Guest Post by Willis Eschenbach (@weschenbach on X, blog at Skating Under The Ice)
Like Don Quixote suffering a coffee overdose, once again I mount my steed, take up my lance, and go tilting at windmills.
Here’s a thought experiment. Consider the Earth with no atmosphere and with the same surface albedo of 12.5% that it has now. How warm would it be?
In this condition, because there is no atmosphere, there are no heat losses from the surface by sensible or latent heat. For the same reason, there’s no reduction in incoming solar power as a result of reflection from clouds or solar radiation absorbed by the atmosphere. Solar radiation absorbed at the surface is radiated straight back to space. So we can calculate the temperature directly from the absorbed radiation using the Stefan-Boltzmann equation. This equation relates watts per square meter of surface radiation to the corresponding temperature of the radiating surface. I used the S/B equation with the average emissivity of the Earth, which is ~ 0.98, to calculate the surface temperature. All values are in watts per square meter (abbreviated variously as W/m2 or W/m2).
Figure 1. Total surface radiation absorption, Earth with no atmosphere, with the same albedo as the current Earth. Total absorbed solar radiation is the incoming 340 W/m2 minus the 43 W/m2 reflected solar radiation. Integer errors are due to rounding. All values are in watts per square meter (abbreviated as W/m2 or W/m2)
Minus two degrees C. What you might call totally chill. Just below freezing, in fact.
Now, over millions of years, let’s slowly add in an atmosphere and when it matches the modern atmosphere in all respects, let it percolate for another few million years.
After adding the atmosphere, there are lots of changes. Atmospheric absorption cuts down on the amount of solar radiation hitting the surface, which cools the surface. Of course, reduced surface radiation also reduces the amount reflected. However, this reduction in solar radiation at the surface is more than compensated for by the atmosphere adding in 340 W/m2 of downwelling longwave (thermal) radiation from the atmosphere to the surface (yellow arrow in Figure 2 below), which provides significant warming.
And back to the chill side, having an atmosphere allows for sensible and latent heat loss to the atmosphere, which cools the surface. In addition, with an atmosphere we get atmospheric absorption of sunlight, emergent phenomena like thunderstorms, and cloud radiative effects, all of which cool the surface.
It’s a mixed atmospheric bag that ends up with about 20°C less temperature change from the no-atmosphere to the with-atmosphere condition than you’d expect based on the large increase of ~ 212 W/m2 in absorbed radiation at the surface.
With that as prologue, here’s the surface radiation absorption with our current atmosphere to the same scale as Figure 1.
Figure 2. Total surface radiation absorption, Earth with an atmosphere. This 510 W/m2 is all the power absorbed by the surface. (Yeah, yeah, I know there’s geothermal heat. It’s on the order of tenths of one W/m2, so it’s always ignored for this level of analysis.) The yellow/black arrow is the downwelling longwave (thermal) radiation from the atmosphere to the surface.
OK. Some simple math.
The amount of radiation absorbed by the surface increased from 298 W/m2 with no atmosphere, to 510 W/m2 with an atmosphere, an increase of 212 W/m2.
Due to surface cooling from reduced solar hitting the surface as well as sensible and latent heat loss to the atmosphere, when the atmosphere was added, the temperature only went from -2°C to 18°C, an increase of 20°C.
This means that for each additional W/m2 absorbed by the surface, including all possible influences and feedbacks from clouds, water vapor, sensible and latent heat loss, etc., in the long term the temperature increased by 20°C / 212 W/m2 = 0.09°C per W/m2
If we use the IPCC canonical value of 3.7 W/m2 per doubling of CO2, this would mean that the equilibrium climate sensitivity at the surface is 0.35°C per doubling of CO2.
“But wait”, I hear you thinking. “Climate sensitivity is how much the temperature changes, not with downwelling longwave radiation at the surface, but how much temperature changes with the “greenhouse effect” radiation (GHE) from the atmosphere and the clouds, as measured at the top of the atmosphere (TOA)”.
And you’d be right. That’s the definition.
However, we can allow for this by understanding the relationship of surface downwelling longwave and downwelling “greenhouse radiation” from the atmosphere and clouds.
When we look at the current global correlation between the poorly-named atmospheric “greenhouse radiation” measured at the top of the atmosphere (TOA) and the surface downwelling longwave radiation, when the TOA-measured greenhouse radiation changes by 1 W/m2, due to internal feedbacks, the surface downwelling longwave radiation changes by 1.26 W/m2. Or vice versa. In either case, they move in synchrony.
Figure 3. Scatterplot, surface downwelling longwave radiation versus TOA-measured “greenhouse” radiation from the atmosphere and the clouds. At the monthly level, there is no temporal lag between the two. Current average GHE radiation is 158 W/m2.
Multiplying the surface sensitivity by 1.26 to convert to sensitivity to greenhouse gases increases the equilibrium climate sensitivity from the surface value of 0.35°C per doubling of CO2 calculated above, to a final figure of 0.44°C per CO2 doubling.
OK, that’s one way to get there. Interestingly, for verification of the sensitivity estimate, we can calculate the equilibrium climate sensitivity in a totally different manner. Consider the same thought experiment as above.
With no atmosphere, the GHE radiation is zero W/m2. Currently, the GHE radiation is 158 W/m2. That gives us 20°C / 158 W/m2 * 3.7 W/m2_per_CO2_doubling = 0.47°C per CO2 doubling … compared to the 0.44°C from the previous calculation using a different method, that’s excellent agreement. So to be conservative, let me call the warming on the order of half a degree C from a doubling of CO2.
This ~ half degree C of surface warming per doubling of CO2 represents a long-term equilibrium calculation, because it includes all known and unknown atmospheric factors and feedbacks involving water vapor, cloud radiative effects, latent and sensible heat losses, atmospheric absorption of solar radiation, all of that. The climate system during the Holocene is basically in a long-term (millennia) dynamic steady-state.
One advantage of using this method to calculate climate sensitivity is that the radiation numbers are large. As a result, small variations or uncertainties in them don’t change the answer much. For example, the current calculated ~ 2 W/m2 increase in “greenhouse radiation” since “pre-industrial” times is not significant because it is only ~ 1% of the change in downwelling longwave radiation from the no-atmosphere to the with-atmosphere condition. Basically, the answer is the same whether it’s included or not.
And while my estimate of the equilibrium climate sensitivity of ~ half-degree per CO2 doubling is well below the three degrees per doubling that the IPCC uses as the sensitivity, it’s not outside historical estimates.
Figure 4. A large number of different estimates of the equilibrium climate sensitivity over the last half century, coded by color. Pale blue lines show the IPCC uncertainty values. My estimate of a half-degree per doubling is shown by the horizontal dotted black line.
Here are the authors, dates, and values of the best estimates shown in Figure 4 that are under 1 W/m2.
Author Year ECS Class
1 Specht et al. 2016 0.37 Theory & Reviews
2 Idso 1998 0.42 Theory & Reviews
3 Lindzen and Choi 2009 0.47 Observations
4 Harde 2017 0.65 Theory & Reviews
5 Lindzen and Choi 2011 0.72 Observations
6 Bates 2016 0.92 Observations
References are in the paper here.
In closing, in my post “Testing A Constructal Climate Model“, I described how the Constructal model estimates a climate sensitivity of 1.1°C per doubling of CO2. However, in that post I said:
Finally, this is a maximum sensitivity which does not include the various emergent thermoregulatory mechanisms that tend to oppose any heating or cooling. This means the actual sensitivity is lower than ~1.1°C per 2xCO2.
This latest result, of ~ half a degree warming per doubling of CO2, which includes not only emergent phenomena but all atmosphere-related phenomena, is in line with that assessment.
In related news, providing this method holds up, this means
“APOCALYPSE CANCELLED! SORRY, NO REFUNDS!”
It’s very unlikely that we will double the current CO2 level. Burning all total known reserves, not just the proven reserves, but all known reserves, will emit about 4,800 gigatonnes of CO2. This will raise the CO2 level in the atmosphere by ~ 280 ppmv. That’s far from doubling the current 420 ppmv atmospheric CO2 level.
This means that about a third of a degree of future warming lies in the ground in the form of fossil fuels.
Sadly, I fear the chance that this analysis will convince any true believers is small. They’re caught in the Upton Sinclair Trap. He famously said:
“It is difficult to get a man to understand something, when his salary depends on his not understanding it.”
It does seem recently, however, that more and more people are seeing through the climate grift. At least the US Government seems to be getting off the climate merry-go-round of endless failed predictions.
And to return to my analysis, what am I missing here? Where are my math mistakes or the holes in my logic? Why isn’t this a solid estimate of the equilibrium climate sensitivity, giving the same answer when calculated in two different ways?
Regards to all on a cloudy fall day,
w.
[UPDATE] In response to comments, a clarification.
A general comment for all, to clarify my position.
I do think that radiatively active gases affect the temperature. But I don’t think that small variations of a few watts per square meter in the resulting downwelling radiation change the temperature anywhere near as much as the conventional wisdom would have you alarmed to believe.
In addition to theoretical arguments and other evidence, my own climate model is able to do a very good job of emulating the real-world temperature using only albedo and the Ramanathan “greenhouse factor”, the percentage of upwelling surface radiation that is absorbed in the atmosphere.
Here is a more recent graphic from my ever-evolving Constructal climate model, showing how those two measurable environmental variables, albedo and greenhouse factor, are all you need to emulate the absolute temperature.
This also makes physical sense. Albedo controls how much energy is entering the system at any time. The greenhouse factor controls how much energy is leaving the system at any time. Any imbalance between those two will be reflected in a temperature change … just not as much as folks think.
This excellent goodness of fit of the Constructal climate model is further evidence that the greenhouse factor, the percentage of upwelling radiation absorbed by radiatively active gases in the atmosphere, does in fact affect the surface temperature.
Regards to everyone,
w.
As Is My Habit: I ask that when you comment, you QUOTE THE EXACT WORDS you are discussing. This is essential to prevent misunderstandings.
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“Total absorbed solar radiation is the incoming 340 W/m2”.
Does this include the 52 – 55% of incoming solar radiation that is in the infrared region?
From Wikipedia
“ In terms of energy, sunlight at Earth’s surface is around 52 to 55 percent infrared (above 700 nm), 42 to 43 percent visible (400 to 700 nm), and 3 to 5 percent ultraviolet (below 400 nm)”
Yes….1360 watts/sq. M at Earth’s average from the sun (shining on a silhouette of area Pi x R^2 where R is the radius of the Earth), then averaged over the surface area of the planet of 4 x Pi x R^2….so average of 1360/4=340….area of a circle being grade 4 math, and area of a sphere grade 5 math….
And the 187 average that gets through according to Fig 2….is highly variable by locality….obviously zero at night, so must be twice as high on average during the day…varies by latitude (nightime at one pole or the other six months of the year)….varies with cloud cover from an albedo of about 0.1 (clear sky over ocean) to about 0.8 for low cumulus cloud cover and clouds of many types cover about 65% of the planet’s surface….varies by actual surface temperature….and the amount of water vapor in the 8-14 micron wavelength atmospheric window.
Using the “Pi x R^2” approach has the unstated assumption that the angle of incidence for all incoming energy is normal (~90 deg). That ignores the high reflectivity of the 71% of the surface covered with water, most notably specular reflectance at the terminator in the absence of clouds.
Actually, even the terrestrial surface, which is generally considered a diffuse reflector, shows a variation in reflectivity with the angle of incidence. If you are fortunate enough to live somewhere where you can observe corn fields, take note of the appearance after harvest, with dirt clods and broken corn stalks in the fields, while looking at the setting sun. At glancing angles, even what are classically considered to be rough surfaces, will have obviously high reflectivity.
