bouncing or re-emitting of that energy does NOTHING to heat anything as that has already happened. Every bouncing or re-emitting loses energy at every encounter.

    Recalculate this process with a rotating disk where the input of energy from the sun side, loss of energy on the dark side. During the day side far more energy is lost to space than this diagram shows. During the night, that loss drops off, until the sun reheats again.

    It is also incorrect that the earth is warmer than the moon. Daytime surface temperature of the moon is 100C. This shows that the rotation of the earth means that the sun side of the earth cannot reach the highest temperature it could. It’s like moving your hand over a flame. Move it fast, and you dont get burned. Move it too slow, and you get burned.

    A 48 hour rotating earth would get much hotter during the day, and much colder on the night side, producing an unlivable planet. A 12 hour rotating planet would not get enough sun during the day making a snow ball earth. Both of those cases would have the same NET energy from the sun, but totally different planets.

    Reply

    • Avatar

      Richard Wakefield

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      “Recalculate this process with a rotating disk where the input of energy”

      Should be

      “Recalculate this process with a rotating sphere where the input of energy”

      Reply

    • Avatar

      Jerry Krause

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      Hi Richard,

      Amen!

      Have a good day, Jerry

      Reply

  • Avatar

    Shawn Marshall

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    What an utterly sensible post – thank you. Even non-technical people should easily understand your points.

    Reply

    • Avatar

      Alan

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      But not the nutty Professor Catriona McKinnon featured in another article or any of the child and adult followers of Greta, and apparently most of the media.

      Reply

  • Avatar

    geran

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    I find this article rather deceitful. The writer rambles on and on, claiming there is something wrong with the IPCC pseudoscience, but then accepting every part of it!

    Just the first 3 examples:

    “The Earth receives net energy of 240 Wm-2 as a difference between the incoming insolation and the reflected shortwave radiation at the top of the atmosphere.”

    Completely wrong! The “240” figure results from averaging solar flux density. You can NOT average flux density!

    “Based on observations, the Earth’s surface absorbs 165 Wm-2 and the missing portion 240-165 = 75 Wm-2 has been absorbed by the atmosphere.”

    Completely wrong! The “165” figure results from averaging solar flux density. You can NOT average flux density! Again, the writer is just regurgitating the IPCC pseudoscience.

    “Satellite observations confirm that the Earth radiates 240 Wm-2 longwave radiation into space.”

    Maybe the writer could link to some of these “satellite observations”?

    The comedy continues.

    Reply

    • Avatar

      Zoe Phin

      |

      Good article.

      Notice there’s a loop between surface and atmo.

      The surface needs that 345, while atmo needs 395. This is a classic chicken-and-egg problem, or lifting yourself by your bootstraps.

      The problem is easily resolved using real science:

      https://phzoe.wordpress.com/2020/02/25/deducing-geothermal/

      From your diagram, we can deduce that geothermal provides the surface: ~510 – ~165 = ~345.

      Isn’t it obvious?

      https://www.routledgehandbooks.com/assets/9781315371436/graphics/fig3_3.jpg

      ^ geophysicists guessed 0C at the surface, but it’s more like 4-5C

      Reply

    • Avatar

      Zoe Phin

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      LOL.
      Even Postma now agrees the sun is not enough, that’s why he needs backgravity heating via N&Z, but N&Z have their causality reversed.

      Reply

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        CD Marshall

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        The Sun is more than enough, Zoe. 960 W/m^2, you do the math. Everything on Earth works to defuse the full potential of sunlight.

        Even at the Equator, potential temperature of full sunlight is reduced.

        Reply

      • Avatar

        Robert Kernodle

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        Even Postma now agrees the sun is not enough, that’s why he needs backgravity heating via N&Z, but N&Z have their causality reversed.

        That’s a pretty big misrepresentation, as I see it, Zoe.

        Reply

    • Avatar

      chris

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      Hi Geran. I think that the point was to point out that if someone were to use only the info as stated in the IPCC pseudoscience that the pseudoscience doesn’t even stand up to its own scrutiny. If it can’t stand up against itself then it has to be abandoned.

      Reply

      • Avatar

        geran

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        Chris, what I’ve learned is there are people who willingly seek to distort the Skeptic message. Whether through incompetence or dishonesty, the effect is the same — Keep the nonsense going. Someone that claims the pseudoscience is wrong, but then accepts every part of it, is not helping to squash the nonsense.

        Reply

        • Avatar

          Zoe Phin

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          I know, right?
          Even though Gay-Lussac’s Law specifically says that Temperature determines Pressure, some skeptics have reversed it.

          Turn off Solar and Geothermal, all the gases in the atmosphere become solids and fall to the ground. Pressure is still exerted, but no atmosphere.

          Some skeptics would rather hold onto pressure raising temperature, which is not true, rather than stop their geothermal denial – which they share in common with alarmists.

          Reply

          • Avatar

            Herb Rose

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            How long did it take for you to learn that Zoe? Do you still believe that the temperature is higher at lower altitudes because gravity has added kinetic energy to the gas molecules?

          • Avatar

            Alan

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            Have you never inflated a bicycle tyre with a hand pump? The pump gets warm and this is because of the gas law. The temperature of a gas increases when the pressure increases. Since you are so enthusiastic about geothermal you should also know about temperatures in deep mines. They are warmer because of geothermal energy, but they cannot be cooled by pumping cooler air from the surface. This is because the higher air pressure in a deep mine raises the temperature of the surface air when it is pumped into the mine. The gas law works, including by gravity compressing the atmosphere.

          • Avatar

            Herb Rose

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            Hi Alan,
            The bicycle pump heats up from friction. The increasing pressure is from the increase in the number of molecules (n) not an increase in kinetic energy (t) (PV =nrt).
            If I have a container that is divided in two by a closed ball valve and one half of the chamber has gas molecules while the other is empty the side contains the gas molecules will have a temperature T. If I open the valve the gas will expand into the whole chamber but nothing has been done to change the kinetic energy of the molecules. Because of the increased volume the molecules will strike the thermometer less often and transfer less kinetic energy to it. The temperature will drop not from a change in kinetic energy but a reduction in mass (n) transferring kinetic energy.
            Herb

          • Avatar

            CD Marshall

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            Actually deep mines are slightly cooled with great effort and cost:

            [Environmental Engineering of South African Mines, 1989, pp 403-404].

            The increase in dry bulb temperature of air due to auto-compression is 9.66C/km, if there is no change to the humidity of the air.

            To keep those super-temperatures from becoming deadly, an ice-slurry mixed with salt is pumped down from the surface; huge fans then blow air over the ice, forming a controlled cold-air system within the mine—its own internal weather system. The above-ground ice-making plant goes through 6,000 tons of ice a day. Ultimately, this means that many tunnels can be kept at an almost bearable 85 degrees

          • Avatar

            Zoe Phin

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            Alan,
            Who’s pumping extra air from some place into our atmosphere?

