Alarmist Study Fails to Measure CO2 Back Radiation
The paper (image, above) by Wong & Minnett is written by climate alarmists. [1] They are attempting to actually measure warming of the ocean from the back radiation from CO2 molecules. They were unable to.
This appears to be the first attempt by anyone to measure potential warming of the Earth’s surface from back-radiation. There were a number of important comments they made.
Before going into those comments, it has to be noted that Wong & Minnett tried to measure warming from back-radiation from CO2 molecules. They were able to measure back-radiation from the atmosphere and from clouds. They could not determine the source of the back-radiation.
They did not even think about measuring back-radiation from water vapor. They did look at water vapor but only as a potential theoretical feed back from CO2 warming. Despite there being 100 times more water vapor than CO2 and despite it being up to 33 times more radiatively active and despite 85{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} to 95{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of all back-radiation was coming from water vapor, they did not consider it.
So despite the vast majority of back-radiation was coming from water vapor, Wong & Minnett stated they did not know whether the back-radiation was coming from CO2 or the clouds. Their paper was built on an assumption that all back-radiation was coming from CO2. A false assumption.
Wong & Minnett tried to separate the warming of the ocean from direct solar radiation in the visible spectrum, from potential theoretical warming from back-radiation. They found that task not possible.
But Wong & Minnett did determine that the back-radiation of IR EMR only penetrated the oceans to “within the top micrometers of the ocean’s surface”. They were able to determine that whatever effect this was going to have on ocean warming was extremely minimal.
Even if there was a difference between solar radiation and back-radiation temperatures, their instruments were not sensitive enough to detect it.
This paper was 2018. The latest and greatest state of the art thermometers did not have the number of decimal places to be able to detect any warming difference. Let us say they eventually invent such a thermometer. The warming difference, if there ends up being one, will be in micro-degrees. But then 85{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} – 95{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of those micro-degrees are coming from water vapor.
Just 5{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117}-15{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} are coming from CO2 molecules. And just 4{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of those CO2 molecules are from humans. So that will be theoretical 4{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of 5{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of micro-degrees. And that will only be in the micrometer surface skin of the ocean. That will do absolutely nothing to heat the oceans. The oceans are 70{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of the Earth’s surface area. So it does nothing to heat the Earth.
The whole experiment is in vain. Back-radiation is the Earth’s radiation coming back on itself. It is impossible for the Earth to heat itself with it’s own radiation. That contravenes the 2nd Law of Thermodynamics.
[1] The Response of the Ocean Thermal Skin Layer to Variations in Incident Infrared Radiation Elizabeth W. Wong Peter J. Minnett (23 March 2018) https://doi.org/10.1002/2017JC013351 https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2017JC013351
About the author: Brendan Godwin served as a Weather Observations & General Meteorology Radio (EMR & Radar) Technical Officer, Australian Bureau of Meteorology (Retired)
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Jan Sevenhans
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Back radiation is TMin that you can measure at night : https://www.linkedin.com/pulse/blue-greenhouse-climate-cooling-red-summer-peak-sevenhans/?published=t
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Al Shelton
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Ask Wong and Minnet to get a tub and fill it with dry ice [99+% CO2]
Now if the Earth’s CO2 [0.04%] can back radiate IR to warm the Earth then the dry ice at close to 100% should quickly heat my beer when I put the bottle into the tub of dry ice.
Well, Wong and Minnet, does my beer get hot??
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Barry
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I think you can just leave the beer any where around the tub and wait a few seconds for it to boil in the back radiation. Once again can’t reproduce the effect of back radiation but for sure it exists.
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chris
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Water vapor, co2, etc are not mirrors. They don’t send the IR back the way that it came. They do radiate their own IR light like all materials do. If co2 could reflect 100% of the IR that got to it back then we wouldn’t get any from the sun. Lets pretend that it can. If the Earth is sending out 340 Wm^-2 times .04% (.0004) = .136 Wm^-2
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Zoe Phin
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lol.
“Backradiation” is actually geothermal.
http://phzoe.com/2020/04/29/the-irrelevance-of-geothermal-heat-flux/
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geran
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I appreciated zany Zoe’s ongoing comedy routines, but just in case she confuses anyone, here is what is going on:
“Back-radiation” is the term used in climate pseudoscience to describe the flux of photons coming from the sky. Since all matter emits photons, molecules in the sky also emit photons. Some of these photons travel to space, and some travel to Earth. The photons that travel to space represent lost energy, resulting in cooling of the Earth.
