Kevin Kilty
I’d say there is no one who writes comments in threads at WUWT who doesn’t believe that radiation provides the beginning and end points of Earth’s climate. Insolation provides energy input to drive the climate, and radiation takes away waste heat at the top of atmosphere (TOA), whatever TOA means for outgoing LWIR (more about this in a moment). Figure 1 shows, using Kelvin’s statement of the Second Law of Thermodynamics, how any heat engine, climate included, must work. It must procure energy from a hot place, whether or not we can call this a reservoir in the sense of thermodynamics, or not. Some of that energy is converted to work. The remainder is discarded into a cold place, with the same statement about whether or not that is a “reservoir”.
Figure 1. An engineered heat engine (Figure 1A) moves heat in the direction it would flow spontaneously but redirects some to do external work. By the first law Work=Qh-Qc, but by the second it can be no greater than Work=Qh(1-Tc/Th) where the temperature factor is Carnot’s theorem. In the atmosphere (Figure 1B) Work is done internal to the system and the dissipation of this work as heat may feed back into the so-called reservoirs so that the Carnot factor no longer really applies. Figure 1C simply shows what occurs in a power plant where flow-work is what runs the turbine.
However, there is endless discussion at WUWT, bordering on argument at times, about what goes on with regard to radiation, particularly the CO2 contribution, in the intermediate air.
The community here at WUWT are largely climate skeptics which means they tend to discount the importance of CO2. Fine. However, whether one seeks to discount or magnify the influence of CO2, one is obliged to use physically relevant arguments to do so. I hope to deep six a number of talking points that occur over and over with as rational an explanation of things as I can muster.
Transmittance measured and calculated
Let me begin with a diagram showing data collected in a laboratory. This is from the NIST website. Figure 2 shows the transmittance of IR radiation through a sample column of CO2 and Nitrogen, measured with an IR spectrometer. Transmittance is defined in terms of power transmitted by a beam of radiation. Thus, the deep troughs indicate energy per unit time that has been redirected from the beam. Where has it gone?
Figure 2. The IR spectrum of CO2. For reference with Figures 3 and 4 the deep troughs are located at 15um (667 inverse cm), 4.26um (2350 inverse cm) and 2.7um (3,700 inverse cm).
First, let’s examine how well a calculating engine, MODTRAN, reproduces this same picture. The pertinent parameters of the path through the sample tube in the spectrometer were 2.6 cm-atm of CO2 (the 200mmHg sample of CO2 was augmented with N2 gas to a total pressure of 600mmHg so that even the pressure broadening is representative). Figure 3 shows how the MODTRAN version at Spectral Sciences, Inc. does with parameters adjusted for 2.6cm-atm CO2.
Figure 3. Source of IR is a blackbody spectrum at 298K which tails-off strongly below 4um wavelength. CO2 at 26ppm and 1 km of pathlength is 2.6cm-atm approximately.
Looks pretty close, does it not? What it says, implicitly, is that while the path length for each plotted wavelength of EM radiation contains the same concentration of CO2, we get hugely different results. Concentration doesn’t matter because the cross-section for absorption varies hugely by wavelength. Claims that CO2 can’t matter much because it is a minor chemical constituent are simply wrong.
In fact, with this version of MODTRAN I can easily check whether or not Beer’s law works. Beer’s law is defined in terms of Absorbance. Absorbance is equal to -Log(Transmittance). At a wavelength of 4.71 um, these three models of atmosphere provide the indicated transmittances:
1700cm-atm H2O, 0ppm CO2, 20km path, T=0.741
0cm-atm H2O, 400ppm CO2, 20km path, T=0.919
1700cm-atm H2O, 400ppm CO2, 20km path, T=0.708
Beer’s Law calculated for this situation is in order 0.741×0.919=0.68
And the resulting transmittance is 0.68. Comparing 0.68 to 0.708 is well within the uncertainty of all things considered. So, MODTRAN looks like a pretty solid calculator when it comes to transmittance. We will examine radiance calculations shortly, but let’s return to the NIST data. Now back to answer the question about where does this energy go?
Where does the redirected energy go?
There are really only two possible responses to the step beyond absorption. The radiant energy is either scattered into other directions other than forward, or it is absorbed by CO2 and stored as molecules with elevated energy. Scattering is nearly zero because of the size discrepancy between the EM radiation and the molecular scale. CO2 strongly absorbs radiation at the wavelengths of those deep transmittance throughs in Figure 2. Now what is the next step in the chain of conversions?
People at WUWT now claim, rightly so, that this absorbed radiation very quickly passes into the kinetic energy of nitrogen in the case of Figure 2, but into Oxygen and Nitrogen in the case of the atmosphere. This is a good thing.
Energy quickly leaves the CO2 molecules into a sea of energy shared among molecular rotations, vibrations, and speed. It is not a one-way evolution of exchanges, because down at the molecular level, what proceeds in one direction goes just as surely in the opposite, maybe at a different rate. So, evolution proceeds toward a distribution of energy by all conversion means until it is a stationary distribution – an equilibrium state. This is the Boltzmann distribution, and the atmosphere arrives there through a detailed balance of energy exchanges. Once we arrive at the Boltzmann distribution it becomes possible to speak of a “temperature” of the ensemble of molecules in near equilibrium with the sea of EM radiation within it.
I never see any reference to Boltzmann distribution, or Maxwell distribution (molecular speeds) or concept of detailed balance, even though these concepts are central to understanding both the concept of temperature (especially LTE) and radiance in the atmosphere.
The background blackbody radiation from the Earth’s surface, along with emitted radiance from nearby atmosphere, simply replaces energy lost from the local atmosphere by various means. This sum total of interactions, and the rates at which they occur, keeps the atmosphere in local thermodynamic equilibrium (LTE).
So, rather than the rapid conversion of absorbed radiation from CO2 to other molecules and other forms of energy preventing back radiation, this conversion makes possible the calculation of radiance of the atmosphere by knowing only its temperature and composition. Some of that radiance goes back to the surface.
Accurate transmission calculations, and radiance calculations from specified chemistry and temperature allow MODTRAN to act as a useful transport modeler.
A couple of other things to discuss
There is often an insistence that molecules in a cold atmosphere can’t radiate to a warmer surface because doing so would violate the Second Law of Thermodynamics. Thus, the greenhouse effect is impossible. Moreover a few people deny that a warmer instrument can measure LWIR originating from a colder place.
This insistence shows a misunderstanding of the concept of “temperature”. Molecules don’t have a temperature; they have only energy. The second law applies to temperature which is only meaningful for a large ensemble of molecules. Energy can travel anywhere, bulk heat can, spontaneously, only travel toward cold from hot, unless one forces its travel the other direction with input of work (refrigerators and heat pumps).
Another insistence by folks who overvalue CO2 is that LWIR cannot travel from the surface directly to space because it gets absorbed and much is returned to Earth. Here is how one set of professional climate scientists phrased it.
“… infrared emission from the surface is mostly absorbed in the atmosphere and cannot radiate directly to space. In turn, the atmosphere radiates both up (to space) and down (to the surface). The surface therefore receives a double whammy of radiation from both the Sun and the atmosphere…”
This represents a belief that cooling of the Earth takes place at a mythical level way up high where the temperature is 255K.
This actual level from which radiation heads to space, the cold “reservoir” that allows the atmospheric heat engine, is exceedingly complex and includes even the Earth’s surface in places and at times. Let’s use MODTRAN once again to show its two end-member clear sky models. Figure 4 shows these.
Transmittance from surface to space averaging between 14% and 32% is not negligible. Yet, there are other instances, not covered by models in MODTRAN, where average transmittance is likely higher. Portions of surface above 2,000 meters elevation, for example, are above the bulk of moist air.
The subtropics, where the furthest reach of the Hadley cell returns to Earth, is composed of air that has been substantially dried by tropical precipitation. It is very transparent to LWIR. It is also an interesting case of where dynamics and radiation meet to cooperate in cooling the Earth. The descending air is kept warm through work performed on it by the surrounding atmospheric pressure, (or by gravity if one insists on that explanation that requires further elaboration), and radiates this work away as heat freely to space.
Even in wavelength ranges where the atmosphere is very opaque (many optical depths from surface to space) to the ballistic passage of IR radiation, there is still a flow of radiant energy that more nearly resembles diffusion, or conductive heat flow, with conductivity proportional to temperature cubed – the Rosseland approximation.
Figure 4A.
Figure 4 A and B. These represent end member clear sky models available in MODTRAN. Note that the Arctic winter atmosphere has ample windows that are highly transmissive and can easily present over 30% average transmittance. Even the moist Tropical atmosphere, though, has 14% average transmittance.
Is CO2 a negligible influence?
This essay began as a project to summarize estimates of climate sensitivity and the ways of arriving at it. However, the number of ways of arriving at that number is amazingly large. The methods have become more sophisticated over time; and more expensive! And definitions expanded the effort further by focusing on either the transient or equilibrium values. And then worrying about minor constituents of the atmosphere expanded it further, and on and on.
Soon I felt like someone auditing a Chicago election – the accounting mattered less than who decided what details went into the accounting. However, the topic of climate sensitivity ties in well, and generates two more comments of note.
A Baseline Value of Climate Sensitivity
The effect of adding a step increase in CO2 from 400 to 800 ppm does the following. It produces a step decrease in outgoing LWIR at the top of the (tropical) atmosphere (TOA) by 3.3W/m2.[1] Obviously, the First Law of Thermodynamics applied to this situation reveals that as output is restricted a little, but insolation remains constant, the amount of stored energy in the surface and atmosphere must rise a small amount too.
Just figuring Stefan-Boltzmann (SB) feedback alone suggests a need to raise surface temperature by 0.55C to restore output at TOA. MODTRAN calculations, though, reveal that balance isn’t restored entirely by this adjustment. Through trials I find that the surface temperature must adjust by 0.76C to restore TOA balance. By this point, though, SB applied to the ground surface shows that it radiates 4.5W/m2 – 1.2W/m2 more than the simple application of SB suggests.
People have commented that this presents a paradox, “Where does the additional radiant energy come from?” The answer is that it comes back to the surface from the atmosphere. There is no paradox. It is simply the reality of dealing with boundary problems when an LWIR active atmosphere occupies the space near a boundary.
It is no different than engineers applying a coating to a surface to make an object behave thermally in a different way. This surface coating for the Earth is its atmosphere.
Now further exploration using MODTRAN is warranted because it is perfectly reasonable to suppose that increasing surface temperature will lead to increased absolute humidity with its attendant enhanced LWIR absorption.
The U of Chicago version of MODTRAN allows one to calculate the effect of a constant relative humidity with increased profile temperature. Modeling as before one finds that the surface temperature has to be increased by 1.21C to restore the original LWIR output at TOA. At this point the surface emitted energy according to SB (with emissivity=0.97) is 7.2 W/m2.
This range of values 0.76 to 1.21 centigrade, I feel, provides a useful baseline of what will happen with 2XCO2, as long as the entire process is radiation bound.
The additional comment I would make takes me back to my original motivation of look at climate sensitivity. It is a rejoinder to the often repeated reference to “Simpson and Brunt from 1938”; that atmospheric dynamics makes prediction of what 2XCO2 will do meaningless.
After Moller in 1963 spooked everyone by concluding that an atmosphere with a variable relative humidity could lead to a surface temperature that was “arbitrary[2], Manabe and Wetherald undertook an effort to model the atmosphere that was sophisticated by 1967 standards.[3] They built a model using a 1-D finite differencing algorithm that included either 9 or 18 atmospheric layers and considered the effects of clouds, CO2, ozone, and water vapor either in constant RH or constant vapor pressure models. Their effort was complicated at this time because in addition to addressing the long-time worry over CO2 warming, there was an additional worry over the SuperSonic Transport (SST) adding water vapor to the stratosphere. What they found, however, was that reasonable distributions of absolute humidity would lead to 1.3 degrees centigrade warming while reasonable assumptions about relative humidity would lead to 2.3 degrees centigrade.
Spencer and Christy [4] (also open access) describe the construction of a 1-D model of vertical heat transport starting with the deep ocean and proceeding through the atmosphere. Their model involves three oceanic layers, three atmospheric layers. Having made this model and verifying it. They then considered estimates of energy imbalance made in various ways (satellite radiometery, Argo floats, borehole measurements, etc) and determined what climate sensitivity values produce a consistent story between energy imbalance, their 1-D model, and observed Earth temperatures since either 1970 or 1850 depending on data source, and also consistent with the two-sigma uncertainties in all quantities.
Their results range from 1.86 to 2.49 degrees centigrade per doubling of CO2. Lewis and Curry (2018)[5] found similar values using similar means. Three dimensional computer climate models produce greater climate sensitivity estimates (up to 5K) still. Yet then using these exact codes within the context of Earth system models produce much less warming (3.3K).[6] It is as though one can’t get a reasonable value of climate sensitivity without including every influence from the biosphere, cryosphere, oceans and atmosphere, because then one misses significant negative feedbacks. But in answer to Simpson and Brunt, despite many attempts to determine climate sensitivity in various ways, nothing seems lower than my baseline from radiation alone.
Notes:
1-The range of values for the decline in outgoing LWIR is 1.8 to 3.5 W/m2 through the full suite of MODTRAN models. Not as great as the Tropical model value, but not insignificant either.
2- F. Möller, On the influence of changes in the CO2 concentration in air on the radiation balance of the Earth’s surface and on the climate, Journal of Geophysical Research, 1 July 1963 https://doi.org/10.1029/JZ068i013p03877
3-Syukuro Manabe and Richard T. Wetherald, Thermal Equilibrium of the Atmosphere with a Given Distribution of Relative Humidity, Journal of the Atmospheric Sciences, Page(s): 241–259, 01 May 1967 DOI: https://doi.org/10.1175/1520-0469
4-Roy W. Spencer and John R. Christy, Effective climate sensitivity distributions from a 1D model of global ocean and land temperature trends, 1970–2021, Theoretical and Applied Climatology (2024) 155:299–308
https://doi.org/10.1007/s00704-023-04634-7
5-Nicholas Lewis and Judith Curry, 2018, The Impact of Recent Forcing and Ocean Heat Uptake Data on Estimates of Climate Sensitivity, Journal of Climate, 6051–6071
DOI: https://doi.org/10.1175/JCLI-D-17-0667.1
6-See for instance at The Geophysical Fluid Dynamic Lab and note the table of models.
For those wishing to pursue some challenging explanation of LTE:
Hermann Harde, Radiation and Heat Transfer in the Atmosphere: A Comprehensive Approach on a Molecular Basis, International Journal of Atmospheric Sciences, 27 October 2013 https://doi.org/10.1155/2013/503727
Please expand abbreviations at first use, particularly those abbreviations may have multiple meanings depending on what readers are used to.
“The community here at WUWT are largely climate skeptics which means they tend to discount the importance of CO2.” Personally I would disagree with this statement as it is written. What I think is that people on here, in the main, look at the relative importance of carbon dioxide compared with other factors and influences within a complex system that is the Earths climate and atmosphere, which is where the disagreement with people like Michael Mann and others arises as they only see carbon dioxide as the main cause of climate change, with the planet Venus being used as the ultimate example of what could happen on earth.
It’s a stretch speaking for any “community”, especially when it is effectively an online diaspora of similarly, yet differently, minded people.
Maybe Charles might like to run a poll with the WUWT community? (Just kidding, Charles).
Having had recent communication with the Met Office it’s safe to say that they are firmly convinced it is CO2 and CO2 alone that is causing all the alleged climate woes they insist on telling us about in colour coded warnings – the weather it has to be said is jolly nice, though.
Now that Burnham is about to take up office even the left have started to wonder, in the absence of any statements, programmes or even interviews, what he is about. Which leads to speculation…
How green is Andy Burnham? Britain’s next PM faces tough climate decisions
“Burnham has been very quiet about the climate [crisis] so far,” says Chris Venables, an environmental campaigner and fellow at the Green Alliance thinktank. “I don’t think [it] is at the forefront of his mind, but that does not mean he will water down this agenda.”
…
while Burnham may be getting away with ignoring the climate for now, the crisis is not ignoring Britain.
…
Rescuing the British economy is top of the prime ministerial to-do list; rescuing the climate, and nature, is inseparable from that, says Craig Bennett
…
His philosophy of “Manchesterism” offers clues. – Guardian
His philosophy of “Manchesterism…
Andy Burnham is leaving Greater Manchester with £1.34billion of outstanding borrowing, the largest debt pile of any combined authority in England, as he prepares to enter Downing Street. – GB News
Buckle up….
Addendum – Forget WhatsApp:
“For some of those who have Andy Burnham’s phone number, a notification:
The open-source encrypted messaging service is a favourite of the privacy-minded.” Some briefing to do Andy?
A word of warning…
Russian government hackers targeting Signal and WhatsApp users, Dutch spies warn
I had also read to “The community here at WUWT are largely climate skeptics” and immediately stopped reading. Because I may not qualify myself as a “climate skeptic”. I’m just a natural scientist by training with solid academic credentials, scientific integrity and common sense. I would not read wannabe scientific essays that begin with labeling audiences.
There are quite loud skeptics commenting at times….not many Luke-warmers make their voice heard. Kevin’s assessment of the audience is likely based on experience with comments, not necessarily readership…
Sir, while I fully appreciate your comment I should like to remind you that the skeptical approach to assumptions is an absolute prerequisite to scientific integrity. This is actually what real scientists are, and this is what allows science to advance. The climate propagandists use the term “skeptics” with derogatory connotation also insulting the scientists per se. Therefore such terminology is unacceptable, at least from my point of view. Thank you.
Venus does not have a greenhouse effect, due to the reasons shown below.
Thank you for the detailed explanation, I knew that it was adiabatic changes that caused the high temperature and not carbon dioxide. Also the planets orbital period is less than its retrograde rotational period.
sorry for the mistake on point 8, it should read 29% is absorbed, not 79%
IIRC, at the altitude where atmospheric pressure is 1 bar, the temperature range is similar to that of Earth, and that’s despite Venus being 24 million miles nearer the Sun. If it were the same distance as Earth, the 1 bar temperature would be cooler if albedo, rotation rate etc stayed the same. Can’t remember where I read that, maybe someone more knowledgeable can confirm or correct.
Yes, that is correct. but that is at a height of 43km, that is where pressure and temperature are similar to Earth. At the surface the pressure in 93 atmosphere.
If Venus was the same distance as Earth, it would be cooler as the energy received by Earth is half that Venus receives, but then Venus would not be the same.
No, when compared at similar atmospheric density, the temperature is not the same, but the only differentiating factor is the distance from the Sun.
I like to believe the WUWT community prefers truth and facts over politics and groupies.
Its interesting that term sceptic which is a mandatory status for science or scientific endeavor has be transformed into a negative status.
That is surely the evidence that its a religion.
Imagine sitting in church and asking for a bit of evidence for stuff being proposed, you would be out and that is the mechanism of our universities and scientific institutions.
OK, regarding Michael Mann you have a fair point. I’ve been on this community for twenty years now and I don’t think it anunfair point that the overwhelming majority of folks work toward limiting the argument about CO2 being a problem. However, some of these arguments are easy to demolish and just serve the efforts of people beyond our community to label us “deniers”. It’s a way of not having to consider any of the reasoned discussion here.
Listen, in the last week I had someone argue with me on this site that energy balance is not a worthwhile consideration — arguing in effect that the First Law is not meaningful regarding climate change. This is an example of what I’d like to temper.
Estimates of 2x CO2 are useful as an intellectual exercise in evaluating its impact but are largely irrelevant to climate because natural variability dwarfs the tiny human forcing. Most of modern warming was due to a cloud reduction which was the main cause of the energy imbalance. The false attribution of CO2 to a climate crisis is the preposterous nonsense WUWT exists to debunk.
Mars has an atmosphere that is 95% CO2 and it’s so cold on Mars it snows dry ice
Mars’ atmosphere is also 95% of the way to the complete vacuum
of outer space…really much more like the moon…
Congratulations, Kevin, that essay seem to cover many things well.
A couple of nits – while I like the limiting case of Rosseland radiation as giving understanding where intuition is hard (reduces to diffusion) it is essentially a gray emissivity concept.You can’t apply it to opaque wavelengths, because while that radiation is rapidly absorbed, the re-radiation is over a range of different wavelengths.
“It is no different than engineers applying a coating to a surface “
It is, because we live under the surface, and want to know what the different temperature will be there.
“whether or not Beer’s law works”
You need to be careful here – it doesn’t theoretically apply. Beer’s law works when radiation that is absorbed stays absorbed. So in aqueous solution, say, absorbed light is not reemitted as light. But that’s not true for thermal IR, which can be reemitted as thermal IR. The right extension of Beer’s Law there is Schwarzschild’s radiative transfer.
But those nits are minor.
But those nits are minor.
I noticed that in response to those minor nits you only got three downvotes (none from me, I prefer to say something, like this).
Could it be that the more nits you manage to pick garners proportionally more downvotes? Needs more research…
But “under the surface” is the issue when engineers apply a coating to make thermal solar panels and tube perform better. Then it is exactly the same.
Good point about Beer’s law, but by permanently absorbed, I think the small perpetual losses from the atmosphere are a pretty reasonable approximation to permanently absorbed. My main point about using Beer’s law was to counter that argument that CO2 can’t influences things that H2O already covers.
MODTRAN uses the transport equation of Schwarzschild as you undoubtedly know.
Ah, since CO2 is NOT a greenhouse gas, discussion of even inclusion of Beer’s Law becomes extraneous.
Nick,
I have thought about the Beer’s comment. Somehow, instinctively I knew to go check Beer’s law where absorbance is small and in the short wavelength region of the spectrum, but didn’t think about it much beyond that. This morning I investigated carefully with a spreadsheet.
Assuming LTE, which MODTRAN does, the emittance for radiance is the product of absorbance and the Planck function. In my example I used a feature in the spectrum where absorbance is small, which limits the emittance issue in the first place; but by focusing on short wavelengths I also limit emittance because the Planck function drops off rapidly below the peak (roughly 10um at 298K). At 4.7um it is only about one-sixth of what it is at 10um.
An interesting exercise this morning…
“absorbed light is not reemitted as light.”
Yes it is – when the excited electronic state returns to ground.
What Nick said was: “Beer’s law works when radiation that is absorbed stays absorbed. So in aqueous solution, say, absorbed light is not reemitted as light. But that’s not true for thermal IR, which can be reemitted as thermal IR”, context matters.
The absorbed electronic energy is not retained.
Beer’s Law works because radiation absorbed from the source is re-radiated isotropically.
I would consider the lack of consideration of the blackbody IR radiation equivalent temperatures of the three CO2 emission/absorption bands, the temperatures of the upper tropical troposphere and the surface, and complete omission of the energy transfer and emissions by the water cycle to be major nits. The author is clearly radiation-focused and not looking at the while picture.
As mentioned above, I am not a skeptic but a realist scientist.
The issue is saturation. Carbon dioxide is effectively saturated at about 300ppm. Thus, the estimated 150ppm increase since the beginning of the Industrial Revolution has had a negligible impact on the average surface temperature of the Earth, and 150ppm more would have even less.
That is what Happer and Wijngaarden concluded, a very small effect from more CO2 from where we are now and the bulk of what it can do, temperature wise, is done within the first 100 ppm. Kevin, are they wrong?
I echo that question.
Please see above.
Completely wrong. “Saturation” of CO2 is a misconception, promoted by Hapless Happer and his Denier cronies.
If so why do we talk about CO2 affect per doubling of concentration?
A meaningless point as CO2 is not a greenhouse gas—nothing is. But, Beer’s law does apply and saturation is real, just not applicable here.
OMG!
CO2 is well proven to absorb IR at specific frequencies, I think you are drinking too much beer.
Most of the CO2 effect is in the first 100 ppm then a rapid drop off thus not a factor in today’s “heat” budget.
Happer and Wijngaarden have authored “peer-reviewed” (hah!) and other scientific papers the credibly argue, with supporting mathematics and empirical data, their position that atmospheric CO2 is now saturated in its ability to absorbed any additional LWIR radiating off Earth’s surface.
No matter what search engine I use, I cannot find any scientific paper authored by a “Warren Beeton” that refutes their claim. Care to cite one—just one will do—that shows they are “completely wrong” (your exact words)?
Just curious if you’re willing to provide your credentials and experience that uniquely make you qualified to to attack Happer, or should we start calling you, Bleatin’ Beeton.
WIll Roy Spencer’s criticism do?
https://www.drroyspencer.com/2024/08/yes-the-greenhouse-effect-is-like-a-real-greenhouse-and-other-odds-and-ends/
“No, the Saturation Effect of Increasing CO2 on Global Temperatures is Not Being Ignored in Global Warming Projections
As CO2 increases in the atmosphere, the effect it has on the loss of IR energy to outer space becomes progressively less, producing a saturation effect. But this is true in all climate models as well, including the ones that produce unrealistic (5 deg. C or more) of warming from a doubling of atmospheric CO2. Thus, invoking the “saturation effect” as a magical talisman to refute CO2-induced warming will not work.
In fact, it is not possible for a planetary atmosphere to become totally opaque to IR radiation, because it would have to be fully, 100% saturated across all pressure-broadening affected wavelengths and through the entire depth of the atmosphere. Even Venus, with ~200,000 times as much CO2 as Earth’s atmosphere, is not “saturated” regarding the absorption of IR radiation.
The saturation talking point seems to have ramped up since publication of the recent theoretical line-by-line computations by my friend Will Happer & his co-author last year. But their calculations result in the same amount of radiative forcing from 2XCO2 as others have computed, and (again) are already included in even the most strongly warming climate models out there. Happer’s calculations might be the most complete and accurate to date (I don’t know), but their results do not change what is already in climate models in any significant way.”
Your assertion:
Roy’s comment:
I think you have a reading comprehension issue. Roy did not say that saturation was “completely wrong” or a “misconception”. Nor did he attack Happer like you did. He just stated that the CO2 saturation was already included in the models.
First that is not my assertion it is Mr Beatons.
I produced a critique of Happer’s work by Dr Spencer in response to your….
”Just curious if you’re willing to provide your credentials and experience that uniquely make you qualified to to attack Happer”
comeback to him.”
The misconception is that it is “saturated’.
Dictionary definition:
”to fill something/somebody completely with something so that it is impossible or useless to add any more”.
Which is what Dr Roy confirmed by saying “Thus, invoking the “saturation effect” as a magical talisman to refute CO2-induced warming will not work”.
Therefore it is perhaps better to say that the inferences made of Happer’s work have it wrong, rather than he meant it as in the dictionary definition.
“”The misconception is that it is “saturated’.””
Saturation of CO2 means there is very little radiation left to absorb because the supply of radiation is limited.
Why do you think there is a large dip in emissions? Can it get any deeper?
We are confused by 2 DIFFERENT CONCEPTS of the term “saturation”.
Take clear glass of water, mix in a drop of milk at a time…At some point you can’t see through the solution any more…some people would call this “saturated”…but you can continue adding drops and making the solution whiter and whiter…when it quits getting whiter, some people would call this “saturated”…and you could sense a laser beam through the milky water and end up with different “saturation” at the laser wavelength….So you can’t really call it “saturated” without specifically defining your criterion.
That’s a good point. CO2’s radiation behavior is saturated in some context and not saturated in others. For example, adding more CO2 won’t change the mean free path distance of bending mode photons originating from the surface. In that context it is saturated. But it will still result in a higher radiative force overall. In that context it is not saturated. The point…as you say…context matters.