With nadir viewing, the surface reflects selected wavelengths that we call the ‘color’ of the material. As the angle of incidence approaches 90 degrees, the light begins to approach the spectrum of the sun, which when it has a long path length through the atmosphere, is enriched in the red-end of the visible light.
There are lots of subtleties that you and Willis are ignoring.
They just simply ignore the laws of optics. The works of Newton, Fresnel, Brewster, and many others are airily dismissed with a wave of the hand! Not only that, it is impossible to arbitrarily determine the surface temperature of anything by merely knowing the amount of “radiation” impinging upon it.
Put a redhot cannonball and a similarly sized ball of ice in the Sun. What is their surface temperature? A “climate scientist” will no doubt tell that one is hotter than it “should” be, and the other is colder than it “should” be.
Preference is given to the ignorant and gullible who write papers titled “Atmospheric CO2: Principal Control Knob Governing Earth’s Temperature.”, and similar fairytales.
Then there’s the obvious problem of removing the atmosphere exposes the oceans to a hard vacuum which boils off the oceans and induces sublimation the polar ice fields altering albedo further. So less terminator reflectance (Arnold will be sad)
Way for insulting our host…
Aaanywaaayyy…
The sun ‘s surface is, appararintly, 500 or 5000 degrees, the corona is a couple of million.
Earth is at 14 degrees, this site has educated me to the fact that our Corona sits around 400 degrees.
A worthwhile education starts by watching THE ELECTRIC COMET on the Thunderbolts Project channel.
Willis is an author, not our host. Anthony is our host.
Mister Esch. served up a doozy, we all partook in the feast, picked at the carcass… I even tried to contribute a bit of intellectual spice, and there you regurgitate the Pope’s nose of supercillious dumfaccery.
Try parsing for information, instead of playing at being a superior DEI teacher.
Oh, I forget, you’re hasbrah task is to make sure we cling to your stupid Dark Matters philosophy full of black holes, gods forbid you actually do something useful, like following the link I suggested. &%?##@&
Others have pointed out that Willis isn’t our host. He’s just another ignorant and gullible poster who believes that adding CO2 makes thermometers hotter.
Blathering about “our Corona” inclines me to sit back with a Corona or two, and laugh at some of the mental gyrations employed by commenters trying to appear intelligent.
By the way, if Willis feels insulted, he needs to take some “Improve your Self Esteem” tablets. If Clyde is the ignorant dimwit you imply, why would any rational person feel insulted by an ignorant dimwit?
I suppose you believe that adding CO2 to air makes thermometers hotter, do you? That sounds like ignorance and gullibility to me, so any attempts to insult me will be disregarded.
Not true in the slightest. In the CERES data that I’m using albedo is highest near the poles at low angles of incidence. Here you go.
w.
===
The CERES dataset calculates surface albedo by combining satellite observations and several modeling steps, leveraging both look-up tables and scene classification from MODIS data. The process involves determining the spectral and broadband surface albedo based on scene type (e.g., land, ocean, ice) and atmospheric conditions such as cloud cover and aerosol content. [1] [2]
Calculation Method
• For each CERES footprint, the algorithm first determines the scene type using MODIS imagery and the International Geosphere Biosphere Programme (IGBP) classification. [2] [1]
• If the area is cloud-free, a look-up table derived from models (such as the Langley Fu & Liou model or the COART model for snow/ice) is used. This table factors in sun angle, water vapor, and aerosol optical depth using MODIS data on the Terra satellite. [1] [2]
• Over cloudy regions, CERES uses a preprocessed history map from previous clear-sky observations for that location and date. [1]
• For snow and ice, the algorithm utilizes scene-dependent spectral shapes and look-up table values that account for variations in grain size and snow condition (fresh or permanent). [2]
• For mixed or transition regions (such as coastlines with both land and ocean), monthly averages and weighting based on water fraction are used to select the appropriate albedo source. [2]
Quality Control and Adjustments
• The CERES team addresses special conditions such as thin or thick clouds and makes corrections for instantaneous surface albedo based on empirical models and observed scene characteristics. [2]
• Updates to the look-up tables and algorithms accommodate seasonal and spatial variability in snow, ice, and cloud effects on surface albedo. [2]
• Albedo values are averaged over time (daily, monthly, annually) to create gridded products with broad spatial coverage. [3]
This multi-step approach ensures that the CERES surface albedo dataset reflects realistic environmental conditions by combining direct remote sensing with physically-based models tailored for specific surface types. [3] [1] [2]
[1](https://ams.confex.com/ams/pdfpapers/112637.pdf)
[2](https://asdc.larc.nasa.gov/documents/ceres/guide/cer_crs.pdf)
[3](https://www.tandfonline.com/doi/full/10.1080/17538947.2025.2547288)
Classically, in physics, the angle of incidence is measured from the normal at the point of incidence, to the source. That means, for polar regions, the angles of incidence should be in the range of about 60-90 degrees, NOT low angles!
Thank you for the links. They were informative, but not in the way that you probably thought.
I get the impression that you have not read my submission at https://wattsupwiththat.com/2016/09/12/why-albedo-is-the-wrong-measure-of-reflectivity-for-modeling-climate/
Link [1] of your links is the most useful. Look at Figure 1., “Clear sky monthly mean surface ‘albedo.'” It shows all of the oceans. The color palette suggests that the oceans have a uniform ‘albedo’ of less than 2%, which indicates that what NASA calls ‘albedo’ is equivalent to the specular reflectance calculated by Fresnel’s Equation for a patch of water viewed at nadir. If the minimum reflectance for an angle of incidence (solar zenith angle) of zero degrees is 2% and at a glancing angle (90 degrees) is 100%, clearly the average has to be higher than 2%. My calculations suggest that the total instantaneous hemispherical specular reflectance should be in the range of 15 +/-3%, varying with latitude and season; the instantaneous specular reflectance varies from 2% to 100%, varying with the solar elevation, or time of day. Inasmuch as Figure 1 does not show a variation with latitude, it appears that they have not calculated how the reflectance varies with the angle of incidence.
A global search for “Fresnel” does not return any matches. Perhaps Fu and Liou don’t realize that specular reflectance varies with the angle of incidence.
In any event, I don’t think it is prudent to assume that NASA is infallible, especially when they are advocating for one-control-knob CAGW.
It employs people like Gavin Schmidt, who believes
Certainly, NASA’s administration has been shown to be woefully incompetent in the past, summed up by Richard Feynman’s comment “For a successful technology, reality must take precedence over public relations, for Nature cannot be fooled“, referring to NASA’s incompetence demonstrated during the report into the Callenger disaster, which Feynman contributed to.
That’s a complete waste of time. As Grok (you may have heard of it) has pointed out, emissivity and absorptivity cancel out, as Baron Fourier (possibly more intelligent that you), pointed out a few hundred years ago, when he said that the Earth loses all the Suns heat it receives to space, plus a little internal heat.
Thus, the Earth cooling over the past four and a half billion years. You don’t have to accept reality, of course. You can keep believing that adding CO2 to air makes thermometers hotter, and I’m sure you will have lots of support by other ignorant and gullible “Climate Change causes Weather” cultists.
A wee-bit, ok way, off topic, but you reminded me and I have to share.
So this family that has lived isolated in deep Appalacia finally sends one of their oldest kid, Verne, to a public school.
After a few weeks, while Verne is washing dishes, Pa asks Verne, “What have you been learnin’ in that fancy school?”
Verne, with a round plate in his hand looks up and says, “Pi*R**2.”
Pa clouds up and says, “Enough, you ain’t goin’ back. What a bunch of hogwash. Everyone with any sense knows pie are round. Cornbread are square.”
I’ve always had a big problem with modeling the Earth’s radiation balance by having the input be uniform sunlight over the entire surface 24/7 (as 1/4 the actual incident radiation). As Clyde Spencer notes, it makes the unstated assumption that all sunlight impinges normal to the surface, whose implications I’ve never seen addressed even inadequately, let alone persuasively. This article’s beginning with an analysis of the Earth having no atmosphere, to provide a baseline of sorts, illustrates the biggest problem nicely.
During long coast periods in the space between Earth and Moon, the Apollo spacecraft were exposed to direct sunlight continuously. If we had assumed that the incident radiation was 340 W/m^2, uniformly arriving over the entire spacecraft, we would have installed heaters on board to keep things from freezing up. But since the actual incident insolation was 1,362 W/m^2 on the Sun-facing side of the spacecraft, we had to do something to prevent one side from heating up beyond its ability to function. That “something” was the Passive Thermal Control Mode, colloquially known as the “barbeque mode.” The spacecraft was put into a certain orientation with respect to the Sun, then slowly spun up about its roll axis to about 1 revolution every 10 minutes. This had the effect of evening out the solar-induced heating. But without the detail of PTC, the blithe assumption of uniform insolation of the spacecraft would lead to disaster.
And, by the way, the Apollo spacecraft looked very shiny, and was. It was made of stainless steel because of the thermal loads during launch and reentry. But as shiny as stainless looks, it only has a net reflectance of about 50% solar. That still would lead to unacceptable heating on the sunward side of the spacecraft, but only if one realized that meant 1,362 W/m^2.
Excellent. Averaging insolation over both bright side and dark side, you lose the impact of what the actual insolation value at any point in time truly is. At any point on the earth, the insolation goes from zero to maximum and back to zero. That is, in a sine function. The average is not 1370 averaged with a point on the dark side that is 180 degrees from the sunlit side. There are a number of angle factors to take into account, but basically at any given point on earth insolation is 1370 x sin(θ) x cos(φ).
None of this deals with soil and ocean temperatures subtracting from the incoming insolation and storing heat for later release. Some of the incoming insolation is stored as in a heat sink, otherwise soil temperatures, for example, would never change. Yet they do over the period of incoming insolation. Here is a graph of soil temps.
Using the average between night and day, or dividing the total hemispheric flux by the total spherical surface area, also runs into the inconvenient fact of the SB T^4 relationship. Thus, one is dealing with totally wrong temperature estimates.
Michael. Thank you for reminding me of this fact. The solution to overheating adopted by NASA relates directly to the divide by 4 argument of climate science. The rotation of the command module does not affect the intensity of solar radiation impacting on the lit side, but it does affect the process of heating of the exposed metal surface.
By the process of rotation (the barbeque mode) cold surface metal is always being brought out of shadow into sunlight and hot surface metal is always being passed round back into shadow. But the spacecraft is losing heat continuously from both the lit and dark sides.
This dynamic process can be modelled dividing the lit surface solar intensity impact by 2, (to account for the module’s spherical can shape) then loosing 1/2 of this flux to space by thermal radiation from the lit surface. The remaining 1/2 flux is rotated to the dark side where 1/2 of 1/2 is also lost to space by surface thermal radiation. The remaining 1/4 flux (1/2 * 1/2 = 1/4) is returned to the lit side as sensible surface heat of the uncooled metal surface.
This process of direct lit side heating & cooling and indirect dark side cooling stabilises at a point where 2/3 of the insolation energy is lost to space from the warmer lit side and 1/3 is lost from the cooler dark side.
Clyde, you are exactly correct. Planck even goes through this in his treatise Theory of Heat Radiation. He even addresses the difference between a smooth surface and a “rough” surface. See Sections 8 – 11 and 36 – 43. The angle of incidence is important because it determines the amount of radiation that is absorbed or reflected.
The “flat” theory of absorbance is shown faulty simply be examining the temperature of the tropics versus the poles. If everywhere has the same insolation, then the poles would have the same temperature as the tropics. That just doesn’t occur. As one proceeds from a perpendicular point beneath the sun to a pole, the reflected amount grows based on (1 – cos φ) such that everything is reflected at π/2.