The confusion starts when the climate clowns claim that the photons from the sky can heat the surface. That’s their bogus “greenhouse effect”. Their problem is that a cold sky cannot warm a hotter surface. The heat transfer is from the surface to space, as demonstrated by the lapse rate.
Zany Zoe’s geothermal is not a significant player in Earth’s energy budget, because it is spread too thin over the surface. But, her comedy is hilarious. Enjoy!
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Zoe Phin
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Backradiation was invented to explain why the surface is hotter than the sun alone could make it.
The actual reason why the surface is hotter than the sun alone could make it, is explained by geothermal.
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geran
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You’re still wrong, Zoe!
Like several others, you’ve been sucked into believing Sun can’t heat Earth. Someone falling for such nonsense is as hilarious as it is pitiful.
You should be more accepting of reality:
”But, Earth gets more than enough energy from Sun. In fact, Earth sheds about 30{154653b9ea5f83bbbf00f55de12e21cba2da5b4b158a426ee0e27ae0c1b44117} of solar energy, so as to not get too warm. Earth and Sun are doing fine. It’s the pseudoscience that is pathetic.”
https://principia-scientific.com/the-sun-is-not-an-ice-cube/
Now, more of your zany pseudoscience, please.
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Zoe Phin
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“you’ve been sucked into believing Sun can’t heat Earth.”
I never claimed anything that stupid.
The sun adds ~165 W/m^2 on top of kinetic energy base provided by geothermal.
geran
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Well I did ask for more of your zany pseudoscience, and you sure provided!
Thanks.
You take the bogus “165” (from IPCC energy budget nonsense) and you add it (fluxes don’t add) to your bogus “340” geothermal (that doesn’t exist).
Nonsense that can’t be added, is added to nonsense!
Perfect Zany Zoe hilarious pseudoscience.
Zoe Phin
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Geran,
How does a commercial flir camera manage to detect weak UV leaks?
The device is at room temeprature and then somehow detects UV and the color changes to show the heat up.
According to you this can’t happen.
Planck’s law states that you can add fluxes (intensity × spectral width) of 8,9,10,11,12,… micron bands. But as soon as someone suggests you can also add the 0-1 micron band, you flip out.
Your ignorance is so incredibly annoying!
geran
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Zany Zoe, a FLIR with “UV leak detection” capability has NOTHING to do with the fact that fluxes do not add.
The “UV leak detection” refers to adding a fluorescent dye to a system to find a leak.
You combine your lack of understanding, with your arrogance, for maximum hilarity. And, you’re doing a great job.
More please.
PS Your lame insults just indicate that you realize your pseudoscience has failed. (I learned that indicator from Norman.)
Zoe Phin
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Geran,
How does a thermal sensor detect that UV leak? Remember, according to you, nothing should be detected by a thermal sensor, because you can’t add UV a sensor with its own kinetic energy.
Why you so stupid?
Zoe Phin
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Geran,
Did you think I meant using a black light to find a UV source? lol.
I mean something like this:
https://youtu.be/_A2kakGc4gw
geran
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Zoe, you can’t understand your own link!
The FLIR mentioned has a selector switch….
Maybe you should stick with baking cookies….
Zoe Phin
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Oh it has a selector switch to turn off the sensor’s room temperature (kinetic energy) so that it can detect UV? lol
You didn’t even try refuting Planck’s law.
geran
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Of course I’m not refuting Planck’s Law, zany Zoe. I’m refuting your zany interpretation of it.
Fluxes don’t add. You can’t bake cookies with ice cubes, no matter how many ice cubes you have.
More zaniness, please.
jerry krause
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Hi Geran,
During winter at over the Arctic Ocean and the surrounding continents, atmospheric soundings show that the temperature of the atmospheric layer,in contact with the surface, ‘normally’ increase with increasing altitude. Hence, according to thermodynamics the downwelling IR from the sky should be able to warm the surface.
It seems most people ignore the observed facts of the common diurnal temperature oscillations during a 24hr cloudless period as compared with the common diurnal temperature oscillation during a 24hr overcast period.