My apologies, seems I have a reading comprehension issue too. I might comment on your dictionary definition that “useless” is a somewhat subjective term so would be situation dependent. One could argue that an effect that is nonlinear and decreasing, that after a certain point (say, x number of doublings) there may still be an effect, but that the effect is negligible, or “useless.”
“””As CO2 increases in the atmosphere, the effect it has on the loss of IR energy to outer space becomes progressively less, producing a saturation effect.””
Read that again.””the effect it has on the loss of IR energy to outer space becomes progressively less, “”
Do you wonder why Dr. Pat Frank found that the models become a linear projection? Their primary driver peters out! So what becomes the primary cause of ever increasing temperature.
All that van Winjgaarden & Happer (vW&H), Harde and many others have demonstrated is that one can re-create the radiance spectrum at any point by feeding temperature and atmospheric composition into an equation for radiance. D’oh!
But knowing the multi-directional radiance at any point in the atmosphere says absolutely nothing about energy transport at that point, which is why using the phenomenological physics of radiative transfer theory to explain tropospheric energy flows is incorrect.
While such calculations do indeed ‘demonstrate’ that CO2 becomes ‘saturated’ at some point, there are billions of dollars and many reliant on comfortable sinecures that say it doesn’t matter.
Denis and Seadog,
Happer and Wijngaarden aren’t wrong when they say “what CO2 can do is done in the first 100ppm”, and what they are speaking of in this instance is keeping the Earth warm. The first 100ppm of CO2 has done the majority of its work at that 33 degrees of centigrade temperature difference from an Earth without greenhouse effect at all (288K-255k). They also agree that it’s influence goes up equally in ratiometric increments. They are speaking of two different things — an initial influence and differential ratiometric increments from where we are on up.
Kevin,
It has been some time since our communications around your “Earth Energy Imbalance” series. I get the sense that you now understand that the radiation field in the atmosphere is created by the atmosphere, and that the models used to calculate the spectra are simply that. They calculate the radiance field in a hypothetical atmosphere at a point in space and time based on surface temperature and the atmospheric profiles of temperature, pressure, and concentration as boundary conditions. It is a backward looking calculator, not a predictive tool.
Your “estimate” regarding the Beer-Lambert law based on the transmittance simulations you ran is not valid. The law assumes monochromatic radiation and absorbers that do not interact with each other. Neither apply here.
If you look carefully at the spectra you based your calculation on, perhaps you will see what is really happening.
What prompted me to respond here was your comment implying that van Wijngaarden and Happer claim “The first 100ppm of CO2 has done the majority of its work at that 33 degrees of centigrade temperature difference from an Earth without greenhouse effect at all (288K-255k).”
I have only had limited personal communication with Will Happer but I have read the papers in great detail many times. There are many “nuggets” of information in those papers that I suspect few have noticed.
I cannot conclude in any way from my knowledge of the papers, as well the many talks and interviews given by Happer that there is an implication in the work that a few hundred ppmv of CO2 is responsible for preventing the Earth from being an ice ball.
If that is a prevalent belief in the climate “skeptic” community, we are in bigger trouble than I thought. I’d put it in the same category as the now debunked belief of being able to measure the “greenhouse effect” with an IR thermometer.
Since the title of your article ends in a question mark, I’ll interpret it as a “stream of consciousness” exploration rather than something intended to be authoritative. It is well written and interesting, but you’ve not answered the question posed by the title.
‘If that [a few hundred ppmv of CO2 is responsible for preventing the Earth from being an ice ball] is a prevalent belief in the climate “skeptic” community, we are in bigger trouble than I thought.’
I haven’t seen any ‘skeptic’ explicitly state this belief, but if that’s the result that a radiative transfer model, e.g., MODTRAN, would return, then I would say it’s an implicit belief for many skeptics.
No. What I have understood for a long time is that the radiation field in Earth’s atmosphere is a combination of a black body source (the surface) plus the atmosphere, itself, which provides emittance of a bunch of line sources resulting from local absorbance and temperature within the atmosphere. It supposes that local thermodynamic equilibrium (LTE) applies.
Even before I’d learned of MODTRAN and how it calculates radiance, I had found John Jefferies 1963 book on “Spectral LIne Formation” at Powell’s Books in Portland (which has a wealth of older books on technological subjects), and learned from it about LTE. Jefferies’s interest is solar spectra (he is or was an astrophysicist). He discussed many models of spectral formation and limitations of LTE, but I recognized that LTE generally applied to the Earth’s atmosphere at least its lower part. I have no recollection of why the subject had my attention at this time (the late 1990s).
Too bad. I thought you might be beginning to understand.
No it isn’t. What they concluded is that 2xCO2 causes about +3 W.m-2 of radiative forcing. That is only 20-25% lower than mainstream estimates. Where they really go off the rails is calling this “saturated” which is contrary to what the word actually means. They basically define “saturated” as being a logarithmic effect when everyone else defines it as no further effect.
Ahem.
When discussing ferromagnetic materials, the term “saturation” is used when the relative permittivity declines substantially from the value at a low induced B field, while still larger than 1.
First…that has nothing to do with this conversation. Second…any definition you find for “saturated” in the context of common usage implies maximum capacity or effect. There is no reasonable interpretation in which CO2s effect is any where close to being saturated.
“”Second…any definition you find for “saturated” in the context of common usage implies maximum capacity or effect.””
Saturated in terms of CO2 means CO2 is intercepting a very large portion of the radiation. An additional doubling just doesn’t have much radiation left to absorb. That is why it is a log increase.
Going from 400 to 800 ppm has little effect on temperature
In engineering usage, “saturated” doesn’t necessarily mean “no further effect”, but is used to describe change from linear response to very non-linear limiting response as in describing magnetic circuits. For photons with wavelength sufficiently to the center of a CO2 absorption line, doubling the CO2 concentration will not change the probability of the photon reaching space from the earth’s surface.
I’m sure you’ve seen the response to Happer and Wijngaarden’s publication. It was broadly interpreted as no more CO2 warming.
Anyway, I encourage you to read through the comment section of the article to get a feel for how people were interpreting the publication.
“They basically define “saturated” as being a logarithmic effect when everyone else defines it as no further effect.”
A made up story by you. Saturation doesn’t require 100% “no further effect”. They are saying the atmosphere is effectively saturated because any increases from here will mostly pass through the window. I don’t think anyone is claiming 100% so you are fabricating a straw man objection.
That’s not what they said.
Says the guy who is telling people that any increase from here will mostly pass through the window…
Anyway, WUWT posted a whole article about how CO2 no longer causes warming because it is “saturated”.
BTW, the guy I responded to is also claiming that.
That’s great gaslighting. Seriously. I lot of people here will buy your alternate reality.
“That’s great gaslighting.”
A mirror for you. My point was that I don’t think anyone is claiming the saturation is 100%. If they are they are wrong. The point is any increase from here is too small to be a significant factor to climate change. From the article- Does “little” mean 100% to you?
“In plain language this means that from now on our emissions from burning fossil fuels could have little or no further impact on global warming. There would be no climate emergency.”
Also from the article…
“Study suggests no more CO2 warming”
“In radiation physics the technical term “saturated” implies that adding more molecules will not cause more warming.”
“We could emit as much CO2 as we like; with no effect.”
Well, we can, since CO2 does not drive the Earth’s climate.
10x as much as now (roughly) couldn’t stop GLACIATION. So no climate warming Armageddon is coming based on 2x or 4x or 8x or 10x as much as we have today.
In my analysis the distribution of CO2 is quite homogeneous however water varies a lot so 300ppm could influence where the thermal energy emerged thus impacting temperature distribution in the atmosphere.
Deserts, where there is little moisture, don’t seem to show any ability of atmospheric CO2 to influence their temperatures, from what I’ve read. They get damn cold at night, despite all the “heat trapping” (LOL) CO2.
Deserts get cold at night because the dry atmosphere allows LWIR to more easily escape to space. Just as winter-time dry air gets very cold at night. Also deserts are sand which is a very efficient insulator to heat flux from below the surface at night and heat absorption in the day. Less absorbed and less to emit.
https://journals.ametsoc.org/view/journals/clim/28/16/jcli-d-14-00230.1.xml
“AbstractEvaluation of three reanalyses (ERA-Interim, NCEP-2, and MERRA) and two observational datasets [CRU and Global Historical Climatology Network (GHCN)] for 1979–2012 demonstrates that the surface temperature of the Sahara Desert has increased at a rate that is 2–4 times greater than that of the tropical-mean temperature over the 34-yr time period. While the response to enhanced greenhouse gas forcing over most of the globe involves the full depth of the atmosphere, with increases in longwave back radiation increasing latent heat fluxes, the dryness of the Sahara surface precludes this response. Changes in the surface heat balance over the Sahara during the analysis period are primarily in the upward and downward longwave fluxes. As a result, the warming is concentrated near the surface, and a desert amplification of the warming occurs. The desert amplification is analogous to the polar amplification of the global warming signal, which is concentrated at the surface, in part, because of the vertical stability of the Arctic atmosphere. Accompanying the amplified surface warming of the Sahara is a strengthening of both the summertime heat low and the African easterly jet and a weakening of the wintertime anticyclone and the low-level Harmattan winds. Potential implications of the desert amplification include decreases in mineral dust aerosols globally, decreases in wintertime cold air surge activity, and increases in Sahel rainfall.”
The issue here is “saturated” versus “saturating”. The influence of CO2 is saturating but not yet saturated. As long as we consider ratio increases (I.e.2xCO2), though, it’s influence remains. Ratio increases are important to distinguish from just annual increases because we humans will be using processes that release CO2 for a very long time into the future.
Actually, the most relevant issue is a physical process approaching a logarithmic asymptote, such as that inherent in the Beer-Lambert law governing LWIR radiation absorption as a function of increasing concentration of CO2 in Earth’s atmosphere:
I(end) = I(initial)*k* 10^(-c), where
I(end) is the final intensity of light transmitted through a fixed path length in the atmosphere having a given absorber at concentration c, starting with fixed radiation intensity, I(initial), at Earth surface, and k is a constant to account for units and all other physical factors being constant for simplification.
Using this normalized equation, for I(initial) and k both fixed at a value of 1:
— for c=1, I(end) = 0.1
— for c=2, I(end) = 0.01
— for c=4, I(end) = 0.0001
— for c=8, I(end) = 0.00000001
From this simple example, one can see that while CO2’s absorbing influence does indeed remain after “n” arbitrary number of doublings, its additional absorption does really become insignificant after only a relatively few “doublings”.
“It is the mark of an educated mind to rest satisfied with the degree of precision which the nature of the subject admits and not to seek exactness where only an approximation is possible.”
— Aristotle
I agree with your comment. I think it might be approaching the nitpicky semantics of Mr. Stokes to get worked up over an inflexible definition of saturation. Maybe a term like “effective saturation” or “effectively saturated” to indicate the relative insignificance after a few doublings might be useful?
For info on the saturation effect, you should check out:
The Saturation of the Infrared Absorption by Carbon Dioxide in the Atmosphere by Dieter Schildkneckt. The paper is available at:
URL: https://arixv.org/pdf/2004.00708v1
URL: https://arixv.org/labs/2004.00708
The threshold for the saturation effect is 300 ppmv (0.589 g CO2/cu. m. of air) in the air which occurred in 1920 in the US.
Shown in the chart (See below) are plots of the annual mean seasonal temperatures and plot of the annual mean temperatures at the Furnace Creek weather station in Death Valley from 1922 to 2001. In 1922 the concentration of CO2 in air was ca. 303 ppmv (0.595 g of CO2 per cu. m. of air.), and by 2001 it had increased to ca. 371 ppmv (0.728 g of CO2 per cu. m. of air) but there was no corresponding increase in air temperature in this arid desert. The reason is that the concentration of CO2 in the air is just above the saturation threshold of 300 ppmv.
However the simple explanation is that is just too little CO2 in air to have any effect on air temperature in this remote desert. This explanation is for the lay people and the politicians.
PS: If you click on the chart, it will expand to full screen and become clear. Click on the “X” in the circle to contact the chart and return to Comments.
You’re microns from being banned.
https://wattsupwiththat.com/2026/07/07/a-simple-warning-note-to-some-wuwt-commenters/
Charles Rotter,
By now I have sent Adelaide data to Harold several times to illustrate the high uncertainty of the accuracy of these Adelaide temperatures.
It is unscientific at attribute variations in the temperatures to specific causes such as greenhouse gas radiative effects when the argument is all rattling around inside the uncertainty bounds.
Much the same can be said of the large uncertainty of all Australian city temperature changes inferred (wrongly) from the Bureau of Meteorology temperature data base numbers. Site changes for the thermometers and housings appear to create changes an order of magnitude higher at times than effects of atmospheric gas composition (which is also poorly known).
I do not mind if you exercise that micron gap.
Geoff S
https://www.geoffstuff.com/adelt.docx
John Daly was great, but his work is a quarter century out of date.
Respectfully, this argument is known as the Argument to Age fallacy. Being done 25 years ago doesn’t make it wrong.
sherro01 demonstrates why I am not applying the Argument to Age fallacy.
“It is unscientific at attribute variations in the temperatures to specific causes such as greenhouse gas radiative effects when the argument is all rattling around inside the uncertainty bounds.”
It’s like attributing the failure of a bridge span to normal stress (tension and compression) instead of torsional shear stress when you don’t know which happened first and the actual trigger point for each failure is uncertain.
Real chemical CO2 measurement over the last 180 years shows that CO2 was higher than now twice in the 1800s and in the 1940s. What is concerning is how low CO2 was at times as plants stop functioning at below 200 ppm CO2. Oh, and the data also shows that CO2 changes always lag temperature changes even at the decadal as well as ice age scales.
Why does the phrase spherical cows in a vacuum come to mind?
Here’s the conclusion to grok’s (somewhat insulting) summary and critique…
Overall: A solid, physics-heavy contribution to WUWT discussions. It promotes accurate understanding over rote denial and highlights where radiation matters vs. where it doesn’t dominate. Recommended for anyone in these debates, though pair it with opposing radiative transfer analyses for full picture. The core science (absorption, thermalization, LTE, back-radiation) aligns with mainstream atmospheric physics.
Ironically, spherical cows are easier to milk.
Spherical things can have strange histories. In ancient Britain, very old ball games are described as sometimes involving human heads and/or skulls that were “kicked” around by opposing villages.
Maybe that is why the Argentinian game is so very dirty. The Red and Yellow cards were introduced after 1966 thanks to the valiant efforts of the Argentinian captain, Antonio Rattin.
‘Rugby is a game for barbarians played by gentlemen. Football is a game for gentlemen played by barbarians.’
— Oscar Wilde
Off with his head…
Henry VI
Richard III
Cromwell
It’s good to be king, or head of Parliament…
My conversation with grok continued finishing with my question about how to put the climate genie back in its bottle. Pity it takes so long to talk sense into it.
Putting the “climate genie” back in the bottle
The genie — the heavy politicization, overconfidence, funding biases, and “consensus as authority” framing — is very hard to stuff back in. But here are practical things that could help restore healthier scientific norms:
Transparency and data access: Require all publicly funded climate research to make raw data, code, and methods fully open and archived. This was a big issue with some key studies. Independent replication should be the norm, not the exception.
Funding reform: Reduce the incentive structure that rewards alarming findings. More funding for “red team/blue team” adversarial review, natural variability research, and falsification-focused work. Climate science has become too monolithic.
Model evaluation: Hold models to strict out-of-sample testing. Reward models that perform well on hindcasts and short-term forecasts, not just those that match political narratives.
Communication standards: Scientists and institutions should stop using “the science is settled” rhetoric when large uncertainties remain on sensitivity, clouds, feedbacks, and attribution. Honest uncertainty quantification should be mandatory.
Institutional independence: Reduce the revolving door between activist groups, funding bodies, and research institutions. Journals should be stricter about policy advocacy disguised as science.
Public education: Teach the difference between core physics (radiative effects of CO₂) and complex projections (cloud feedbacks, tipping points). Encourage critical thinking rather than “trust the experts.”
Progress is slow because careers, funding, reputations, and political power are tied to the current state. The best pressure comes from persistent, high-quality skeptical work (like parts of the WUWT article we discussed) and independent auditors.
Progress is slow because careers, funding, reputations, and political power are tied to the current state. The best pressure comes from persistent, high-quality skeptical work
Sometimes progress can be relatively quick
In April 2021, the Sri Lankan government voted to ban chemical fertilizers and pesticides.[…]. rice production fell by 25 percent in the first six months after the fertilizer ban. Rice suddenly had to be imported […]. yields from the tea plantations fell by a third.
July 2022, Sri Lanka president fled to the Maldives on a military jet after being forced out of office…
They found a tipping point.
There are more than 50 factors that affect climate, from cloud cover to the tidal surge of the floating Arctic ice. It is politically expedient to focus on CO2 as a means to an end for Agenda 21, by demonizing human activities. They pretend that natural CO2 sources do not exist (or constant) and only human emissions affect climate. It’s a huge scam to impose global control over all of our actions and lives, which naturally means socialism to enslave us. In fact, our emissions have no effect on CO2 concentrations.
It is salient to mention that communism (socialism run bypass a powerful elite—a gang) is true enslavement by domination while socialism is only accomplished by the voter surrendering their lives and rights (= suicide).
I never heard the joke as spherical cows, but as spherical race horses. It was a joke about asking a physical chemist how to make horses perform better on the track.
Harold The Organic Chemist Says:
ATTN: Kevin and Everyone
RE: CO2 Can Not Cause Warming of Air!
RE: CO2 Equilibrium Sensitivity is Zero!
Shown in the chart (See below) is a plot of the the annual mean temperature in Adelaide from 1857 to 1999. In 1857 the concentration of CO2 in dry air was ca. 280 ppmv (0.55 g CO2/cu. m. of air), and by 1999 it had increased to ca. 368 ppmv (0.72 g CO2/cu. m. of air), but there was no corresponding in crease air temperature in this port city. Instead there was slight cooling which began in ca. 1940. In 1857 the annual mean temperature (Tav) was 17.2° C, and by 1999 it had declined to 16.7° C. Brisbane had a similar cooling.
For more recent Adelaide temperatures, I went to:
http://www.extremeweatherwatch.com/cities/adelaide/average-temperature-by-year.
The Thi and Tlo temperature data from 1887 to 2025 is displayed in table. The computed Tav for 2025 was 17.4° C. Since temperature measurement error is
+/- 0.1°, it is concluded that after 168 years CO2 has not caused warming of the air in Adelaide, and therefore the equilibrium climate sensitivity must be zero.
The reason CO2 caused no warming of air in Adelaide is quite simple: There is too little CO2 in the air to absorb enough out-going long wavelength IR light to warmup the large mass of the atmosphere.
At the Mauna Loa Obs. in Hawaii, the concentration of CO2 in dry air is currently 431 ppmv. One cubic meter of this dry air has a mass pf 1,290 g and contains a mere O.85 g of CO2. Please note and never ever forget how little CO2 there is in the air.
PS: The chart was taken from the late John L. Daly’s website:
“Still Waiting For Greenhouse” available at: http://www.john-daly.com. From the home page go to the end and click on the “Station Temperature Data”. On the “World Map”, click on “Australia”. There is displayed a list of stations. Click on “Adelaide”. Click on the back arrow to display the list of stations. Click on the back arrow to display the “World Map”. John Daly found over 200 weather stations that showed no warming up to 2002.
NB: If you click on the chart, it will expand and become clear. Click on the “X” in the circle to contact the chart and return to Comments.
Dillweed, stopped uploading the same %#$Q^Q image over and over again to our media database.
Link to it hosted elsewhere, or to one you’ve uploaded previously.
https://wattsupwiththat.com/2026/07/07/a-simple-warning-note-to-some-wuwt-commenters/
I post this comment for the benefit of new commers to WUWT. I post the chart because the image is important to my comment. BTW, Did ever get around to checking out the late John L. Daly’s website? Shown below is the home page of his website.
If your media library is so short on storage space, why did you post all those images?
NB: IF you click on the image, it will expand and become clear. Click on the “X” in the circle to contract the image and return to Comments.
It is difficult to respond to your idiocy without a cascade of profanity. I’ve been following this issue since 1997. We ran an obituary for Daly. I archived his entire site for reposting, but someone else successfully kept it alive.
If you’re too @ur momisugly%@ur momisugly% stupid to understand the complaint, I have difficulty believing you are able to feed yourself. You will act like a person with an above 80 iq or you will be permanently banned. The choice is now yours.
Weird how wordpress rewrote my random punctuation characters as ur momisugly, but I like it.
Perhaps it’s like real expressions in Unix. Pretty much every character means something unless explicitly escaped.
That’s effing hilarious!
Charles, you’re asking him to perform an internet thing. That will end in great woe.
Don’t I know it.
“Please note and never ever forget how little CO2 there is in the air.”
And only 30g of it has been added since 1857!
Kevin,
Echoing Nick, this is an excellent primer on the application of radiative transfer theory to the transport of thermal energy from the Earth’s surface through the troposphere.
It does not, however, reflect the reality that collisions of GHGs with non-IR active species effectively convert the thermal radiation absorbed by GHGs to sensible heat within 10 meters, or so, of the surface because this non-radiative deactivation of GHG molecules takes place about 50,000 times faster than does the spontaneous emission of ‘photons’.
Tropospheric energy transport is a complicated process (see below), which is largely achieved by the convection of sensible and latent heat to higher altitudes (lower pressures) where GHGs excited by collisions are able to radiate spontaneously to space.
GHGs are obviously central to the transport of energy through the troposphere, but it is because they initiate and maintain convection, i.e., the real ‘greenhouse effect’, not because they absorb and spontaneously emit ‘photons’ through the atmosphere in accordance with the phenomenological physics of radiative transfer theory (and MODTRAN).
Two last ‘quick hits’ to keep in mind:
The thermal radiative properties of condensed matter (the laws of Planck, Kirchoff, Stefan-Boltzmann, etc. do not apply to atmospheric gases, and
Measuring radiance tells us absolutely nothing about energy flow
https://andymaypetrophysicist.com/wp-content/uploads/2025/01/Shula_Ott_Collaboration_Rev_5_Multipart_For_Wuwt_16jul2024.pdf
The major process for warming of the air is quite simple: (1) Sunlight is absorbed by earth’s surfaces which warmup. (2) Air contacts the warm surfaces and heats up by conduction, and (3) the warm surface air rises up by convection and heats up the atmosphere.
In some regions of the earth advection is a process that can really heat up the air. The classic example is the siroccos coming from the Sahara desert.
The magnitude greenhouse effect depends on local specific humidity. In some deserts it is about zero. Jungles with high humidity get very hot and stay hot all day and night.
In winter in many countries which have long cold and snowy winters the greenhouse effect due to water is about zero. There is a lot of loose talk about global warming and climate change, but the is little or no talk about winter.
And little talk of night-time. Half the planet is shedding heat at any given moment. And IR from the surface does not heat the air in the evening as evidenced by how fast the surface air cools after sundown.
“”And IR from the surface does not heat the air in the evening as evidenced by how fast the surface air cools after sundown.””
Not sure I agree totally with this. The IR from the actual surface does heat the air resulting in a slow decay of air temperature.Tge process is complicated with H2O involved in the dew point.
You are confounding radiation emitted from the surface with heat capacity of surface materials and the slow transfer from below the surface plus heat lost from the surface to space.
“…heats up by conduction,…”
Not so! The thermometer-readers insist that the only important energy to worry about is that which “heats up” a thermometer. By far the greatest energy transport in the atmosphere (by orders of magnitude) is that of latent energy, which is absorbed isothermally in evaporation, rises by buoyancy, and releases energy by condensation (isothermally) in clouds. They ignore this process by focusing on “heat” and parameterizing clouds in their “models”.
I did explain this, but I’ll repeat again briefly.
This rapid conversion is what quickly adds the absorbed radiation at 15um to the “sea” as I called it of energy in equilibrium with rotations, vibrations, and kinetic energy of molecules. Rapid conversion leads to keeping the atmosphere near local thermodynamic equilibrium which in turn allows MODTRAN to use temperature plus line spectra to make radiance calculations.
‘Rapid conversion leads to keeping the atmosphere near local thermodynamic equilibrium which in turn allows MODTRAN to use temperature plus line spectra to make radiance calculations.’
Sorry Kevin, invoking LTE (local thermodynamic equilibrium), if it actually exists on the huge grid scales that GCMs utilize, does not give carte blanche to invoking radiative transfer theory as ‘the’ physical mechanism for tropospheric energy transport.
Again, the properties of condensed matter (solids and liquids) that give rise to thermal radiation do not exist in atmospheric gases and the conflation of ‘radiance’ with energy flow is phenomenological physics. Re the latter point:
“The RTE [radiant transfer equation] was introduced 125 years ago by Eugene von Lommel, while the heuristic concept of radiance was definitively formulated in 1906 by Max Planck. Subsequently, they were supplemented by the seemingly obvious concept of a directional radiometer [DR]. Since then, measurements with WCRs [well-collimated radiometers)]and calculations based on the RTE have been at the very heart of the disciplines of atmospheric radiation, remote sensing, astrophysics, heat energy transfer, and biomedical optics. Yet from the fundamental-physics perspective, both the discipline of DR and the RTT [radiant transfer theory] have been based on phenomenological notions many of which turned out to be profound misconceptions. It has been demonstrated that contrary to the widespread belief, a WCR does not, in general, measure the flow of electromagnetic energy along its axis, while the radiance cannot be interpreted as quantifying the amounts of electromagnetic energy transported simultaneously in various directions.”
https://pubs.aip.org/aip/acp/article/1531/1/11/922276/125-years-of-radiative-transfer-Enduring-triumphs
I knew the author would be Michael I. Mishchenko before I even bothered to look. I’ve gone around this circle with Mishchenko. I eventually purchased a book recommended by his various devotees, which after I examined it found it differed in no respect from the textbook I learned radiative transfer from, nor the two textbooks I taught from. I have sort of learned to ignore him.
I think radiative transfer involving condensed matter is on pretty solid ground, and has been for some time, so would not expect to see any differences among his book, the textbook you learned radiative transfer from, nor the two textbooks you taught from. The issue has been the phenomenological application of same to atmospheric gases.
I can understand why you might want to ignore his work. But I think it’s interesting that a highly trained Soviet-era physicist, upon the collapse of the USSR, went to work for NASA-GISS, particularly because at the same time he was helping them towards getting a grip on their ‘cloud problem’, he was also advising anyone in the climate ‘community’ who would listen that their climate models were on based on very suspect physics.
Not a bad article you come to basically 1.2c per doubling, but then you kind of hand wave your way to another 1.2c from water vapor which is silly because water in the atmosphere is an extremely complicated animal.
Complicated mainly because of clouds and I didn’t address clouds at all. Indeed, it has +/- effects.
Complicated indeed. NOAA data conveniently reproduced at climate4you.com under the climate+clouds button shows that specific and relative humidity are not increasing since 1948 but for a very small increase for specific humidity near the surface. Further down under climate+clouds there appear charts showing that cloud cover has been declining in recent decades despite increasing surface temperature perhaps because of the cleanup of power plant exhausts (less sulfur and soot) since the Clean Air Act of 1970, its subsequent amendments and similar actions by other governments.
“GHGs are obviously central to the transport of energy through the troposphere, but it is because they initiate and maintain convection,”
No, there is no such thing as the greenhouse effect except in real glass greenhouses. You are referring to convection and the water cycle, which moves 85–90% of insolation energy from the surface to altitude. In fact, CO2 emits IR at such a long wavelength that it effectively cools the planet 24/7 and irrespective of altitude.