          • Avatar

            Richard Wakefield

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            Alan: “The bicycle pump heats up from friction. The increasing pressure is from the increase in the number of molecules (n) not an increase in kinetic energy (t) (PV =nrt).”

            It heats up because the molecules are closer together hitting each other more often, which is EXACTLY what temperature is.

          • Avatar

            Herb Rose

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            Hi Richard,
            People keep telling me that temperature is the mean kinetic energy of the molecules being measured. The number of molecules is irrelevant. The molecules in a 100 C oven have the same kinetic energies the molecules in 100C water. 0 C liquid water molecules have the same kinetic energy as 0 C ice molecules. !00 C water molecules have the same kinetic energy as 100 C steam molecules.
            Herb

        • Avatar

          Chris

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          I understand your point. But when a person holds onto a false belief they will even in the face of a good opposing points. So in order to break them of their belief, their belief must first be shown wrong. Then they will be willing to learn something new. I recognized the fallacy immediately, but someone without any science training can be easily fooled.

          Reply

  • Avatar

    Barry

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    I am uneducated but the earth is not a flat disc it is a rotating sphere that gets its warmth from the sun that is thankfully moderated by our atmosphere. This whole article seems to be about trying to make the IPCC model work according to physics. If you want to understand how the climate operates you first have to base a model on reality,the earth is not a closed system.

    Reply

  • Avatar

    Tom O

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    Just a thought, but when I use the word “average,” I tend to think of it as just what it says – the average value of a collection of values. If I have a rotating sphere and take the readings from all possible points, I actually can use the term average and pretend that it covers all possible readings on that surface, thus treat it as a flat disk, if that is my choice. Now if I am looking at the energy influx as an instantaneous value on only one point, and not the average of those instantaneous values, then I will agree that you have to consider the entire rotating surface. So if the influx value IS an average, what’s the problem?

    Reply

    • Avatar

      Richard Wakefield

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      They only way to know if the average is a real representation of the over all energy flow, one would have to do an analysis of each sqr mtr of surface during the rotational day and throughout the seasons. I think you will dint the average is nonsense and doesnt represent what is going on. If you want to study what the planet’s system is doing, you dont make a model that isnt what the planet is, a flat non rotating disk.

      Reply

      • Avatar

        Richard Wakefield

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        Must be tired today, typing badly.

        I think you will FIND the average…

        My point about different rotation rates having the same average shows I’m right about this.

        Reply

    • Avatar

      geran

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      Flux density is NOT a normal scalar. It cannot be added, subtracted, averaged, etc. People try to treat flux density as energy, but the two are not the same. Energy is a scalar. Energy is conserved. Flux density is NOT conserved.

      A good analogy is “temperature”. Temperature is not a normal scalar. A liter of water at 25º, poured into a larger container with another liter of water at 25º does NOT result in 50º water. The mixture is 2 liters of water, because mass adds. Mass is a scalar. The energy adds, because energy is a scaler. But temperature does NOT simply add.

      Reply

      • Avatar

        Brett Keane

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        True, Flux is a Vector Force. Either it can all move in a certain direction because it has more Power, or it cannot. Because of the operation of Quantum Oscillators (Einstein 1917, from Maxwell 1865 and Poisson’s Ideal Gas Laws etc.).

        Reply

    • Avatar

      Alan

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      Averages are never used in science and engineering. How can a bridge be designed by using the average load? How could a jet engine be designed using an average thrust for the journey? How can a thermodynamic calculation be done using an average temperature when it is temperature differences that determine heat transfer? Those who understand thermodynamics know that temperatures in a system can change without any extra energy being added. The Gulf Stream is a good example. It is energy flowing in ocean currents and it raises the temperature along the west coast of the UK, but this is not additional energy, it is heat that is flowing from a location where the energy was added. The El Nino effect is said to produce warmer periods, but it is not new energy added to the earth’s system when it happens. It is about heat flows. In winter locations near the sea are warmer because of the energy retained in the sea.

      Reply

      • Avatar

        Jerry Krause

        |

        Hi Alan,

        Thank you for illustrating how simple things can be if we only ‘actually’ observe a little bit. That is actually what Galileo taught in his famous book that so few of today’s scientists have likely ever read.

        I am going to illustrate a little more. Some know the story of how Archimedes (287?-212 B.C.) was in his bathtub and shouted: Eureka! We have learned that he did this because he knew he had ‘just discovered’ the principle of buoyancy. But is this what he first discovered?

        As I pondered this situation I concluded it may have only been he had seen how to measure the volume of an irregularly shaped body relative to the volume of water that would spill out of the tub if the tub had been filled to the brim with water. For he could weigh the weight of the crown in air. And he could weigh the weight of the water which spilled (if he was able to devise a method of collecting this spilled water). So if he divided the weight of the irregularly shaped body by the weight of the spilled water he would have calculated the density of the body relative to the density of water. We would say, given common words we use today, that he would have measured and calculated the specific gravity of the crown.

        The next step to the principle of buoyancy was he could have place chunk of wood which floated on water to see how much water it actually spilled out of the bathtub to find that its specific gravity was less than than that of water.

        Can you see it was far easier to actually do these experiments than try to reason something without having measured the weight of the body and the weight of the spilled water? Can you see how difficult it would have been to reason geometry without ever drawn a circle with some instrument and then actually have done the experiment of ‘walking’ the instrument about the circle just drawn with it?

        Have a good day, Jerry

        Reply

      • Avatar

        Richard Wakefield

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        EXACTLY!!! Which is why using averages by the AGW High Priests is completely bogus.

        A simple analogy to this is take a picture of a mountain range, with a lake in front, and partly clouded blue sky. Add up the pixels, then divide by the number of pixels to get an average. Apply that average to every pixel. What do you get? A completely gray image with zero detail. No way to know what the actual photo was.

        Reply

      • Avatar

        Jerry Krause

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        Hi Richard,

        Your analogy provides me the opportunity to give you a puzzle. It begins with a story (O Americano, Outra Vez!) about which Richard Feynman wrote (“Surely You’re Joking, Mr. Feynman!”). I do not give the introduction he gave to keep this brief.

        “I said, “Look at the bay outside, through the polaroid. Now turn the polaroid.” “oh, it’s polarized!” they said.”

        I have seen many reflections of solar radiation, mountains, clouds from a smooth water surface. But, if I was wearing polaroid sun glasses, I never took them off and rotated it to see if the reflected image would disappear at some angle.

        And until I had read Feynman’s story multiple times did I see the probable consequence of what the students saw.. Which is that 50% of incident visible solar radiation is reflected from the water surface and the other 50% passes though the surface into the colorless liquid water beneath the surface.