Some folks that I have argued with want to use the theory that a “ray” covers more and more area as you proceed to a pole and is responsible for less insolation per square meter. The problem with that is that it assumes the insolation is from a point source close to the earth. The fact is that the earth is distant enough from the sun that insolation is a plane wave. A plain wave has a constant value at each point. If the value is 1370, then every point in the wave that intercepts the earth has that same value. That means when near the poles, the earth is receiving 1370 at each square meter, but the amount absorbed has been reduced by cos φ, which at π/2 becomes 0.
Lastly, an average insolation value is not appropriate for determining anything. The SB equation has a T⁴ term. One cannot simply ignore the fact that [(240 + δ)/σ]¼ has a much different value than [(240 – δ)/σ]¼ where δ is something like 1000. Each and every point on a latitude receives insolation based upon a sine function with a minimum of zero and a maximum of 1370 as the earth rotates.
I would download the following book:
Copyright © 2017 by Roland Stull. Practical Meteorology: An Algebra-based Survey of Atmospheric Science. v1.02b. Read Chapter 2 thoroughly. It is available here, Practical Meteorology An Algebra-based Survey of Atmospheric Science – version 1.02b. isbn 978-0-88865-283-6
Jim,
Why do you continue to spout such nonsense? You say:
“A plain [sic] wave has a constant value at each point. If the value is 1370, then every point in the wave that intercepts the earth has that same value. That means when near the poles, the earth is receiving 1370 at each square meter…”
I responded to this assertion of your last week, but never got a response:
https://wattsupwiththat.com/2025/11/19/just-how-good-were-the-early-climate-models/#comment-4133626
I stand behind all my arguments in that comment. I’ll try another tack here.
First, consider a photvoltaic panel 1 meter square directly facing the sun. We agree that there will be 1370 watts incident on this panel.
Compare this to two of these 1m x 1m panels each tilted 60 degrees off from directly facing the sun, joined at the far end in a V configuration. The opening to this system is the same 1m x 1m square that the single panel in the first system occupies.
Your analysis says that there is 1370 watts of radiative flux incident on each of these panels, for a total of 2740 watts. Your argument is that half (cos 60 = 0.5) of the flux is reflected from each panel, so there is only 1370 watts that could be absorbed.
But the reflected flux from each panel is largely directed toward the other panel, which will absorb some of it, increasing the total absorption to a value greater than 1370 watts. And this is from radiation that passed through the 1 m^2 opening to the “V”, which only transferred 1370 watts there.
Two questions for you:
1. How is this not a violation of conservation of energy?
2. Why are people not building systems to take advantage of this supposed phenomenon?
Ed, because you believe that adding CO2 to air makes thermometers hotter, I’ll type slowly.
You say –
Well, no, there won’t – and neither you nor anybody else has ever measured that. Oh, unless you really mean that the solar panel is in space, with no atmospheric attenuation taking place.
Maybe you are just trying to perform an illusion, being a fraud rather than a fool? If so, I’ve seen through it, haven’t I?
I would suggest that you hop on over to https://en.wikipedia.org/wiki/Fresnel_equations and try to absorb what it says. Glass also exhibits the phenomenon of varying the reflectivity with the angle of incidence and the angle of polarization. The original direct light from the sun is unpolarized. However, after the first reflection, it will be partially polarized and behave differently on the second reflection for your suggested V-configuration.
Clyde,
That is irrelevant to the point I am making. In my configuration, assuming Jim’s argument that the tilted panels receive the same power flux density as a panel directly facing the flux direction, if ANY of the solar radiation reflected off one panel is absorbed by the other, there is a 1st Law conservation of energy violation. Differing absorption for s and p polarization doesn’t matter.
You might want to explain why you leave out the fact that the absorption by the other panel also reflects back to the first panel. In other words, there is no energy gain, only equilibrium.
Jim: You are analyzing a different problem here – the power exchange between panels. I am analyzing the exchange between the system containing both panels and the rest of the universe.
Each additional absorption by either panel adds to the total absorbed by the system in your analysis, increasing it further from what entered the system, and making the violation of the 1st Law worse.
Each “additional” absorption by one panel is offset by that same panel emitting the same back toward the other panel.
Why are you continuing to press your straw man that does not by any stretch of anyone’s imagination meet the physical structure of the surface of the earth.
“Compare this to two of these 1m x 1m panels each tilted 60 degrees off from directly facing the sun, joined at the far end in a V configuration.”
You are confusing your configurations. Point A on the surface of the Earth does not face Point B on the surface of the earth. Reflected energy from Point A does not impinge of Point B.
Tim,
I’m afraid you miss my point entirely. I was saying that IF a surface tilted with respect to the flux direction received the same power flux density as one directly facing the flux direction, as Jim asserts, then a configuration such as the one I suggested COULD be set up.
In such a case, more power would be absorbed than what entered the top of the “V”, a blatant violation of 1st Law conservation of energy. So Jim’s claim cannot be true.
You are ignoring that if the panels are similar, then any reflection from one, will also be offset by reflection from the other essentially meaning equilibrium.
That is why “back radiation” is real problem with radiation. Planck discusses this very scenario and shows how compensation occurs by immediately radiating away any reflected radiation.
No, each subsequent absorption by either panel adds to the total absorbed by the combined system of energy from outside the system, increasing it further beyond the power input to the combined system.
Have you ever done any formal thermodynamic analysis of systems?
Why yes I have. While in college I worked on a team formed by a professor consulting on pump houses and stilts for the Alaskan pipeline. I have also done design work on heat sinks for HF RF amplifiers that required analysis of conduction, convection, and radiation components. Many required forced air cooling due to size restrictions.
I have a BSEE and have studied high temp steam boilers and cooling down to hand held devices requiring heat dissipation. I have spent the last four years studying Planck’s thesis on heat radiation in detail.
From Planck:
Your assertion requires that a mirror absorb and store that energy. That just doesn’t occur.
Show a drawing of your experimental system along with the math that suppots your assertion.
I suggest that you go back to the drawing board — literally. Sketching your suggested configuration shows that only half of the light rays reflected off one panel will impinge on the adjacent panel. The first reflection, at an angle of incidence of 60 degs, looks to reduce the transmitted light (to the photovoltaic layer) to about half of what it would be at normal incidence.
It doesn’t appear to me to be a violation of conservation of energy. It looks like you ignored how much light is reflected at each bounce and maybe didn’t realize that only half of the rays hit the adjacent panel. So that answers your second question of “why” people aren’t taking “advantage of this supposed phenomenon.” It doesn’t exist.
I have sketched the system in some detail – WUWT is not cooperating in attaching my diagram. So I will try to explain verbally here.
Using Jim’s analysis, if the reflection is specular (mirror-like), the reflection off Panel A will be directed straight at Panel B, striking it at a 90 degree angle. (The same for reflection off Panel B directed at Panel A.)
Even with more diffuse reflection, the bulk of Jim’s reflected radiation will strike the opposite panel at a high angle, where Jim’s analysis says there will be high absorption. And virtually all this reflected radiation will be directed toward the other panel.
But the point remains. Under Jim’s analysis, ANY subsequent absorption of radiative energy reflected from the solar radiation entering the system means that the two-panel system is absorbing more power than enters the system, which is a physical impossibility.
Read this carefully. What you are describing is a situation where equal amounts are being reflected from one panel to the other.
As panel A absorbs 1/2 of the insolation while sending 1/2 toward panel B, guess what, panel B is absorbing 1/2 of the insolation while sending 1/2 towards towards panel A.
Lastly, as you’ve been told, your arrangement DOES NOT MIMIC the surface. It is a straw man that proves nothing about insolation intercepting a spherical surface such as the earth.
If you wish to perform a better arrangement, lay two 1 m^2 panels flat one the earth. Show how the reflected insolation from one can be absorbed by its neighbor.
Jim: You continue to completely miss the point. I am evaluating the energy balance of the combined dual-panel system. Yes, the reflected insolation from panel A to panel B is the same as from panel B to panel A. But both of these act to increase the total aborption of the dual-panel system.
Going carefully step by step. The open top of the “V”, 1m^2 in area is receiving parallel insolation flux of 1368 watts (to use your number; the particular value is not important). Each panel is tilted 60 degrees away from normal to this flux, forming the V.
You claim that each panel has the full 1368 watts incident upon it, but only can absorb half of it (684 watts), because cos60 = 0.5, reflecting the other 684 watts away. So far, we have 1368 watts in, 2 times 684 watts = 1368 watts absorbed. So far, so good.
But because of the geometry of the V, some (probably most) of the reflected flux from Panel A will hit Panel B, where some of it will be absorbed. And yes, some (probably most) of the reflected flux from Panel B will hit Panel A, where some of it will be absorbed. In both cases, the energy absorbed by the full system is increased, creating the case where the system absorbs more energy than it receives. This is the 1st Law violation I keep noting.
But, you protest, my arrangement “DOES NOT MIMIC” the earth’s surface. Irrelevant! You have made a very general argument about what happens to radiation when it hits a non-normal surface. You have deluded yourself into thinking that because in the earth’s case, the reflection from the surface is (mostly) directed back into space, that your argument can work.
But I am simply exploring the ramifications of your argument. The properties of this reflection CANNOT BE dependent on what there is “downstream”. I only need to show a single case where your analysis yields a physical impossibility to invalidate that analysis.
Let’s say Panel A is lying on flat ground a 60 degrees north latitude. Joined at the north end of this panel, we add Panel B, whose southern end is tilted up 30 degrees, so the combined system has an included angle of 60 degrees. This is my “V” dual-panel system. At noon on the equinox, the parallel “plane wave” flux from the system flows directly into the mouth of this V. The reflected flux off Panel A, which you have implicitly assumed to travel off back into space, instead encounters Panel B, where some of it is absorbed.
Surely you cannot be arguing that what happens to the radiation striking Panel A directly from the sun is dependent on whether there is a Panel B present or not! Can you?
You are trying to begin a debate about a straw man you have made up. No thanks, find someone else.
You keep trying to describe the Earth as being the front of a parabolic antenna. It isn’t. It’s the *back* side of a parabolic antenna. If you have a plane wave hitting the back side of a parabolic antenna then just how does Point A on that back side reflect any energy to Point B on the backside of the antenna? The front side concentrates the energy received. The backside disperses it!
What parabolic shape am I describing? I think you are probably trying to distinguish a system that is concave facing the sun (my example), versus one that is convex (a smooth globe) or at least flat.
As I explained to Jim above, this is irrelevant. His argument for what happens to radiation hitting a surface CANNOT be dependent on what subsequently happens downstream of that event, whether the reflected radiation encounters something that absorbs it or not.
First, this is a straw man argument that has nothing to do with the geometry of the earth. The earth is a sphere and the surface varies both north to south and east to west. The surfaces will never face each other directly. This obviates your example arrangement of solar panels.
Second, a plane wave has both a magnitude and direction. You need to review your vector math and see how your straw man setup will work when the edges are aligned with a longitudinal line. Show us your geometry with the vectors appropriately drawn.
You obviously did not download and read the textbook section I posted above. Here it is again.
https://www.eoas.ubc.ca/books/Practical_Meteorology/
Here is an image from the book that outlines the geometry and equations necessary to calculate the absorbed insolation at a point on the earth.
Here is a prompt you can give to CoPilot (think deeper) to obtain the same information that the textbook discusses.
Perhaps you can discuss how these resources are incorrect. Beware 99% percent of the internet references that use the sun as a point source rather than a plane wave whereby a ray is spread over larger and larger areas as you move around the sphere toward the poles.
I forgot to add this image of insolation at my location. As you can see, this shows what is basically a sine wave function from sunrise to sunset. You might want to explain this using your interpretation of the process that creates this and how an average is not actually a good representation of the temperature considering that there is an exponential factor of 4 involved.
Good grief, Jim! The sources you cite support MY arguments, not yours! You don’t even understand the basic concepts here.
The textbook example you cite asks, “During the equinox at noon at latitude ϕ =60°, the solar elevation angle is Ψ = 90° – 60° = 30°. If the atmosphere is perfectly transparent, then how much radiative flux is absorbed into a perfectly black asphalt parking lot?”