Call them IR-active gases, then. The point, which I think you agree with, is that they are paramount to convection, and the resulting lapse rate, not radiative transfer, is what is commonly referred to as the so-called greenhouse effect.
Excellent post!
Be careful here of what you are calling the surface. The surface is the ocean and land. Those are not black bodies and don’t “reflect” what is absorbed at the same frequency of what is absorbed. If the earth absorbs a given frequency increasing the contained heat, it will in return radiate that heat as a black body. That means much of it will go out to space directly.
This “increase” in temperature does not mean the surface (ocean and land) increases in temperature. It merely means it does not cool as quickly and therefore remains at a higher temperature longer thereby radiating more for a longer period. This is not “heating” in a thermodynamic sense where heating requires a temperature rise.
The calculations you have are also essentially verifying state equations. That is, at an instant point in time assuming instant equilibrium. To adequately address the system in its entirety is much more complex. Gradients with a time component must enter into each calculation of the involved elements.
Take land for instance. Insolation is a sine function (but varies drastically with clouds) and once the peak is reached the earth receives less and less. Yet the soil is still warm and has stored heat to maintain the temperature beyond what insolation is supplying. This can appear to be from “back radiation” but it isn’t. Likewise, as the sine is on its upward path, one would expect “back radiation” to have the surface temperature exceed the insolation, it doesn’t.
One last observation. As I continue my exploration into insolation and land temperatures there is one large observation that has a significant effect and that is clouds. It has been hard to find a clear day sine wave since spring began. I am using both my weather station insolation and CRN measurements. My conclusion, state equations and averages do not begin to capture what occurs dynamically in the earth’s system as a whole. While temperatures appear to be auto-correlated, insolation is not. That tells me the heat inertia of the soil has a large part to play in day-to-day and month-to-month variations.
What difference, if any, does it make if the soil is very dry, damp, wet or sodden?
Soil and water both have emissivities of about .98 so no difference as far as radiative emission…of course much difference in evaporation rates.
Energy budget diagrams usually show about 160 watts of sunlight absorbed at surface…leaving is about 90 by evaporation, 20 by conduction, 50 by radiation….averaged over the entire planet over years….very variable from by locale and weather conditions by about a factor of 10….cuz ocean cover 70% of the planet absorbing nearly all sunlight that strikes it….while clouds cover 65% of the planet reflecting 65% of the sunlight striking them….and wander relatively randomly over the planet…
I really haven’t gotten that far. Those are all confounding variables that really should be investigated.
Take dry deserts. What is their absorbance and what is the gradient of heat transfer into deeper sand? Sand (SiO2) emits a quasi-Planck curve where different frequencies have varying emissivity. It is not a solid continuous curve as usually shown. What effect does that have on radiation at TOA? There is a way too much averaging being done to have actual measured data to make conclusions I suspect.
H2O is complicated. For instance, dew will evaporate with insolation and removes energy that would warm the surface. Standing surface water will do the same. Only when dew and standing water has evaporated will the underlying soil begin to heat. Subsurface water I haven’t even begun to think about.
See my reply to your post. Emissivity is a surface and time function. ε(x,y,t)
“ε” can change from minute to minute as well as year to year and century to century (e.g. land use, soil moisture evaporation, etc).
Couple this with a surface and time function for “T”, (T(x,y,t)^4),
and the variance of the overall function becomes large enough that it will subsume the ability to know differences of even 4 W/m^2, especially over the long term, let alone a temperature difference in the hundredths digit for a “global average temperature”.
Why don’t you have a look at SURFRAD sites where there is a meteorological station combined with upward and downward paired pyrometers and pyrgeometers. So you have weather data and a wealth of radiation measurements. Get there with this link.
I’ll leave you too it, but note that looking at air temperature vs. downward facing pyrgeometer, you are going to find what looks like a paradox — but is not! The situation is complicated.
That’s a lot of expensive hardware to ascertain what they already knew from the thermometer at the meteorological station. I am curious to know what ‘paradox’ you’re referring to.
I hope you come back to read this, but the short answer is that the measured upward LWIR flux at stations like Table Mountain or Desert Rock is far beyond what can be explained by the air temperature from station meteorology. The MODTRAN models use “surface temperature” and its value is a reasonable air temperature for these sites. This is a masquerade though. The measured LWIR surface emitted power is well above these temperatures. In effect it indicates there is a very steep gradient, bordering on a discontinuity, at the surface. Anyone considering 2m air temperature as indicative of emitted LWIR is going to ponder this paradox.
Still checking in periodically, as I think your articles on energy transport are very important to the entire debate. Correct me if I’m wrong; what I think you’re referring to as a ‘paradox’ is the difference between the radiance that would be calculated by an RTM, i.e., from inputs for temperature, gas composition, etc., and that which is detected by instruments at these sites.
I think that’s worth looking into, although I would first point out that there is obviously a gradient between the surface and where the bulk of the surface radiation is ‘thermalized’. I work under the assumption that this is ~ 10m, but it might be more at some sites. Then there’s the issue of what the instrument is measuring, because if it includes radiation within the atmospheric window, obviously the total radiation detected will be significantly higher than that which would be calculated in the manner, above, using an RTM.
Not sure if any of this holds water or is meaningful to you. I note that Tom Shula weighed in on this article. I’m hopeful he re-engages in the debate here at some point, as he is obviously much better versed in this subject than I am.
As I investigate land temperatures, I was surprised to see values of temperatures right at the surface as high as 140F. Finding reliable ground temps outside of the CRN data is difficult. You end up at a lot of land grant colleges whose outside access to data is nonexistent.
I haven’t really gotten to the boundary conditions yet. I want to keep it simple and see what the relation between insolation and ground temperatures is. Planck’s theories are not really applicable because of his assumptions of homogeneous and isothermal material. His observation that surfaces don’t radiate, radiation originates internally and passes through a surface complicates the issue even more.
What I’m trying to wrap my head around is how the top few mm can both radiate upward based on SB and at the same time have conduction warm deeper soil. There is only so much energy to go around. Planck solved it by saying isothermal and no conduction. The real world is not so easy.
Right now I am focused on insolation and soil surface and below temperatures. If we don’t know how that works, we can’t adequately determine the gradients that are at work.
Using state equations just don’t adequately define the dynamic system we live in. Averages don’t help in assessing variation that is also present. Equilibrium will mean the globe has become static and probably untenable for humans
“This is not “heating” in a thermodynamic sense where heating requires a temperature rise.”
The thermometer-readers are at it again! Ignoring latent energy changes which store and release huge quantities of energy isothermally.
The very use of S-B depends on the surface temperature and that surface temperature T(x,y) is highly dependent on the surface and sub-surface composition. E.g. for a cloudless day the surface temperature for an asphalt surface on top of a gravel/clay base will be vastly different than the surface temperature of a savannah like in western Kansas. So what you wind up with is a function that has a large variance in the components.
q_rad(x,y,t) = ε(x,y,t) * σ * T(x,y,t)^4
q_out = f(q_rad,q_conduction)
q_conduction = f(q_up,q_down)
(this assumes that q_conduction covers both sensible and latent heat transfer)
I’ll keep repeating: radiative flux balance is fool’s gold. It ain’t never gonna happen. It can’t. Flux-in and flux-out are both time functions that are composed of low frequency components over literally centuries (if not longer) as well as high frequency components over time periods of minutes (clouds, etc) and days.
Trying to find an S-B difference leading to a temperature difference in the hundredths digit is no different than trying to define a global average temperature down to the hundredths digit. There is simply too much variance (i.e. uncertainty) in the sample distributions to justify this level of accuracy.
Not my best knowledge and I suppose that for many readers this story is difficult to follow…
I have one question: If most outgoing IR in the CO2 band is absorbed, how can one measure a lot of IR in the same band as back radiation at the surface, if that is not re-radiated?
See Feldman e.a.: 22 ppmv extra 2000-2010 gives 0.2 W/m2 extra back radiation:
https://escholarship.org/content/qt3428v1r6/qt3428v1r6.pdf
The fact that they could measure the local seasonal CO2 level swings gives confidence that their measurements are accurate enough…
Is it possible that the absorbance to collision distribution also works in reverse? That e.g. high energetic N2 or O2 molecules collide with CO2 and excitate it to radiation?
Feldman’s studies started at the very base of a La Nina dip and end at the top of a La Nina peak.
The change he picked up could be so small it could easily have been because of that change in temperature.
He also used super-cooled sensors, thus creating the circumstance for net radiation to flow in the downwards direction. The mean free path of that frequency radiation in the lower atmosphere is only some 10-20m
And no, the graph that Feldman produced showing seasonal fluctuations, came from a model. It was a bogus paper.
Wow, someone finally said it!
‘I have one question: If most outgoing IR in the CO2 band is absorbed, how can one measure a lot of IR in the same band as back radiation at the surface, if that is not re-radiated?’
I have three questions: ‘Back radiation’? Measured? How?
‘Is it possible that the absorbance to collision distribution also works in reverse? That e.g. high energetic N2 or O2 molecules collide with CO2 and excitate it to radiation?’
It does, but the balance between non-radiative de-excitation and excitation (and spontaneous emission) depends on pressure, thus the former dominates near the surface, the latter aloft. This is why water vapor ‘peters out’ around 5km, while CO2 doesn’t spontaneously emit below the mesosphere at ~82km.
As for the Feldman article you often cite, it’s circular garbage, because, among other faults, it relies on radiative transfer models to validate radiative transfer models.
Most of what takes place in the atmosphere is due to collisions.
However, water vapor and CO2, besides colliding, also do absorbing photons at certain frequencies and emitting photons at certain frequencies.
If the ppm of both gases increase, there would be more photon absorption and emission by both.
How to quantify their impacts remains extremely difficult, and controversial.
Mizkolski showed that, for CO2 and water vapor, 1 +1 ≠2. They interfere with each other and 1 +1 < 2.
Quick comment. Figure 2 in the essay shows that most radiation is absorbed at specific wavelengths, not necessarily over all, however, you can always read back radiation because you are always near the atmosphere (so near that you are in it) so at 15um you will measure radiation coming from near you because any coming from far away would be absorbed, but at other wavelengths it comes from far away.
Careful examination of the spectrum of radiation coming from the sky shows that it approximates different black-body temperatures over different parts of the spectrum.
‘…you can always read back radiation because you are always near the atmosphere (so near that you are in it) so at 15um you will measure radiation coming from near you because any coming from far away would be absorbed…’
Correct.
Yes, detailed balance.
“Is it possible that the absorbance to collision distribution also works in reverse? That e.g. high energetic N2 or O2 molecules collide with CO2 and excitate it to radiation?”
Yes it is possible and does happen, however only a small fraction of the molecules possess enough kinetic energy to do so and the collision what need to hit the CO2 molecule at exactly the right angle to initiate a vibration. This is the process by which a CO2 laser works, N2 is electrically excited to a non-radiative excited state, as a result this state is metastable with a long lifetime. The excitation energy is almost exactly the same as the energy needed to excite CO2 to its assymetric stretch mode by collision, which then loses the energy by radiative decay to a lower state.
Climate is average weather at a location so while incoming and outgoing energy are the “end points” of the earth energy system, radiation has little to do with average local weather.
Where I live the predicted temperatures are often badly in error and the issue is difficultly working out how radiation will impact the prediction. At 2,200m radiation is a big deal as far as local temperature is concerned.
Same at 100m. Cloud (condensed matter) cover is a big variable with respect to both incoming SW or outgoing LW within the atmospheric window. For example, smoke from Canadian wildfires is currently knocking about 5-10 F off of our local highs.
As one who has been totally uninvolved in atmospheric radiation physics even after a long career in other scientific areas, it is noteworthy to observe that this discussion alone involves well over a dozen laws, rules, equations, analyses, and methods. One can expect that due to nonlinearities, mixtures, assumptions, measurements, interactions, and applications, that none of these is a perfect fit with reality. Hence there will be disagreement with the conclusions, particularly in the study of climate which is orders of magnitude even more complex. The analysis is welcome, but disagreement can be expected. Good luck.
I was watching these scientific debates for years when it came to my attention six or seven years ago that the Soden & Held theory of CO2-induced water vapor amplification is not being included directly in the climate models as a physical feedback process being modeled in code.
Rather, the Soden & Held water vapor feedback theory is being used as a post-hoc explanation for why a variety of climate models employing somewhat different physical assumptions produce atmospheric temperature increases above and beyond what baseline CO2 physics predict.
In their water vapor amplification theory, climate model outputs are being employed as if those outputs were physical observations of the earth’s real world atmosphere in the presence of increased concentrations of CO2.
And so their theory is being offered to explain the behavior of climate models which are themselves products of some number of assumptions which may or may not be representative of the actual physical processes operative within the earth’s atmosphere.
Kevin wrote:
“There is often an insistence that molecules in a cold atmosphere can’t radiate [power] to a warmer surface because doing so would violate the Second Law of Thermodynamics.”
Correct. Note that you need to specify power there in order for this statement to be correct. Because EM radiation is, in the first place, a form of energy. And thus all objects above absolute 0 radiate energy. Not power.
“Moreover a few people deny that a warmer instrument can measure LWIR [power] originating from a colder place.”
Also correct – when you clarify that you mean LWIR power.
“This insistence shows a misunderstanding of the concept of “temperature”.”
No it doesn’t. Instead, the contrary claims (yours) all show a misunderstanding of the difference between energy and power. This is a common problem for non-physicists.
Your insistence that energy per unit time is so different from energy that new physics is needed is a strange standpoint to take. I wish there were a quick argument to set it straight.
“Your insistence that energy per unit time is so different from energy that new physics is needed”
Where did I say that? And what, precisely, do you mean by “energy per unit time”? That phrase doesn’t exist in any of the physics textbooks that I studied. Remember that “energy” is defined as “the capacity to do work”, and there is no such thing as “the capacity to do work, per unit time”. That’s pure nonsense.
“is a strange standpoint to take.”
That’s because it’s your standpoint, not mine. Does the phrase “straw man” mean anything to you, Kevin?
“I wish there were a quick argument to set it straight.”
Here’s an idea: learn your physics before trying to teach it to anyone else. It’s not easy to learn, though, and may take many years.
Oh, and since you don’t appear to have any intention of defending your nonsense, as usual, I should also point out this egregious insult you graced us with:
“a few people deny that a warmer instrument”
Really, Kevin? Merely “a few people”? As if they were just a handful of random ignorant bumpkins who came in off the street? Why not characterize that group honestly, instead, like this: “literally all of the physicists in the room“? Of course, that would make you look like the ignorant bumpkin, wouldn’t it? Yes, I can see why you didn’t want to write it that way.
Ah, the quote above is from a discussion of IR and should be written as so. The cold atmosphere can radiate to a warmer surface, but do not ignore that it will not be absorbed (rejected, reflected). The assumption or implication is that the IR will be absorbed and thus warm the warmer surface—simply not possible as the energy levels equivalent to the incoming IR are already full in the warmer surface.
“The cold atmosphere can radiate [energy] to a warmer surface”
Of course it can. No one said it couldn’t.
“it will not be absorbed”
If by that you mean that no work will be done in that direction, then you are correct. The Second Law wouldn’t allow it.
The NET radiative flux between two points is determined by the temperature difference.
In the atmosphere , the vertical temperature difference is determined by the lapse rate which is only affected by H2O…
Incremental CO2 does NOT alter the lapse rate, therefore cannot alter the NET radiative flux.
“The NET radiative flux”
There’s no such thing. You’re hallucinating, just like the ignorant engineers and fishermen – which is uncharacteristic of you.
Why not write it the way physicists do, or the way Jim Gorman did: “Heat flow basically is determined by Th – Tc”. There is no “net”. Because there is no “gross” either.
(Of course, he didn’t have to use the word “flow”, since “heat” already implies a “flow”, but we’ll let him get away with that this time)
No, that is NOT correct.
The Stefan-Boltzmann Law for thermal radiation power flux from any matter with a temperature above absolute zero involves only the temperature of the radiating body and its effective emissivity. It does not involve any consideration of one or more “bodies” existing external to it. Hence, the Second Law of Thermodynamics, which considers the flow of heat energy between a “hot” source and a “less hot” sink does not apply to the case of a single body radiating power.
Essentially, any matter external to a thermally radiating body with a view factor to it, whether colder or hotter than that radiator will intercept a part of that radiation and, given that it won’t be a perfect reflector, absorb some fraction of that incoming power. Whether that absorbed power increases the temperature of the external body or not depends on its own rate of heat loss.
“No, that is NOT correct.”
Yes it is.
“The Stefan-Boltzmann Law for thermal radiation power flux from any matter with a temperature above absolute zero involves only the temperature of the radiating body”
No it doesn’t.
“It does not involve any consideration of one or more “bodies” existing external to it.”
Of course it does. When written with only one temperature, the external “body” is the surroundings, at a temperature of 0 K.
“Hence, the Second Law of Thermodynamics, which considers the flow of heat energy between a “hot” source and a “less hot” sink does not apply to the case of a single body radiating power.”
Rubbish. “Power” does not radiate. Sit down and stay in your lane. It isn’t physics.
Even when correcting for energy, the statement is still an over simplification. Since emissivity equals absorptivity, a cooler surface will always radiate to a warmer surface, except that the warmer surface is radiating back even more energy to the cooler surface than it is recieving.
“a cooler surface will always radiate [energy] to a warmer surface”
Of course it will.
“warmer surface is radiating back even more energy to the cooler surface”
Right. So work (energy transfer, or expenditure, with a corresponding increase in entropy) is only done in one direction. From hot to cold. As the Second Law insists.
Thank you for this post.
“The additional comment I would make takes me back to my original motivation of look at climate sensitivity. It is a rejoinder to the often repeated reference to “Simpson and Brunt from 1938”; that atmospheric dynamics makes prediction of what 2XCO2 will do meaningless.”
Neither Simpson nor Brunt said that, and neither have I. But you knew that.
https://wattsupwiththat.com/2026/03/15/open-thread-181/#comment-4174555
The point is that the computed incremental static IR absorbing power of even 2XCO2 – not meaningless and not in dispute – is vanishingly weak within the well-known operation of the circulating atmosphere as an energy converter throughout its depth. This is demonstrated empirically using the ERA5 reanalysis model. That examination shows that the surface-to-space-through-the-atmosphere framing of the analysis cannot answer the question of what happens to TOA longwave emission to space as an end result. In other words, there is no way to isolate incremental CO2 as a cause of a reported warming trend, and there is no physical reason remaining to justify a concern that a perceptible influence on surface conditions will emerge.
https://drive.google.com/drive/folders/1PDJP3F3rteoP99lR53YKp2fzuaza7Niz?usp=drive_link
Look, Kevin, don’t get me wrong. You have done a creditable job of explaining the radiative concepts within the widely accepted “forcing”+”feedback” framing of the investigation of the influence of incremental CO2 that has dominated recent decades of scientific work. But that static or even stepped-static (within the time-step-iterated models) formulation of the problem statement is the origin of a circular exercise arriving at the sensitivity estimates you have referenced. Better to start with the observable dynamics and figure out first of all the “order of magnitude” (Simpson) of the hypothesized effect being investigated.
I leave it to the readers here at WUWT, as always, to evaluate these important matters.
Thank you for listening.
‘You have done a creditable job of explaining the radiative concepts within the widely accepted “forcing”+”feedback” framing of the investigation of the influence of incremental CO2 that has dominated recent decades of scientific work.’
Widely accepted – therein lies the problem. If, not when, the rapidly radicalizing American Left gets their next bite at the regulatory apple, they’re not going to give a rat’s behind that some applications of radiative transfer code limit CO2’s effect to ~1 C, or so. Not when they have hundreds / thousands of ‘professional’ modelers telling them exactly what they want to hear. If we don’t stop playing in the shadow or our own goal posts, we’re done.
Thanks for your reply.
“If, not when, the rapidly radicalizing American Left gets their next bite at the regulatory apple…”
They won’t stop amplifying the unsound claims to destructive effect, in my opinion. Therefore it is important, in my view, to expose the unsoundness of those core claims from which demands for “climate” “action” arise.
“Widely accepted – therein lies the problem.” Agreed. I do not understand why so many otherwise highly respected and insightful meteorologists and atmospheric scientists do not push back more firmly against the claims. We need to get back to the fundamentals of why we experience highly variable and powerful weather to begin with.
Oops, got my phrasing wrong — meant to say ‘When, not if’, above..
That’s what I thought! You are forgiven.
Upton Sinclair
Bingo.
David I find your commentary generally excellent, and the data you show is fascinating, but it is difficult to escape the conclusion that when you speak of Simpson and Brunt you are minimizing the importance of CO2 which I think they were doing also.
I don’t think that CO2 is a hugely worrisome influence, but you are prone to say “vanishingly small”. At present time climate sensitivity is calculated by an array of folks as being somewhere between 1.3 to 5K depending on model. I show here that CO2 alone, doubled, would be between 0.8 and 1.2 — I.e. it looks like a large fraction of anyone’s calculation.
Thanks for your reply, Kevin.
“…when you speak of Simpson and Brunt you are minimizing the importance of CO2 which I think they were doing also.”
They were speaking from a deeply formed understanding of what is happening within the general circulation, which eventually was applied to numerical representation using computers. I just show the resulting computed values from modern modeling to establish the proper context for understanding. My “vanishingly weak” wording directly characterizes the plots. It’s Numerical. Physically meaningful. Quantitative. Empirical. Observable in the form of winds and weather. It’s important not to minimize the motion of the compressible working fluid in analyzing whether to expect the disposition of absorbed energy to be perceptibly influenced by incremental CO2.
All the best to you.
What is a missing here is that the temperature of the IR emitter (-17 deg C) and the absorber (15 deg C) are ignored. The models also use a flat Earth with no night-time. Huge errors thus come from these omissions.
“”The models also use a flat Earth with no night-time. Huge errors thus come from these omissions.””
Exactly. The assumption is that the surface area that is absorbing is only πr². As you point out, that means one must also assume πr² only is emitting and not 2πr². Somehow half the earth doesn’t emit.
No the assumption is that 2πr² is absorbing and 4πr² is emitting.
David, not dissing either you or Kevin….you are both on the same page, just different paragraphs…
If line by line absorption calcs say 2X CO2 is going to absorb 3 watts more energy calculated at TOA….and present 240 watts gives us 288 C here at surface, call it 1:1…a first approximation is that 3 watts would give about 3 degrees more warming…of course that’s wrong because of the T^4 nature of radiative emission, so you end up with around a degree of warming for 2XCO2 taking that into account as Modtran does, and others can make the number higher by other considerations as Kevin has described.
Your rather good evaluation of GOES sat info shows that there should be even less warming than predicted radiatively because column kinetic energy can dissipate those watts so easily to a slightly higher altitude where…umm…complicated…best look at your vid…keeping in mind that the watts have to go somewhere, and radiation is the only exit for heat from Planet Earth….
https://youtu.be/I0OCzxUyMqQ
Thanks for linking to that video.
In my opinion, everything one thinks about the influence of incremental CO2 on TOA longwave emission to space must be checked for consistency with what we “see” from space in those visualizations.
All the best to you.
You will get no argument from me involving the need for consistency with observations. It’s just that observations do not make the contribution of CO2 vanishingly small and I don’t see that convection and latent heat, the stuff of atmospheric dynamics, negates the contributions of CO2 to LWIR radiation. Maybe if we had specific instances where you see the dynamics as making CO2 vanishingly small.
Now, I do not dispute that clouds can make the contribution of CO2 a shorter range phenomenon than in clear air.
“…I don’t see…”
Not yet. But I’m not giving up on you.
“…that convection and latent heat, the stuff of atmospheric dynamics…”
The stuff of atmospheric dynamics also includes energy conversion, which I encourage you to explore more deeply for yourself.
“Maybe if we had specific instances where you see the dynamics as making CO2 vanishingly small.”
OK. At specific latitude rings (10N/S, 23.5N/S, 45N/S, 66.5N/S), for all hours at all longitudes for 2022, I plotted modified histograms of the ERA5 “vertical integral of energy conversion” hourly parameter, which is in units of W/m^2. The maximum influence of the 2XCO2 case (say ~4 W/m^2) is visualized on the plots to fit within a fraction of the width of the “0” index mark on the horizontal axis. Vanishingly weak. See the “Histograms” folder here.
https://drive.google.com/drive/folders/1PDJP3F3rteoP99lR53YKp2fzuaza7Niz?usp=drive_link
(Bonus: You might also be interested to note that at each of these specific latitudes, the mean of all the values is not zero. This relates to regions where there is a net generation of kinetic energy and regions where there is a net conversion to internal energy + potential energy. But that’s not the main point I am making right now.)
Thanks again for your post and your replies.
‘Your rather good evaluation of GOES sat info shows that there should be even less warming than predicted radiatively because column kinetic energy can dissipate those watts so easily to a slightly higher altitude where…umm…complicated…’
So close! Just substitute ‘non-radiatively excited GHGs can then spontaneously emit energy to space’ for “umm…complicated”. The main takeaways are that (outside the atmospheric window) the link between radiation from the surface and radiation to space is convection and the so-called GHE exists because convection scales linearly with T, not T^4.
Frank, I think you mostly missed some nuances of the “complicated” part…but 7 out of 10 marks for you and no down vote today…
High praise indeed! I’m sure we’ll have opportunities to exchange thoughts on some of the nuances down the road.
Once I had a degreed engineer pitch an ‘invention” of his that would use the waste heat from a manufacturing plant (rejected into a river) to make electricity. It was hard to wrap my head around because it his was very complicated, he presented heat tables with enthalpy values to support his case. I finally realized he was proposing an isothermal cycle to produce electricity.
I told him his cycle violated Carnot’s Law and he happily said “Yes!” We continued to listen and then politely showed him the door.
I like that you were “polite” about this…Donald Simanek, a retired physics professor from a small college (if he is still with us) and I used to correspond about perpetual motion machines. You might look for his “museum of impossible machines” online.
Thanks for that reference. It has been very entertaining.
Did Donald explain to you the difference between “energy” and “power”, Kevin? Maybe you should ask him about that. Then ask him if that difference is an important one, and also whether it constitutes “new physics”.
You could also ask him whether he thinks a measuring instrument can measure power coming from a colder object. Let us know what he says.
“I like that you were “polite” about this”
Just out of curiosity, does spouting ignorant nonsense, insulting all of the physicists in the room, and then running away count as “polite” where you come from? Who taught you to behave like this? And what exactly would you call “rude”?
Or are people only considered to be “polite” when they agree with you?
Excellent write-up Kevin. It was satisfying to find that my personal Modtran efforts and impressions of its usefulness match someone else’s ! The “Back-Radiation-Impossible” gang will likely be waiting for you in comments alley….
Waiting for me and not so “polite” as “oeman50” above, maybe. I appreciate your one-man efforts to pin-down NS!
NS comments are a good teaching situation…if one has time to answer Gish Gallops of incorrect claims….
“NS comments are a good teaching situation”
That may be the only correct thing you’ve ever said!
“The “Back-Radiation-Impossible” gang”
You mean the physicists? Perhaps you should sit down, DMac. You’ll sound a lot smarter that way.
Kevin, Thanks for injecting some science for discussion into this forum,
It’s my understanding that Modtran calculations presuppose constant thermal gradients. Temperature is introduced via Boltzmann’s Constant (kT) which presumes an isothermal profile of maximum entropy and zero dissipation. Local thermodynamic equilibrium arguments assume constant values for extensive parameters, but gradient-dependent complications remain unaltered.