        Now, I go back to his introduction. “Then I said, “Have you ever heard of Brewster’s Angle!” “Yes sir! Brewster’s Angle is the angle at which light reflected from a medium with an index of refraction is completely polarized.” ”

        Until I read Feynman’s story, I had never heard of Brewster’s Angle and I do not know what I would see through a single lens of my polaroid glasses. But I have long considered that the oceans albedo cannot measured from space for a different reason. Now I question what is the relationship of the water’s Brewster Angle to a water’s surface reflection of solar radiation including the IR portion of solar radiation which at some wavelengths is opaque (not transmitted through water).

        Any comments?

        Have a good day, Jerry .

        Reply

  • Avatar

    Barry

    |

    Averages are a very misleading term, as they only pertain to an outcome that can be used in specific ways. For an example if I am using heat to bend a half inch steel bar that is 10′ long I will want to get about one inch of the bar red hot or about 900degrees F . It is a useless calculation to say that if I get the bar to an average temp of 900 Divided by 120″ or 7.5 degrees that I will now be able to bend the bar. If how ever I’m trying to determine how much heat I have used per cubic inch of that particular bar it may be useful. Averages are really only good for certain operations and do not apply to much in the real world.

    Reply

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      Richard Wakefield

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      Excellent, and quite correct.

      Reply

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      Jerry Krause

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      Hi Barry,

      One cannot avoid averaging varying variables that can rapidly change. NOAA’s SURFRAD project records and reports radiations and basic meteorological factors like air temperature every minute. Which allows one to see how rapidly some of these variables can change but contributes little to a general understanding what has occurred during the 60 minutes of an hour. To repeat myself, the 60 times greater number of specific numbers do little to produce a greater understanding. The other NOAA project, USCRN, calculated the hourly mean of the previous hour plus the maximum and minimum values (actual measurements) measured during the previous hour. An analysis of these three numbers, given some experience, can really show one what was occurring during that previous hour. Like how rapidly the temperature can naturally increase, or decrease, during certain hours of the day.

      Just some information which I find useful to know about. Just learning about the Arctic Research Expedition opened a window of the vast data of which I was totally unaware. Of course, given my attempt to possibly explain how prehistoric glaciers could have been formed. I should have searched out the critical data which already existed; but I had not.

      Maybe now I can compose Part 5 of this series and not embarrass myself by not being aware of what I should have known.

      Have a good day, Jerry

      Reply

      • Avatar

        Barry

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        Point taken Jerry that is one of those averages that is useful . My point was that you need to know what you are trying to accomplish with the average nearly before you decide to do an average as they sometimes give misleading results
        Thanks Barry

        Reply

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    Herb Rose

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    Hi Dr. Ollila,
    What makes you think that the short wave radiation from the sun is being reflected? What causes the O2 molecules in the upper atmosphere to spit into oxygen atoms? If the uv light from the sun spits O2 in the stratosphere, creating the ozone layer, why wouldn’t O2 molecules throughout the atmosphere absorb uv? What do you believe is creating the ionosphere reflected energy from the Earth’s surface?
    At an altitude of around 100 km the concentration of N2 begins to decrease and the concentration of oxygen atoms increase. This is from the absorption of uv by O2. N2 also absorbs uv but because it takes 940 kjoules/mole to split the N2 triple bond compared to 490 kjoules/mole to split O2, that energy is being converted into kinetic energy of the molecule.The ionosphere is a result of atoms absorbing energy from the x-rays being emitted by the sun.
    The short wave radiation (which has more energy than the visible spectrum) is not being reflected by the atmosphere but absorbed and converted to kinetic energy.
    The problem with all the models is they ignore the majority of solar energy being absorbed by the N2 and O2 in the atmosphere and only count the solar energy reaching the troposphere.
    Herb

    Reply

    • Avatar

      Antero Ollila

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      A direct question: What makes you think that the shortwave radiation from the sun is being reflected? I trust on direct observations by satellites orbiting the Earth. They show consistent results on a daily and monthly basis. There were comments that there is no such thing as an average of fluxes. Of course, there is. Fluxes do not deviate from other physical observations in this sense. NASA calculates average flux values all the time and the results are convincing.
      Contrarian scientists will not win this battle against the IPCC using strange principles of physics. Ordinary but correctly applied physics is enough.

      Reply

      • Avatar

        Herb Rose

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        Hi Antero,
        I am curious about what you think is creating the ionosphere and the elemental atoms in the upper atmosphere if shortwave radiation is being reflected. N2 and O2 do not absorb visible or infrared light so what causes them to split into atoms? Some uv light makes it to the surface why isn’t it being reflected? The ozone layer is a result of uv splitting O2 and the atoms combining to form O3. Why wouldn’t the uv be absorbed by O2 throughout the atmosphere?
        Since the x-ray wavelength emitted by the sun does not make to the surface of the Earth do you really believe that the gases in the atmosphere can reflect x-rays? X-rays are absorb by atoms creating ions that form the ionosphere. How do satellites distinguish between short wave radiation being emitted as atoms absorb the x-rays and re-emit radiation from reflected radiation?
        Why is the thermosphere so “hot” if energy is not being absorbed and converted to kinetic energy? All objects absorb radiated energy depending on bond length. How can people believe that O2 and N2 are not absorbing radiated energy just because they don’t absorb visible or ir light?
        Herb

        Reply

      • Avatar

        Jerry Krause

        |

        Hi Herb,

        You asked a very important and fundamental question. “Why wouldn’t the uv be absorbed by O2 throughout the atmosphere?” But you didn’t answer it for Antero.

        The simple answer is that the number of photons capable of dissociating the oxygen molecule is finite and gets consumed in passing down through the atmosphere where the oxygen atom density is ever increasing. So there are optimum ranges of elevation, at which these photons are consumed by this reaction of photons with oxygen molecules, which depend upon the latitude and the season.

        Have a good day, Jerry

        Reply

        • Avatar

          Herb Rose

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          Hi Jerry,
          The entire atmosphere is equivalent to a layer of water 10 meters deep. Wherever in the atmosphere uv light intercepts a O2 or a N2 molecule it will be absorbed.
          Did you look at the composition of the atmosphere with altitude? At just under 100 km the percentage of N2 decreases and the percent of oxygen increases until nitrogen disappears and the atmosphere is made of oxygen and helium. The reason oxygen replaces nitrogen is because it becomes less dense as it absorbs uv. This is a result of it having two bonds (which require 450 kjoules/mole to break) while N2 has a triple bond holding the atoms together and requires 940 kjoules/mole to break. The last bond requiring the most energy.
          Have a good day,
          Herb

          Reply

          • Avatar

            James McGinn

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            Herb:
            The reason oxygen replaces nitrogen is because it becomes less dense as it absorbs uv. This is a result of it having two bonds (which require 450 kjoules/mole to break) while N2 has a triple bond holding the atoms together and requires 940 kjoules/mole to break. The last bond requiring the most energy.

            James:
            Interesting stuff. Makes sense. Thanks for this.