In radiative physics, “perfectly black” means “absorbs all incident radiation”, that is, an idealized blackbody. (This includes all angles.) The reduction in absorbed flux comes not from reflection, as there is none in this example, but from the basic geometry, as I have argued.
You asked me to consult Copilot, so I put in your exact prompt.
Copilot replied with “If the Sun is 45° above the horizon, the incoming insolation strikes the Earth’s surface at an angle, causing the rays to be spread out over a larger surface area, reducing the intensity of the radiation.”
This is EXACTLY what I have been arguing, and what you have been denying!
I’m not sure if I will be able to break loose much more time soon for more replies, but I will try.
“n radiative physics, “perfectly black” means “absorbs all incident radiation”, that is, an idealized blackbody. (This includes all angles.) The reduction in absorbed flux comes not from reflection, as there is none in this example, but from the basic geometry, as I have argued.”
None of these assumptions by Planck apply to the Earth.
Tim: You still miss my points completely.
1. Jim pasted a textbook example showing the “cosine” effect of radiation hitting a non-normal surface. He claims that this cosine factor is due to reflectivity difference with angle. But this textbook example includes this cosine factor for a ssurface with ZERO reflectivity, which doesn’t agree with Jim’s argument at all. The fact that the earth’s surface is not a perfect blackbody in no way contradicts this.
2. My analysis uses perfectly parallel rays of radiation, which is only a slight oversimplification. The “spread” on the non-normal surface is NOT due to lack of parallelism in these rays. To build on your laser example, imagine a group of perfectly parallel laser beams with 1 cm spacing. When these hit a surface tilted 60 degrees from normal, the spots on that surface will be 2 cm apart, not 1 cm. Nothing to do with non-parallel rays or reflectivity of the surface.
3. As you say, the “real world of surface irregularities” is complex, nowhere near the simple cosine effect that Jim asserts. You say that Planck “accounts for angle of incidence by assuming a black body has zero reflectivity”. Yet the cosine effect is still here when there is zero reflectivity, which is my point exactly.
If the asphalt surface only absorbs 1/2 the insolation, then the question you need to answer is:
Where does the other 1/2 of the insolation go when there is no reflection?
I want to see a reference that describes what you think happens to the remaining energy.
There is no “remaining energy”. There is no “other 1/2 of the insolation”.
You evaluate how much insolation reaches a surface by performing a surface integral calculation. The surface integral of a vector flux hitting a surface involves first taking the dot product of the flux vector and the unit surface normal vector. The dot product includes the cosine of the angle between these two vectors.
In the example we have been using, where this angle is 60 degrees, the cosine is 0.5. So the flux of 1368 W/m2 (to use your number) hitting a surface 60 degrees off of normal to its direction results in a 684 W/m2 flux density on that surface.
This is completely standard math and science that you should have gotten in the first year of a college science or engineering course of study. It’s been around a lot longer than any “climate science”.
For reference, these Wikipedia pages are decent:
https://en.wikipedia.org/wiki/Surface_integral
https://en.wikipedia.org/wiki/Divergence_theorem
Both show the use of the dot product of these vectors, which of course includes the cosine term.
Where does the other 1368 – 684 = 684 W/m2 disappear to? You were interested in conservation of energy. Is the remaining 684 somehow destroyed?
It was never there in the first place. When tilted by 60 degrees, the panel only intercepts 1/2 of a m^2 of insolation flux, not a full m^2. So if the flux is 1368 W/m^2, the 1 m^2 tilted panel only intercepts 1368 * 0.5 = 684 watts.
Come on dude. The panel DOES intercept the entire ray. However it only absorbs the perpendicular component of the vector.
You need to show a reference with a diagram and math supporting your assertion.
“ssurface with ZERO reflectivity”
The Earth is neither perfectly reflective nor is it have perfect absorption.
“ When these hit a surface tilted 60 degrees from normal, the spots on that surface will be 2 cm apart, not 1 cm.”
The rays will *still* be 1cm apart.
It does *NOT* matter how far apart the spots on the surface are. Those laser beams will still be 1cm apart. The radiation intensity associated with each of those rays will still be the same.
The difference will be the angle of incidence on each spot.
Heat transfer is accomplished only by the proportion of the ray that is perpendicular to the surface, dA is the incremental surface perpendicular to the ray. On a spherical surface that perpendicular area is defined as dA = Acosθsinφ. A larger subtended spot will still only project a certain area perpendicular to the plane wave. The larger the spot (i.e. the further around the sphere it is) the smaller Acosθsinφ becomes. Meaning the less heat transfer there will be. At 90deg there won’t be any heat transfer because cosθ = 0.
It isn’t a matter of the spot becoming bigger, it’s a matter of the area perpendicular to the ray becoming smaller because of the angle of incidence.
Think about it. If the laser beams are 1cm apart and the spot grows to 2cm then why doesn’t 2 beams hit that spot? If it grows to 3cm then why doesn’t 3 beams 1cm apart hit it? if the influence of laser_beam1 goes down by half as the area of the spot doubles then what happens to the toral when the laser_beam2 joins in on action?
Tim: I can’t believe I need to explain this in more detail. Let’s build on the parallel discrete lasers example.
We have a grid of 100 by 100 parallel lasers, 1 cm spacing both axes, all directed perpendicular to the plane of origin, and each transmitting 0.1 watt. The total radiative flux is 100 x 100 x 0.1 = 1000 watts, and since this grid covers an area 1m x 1m, the flux density is 1000 W/m2.
If these beams hit a 1m x 1m surface that is perpendicular to the beams, there will be 1000 watts of radiative flux onto that plane (100 x 100 x 0.1). The spacing of the laser spots on the surface will be 1 cm in both axes.
Now we take this 1m x 1m plate and tilt it 60 degrees in one axis. In one direction, only 50 of the rows of lasers will hit the plate, with the rows space 2 cm apart along that axis. (In the other direction all 100 colums will hit the plate.) So the radiative flux hitting the tilted plate is 50 x 100 x 0.1 = 500 watts. Since the plate is still 1 m^2 in area, the flux density hitting it is 500 W/m^2.
Of course, the same effect holds true for more continuous radiative fluxes.
This difference is before any possible reflection. I am only talking about the radiation incident on the surface in the first place.
Tim – You say:
“Heat transfer is accomplished only by the proportion of the ray that is perpendicular to the surface, dA is the incremental surface perpendicular to the ray. On a spherical surface that perpendicular area is defined as dA = Acosθsinφ. A larger subtended spot will still only project a certain area perpendicular to the plane wave. The larger the spot (i.e. the further around the sphere it is) the smaller Acosθsinφ becomes. Meaning the less heat transfer there will be. At 90deg there won’t be any heat transfer because cosθ = 0.”
This is exactly what I have been arguing to Jim these past few days, and he has argued vehemently against. Perhaps you can succeed in convincing him where I have failed.
Then you say, with regard to our example of discrete lasers:
“It isn’t a matter of the spot becoming bigger, it’s a matter of the area perpendicular to the ray becoming smaller because of the angle of incidence.”
I never argued about the size of the spots – I talked about the spacing. We are actually in agreement here.
Now we just have to convince Jim that this has nothing to do with reflectivity!
“This is exactly what I have been arguing to Jim these past few days, and he has argued vehemently against. Perhaps you can succeed in convincing him where I have failed.”
If you think that is what Jim has been doing then you have a reading comprehension skill problem.
“I never argued about the size of the spots – I talked about the spacing. We are actually in agreement here.”
How does the spacing change? The spots don’t move in relation to each other. The only thing that changes is the perpendicular area presented to the plane wave. And that perpendicular area is related to the cosine of the angle of incidence.
Tim:
Jim’s original comment that brought me in said:
He is very clearly arguing that the lower absorption at low incident angles is due solely to increased reflectivity. I have been arguing that this is wrong, that it has nothing to do with reflection. This cosine effect will occur even on a perfectly absorbing surface.
He has twice presented me with a textbook example that explicitly uses a perfectly absorbing surface (i.e. no reflectivity) to try to argue that the effect is due to reflectivity!
In your most recent comment, you say, “The only thing that changes is the perpendicular area presented to the plane wave. And that perpendicular area is related to the cosine of the angle of incidence.”
That is exactly what I have been arguing, and against what Jim has been arguing.
The textbook very explicitly states that the surface is not a perfect absorber.
Your assertions would be much more convincing if you would include the texts and graphics that support you.
That’s funny… In all the equations you have presented, your own and textbook, the albedo factor A has never appeared once. How could this be the cause if it does not appear in the equations?
In the diagram you have presented twice, it states clearly that the surface is “perfectly black”, which means no reflectance. Yet the cosine term is there, and A is not.
What do you think albedo is?
I’m really beginning to wonder if you are just an AI troll.
My points about albedo (reflectivity) are perfectly valid, A “perfectly black” surface, as in the example Jim provided twice, has zero albedo. And yet, the cosine factor appears in that example.
That is *NOT* what a “perfectly black” surface means. It does *NOT* mean a zero albedo.
What is the characteristic of an object that is black vs an object that is white?
“Black” objects absorb incident radiation of all wavelengths, reflecting none. This means that, by definition, they have zero albedo. “White” objects reflect incident radiation of all wavelengths, absorbing none. They have 100% albedo.
In non-technical usage, people are talking about visible wavelengths. In technical contexts such as that textbook, the term refers to all wavelengths of interest.
So yes, Jim’s example was talking about a surface with zero albedo, yet still containing the cosine term.
Heat transfer and reflectivity are *not* the same thing. Heat transfer is only based on the normal vector to the surface.
Here is what copilot says:
When you imply that the warming is less because the area impacted on the surface changes the implication is wrong.
The area impacted doesn’t change. The radiation intensity doesn’t change. Remember back to the radiation being made up of laser beams, that laser beam doesn’t change intensity nor does the area it hits go up. If the area *did* change then more laser beams would hit that increased area keeping the total radiation the same.
What changes is the angle of incidence which determines the amount of radiation aligning with the surface normal vector.
It’s why a radiation intensity measurement device that depends on the heating of a sensor must be perpendicular to the radiation in order to get an accurate measurement. The area of the sensor doesn’t change. The collection aperture doesn’t change. The beam width doesn’t change. What changes is the angle of incidence compared to the normal vector of the sensor surface.
For a true black body the angle of incidence doesn’t apply, it’s a specific assumption that Planck made in his treatise. That is *not* true for a black object that is not a black body.
This is unbelievable! You keep citing sources that agree with me, and undercut your whole argument!
Your latest copilot response:
“What changes is the projection of the sunlight beam onto that area.”
“So yes: what I described as “projection” is precisely the cosine dependence on the surface normal vector. Asphalt heating depends on how closely the Sun’s rays align with the surface normal.”
Then you say: “When you imply that the warming is less because the area impacted on the surface changes the implication is wrong.”
What do you think this “projection” copilot is referring to? If you are correct, why doesn’t it mention anything about reflection/reflectivity?
Let’s see if I can get WUWT to accept this drawing, or at least the link to it. The top diagram shows lines of parallel flux hitting a perfectly absorbing perpendicular plane. All ten of the flux lines hit the surface (lets call it 1 m^2), with the spacing between points on the surface the same as between the flux lines.
The bottom diagram shows these same lines of parallel flux hitting this same perfectly absorbing plane of of 1 m^2 that has been tilted 60 degrees (cos(60)=0.5). Only half of the flux hits the plane now, and the points on the plane are now twice as far apart. In this case, the absorption of the perfectly absorbing 1 m^2 target plane is one-half of that of the perpendicular case.
https://docs.google.com/document/d/1aVS9hSle0I3HI4vsxbs-dIyGQiq_B_rQGXmNfd7domY/edit?tab=t.0
Let’s try this link for the diagram:
I don’t care to have an account at drop box.