The elementary linear dissipation model, d J/d T1 = J * [(1/T1) + (1/(T1 – T2))], depends only on measurable boundary parameters, encompassing internal complications of greenhouse gases, oceans, clouds, etc. (An alternative baseline?) Why the Stefan-Boltzmann constant for black body radiation should apply to thin atmospheric layers is an interesting exercise for the mathophilic.
Why the physics
Stefan-Boltzmann constantfor black body radiation should apply to thin atmospheric layers is an interesting exercise for the mathophiliccommon, but problematic, conjecture.Black body—even grey body—radiation does occur with thin or even thick atmospheric layers.
Gases, due to their relatively low density, emit and absorb radiation at highly specific, discrete wavelengths based on their molecular structure, as compared to condensed matter such as solids and liquids that emit radiation continuously across a portion of the EM spectrum as a function of temperature.
The distinguishing difference is that solids and liquids have their constituent particles (atoms or molecules) so closely packed together—for solids, typically in a “lattice”/crystal structure—that they experience constant collisions and overlapping interatomic/intermolecular fields, which “smear” allowable energy levels into a broad, continuous range.
You give a great summary of the properties of condensed matter that allow it to absorb and emit black body radiation, but then go on to say that gases, which completely lack these properties, can do so as well.
“The reason this matters is that the radiation mentioned in the quote, black body radiation, is mainly produced by oscillations in the electron density. Thermal motion makes the atoms oscillate and this produces changes in the electron density. This in turn produces oscillating electric dipoles, and those dipoles emit the black body radiation.
In a gas at everyday temperatures and pressures this can’t happen because the gas atoms or molecules are too widely spaced. There is no continuous distribution of electrons to oscillate, and as a result gases do not emit black body radiation. The radiation emitted by gases generally consists of sharp lines related to rotational or vibrational transitions.”
difference-in-thermal-radiation-between-condensed-matter-and-gases
Well, I simply never stated such. In fact if you bother to read (re-read ?) my post immediately above, you may note my statement: “Black body—even grey body—radiation does occur with thin or even thick atmospheric layers.”
I did note your statement, I just respectfully disagree that gas layers can be treated like black bodies.
You are correct.
It is very common in engineering work to assign an “effective emissivity” to a “depth” of CO2 molecules of known temperature and concentration so that you can use your T⁴ formulae to find out how hot your furnace wall is going to get (see Hottel charts)…that’s just one example…many more examples if you are calculating the effects of CO2 and water vapor on CCD sensors…you aren’t treating them as “black bodies”… you are expressing how far they deviate from a black body…
Alfred Schack?
Sorry, my mistake. I meant to state in my above post that thin or even thick atmospheric layers, being gases, cannot emit black body radiation.
It’s my own inattentiveness to carefully re-reading what I type out versus what I think I’m saying BEFORE posting.
You hit the nail squarely!
MODTRAN is a more general calculator than the free versions we get to work with, but our limited versions presuppose temperature profiles in a number of models one can choose. Thus radiance is dependent on these models. The only parameter we have at our disposal regarding temperature is a “surface temperature” and adjusting this adjusts the entire column similarly. Thus, MODTRAN does not presuppose gradients, but rather temperature models and then enforces gradients.
LTE as you might imagine intersects with a number of scientific and engineering topics; spectral lines of stellar atmospheres, planetary atmospheres, ovens, furnaces, and other engineering systems that involve atmospheres resulting from combustion, and finally in the design of remote measuring systems dependent on the schwartzschild transport function in Earth’s atmosphere. LTE is a hugely simplifying assumption in this as it presents a simple source function for radiance. Is it a good approximation? Well, emissivity and absorptivity are related to good accuracy by
where B is Planck’s function as long as a gas is enclosed in a cavity with specified boundary temperature. Now you can argue that the open surface of the Earth is hardly a cavity, but that quarrel applies to all surfaces, and just applying an emissivity less than one is maybe not a great approximation to non-cavity behavior. In particular it applies to the surface of the Earth itself where we use a value of emissivity so close to 1.0 that you’d say it approximates cavity radiation really well. So, as long as we know the absorptivity, which we do having measured it, and the temperature which we suppose is described by assigned model temperatures, then the LTE approximation lets us calculate radiance with accuracy.
With regard to your dissipation function, I have regarded it at several times, but I struggle to see how to apply it. One stumbling point is the boundary issue. I have described the upper boundary of this atmospheric analysis as complex, involving the surface in places, and not at a constant T.
Apparently, that includes you and your pages of handwavium.
Your diagrams are Carnot type heat engine cycles which describe converting energy to work, refrigeration, HVAC loops. Carnot’s kinetic theory of heat overthrew caloric and made many people mad. These graphics are not about radiation. They describe a Rankine cycle for steam turbines, Brayton cycle of combustion turbines. Ammonia and mercury have even been used as working fluids.
Attached is a graphic describing how heat transfer works and how Earth’s surface cannot be BB and “back” radiation does not exist.
As demonstrated by experiment, the gold standard of classical science.
For the experimental write up see:
https://principia-scientific.org/debunking-the-greenhouse-gas-theory-with-a-boiling-water-pot/
Search: Bruges group “boiling water pot” Schroeder
The “back radiation” derives from the fact that emission is isotropic, that is, equally in all directions. That means 20% in all directions, up, down, sideways and front to back.
One of the complicated issues that is always ignored is the inverse square law. If a cubic meter of CO2 radiates 10 W/m² at 10 meters of height, 10/10² = 0.1 W/m² at the surface of the earth. This is the same law that explains why the earth only receives 1360 W/m² from a very hot sun.
“One of the complicated issues that is always ignored is the inverse square law. If a cubic meter of CO2 radiates 10 W/m² at 10 meters of height, 10/10² = 0.1 W/m² at the surface of the earth. ”
Not trying to start another argument, but I think this is misleading. What matters is the total energy hitting the Earth.
Any CO2 emitting energy in all directions will hit a specific point on the surface, following the inverse square law. But it will also hit every other point on the surface within it’s line of sight. As the Earth takes up almost 50% of visible space from the CO2 moleclue, then around 50% of its energy will be absorbed by the surface.
Looked at from the other direction, if you are a square metre of the Earth’s surface, the amount of energy you get from a given molecule diminishes the further that molecule is from the surface, but you are also receiving energy from the entire visible atmosphere at the same time. The higher up the atmosphere, the more of it is in your line of sight. And as the atmosphere surrounds the earth the total energy hitting the surface is going to be close to 50% of the energy emitted by the atmosphere.
The inverse square law only applies to energy dispersing from a single point, and hitting a constant area.
“Any CO2 emitting energy in all directions will hit a specific point on the surface, following the inverse square law. But it will also hit every other point on the surface within it’s line of sight. As the Earth takes up almost 50% of visible space from the CO2 moleclue, then around 50% of its energy will be absorbed by the surface.”
Huh? 50% of the visible space would be half up and half down! At best it would be 1/2 of the downward half.
Nor does this actually make any 3D sense. That CO2 molecule actually emits a flux of roughly about 10^-32 W/m^2. Pretty damn small.
And if that doesn’t hit the surface of the earth at 90deg then the amount that gets absorbed is (10^-32)cos(Θ).
Flux is usually defined at a SURFACE, not by molecule. That means that you need to set his up as a 3D function which itself has CO2 density per volume per height as a component. And then integrate that over the volume involved. And to complicate it further, that flux will be per steradian so the height also determines the amount per m^2 that is received.
And what you are going to find is that climate science has made one more garbage assumption that CO2 is well-mixed globally so that the variance of CO2 per height-latitude-longitude can be ignored and some guess at an “average” value can be assumed.
Good luck in figuring this all out. I’m not going to hold my breath because I doubt you understand the 3D math involved.
Note: I still haven’t figured out how climate science has separated out the LWIR at the earth’s surface between that from H2O and that from CO2. That seems to be just one more guess based on a whole lot of “simplifying” assumptions – i.e. guesses. At least I can’t find anywhere how one piece of radiation is “tagged” so that the earth can differentiate it from another piece.
Bellman is correct. The inverse square law does not apply in the way you are suggesting because the receiving surface is itself in view of another surface; not a point source. Consider two concentric spherical surfaces with the inner shell analogous to the surface of Earth and the outer shell analogous to the atmosphere surrounding the Earth. Because the view factor of the outer shell to the inner shell is 1 that means whatever flux (in W.m-2) the outer shell emits towards the inner shell is what the inner shell will receive.
[1] The astute reader may initially question the above statement with the argument that the inner shell has a smaller surface area and so to conserve energy it must receive a higher flux (in W.m-2) than what the outer shell emitted. But that way of thinking, while admirable, is wrong because some of the emittance of the outer shell misses the inner shell and is received back by the outer shell itself. If the inner shell has the same radius as that of Earth and the outer shell is 10,000 m “above” it then the outer shell will receive approximately 0.3 % from itself and the inner shell will receive 99.7 % if the outer shell is an isotopic diffuse radiator. The total energy thus received by the inner shell is 49.8 %.
[2] If we consider F_OI the view factor from the outer shell to the inner shell and F_OO being that from the outer shell to itself then F_OI + F_OO = 1 because of the summation rule. With some geometry it can be show that F_OI = A_I / A_O where A_I and A_O are the areas of the inner and outer shells respectively. And with some trivial algebra F_OI = (R_I/R_O)^2.
[3] Because the radius of the Earth is significant compared to the altitude (above Earth’s surface) of the effective emission height we are usually justified in stating that 50% of the radiation from GHGs has a surface directed vector at least for first order approximations.
He is not correct and neither are you. If you would study Planck’s Theory of Heat Radiation, you would know what occurs.
From Innovative Tool to Determine Radiative Heat Transfer Inside Spherical Segments:
From this you can see that a perpendicular ray between the surface would have both angles equal to cos(0)=1 and that r² will be the minimum. It is possible to have the radiating surfaces be small and the distances between the two surfaces also be small. That allows one to assume a factor of 1 for the viewing factor.
You just forgot about the distance between the shells. Look at the diagram I showed. If we are talking about an area of 1 square meters radiating the distance of 10,000 meters becomes significant and one must deal with solid angles. The r² (inverse square law) becomes a factor.
r² = (10,000 m)² = 1 × 10⁸
That is a big number to be dividing by. Radiated energy from that height won’t have much effect on the inner shell.
The magnitude is also reduced by the Optical Path Length (OPL). My research indicates the OPL for the radiation at the line center is about 20 meters. That means at 20 meters the Intensity would be:
Iᵤₚ = I₀e⁻ᵗ = I₀e⁻¹ = 0.37 I₀,
then, on the back to the surface,
Idown = Iᵤₚe⁻ᵗ = Iᵤₚe⁻¹ = 0.37 Iᵤₚ
substituting, you obtain the amount reaching the earth,
Isurface = I₀e⁻² = 0.135I₀.
In addition to the path loss, after absorption by the land surface, the energy will be reradiated using the original spectrum which means a lot goes out the atmospheric window. That will make any effect back radiation has on the atmospheric temperature pretty small.
Says the guy whose understanding of geometry causes him to believe that the solar flux at TOA is 552 W.m-2 as opposed to the correct value of 340 W.m-2.
Your hubris often causes you defend one incorrect claim by stating a different incorrect claim. Perhaps you should look in the mirror before insinuating that the rest of us lack the level of study you have achieved.
Your own source is consistent with Bellman’s statement and inconsistent with yours.
No I didn’t. Any half reasonable reader would see that I explicitly stated the view factor as the square of the ratio of the two radii for the concentric spherical surface scene.
No it doesn’t. At least not for concentric spherical surfaces in which one is enclosing the other. A
Note that the scene of the Sun emitting toward the Earth is different than the scene of the atmosphere emitting toward to the surface. The more general application of view factors handles both scenes only one of which (Sun to Earth) simplifies to the inverse square law. The other (atmosphere to surface) simplifies to an equivalence such that the flux upon the inner surface is equal to the flux departing the outer surface regardless of the distance between them. That’s just how the geometry works out.
[1] Again…as I said above the inner surface area is smaller than that of the outer. If the fluxes are equivalent that means the inner is receiving less power and thus energy than the outer sent. This is compensated by the fact that the outer surface partially views itself so some of the power/energy it sends is received by itself.
Here is the reference I gave again. https://www.mdpi.com/2076-3417/13/14/8251
Look closely at what this is showing for concentric spherical surface. You will notice that the radii of the spheres is not the determining factor. The distance that a ray follows from one to the other is the determining factor and is presented as “r”, the distance between them.
Let me add some text from Planck’s Theory of Heat Radiation also.
Funny how the term of r², the distance between the two surfaces keep showing up in all of these references. Do you have any references to support your assertions?
The graphic is not showing concentric spherical surfaces. It’s not even showing spherical surfaces. And it is meant as a schematic for visualizing a general equation that can be used for any two surfaces regardless of shape or configuration.
BTW…notice that the authors of that publication do not analyze the concentric spherical surfaces scenario. That’s probably because it is a trivial scenario. Instead they focus on far more complex scenes.
And what happens when you do the full integration for the concentric spherical surfaces scene?
The r^2 term in general form depicted in the schematic you posted does not survive in the final solution for some of the scenes analyzed by the authors cited source.
Your own source is adequate enough. Follow the procedure. Do the full integration of the concentric spherical surface scene and see what happens with that r^2 term depicted in the schematic you posted.
“No I didn’t. Any half reasonable reader would see that I explicitly stated the view factor as the square of the ratio of the two radii for the concentric spherical surface scene.”
Except the radiation doesn’t begin and end at the center of the earth!
It begins on one sphere and ends on the other! The path loss is the distance between the two points, not the two points to the center point of the system!
“That is a big number to be dividing by. Radiated energy from that height won’t have much effect on the inner shell. ”
Just explain what you think happens to all the energy lost.
ONE MORE TIME!
Go look up the Divergence Theorem!
Received flux is a vector that has two components, a normal component and a tangential component. Only the normal component gets absorbed.
The tangential component skims the surface and is never absorbed by the volume below the surface.
Since the ideal earth we are considering is a sphere that tangential component never encounters any other piece of the surface and, therefore, can never be absorbed by the earth. It continues on into the atmosphere headed toward space.
There is no “lost” energy.
You are talking about energy hitting at a tangent. I’m asking about the energy “lost” due to the inverse square law.
If a piece of the atmosphere 1m directly above your head hits you with X amount if energy, and the same size of atmosphere 10m above your head hits you with X/100 energy, what happens to the other 99% of the energy? It can’t all be hitting the earth at a tangent.
As an aside, energy skimming of the Earth doesn’t go straight out to space, it has to go through the atmosphere again, with a good chance if being re-absorbed.
No energy is LOST! You simply can’t grasp physical reality at all, can you?
Put a gallon of water in a 8″ round pan. Measure the height of the water in the pan. The height will be proportional to the pressure being imposed on the bottom of the pan at any dA area.
Now, pour that water into a 16″ round pan and measure the height of the water in the pan.
Did you lose any of the water or do you still have a gallon in the 16″ pan?
Did the height of the water in 16″ pan become less than in the 8″ pan? Does that mean that the pressure on a point of dA in the 16″ pan is less then it is in the 8″ pan?
That gallon of water is similar to the intensity of the radiative flux. Spreading the water out over a larger and larger area giving less pressure at any dA point is similar to what happens to an emission ray of radiative flux and its intensity. As the energy in that steradian gets spread over a larger and larger area, the intensity at any dA point goes down. NO ENERGY LOSS!
The *total* energy when integrated over the spherical EM wave never changes. The intensity of that EM wave at any dA point goes down, however, as the radius of the sphere goes up.
NO ENERGY LOSS!
“If a piece of the atmosphere 1m directly above your head hits you with X amount if energy, and the same size of atmosphere 10m above your head hits you with X/100 energy, what happens to the other 99% of the energy? It can’t all be hitting the earth at a tangent.”
Nothing happens to the other 99%! Radiative flux between surfaces is calculated using this formula.
F(i->j) = (1/A_i) ∫_A(i) ∫_A(j) [cos(θ_i) cos(θ_j)]/πr^2 dA_i dA_j
“i” is a point on the originating sphere. “j” is a point on the receiving sphere. Break it down into pieces.
cos(θ_i) gives the dependence on emission angle of the radiation at the originating point. cos(θ_j) is the dependence of the emission angle at the receiving point. (1/πr^2) is how the intensity of the radiation changes over a distance (i.e. how the radius of the spherical wavefront changes).
So what? How does that affect the surface of the earth? Are we going to return to you claiming that the reabsorbed heat gets retained forever? And that somehow means the surface of the earth climbs back up the temperature gradient?
“NO ENERGY LOSS!”
Your usual shouty distractions. By lost, I clearly meant lost to the earth, because that’s your claim.
If a particle of the atmosphere is at 10m it will only have 1% of the energy hitting the Earth than the same particle at 1m. You seem to think that this means any energy from the atmosphere is irrelevant. You imply that 99% of the energy has been lost to space, and I’m telling you that it is not lost, it is the same amount of energy spread over a larger area.
Either you are claiming that most of the energy is lost to space, or you accept that half is absorbed by the earth. It can’t simply disappear.
Look at this diagram. See how the energy expands as the surface area of each larger sphere expands. That is the inverse square law. If the first “square” shown is 1 m², then a 1 m² one the next “sphere” will have less energy. A 1 m² on the 3rd sphere will have less. The factor is 1/r².
Let me disabuse you about discussing concentric spheres as if they are identical. The inner sphere is the earth and it radiates based on temperature as a Planck curve. The power radiated under that curve is calculated using the SB law which is derived by integrating the Planck energy at each frequency of the curve. The outer sphere you are discussing is actually a gas that radiates. Gases do not radiate power as an integrated Planck curve. Gases radiate at specific frequencies. You cannot add the energy at one frequency, say for CO2, to the entire Planck curve for the earth. In essence, that invalidates the use of the SB equation to analyze the system.
Here is a quick and dirty image I had CoPilot throw together to illustrate how small the CO2 back radiation is in the scheme of things.
” That is the inverse square law. ”
This is just getting embarrassing. How many more times do I have to explain that I understand the inverse square law, and am not saying it’s wrong. Just stop making these strawmen arguments and try to address my argument.
You called me out on my use of the inverse square law dude.
My reference is far from a straw man.
That’s because your use is wrong. It doesn’t mean the law itself is wrong.
Your own reference is contrary to your argument.
I said it was misleading. I explained why it was misleading. I did not say the law was wrong. It’s just not relevant to a discussion of how much energy is radiated to the earth from the atmosphere.
Let’s see what your artificial authority has to say:
Me: “Is the inverse square law applicable when considering energy radiating from the atmosphere to the earth’s surface?”
Copilot:
Short answer: Not in any simple or useful way. The inverse square law does not apply to energy radiating from the atmosphere to Earth’s surface because the atmosphere is not a point source, and radiative transfer in the climate system is governed by emission, absorption, and re‑emission within a dense, extended medium, not geometric spreading in free space.
Why the inverse square law doesn’t apply to atmospheric back‑radiationThe inverse square law requires three conditions:
Radiation from the atmosphere violates all three:
Inverse square behavior is only valid for point sources in a vacuum , and radiative transfer in media requires emission and absorption coefficients instead .
What does govern downward atmospheric radiation?Downwelling longwave radiation at the surface is determined by:
Each layer emits according to the Planck function, and the surface receives the integrated contribution from all layers weighted by their transmissivity.
This is why the surface receives ~300–350 W/m² of downward longwave radiation—comparable to solar input—despite the atmosphere not being a point source.
Why geometric spreading doesn’t matter hereRadiation from a nearby extended surface (like the sky) does not dilute with distance the way radiation from a distant point source does. In heat‑transfer terms, the atmosphere and surface have a view factor near 1: each sees almost all of the other. View factors replace inverse‑square geometry for extended surfaces .
Bottom lineAtmospheric radiation to the surface is a radiative‑transfer problem, not an inverse‑square problem. The inverse square law applies to point sources in free space, not to extended, absorbing, emitting media like Earth’s atmosphere.
I’ve had this argument with CoPilot before. It is mistaken because it gathers its information from incorrect sources.
Technically a “point source” would be a single atom or molecule. That is, a point with no size. Even Planck admits that this would invalidate any macro analysis of heat transmission. Consequently, he and all my thermodynamic books deal with AREAs. Specifically, dA (small area) in math terms. Planck defines dt as a small volume that is large enough to contain sufficient matter that atomic implications can be ignored in a macro analysis.
If you examine my image of a spherical surface it defines the AREA as dA₁ and dA₂.
Again, CoPilot assumes this from GHG consensus sites. It took me a while to get CoPilot to actually research sites that deal with the emittance of gases. Gases such as CO2 DO NOT emit a Planck function like a black body. Why do you think the folks that know something about radiation make the distinction between solid/liquid and gases when it comes to using the SB equation?
Here is a graph of the absorbance/transmittance of CO2.
As you can see CO2 has basically zero transmittance outside of the 15 um band. That is what the surface absorbs, but the surface DOES emit as a black body, consequently, the CO2 “back radiation’ is diluted to a very small change in the Planck distribution of the surface radiation.
CoPilot did not inform you of the Optical Path Length decay that CO2 radiation encounters on its way back to the surface also. Tim has mentioned this constantly to you. CO2 can’t distinguish between upward and downward radiation. It absorbs downward radiation just as easily as upward. This causes a decay to occur. You choose the OPL length you wish. Many folks talk about 10 to 20 meters. That means CO2 radiation at that height and above will never reach the surface. Why don’t you investigate that and see what you find. CoPilot can help you with that. It did me once I had it investigate the issue (it didn’t do it on its own).
I tried it with Gemini 3.5 Flash Thinking, Gemini 3.1 Pro Thinking, and Claude Fable 5 Effort Max (that ate through a significant portion of my allowance BTW). All of them said the flux on the inner surface was equal to the flux emitted by the outer surface. All of them did the full surface quadruple integration. All of them were consistent with each other. All of them said that the inverse square law does not apply to a concentric spherical surface scenario.
Maybe it is you that is mistaken?
The original point was that the surface is the source. The outer sphere receives less per unit area due to the inverse square law. As that lower value is radiated back from a unit area, the radiation expands in a spherical fashion. What strikes a unit area on the inner sphere at a normal angle is thereby less by 1/r².
Secondly, you haven’t addressed the fact that there is decay over the path from absorption and the fact that CO2 does not emit as a Planck curve, but only in a small band.
“The original point was that the surface is the source.”
Really? The comment I was responding to was:
You are still wrong, for the same reason. Energy radiated by the surface is radiating into an atmosphere that completely surrounds it. The only question is what proportion of emitted energy is absorbed by the atmosphere.
” What strikes a unit area on the inner sphere at a normal angle is thereby less by 1/r². ”
Again what do you think happens to all the energy not striking the Earth? You are either suggesting it accumulates in the atmosphere indefinitely, which is obviously not true, or that it all radiates into space. But why would more radiate into space because of the inverse square rule?
“you haven’t addressed the fact that there is decay over the path from absorption”
Operative word is “absorption”. It isn’t lost, it just moves from one molecule to another. At some point it has to leave the atmosphere, either to the surface or space. All you are doing is illustrating why the inverse square rule does not apply.
“You are still wrong, for the same reason. Energy radiated by the surface is radiating into an atmosphere that completely surrounds it. The only question is what proportion of emitted energy is absorbed by the atmosphere.”
Energy absorbed by the atmosphere is *NOT* the energy absorbed by the surface.
Did you *really* think that you could change the goalposts from the surface of the earth to the atmosphere and get away with it?
“Again what do you think happens to all the energy not striking the Earth? You are either suggesting it accumulates in the atmosphere indefinitely, which is obviously not true”
Stop making up things. No one *EVER* said that heat is retained by the atmosphere. YOU are the only one that ever thought “back radiation” represented retained heat causing a reversal of the temperature gradient instead of just slowing the cooling rate.
“or that it all radiates into space”
There’s that “retained heat” you are so fond of! If it doesn’t radiate back to space then it must be retained. So which is it? Pick one and stick with it!
“Did you *really* think that you could change the goalposts from the surface of the earth to the atmosphere and get away with it?”
I was responding to Jim doing just that, when he said “The original point was that the surface is the source.”
Try reading the context of my comments before accusing me of your own sins.
“Stop making up things. No one *EVER* said that heat is retained by the atmosphere”
Firstly, it’s energy being retained, not heat. And secondly, that’s the implication of what you are saying. That’s why I keep asking you to explain what you think is happening to the energy. Energy is conserved. If it isn’t being absorbed by the Earth it is either staying in the atmosphere or radiating into space.
“There’s that “retained heat” you are so fond of! If it doesn’t radiate back to space then it must be retained.”
Or it gets absorbed by the Earth. What else is there?
“All of them said the flux on the inner surface was equal to the flux emitted by the outer surface.”
Then they are all wrong.
Radiation from the inner sphere will always hit the outer sphere. Radiation from the outer sphere will *NOT* always hit the inner sphere.
You apparently just blew off my example to you of a golf ball inside a basketball. Some of the radiation from the inside of the basketball will miss the golf ball and just light up the far side of the basketball.
In addition, only the NORMAL portion of that radiation from the inner surface of the basketball will penetrate the boundary of the golf ball to warm it.
Whatever source you are looking at is treating the radiation as “bullets” that are only emitted normal to the emitting surface, i.e. the outer sphere, and striking the surface of the inner sphere at only 90deg.
That is *NOT* how an EM wave works.
What in Pete’s name has happened to the higher education system today that it doesn’t teach Maxwell theories on electromagnetics and 3D vector calculus any longer?
Put a marble on the lens of a flashlight pointing up to the ceiling. Does it block *all* of the light bouncing off the inner reflecting sphere of the flashlight? That inner reflecting sphere is just like the inner half of an outer sphere and the marble is like an inner spherical surface. Does *all* of the light being emitted from the flashlight hit the marble?
I point you once again to the general equation for the flux between two surfaces. The equation only becomes 1 for two flat, infinite planes.
*NOT* for two spherical surfaces. The 1/r^2 and the cos(A_i) and cos(A_j) factors in the equation prevent that from being the case.
Show us how that general equation becomes 1 for two concentric spheres. If you can’t then you need to ask those sources of yours to try and do it!
Lol.
“I’ve had this argument with CoPilot before.”
That’s sad.
“Technically a “point source” would be a single atom or molecule.”
Technically a point source would be a single point – smaller than any sub atomic particle. But it’s reasonable to treat any relatively small object as a point. It’s a reasonable approximation in most cases, but not when the object covers a large area close up area, or in this case when the object is completely surrounding you.
“If you examine my image of a spherical surface it defines the AREA as dA₁ and dA₂.”
What image? The one you keep showing is just showing two arbitrary surfaces, not specifically spherical. And the dA’s are not “the AREA”. They are a vanishingly small area of each surface – i.e. something that has a point as a limit.
“Again, CoPilot assumes this from GHG consensus sites.”
Naturally as no-one with any sense disagrees with it – but it’s still irrelevant to the question of how you are using the inverse square law.
“CoPilot did not inform you of the Optical Path Length decay that CO2 radiation encounters on its way back to the surface also.”
What has any of this to do with the inverse square law. This is just your attempt at a distraction. It doesn’t matter how long it takes for energy to move through the atmosphere – the point is that half of it will eventually hit the surface and half will go into space.