            James McGinn
            Solving Tornadoes: Woke Meteorology
            https://anchor.fm/james-mcginn

  • Avatar

    Chris

    |

    Geran, I understand your point. But when a person holds onto a false belief they will even in the face of a good opposing points. So in order to break them of their belief, their belief must first be shown wrong. Then they will be willing to learn something new. I recognized the fallacy immediately, but someone without any science training can be easily fooled.

    Reply

  • Avatar

    lafargue

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    I think that the atmosphere cannot be considered to be homogeneous.
    On one side, there is air, trace gases, on the other, clouds and solid aerosols!
    Gases which are 500 to 1250 times less dense do not have the same properties as liquids or solids, water or ice from clouds and aerosols.
    That the upper clouds at -31 ° C (242 ° K) intercept the radiation of the ground at 390W / m2, it is physically possible because at 242 ° K they can radiate 195W / m2 on each side. So they transmit 50% to space and return 50% to the ground and do not accumulate energy. It is the only thermally stable situation that can warm the climate.

    A cloudless sky cannot intercept anything otherwise the air temperature would rise very quickly (see the Sahara).
    As for lower clouds, their temperature at the bottom at 2000 m is around 0°C, 273°K and therefore they radiate 315W / m2, towards the ground transmitting 75W / m2 upwards while receiving 390W / m2 of floor. In this case, the top of the low clouds is subjected to space cold and therefore tends to cool strongly which cools the air by convection.
    With a flux of 75W / m2 we can expect temperatures from -82 ° C to 4 or 5000 m on the top of the lower clouds
    In summary, with an overcast sky at 65% clouds (70% high and 30% low)
    ground radiation through the atmosphere would be
    0.35 * 390 + 0.3 * 0.65 * 75 + 0.7 * 0.65 * 195 or 240W / m2
    what else ?

    Reply

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    Barry

    |

    Point taken Jerry that is one of those averages that is useful . My point was that you need to know what you are trying to accomplish with the average nearly before you decide to do an average as they sometimes give misleading results
    Thanks Barry

    Reply

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    Barry

    |

    Point taken Jerry that is one of those averages that is useful . My point was that you need to know what you are trying to accomplish with the average nearly before you decide to do an average as they sometimes give misleading results
    Thanks Barry

    Reply

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    Jonas

    |

    Planck did not derive any radiation law.Planck derived the equilibrium field inside a closed box.
    Boltzmann did it. His analysis is based on classical thermodynamics.

    My understanding is that he showed that there will be a radiation pressure in a warm body, This pressure is proportional to T^4.
    The radiation pressure is the driver for emission (my thinking). Emission is proportional to the difference in radiation pressure (just like a gas). I think this conceptual model makes it much easier to understand thermal radition.

    E.g. if there is incoming radiation – sun shining on earth – that radiation will increase the radiation pressure outside the body, why it will radiate less.
    If one just apply T^4 emission independent of external radiation – the result will be wrong. You overestimate the radiation losses, since the external field will reduce radiation losses.
    No need for Green House effects – just avoid estimating emission as if there are no external field.

    Reply

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      JDHuffman

      |

      The Stefan-Boltzmann Law states that the emission from a body is proportional to the fourth power of the absolute temperature. Stefan developed it empirically and Boltzmann proved it mathematically. “Radiation pressure” is a physical force, unrelated to the S/B Law. An object emits based on its own temperature, not the temperature of its environment.

      Reply

      • Avatar

        Jonas

        |

        The radiation pressure is proportional to the 4:th power of the temperature. That is what Boltzmann showed.
        Emission proportional to pressure will give a T4 law. I think it is very logical that the radiation pressure in a body is the driver for emission.

        Reply

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          JDHuffman

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          Jonas, it may be the concept of “radiation pressure” that is confusing you. Radiation pressure is an insignificant effect. Check your source that says “what Boltzmann showed”. You are being misled.

          Reply

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            Jonas

            |

            Hi JD,
            I do not think that I am misled. Please explain your view. What do you think Boltzmann showed?

            By the way, radiation has a momentum. You need a force to generate momentum. Temperature does not have momentum.

            Yes, radiation pressure is low, but the flow resistance can also be very low.

          • Avatar

            JDHuffman

            |

            The S/B Law has NOTHING to do with “radiation pressure”, as I explained. So if you believe it does, you have been misled. Where is your source that misled you? What is your source that makes you claim: ”The radiation pressure is proportional to the 4th power of the temperature. That is what Boltzmann showed.”

            The only momentum photons have is due to relativistic considerations, since a photon has zero rest mass. You don’t need relativistic considerations to measure temperature. Someone is misleading you.

          • Avatar

            Jonas

            |

            E.g /
            Knizhnik, Kalman. “Derivation of the Stefan–Boltzmann Law” (PDF). Johns Hopkins University – Department of Physics & Astronomy

            You can easily find the derivation if you search the net for some minutes. Boltzmanns derivation (based on classical thermodynamics) shows the the energy density and the radiation pressure is proportional to the 4:th power of temperature

          • Avatar

            JDHuffman

            |

            So you can’t provide the exact link as to how you got misled?

            That might tell you the problem….

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    Jonas

    |

    Additional comment – in a box (room) the walls will radiate. That will give rise to an radiation field in the box. Equilibrium will be reached when the radiation pressure in the box equals the radiation pressure in the walls. At that stage there will be no radiation from the walls (no gradient). If you apply the “always emitting T^4 idea- there will be a mess of fluxes going back and forth in the box.

    The radiation pressure in the walls is proportional to T^4, why the equilibrium field in the room also will be proportional to wall temperature ^4

    Reply

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    Kevin Doyle

    |

    Dr. Antero Ollila,
    The illustration you posted and your article are complete nonsense, and no reflection of reality.
    It is embarrassing that anyone would attempt logical reasoning and math with completely false mathematical assertions.
    Incoming energy from the sun upon half of the Earth is balanced by energy emitted from the entire globe. Why is this even subject to debate?
    The real question is how can cold CO2 in the atmosphere ‘warm’ the surface of the Earth? Thus far, no one has been able to demonstrate this in a laboratory.
    Perhaps, you should refrain from theoretical masturbation, and actually do some real science?
    I find it interesting the people who design and build nuclear reactors still haven’t discovered this miracle gas CO2. Yet, theorists have it all figured out.
    KD

    Reply

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    Antero Ollila

    |

    I am sorry that I became aware of my post on this webpage so late because this post was published in the first place on Climate Change Dispatch pages. A general observation about the comments is that the most active commentators are those who want to represent their own physics because these comments are the only opportunity for them. They could never get through their ideas on any webpages as a story.

    Just a few general pieces of evidence.