I borrowed an iphone. Keep in mind a laser as you have drawn them is not an exact replication and neither is a flat surface.
In a plane EM wave there is no separation as one would get with separate lasers. Each m² gets the same insolation. Every μm² receives the same insolation.
Nothing you have shown refutes that only the perpendicular component of a ray is absorbed.
Replace each of the discrete lasers with a million lasers each of one-millionth of the power, and the result is the same. Continue the process to the continuous case (or at least quantum photon case) and the result is the same.
In all these cases, the 60-degree tilted panel intercepts only half of what the normal plane does. What it does not intercept, it cannot reflect. If the parallel flux density is 1368 W/m^2, the tilted 1 m^2 panel can only intercept 684 watts.
You claim that this tilted 1 m^2 panel intercepts the full 1368 watts, but reflects half of it. Let’s look at the implications of that. I have uploaded a diagram of the case I was verbally describing earlier. (Dropbox is the only method I have gotten to work.)
https://www.dropbox.com/scl/fi/wan9el98my8yy84qrjyzo/V-Shaped-Receiver-Rotated-2025-11.vsdx?rlkey=9s3ho3tmqpp6pfo6qe8qhppng&st=mo1pb3b0&dl=0
Like the other diagram, this shows a side view of the system, 1 meter in “depth” (into the page).
The diagram shows discrete flux lines, but the continuous case is the same. The bottom half is the equivalent of the diagram you have already looked at. Your claim is that the bottom 1 m^2 panel has incident upon it the full 1368 watts, but because of the angle, even a “perfectly black” surface will reflect half of the flux. This reflected half is shown by narrower arrows in the diagram. They are shown for specular (mirror-like) reflection, but more diffuse reflection does not affect the argument.
The upper panel will intercept at least some of the radiation reflected from the lower panel. (If the reflection off the bottom panel is specular, this reflection would be perpendicular to the top panel, and potentially all of it could be absorbed.)
The same is true for radiation hitting the 1 m^2 upper panel first. You claim that this panel would have incident upon it the full 1368 watts, reflecting half of it toward the lower panel, which will absorb at least some of it.
So let’s look at the full system now. We have 1368 watts entering the 1 m^2 opening on the left. I think we are agreed that if there were a panel there, perpendicular to the flux, it would have 1368 watts incident upon it.
You claim that the lower panel has 1368 watts incident upon it, absorbing 684 watts and reflecting 684 watts. Similarly, the upper panel has 1368 watts incident upon it, absorbing 684 watts and reflecting 684 watts.
So far, we have 1368 watts entering the system, and 684+684=1368 watts absorbed. So far, so good. But we haven’t considered the reflected radiation you claim. At least some (and probably most) of this is absorbed by the opposite panel. And if ANY of this is absorbed, the system is absorbing more power than is entering the system.
This is a violation of both the divergence theorem and the 1st Law of Thermodynamics (conservation of energy). Therefore, it cannot be so!
I think you are confusing yourself by only considering the wavefront, when you really need to be looking at the power flux. My points are painfully obvious to anyone examing the radiative power flux vectors.
The 60 degree plane intercepts the entire wave. Some is reflected and some is absorbed. You have yet to tell us where the half that is reflected goes. Maybe this will help.
Light Reflection, Absorption and Transmission in Remote Sensing – GIS Geography
“The 60 degree plane intercepts the entire wave.”
Don’t be absurd! The diagram clearly shows that the 60 degree tilted plane only intercepts half of the flux that the perpendicular plane does. What is not intercepted cannot be either absorbed or reflected.
“You have yet to tell us where the half that is reflected goes.”
That’s because it DOESN’T EXIST!! I went into great detail showing that if the tilted plane intercepted as much as the normal plane did, but reflected half of it, there would be blatant 1st Law violations. You have presented no counterargument.
The page you link addresses nothing on this point. All it says it that reflection can occur.
I want to emphasize that I am not arguing that reflection cannot occur. It’s just that your concept that a non-normal surface has as high a flux density incident on it as does a normal surface, but it reflects an amount purely based on the angle, cannot possibly be true.
That is absurd. Of course it exists. The insolation has an intensity of 1370 W/m^2.
The energy exists! If it hits at 90°, you have already said 1370 would be absorbed. If the plate is turned to 60°while the same ray is incoming, the ray still has 1370. Where does the “other half” disappear to? Simply saying the non-absorbed half no longer exists is bizarre.
Ed obviously believes that your bathroom scale would read differently if you lay on it or stand on it or even if you lean forward or backward!
You never took AP Physics in high school or calculus based physics in college, did you?
This entire subject is no different than the inclined plane and gravity in physics. Gravity is a wave, it is a vector with direction and magnitude. It has intensity.
If you inclined plane is horizontal to the gravity wave, you get maximum downward force on a car placed on the plane, i.e. maximum friction. You get zero horizontal force so you have no acceleration of the car.
Tile the plane with respect to gravity and you then have to break the gravity wave into two vector components. One component is parallel to the inclined plane and the other is vertical to the inclined plane.
The parallel vector accelerates the car. The perpendicular vector produces a friction force on the car. Thus the acceleration goes up while friction goes down. Add the parallel vector to the perpendicular vector and you get the total force, the same total force you get if the inclined plane is horizontal.
For a parallel wave like that from the sun the same thing happens. The vector parallel to the surface goes up and the perpendicular vector goes down. For the EM wave, the parallel component produces no heating. The perpendicular component does.
The problem with your diagram is that an EM wave is continuous. It’s intensity is continuous. It’s not made up of discrete arrows. It’s no different than gravity. The same continuous intensity hits the inclined plane and the car no matter its orientation. The difference is the vectors produced, the parallel vector and the perpendicular vector. The total force on that car on the inclined plane is the same no matter its orientation. Only the parallel and perpendicular vectors change. One goes down, the other goes up. If the car intercepted less of the gravity wave then the total parallel and perpendicular vectors would not add to the same magnitudes as you increased/decreased the tilt of the inclined plane.
Think of it this way, the total watts is the integral of the two vectors over the area illuminated. So you get
I[cos(Θ)] ∫ dA from 0 to 1 + I[sin(Θ)] ∫ dA from 0 to 1
(assuming you have a square area that is 1m on each side)
This basically evaluates to { I [cos(Θ)] + I[sin(Θ)] } * 1 where 1 is the area of one square meter.
In your example the total intensity over the area of the plane goes down as you rotate it. If that wave was gravity then the weight of the plane would go down, a physical impossibility. Do you weigh differently if you lay down on your bathroom scale than when you stand on it?
For the record, I did study AP Calculus and Physics in high school, getting a 5 on both exams. I continued my studies of both at MIT (heard of it?), in the advanced track that only a minority of students there attempted. I assure you that you would have been laughed out of those courses!
You have a completely erroneous concept of flux and flux density. My basic diagram shows a rightward parallel uniform flux of a cross-sectional area of 1 m^2. I represent it by discrete lines, but you can consider it a representation of a uniform flux (like we draw “lines” of magnetic flux to represent a continuous flux).
If the flux density is 1370 W/m^2, as you suggest, the total power in this flux is 1370 W/m^2 * 1 m^2 = 1370 watts. (Because it is uniform, we don’t have to do the formal integral.) A planar panel of 1 m^2 area with surface normal to this flux has 1370 watts incident upon it.
The lower diagram shows the same parallel flux and same 1 m^2 panel, but with the panel tilted 60 degrees. Now only the lower half of the flux, with a cross-sectional area of 0.5 m^2, hits the panel. So the radiant power incident on the panel is 1370 W/m^2 * 0.5 m^2 = 685 watts. What happens to this flux is dependent on the surface properties of the panel, but the key point is that there is only 685 watts hitting the panel.
The top half of the flux, also transmitting 685 watts, misses the panel completely. It cannot be absorbed or reflected. You seem to think that this somehow hits the panel as well. I’d like to know why!
As to your scale analogy, if I put a 10-pound weight on a bathroom scale, then tipped the scale 60 degrees (with the weight glued on), the scale would only report 5 pounds.
“You have a completely erroneous concept of flux and flux density.”
How then do you explain the inclined plane experiment? Gravity *is* a flux and has a flux density.
“Now only the lower half of the flux,”
You’ve been asked this at least a dozen times – what happens to the other half of the flux? Does it just disappear somehow? You *HAVE* to account for it in some manner since just rotating the plane doesn’t change the flux density. Your “diagram” doesn’t account for the other half of the flux!
“The top half of the flux, also transmitting 685 watts, misses the panel completely.”
So you *do* think that the other half just disappears! Do you *really* believe that the surface of the earth is just 1m^2? And that if you tilt the earth that half of the sunlight just disappears by missing the surface of the earth?
And you *do* have to integrate over the entire area to get total energy delivered by the impinging power.
“It cannot be absorbed or reflected. “
Does tiling the inclined plane with the car on it get lighter when you tilt the plane? Does rotating the inclined plane to 90deg suddenly cause the car to become weightless because all of the gravity wave misses it?
Of is part of the gravity wave change the friction between the inclined plane and the car and another part of the gravity wave accelerate it? What happens to the friction when the plane is tilted to 90deg?
Again, the plane wave from the sun has DIRECTION and MAGNITUDE. Tilting the receiving area does *NOT* decrease the magnitude nor does it decrease the intensity of the sun’s insolation radiation. Since it has direction and magnitude you *have* to treat it as a vector and not as a scalar quantity the way you are doing. Only the perpendicular component of the vector transfers heat, the parallel component does not.
Think of it this way. If you take an area of dA, i.e. the area as x and y approach zero, does NO heat get transferred into dA because all of the sun’s insolation misses it? For if that is true then the heat summed over A would be zero because the contribution to the integral by each dA element would be zero. I.e. Σ dA from 1 to n would be zero. 0 + 0 + 0 + … + 0 = 0.
Just how much sun insolation does your example provide for each dA element? And, one more time, what happens to the insolation that misses the dA?
This *is* simple vector math. The same vector math used with the inclined plane. Why don’t you try to explain how the inclined plane works with the gravity insolation using your example. Show us how the gravity insolatin “misses” the car as you rotate the inclined plane.
“As to your scale analogy, if I put a 10-pound weight on a bathroom scale, then tipped the scale 60 degrees (with the weight glued on), the scale would only report 5 pounds.”
You missed the WHOLE POINT of the example! What happens to the rest of the mass glued to the scale? Does it disappear? Does the gravity wave “miss” part of the mass? Did you think everyone would miss your moving the goal post by “gluing” the mass to the inclined plane? What happens if the mass is *NOT* glued to the scale?
Yep! You have confirmed it. As I said, you have a completely erroneous concept of flux and flux density.
I provide an example where only the lower half of the 1 m^2 of flux hits the plane, now that it has been tilted 60 degrees. You insist I answer the question “what happens to the other half of the flux? Does it just disappear somehow?
That has to be one of the stupidest questions I have been asked in a long time! What happens to the solar insolation that just misses the earth? It just continues on and is no longer relevant to the problem!
You say my diagram “does not account for the other half of the flux.” It certainly does – it shows it continuing past the plane so it cannot affect the plane, just like the solar flux that passes over the earth’s poles and continues on without further affecting the earth. Why is this so hard to grasp?
You claim “You *HAVE* to account for it in some manner since just rotating the plane doesn’t change the flux density.” Here is where you confuse flux and flux density. If the density doesn’t change, but the area perpendicular to the flux is reduced by half, the flux is reduced by half.
Let’s reduce it to the most basic case. In my lower diagram, consider the trapezoidal area in the bottom half. 685 watts enter this area from the left. You claim that 1370 watts exits this area on the right, either from absorption by the panel, or reflection off it.