“This causes a decay to occur.”
A decay in what? All you are saying is that it takes longer for energy to move through an atmosphere then it does through a vacuum.
“That means CO2 radiation at that height and above will never reach the surface.”
Then the atmosphere is just going to keep getting hotter indefinitely, or else you think the energy is decaying, in violation of the first law of thermodynamics.
“Why don’t you investigate that and see what you find.”
Or maybe you could provide a reference – and then explain what this has to do with the inverse square law.
“Each layer emits according to the Planck function,”
Ask copilot how CO2 radiates vibrational energy as a Planck function.
The amount absorbed is a different topic.
We’re discussing your quote…
“One of the complicated issues that is always ignored is the inverse square law. If a cubic meter of CO2 radiates 10 W/m² at 10 meters of height, 10/10² = 0.1 W/m² at the surface of the earth. This is the same law that explains why the earth only receives 1360 W/m² from a very hot sun.”
Bellman and I are responding to what is being received because that is what you originally proposed. The inverse square law cannot be strictly applied to the atmosphere-surface scene like it is for the Sun-Earth scene.
BTW…check your units. Your quote states 1360 W.m-2 (which I happen to agree with), but since that is an annual average of an intensive property and since you believe those automatically inherit the units of the time period being averaged I think you meant to say 1360 W.m-2.year-1 like what Pat Frank insists on. Even though I disagree with it I still think it worthwhile to remind you to be consistent with the application of your rules if you are truly as confident of them as you let on.
“The amount absorbed is a different topic.”
NO, it *is* the topic at hand. It is only the absorbed part of the incidence ray that crosses the surface boundary and creates effects on the volume below the received dA.
1360 W/m^2 is the annual AVERAGE. That’s the assumed instantaneous value at any point in time during the year. The annual TOTAL would be 1360 joules/sec-m^2 * 31,557,600 sec = 43,000,000,000 joules/m^2 entering the system.
The *issue* is that each time that 1360 joules/sec-m^2 is used the total measurement uncertainty gets larger – since measurement uncertainty adds, it doesn’t cancel.
You keep wanting to say the annual average is 1360 +/- 0 W/m^2. It isn’t. It’s 1360 +/- u W/m^2. And “u” is *NOT* zero. It all stems from the fact that you can’t tell sampling uncertainty from measurement uncertainty. If “u” is not zero then the measurement uncertainty of the total goes up.
The sampling error tells you how precisely you have located the population mean. IT DOES NOT TELL YOU HOW ACCURATE THAT PRECISELY LOCATED MEAN IS. Each time you add another random variable into the distribution for which you are calculating the average using the estimated values of the components, the variance of the distribution goes UP. It is the variance of the distribution that is the measurement uncertainty. It is the variance of the distribution that describes the interval of values that can reasonably be assigned to the measurand. The measurement uncertainty is *NOT* the sampling uncertainty, the sampling uncertainty tells you nothing about the accuracy of the mean.
First…that’s not how your brother started the conversation.
Second…how are you going to correctly figure out how much is absorbed if you can’t even get the amount received correct?
I never said the uncertainty was 0 W.m-2.
Wait…why did you not state it as W.m-2.year-1? Afterall that is an average over a 1 year period you have vigorously defended the notion that averages inherit the time units in the denominator previously.
I don’t get me wrong. W.m-2 is the correct units. It would be awesome if you’ve changed your position in this regard.
“I never said the uncertainty was 0 W.m-2.”
You sure as all git out didn’t quote what it was! That only makes sense if it is zero!
“I don’t get me wrong. W.m-2 is the correct units. It would be awesome if you’ve changed your position in this regard”
The ISSUE is that if that 1360 estimate value has an uncertainty, e.g. the variance of the value over the year, then every time you use it measurement uncertainty grows.
You are standing on your head trying to justify that it doesn’t. And when you are called on that you just say “I never said it was 0”.
So if I state S = 1360 W.m2 the significant figure rules say that the uncertainty is 0 W.m-2?
The issue is that your statements are inconsistent at best. You tell everyone that an average must include the time dimensions of the period in the denominator, but then don’t include time dimensions in the denominator yourself when using the same average Which is it? Do they include time dimensions in the denominator or not?
It is not lost. You still don’t understand why the radiation intensity over a constant area of 1 square meter from the sun diminishes as it travels to earth do you? Radiation expands spherically. As it expands, the energy is spread over larger and larger surface areas.
If you look at a sphere 1 meter from the source and an area of 1 square meter on that sphere is receiving 100 W/m². But, guess what 1 square meter on a sphere two meters away from the source will receive? 25 W/m²! How about 3 meters away? Again, one square meter on a sphere 3 meters away will only receive 100/9 W/m².
Golly, this is basic stuff. It applies to all isotropic EM waves from 1 Hz to 10x Hz.
Energy isn’t lost, it is simply spread over a larger and larger area.
“You still don’t understand why the radiation intensity over a constant area of 1 square meter from the sun diminishes as it travels to earth do you?”
If you stopped this patronising spiel and tried tried to figure out what I’m saying, you would realise that I’m not in any way disagreeing with the inverse square law.
“Energy isn’t lost, it is simply spread over a larger and larger area.”
Which is the point. When you have an atmosphere surrounding a sphere, the are that is being spread over is still all the Earth.
If you want to imagine a single point of atmosphere emitting a sphere, then what is relevant is how much of that sphere intercepts the surface. When you are looking at the Earth from the sun, then this diminishes with the square of the distance. Move the Earth twice as far from the sun, and it’s diameter as seen from the sun, halfs and so it’s surface is only a quarter, hence it only gets a quarter as much energy. This is the same weather you look at energy per square meter, or total energy over the whole surface.
But when the energy source is very close to the surface, this is not the case, because of the curvature of the earth.
Say your source is 1m from the surface, and say directly under it the surface is receiving 100W/ m². Now move the source up to 10m above the surface, and now the surface directly under it is only receiving 1W/m². This reduction is because the same amount of energy is being spread over a wider area. But obviously the total energy received by the Earth is the same, because in both cases the Earth takes up half the sphere of the emissions. Any energy radiating in a slightly downward direction is going to hit the Earth.
Therefore you have to conclude that the area of the surface that the 10m source hits is a lot larger than the area hit by the 1m source. And this is what you should expect from the horizon effect. The surface of the Earth visible from 1m above ground is much less than the surface from 10m. And the same is true even if you you are 10s of kilometres. The energy is spread very thin, but over a very wide area. If almost half your field of view is taken up with the Earth, then almost half the total energy must hit the Earth. It can’t just vanish.
And somehow my statement about 1/r² is wrong?
Be careful about how you spread that extra energy around the inner sphere or you will have more, much more, energy entering every point on the inner sphere than the outer sphere is actually radiating. Remember, even two spherical bodies side by side will also spread the radiation from a dA area around to the other sphere. How do you handle that?
“And somehow my statement about 1/r² is wrong? ”
This is just getting ridiculous. No, the inverse square law is not wrong. What’s wrong is you claiming it effects the total energy absorbed by the earth.
“energy entering every point on the inner sphere than the outer sphere is actually radiating.”
No you won’t. You can’t increase energy by spreading it over a wider area.
” How do you handle that?”
Perhaps if you explained exactly what your problem is I could help.
“The energy is spread very thin, but over a very wide area. If almost half your field of view is taken up with the Earth, then almost half the total energy must hit the Earth. It can’t just vanish.” (bolding mint, tpg)
See the bolded part. It may *HIT* the earth but only the normal component can cross the boundary! The rest skims along the tangent to the sphere that is the Earth AND DOES NOT GET ABSORBED! And since it is a sphere the tangent part jut heads back out toward space!
As usual you STILL HAVEN’T BOTHERED TO LOOK UP THE DIVERGENCE THEOREM!
You keep hanging on to the garbage assumption that the atmosphere and the earth are two parallel, flat, infinite plains. Reality is just not in your in your wheelhouse at all.
“ It may *HIT* the earth but only the normal component can cross the boundary! The rest skims along the tangent to the sphere that is the Earth AND DOES NOT GET ABSORBED! And since it is a sphere the tangent part jut heads back out toward space!”
Utter rubbish!
You have *still* not bothered to look up the Divergence Theorem, have you?
In words: “The total outward (inward) flux of a vector field through a closed surface equals the volume integral of the divergence of that field inside the surface”
Heat flux *is* a vector field.
∫∫∫_v (V · F) dV = ∫∫_∂V F ·n dS
(bolding indicates a vector, n is the normal vector)
F ·n gives you the normal component of the vector field.
In other words, the normal component of a vector field determines what crosses a closed boundary; the tangential component never contributes to net flow across the boundary.
A closed surface can only *see” the normal projection of a vector field. It cannot “see” the tangential component. The tangential component just slides along the boundary and never crosses the boundary.
This is no different from Gauss’s Law for an electric field.
∫∫_∂V E· n dS = Q_enclosed/ ε_0
I could go deeper but you really need to study this on your own.
(hint: F · n = |F| |n| cos(θ), since n is the unit normal vector |n| = 1.
“But when the energy source is very close to the surface, this is not the case, because of the curvature of the earth.”
“Say your source is 1m from the surface, and say directly under it the surface is receiving 100W/ m². Now move the source up to 10m above the surface, and now the surface directly under it is only receiving 1W/m². This reduction is because the same amount of energy is being spread over a wider area. But obviously the total energy received by the Earth is the same, because in both cases the Earth takes up half the sphere of the emissions. Any energy radiating in a slightly downward direction is going to hit the Earth.”
It’s the curvature of the earth that causes the radiation to miss it! As the surfaces move apart the subtending angle that the earth presents gets SMALLER. Not only is the emission sphere diluting by 1/r^2 but the amount of the EM wave intercepted by the earth gets smaller too!
The subtending angle is proportional to R/d where R is the radius of the earth and “d” is the distance between the centers of the spheres. As “d” increases the subtending angle gets less and the area presented to the spherical EM wave gets less. As altitude of the CO2 that is radiating gets larger the less of its radiation hits the earth and the radiation that does hit the earth is more diffuse. It’s a double whammy.
You *really* have no grasp of 3D geometry do you?
” As the surfaces move apart the subtending angle that the earth presents gets SMALLER.”
Not when you are close to the surface. Have you ever been up a tall building or a plane? The higher up you are, the further you can see. The angle you can theoretically see is close to 180°.
The Earth’s diameter is ~12750 km, the troposphere is on average ~13km high, so the effect of the altitude of the CO2 emitter is insignificant!
Path loss is calculated from originating point to destination point.
Radiation from CO2 does not flow to the center of the earth and then back to the surface. The center of the earth is irrelevant to the EM spherical wavefront.
Radiation from CO2 travels the 13km path to the surface with no excursions in between!
That spherical wavefront from CO2 in the atmosphere has an increasing radius and, therefore, an increasing surface area. The intensity per m^2 gets less as the surface area increases. It gets less by 1/r^2. Go look up the term steradian. The intensity per steradian stays the same, but not per m^2.
The center of the spherical wavefront is at the emitting particle and the radius of the spherical wavefront gets larger as it travels from CO2 to the earth’s surface.
The flux received by any point on the earth’s surface is 169 x 10^-6 less than when it started out from the emitter.
ROFL! A single CO2 molecule is not a point source?
You are no better at 3D surfaces than Bellman.
Being received does *NOT* mean that it will all be absorbed!
1.Radiative flux from a surface is W/m^2. That radiative flux when received at a distance is based on the steradian, in other words the fact that the spherical area covered by the ray increases with distance. That is the whole basis of the inverse square law. It’s why the radiative flux from the sun doesn’t hit the earth at the intensity at which it is emitted.
2.As expected, you’ve never bothered to go look at the Divergence Theorem or if you did you didn’t understand it. According to the Divergence Theorem only that portion of the radiative flux vector that is normal to the surface is absorbed, the rest just skims the surface tangentially and does nothing.
3.The TEMPERATURE rise that the absorbed flux causes is based on the specific heat capacity of volume intercepting the flux. The SPECIFIC HEAT CAPACITY is *not* a constant. It is a highly variable function, f(x,y,t) over space and time.
Every single thing I said is correct. You’ve refuted nothing. All you’ve done is parrot the garbage simplifying assumptions made by climate science. “you can ignore distance for radiative flux”. “Everything received is absorbed” “the earth is a homogenous mass with a constant absorptivity”
Says the guy whose 3D geometry understanding leads him to believe that solar irradiance at TOA is 552 W.m-2 as opposed to the correct value of 340 W.m-2.
Lay it out for me straight. In the concentric spherical surface scenario where the inner and outer surfaces have radii of 6371 km and 6381 km respectively and where the outer surface is an isotropic diffuse radiator with an inward radiant exitance of 400 W.m-2 how much flux is received by the inner surface?
“how much flux is received by the inner surface?”
I gave you the integral to figure that out. It’s a complex double integral where the flux intensity is based on 1/r^2 = 1/10km^2 and the vector value normal to any point on the inner sphere is dependent on cos(A_i) and cos(A_j).
Don’t blame me if you can’t figure the integral out.
For a point on the inner sphere receiving the flux vector from a point on the outer sphere that is perpendicular to the point on the inner sphere, i.e. the entire vector is normal to the surface to the point on the inner sphere, the flux intensity received is 400 W/m^2 / 10000meters^2. The flux intensity at that point would be 4×10^2 * 1×10^-4^2 = 4 x 10^-6 W/m^2. That’s .000004 W/m^2. The flux intensity received from any other point on the outer sphere (that can see the point on the inner sphere) by the same point on the inner sphere will be even less because the path from outer sphere point to the inner sphere point is longer. What do you think the cos(A_i) is for?
3d Vector Analysis just seems to escape you.
I already figured it out. Thanks. It is actually the reason I challenged your brothers statement to begin with.
Is that your answer to my question…” In the concentric spherical surface scenario where the inner and outer surfaces have radii of 6371 km and 6381 km respectively and where the outer surface is an isotropic diffuse radiator with an inward radiant exitance of 400 W.m-2 how much flux is received by the inner surface?”
Do you want to think about it some more before committing to 0.000004 W.m-2?
“ According to the Divergence Theorem only that portion of the radiative flux vector that is normal to the surface is absorbed, the rest just skims the surface tangentially and does nothing.”
Nonsense!
How many times must the divergence theorem be explained to you before it sinks in?
∫∫∫_v (V · F) dV = ∫∫_∂V F ·n dS
F ·n is the portion of the flux vector that can cross the receiving boundary.
It’s EXACTLY the same as Gauss’s Law for electric fields.
∫∫_∂V E· n dS = Q_enclosed/ ε_0
In essence, the two equations say that the component of the field vector that is parallel to the surface never crosses the closed boundary. Only that portion that is normal to the boundary can cross the boundary.
E is the electric field vector, F is the heat field vector. No difference in the math.
It is *not* nonsense.
“50% of the visible space would be half up and half down! At best it would be 1/2 of the downward half. ”
Show your workings. How high up would you have to be for the Earth to only occupy 25% of your field of view?
“Good luck in figuring this all out.”
As so often you can’t see the woods for the trees. All you need to ask yourself is what happens to all the energy lost due to the inverse square rule. Does it just vanish – breaking the conservation of energy, or does it all just go into space?
“Show your workings. How high up would you have to be for the Earth to only occupy 25% of your field of view?”
You’ve still not figured out how radiative flux works, have you? Nor do you understand 3D geometry at all!
It simply doesn’t matter what angle the earth subtends when trying to calculate the joules entering a dA piece of the earth’s surface based on its distance from the source and the angle of incidence.
The temperature rise caused by the absorbed flux is based on the composition of the volume below each dA piece of surface. It is *not* a constant. Even the surface of the sea has a variable specific heat capacity associated with the dA piece you are looking at. All kinds of things make up the specific heat capacity of that piece of dA, things like salinity, contaminants like microplastics, plankton, even wave action. And that specific heat capacity is a complex function of (x,y,t).
You are trying to justify the same kind of garbage climate science assumption that the earth can have an “average” temperature that determines the radiative flux at any point in space and time only here you are assuming that the earth is a homogenous volume with an average specific heat capacity that will determine the “average global temperature” based on it absorbing everything it receives (i.e. a black body).
Reality just doesn’t seem to be where you and bellman live.
Put a sphere around the middle of a cube. The 1/r² applies to any “ray” that strikes the earth. Certainly, a ray that is perpendicular, i.e., the shortest distance, will have the highest intensity. However, as the length of the other rays that encounter the earth, simple geometry tells you that the distance increases, and therefore, the 1/r² factor reduces the intensity.
Your statement about 50% indicates you still do not understand how a body radiates. It does so at full power in ALL directions. Up, down, sideways, front, back. A body that radiates 100 up also radiates 100 down. A body that radiates 100 directly north also radiates 100 south. It is an ever expanding sphere. You pick an x,y,z direction and the same amount is radiated in any other x,y.z direction.
Put the half down and half up out of your mind. It is 100% percent in ANY AND ALL directions from a point.
And before you go off on a tangent, this is predicated on having a body that is large enough to contain sufficient molecules such that random directions of radiation from each molecule all add up to all directions. In other words, it is a macro solution, not a micro solution. Planck discusses this in the first few paragraphs of his Theory of Heat Radiation. You should download it and study it.
You’re still discussing the behavior of radiation from a point source upon the receiving surface. The inverse square law does not apply to a surface source emitting toward a receiving surface in complex scene. Instead you have to use the concept of view factors.
[1] You can actually derive the inverse square law as a special case of view factors for a point source. You can do this by constructing the scene as concentric spherical surfaces where the limit of the radius of the inner surface approaches zero. When you do this you’ll get a view factor F = A * cos(θ) / 4πr^2. A few more steps conserving energy results in the received radiation q on the outer shell to be proportional to inverse square of its radius such that q ∝ 1/r^2.
“ou’re still discussing the behavior of radiation from a point source upon the receiving surface. The inverse square law does not apply to a surface source emitting toward a receiving surface in complex scene. Instead you have to use the concept of view factors.”
View factor is irrelevant. The surfaces are not flat, neither the expanding sphere of radiative flux or the receiving surface. The angle of incidence at the receiving surface determines the normal component of the radiative flux vector which is what is absorbed and the distance from the originating point to the receiving point determines the intensity of the radiative flux per square meter at the receiving point.
All the view factor provides is the portion of the radiation that is leaving surface “i” that reaches surface “j”.
F(i->j) = (1/A_i) ∫_A(i) ∫_A(j) [cos(θ_i) cos(θ_j)]/πr^2 dA_i dA_j
It is a function of the angle of incidence leaving and the angle of incidence receiving and is based on the inverse square law (1/πr^2). Please note carefully the [cos(θ_i) cos(θ_j)] term. For spherical surfaces this makes contributions at any point on the receiving surface decrease very fast as you move around the originating sphere.
The subtending angle only determines the limits of integration for A(i) and A(j). As the subtended angle goes down the area of the originating source that can “hit” the receiving source gets smaller and smaller. If you will it is the “shadow” the receiving surface casts on the originating surface. As the distance between the surfaces increases the radius of the shadow on the originating source shrinks.
The temperature rise at dA_j is based solely on the vertical component of the received flux and the specific heat capacity of the volume below dA_j. There is no simple linear “average” value. That also means that the radiative flux *out* from the surface of the earth is not a simple linear average either. Especially when considering it is a function of time as well as space. To complicate it further the amount of radiating CO2 in any volume is also a function of (x,y,t). Assuming a well-mixed CO2 is just one more garbage simplifying assumption by climate science.
You keep wanting to present radiation from CO2 in the atmosphere as being from a flat, infinite plane and being received by a flat, infinite plane representing the earth’s surface. For two flat, infinite planes F(i->j) = 1.
I’m sorry, it just doesn’t work that way in reality. It’s just one more garbage assumption by climate science.
Here is a quote from Planck in case believe what you have said.
I’m curious…what possible rationalization are you using to state that the portion of radiation leaving the source surface and arriving at the destination surface is not relevant?
I’m curious…when discussing two concentric spherical surfaces with radii of 6371 km and 6381 km how are you visualizing that in your mind that makes you think those surfaces are flat and infinite?
You seem to have the same blind spot for radiative flux that you have for photons. Neither are bullets, they are EM waves with magnitude and direction. Nor is the earth a black body. It simply does *NOT* absorb everything it receives regardless of the angle of incidence. Because the earth is not a black body it is *not* isothermal either. Meaning the temperature at any point on the surface is dependent on the conditions at that point and can be different than the temperature at any other point. T is a function of (x,y,t).
*I* am *NOT* visualizing those as flat and infinite planes, YOU ARE.
Any point on the outside sphere emits at a rate of W/m^2 IN ALL DIRECTIONS. The amount of that radiation that gets absorbed at any point on the inside sphere depends on the angle of incidence on the inside sphere. The temperature that is engendered at that point on the inside sphere is a complex function based on the composition of the volume below that point.
I gave you the integral needing to be solved for just PART of the system. It is a complex function of cos(θ_source), cos(θ_destination), r^2, the area of the originating source involved A_i, and the area of the destination source involved A_j. It doesn’t even include the specific heat capacity of the destination point or its absorptivity.
If you’ll just think about it for even a second it should be obvious. Put a golf ball at the center of a basketball. Pick a point on the surface of the basketball. It will radiate all the way from 0deg to 180deg away from that point, say at a flux value of 100W/m^2 for any angle you pick. How much of the radiation at 0deg or 180deg (i.e. horizontally away from the point) will hit that golf ball? How much of the radiation at π/2deg (i.e. vertically away from the point) will hit the backside of the golf ball?
Much of the radiation from that point on the basketball is going to miss the golf ball and just exit through the surface of the basketball somewhere.
Now stick a volleyball inside the basketball or even a deflated basketball that is just a little smaller than the inflated one. How much heat energy is going to be absorbed by the inside deflated basketball? Certainly not all of it. Rays emitted at 45deg from the point on the outside basketball are going to hit the inside basketball at what angle? 45deg you say? What is the normal vector at 45deg? Could it be about 0.7 of the total magnitude? So only 100 W/m^2 * 0.7 is going to get absorbed! The tangential component just skips right over the surface of the inside basketball and exits through the outside basketball.
“6371 km and 6381 km ‘
Where in Pete’s name do you get these values? The radiation from a CO2 molecule doesn’t start at the center of the earth. And the distance from the surface of the earth to the center of the earth is not the path the radiation travels.
The path travelled by the radiation is 6381km – 6371km = 10km! = 10,000meters. 10 x 10^3. (10 x 10^3)^2 is what? Could it be 100 x 10^6?
So the flux intensity at a point on the surface of the earth from a point at 10km altitude is going to be x/(1 x 10^8). And the maximum that can be absorbed will be [x/(1 x 10^8) * cos(θ).
Do you want to have a genuine conservation or do you want to continue with your typical gaslighting?
Again…the atmosphere is not a point source. It is a surface radiating from its entire surface area; not a single point. The amount of radiation a point on the surface is receiving is the integration of all points radiating toward it from the atmosphere; not a single point.
Again…the atmosphere is a finite spherical surface enclosing the surface which is also a finite spherical surface. Neither of those are flat or infinite. So whatever bizarre thought process you had that made you believe they were flat and infinite is wrong.
No, I am discussing it from the standpoint of a surface radiating.
Here is how Planck describes it.
Look at the other reference. dA1 and dA2 are not point sources. They are small areas on each surface.
Look at what Planck said about absorption or scattering caused by the intervening space. It should be treated as essentially a vacuum. Yet “r” still applies.
Show us some resources that describe how distance between emitter and absorber does not affect the value of intensity. I’ve shown you mine, now show me yours.
Your own source. Just apply the concept of view factors (which your source uses) to the scene of concentric spherical surfaces in which the outer encloses the inner. In that scene the distance between the inner and outer surfaces has zero effect on the flux the inner surface receives from the outer surface assuming the outer surface is an isotropic diffuse radiator. The received flux on the inner surface equals the emitted flux from the outer surface. This is true regardless of the distance between the inner and outer surfaces.
Right…so why are you and your brother hyper focused on the radiation received from a single point on the source surface and ignoring all of the other points that must be integrated from the source surface that are also emitting toward the receiving surface?
Because that is how it is analyzed. I am not harping on anything. I am showing you references that explain how to do so. You have failed to provide any reference to support your position.
My position is that the distance is important to determining the loss. Both sources show basically the same equations that include a distance “r” that is squared. That agrees with both my sources.
If you want to maintain that the EM wave intensity does not diminish with equal areas at both the emitter and the absorber, that is your prerogative. It should move you to show the math of how the r² disappears rather than just claiming that it does.
The folks here can judge for themselves if my sources are correct or if you are. End of story.
Does anything you said address my point?
“The 1/r² applies to any “ray” that strikes the earth.”
No. “A” ray won’t. It’s the density of rays that gives you the inverse square law. The rays are spreading out in all directions. The further they go the fewer there over any area.
“Put the half down and half up out of your mind. It is 100% percent in ANY AND ALL directions from a point.”
Yes, that’s the point. The emissions go in all directions, almost half of them hit the earth. Even from a high altitude, the Earth will have a large angular diameter, meaning a high proportion of the total energy emitted will hit the Earth.
Using an online calculator I estimate that at 50km the Earth will have a diameter of 166°. That’s over 45% of the total sphere, so a random ray has a 45% chance of hitting the Earth.
Most CO2 is going to be well below 50km, and so will have a slightly larger target.
“No. “A” ray won’t. It’s the density of rays that gives you the inverse square law. The rays are spreading out in all directions. The further they go the fewer there over any area.”
The intensity of the ray is based on steradians, not radians. It isn’t a matter of “fewer” rays.
From any point source, the EM wave is a spherical wavefront. The intensity the EM wave presents at a distance is constant for a steradian. It is *NOT* constant for a per m^2 area. The surface area of a sphere is 4πr^2. Note carefully the r^2 term. The surface area of the EM wave from that point source that is subtended by a steradian is different at 1m than it is at 10m. If the point source is emitting at 100 W/m^2 then the intensity of that SINGLE RAY at 10 meters distant will be (100 W/m^2)/ 10^2 = 1 W/m^2. That happens solely because of the expanding sphere that single ray (i.e. EM wave) represents.
The *total* flux received at a point on the receiving surface is a complex integral.
F(i->j) = (1/A_i) ∫_A(i) ∫_A(j) [cos(θ_i) cos(θ_j)]/πr^2 dA_i dA_j
Note the 1/πr^2 term, that is the inverse square law. The limits of integration for A(i), the originating source, is the area that can “see” the receiving source. I.e. bdgwx’s viewing angle. It is related to R_j (the radius of the receiving surface) divided by D, the distance apart the surfaces are. In essence it is the shadow the receiving surface casts on the originating surface, that determines the horizon for a point on the originating source. Points on the originating surface that are “below” the horizon (i.e. outside the shadow) can’t “see” the receiving surface. Conversly, only the receiving source area that can see the originating source are able to receive radiation from the source. But at high angles of incidence, i.e. cos(θ_j) -> 90deg, the normal vector -> 0.
And this *all* ties into the temperature of the earth at any point in space and time. It’s all a complex function of ε(x,y,t) and T(x,y,t)^4. No amount of measurement today gives enough granularity to come even close to an accurate picture, meaning the uncertainty of radiative balance is unknown probably to the tens digit let alone the units digit!
This whole thing of radiative balance is a mess, chock full of guesses at “average” values just like with the “average global temperature”.
Since you obviously don’t understand what I posted, here is the photo again and its explanation. Note the phrase “Spherical Segments”.