How about this observed actual data???
Have a good day, Jerry
Have a good day, Jerry
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geran
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Hi Jerry,
The lapse rate also applies to the Arctic. You may have chanced upon some temperature inversion. I would have to see the data to know.
Remember, lapse rate is -dT/dz. So as z increases, T decreases. But lapse rate is just a general value. Local conditions can alter that value quite rapidly.
Have a great day.
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Herb Rose
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There are two ways an object loses energy,: by collisions (the fast way) or by radiation (the slow way). The more molecules there are, the more collision occur and the faster the energy loss. The fewer the number of molecules the slower the energy loss. On a clear night (anywhere) there are fewer molecules in the atmosphere than on an overcast night. This would mean that if the surface of the Earth is heating the atmosphere the Earth would cool slower on a clear night than on a cloudy night where there are more molecules to conduct energy. Overcast nights are warmer than clear nights because during the day the uv light from the sun is heating the N2 and O2 in the atmosphere, NOT the surface of the Earth. When there is water in the atmosphere it is better able to conduct energy from the atmosphere to the surface of the Earth than when those water molecules are absent.
Herb
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tom0mason
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Herb Rose,
When a molecule looses energy by collision or re-radiation does it necessarily mean that this or that is heated?
Surely the transfer of energy and subsequent re-radiation, that while moving down the EM spectrum, may actually fall below the IR band? However, also there is the transfer of energy that while moving down the EM spectrum and originating much above the IR band, may actually fall inside the IR band?
My point is that radiant energy when translated through interactions with matter do not always give heat.
Or maybe our florescent bulbs don’t really work as many consider they work (UV to visible light).
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Herb Rose
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Hi Tom,
When molecules collide energies equalize. The internal energy of a molecule (vibration of atoms across a bond) converts to kinetic energy of the molecule.
What wavelength an object absorbs depends on its electric fields. Atoms will absorb x-rays, O2 and N2 absorb uv. The larger molecules can absorb wavelengths over multiple bonds then radiate the energy in even longer wavelengths This is why even though longer wavelengths have less energy they travel further in space (hence ir telescopes in satellites above the atmosphere) than shorter wavelengths which are absorbed. Matter converts shorter wavelengths into longer wavelengths.
In the florescent bulb high frequency electric waves are absorbed by mercury vapor which radiates uv light which is absorbed by the white coating inside the bulb and converted to visible light. So while matter is able to re-radiate energy in various ways the shorter wavelengths are absorbed by other molecules with the right bond length and converted into longer wavelengths.
Herb
jerry krause
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Hi Geran,
Try this link (http://weather.uwyo.edu/upperair/sounding.html) You seem totally unaware of it .
Have a good day, Jerry
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geran
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That’s a great link, Jerry. Thanks.
Except for a few weather events, you can clearly see the lapse rate–temperatures decrease with altitude.
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jerry krause
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Hi Geran,
You have just demonstrated that you practice ‘minute science’. I practice ‘Einstein Science’: “It’s not that I’m so smart it’s just that I stay with problems longer.” (A. Einstein)
And clearly you are a ‘cherry picker’.
Have a good day, Jerry
Have a good day, Jerry
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geran
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Jerry, that’s really impressive that you practice “Einstein Science”.
I didn’t realize that “Einstein Science” required insulting people.
We live to learn.
Have a great day.
(Since you love duplication.)
Have a great day.
Joseph Olson
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The first published experiment on phantom “back radiation” was by Dr Nasif Nahle, posted at PSI in 2011, under “Publications”
“Earth’s Missing Geothermal Flux” at FauxScienceSlayer(.)com > sets base surface temperature
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Brian James
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Apr 29, 2020 UFOs, Comet ATLAS, 1998 OR2, Magnon Repulsion | S0 News
https://youtu.be/IQx2E6deMuE
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Jonas
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If IR is absorbed in the top micrometers, then all emission should also come from those micrometers ?
I do not get it. Suppose you hav a pond with water (at 20C). The top micrometers emitt about 400 kW/m2. Why does not that layer freeze to ice immedeatly ??
Or why does it not boil immedeatly when exposed to IR radiation ?
Convection ? If convection can work on a fast time scale it should be part of the absorption/emission process.
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J Cuttance
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‘Or why does it not boil immediately when exposed to IR radiation ?’