    The energy balance of the Earth is based on the observations. The incoming insolation and the reflected SW radiation have been continuously measured by satellites and they show the average flux values as shown in my figure. The only counter-argument that I have received is that they cannot be measured and there are no instruments capable of doing it. This means that these people believe in conspiracy theories that the USA government is using a huge amount of money for nonsense. Probably a man was never in the moon…

    Also, the emitted LW flux by the Earth’s surface have been measured – and surprise, surprise – its value is in line with its temperature according to Planck’s law. This time I did not notices a single comment that the atmosphere cannot reradiate LW radiation on the surface, because “heat cannot be transferred from a colder object to a hotter object”. Also, this radiation has been measured – it is not fabricated.

    Here I explain what happens when the atmosphere emits LW radiation according to the second law of thermodynamics:

    The radiation is energy transferred radially away from the bodies, and its magnitude is given by the Stephan-Boltzmann law. The radiance (watts per square meter per steradian) is proportional to the fourth power of the temperature. Thus, a colder body will radiate to the hot body much less than the hot body to the cold. As the hot body will be cooling by its radiation, the rate of cooling will be less than it would be if there were no cold radiator close to it. On the other hand, as the incoming radiation from the hot body is much larger than the loss of the cold body’s black body radiation, the cold body will get heated. A cold body emits photons as well as the hot body but less. When a photon from the cold body hits the hot body, it will be absorbed, and a part of its energy will increase the temperature of the hot body: energy cannot be destroyed but it can change its form.

    A colder body has no idea that a hotter body would come into its vicinity. Therefore, it emits radiation despite a hotter body.

    I am sorry but I could not notice a single comment based on real physics, which would question my presentation. If I missed these comments, please response.

    Reply

    • Avatar

      geran

      |

      Antero, the measured values do NOT “show the average flux values as shown” in your figure. The incoming solar has been divided by 4. They take the actual measurements, then divide it by 4. That makes your subsequent conclusions WRONG.

      But, it gets worse.

      You state: ”When a photon from the cold body hits the hot body, it will be absorbed, and a part of its energy will increase the temperature of the hot body…”

      That is clearly a violation of the Second Law of Thermodynamics. A cold body can NOT increase the temperature of a hot body. And just to casually state that the photons from the colder body “will be absorbed” means that not only don’t you understand thermodynamics, but you don’t understand quantum physics either.

      And I notice you did not include any links to support your claim: “Satellite observations confirm that the Earth radiates 240 Wm-2 longwave radiation into space.”

      You seem to be lacking in both a knowledge of the relevant physics and the actual data to support your words. But, you appear very well versed in the IPCC nonsense. We see this a lot.

      As to your claim that you did “not notice a single comment based on real physics”, obviously you overlooked several, including mine:

      https://principia-scientific.org/how-the-ipccs-greenhouse-definition-violates-physical-laws/#comment-32810

      https://principia-scientific.org/how-the-ipccs-greenhouse-definition-violates-physical-laws/#comment-32841

      Reply

      • Avatar

        Antero Ollila

        |

        This is a blog story and not a scientific publication. If you really want to find all the details they are available here:https://doi.org/10.9734/psij/2019/v23i230149

        ​I did not show that the incoming insolation flux must be divided by to get 340 and that the reflected flux of 100 must be subtracted to get 240 W/m2 because they are not in focus when we talk about the GH effect. The story is long enough already and I have to make choices to keep it concise.

        Reply

        • Avatar

          geran

          |

          Antero, if you rely on pseudoscience, making excuses doesn’t help.

          You must stop relying on pseudoscience. That’s what helps.

          Reply

  • Avatar

    Antero Ollila

    |

    Greenhouse effect. There are a few people who try to show that there is no greenhouse effect needed. A simple fact is that the net energy received from the Sun is only 240 W/m2 corresponding to temperature -18 C degrees. Therefore, there must be a phenomenon we call a GH effect. SW radiation of 240 W/m2 cannot maintain the temperature of + 15 C degrees. It is just simple like that. Otherwise, we can forget the laws of physics. The direct observations show that a surface gets much more radiation: 165 SW radiation and 345 LW radiation.

    Figure 1 shows to recycling energy flows:
    1) The Earth receives net energy of 240 W/m2 and the same amount is recycling back into space. If it would not happen, the Earth would cool down or warm-up.
    2) Another recycling of energy happens between the atmosphere and the surface. The atmosphere receives net energy from four sources: SW radiation 75, LW absorption by GH gases and clouds 155, latent heating 91 and sensible heating 24; totally 345. If this much of energy flows into the atmosphere, it must get rid of it. Otherwise, it would warm up. This energy is in a kind of trap in the atmosphere. Only one heat transfer process is possible in this case, and it is radiation. The atmosphere radiates LW radiation to the surface 345 W/m2 and this way also this energy recycles continuously between the atmosphere and the surface and we call this phenomenon a GH effect.

    Reply

    • Avatar

      geran

      |

      Antero, there is no “greenhouse effect”! The “-18 ºC” is based on the calculation for a black body. It has no application to Earth. Earth has oceans and an atmosphere. An imaginary black body doesn’t. GHE believers live in a fantasy world of make-believe science.

      Your figures are confused and invalid. You claim that the surface gets 165 W/m^2 (SW) plus 345 W/m^2 (LW). You claim that those different fluxes add, for a constant 510 W/m^2. If that were true, the equilibrium temperature of Earth’s surface would be 35 ºC (95 ºF). Obviously you can’t even figure that out.

      Learn some physics.

      Reply

      • Avatar

        Antero Ollila

        |

        I think that you do not understand what the energy balance means. In my figure, the Earth’s surface is in the balance which means that the incoming energy 165 SW radiation and 345 LW radiation make totally 510 W/m2. This means the energy fluxes leaving the surface must be totally the same as they are: 395 W/m2 LW radiation according to Planck’s law and two other energy fluxes namely latent heating 91 and sensible heating 24; totally 510 W/m2. I think that you make a very simple error and it is that there are other energy fluxes leaving the surface than the LW radiation flux. I think that you have not too much understanding of physics, and if I were you, I would shut up my mouth and make some courses of physics.

        Reply

        • Avatar

          geran

          |

          Antero, I showed you why the 510 is WRONG. But, you can’t accept reality. You’re too far gone into your false religion. You’re opposed to learning.

          We’ve seen it all before.

          Reply

  • Avatar

    Antero Ollila

    |

    It seems like some commentators did not notice that my story does not confirm the GH effect definition of the IPCC. The final result of my research study is that climate sensitivity TCS of the IPCC cannot be fitted into the real GH effect. TCS value of the IPCC is 1.8 C degrees but my value is only insignificant 0.6 C degrees.