You questioned my physics background, so I am questioning yours. In any decent introductory E&M physics course, you will learn about the divergence theorem, which states that the net flux across a system boundary (calculated in general by the surface integral), is equal to the amount generated inside that system, which in this case is zero. But you have 685 watts more leaving than entering. Did you sleep through the lecture on the divergence theorem?
You say ” Since it has direction and magnitude you *have* to treat it as a vector and not as a scalar quantity the way you are doing.”
WTF??? I have been saying from the very beginning that you have to take the dot product of the solar flux VECTOR and the surface normal VECTOR to compute the flux incident on the surface. Were you even paying attention?
As to the gravity analogy (be careful of analogies, BTW), if I used your analysis, I would think that the 10-pound weight on the 60-degree tilted scale would put 10 pounds of force on the scale’s spring. My analysis says that there will be only 5 pounds of force on the spring. (And there is no reflection in this case.)
You ask “What happens to the rest of the mass glued to the scale?” Technically, you should be asking about the weight force, but we’ll let that slide. That creates a perpendicular force that does not compress the spring (hence the need for glue…)
I could go on and on answering your other ridiculous arguments, but there is no point.
Ed, it is *this* simple.
In your example you say the flux becomes (1/2)I because less flux hits the surface with a 60deg tilt. The other half “misses” the surface.
The example then shows that (1/2)I hitting the 60deg surface at a 60deg angle.
Now, in order for that (1/2)I hitting at a 60deg angle to warm the object either 1. it is *not* true that heat is only transferred from the perpendicular component of the flux OR 2. all objects are black bodies where the angle of incidence doesn’t apply.
If it is *true* that only the perpendicular component of the flux can transfer heat then what you wind up with in your example is one half of (1/2)I is the warming component.
Thus the final warming flux is (1/2)(1/2)I = I/4.
So which is it? It *has* to be one of these in order for your example to work. Pick one and stick with it.
“What happens to the solar insolation that just misses the earth?”
Flux is flux. It is not cumulative as far as intensity. It doesn’t matter how small you make the receiving surface, that surface will receive the same amount of radiant flux intensity. That means that dA will get 1370 W/m^2.
The earth is a sphere. Rotating it presents the same amount of surface to the plane wave from the sun, it does *not* make the receiving area smaller. So what part “misses” the earth when you rotate it? The same amount that hits it before you rotate it?
“but the area perpendicular to the flux is reduced by half, the flux is reduced by half.”
What changes is the perpendicular component of the radiant flux. The radiant flux remains the same, it is *NOT* reduced by 1/2.
The total flux is I_total = I[cos(Θ)] + I[sin(Θ)]. What changes is Θ, not I_total. When Θ changes, the perpendicular and parallel components change values but together they still add to the same total!
I pointed this out to you but you ignored it – as usual.
“WTF??? I have been saying from the very beginning that you have to take the dot product of the solar flux VECTOR and the surface normal VECTOR to compute the flux incident on the surface. Were you even paying attention?”
You don’t seem to realize what you are saying here! You are admitting that it is the perpendicular component of the solar flux VECTOR that transfers heat. But your example shows the solar flux VECTOR being (1/2)I being incident on the surface. At 60deg the perpendicular component of the solar flux VECTOR is (1/2) * (1/2)I. So in your example you wind up with I/4 being the heating component, not I/2.
“f I used your analysis, I would think that the 10-pound weight on the 60-degree tilted scale would put 10 pounds of force on the scale’s spring.”
You can read very well. I said *nothing* about what would happen if you LAID on the scale or if you stood on the scale. I asked you what would happen.
YOU created a red herring of addressing what would happen if you glued a weight onto the scale and then tilted the scale! In other words you didn’t answer what I asked and then you ignored what happens to the parallel force applied to the weight on the scale by gravity. You basically tried to modify the inclined plane experiment so you could ignore the movable car on the inclined plane!
That tells me that you either can’t read or you are frantically searching for a rational to support your unphysical concept of how radiant flux works.
“Technically, you should be asking about the weight force, but we’ll let that slide.”
Malarky! The point is that the total force generated by gravity doesn’t change. What changes are the perpendicular and parallel vectors. Just like with radiant flux impinging on a surface. If you ignore the force on the weight in the parallel direction, then you are ignoring the total system so you can rationalize your concept of how forces interact, being it a rate of transfer of energy associated with moving a weight or a rate of transfer of energy associated with the transfer of heat.
“What do you think this “projection” copilot is referring to?”
Again, the size of the illuminated spot doesn’t change. That would require the surface to expand like a balloon!
The projection is the I * cosΘ. That does *NOT* indicate in any way that the size of the illuminated area changes. It only indicates that the it is the normal component of the incident ray that transfers heat.
“ Only half of the flux hits the plane now,”
I can’t see the diagram you have referenced. Google Doc’s requires me to have authorization.
The ENTIRE FLUX is hitting the area. The only component of that entire flux that transfers heat is the normal component. At 60deg the normal component is 1/2 of the total flux (cos(60) = 1/2). That means that the non-normal component is also 1/2 of the total flux. But the total flux hitting the surface is 1/2 + 1/2 = 1.
That non-normal component does not go to 0 (zero). It remains there physically. It just doesn’t transfer any heat!
Neither does the size of the illuminated spot change. It remains exactly the same.
You are trying to make another strawman. So what if a body is considered “black” and absorbs and emits 100% of radiation that impinges on it? That is just one item.
You keep saying a diagram I have shown twice clearly states the surface is perfectly black. You need to post the date and time of those posts so I can view them.
I don’t remember any of them saying that a black body is reflective. By definition, an ideal black body is non-reflective. However, real world “black bodies” can only minimize reflection, thus the cosine factor is applicable.
I suspect you are misinterpreting diagrams like the following that don’t show the reflected component in the graphic because it is not pertinent to the subject of absorbance.
Notice that it does not show the reflected portion of the insolation K↓ in the graphic. However, the reflected wave K↑ is defined using both albedo and cosine.
Here are the comments where YOU included an example off the cosine effect reduction with angle on a “perfectly black” (no reflectivity) surface. Of course, this completely undermines your own arguments.
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4135915
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4136299
In your latest example, with a surface that is not “perfectly black”, you should notice that it first multiplies the solar flux that gets through the atmosphere by the cosine factor (0.3 in this case), and THEN applies the albedo reflectivity value.
Show us the heading of the posts you are referencing.
If you put your cursor over the date and time you will see the “link symbol” displayed. You can click on it and the line will be copied to your clipboard.
You don’t remember your own comments? All right, here goes:
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4135915
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4136299
A “perfectly black asphalt” parking lot is NOT* a black body! You are not demonstrating any reading skills at all! The example does *NOT* say “black body” anywhere in it.
Do you *really* think a perfectly black asphalt parking lot absorbs as much heat from the sun at sunrise and sunset as it does at noon? If it does do so then the elevation angle in the example becomes useless. But apparently the author of the example didn’t believe that a “perfectly black asphalt” is the same as a black body!
I think I asked you this once before and you didn’t answer!
You have to read for *MEANING*. What do you think the author of the example meant by using the term “perfectly black asphalt”? What does that tell you? (hint: think “black”)
The example very clearly refers to a blackbody with its use of “perfectly black” (zero reflectivity). What do you think it refers to? A body that reflects red?
You ask: “Do you *really* think a perfectly black asphalt parking lot absorbs as much heat from the sun at sunrise and sunset as it does at noon?”
You haven’t understood a single word I have said through this entire thread! I have said over and over again that lower angles of incidence cause reduced radiative flux onto a surface (and therefore maximum potential absorption), but that this is a purely geometric effect, without considering reflectivity effect.
So, no, I do NOT think a perfectly black asphalt parking lot absorbs as much heat from the sun at sunrise and sunset as it does at noon.
Four days ago, I used the example of parallel laser beams space 1 cm apart and how they fell on a perpendicular plate, and one tilted 60 degrees:
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4136256
I built on that example to make it a 100 x 100 grid of parallel lasers in a 1 m^2 area:
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4136361
I explained how the 1cm spacing of the beam would create a 2cm spacing between the resulting spots on the tilted plane. While you had trouble accepting that (very elementary) point, you did acknowledge that the ” thing that changes is the perpendicular area presented to the plane wave. And that perpendicular area is related to the cosine of the angle of incidence.”
Geometry, not reflectivity! My point exactly!
In the example of the 100 x 100 parallel laser beams, the tilted 1m x 1m target plane only intercepts 0.5 m^2 to the grid of beams, so only has half of the incident radiation upon it. Nothing to do with reflectivity.
Once again, here are the examples you posted with a perfectly absorbing surface :
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4135915
https://wattsupwiththat.com/2025/11/24/a-world-without-air/#comment-4136299
“Once again, here are the examples you posted with a perfectly absorbing surface :”
Once again, the term “perfectly black asphalt” does *NOT* imply a perfectly absorbing source, i.e. a black body.
Think about it.
I have thought about it. There is nothing else it could conceivably refer to.
“I have thought about it. There is nothing else it could conceivably refer to.”
So a black asphalt parking lot is a black body? It warms as much at sunrise and sunset as at noon.
I worked for two years as a highway maintenance crewman for the State Dept of Transportation. Trust me, that asphalt gets a LOT warmer at noon than it does at sunrise! But it still looks as black at sunrise as it does at noon!
If you can’t understand that then I suggest you get out of your basement once in a while and experience the real world.
“ This cosine effect will occur even on a perfectly absorbing surface.”
Did you read this after you wrote it?
The incident power has three components: 1: absorbed power, 2. reflected power, and 3. transmitted power (i.e. transparency factor). A perfect absorber means reflection = 0 and transmission =0.
The cosine factor does NOT apply to a perfect absorber since it is a black body. Read Planck.
The cosine factor only applies to a surface that has a combination of reflectivity and transmission. The Earth is opaque so transmission piece is zero leaving only a combination of absorption and reflectivity.
Climate science likes to pretend that a parameterization of “albedo” (i.e. some kind of average value) has no measurement uncertainty and the albedo parameterization is 100% accurate so it can be used as a 100% accurate reflectivity factor. The *real* issue is that the actual measurement uncertainty will be larger than any difference trying to be identified.
If *anyone* in climate science ever identified a measurement uncertainty budget I’d probably have a seizure!
The cosine factor does NOT apply to a perfect absorber since it is a black body. Read Planck.
Tell that to Jim. He has twice presented a textbook example of a perfect absorber using an equation with the cosine factor. Why doesn’t Jim understand your point?
And yes, I have read Planck…
“Tell that to Jim. He has twice presented a textbook example of a perfect absorber using an equation with the cosine factor. Why doesn’t Jim understand your point?”
A “perfectly black asphalt” parking lot is *NOT* the same as a perfect absorber, i.e. a black body.
If it was a perfect absorber then it would get as hot at sunrise and sunset as it does at noon!
What is the prevailing characteristic of a “perfectly black asphalt parking lot”?
Think about it for a minute.
At low angles of incidence, the radiation impinging on each unit area of the (perfectly absorbing) surface is less than when the sun is directly overhead. Simple, simple concept, but evidently entirely beyond you!
You have yet to discover the difference between a spherical wave and plane wave.
A spherical wave has decreasing intensity at any given point based on distance. The “rays” contained in it are diverging.
A plane wave has a uniform intensity at any point on the wave. The “rays” contained in it are coherent.
Baloney! I have been talking explicitly about completely parallel flux, what you call a plane wave, this entire discussion. (It’s a very slight oversimplification, as there is ~1/2-degree divergence in the radiation due to the subtended angle of the sun.)
When this parallel radiative flux hits a surface that is not perpendicular to the direction of the flux, the intensity of the flux on that surface is less than that on a surface that is perpendicular to the flux, even if all of the radiation is absorbed. Basic geometry.