Exactly! What the hell is this 250W/m2 downwelling IR doing? If it penetrates a few microns then why aren’t the oceans steaming away all night?
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jerry krause
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Hi Jonas and J,
You have only considered one mechanism of energy transfer by which a surface of condensed matter can be warmed or cooled. But there is another: the thermal conduction between the atmosphere and the surface and thermal conduction between the condensed matter beneath the surface layer and the surface layer. So while the focus is upon a very thin surface layer there is the fact that far more matter is involved than that of the surface layer via simple thermal conduction which only requires a temperature difference (gradient)..
Have a good day, Jerry
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Henrik Rasmus Andersen
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Actually Wong and Tennett did measure the back radiation and included the spectra as figure 3 in the mentioned paper. They were not the first to do this as this was achieved already about 80 years ago. The discussion section of the paper makes clear which part of the IR spectra is dominated by water so the claim by Godwin that they assume it is all CO2 is fantasy.
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JDHuffman
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Henrik, Fig. 3 indicates the back-radiation, not the “warming of the ocean from the back radiation from CO2 molecules.” So either they are innocently confused, or purposely trying to confuse the reader.
Whatever the case, their Fig. 3 results are incorrect. The graph indicates the same for both “with clouds” and “without clouds”, at 15μ, which is clearly wrong for low clouds.
So Godwin’s concluding sentence stands “The whole experiment is in vain. Back-radiation is the Earth’s radiation coming back on itself. It is impossible for the Earth to heat itself with it’s own radiation. That contravenes the 2nd Law of Thermodynamics.”
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jerry krause
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Hi JD,
You and I simply measure the downing welling IR from the atmosphere with our IR thermometers. These IR thermometers do not distinguish from where these IR photons originate But you have measured temperatures of neg 40F and I of neg 30F when the atmosphere appeared, during the time of my measurements, to be exceptionally blue (clear, cloudless).
R. C. Sutcliffe, my favorite meteorologist wrote (Weather and Climate, 1966): “The natural atmosphere, however clean it may appear to be, is always supplied with a sufficient number of minute particles of salts, acids or substances.”
I question: Why are not these minute particles of condensed matter (liquid or solid) emitting radiation according to the temperature of the supporting atmosphere?
Have a good day, Jerry
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JDHuffman
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Hi Jerry. I would offer that the minute particles ARE emitting, since everything above absolute zero emits.
Latest readings:
Directly overhead, blue sky surrounded by puffy clouds ==> 6.5 ºF
Ground (shade) ==> 79.7 ºF
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jerry krause
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Hi JD,
Since it seems we (you and I) agree that all condensed matter (liquid or solid) emits if its temperature is above zero Kelvin, I want to sidestep the issue of whether gas atoms or molecules can emit while I consider something else my favorite meteorologist, R. C. Sutcliffe, wrote.
“Long-wave radiation from the earth, the invisible heat rays, is by contrast totally absorb by quite a thin layer of clouds and , by the same token, the clouds themselves emit heat continuously according to their temperatures, almost as though they were black bodies.”,
I have to admit that Sutcliffe has written something which cannot stand reasonable analysis. For if the earth emits according to its temperature, say 288K, and the thin, high, cold cloud totally absorbs the radiation that the earth emits and then emits according to its temperature, say 260K. There must be a serious energy balance problem.
However, it seems it would be possible to get a plane capable of flying above this thin cloud top with one of our IR thermometer pointing downward to measure the temperature of the cloud top. And then drop down into the thin cloud and measure its temperature with a conventional thermometer. And then drop below the thin clouds and measure the temperature of the earth with the IR thermometers. And then, while flying below the cloud and point the IR thermometer and measure that the temperature of the cloud bottom.
And I expect (predict) that the upper two measured temperatures would be near 260K and the bottom two measured temperatures would be near 288K.
For instead of the bottom of the thin cloud totally absorbing the upward radiation from the earth,, the bottom of the cloud is scattering the most of the earth’s radiation back toward the surface earth.
This is not a thought experiment; it is an actual experiment which could be done now that we have the modern inexpensive IR thermometer. JD, what do you think?
Have a good day, Jerry
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Herb Rose
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Hi Jerry,
Every object absorbs radiated energy and every object above absolute zero emits radiated energy.