From Innovative Tool to Determine Radiative Heat Transfer Inside Spherical Segments:
You haven’t shown how distance traveled is removed. In fact, I don’t understand how your calculation has anything to do with the inverse square law at all.
” emission is isotropic”
Emission of what, though, Jim? Remember, you told us that “radiation is energy”. As did your idiot clone Tim, and also Willis, and even Andy May for that matter. So you wouldn’t be trying to describe that emission in Watts, would you?
“CO2 radiates 10 W/m²”
Ah, yes, of course you would. Do we measure energy in Watts, Jim?
MODTRAN assume that all energy transfer is occurring via radiation. (Not an issue, that is what it was designed to do.)
However, in the real world “energy” reaches TOA and from there space, via two mechanisms. Radiation and convection. As the “surface” temperature increases compared to “space”, the amount of convection will also increase.
Convection is important. At the surface, in fact, it may carry a majority of heat upward. Yet, at some point the majority of heat is carried by radiation as bulk heat transport depends on mass and mass is rapidly declining with altitude. The cross over in mid-latitudes appears to be around 2km.
You can see by my essays over the years that I have consistently criticized local radiation arguments for ignoring vertical transport by mass motion, and also ignoring horizontal transport from tropics to poles.
Kevin, I can’t help but think that you and I are on the same page, but are somehow talking past each other. For example, I agree with what you say here, but would kindly point out that declining mass with altitude is no less a factor for radiative transport than it is for ‘bulk heat transport’. In other words, while non-radiative excitation is still prevalent at altitude, the lower mass at that level greatly increases the relative amount of radiative (spontaneous) energy emission that can escape to space.
I’m not sure what MODTRAN is but suspect it’s a hocus pocus box nobody understands. Met too many young engineers who trusted complex algorithms/spreadsheets they could not explain.
255 K is not measured, it’s S-B calculated average balance of Earth’s OLR surface w 30% albedo.
1,368/4=342*.7=240=255K
Remove GHE water vapor and Earth bakes in 250F solar wind. Core punches through empty ocean floors flooding basin w dark lava and albedo goes to .05 and temp to 394 K.
“Their results range from 1.86 to 2.49 degrees centigrade per doubling of CO2. Lewis and Curry (2018)[5] found similar values using similar means.”
Ceteris paribus.
And there goes the whole ball of calculational wax.
Such precision, too! Nothing wrong with those models…
/s
But the ability to extract a pertinent point and work with it, Ceteris paribus, is the essence of science, isn’t it?
Exactly. The concept of ceteris paribus is so important to the scientific method that scientists go to great lengths to setup an appropriate control group as a baseline during experimentation specifically to achieve it. Pat Frank is not the first person I’ve seen question a core principal of the scientific method here nor will he be the last.
The climate state is not static.
A ceteris paribus model of a dynamic physical system is not predictive of future states.
“Pat Frank is not the first person I’ve seen question a core principal of the scientific method here nor will he be the last.”
Yet another example of you not knowing what you’re talking about, bdgwx.
A projectile is part of a dynamic physical system. I can still invoke the concept of ceteris paribus to predict how a change in launch angle will change the distance traveled. Just because you are incapable of doing it doesn’t mean everyone else is as equally incapable.
Says the guy who thinks TSI is 1360 W.m-2.day-1 (with the extra day-1 unit) as opposed to the actual value of 1360 W.m-2.
“A projectile …”
So for you, a projectile equates to the climate. Wonderful. The acuity of thought I’ve come to expect of you, bdgwx.
“… guy who thinks TSI …”
I don’t think that.
It’s what you said here. And it wasn’t some knee-jerk comment to be argumentative or a typo. You defended it repeatedly and vigorously. You accused me of being disingenuous and not understanding how averages work because I accept that TSI is 1360 W.m-2 (without the extra day-1 or year-1 unit you insisted must be there). You even insinuated that NASA’s stated value of TSI was wrong because they used units of W.m-2.
So if you truly believe that is TSI is 1360 W.m-2 (without the extra day-1 or year-1 units) then why did you go to great lengths to convince everyone otherwise?
“It’s what you said here.”
That discussion concerned the mean of TSI every day, noting that the TSI orbitally varies ±3% over the year.
Noting the mean of the TSI is 1360 Wm⁻² d⁻¹ is neither the constant of your “TSI is 1360 W.m-2.day-1” nor the per-day irradiance total your rendition implies.
Leave it to you to misrepresent the discussion. Or, to be charitable, maybe you didn’t understand what was being argued, and still don’t.
Funny…you originally said and I quote “Solar flux is a constant at 1360 W/m^2 every single second.”
I had to explain to you that daily flux values are not constant to which you responded “Whoop-de-do.”
So you do still think the annual mean of TSI is 1360 W.m-2.day-1? Yes? No?
You can see quite clearly that the annual average TSI commonly referred as the solar constant S was how the discussion started.
Specifically ~1360 W.m-2 (without the extra day-1 or year-1 units) is the result when you average measurements of zenith solar flux over the course of a year. The example I gave was of daily measurements that are aggregated such that S = Σ[Fi, i:{1 to 365}] / 365 or if over a 20 year period it would be S = Σ[Fi, i:{1 to 7305}] / 7305. I also mentioned that you can use the mean value theorem of calculus to compute it using a full integration of Earth’s orbit.
That’s when you stated it was actually 1360 W.m-2.day-1.
Maybe it is you that didn’t understand the discussion. This is your out. If you don’t want people to think you believe that the annual average TSI is ~1360 W.m-2.day-1 then now is your chance to clarify your position.
“… to which you responded “Whoop-de-do.””
To which I responded, “Whoop-de-do. That doesn’t change the fact that you grafted ‘-day’ onto your numerator to serve your specious argument.”
Let’s see: your rendition of the exchange includes a lie of omission.
The whoop-de-do refers to the irrelevance of your TSI argument to your antecedent opportunistic inclusion/exclusion of the denominator dimension in a mean, as in “N is unitless.” , or not, whichever momentarily suits you.
You wrote, “That’s when you stated it was actually 1360 W.m-2.day-1.”
What I wrote was, “When the solar constant is given by specifying the information is its mean, 1360 W/m^2/day, it is correct every day.”
You crafted an obvious misrepresentation of my statement, bdgwx.
Your argument here is hopelessly dissimulated, whether by intent or by inadvertence.
First…I never used units of W.m-2.day for solar flux. The first mention of it was…wait for it…by you in this post. And as you can see from the conversation you did so because you did the integral wrong.
Second…given that you arbitrarily added time units in the denominator of other flux values and that you said “When the solar constant is given by specifying the information is its mean, 1360 W/m^2/day, it is correct every day.” and your vigorous challenges of any claim that it is 1360 W.m-2 (without the extra year-1 or day-1 units) how else is any supposed to take your commentary as you believing the units really are W.m-2.day-1 as you said?
If you don’t actually think an average flux over a period of time inherits those time units in the denominator then 1) step up and say so and 2) go back correct all of the times in which you did it. Ya know what…just doing #1 alone would be a massive contribution to the discussion and all it takes is a single sentence.
Bdgwx, you’re clearly unable to distinguish a statistic from a physical variable.
Though the difference has been discussed repeatedly in these conversations, your ignorance in this matter appears irrefragable.
Climate modelers display the same immovable inability.
You also show no understanding of the need for, or method of, dimensional analysis, or of retaining dimensional units through a calculation.
You seem determined to retain this state.
You’re welcome to continue blowing hot over the retention of dimensional units in the denominator of a mean.
You’re further welcome to continue accusing me of egregious errors in my continued use of standard methods.
But you’re wrong. You’ve been wrong all along, and you appear determined to remain so.
Finally, a word of advice – ignorant hostility is no recipe for a constructive outcome.
He’s a troll, Pat.
(I know, stolen quote, but apropos.)
I don’t know what to tell you Pat. You say the annual average TSI has units of W.m-2.day-1. I gave you multiple opportunities to reconsider your position and you chose not to. At this point I don’t really have any other choice but to accept that when you said W.m-2.day-1 you actually believed it to be W.m-2.day-1. And if you don’t then you’ve done everything you possibly could to convince me otherwise.
You fail to deal with physical quantities. 1360 W/m² tells you what from a dimensional analysis standpoint? Is that determined by averaging what, seconds, hours, days, months, years, decades? How you determine a value is important. Even more important is the variance that goes along with the time frame from which you determined the value. 1360 W/m² ifs fine for the portion of a day, that is in view of the sun. What is it for periods when the sun isn’t visible? How do you denote between them.
This whole argument perspective of a dimension is not very useful in the scheme of things. More important is how the value is used in a calculation. Is it wrongly used in a calculation? If it is, then show how it affects the mathematical outcome. You should be dealing with the value and its uncertainty, not how it is labeled.
“You say the annual average TSI has units of W.m-2.day-1.”
What I wrote was, “Note that without the per-day unit in the denominator, a claim that the solar constant is, e.g., 1360 W/m^2 would be wrong every day of the year, but one.
“When the solar constant is given by specifying the information is its mean, 1360 W/m^2/day, it is correct every day.”
You crafted an obvious misrepresentation of my statement, bdgwx.
Your argument is specious, no matter whether tendentious or from ignorance.
That is, you’re either unable or unwilling to distinguish a statistic from a physical variable.
Honestly, bdgwx, I’ve no interest in convincing you of anything.
Why the equivocation? The annual average TSI or solar constant S has a single defined set of units. Stop beating around the bush and just state what you think it is.
No equivocation, bdgwx. When the solar constant is given by specifying the information is its mean, 1360 W/m^2/day, it is correct every day.
Try something new: stare at that sentence until you figure out what it means. Clearly, it’s escaped you for 10mos, 9 days.
I see units of W.m-2.day-1. Is that what you’re sticking with?
So, you don’t see anything else there, Got it.
No I don’t. I certainly don’t see W.m-2. And anytime I say the annual average TSI or solar constant S has units of W.m-2 you argue with me.
“No I don’t.”
So, then, you’ve been arguing a chimera.
Evidently, you’re so hot to find a mistake that you’ve become unable to read for meaning.
The original point centered on the daily transition of TSI across a year, and the general relevance of its daily mean value. Not the annual mean.
All in the larger context of you persistently denying the necessity of denominator units in a calibration uncertainty mean.
All this to, one hopes, illuminate the meaning for you. Please do not endeavor to litigate denominators again.
I’m hot about your claim that averages over a period of time inherit time dimensions in the denominator. I don’t know how to make that any more clear.
I’ve never mentioned a daily mean. I’ve only ever talked about it’s long term mean which is ~1360 W.m-2. I gave 3 different examples of computing the mean: 1) by summing up 365 measurements taken daily over 1 year, by summing up 7305 measurements taken daily over 20 years, or by doing an integration of TSI over one orbital cycle. All of these yield units of W.m-2. You argued with me about it saying I was wrong.
And this all came up because you took an annual average mean of another flux and arbitrarily added year-1 units to it. As a way of demonstrating the absurdity of that I brought the annual average TSI into the discussion.
Do you or do you not think the annual average TSI or solar constant S has units of W.m-2?
“I’m hot about your claim that averages over a period of time inherit time dimensions in the denominator.”
Well, then, I suggest you take a high-school science class; at a level certain to include dimensional analysis in its curriculum.
Conversation beyond that is pointless.
The rest of your argument stems from your apparently irrefrangible inability to understand that dimensional units are always retained.
It is all about a mathematician’s in ability to distinguish between numbers and physical quantities. I suspect he never learned how to find the limiting reactant in a chemical reaction.
Honestly, Jim, I don’t understand the difficulty. But by all evidence, it afflicts pretty much all climate modelers, too.
Not a single one of them tells you to put the time dimensions in the denominator of an averaged value.
You and the Gorman’s are the only people I’ve ever seen make this claim…ever. And in the case of the Gorman’s even they don’t seem eager to abide by your rule consistently as evidenced by posts in this very article.
The issue is – did using a per day label affect the outcome? If not you are nitpicking something of no value in judging the value of the study to dismiss it.
Like it or not, iterative processes add uncertainty at each iteration. It is no different than laying boards end to end. The uncertainty adds, maybe in quadrature or maybe not.
Likely, you’re not looking in the right place. Retaining dimensions is absolutely critical in scientific calculations
Maybe this will help: How math in science is different from math in math (2013)
Followed by, Symbols in science. Specifically, this bit:”The population density of trout in a stream is
f(x)=20[(1+x)/(x²+1)]
“where f is measured in trout per mile and x is measured in miles.
x runs from 0 to 10. Write an expression for the total number of trout in the stream. Do not compute it.
“Although this is a perfectly reasonable problem for a math class (to see if you know how to write an integral), in a physics class, any student writing an equation like this would lose significant credit for failing to be consistent about dimensions (and we’ve shown this with an “unhappy” emoji).
“If x is measured in miles, this means that the “1” in the numerator has to mean “1 mile”, not just “1”, and the “1” in the denominator has to mean “1 mi2“, not just “1”, So in this equation “1” is used twice, but means two different things, neither of which is “1”.”
Dimensional units are always retained in scientific calculations. A physical mean must always retain the dimensional unit in the denominator, so as to retain the original physical meaning.
Your own source is inconsistent with your claim. No where in either of those citations does it say to arbitrarily inject time units in the denominator when computing an average over a period time or some other domain. In fact, neither of them even discuss averages. Nonetheless, if you follow the procedure set forth in those references then you’ll see that an average retains the same units as the quantity itself. The most convincing way to prove this for yourself is to use the mean value theorem for integrals.
“arbitrarily inject time units”
It is to laugh.
“you’ll see that an average retains the same units as the quantity itself.”
Not when the divisor has dimensional units.
Maybe this will help. If not, then your case is hopeless.
It does help. I can now see how some of your confusion arises.
The example you presented is one of a rate. When you divide a change in quantity by a change in time you get a rate. In this particular case the result happens to represent an average making it an average rate. This brings time units in the denominator because it is a rate
The context we are discussing is one of state. When you integrate a quantity of state (ie. intensive property) and divide by the domain you get an average state by way of the mean value theorem of integrals. This does not bring time units in the denominator because it is a state.
You have conflated states with rates. The annual average TSI is a state. The annual average RMSE of the flux provided by Lauer & Hamilton is for an state. The monthly average temperature is a state. This is why those qualities do not have time dimensions in the denominator. Like I said… you can prove this for yourself using the mean value theorem for integrals.
You still don’t get it.
Dimensional units must be retained. End of story.
“When you divide a change in quantity by a change in time you get a rate.”
Averaging a time-varying state quantity per unit time, yields mean/time-unit. Mean/time-unit is is a statistic. Not a rate.
A rate is a velocity; change with time; a vector. Not a statistic. A quantity time-average is not a vector. It is a statistic.
Dividing a series of measurement errors by number of measurements, yields (mean error)/measurement. That quantity is a statistic, not a rate. And “measurement” is retained as the denominator.
Density is an intensive property. But it varies with temperature. Suppose one wanted a daily average density over a seasonal year. Following 365 measurements of density, taken at a varying ambient temperature, the mean is (1/365 days)*Σ(density)ᵢ = ((mean density)/day. It’s not a rate. It’s a statistic.
Not that I want to get involved in another endless argument, but bdgwx is correct – the units if an average have to be the same as the thing you are averaging.
“Dimensional units must be retained. End of story.”
Yes, but if you are dividing by units with a dimension to get an average, it’s because you are dividing a sum or an integral that has dimensions of that quantity multipled by the dimensions of the divisor.
E.g. if you want to know the average velocity of an object, you can integrate it’s velocity over a period of time and then divide by.time. But that integral is velocity × time, i.e. distance. And the average is velocity × time / time = velocity.
“the mean is (1/365 days)*Σ(density)ᵢ = ((mean density)/day.”
If you are summing densities then you are not dividing by time, just by the number of measurements. Using a density/day as the units if the average makes no sense. It implies the longer you left the object the denser it would get.
“that integral is velocity × time, i.e. distance. And the average is velocity × time / time = velocity.”
Velocity is distance/time; a vector quantity. Velocity×total time = total distance; a unit quantity. Velocity×total time/total time = (total distance)/(total time) = (distance/time) mean.
Your first-step cancellation of time violates the correct operational order.
The final (distance/time) mean is a statistic of mean displacement per unit time. It’s not a velocity, even though, as a statistic, it has the same units as velocity.
Again, the vdt integral yields distance. Total distance/total time = mean distance/unit time; a statistic, not a velocity.
“It implies the longer you left the object the denser it would get.”
It implies no such thing. It provides the mean density one may use on any given day. One must have no sense of the physical reasoning that frames the statistic to reach your mistaken interpretation.
You’re right. It is another endless argument. But the endlessness obtains because you and bdgwx persist in re-raising it.
bdgwx is incorrect. So is your analysis, which is careless about physical meaning.
For whatever reason, neither you nor bdgwx evidence any grasp of how to reason as a physical scientist.
“Velocity is distance/time; a vector quantity.”
I should have said displacement rather than distance, but it makes no difference if we are talking about velocity or speed, the point is the units.
“Velocity×total time = total distance; a unit quantity.”
I’m not sure what you mean by a “unit quantity”. Displacement is a vector, distance is a scalar.
“Your first-step cancellation of time violates the correct operational order.”
What? You are multiplying and dividing, you can do that in any order.
“It’s not a velocity, even though, as a statistic, it has the same units as velocity.”
As I keep saying, an average velocity is not a velocity. But it should have the same units.
“It provides the mean density one may use on any given day.”
Would that be different if you gave the result as “(mean density)/year”, or “(mean density)/second”?
“For whatever reason, neither you nor bdgwx evidence any grasp of how to reason as a physical scientist.”
If being a “physical scientist” means not understanding dimensions, then I’m glad I’m not one. Could you provide a reference in any physical science document (excluding your own works) that give average over time in units that include dividing by time?
In everyday speech, people will sometimes say thing like, the average height is 1.8m per person, which is sloppy but can help to emphasize exactly what’s being averaged. But I would hope “physical scientists” wouldn’t confuse that with an actual physical unit.
“What? You are multiplying and dividing, you can do that in any order.”
Actually the standard order of operation taught in the U.S. is called PEMDAS. It stand for:
Parentheses,
Exponents,
Multiply (left to right),
Divide (left to right),
Add (left to right),
Subtract (left to right)
In many cases multiplying and dividing can be done in any order, but not always. Follow PEMDAS and you will have support for your answer when the order does matter.
“Actually the standard order of operation taught in the U.S. is called PEMDAS.”
Use whatever mnenomic you want, but it doesn’t mean you do M before D or A before S.
A quick search gave me this:
P Parentheses first
E Exponents (powers, square roots, and so on)
MD Multiplication and Division (left-to-right)
AS Addition and Subtraction (left-to-right)
Divide and Multiply rank equally (and go left to right).
Add and Subtract rank equally (and go left to right)
https://www.mathsisfun.com/operation-order-pemdas.html
That’s why x/yz = (x/y)z and not x/(yz).
And this is only really about the convention for writing equations. It has nothing to do with the question of whether multiplying by time before dividing by time makes any difference.
In general you are correct. I should have prefaced that with the addition that proper use of parentheses is needed to properly indicate correct grouping. This becomes necessary as students in stem begin doing experiments using formulas that require proper grouping.
The main point of having a defined order of operation is to avoid having too many parentheses.
It still has nothing to do with there being a correct order for multiplying and dividing by time.
“What? You are multiplying and dividing, you can do that in any order.”
Not if you’re calculating a physical quantity. Order of operation is critical.
“Could you provide a reference in any physical science document (excluding your own works) that give average over time in units that include dividing by time?”
Here you go.
Go away, Bellman. You’ve got no clue.
A chat bot is your your scientific source? Fair enough, it isn’t saying the units of an average capacity are different than the actual capacity. In each case the units are number divided by time.
But to be sure I asked it about getting the average velocity using integration and what units the average would have.
It concludes
———-
The integral represents the total displacement of the object during the time interval.
Dividing this displacement by the total time interval yields the average velocity.
Average velocity is a vector quantity with units of displacement per time, commonly meters per second (m/s).
———————-
https://www.studocu.com/en-us/ai/project/019f9dd9-b121-72d8-9318-1572f37b678a?sc=4qotca3n&view=creation&id=612649ee-53b9-4e7e-96bb-264b222d132c
“The final (distance/time) mean is a statistic of mean displacement per unit time. It’s not a velocity, even though, as a statistic, it has the same units as velocity.”
They still don’t buy that an average is a statistical descriptor of a distribution of data. It is *not* a measurement. The average cannot be measured, it can only be calculated as a statistical descriptor. And if the distribution is not Gaussian, the average value is not as useful as the median and mode statistical descriptors!
The only way that “average” value actually makes sense is to include the variance and/or the 5-number statistical descriptors if the distribution is not Gaussian. All of these thing are STATISTICAL DESCRIPTORS of a distribution of measurements, they are *NOT* measurements of anything physical.
An “average” of distance per time is *NOT* an actual rate. It is a statistical descriptor of the distribution of velocities (i.e. rates) over the distance and time.
If you travel 2 miles in 2 minutes you have averaged 1 mile/min. If you travel the first mile in 0.5minutes and the second mile in 1.5 minutes you still averaged 1 mile/min. But where would you measure that average velocity? Again, that 1 mile/minute is a statistical descriptor of the data distribution, it is not an actual physical rate that can be measured. (1,1) vs (5,2/3) – two different data distributions with the same average but one has a variance of 0 and the other a variance of 9. The average and variance together describe the data distribution, but are *NOT* measurements.
Every time you add a measurement into the distribution, e.g. from a step-by-step model, the measurement uncertainty increases. Unless you are a climate scientist that believes that all measurement uncertainty is random, Gaussian, and cancels.
Does any of this have anything to do with topic of the units of an average? It the average velocity m/s or m/s²?
“ the units if an average have to be the same as the thing you are averaging.”
The average is a statistical descriptor – *NOT* a measurement of a measurand existing in physical reality.
It only describes the distribution of the data, it does *NOT* represent an actual measurement of anything.
“Using a density/day as the units if the average makes no sense.”
And using temperature/day as the units makes sense?
density – intensive property
temperature – intensive property
If the average density/day makes no sense then why does the average temperature/day make sense?
As Pat noted, density (D) is temperature dependent. And temperature is time dependent. D(T) –> D(T(t))
Daily temperature, D, is no different. D(T) –> D(T(t))
Either density/day and temperature/day makes sense or neither do.
Which is it?
“The average is a statistical descriptor ”
Yes, it’s a statistical descriptor that had the same dimensions as the thing being averaged.
“And using temperature/day as the units makes sense?”
Nope. If you are averaging temperature the unit of the average is temperature.
“If the average density/day makes no sense then why does the average temperature/day make sense?”
It doesn’t.
“Either density/day and temperature/day makes sense or neither do.”
Neither do.
You do realise you are agreeing with me and bdgxw in this, and disagreeing with Pat Frank?
“Yes, it’s a statistical descriptor that had the same dimensions as the thing being averaged.”
So what? It is a statistical descriptor that describes the distribution of the measurement values. It is *NOT* itself a measurement.
“Nope. If you are averaging temperature the unit of the average is temperature.”
You just contradicted yourself. The dimension of the thing being averaged is collected over a period of time. So you have the measurement AND the time period involved in the average. e.g. ΣT/hour, etc.
If the thing being averaged is temperature, you divide it by the time period over which the temperature is measured.
ΣT/#seconds, ΣT/#hours, ΣT/#month, etc.
If you do not provide the dimension for the time period being used then you have no way to know if 70F is a daily average, an hourly average, a monthly average, etc.
Even worse, without the variance associated with the average over a time period you have no idea of what the distribution of temperatures actually looks like. The average by itself is not a sufficient descriptor of the distribution – unless you are a climate scientist or a supporter of climate science today where it always assumed everything is Gaussian.
Your grasp of physical science has not improved over the time you have spent on WUWT.
“Neither do.”
Ahhhh! So now you have joined those who say the global average temperature is a scientific hoax. The daily mid-range temperature used in formulating the global average temperature is of the dimension T/day.
The mid-range temperature is actually of the form (Tmax/day + Tmin/day) since the temperatures are collected over the diurnal period of a day. Thus the average is (Tmax/day + Tmin/day)/2 –> T/day.
If you don’t know the dimension of “day” then you can’t distinguish whether the average is taken over a day, a week, a month, a year, etc.
My guess is that your cognitive dissonance will now say that density/day is illegitimate while temperature/day is legitimate.
“You just contradicted yourself.”
Only in your delusional mind.
“If the thing being averaged is temperature, you divide it by the time period over which the temperature is measured.”
No you don’t. If you are just summing things you divide by the number of measurements, not the time. Think about it if you are capable, suppose you take suppose you take 10 measurements over a 5 hour period. Do you think dividing by 10 hours is going to give you a correct average?
The only time you consider the time is if you are doing a weighted average based on time, and the you are dividing temperature multiplied by time. Or you are using an integral.
Look at your favorite example in TN1900. You are dividing the sum of daily measurements by the number of days, but the result is an average in °C, not °C/day.
“If you do not provide the dimension for the time period being used then you have no way to know if 70F is a daily average, an hourly average, a monthly average, etc.”That’s why you specify the time interval when relevant. E.g.the annual average temperature – not temperature/year.
“If you do not provide the dimension for the time period being used then you have no way to know if 70F is a daily average, an hourly average, a monthly average, etc.”
You are either incredibly stupid or are trolling. Try to argue the point without lying about what I’ve said.
“My guess is that your cognitive dissonance will now say that density/day is illegitimate while temperature/day is legitimate.”
Your guesses are as accurate as everything else you write. I don’t know how many more times I can say that temperature/day is not the correct units for an average of temperature.
Here is an average temperature: 70F.
How will you use that temperature in any functional relationship? How about a functional relationship describing average enthalpy for a week? For a month? For a year?
Is that 70F average temperature a constant that can be used for enthalpy regardless of what the functional relationship is?
“Do you think dividing by 10 hours is going to give you a correct average?”
The average is a STATISTICAL DESCRIPTOR of a distribution of measurements. That means you HAVE to know what interval the measurements are taken over. Forget temperature. If I tell you that the average density of topsoil is x kg/m^3, do you need to know what area was measured in order to use it? Can I go grab a pickup load of topsoil out of my backyard and use “x” to determine the load weight on the pickup? Can I go grab a pickup load of topsoil from a soybean field a county away and use the same average figure to determine the load weight on the pickup?
If I tell you that the average height from the table to the culet for an 8mm brilliant cut sapphire is 6mm does that apply to *all* 8mm brilliant cut sapphires or only to some of them? Do I need to know which ones it applies to in order to size the mounting for a brilliant cut sapphire I might order from LapidaryA as opposed to LapidaryB?
If I am sizing the lateral field for a septic system at a new house can I just use the average percolation for the county? Or do I need to know what the average rate is for the pasture where the house is going to be built?
If I tell you the average temperature is 70F do you need to know what interval of time that was measured over? A week? A day? A year?
Total and utter malarky! It has NOTHING to do with a weighted average. You are throwing crap against the wall hoping something will stick!
READ THE TN FOR MEANING AND CONTEXT! The measurement average is being used as the BEST ESTIMATE FOR THE VALUE OF THE MEASURAND – Tmax! The assumptions in the TN make the exercise into multiple measurements of the same thing, Tmax, and not multiple measurements of different things over a period of time! THERE IS NO TIME INTERVAL INVOLVED IN THE EXAMPLE! It’s a Type A uncertainty example using experimental observations in a repeatable environment!