“When this parallel radiative flux hits a surface that is not perpendicular to the direction of the flux, the intensity of the flux on that surface is less than that on a surface that is perpendicular to the flux”
The intensity of the flux is *NOT* less. It remains the same. The size of the spot being illuminated does not change. It remains the same.
You keep making these unphysical assertions. Radiant intensity is radiatt intensity. The receiving surface can’t change that value. The size of the illuminated spot doesn’t grow as the surface moves. That would require the surface to expand like a balloon.
What changes is the angle of incidence. Only that component of the radiant flux that is normal to the surface transfers heat. It’s just that simple. It’s not complicated. There is no reason to make anything “change”, not the radiant intensity or the size of the surface.
Did you ask it to “think deeper”? Did you prepare it previously with the instructions needed to obtain factual, objective, scientific answers? It appears you did not do this, so you simply got the “consensus” answer that is shown in 99% of the climate science documents. Those documents are based on the sun being a point source where the earth would be covered with a “cone” of rays, not on a plane wave.
Here is what I received from CoPilot for the question,
Instantaneous flux on a horizontal surface
There is more, but I think this explains the theory pretty well. It DOES NOT indicate that “causing the rays to be spread out over a larger surface area, reducing the intensity of the radiation.”is a correct answer to the question. You should have been leery when there was no formal mathematics shown.
You might want to look closely to see how well it agrees with the textbook I showed above and the page from it.
You need to read W.E.’s posts closer and learn how to set up instructions that an AI will follow to get the correct scientific answers. I have used CoPilot enough that those instructions are remembered every time I ask it a question.
Jim: You keep digging yourself deeper! You gave me an AI prompt that you expected to back your argument. I used that exact prompt, and the reponse did exactly the opposite!
I’ve done some variations now, and cannot find anything to back your viewpoint. In one response today, I get:
“For a horizontal surface, that angle is the solar zenith angle 𝜁, so the instantaneous absorbed flux is proportional to cos𝜁 (before considering atmosphere and albedo). This is just the geometric projection of a plane wave onto the surface, often written 𝐼=𝐼0cos𝜃 for a beam on a tilted collector.”
Note carefully the phrase “before considering atmosphere and albedo”. That is, the cosine effect is there even without any reflectivity.
You seem to think that I am arguing that there is no trigonometric effect. No! I am arguing, and this is backed up by all the sources YOU cite, that the trig comes from the pure geometry and is independent of any surface reflectivity effects.
You say: “You might want to look closely to see how well it agrees with the textbook I showed above and the page from it.” Why do you keep citing an example that contradicts your argument?? The example states very clearly that it has nothing to do with reflectivity.
In any 1st year college introductory E&M course, you learn how to compute the flux transfer using surface integrals. The expression inside the integral is the dot product of the flux vector and the unit surface normal vector. The dot product includes the cosine of the angle between these two vectors. THIS is fundamentally where the cosine factor comes from — and it has NOTHING to do with the properties of the surface. And I don’t need an AI response to tell me this; I can go to any of my old texts.
The fact that you seem not to understand this very basic math and science that you would have gotten in the 1st year of an EE major is why I question whether you really studied this.
I understand the math perfectly well. I also know how to research a subject and learn from experts.
Ok, so let’s explore some of the resources that CoPilot used to develop its answers.
Insolation
Lecture11 — Insolation
ESCI 340
Have you downloaded and studies the book I referenced. It agrees with these resources also.
Have you studied Planck’s thesis, particularly Part II, Chapter 1
As usual, you have provided no references that support your assertions, only your assurance that you know what you are talking about. You haven’t even attempted to deal with the references I have shown you by showing where they are incorrect or where I have misinterpreted what their equations say.
Please show where this example is wrong if you want to sustain your assertions.
Jim: You have provided exactly ZERO references stating that the reduction in incident radiation with angle is due to reflectivity.
While you have cited many sources that show the trigonometry involved with the angle of incidence, NONE of them claim this is due to increased reflection with angle.
Comically, you cite for a second time the example of insolation at 60 degree latitude, when I have already pointed out that it states that the reduction occurs on a surface with no reflectivity!
I only need to cite back to you the sources you provide to show you that you are wrong.
Elsewhere in this thread, I have linked the Wikipedia pages on surface integrals and the divergence theorem, which are straight out of any standard E&M or calculus text. These explain how the dot product involved in calculated the surface integral introduces the trigonometric functions that appear.
Comically, you state the the surface has no reflectivity without describing where the energy that isnt absorbed disappears to.
Wikipedia page, don’t make me laugh!
The Wikipedia pages I read about divergence never deal with absorption and transmission, only diverse reflection with no absorption, i.e., no heat transfer at all.
Those Wikipedia pages agree exactly with my college math and physics textbooks. The equations there are completely relevant to the heat transfer topic at hand here. The fact that you cannot begin to understand this tells me that you are completely out of your depth here.
They may agree but your interpretation is incorrect. Here is a pertinent image from one of your references.
Take a close look where it describes the “normal (perpendicular)” component of F(x).
Fn(x) = F(x) cos θ
There are typically three components involved: 1. Absorbed, 2. Reflected, and 3. Transmitted. The Earth is pretty opaque so the transmitted part is pretty small as well. If the Earth is not a black body then you don’t get 100 % absorption. That leaves reflection as a major component. If absorption only happens with the normal vector then what is left over *has* to be reflected. It can’t just “disappear”.
Jim – Seriously??? You think that this is an argument against me?
I have been arguing explicitly from the start that this basic geometric effect is the source of the cosine term. A search of the comments for the term “dot product” shows that I have used that term 6 times already in my responses to you.
In one of these comments, I said: “The dot product includes the cosine of the angle between these two vectors. THIS is fundamentally where the cosine factor comes from — and it has NOTHING to do with the properties of the surface.”
What you just posted agrees exactly with my arguments. It states that “the flux through each patch is equal to the normal (perpendicular) component of the field Fn(x)=F(x)cos(Theta)…”
Note the wording about “the flux through each path”. This is about the incident radiation, before any possible reflection.
Again, you cite an example that has nothing to do with reflectivity to try to bolster your argument about reflectivity.
Look closer at the diagram from your wiki reference. It does show a dot product TO CALULATE one component. “Fn(x) = F(x) cos θ”. What happens to the component calculated from “F(x) cos(90°-θ)”? You just ignored it by saying it doesn’t exist!
Look carefully at the red square. There are two vectors, vertical and horizontal. Only the vertical is pertinent to absorption.
Jim, thank you for the vote of confidence. It is like a breath of fresh air to see that someone I respect understands the phenomenology. That goes for you too, Tim.
Jim, thank you for the link to the meteorology book. So far, I have only looked at the chapter on atmospheric optics. Based on that, I’m sure that it will be a useful reference in the future.
You are most welcome. I have spent many years now searching for resources on atmospheric processes. There is a plethora of information on the internet if you can find it. Same with measurement uncertainty. Yet most people, even here, fail to post the resources that support their assertions. Climate science seems to treat radiative effects as isolated from all the other processes along with using arithmetic averages of processes that are exponential and that have gradients. My training in designing heat sinks and cooling towers tells me that just doesn’t work in the real world.
It is interesting that the LLMs don’t seem to be aware of either bi-directional reflectance distribution functions (BRDF) or Fresnel’s Equation for calculating the amount of radiation specularly reflected from a smooth surface.
Yes, the Total Solar Irradiation (TSI) figure of 340 W/m2 includes all of those wavelengths of solar radiation.
w.
Which is a “calculated” or “guessed” figure, as no present radiometric equipment is capable of measuring Total Solar Irradiation. Completely pointless – over the past four and a half billion years or so, the surface has cooled.
You don’t have to believe it, of course.
But the sun has an average temperature and the earth has a mean orbit radius. That means you can calculate the nominal. Who needs variable sunlight.
Sparta, as I said – calculations, which are completely meaningless.
What’s the temperature outside at the moment?
Michael,
I apologize for omitting the classical /sarc annotation.
Sparta, sorry. Sometimes, when I assume sarcasm, someone is being deadly serious. I feel your pain!
Then how does 340 W/m2 turn into 346 W/m2 of downwelling radiation plus 23 W/m2 of reflected solar radiation?
David, you have to believe in the Magic of CO2 – practised by the Climate Wizards.
They can turn anything they like into anything at all! Slow cooling turns into heating! Celsius degrees turn into Watts per square meter! Dimwits turn into Climate Scientists!
See how easy it is?<g>
Oh well. I would always encourage to reflect on basic climate physics. It is just that there seems to be an endless number of pitfalls.
First we have the surface albedo/emissivity issue. Earth, the surface, is pretty dark. About 1/2 the albedo is due to clouds, and another quarter is due the air! It is something many people forget, but the air does reflect a lot of light. Technically of course it is not reflecting but scattering, and you can see that with your own eyes. The same bluish light the clear sky shines down on you, it also shines into space.
The surface albedo is pretty low, just about 0.07 or so. The emissivity (=1-reflectivity) however is much lower than usually assumed. With water it is 0.91, with the global average not deviating a lot from it (wavelength on x-scale, dimensionless Planck functions solar/terrestial for the sake of weighting) ..
Left on its own, in theory, the surface would take on a temperature of ~280K, just 8K less than we have. And while this might even promote the argument the article tries to make, we have other issues..
There just is no “TOA-measured “greenhouse” radiation“. You must neither relate dOLR (the 3.7W/m2 are not dOLR, but the sum of “fluxes” at the tropopause) to “back radiaton”. The latter is just irrelevant – and would grow just because of increasing temperatures.
All further conclusions are baseless..
I specified how it is measured, using the method of Ramanathan. Upwelling surface longwave (thermal) radiation is on the order of ~ 400 W/m2. TOA measurements show ~240 W/m2 emitted to space. What happened to the other 160W/m2?
Since it’s not going to space, it must be returning to the ground.
This is basic stuff.
w.
The term “radiative (energy) flux” has corrupted lots of minds. “Climate science” is a stupid science. Just ask the AI how much radiation all the molecules in your body emit. Spoiler: ~1MW! And then try to figure out where that “energy” goes..
The average skin temperature of the human body is ~34°C. Per the S/B equation, this radiates about half a kilowatt per square meter.
The average human body has about 1.7 square meters of skin, so that’s about .85 of a KILOWATT, nowhere near the one MEGAWATT figure you provide.
Finally, your claims that
“The term “radiative (energy) flux” has corrupted lots of minds. “Climate science” is a stupid science.”
are certainly your opinion, but you have not supported them with evidence.
Regards,
w.
For one, you have no special “skin molecules” that emit radiation exclusively, rather they all do it. That is why I suggested you ask AI..
If you want to see a true and substantial falsification of “the science”..
https://greenhousedefect.com/the-holy-grail-of-ecs/how-to-flip-the-sign-on-feedbacks
E:
One of the first things you learn in an introductory thermodynamics course is do define a “control volume” and then consider only the energy transfers across the boundaries of that volume. Any internal transfers do not affect the energy of the overall volume.
Two control volumes under discussion here are the earth (solid and liquid portion) and the human body. In the analysis of the human body, skin molecules are “special” because they are at the surface of the volume, and therefore can transfer energy to and from the control volume. Compare this to, say, molecules of the lining of the intestine, which cannot do this. (Because of their location, not the type of molecule.)
So the fact that molecules internal to the human body radiate is not relevant to the thermodynamic argument made here. The same goes for molecules internal to the earth. In these cases, one internal molecule radiates some energy, and another absorbs it. Netting out to zero.
Again, this is the most basic thermodynamic analysis.
Dude, the person you are talking down to are my feet.
As I expected, not even an attempt to defend your arguments.
No, he accepts reality. Arguments of this sort are for the ignorant and gullible who wrongly believe that adding CO2 to air makes thermometers hotter.
It doesn’t.