Objects will equalize their energy with the energy field they are in.
Energy decreases with distance from its source and as a result of energy being absorbed by objects.
At an altitude of 100 km the concentration of matter is .00005 kg/m^3. This matter will equalize with the energy field. The energy field will decreases as it descends further in the atmosphere from both the distance from it source (the sun) increasing (small change) and because more molecules are absorbing energy converting it to kinetic energy.
At sea level the concentration of molecules in the atmosphere is 1.1 kg/m^3. This matter will equalize with the reduced energy field.
The kinetic energy of the gas molecules at sea level will be less than 1/22,000 the kinetic energy of the molecules at an altitude of 100km. The kinetic energy of the molecules in the atmosphere increasing with altitude as the strength of the energy field increases because of less energy being lost to matter.
Have a good day,
Herb
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jerry krause
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Hi Herb,
You wrote: “Energy decreases with distance from its source.” Here I though there was a observed scientific law termed the Conservation of Energy. Guess I must be wrong as usual.
Have a good day, Jerry
JDHuffman
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Hi Jerry.
Earth’s surface at 288 K would be emitting about 390 W/m^2, allowing for perfect black bodies. If the cloud were emitting at 260 K, that would be about 259 W/m^2, again allowing for perfection. So since the thin cloud has 2 surfaces, top and bottom only, being very thin, it would be receiving 390(A) Joules/sec and emitting 259(2A) Joules/sec. It appears that the cloud would be losing energy faster than it is gaining energy, but remember it is also receiving energy from the atmosphere (and from Sun during daytime). So, I have no problem with Sutcliffe’s statement.
Your proposed experiment is realistic, and would likely further verify Sutcliffe.
This morning, Sun is not fully above horizon, heavy cloud cover, no clear sky, rain predicted:
Directly overhead ==> 36.0 ºF
Ground ==> 65.1 ºF
Even the “warm” clouds would be unable to heat the surface.
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jerry krause
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Hi JD,
You wrote: “So since the thin cloud has 2 surfaces, top and bottom only, being very thin, it would be receiving 390(A) Joules/sec and emitting 259(2A) Joules/sec.” Then you wrote: “It appears that the cloud would be losing energy faster than it is gaining energy.” Not according to the words you had just wrote. I had to assume that because the clouds had two surfaces (at the temperature of the top surface it was emitting 2 X 259 joules/sec. Which I now see what I just wrote is what you wrote (2A).
So we (you and I) agree there is still an energy balance problem. But I need to ask you: Do you believe that the bottom temperature of the cloud bottom is actually 36.0 ºF? For we expect there is an atmospheric temperature gradient where the atmosphere’s temperature decreases with increasing altitude even if there is no vertical convection actually occurring at the time you measured its bottom temperature. Certainly the ground surface is being warmed by conduction from the soil beneath the surface and the bottom of the cloud is being cooled by conduction by the colder atmosphere above it (due to the temperature gradient of the atmosphere).
So to know what the temperature of cloud bottom is, we need to fly the plane up into the ‘bottom layer’ of the cloud and directly measure the atmosphere’s temperature with a common thermometer.
I admit I have trouble explaining (understanding) the difference between the two temperatures you have measured but do not question their validity. For I can remember measuring similar temperatures. But I have to ask: What were the conventionally measured air temperature temperatures during the previous several hours if the cloud deck was present earlier? For my experience is that there is very little cooling at the earth surface if there is such a cloud deck during the nighttime.
We must accurately define the system we are studying. And maybe we haven’t. But we also must make actual observations (measurements) to have a clue as to what there is to actually define. Which hasn’t yet been done.
Have a good day, Jerry
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jerry krause
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Hi JD,
Directly overhead ==> 37ºF
Ground ==> 54 ºF
Light rain at 6.36am
Confirming valid measurements even if we (at least I) do not totally understand (can explain) them yet.
Have a good day, Jerry
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jerry krause
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Hi JD,
Forgot, one degree cooling during past 5 hours. This is critically important.
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jerry krause
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Hi JD,
Most important is that we have an airport about 3 miles from our home from which atmospheric soundings are launched at 4am and cloud bases are automatically measured from the ground on the hour. So the measured temperature of the cloud base a little after 4am was about 32 ºF and the air temperature was 50 ºF.
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