    Water is the most important GH gas if we mean the absorption of infrared radiation by GH gases and clouds. Its portion is about 80 % and the same of CO2 is about 13 %. But if we calculate the contributions of the GH effect according to the correct GH effect definition the portions are as follows:
    – water 33.6 %
    – latent heating 33.6 %
    – clouds 13.3 %
    – sensible heating 8,9 %
    – carbon dioxide 7.4 %
    – ozone 2,6%
    – methane & nitrogen oxide 0,7 %.
    GH effect is a very important factor in the climate and CO2 has only a minor role in that phenomenon. Water controls in many ways. Oceans control the dynamics of climatic changes because of its long-residence time. Water has also the most important factor in the GH effect: Absorption 33.6 + latent heating 33.6 + clouds 13.3 = 80.5 %.

    Reply

    • Avatar

      geran

      |

      You want to believe in the GHE, while trying to claim it’s not as bad as the Alarmists claim. You’re trying to live in the middle because you don’t know the relevant physics. People like you are called “Lukewarmers”.

      Your claim of a TCS of 0.6 ºC is as bogus as the 1.8 ºC claim. The TCS of CO2 would be negative. Add more CO2 to the atmosphere, everything else being the same, and the temperature drops slightly.

      Again, learn some physics.

      Reply

  • Avatar

    Antero Ollila

    |

    I do care about which names you use for classifying people. My opinions are based on scientific studies and not on the physics of your own.

    Reply

    • Avatar

      geran

      |

      Antero, I’m glad you admitted that you are relying on your “opinions”. But sadly, “opinions” are NOT science. And layers upon layers of pseudoscience papers are NOT “scientific studies”.

      If one of your “scientific studies” violates any law of physics, it is NOT scientific. That means any future “scientific study” that relies on a discredited “study”, is also discredited. That’s why IPCC/AGW/GHE/CO2 nonsense is pseudoscience, it is all based on major violations of the laws of physics.

      Reply

      • Avatar

        Antero Ollila

        |

        I did not say that I am relying on my opinions. I said that my opinions on climate change issues are based on scientific studies. You try to twist what I say.

        Reply

        • Avatar

          geran

          |

          Antero, do you rely on your opinions, or do you question them? It appears as if you believe in your opinions. “Beliefs” and “opinions” are NOT science.

          And, again, your opinions are based on pseudoscience.

          I stand by my words.

          Reply

  • Avatar

    Antero Ollila

    |

    Your comment that increasing the CO2 content decreases the surface temperature means that you are living totally on your own planet, with your own physics. It looks like you have no understanding of what energy balance means as I showed you earlier. It is usually hopeless to make any reasonable conversation.

    But try me. What is is your energy balance for the Earth. How much energy Earth receives to the atmosphere and to the surface and what are the energy fluxes leaving the surface and the atmosphere.

    Reply

    • Avatar

      geran

      |

      Antero, you do NOT have an “energy balance”. You have a false, incorrect, inaccurate attempt to balance fluxes. But, fluxes do NOT balance!

      Take your imaginary blackbody sphere, absorbing 960 W/m^2. At equilibrium, the sphere is emitting 240 W/m^2. Do you believe that 960 = 240?

      Flux does not “balance”. Flux is NOT conserved. Energy is conserved. Energy balances.

      Learn some physics.

      Reply

      • Avatar

        Antero Ollila

        |

        It looks like that you have no understanding what energy balance means. Hopeless.

        Reply

        • Avatar

          geran

          |

          Another clown trick, Antero?

          Reply

  • Avatar

    Antero Ollila

    |

    I try to continue the conversation.

    As I wrote before, this statement that “heat cannot be transferred from a colder object to a hotter object” you do not understand. You did not answer what is the situation:
    – Is a colder body still emitting photons when a hotter body is somewhere nearby?
    – What happens to a photon hitting the hotter body from a colder body?
    – What is the physical law which says that above a certain temperature a hot body will not absorb a photon?

    Can you give precise answers to these questions?

    Reply

    • Avatar

      geran

      |

      Yes
      It is reflected

      A hot body can absorb photons from a hotter body. But, not from a colder body. Photon absorption is based on wavelength compatibility. A photon that has a wavelength too long for the surface gets reflected.

      Precise enough for you?

      Reply

      • Avatar

        Antero Ollila

        |

        Not even close to. It looks like that you mix the absorption by the black surface to the absorption of a photon by GH gases. Try again.

        Reply

        • Avatar

          geran

          |

          No Antero, I answered your questions. Trying to misrepresent me is a trick used by climate clowns.

          Are you just another climate clown?

          Reply

  • Avatar

    JDHuffman

    |

    Even if a person does not have a background in physics, they can quickly learn that the “greenhouse effect” is pseudoscience. Purchase an inexpensive IR thermometer. A clear sky, directly overhead, typically has a temperature -40 ºC, or lower. For someone familiar with temperatures, it should be clear that -40 ºC is not going to “heat the planet”.

    The only time sky temperatures get closer to surface temperatures is when there are clouds. Cloud temperatures are affected by the lapse rate, so low clouds are warmer than high clouds. But, clouds are never warmer than the surface. Currently at my location we are expecting rain, so the sky is covered with low clouds. Directly overhead is very warm, 34.4 ºF (1.3 ºC), but the ground was even warmer, 50.5 ºF (10.3 ºC).

    The sky isn’t warming the surface. The surface is warming the sky.

    Reply

    • Avatar

      Jerry Krause

      |

      Hi JD,

      Almost missed your comment. “Purchase an inexpensive IR thermometer. A clear sky, directly overhead, typically has a temperature -40 ºC, or lower.”

      Your advise to “Purchase an inexpensive IR thermometer” and to use it is right on. But from my now experiences I have to question “A clear sky, directly overhead, typically has a temperature -40 ºC, or lower.” For either my IR thermometer is inferior to yours or the temperature of a cloudless sky varies from one location, or season, to another.

      And based upon my experiences of making measurements of different variables with various instruments, I believe my less than $30 US IR thermometer is making measurements with a precision that far exceeds the manufacture’s specifications.

      If Herb would purchase such an IR thermometer which many people beside him do not believe can so simply measure temperature., and compare its ‘reported’ temperatures with those devices which measure meteorological air temperatures and water temperatures etc. Herb might become a believer there is this thing called temperature and these other doubter might begin to believe the IR thermometer is a devices which too actually measures temperatures with a precision better than Galileo’s thermometer.

      Have a good day, Jerry

      Reply

      • Avatar

        Herb Rose

        |

        Hi Jerry,
        I believe in temperature I just don’t believe a thermometer gives an accurate measurement of kinetic energy in a gas. At sea level a cubic meter of air has 1.1 kg of molecules. A Cubic meter of water has 1000 kg of water. If the temperature of the water and the air is the same it means that for every gram of air transferring energy there are 1000 grams of water transfer energy I will go outside on a 50 F day and jog for 45 minutes. You go swimming or exercise in 50 F water. You will die from hypothermia I will not.
        Have a good day,
        Herb

        Reply

      • Avatar

        Jerry Krause

        |

        Hi Herb,

        “In the English Channel the water temperature can vary from 59°F (15°C) at the end of June, increasing to 64/65°F (18°C) by the beginning of September. It is extremely important that you are accustomed to swimming in temperatures of 59°F and below comfortably before contemplating a channel swim.”