Judas Priest man! When are you going to stop cherry picking stuff that you haven’t studied for meaning and context?
I’m not lying about *anything* you’ve said. I’ve replied to direct quotes of what you post.
Then how do I know when you post an average temperature if it is from measurements taken over a day, a month, a week, a year?
If you tell me that the average temperature for Latitude X/LongitudeY is 70F what can I use it for? As an annual average? A millennial average? A daily average?
All you are doing is expressing your belief in the meme of “numbers is just numbers”. They have no relationship to physical reality.
Just a few, hopefully, final comments as you are in full rant mode now.
“I’m not lying about *anything* you’ve said. I’ve replied to direct quotes of what you post. ”
My direct comment was that neither Density/Time, or Temperature/Time made sense as units of the average. Your reply was
If you think that’s a reasonable deduction, then I really don;t want to know what’s going on in your head.
“READ THE TN FOR MEANING AND CONTEXT!”
Stop shouting – if you think you have a reasonable point to make try making it without the caps lock key.
I did read the TN for context – the context being what was the correct unit for an average of temperature. You claimed that dividing by a number of days meant the average was Temperature/Time. The example makes it clear that you are wrong, as do every other source I can find. An average is always in the units of the thing being averaged.
Somehow you’ve convinced yourself that finding an monthly average is not finding a monthly average just becasue you’ve treated it as a repeatable set of measurements. The mental twisting you must go through in order to avoid the obvious is painful to watch. And you have the gall to accuse overs of cognitive dissonance.
“Then how do I know when you post an average temperature if it is from measurements taken over a day, a month, a week, a year?”
Because the description of the average should tell you. An average temperature with no context is meaningless. It’s always an average temperature of something.
Look at any of the UAH updates.
That tells you that it’s the average for the month of June 2026, along with additional information, such as it being global and for the lower troposphere. What you do not do is say that it was +0.46°C/month.
If you have the anomaly for 2025 you say it’s the annual anomaly for 2025, not claim that it’s units are °C/year. It it’s for a specific region you say what that region is, not claim it’s units are °C/year/m².
“My direct comment was that neither Density/Time, or Temperature/Time made sense as units of the average. Your reply was”
Hmmmm……
Here is the exchange:
—————————-
tpg:“If the average density/day makes no sense then why does the average temperature/day make sense?”
bellman:It doesn’t.
tpg:“Either density/day and temperature/day makes sense or neither do.”
bellman: Neither do.”
——————-
Not much deduction needed.
you then go on to say:
————————-
Think about it if you are capable, suppose you take suppose you take 10 measurements over a 5 hour period. Do you think dividing by 10 hours is going to give you a correct average?
————————
You are admitting that you need to know the interval over which the average is taken in order for it to make any sense.
If you don’t include the interval then how do you know it was a 5 hour average vs a 10 hour average?
If I tell you the average measurement is 70F how do you know if that is a daily average, a weekly average, a monthly average, etc?
It’s why you have to say the average is 70F/day, 70F/week, 70F/month, etc.
But you just keep saying that 70F is the proper way to state the average – no interval needed.
No lying about what you said. No deduction about what you said is needed. It’s all there in plain English.
In one spot you say the dimension of the average doesn’t need to specify the interval over which the average is calculated and then in another spot you say you need to know the value of the interval over which the average is calculated – i.e. the interval dimension as in 70F/day, etc.
You just say what you need to say in the moment with no regard for being consistent.
“Not much deduction needed. ”
I said that temperature/time is not the correct unit for an average temperature, and you deduced that I’m saying global average temperature is a hoax. This says more about your abilities of deduction then it does about me. As always you filter everything through your own warped mind -it’s just so much easier for you to debate your own phantoms than what people actually say.
“You are admitting that you need to know the interval over which the average is taken in order for it to make any sense.”
There you go again. That’s not what I’m saying at all.
“If you don’t include the interval then how do you know it was a 5 hour average vs a 10 hour average? ”
All the answers would be found if you actually read what I said. If it matters if you averaged over a 20 hour period rather than a 10 hour one, you should state that in your description. I’ve no idea how you think quoting a figure per hour would help.
“It’s why you have to say the average is 70F/day, 70F/week, 70F/month, etc.”
Could you provide a single example if someone saying that in a scientific document?
No, it’s the average of Tmax! The value of a measurand. The assumptions make those measurements into a repeatable set – i.e. taken over a short period of time, basically meaning no interval is involved. The environment in which the measurements were made remained the same.
Again, READ THE TN FOR MEANING AND CONTEXT!
Stop cherry picking stuff you refuse to understand.
No, I did not claim that for TN1900. I claimed the exact opposite!
tpg(bolding mine, tpg): “The assumptions in the TN make the exercise into multiple measurements of the same thing, Tmax, and not multiple measurements of different things over a period of time!” )
READ THE TN! The assumptions make this into multiple measurements of the same thing (Tmax) under repeatable conditions!
Why do you absolutely refuse to read the TN rather than just cherry picking from it?
TN1900, Ex2:
——————
“The daily maximum temperature r in the month of May, 2012, in this Stevenson shelter, may be defined as the mean of the thirty-one true daily maxima of that month in that shelter” (bolding mine, tpg)
“The equation, ti = r+Ei, that links the data to the measurand, together with the assumptions
made about the quantities that figure in it, is the observation equation.”
——————
The measurand is Tmax. No time interval involved.
READ THE TN!
“No, it’s the average of Tmax!”
For the month of May.
You can witter on about repeatability and assumptions all you like – the fact is you are dividing the observations by the number of days, and so by your logic the units should be °C/month.
Do you really think the average is the one and only temperature you will ever get in May, rather than the average of all days in that month?
“No, I did not claim that for TN1900. I claimed the exact opposite!”
Read what I said for context and meaning. I said you claimed that whenever you divided by the number of days you had to include time in the average. I know you did not claim that for TN1900 becasue you would then have to admit it’s a nonsense claim. It’s the very fact that as soon as you are provided with evidence you are wrong you twist and turn trying to find a reason why it’s a special case. You just don;t make a convincing argument for why it’s different.
Your logic is that you need to say this is °C/month or else how do you know the measurand is for one month rather than a year. Why do you think this is different in this case? How are you supposed to know what 25.59°C means, if you don’t have the units telling you it’s for one month?
“The measurand is Tmax. No time interval involved.”
You’ve literally just quoted them saying it’s the daily temperature in the month of May 2012 and it’s defined as the mean of 31 daily maximums. What do you think month and daily mean if they are not time intervals.
Still – you could stop this argument by actually providing a reference that quotes average temperature over a time interval in units of °C/time.
Even Pat Frank doesn’t do that. Throughout his paper he describes temperature uncertainties and anomalies in terms of °C. E.g. in the conclusion:
https://www.mdpi.com/1424-8220/23/13/5976
You still can’t distinguish between 1.calculating the “best estimate” of the value of a measurand and 2. averaging the distribution of a set of measurands over time.
It’s being divided by the number of observations! It’s no different than taking 30 measurements of a crankshaft journal using a micrometer. It’s a Type A experimental situation where the best estimate is the average of the number of measurements and the measurement uncertainty is the variance of the distribution of the measurements. NO TIME INTERVAL involved.
“You can witter on about repeatability and assumptions” just about says it all. You simply refuse to read for meaning and comprehension. Your reading comprehension skills are basically zero.
It is the repeatability and assumptions that make the Example what it is! You want to redefine the assumptions in order to fit the example into your limited understanding of metrology!
I can’t even tell where you are getting this from. It has nothing to do with TN1900. And you have absolutely no understanding of averaging the daily temperature profile over time, You believe that (Tmax + Tmin)/2 is a good representation of the average temperature for each day!
I quoted you directly. The context and meaning is right there in the quotes.
“I said you claimed that whenever you divided by the number of days you had to include time in the average.”
I know that is what you said. And then you said you don’t have to let the reader know what the time interval involved is. Just quote the 70F with no time interval shown as a dimension.
TN1900 is an example of how to do a Type A experimental calculation for measurement uncertainty. It is using a set of observations to determine this. Those observations have to meet repeatability conditions and that is what the assumptions in TN1900 are doing. Repeatability requires taking the measurements over a short period of time, not over 30 days.
For some reason you can’t admit that.
I’ve given you the assumptions in TN1900. You refuse to admit they are there in order to make the example into showing how to do a Type A measurement uncertainty calculation. It *is* an EXAMPLE, not a special case.
Asked and answered. Your lack of reading skills is showing again.
“The compilation of land- and sea-surface LiG uncertainty yield a 1900–2010 global air-temperature record anomaly of 0.86 ± 1.92 °C (2σ)” (bolding mine, tpg)
ROFL! This is not specifying the time interval? Unfreakingbelivable!
“ROFL! This is not specifying the time interval? Unfreakingbelivable!”
You don’t think 1900-2010 is a time interval?
“You don’t think 1900-2010 is a time interval?”
*YOU* were the one that said:
“Even Pat Frank doesn’t do that. “
not me.
Sorry. Admits your usual hysterical act I missed the question mark.
So you accept the average is over a time interval, but don’t think he should have used units of °C/century, or whatever.
You really need to try to explain exactly what you want to say. Are you saying temperature/time is the units that have to be used, or are you saying the “/time” is a shorthand for specifying the time period, and the units are just temperature?
“Because the description of the average should tell you. An average temperature with no context is meaningless. It’s always an average temperature of something.”
The description is given by the dimension!
70F/day, 70F/month, etc.
You are trying to imply that:
are two different things!
Unfreakingbelivable!
What does 70F/day actually mean?
Temperature does not accumulate over time. Temperature/time implies a rate of warming. E.g. global temperatures have increased by 0.16K/decade.
Still, as you think a daily temperature per day is a meaningful unit, you should have no problem finding any number of examples of such a unit being used in the scientific literature. I’ve given you two examples where it is not used, TN1900 and the UAH monthly updates, so why don’t you provide an actual example demonstrating what you are claiming.
Really? There are lots of HVAC engineers that will disagree with this.
What in Pete’s name do you think a “degree-day” is?
Here are three references:
1.Degree‑days — Encyclopedia of Snow, Ice and Glaciers
2.A global degree days database for energy‑related applications, Energy, 143, 1048–1055
3.Historical global gridded degree‑days: A high‑spatial resolution database of CDD and HDD. Geoscience Data Journal, 6(2), 214–221.
“Really?”
Yes, really.
“There are lots of HVAC engineers that will disagree with this.”
They think that if it was 20°C yesterday and 25°C today, then it was 45°C in total?
“What in Pete’s name do you think a “degree-day” is?”
A degree day is multiplying a temperature by time, not dividing it. It has units of degree day, not degree/day. There’s a clue in the name.
We discussed degree-day’s extensively at least two years ago.
You apparently learned nothing and remember nothing about it. Typical.
Did you even bother to obtain the references I gave you and read them for meaning and context?
“A degree day is multiplying a temperature by time, not dividing it. It has units of degree day, not degree/day. There’s a clue in the name.”
It’s *NOT* multiplying temperature by time. It is *adding* temperatures over an interval. It is the area under the temperature curve! It is an integral!
Just like the average, the degree-day is *NOT* a measurement. It is a descriptor of the temperature distribution.
HDD = ∫ (T_b – T(t) dt from t_a to t_b.
CDD is similar.
Do you *ever* actually research anything before putting forth assertions you *think* are true?
“It is the area under the temperature curve! It is an integral!”
Yes, which has dimension temperature × time.
All your patronising ad hominems are just you trying to deflect from that simple point.
“That tells you that it’s the average for the month of June 2026, along with additional information, such as it being global and for the lower troposphere. What you do not do is say that it was +0.46°C/month.”
ROFL!!
It wouldn’t be +0.46C/month, it would be
+0.46C/june2026
Again, you are trying to imply that
are somehow different. Do you not see how idiotic such an assertion is?
“ROFL!!”
Your usual response whenever I demonstrate why you are wrong. Show me where Spencer gives the units as °C/month.
“Again, you are trying to imply that
saying 70F per month, and
writing 70F/month
are somehow different. ”
No. I’m telling you they are both wrong. Unless you are talking about a rate of change, an average temperature per month is meaningless.
WTF is a june2026 unit?
No one is saying or imply that. What we’re saying is that they are both wrong. An average temperature for a month is just C. It’s not C/month or C per month. Just C. Burn this into your brain so that you don’t have keep asking about it.
Both assertions are idiotic.
It is both an interval of time and a descriptor of what interval of time.
How is this any different than speaking: X deg average over June 2026?
Discussing physical science with you and bellman is like discussing it with a 1st grader just learning how to read!
Have you *ever* kept an experimental journal? Did you write out all your measurement entries in long hand script?
Then why are you asking WTF does june2026 mean? What it means is pretty obvious to anyone that can read!
No, it isn’t. It’s for a SPECIFIC INTERVAL AT A SPECIFIC POINT ON THE UNIVERSE’S TIMELINE!
What is June, 2026 if it isn’t a month? What does it mean if you don’t know what time interval is involved and when that time interval occurred?
If I give you this distribution (70C, 71C, 69C, 65C) and tell you they are average values can you use the distribution for anything meaningful in physical reality? Is the 70C average from June? July? August? And what year do the components of the distribution apply to?
Is it 70C for June, 2026, 69C for September, 2020?
If I give you *this* distribution (70C/june2026, 71C/july2024, 69C/sept2021, 65C/oct2021) what information can you glean from the values in the distribution? What if the distribution is (70C/june2026, 71C/july2026,69C/aug2026,65C/sep2026) would the distribution be more useful?
Or would you just write this all out in longhand script?
(70C in June, 2026; 71C in july, 2026; 69C in September, 2026; 65C in October, 2026)
Again, the fact that these are AVERAGES, you know they are statistical descriptors and not actual measurements.
If you were trying to use this data what is the first question a real physical scientist would ask about this distribution?
“Have you *ever* kept an experimental journal? Did you write out all your measurement entries in long hand script?”
Congratulations. You seem to have finally admitted what I was suggesting a few days ago. These are not units of measurements, just a short hand way of describing what this is an average of.
Like when I said you might say the average height of a population is 1.8m per person. That’s fine, but you should not confuse this with the actual units of an average.
Unfreakingbelievable!
The interval of time over which measurements are taken is not part of the measurement.
GUM:
—————–
3.1.1 The objective of a measurement (B.2.5) is to determine the value (B.2.2) of the measurand (B.2.9), that is, the value of the particular quantity (B.2.1, Note 1) to be measured. A measurement therefore begins with an appropriate specification of the measurand, the method of measurement (B.2.7), and the measurement procedure (B.2.8).
——————–(bolding mine, tpg)
You have *NEVER*, not once, actually read the GUM.
Yet here you are, trying to lecture everyone on the fine points of metrology.
The time interval involved is part of THE METHOD OF MEASUREMENT and part of the MEASUREMENT PROCEDURE. The time interval is, therefore, part of determining the value of the measurand (see the GUM above). The location of the measurement on the timeline is the very same thing.
You didn’t answer if you’ve ever kept an experimental journal. It’s pretty obvious that either you haven’t or that whoever taught you how to do one missed getting some very important parts of the information requirements into your head (and based on what we see from you continuously that is quite likely).
Measurements have dimensions. One of those dimensions is the time interval involved and its location on the time line. Since the dimensions of the measurements carry over to the average (as you and bdgwx have said repeatedly), the time interval and timeline dimensions carry over to the average as well.
That means that all you are doing here is whining about whether that dimension has to be written out in longhand script or not. You are saying that it has to be written out instead of being specified the way measurement dimensions are usually given.
You can just keep on whining but no one will be listening.
Could you for once answer a question and explain what you mean without these pathetic personal attacks. Are you or are you not saying the dimensions of an average temperature, taken over a period of time is temperature/time?
Exactly. You cannot arbitrarily inject them where they don’t belong.
n is unitless so the formula is (1/365)*Σ(density) and thus the final units are still kg.m-3.
This should be intuitive and obvious. Think about a sample of size n of lengths of boards. If you lay the boards end to end then to get the total length of the boards you would do Σ[x_i, i:{1 to n}]. You intuitively know that the sum must have units of meters (m) so n must be unitless to make that happen. It might be even more convincing if you let x = x_i for all i in which case you’d get Σ[x_i, i:{1 to n}] = n * x. Since x is meters (m) then only way for n * x to also be meters (m) is if n is unitless. And of course this is true regardless of the units that x actually is including W.m-2.
Another way to intuitively reason that n is unitless is to consider what happens in a standard deviation if n actually had units watch it blow up.
If your intuition fails you here I can walk you through the mathematical derivation of how 1/n comes into existence why it doesn’t have units.
Exactly. That means you cannot arbitrarily inject time in the denominator and pretend like it is a rate when it is not.
Dimensional units: you cannot arbitrarily remove them where they belong. As you have repeatedly done.
“n is unitless” An arbitrary removal.
Wrong, n is of dimension day.
The mean length in meters of a set of boards is m/board.
“That means you cannot arbitrarily inject time in the denominator and pretend like it is a rate when it is not.”
There’s nothing arbitrary about my use of denominator units. A statistic of denominator unit time is not a rate.
See my reply to Bellman. You’re wrong. He’s wrong.
And what happens in the standard deviation formula when you arbitrarily give n dimensions?
Unit dimensions are not arbitrary, first.
To your question, the RMS of the density measurements has no physical meaning because there is no negative density.
Setting that aside, it’s the physical meaning thing again.
Each density measurement is dᵢ (kg-m⁻³). Σ(dᵢ)² (i=1-365) = D (unit kg²/m⁶); D×(1/365 days) = D’ day⁻¹.
D’ day⁻¹ is the nominal density variance per unit day. The RMS is√D’ day⁻¹ = ±d’ kg-m⁻³ day⁻¹. ±d’ (±kg-m⁻³) has no physical meaning in this setting, but there it is.
The physical reasoning for exclusion of day⁻¹ is that, in this frame, ‘day’ is just a time index. D’ day⁻¹ is the mean density variance per unit day. The kg²/m⁶ variance is the quantity of interest; day⁻¹ is not a part of the variance.
That is, the variance index of day⁻¹, merely indicates the unit time across which (dᵢ)² was summed. The variance itself is kg²/m⁶, indexed as ‘per unit day’ (which allows one to recover the original sum). The variance is not kg²/m⁶-day⁻¹, i.e., all one unit.
The root is, of physical necessity, also indexed per day. The pertinent root is then √(kg²/m⁶) day⁻¹
The ‘day⁻¹’ remains outside the root (the √) because it is an independent index of time and not an intrinsic dimension of the measurement or of the variance.
How math in science is different from math in math
First…RMS is different than SD. Whatever. It demonstrates the problem in a similar way.
Second…n does NOT exist outside the square root. The formula for RMS is sqrt[ Σ[x_i^2, i:{1 to n}] / n ].
Third…there is an obvious problem with RMS if you give n some units. For example, if you assign day units to n then a density RMS would compute as sqrt[ (kg.m-3)^2 / day] which reduces to sqrt[(kg.m-3)^2] / sqrt[day] = kg.m-3.day^(-1/2). What is a day^(-1/2)?
Fourth…the reason I’m presenting this to you as an exercise is so that you can see for yourself that there is an obvious problem when you give n units and I’m hoping this provides the epiphany you need to see that there is a problem with your understanding.
Anyway, correct the math mistake of pulling n outside of the square root and resubmit for review.
Excluding the time index not a mistake.
See, your invariable limitation, bdgwx, is that whatever you don’t understand is wrong.
The formula for RMS is sqrt[ Σ[x^2] / n ]. The n is inside the square root. That means you have take the square root of whatever dimensions you gave to n. If n has units of day then RMS computes to having units of day^-(1/2) in it. That’s just how math works.
And for SD it is sqrt[ Σ[(x_i – x_avg)^2] / n ] where x_avg = Σ[x]/n . The only ways this works is if n is unitless otherwise x_i – x_avg would be using different units which is disallowed.
Pat, the fact that SD isn’t even computable when n has dimensions should an epiphany that something isn’t right with your understanding.
“And what happens in the standard deviation formula when you arbitrarily give n dimensions?”
The standard deviation is a STATISTICAL DESCRIPTOR. It is *NOT* a measurement. The standard deviation of a set of measurements in a Type A experimental situation tells you the interval of values which is reasonable to assign to the measurand based on the distribution of observed values. But it is *NOT* itself a measurement.
The SD in a Type A experimental situation is calculated using observations that have no given measurement uncertainty. Therefore the SD is found using only stated values of the observations.
If the observations have measurement uncertainty, then the SD of the stated values is *NOT* the measurement uncertainty, the propagated sum of the measurement uncertainties becomes the measurement uncertainty. The “average” of the stated values may be used as a “best estimate” for the measurand — IF IT CAN BE ASSUMED THE MEASUREMENTS ARE GAUSSIAN. If the measurements are not Gaussian, then the average may *not* be the best estimate of the measurand, it may be the mode or the median.
*YOU* never bother with giving a measurement uncertainty for the 1360 stated value. Why is that? Is it so you won’t have to propagate that uncertainty over each iteration in a step-by-step model?
I invite you to do the unit analysis with SD when n is arbitrarily given units and see what happens as well.
The silence is deafening.
“n” is *NOT* arbitrarily given units.
Nor does the data have to be measurements with uncertainty.
If I tell an investor that the average number of shares traded for Nvidia is 15,000 at $1000 each what is the first question an astute investor should ask? What is the second question?
If you want to talk about deafening silence I’ll ask this again:
“*YOU* never bother with giving a measurement uncertainty for the 1360 stated value. Why is that? Is it so you won’t have to propagate that uncertainty over each iteration in a step-by-step model?”
Why don’t you want to discuss the SD and what happens when n is given units?
Can you read? The SD is a statistical descriptor. It is *NOT* a measurement. The SD of a population has no measurement uncertainty itself!
A stated value for a measurement *should* be given as x +/- u “units”.
“u” itself describes the interval around “x”. It is the value “x” that has the dimension.
See Taylor, Eq 2.6 for an example.
(measured g) = 9.82 +/- 0.02 m/s^2
It is *NOT* written as 9.82 m/s^2 +/- 0.02 m/s^2
You are deflecting and diverting again. Uncertainty has nothing to do with this conversation. I’ll ask you again…what happens to an SD when you give n dimensional units?
“Wrong, n is of dimension day.”
He’ll never get it.
Agreed.
The debate here has descended into being pointless.
Getting the units correct is of the upmost importance. One wrong move can spoil the whole calculation as evidenced in your publication in which you took the 4 W.m-2 value from Lauer & Hamilton and arbitrarily added year-1 units to it thus fooling you into believing that you then had to multiple their value by the number of years.
“Getting the units correct is of the upmost importance”
If density/day is wrong then so is temperature/day.
Density, D, is temperature dependent. D(T). But T is time dependent –. D(T(t))
Daily temperature, D, is temperature dependent, D(T). But T is time dependent –> D(T(t))
Both are equivalent.
If density/day makes no sense to you then how does temperature/day make any sense.
Either they both make sense or neither do.
Which is it?
Neither. As I keep saying none of it makes sense. It doesn’t matter if it is density, temperature, flux, or some other intensive property. When you average it over a period of time it doesn’t change from a descriptor of state to a descriptor of rate. The units remain as-is. Burn this into your brain so you don’t have to keep asking.
Then why do you apparently support the climate models outputting a global average temperature?
You simply can’t seem to grasp that a statistical descriptor is *NOT* a measurement. A “rate” is a measurement. An average of rates is *NOT* itself a measurement, it is a statistical descriptor of a distribution of measurements compiled from a collection of rates.
miles/hour is a measurement of a physical phenomenon. An average of a collection of miles/hour data is a description of the distribution associated with those measurements.
The dimensions of both may look the same but they serve a different purpose.
If I give you a value of 70F can you tell if that is a measurement of a physical attribute or if it is an average calculated from a collection of measurements forming a distribution?
Can you even tell if 70F is an average temperature over a day? A week? A month? A year? A temperature of 70F over a week tells you nothing about the temperature over a year.
The use of the dimension of “time” in an average tells you the time period involved in the measurements forming the distribution.
If I tell you that you are traveling at a mile/minute that describes a physical reality, a measured attribute. If I tell you that you averaged a mile/minute how do you know over what time period that average describes?
The only way to describe the distribution of measurements is to add a dimension describing the distribution.
Average *has* to become:
(miles/minute) / hour, (miles/minute) / day, (miles/minute) / week, etc
in order to understand the distribution shape.
miles/minute is a RATE. miles/min = v
v/day, v/week/ v/hour is a STATISTICAL DESCRIPTOR OF A COLLECTION OF “v” values over a period of time.
It’s no different with flux. F = joules/sec-m^2. It’s a rate. I can measure the rate. It’s a physical reality.
The average of a collection of F values is *NOT* a measurement, it has no physical reality. It is a statistical descriptor. But an average of joules/sec-m^2 is lacking a needed dimension that describes over what time period the collection of measurements were made.
is it F/day? F/minute? F/year?
Can you substitute F/day for F/year? Will the daily average *always* be equal to the annual average? If they cannot be assumed to always be the same then it is REQUIRED that a dimension is needed to differentiate them.
All Pat has done is added the needed dimension to DESCRIBE the time period over which the values used for the calculation covers.
In physical science that is NECESSARY piece of information in order to judge the appropriateness of further calculations.
You just keep trying to validate the typical statistician meme of “numbers is just numbers” that is so prevalent in climate science. You are trying to justify the view that the time period over which an average applies is not needed, 70F is 70F and who cares if it is an actual physical measurement or an average over a year? A week? A day?
Numbers is just numbers, right?
“Mean/time-unit is is a statistic”
Statistics world is not the real world. Some statisticians, mathematicians, and climate scientists can’t seem to understand that.
A statistical descriptor is *NOT* a measurement of a measurand existing in physical reality.
JCGM 100:2008 says it is.
Let’s see what JCGM says about a Type A analysis.
All observations are estimates. A Type A analysis can use a statistical parameter to obtain an an estimate. However, the downside is that there is ALWAYS uncertainty associated with that estimate that must also be quoted.
End result? A mean is not a measurement, it is an estimate calculated from a number of observations.
You do realize that there is no guarantee that the mean is THE true value, right?
You have never accepted the physical reality of measurements. They are not just numbers to be played with.
“JCGM 100:2008 says it is.”
No, JCGM 100:2008 says it is the BEST ESTIMATE of the value of the measurand. It is *NOT* a measurement, it is an ESTIMATE!
The BEST ESTIMATE is a statistical descriptor of the distribution of actual measurements – it is *NOT* a measurement.
And even then, the GUM assumes that the measurements are Gaussian with no systematic effects – which is *NOT* a good assumption for real world measurements of different things using different instruments under different conditions.
GUM:
———————-
4.1.4 An estimate of the measurand Y, denoted by y, is obtained from Equation (1) using input estimates x1, x2, …, xN for the values of the N quantities X1, X2, …, XN. Thus the output estimate y, which is the result of the measurement, is given by
y = f(x_1,x_2, …, x_n)
—————— (bolding mine, tpg)
x_1, x_2, etc are MEASUREMENTS.
y is *NOT* a measurement. It is an estimate of the value of the measurand.
“The context we are discussing is one of state”
If you are discussing something that has time in the denominator then you are discussing a RATE.
W/m^2 does not appear to have time in the denominator because you have not broken down Watts into its fundamental components.