        I understand that 59°F is not 50°F but I doubt anyone who actually swims the Channel does it in 45 minutes. I run about 32 minutes, in the same clothing as I run in a 79°F temperature except I wear gloves, when the grass beside sidewalk on which I run has frost on it. Do you not know that a heat engine (the body produces several times the waste heat energy than it uses to produce useful work. So if your body is capable of doing long term vigorous physical activity if seems one does not need to worry of hypothermia when swimming 45 minutes in 50°F water.

        Herb, you just ‘state’ so many things which are not true if you would test them or research what others have done.

        Have a good day, Jerry

        Reply

        • Avatar

          Jerry Krause

          |

          Hi Herb and also James if he reads this,

          I forgot a very important fact. The heat engine doing useful work must be cooled or it will destroy itself.

          I am so sure that I am willing to be wrong by not looking at the results of the data the atmospheric soundings done near the line of tornadoes which occurred in Tennessee two nights ago. For I am positive if you study the atmospheric sounding data you will find there was a jet stream over the path of these multiple tornadoes. For it is my idea that the jet steams cool the waste heat of the heat engines known as thunderstorms which are short lived if there is nothing to remove the waste heat energy which destroys the short lived thunderstorm. I am sure others have made this observation.but question if it is common knowledge.

          So, yes, I am trying to make it common knowledge. Whatever ‘it’ is.

          Have a good day, Jerry

          Reply

        • Avatar

          Herb Rose

          |

          Hi Jerry,
          People swimming the channel wear wet suits to prevent heat loss. The rate of heat loss in water is so great that your body cannot produce enough heat to compensate for the loss no matter how good a swimmer you are.
          Have a good day,
          Herb

          Reply

        • Avatar

          Jerry Krause

          |

          Hi Herb,

          Yes, I am trying to destroy your credibility as to fundamental scientific knowledge. This so the less informed accept the validity of ‘everything’ you write. For a fact is that you often (even) do write about commonly accepted fundamental scientific ideas. But even these commonly accepted ideas should not be accepted as a ‘proven’ truth as Jim Allison (a Nobel Prize winner) attempted to alert PSI readers in a December 12, 2019 PSI posting.

          But I believe the following is a historical fact: “Captain Matthew Webb (19 January 1848 – 24 July 1883) was the first recorded person to swim the English Channel for sport without the use of artificial aids. In 1875, Webb swam from Dover to Calais in less than 22 hours.”

          It would seem ‘modern’ wet suits would be considered an artificial aid if they had been available in 1875. Research what you claim before you put your foot in your mouth.

          Have a good day, Jerry

          Reply

      • Avatar

        JDHuffman

        |

        Jerry, you may need to “calibrate” your IR thermometer for sky temperatures. Find a dark night where the stars are very clear, that is, no clouds, no haze — very clear. Point your thermometer up and see if it indicates at least -40 ºF. If not, adjust the emissivity until you get at least -40 ºF.

        Reply

        • Avatar

          Jerry Krause

          |

          Hi JD,

          I understand we do not know the emisstivy of the clear sky but I am not going to change the emissivity of my IR thermometer because the other temperatures it measured seem to be very consistent with the temperatures measured more conventionally.

          You and I both know the temperature of the atmosphere is expected to decrease with altitude, Hence it is impossible to claim to measure the temperature at a certain altitude. What we are measuring with the IR thermometer pointed upward is the temperature that a a condensed matter surface would have if there was such a surface, which there is not.

          But a temperature much less than the ambient air temperature near the earth’s surface can only slightly slow the cooling of the nighttime air temperature. Hence, you and I and others (who need to buy an IR thermometer can compare the difference between the earth’s surface temperature (which we can reasonably measure with the IR thermometer to find the rate of the surface’s cooling during the nighttime when there are no apparent cloud relative to the difference between the two temperatures which can be measured with the same instrument.

          And then compare our results to see if there is any consistency.between what is measured at different locations at different times by difference ‘scientists’. For if one begins such an observation project one instantly becomes a scientist.

          Have a good day, Jerry

          Reply

    • Avatar

      Jerry Krause

      |

      Hi JD,

      You began the previous comment: “Even if a person does not have a background in physics, they can quickly learn that the “greenhouse effect” is pseudoscience. Purchase an inexpensive IR thermometer. A clear sky, directly overhead, typically has a temperature -40 ºC, or lower. For someone familiar with temperatures, it should be clear that -40 ºC is not going to “heat the planet”.

      I quote this because it seems you and I are the only ones here at PSI who are interested in actually doing some critical SCIENCE. Critical because the fundamental basis of SCIENCE is observations. Observation that need to be explained or observation that test the validity of ‘theoretical’ scientific ideas. Which theoretical idea to which you refer is that if the atmosphere did not contain carbon dioxide molecules and other molecules with similar properties the atmosphere’s temperatures would be about 33C (58F) lower.

      If this theoretical idea were only a wrong scientific idea and no one was paying it any attention, as is the case with the theory of black holes or why the surface temperature of Venus was what it is observed to be, there would be no problem. But the economy of our world is being changed on the basis of this wrong theoretical idea. And you are absolutely correct that everyone should purchase an inexpensive IR thermometer and observe for themselves the measured temperature of the clear sky directly overhead.

      But you cast doubt upon what I see with my inexpensive IR thermometer (IR-T). Does it really matter if the temperature indicated by my IR-T is -25F instead of -40F when the temperature of the atmosphere in which I am standing is +40F or some other temperature very significantly greater?

      You suggested I modify (adjust) my IR-T so that the temperature it measures is maybe 15F lower. Because you haven’t replied to my comment I am not sure if you realize how serious your suggestion was relative to the people who you suggest need to purchase an IR-T.

      It is too easy to argue (state) that one cannot measure (observe) anything with any device. I have directly observed by reading (seeing) or hearing (listening) that very few scientists these days have read Galileo’s famous book or Newton’s famous book. And I know that I had not read either book until I was about 55 years of age.

      So I did not know, that Newton had written in his preface (as translated by Motte): “Therefore geometry is founded in mechanical practice, and is nothing but that part of universal mechanics which accurately proposes and demonstrates the art of measurement.” I only quote this statement because of the words: “proposes and demonstrates the art of measurement”.

      You only focus on pointing your IR-T directly upward (vertical). I do this but I also point my IR-T down toward the horizon. Where I expect any radiation from the atmosphere is coming from a lower elevation of the atmosphere which I, understanding that since the temperature of the atmosphere is expected to decreases with increasing altitude, reason the temperature of this lower atmosphere should be greater. Hence, when I observe that the temperature of this less ‘vertical’ atmosphere is greater as I had reasoned, I believe my IR-T is maybe accurately measuring to the temperature of the atmosphere from which the IR radiation is coming. And I know I reason this because of have learned the ‘art of measurement’ by my experiences of actually measuring with various devices (instruments).