W/m^2 ==> joules/sec-m^2. Joules per second per square meter.
Integrating joules/sec-m^2 over time gives you the total joules/m^2.
[joules/sec-m^2] * sec ==> joules/m^2
∫ f([joules/sec-m^2) * dt where t is in seconds –> joules/m^2.
But somewhere you have to define what f(joules/sec-m^2) actually *is* using a functional relationship to relate the components of joules, seconds, and square meters in order to actually compute the integral.
You then divide the total (joules/m^2) by the total time
total(joules/m^2) / total(time) –> joules/sec-m^2
in order to find the average joules/sec-m^2.
You *HAVE* to follow your dimensions all the way through.
If you find the average per month then you wind up with
∫ f(j,t,a) dt from 1 to 30 to get total joules/m^2 for the month and divide by the number of days to get the average.
∫ (joules/sec-m^2) dsec –> (joules/m^2)
(joules/m^2) /30 days –> joules/day-m^2.
The biggest problem you have here is defining f(j,t,a) as a functional relationship. It appears that you want f(j,t,a) = K, a constant. In other words, a garbage climate science assumption. And then you want to use the garbage climate science assumption of “numbers is just numbers” to claim “n” in an average has no dimension. It’s not “seconds in a minute” or “minutes in a year” or anything other than it’s just a number.
Another major problem is that you are simply ignoring the measurement uncertainty throughout the function.
If F = Asin(t) +/- ε(t) and t varies from 0 to pi/2 then the integral of the function (needed to calculate the average) becomes
∫ A sin(t) dt +/- ∫ε(t) dt, evaluated from t = 0 to t = pi/2
If ε(t) is a constant ε, say 0.1, then the uncertainty of the total becomes (pi/2) * ε –> (1.57)(0.1) = 0.157. The uncertainty of the sum has grown. And the sum is A(1) since ∫sin(t) = 1. And that carries over onto the average value of A/(1)/(pi/2) +/0 0.157.
When are you going to learn to 1. identify measurement uncertainty right at the very beginning and, 2. propagate it correctly?
Your posts reads like you are challenging the mean value theorem for integrals or at least the inevitable result that it yields units of W.m-2 when f(x) is W.m-2 and x is seconds (s). If that’s what you are intending then that is sad because this is probably the easiest integral you’ll see.
I’m not challenging the mean value theorem.
I’m challenging your claim that the mean value theorem can give you a value for the mean. The theorem just says that it exists. It doesn’t tell you what it is!
In order to know what it is you *have* to have a functional relationship to integrate. You have yet to define what that functional relationship actually *is*!
First…it is the mean value theorem for integrals.
Second…I’ll ask again…what do you think integral[f(x) dx, x:{a to b}] / (a-b) is if not the mean value?
That is a fine generalization, but it lacks specificity. Let’s say you are averaging a month’s worth of Tmax temperatures, an example would be in TN 1900.
What exactly is the functional relationship that you are integrating to find the mean? Is the relationship Gaussian? Is it skewed (not symmetric)? Is it a t distribution? Lorentzian?
Just saying f(x) is fine for math, but for science you need an actual relationship to study. Measurement distributions can have quite different equations defining the PDF’s.
What we are trying to explain is that real world physical measurements require a mental change from just thinking about numbers as numbers with abstract definitions, to one that understands the length of a pushrod in an ICE engine will never be known exactly. Yet one has to decide if the uncertainty interval (not the mean) is such that the engine will meet design criteria over the needed rpm range regardless of where in the interval the length actually lays.
You didn’t answer the question. You say integral[f(x) dx, x:{a to b}] / (a-b) is not the mean value of f(x) from a to b. So what do you think it is?
Stop being so willfully ignorant.
He did *NOT* tell you that the mean is is not the mean.
He told you that the theorem tells you *what” exists, the integral tells the value of the “what”.
THEY ARE NOT THE SAME.
What the hell does that mean? The integral is literally the salient part of the theorem.
I’ll repeat again…the mean value theorem for integrals tells you what the mean value is.
I didn’t say that. I said that to do science you must move from a blackboard f(x) to an actual functional relationship definition for f(x).
Here is a symmetric Gaussian function.
f(x) = {1/(σ√(2π))}e⁻⁽ˣ⁻ᵘ⁾^²/²σ^²
Show the math for finding the average value of that function from (-∞, μ), that is, the bottom half. Then do the same for the top half.
Maybe using medians and a 5-number description would be easier.
Tim said and I quote: “I’m challenging your claim that the mean value theorem can give you a value for the mean. The theorem just says that it exists. It doesn’t tell you what it is!”
Do you agree with Time that the mean value theorem for integrals does not tell you what the mean value is?
The integral itself is *NOT* the mean value of integrals theorem.
Again, the theorem tells you *what* exists. The integral tells you the *value* of the “what”.
They are *NOT* the same thing.
Stop trying to conflate the two.
I’ll ask *YOU* AGAIN.
WHAT IS f(x)? What functional relationship did you actually evaluate?
WTF?
The integral is part of the theorem…the salient part of it I might add.
Here’s what I think…you and/or brother made absurd claim that the MVTI does not tell you the mean and to defend it you have to make yet another absurd claim by insinuating that the integral is not part of the theorem.
If you don’t want people to think you’re challenging the MVTI then now is your chance to say that you accept the MVTI and all its consequences including the fact that it tells you what the mean is.
That’s like saying ∫f(x)dx will give you a value. IT WON’T!
And you still refuse to say what f(x) *IS*! x certainly isn’t “seconds” in this context. Flux is not given in “seconds”. It’s given in joules/sec-m^2.
It does *NOT* give you the mean.
Tell me what value the integral ∫f(x)dx evaluates to!
No one is saying that! One more time: what does ∫f(x)dx evaluate to? If you can’t answer that then the only possible conclusion is that the theorem only says the mean exists, NOT WHAT IT IS.
For at least the fifth time: WHAT VALUE DOES ∫f(x)dx EVALUATE TO?
For at least the fifth time: What is functional relationship represented by f(x)?
If you continue to refuse to answer these questions then the only conclusion can be that you *KNOW* that the MVT only tells you that something exists but does *NOT* give what the value is for that “something”.
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.
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How am I supposed to have a conversation with you when you act like this?
The area under the curve. And then when you divide by (b-a) it gives you the mean value of f(x) over span from a to b. That is the mean value theorem for integrals.
And exactly what functional relationship defines that curve? “a” and “b” are used to calculate points on that curve. The relationship between “a” and “b” and the corresponding values of the dependent variable must be known to evaluate the integral, right?
f(x) is not a mathematical description of a relationship. It is a place holder that signifies a relationship exists. Therefore, in this case, it only tells you that a mean value exists (in general), but f(x) does not allow the calculation of the value of the mean.
PS: There are cases where f(x) does not adequately indicate discontinuities in relationships. That means f(x) must be defined in order to recognize those discontinuities. How many means does “1/x” have?
“The example I gave was of daily measurements that are aggregated”
What is the measurement uncertainty of that aggregate? 0 (zero)?
“I also mentioned that you can use the mean value theorem of calculus to compute it using a full integration of Earth’s orbit.”
Tell us what you think the “mean value theorem” gives you. Is it the “average” value, the”median value”, or is it something else entirely?
I suspect you are confusing the “mean value theorem” and the “average value theorem”.
Even the average value theorem only tells you that there is at least one spot in the function where the function equals the average. You can’t calculate the average value from the “average value theorem”. It’s the same with the “mean value theorem”, it doesn’t allow identifying the average value of the function, it only identifies that somewhere along the function, the slope of the function will be the same as the slope of the line connecting the endpoints. That may or may not be the “average” value.
First…to be clear I’m talking about the mean value theorem for integrals.
Second…it gives you W.m-2. Ya know…the units of flux for the annual average TSI or solar constant S. There is no extra day-1 or year-1 units that appear in the final result.
Third…before you start your typical gaslighting know that I explicitly stated “mean value theorem for integrals” each time I referred to it. It was Pat that shortened it to just “mean value theorem”. So if you think you have a “gotcha” on me you can direct it straight to Pat.
No I am not.
That is absurd. Of course it does.
“First…to be clear I’m talking about the mean value theorem for integrals.”
The mean value theorem for integrals just says that there is a point on a continuous function that equals the average. The average still has to be calculated.
[1/(b-a)] ∫ f(x) dx, from a to b.
“Second…it gives you W.m-2. Ya know…the units of flux”
If f(x) is in W/m^2 = joules/sec-m^2 and “dx” is in sec then the integral gives you joules/m^2, i.e. the total joules received over the time interval, not the flux value.
You then divide by the time interval to get an average value of the flux.
What did you use of f(x)? What did you use for a and b?
“Third…before you start your typical gaslighting know that I explicitly stated “mean value theorem for integrals” each time I referred to it. “
It *still* isn’t obvious that you know what the “mean value theorem for integrals is.
It does *NOT* reduce the measurement uncertainty of the average value to zero.
And, in fact, since you have to calculate the total amount of joules to start with, and it is joules that have to balance (not the flux) why don’t you just then calculate the total joules-out in order to see what the balance is? THE FLUX IN AND OUT WILL NEVER BALANCE since they occur over different time intervals.
“That is absurd. Of course it does.”
No, finding the total joules/m^2 is what the integral gives you. That is not the average joules/sec-m^2. Since it is the joules-in and joules-out that have to balance why even bother with the some kind of normalized average values?
I’m curious…if you don’t think integral[f(x) dx, x:{a to b}] / (a-b) is the average of f(x) over x:{a to b} then what do you think it is?
What I said I did. f(x) is W.m-2 and x is s. Therefore the mean value theorem for integrals evaluates to W.m-2.
It is as I stated. f(c) = integral[f(x) dx, x:{a to b}]. I can’t make it any more obvious than that.
The mean value theorem for integrals does NOT yield units of j.m-2 when f(x) is W.m-2 and x and is s. It yields units of W.m-2.
” integral[f(x) dx, x:{a to b}] / (a-b)” IS THE AVERAGE. It’s why I said the mean value theorem does *NOT* give you the average. The average gives you the average, not the mean value theorem.
I said: “The average still has to be calculated.”
Can you read?
W-m^2 is a dimension. You don’t integrate dimensions. You integrate functions whose components have dimensions. A function gives you the value of the radiation at a point in space and time. What FUNCTION did you integrate?
If F = f(x) where F is the flux then what is the functional relationship between F and x? Is it F = K, a constant? F = Ksin(x)? F=Kcos(x)? F = Kx^4? What are you integrating? What *is* f(x)?
If f(x) is in W/m^2 and “s” is seconds, exactly what are you doing?
f(x) should actually have the dimension of joules/s-m^2. Then the integral is
∫f(x)dx –>
∫ (F_joules/sec-m^2) dsec
And you get joules/m^2 for the integral’s dimension But you *still* can’t calculate anything until you define what f(x) is in the equation of F = f(x) .
If a = s0 and b = s1 then f(s1) – f(s0) gives the total joules/m^2 and you divide by the total time interval, s1 – s0, to get the average joules/sec-m^2 –> W/m^2.
Like I said, the mean value theorem doesn’t tell you what the average *is*. You have to calculate the average.
BTW, it can’t be f(x) where x is seconds. It has to be f(x,j,a) where x is seconds, j is joules, and a is area – if your dimension is going to be joules/sec-m^2.
I’d love to see bdgwx make a 1000 yard shot to target from the top of a hill down a valley.
All the world’s a fused globe of plexiglass.
Actually you can’t. The launch angle and other variables are all interrelated. Changing one thing, changes other variables. That is a functional relationship. For example, a different launch angle modifies the vectors. Friction can vary because the speed may change. The aerodynamics will be different because of the path through air at a different angle. Different heights may encounter varying wind speed and direction reducing accuracy.
In other words ceteris paribus can’t be used because not everything except launch angle remains the same. You’ve obviously never shot high power rifles at distances of 300 – 500 yards. Lots of variables change and nothing ever stays the same.
“I can still invoke the concept of ceteris paribus to predict how a change in launch angle will change the distance traveled.”
How? In the REAL WORLD, the Coriolis Force will change the destination point as will any wind. Even the powder load, which has a measurement uncertainty, will impact the distance. You can’t just hold the powder load “constant”.
It’s the same with climate, you *can’t* hold all factors the same IN THE REAL WORLD. Doing so makes your predictions nothing but a best estimate with an uncertainty dependent on the uncertainty of the factors you tried to hold constant.
No.
The essence of science is a deductively predictive theory, and the accurate requisite observation. Climate models are not deductively predictive. Climate is not static.
Control of variables is the essence of laboratory experiments.
Earth is cooler not warmer.
GHE balance graphics are trash.
Earth cannot radiate “extra” energy as a BB.
GHE = bogus & CAGW = scam.
What CO2 does and how it does is moot.
Still missing the process of kinetic energy at the surface becoming potential energy within an atmosphere. The former radiates but the latter does not.
Thus one must consider the delay in energy loss to space that arises during the process of conversion and reconversion (convection up and down).
That delay is what fixes the surface temperature rather than the composition of the atmosphere.
If the composition changes then the rate of turnover within the atmosphere changes to neutralise any effect on surface temperature.
‘Still missing the process of kinetic energy at the surface becoming potential energy within an atmosphere.’
And not a minor process, either. The reason is because once a near-surface parcel is kinetically warmed by thermalization, it doesn’t just sit there in ‘LTE’, it lifts. And in so doing, it does work with an accompanying increase in entropy, hence the process is irreversible.
It matters not that LWIR is sent downward toward the surface. The IR from above is from gases much colder than the surface. Thus the energy levels of the surface, equivalent to the gases, are full and the IR is reflected (rejected). No warming can occur (= no greenhouse effect).
For the three CO2 absorption bands, they are equivalent to blackbody radiation of objects at 800, 400, and -80 deg C. Nothing in the atmosphere (upper tropical troposphere, “hotspot,” at about -17 deg C) is hot enough to emit the first two and nothing on the surface (about 15 deg C) is cold enough to absorb emissions at -80 deg C. Thermodynamics, again.
Also, you pointedly ignore the water cycle which carries 85–90% of surface energy to altitude where the energy is released by condensation through the latent heat of water. Trenberth made the same omission and then agonized over the missing heat hiding in the oceans, waiting to jump out at us—what a rube).
With the temperature differences between the upper tropical troposphere and the surface, there is no greenhouse effect and there are no greenhouse gases (all this from failed hypotheses by Arrhenius in 1896 and thoroughly disproven since many times). However, your taking this for granted and then discussing winter water vapor is a joke, as CO2 is not a factor.
In fact, physicists have been unable to make CO2 warm anything and have realized that it is the world’s best refrigerant. For one thing it is constantly emitting the -80 deg C IR which nothing in the atmosphere can absorb, being too hot, and works 24/7 to cool the planet by sending this IR out to space. It is stable, abundant, cheap, nontoxic, and nonflammable, a win-win. Mercedes car A/C units and new skating rinks are using CO2 as the primary refrigerant. More CO2 means more cooling and a greener planet, as it is PLANT FOOD AND WE NEED MORE NOT LESS.
Nice treatment but missing some key facts. NASA has been looking for the “hotspot” for decades and in fact cannot find it. However, they observed that this region had been gently cooling for decades despite increases in CO2 concentration.
“It matters not that LWIR is sent downward toward the surface. The IR from above is from gases much colder than the surface. Thus the energy levels of the surface, equivalent to the gases, are full and the IR is reflected (rejected).”
Absolute nonsense, while the emissions from a gas molecule such as CO2 are the result of a decay from a specific molecular energy level the absorption at the surface is by a solid or liquid effectively a black body, the IR is absorbed.
The earth is not a black body. It is not isothermal nor does it absorb all LWIR incident on the surface.
The infrared (IR) absorbance of Earth’s surface is extremely high, meaning the surface absorbs and emits thermal infrared radiation almost like a blackbody. By Kirchhoff’s Law of thermal radiation, a surface’s absorptivity equals its emissivity ( 𝛼=𝜀) at thermal equilibrium, meaning natural surfaces with high infrared emissivity also have high infrared absorbance.
Earth’s surface emissivities (εs) have been inferred with satellite-based instruments by directly observing surface thermal emissions at nadir through a less obstructed atmospheric window spanning 8-13 μm.[29] Values range about εs=0.65-0.99, with lowest values typically limited to the most barren desert areas. Emissivities of most surface regions are above 0.9 due to the dominant influence of water; including oceans, land vegetation, and snow/ice. Globally averaged estimates for the hemispheric emissivity of Earth’s surface are in the vicinity of εs=0.95.[30]
29 “ASTER global emissivity database: 100 times more detailed than its predecessor”. NASA Earth Observatory. 17 November 2014. Retrieved 10 October 2022.
30 “Joint Emissivity Database Initiative”. NASA Jet Propulsion Laboratory. Retrieved 10 October 2022.
“The infrared (IR) absorbance of Earth’s surface is extremely high, meaning the surface absorbs and emits thermal infrared radiation almost like a blackbody”
Silica minerals make up a significant portion of the land surface on the earth. It’s absorptivity in the LWIR bands range from 0.90 to 0.95. That is *not* black body.
Nor does the earth immediately radiate absorbed LWIR which a black body does. Nor is the earth isothermal which a black body is.
These all have a significant impact on the thermodynamics of the earth – meaning it is *not* a black body. Climate science assuming that it is a black body means climate science totally ignores thermodynamic issues like the specific heat capacity variance of the earth components just below the surface, the varying thermal conductivity of the components of the earth just below the surface, and the density of the components of the earth just below the surface.
These factors all go into determining the thermal inertia of the various surfaces of the earth – i.e. the speed at which surface temperatures respond.
Bottom line? Radiant flux balance is a joke. The emission spectra of the earth will vary widely over the surface and the emission spectra will vary widely over time. High values of variance means any “average” value of outgoing flux at any point in time and space will have a large uncertainty.
4 W/m^2 is a relative uncertainty of 0.2%. The daily variance of incoming flux is over 450,000. (12 hours of 1360 and 12 hours of zero). Even if the 1360 value has an uncertainty of zero the variance of the daily average will be huge meaning its uncertainty will also be large, far larger than the difference attempted to be identified of 4 W/m^2. If the measurement uncertainty of the 1360 W/m^2 over an hour is just 1%, the total uncertainty over 12 hours will be 4W/m^2. Add this to the measurement uncertainty of the earth’s emissions over 24 hours and there is simply no way to know if the flux imbalance is 4 W/m^2.
“For the three CO2 absorption bands, they are equivalent to blackbody radiation of objects at 800, 400, and -80 deg C. Nothing in the atmosphere (upper tropical troposphere, “hotspot,” at about -17 deg C) is hot enough to emit the first two and nothing on the surface (about 15 deg C) is cold enough to absorb emissions at -80 deg C.”
Incorrect, the temperatures you refer to are the temperatures at which a black body emission would have a maximum at that frequency. BB emission/absorption will still occur at other temperatures.
Here’s a graph of BB emission at two temperatures, notice that the emission is at all frequencies.
Both temperatures have emission at 15 microns but the higher temperature emits more. For example, at a temperature of -80ºC a BB would have a peak emission of ~1.1 W/m2/sr/μm at 15μm, whereas at 25ºC it would be ~9.6 at 9.5μm and the emission at 15μm would be ~6.5, about 6 times more than at -80ºC!
Several problems with just the first paragraph of the above article, particularly with the lead-in graphic that is also the Figure 1 presented within the article itself.
1) This simplified set of diagrams showing “heat engines” ignores the HUGE energy flows in/out of the phase changes and heat capacities of water on Earth (ice/snow<—>liquid and liquid<—>water vapor) where no work is done and the associated temperatures at phase change are ~constant. Hence, Earth and its hydrologic cycle is far more than just a simple “heat engine”.
2) All three figures in those graphics do not reflect the inherent inefficiency of all “heat engines” (the entropy) that occurs in actual practice . . . such is not part of the output work. Although this is mentioned-in-passing in the description underneath Figure 1 in the body of the article, note that entropy is NOT properly accounted for in Figure 1C where there is energy (not work) loss that inevitably appears as “waste heat”.
3) The description of Figure 1B states “In the atmosphere (Figure 1B) Work is done internal to the system”. But note that the control volume shown around Figure 1B includes both the the “hot reservoir” (assumed, in context, to be the Sun) and the “cold reservoir” (assumed, in context to be deep space). The atmosphere does not contain either of these “reservoirs”.
Separately, within the the subsequent body text of the full article there are these statements:
“Energy quickly leaves the CO2 molecules into a sea of energy shared among molecular rotations, vibrations, and speed. It is not a one-way evolution of exchanges, because down at the molecular level, what proceeds in one direction goes just as surely in the opposite, maybe at a different rate. So, evolution proceeds toward a distribution of energy by all conversion means until it is a stationary distribution – an equilibrium state.”
(with my bold emphasis added)
The bolded text is not correct, and it even conflicts with the sentence that follows it.
The Second Law of Thermodynamics governs the overall flow of energy distribution within a control volume and clearly states that, in an overall interacting ensemble state, energy will tend to flow from “hot” (energetic) molecules to “cold” (less energetic) molecules. This is sometimes referred to as the “Zeroth Law of Thermodynamics”, which dictates that heat naturally flows from regions of higher temperature to lower temperature until thermal equilibrium—a uniform temperature distribution per Maxwell-Boltzmann statistics—is achieved.
Yes, entropy is why thermalization is effectively a non-reversible process for purposes of ‘back radiation’.
Thermalization is the reaching of the Boltzmann distribution and involves both molecular speeds or kinetic energy and other energy forms as well. So, by the Boltzmann distribution there is always some populations of CO2 molecules that are in an energy state equivalent to having absorbed 15um radiation — 0.036 or so at 300K — and though a small fraction, there are still so many per unit volume that they produce a measurable flow of 15um radiation into all directions, including toward the surface.
In effect, thermalization does not prevent this radiation — it enables it.
I don’t disagree with much of what you say in this comment, but let’s talk about CO2 specifically. As you mention, there is always a ‘small fraction’ of molecules that can spontaneously emit ‘photons’, but we’re talking about 50,000:1 in terms of the incidence of non-radiative to radiative emission at sea level.
Having said that, I have no doubt that if I had 15 micron eyes, the air around me would be glowing, not just from CO2, but mainly from water vapor, which is much more prevalent and overlaps CO2 at this frequency. And, as noted elsewhere, I wouldn’t be able to see beyond 10-20 meters, but more importantly, I would not be able to discern the actual direction of energy flow from this radiance using any instrument currently in existence.
So here’s the real problem we should all have with the ‘back radiation’ shown on a typical Earth Energy Imbalance (EEI) diagram. Specifically, what is it, from whence in the ‘wild blue yonder’ does it arise, what does it actually mean in terms of energy flow, surface temperature, etc.?
Because for my money, you tell me the radiance and atmospheric composition and I, or someone more adept at doing physics calculations, will tell you what the local temperature is. Or alternatively, I can easily figure this out myself by looking at a cheap thermometer.
And I can also explain how GHGs generally cause the troposphere to convect, thereby making the lower troposphere warmer than the upper troposphere, aka, the greenhouse effect, without invoking a phenomenological physics conjecture from over a century ago that over recent decades has left us economically and politically at the mercy of some seriously bad actors.
But the Boltzmann distribution refers to the kinetic energy not the vibrational energy which is the source of the radiation.
Not quite correct.
Vibration of molecular bonds between atoms in molecules is one of the means in which mechanical energy is stored. According to the equipartition theorem, each degree of freedom that is active for a molecule at a given temperature for a gas in thermal equilibrium tends to account for (1/2)*k*T of the total mechanical energy, where k is the Boltzmann constant and T is absolute temperature.
At the temperature range across Earth’s atmosphere CO2, nitrogen and oxygen have (5/2)*k*T of total energy . . . that is, they each have five active degrees of freedom, with only 3 being associated with x, y and z translational degrees of freedom; the other two involve mechanical energy present in vibrations of molecular bonds.
During molecule-molecule collisions in the atmosphere—particularly between LWIR-energized CO2 and N2 and/or O2—energy is exchanged using all five degrees of freedom.
CO2 has 9 degrees of freedom being a triatomic molecule, there are 4 vibrational degrees of freedom. For the 15μm emission the bending mode vibration has to be excited to its first level.
You simply don’t understand that the other possible degrees of freedom for CO2 are NOT active in the temperature range of CO2 in Earth’s atmosphere . . . they need much higher temperatures to become active.
As I stated, at the range of temperatures only two modes for CO2 molecular vibration are available to “store” mechanical energy. Any competent Web search AI will confirm this.
“You simply don’t understand that the other possible degrees of freedom for CO2 are NOT active in the temperature range of CO2 in Earth’s atmosphere . . . they need much higher temperatures to become active.”
Not true, all 4 vibrational modes are active in the Earth’s atmosphere, one of them is not radiationally active but the other 3 are. The Asymmetric stretch mode lies in the frequency range between the solar radiation and the Earth’s IR emission so is not a major absorber however it can be collisionally activated. Also for some reason you neglect the rotational degrees of freedom.
From Google’s AI bot:
“At room temperature CO2 has five active degrees of freedom (3
translational and 2 rotational). The vibrational modes are mostly ‘frozen’ and do not contribute to the heat capacity.
At high temperatures: Vibrational modes are activated, adding
additional vibrational degrees of freedom.”
So, yeah, I mistakenly said the active mode for CO2 in Earth’s atmosphere included 3 for molecular translations and 2 for vibration of intermolecular bonds, whereas I should have stated beyond translational DOFs there are two rotational DOFs, not vibrational DOFs . . . the main point being that the CO2 molecule does not have nine DOFs active under its range of atmospheric temperatures.
Do you suppose that the heat engine of Figure 1, when applied to the turbine in a power plant, ignores phase changes and heat capacity?
The Zeroth Law involves the transitivity of temperature measurements and is essentially a justification for using thermometers.
The direction of spontaneous heat transport is the Second Law. Both the Kelvin and Clausius statements of the Second Law dictate the flow of heat. You have the zeroth and second laws mixed up.
The Boltzmann-Maxwell distribution involves molecular speeds and is distinct from the Boltzmann distribution.
“Do you suppose that the heat engine of Figure 1, when applied to the turbine in a power plant, ignores phase changes and heat capacity?”
Yes, it does when the steam turbine is run properly. A well run steam turbine will, at all cost, avoid reducing the steam quality below 100% (that is, risk condensing water droplets) which can cause rapid erosion of the turbine blades and turbine vibrations/aerodynamic stalling.
As for the heat capacity, no, because the Rankine cycle that is fundamental to steam turbine power plants is based on the total “heat capacity” changes of water with such mostly involved with phase change between liquid water on the condenser output, pump, and boiler input segments of the cycle and superheated steam on the boiler output, turbine and condenser input segments of the cycle.
In fact, it is the heat capacity (enthalpy) change between high pressure, superheated steam input to the turbine and the lower pressure, lower temperature dry steam leaving the turbine that provides the output power (represented as torque and rpm) that the steam turbine generates.
Does WaveTran work properly in the Stratosphere, with the temperature profile inverted?
I am unfamiliar with WaveTran, so I can’t say.
Looks interesting & detailed . Radiation to space forms the envelope . And it is crucial to understand that a gray , ie: flat spectrum , comes to a radiative equilibrium of ~ 278.7+-2.3 around our orbit . https://cosy.com/y26/NL202603.html#278.7+-2.3 .
Any further calculations start with that .
BTW : CoSy Sat 1100-6 Zoom tmro . em for invite .