      Therefore, I hope some readers will purchase an IR-T because they can trust what this device measures. And when there are scattered clouds in a clear sky, I would hope they would point the IR-T at the clear sky and at a scattered cloud. For I know from experience how different the observed temperatures will be. Which is further ‘proof’ that the IR-T is reasonably accurately measuring which is claimed it should be measuring.

      Yes, another fact is that to accurately describe ‘what is’ requires many words. Many words which some might naively consider to be rambling.

      Have a good day, Jerry

      Reply

      • Avatar

        JDHuffman

        |

        Jerry, I only meant that adjusting your device’s emissivity was an option. If you’re happy with it the way it is, that’s no problem. You’re correct, even if it doesn’t read all the way to -70 ºF, it still demonstrates that the atmosphere is not warming the surface.

        This morning, very clear blue sky, bright sun:

        Directly overhead ==> – 57.6 ºF
        Ground (in shade) ==> 44.7 ºF

        Reply

      • Avatar

        Jerry Krause

        |

        Hi JD,

        You report two temperatures that you measure with your IR-T. Why not compare four temperatures: Directly overhead ==> – 57.6 ºF, Ground (in shade) ==> 44.7 ºF, Ground (in direct sun) (?? ºF) and Conventional Air (?? ºF)?

        Why not measure and report the measurement of IR from clouds?

        Advice from Galileo: “Measure what is measurable and make measurable what is not so.”

        Have a good day, Jerry

        Have a good day, Jerry

        Reply

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    Antero Ollila

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    I have lived on the countryside in Finland with a constant threat of frost during the growing season. Everybody knew that if the sky is turning cloudless during the late August night, there is a risk of frost destroying the harvest. Why this? Because the surface radiated into the black sky having a temperature well -200 C degrees. If there were clouds in the sky, there was no risk. What is the reason? It is simply that the cloudy sky radiates much more LW radiation than the clear sky. The surface has the same temperature in the beginning but the radiation by clouds changes the situation. Every material emits radiation above absolute zero. The photons are emitted by the clouds and these photons are absorbed by the surface. Photons mean energy. Energy cannot disappear. If you think otherwise, do you know what is the temperature when a black body stops absorbing photons?

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      geran

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      Well you finally got something right, Antero. Of course it was because I pointed clouds out to you. But, the point you are missing is that clouds are NOT CO2. And, clouds do not raise the temperature of the surface. The CO2 GHE is bogus, clouds are real.

      But photons emitted by a colder surface can NOT warm a hotter surface. You’re still making that mistake.

      Reply

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      Jerry Krause

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      Hi Antero,

      You wrote: “Everybody knew that if the sky is turning cloudless during the late August night, there is a risk of frost destroying the harvest.. .. If there were clouds in the sky, there was no risk.”

      I believe everybody in Finland did know this. And I believe that Svante Arrhenius and everybody in Sweden did know this also. But today I doubt if everybody, who live at high latitudes, know this because majority of the populations of these countries have no crops that could be destroyed by an early frost.

      But I have to ask you, Antero, if you expect that the air temperature at sundown was such that everybody, from experience, knew that if the sky was cloudless that frost was likely. But if the sky was cloudy, they knew from experience that frost was not likely even though the air temperature at sundown on both evenings was the same.

      Now, I’ve set you up. Everybody back in that former time, probably did not know what the temperature of the clouds might be. And I will continue to set you up by asking: Did you or anyone else worry about what the elevation of the clouds might be? I doubt if you or anyone else did.

      But you rave just written: “It is simply that the cloudy sky radiates much more LW radiation than the clear sky. The surface has the same temperature in the beginning but the radiation by clouds changes the situation. Every material emits radiation above absolute zero. The photons are emitted by the clouds and these photons are absorbed by the surface.” You started with a focus upon temperature and then you ignore what the temperature of the matter which emits actually is.

      Do you see your problem and Arrhenius’s problem that is different from your problem? For he totally ignores any possible influence that clouds might have upon the radiation emitted by the earth’s surface and the crops’ surfaces.

      Have a good day, Jerry

      Reply

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      Herb Rose

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      Hi Antero,
      As you are new to this site I will explain something I have been arguing for some time.
      You cannot use the temperature as recorded on a thermometer as an accurate indication of kinetic energy. The atmosphere exists because energy converts oxygen and nitrogen into gases and additional energy cause these gases and atmosphere to expand in volume. According to the universal gas law increasing kinetic energy results in lower density of a gas. Since the atmosphere’s density decreases with increasing altitude, the kinetic energy of the molecules are increasing. (P in the universal gas law is gravity which confines the atmosphere and resists expansion, not atmospheric pressure.) The clouds and all the atmosphere is “hotter” (the molecules have greater kinetic energy) than the surface of the Earth and the reason cloudy nights are warmer is because the liquid water droplets contain stored energy (which gases do not) and radiate that energy to the surface of the Earth. “super cooled water” at -50C is not good at reflecting heat but very good at absorbing heat.
      Herb.

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    Jerry Krause

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    Hi Professor (emeritus) Antero Ollila,

    Finally looked up who you were. Isn’t PSI a good forum to discuss scientific ideas that need correction? But we must Focus, Focus, Focus!!! And not allow ourselves to get distracted.

    You have not replied to my question of March 3, 2020 at 7:01pm. Which is relative to your comment of March 3, 2020 at 5:41pm. People who live at latitudes greater than 45 degrees have an advantage over those who live only a little lower than 45 degrees. And maybe those who live at very near 45 degrees have an advantage over both of the those I have grouped. For I know how dramatically the winters change a little above 45 and a little below. And you had a somewhat advantage over Svante Arrhenius because Finland’s weather and climate is maybe not influenced as much as Sweden’s by the Gulf Stream.

    My effort here (this comment) is to get you involved in a conversation in which you and I do not allow others to distract us from what we know by experiences and pondering..

    And hopefully Rosie Langridge might join us so we can better focus her attentions, and not be distracted by her critics.

    While I wait for your desired response I will focus my attention better upon what you have already written here at PSI.

    Have a good day, Jerry

    Reply

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      Jerry Krause

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      Hi Zoe,

      Any calculation which uses an averaged over a period of more than a hour has no validity for the sphere of the earth clearly does not have a homogeneous surface and therefore is not warmed uniformly by either solar radiation or geothermal energy..

      Have a good day, Jerry.

      Reply

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        Zoe Phin

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        True, and gravity varies, and distribution of night time sky varies, and biology varies, and humam contruction varies, and this and that varies and now we can’t model anything?

        An hour is also arbitrary. Why not take it down to Planck time?

        It’s not that you’re wrong, it just appears as evasion to many.

        Reply

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