Not quite – the Earth is losing about 44 TW. I know you don’t believe it, but it’s true. To save you embarrassment, I’ll let an AI apologise on your behalf –
You can “argue” all you like. Adding CO2 to air does not make thermometers hotter. Anybody who believes that is obviously ignorant and gullible.
Wouldn’t you agree?
Why? I suppose you believe that a colder atmosphere can raise the temperature of a hotter surface. Step outside at night – the surface cools.
So much for “it must”.
Untrue. You are confusing the flow of heat (hot to cold) with radiationall energy transfer.
You don’t believe that the surface cools at night? No wonder that you can’t bring yourself to actually quote me!
I’ll laugh as I point out that you misspelled “radiational” – unless “radiationall” is another example of pseudoscientific jargon.
Perhaps you should consider how a cold laser can melt steel.
Perhaps you should learn about coherent light, lasers, energy, heat, quantum physics, energy conversion, and so on.
Turn the source of power off, and your “cold laser” won’t warm anything hotter than itself, far less “cut steel”.
You’re not alone – here’s something from a fairly recent biophysical paper –
The author is confused and ignorant. His claim that the observations “ cannot be explained currently by physics laws, even at nanometer scales” is breathtakingly presumptuous.
He seems to know about as much about physics as you – not everything. Nor do I, but I know more than either you or him, obviously.
You believe that adding CO2 to air makes thermometers hotter. It doesn’t, and no amount of wishful thinking can obliterate reality.
The flow of heat is controlled by the net flux flow. If you are using the SB equation to determine the amount of heat transferred between bodies it is important to determine the temperature difference, i.e. (T1 – T2). When thermal equilibrium is reached there is no NET flux, but there is still energy transmitted between the two bodies.
Think about this, two bodies, Body A and Body B. Body A has a higher temperature than Body B. If Body B’s flux can be added to Body A thereby raising its temperature, then Body A will begin radiating a higher flux which will raise Body B’s temperature, which then raises Body A’s temp, which then raises Body B’s temp and so on. Where does it stop?
Planck explains this in his Theory of Heat Radiation. Read Section 102 about entropy. Pay attention to this.
Think about what this says. Back radiation is absorbed and immediately emitted. If it is immediately radiated, then its temperature can not raise.
“net flux flow”
You have literally no clue what that phrase means, Jim. Nor does anyone else. It’s gibberish. Not physics. I’m sorry that your engineering professors did that to you, but then, they weren’t being paid to teach you physics, were they?
Note that two of those words are synonyms, brought to you by the Department of Redundancy Department, and the third one has no relationship to them.
Can you tell us what “gross flux flow” means, to start with? And how do we measure it?
You know what. You are a troll and don’t really deserve a response. But, here is one,
Albert Einstein is often quoted as saying, “If you can’t explain it simply, you don’t understand it well enough.”
There are many on here that appreciate a simple down-to-earth explanation of complicated phenomena. The fact that you continually denigrate those who try to explain things in a simple manner makes you a TROLL that adds little to any discussion. GO AWAY and bother others.
“denigrate those who try to explain things in a simple manner”
No, Jim, I am pointing out that you are wrong. And have literally no clue what you are talking about. That is only “simple” in the sense of “simpleton”. Einstein would not have approved.
Can you answer my question and tell us what “gross flux flow” means, please?
If you are unable to answer your own question from the context, then you are dumber than a real TROLL.
You are looking to start a demonic back and forth so you can apparently receive more renumeration. I won’t play your game. Go bugger off elsewhere.
“If you are unable to answer your own question”
No, Jim, I am not teaching me, I am teaching you. Please answer the question.
“demonic”
Where did that come from? Are you a psychopath? Apart from anything else, you are in gross violation of the Engineering Ethics you theoretically signed up for, never mind the site’s policy regarding basic civility and courtesy. Moderators? Moderators?
Please just stick to the science and politely answer the question. If you don’t know the answer, just say “I don’t know”. Remember, there is no shame in admitting that, but plenty in calling your physics teachers “demonic” and telling them to “bugger off”. Is that how you treated your engineering professors too? Who taught you to behave like this?
No, you are not teaching anyone on this thread. Your only pastime is telling people they know nothing compared to you. I have no desire to enter into a demonic back and forth with you. If that upsets you, so be it. Find someone else to argue with.
“Find someone else to argue with.”
No, Jim, I’m teaching you physics, and I’m not going away until you learn how to do it properly. If you want to spout BS without being called out on it, try The Conversation or Skeptical Science. They love that stuff over there.
“demonic”
You sound like this wretched stereotype, penned by The Copernican, and highlighted recently by Jo Nova:
“The [urban] bugman’s neurological model of reality is divorced from reality. They hallucinate truths that make no sense, and they delude themselves into provably false ideas, and violently attack anyone with a model of reality more accurate than their own.”
Are you an “urban bugman”, Jim Gorman? You behave like one.
Calculating the upwelling EM radiation based on surface temperature. Ok. What about the energy given up by the ground due to latent heat effects, conduction (surface) and convection (vertical). Those energy paths deplete the EM energy emitted.
As the surface warms, some of the energy goes into the ground. Water is different than granite is different than asphalt. But there is a storage phenomenon that has to be included.
EM fields are spherical phenomenon. The earth’s surface is not flat. Is it not possible that some of that energy went in a direction that was not captured by the relatively small focal plane array in the satellite sensor?
Concluding that it must be returning to the ground ignores the alternative explanations.
I suggest you discontinue using the flat earth model. The average of 60C and -20C is 20C but the EM energy emitted at 20C is not the average of the EM emitted at (60C + -20C)/2. This is a big reason why the so call GAT is bogus.
You accurately described the inability of arithmetic means to adequately address exponential and trig functions.
Be careful of using arithmetic averages when dealing with a function that has 4 as a power/root and is related to trig functions. For example, what is the average insolation for a point directly under the sun from sunrise to sunset? What is the resulting temperature. What is the average temperature at that point from sunset to sunrise?
“Upwelling surface longwave (thermal) radiation is on the order of ~ 400 W/m2.”
No it isn’t. Measurements show 0 to 100 W/m^2 depending on humidity. When are you planning to stop lying, Willis?
“returning to the ground.”
You are hallucinating.
It is about time to give atmospheric pressure a bit more warming respect.
While it is hard to measure a temperature difference after a 10% increase in moisture or co2,
atmospheric pressure has been quite reliable and constant with 2/3 of a degree change for every 100 meters of change in altitude.
IMO it deserves the warming control knob award a bit more than the “It’s Co2 or Nazi” – narrative.
Nope. Atmospheric pressure alone can’t warm the surface. See my post entitled “A Matter Of Some Gravity“.
w.
Neither can adding CO2 to the atmosphere. You are delusional and flying in the face of four and a half billion years of history if you believe otherwise.
Before I respond to this comment by Willis, I want to highly recommend this lecture by Michael Mishchenko (1025) Michael Mishchenko Maniac Lecture, January 26, 2015 – YouTube
Pressure is a function of gravity working on mass. As Rick Will has pointed out many times, the maximum ocean temperature of 30C is dependent on surface pressure. Increase the surface air pressure and the maximum ocean surface temperature increases and the earth’s surface warms. So if atmospheric mass is increased by non greenhouse gases, which by definition increases surface air pressure, it will have an effect on the Earth’s surface temperature. The question is how big is the effect.
Here is my question for Willis: Do you believe in the conservation of photons? Energy is certainly a conserved quantity. Where I think we differ in opinion is that I believe near the Earth’s surface where greenhouse gases have absorbed photons, the kinetic energy created is pasted on to other molecules by collision. This creates convective forces. Since temperature is just an intensive property of kinetic energy, the collisions increase the temperature locally. This process weakens the higher in the atmosphere you go as excited CO2 molecules experience fewer collisions, so they return to their ground state through reradiation of absorbed photons.
This is all complication stuff. I’m not sure that analyzing made-up worlds provide much insight into the Earth we inhabit. I think my issue with your model is that it is a pure radiation model where convection/advection play no role and no atmospheric mass other than greenhouse gases matter.
Good catch! Willis’ model assumes a completely static atmosphere, ignoring the fundamental roles of convection and the phase changes of water.
Not true, either of you. Every single phenomenon is included, moving atmosphere, phase changes of water, everything known and unknown.
With all of those other phenomena, the increase in “greenhouse” radiation increases the temperature by only ~20°C. That’s my point. Without all that stuff, if it were pure radiation only, the surface temperature per S/B of 510 W/m2 as shown in Fig. 2 would be 37°C, not 18°C.
That’s why the equilibrium climate sensitivity is so small—because the surface is cooled by the convection, the phase changes of water, and all the rest of the thermoregulatory emergent climate phenomena.
w.
Willis, I’m not sure what I said is untrue. You believe that the only way for an excited CO2 molecule to return to its ground state is to reemit the photon, Thus, you believe in the conservation of photons. I believe that in the lower atmosphere, an excited CO2 molecule almost always returns to its ground state by passing its vibrational energy to other molecules through collisions.
I have never seen any measure of downwelling radiation at the Earth’s surface of 510 W/m2.
I believe that convective forces are created throughout the atmosphere due to collisions. You seem to believe that all convective forces start at the Earth’s surface, which is a direct consequence of your belief in the conservation of photons.
I think your end results are basically correct, I just think you get the dynamics involved incorrect.
Was that the one where you demonstrated that you had no clue what the Stefan-Boltzmann equation means, Willis?
“The Stefan-Boltzmann equation specifies how much radiation [energy] is emitted at a given temperature”
No it doesn’t. The S-B equation gives you Watts, but energy is not measured in Watts, is it, Willis?
Steve, replying to you is generally a mistake … but I’ll try anyhow. I can’t find that quotation anywhere in this thread. And why did you stick the word “energy” into the middle of the purported quotation?
w.
“replying to you is generally a mistake”
Please cease the endless insults and just stick to the science. Thanks.
“I can’t find that quotation anywhere in this thread”
It was in the article you told us to read, in the comment that I replied to. The article you wrote 14 years ago, entitled “A Matter of Some Gravity”. That’s why I put the quotation after my phrase “Was that the one [article] where you demonstrated …” So yes, before the quotation itself I could have written, for clarity, “In that article, you said”. Sorry if I confused you about where your own words came from. I know you’ve written a lot of them and it’s hard to keep track of them all.
“And why did you stick the word “energy” into the middle”
Because you told us that “radiation is energy”, remember? I’m simply reminding you of your own definitions, because you seem to keep forgetting them.
“of the purported quotation?”
It’s not “purported”. It’s right there in black and white, with your name on it.
You are incorrect. The S-B equation gives radiant exitance (intensity) and has units of Watts/m².
“radiant exitance”
As usual, Jim Gorman, you have literally no clue what that phrase means. Nor does anyone else. It’s undefined gibberish, much like your previous phrase “net flux flow”, which you still haven’t defined for us. You should probably just sit down and stay in your lane. This is “demonic” physics, remember? It’s a complex topic and definitely not suitable for undereducated numpties. Approach with caution lest your soul be sucked out through your eyeballs.
Note that the “per square meter” part is totally irrelevant here, which is why I left it out. The important part is the Watts, like I said. What do you think a “watt” is?
No, it is absolutely relevant. Without knowing the area radiating there is no way to know the total watts being radiated.
“No, it is absolutely relevant.”
It’s not. Can you tell us what a Watt is, please?
Steve, you say:
From the web.
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But hey, why let facts get in your way?
w.
Willis quotes some equally uninformed engineer on the web:
“exitance is the radiant flux (power) emitted”
Just because someone else wrote it doesn’t mean it isn’t gibberish. You have no idea what any of that means. Nor does anyone else. Your entire quote is a hallucination. Because none of it can be measured. And according to you, if it can’t be measured, it is a fantasy. (That’s a paraphrase, but I can find your exact quote if you need it)
No, power is not something that can be “emitted”, Willis. You know that very well. Or at least, part of your brain does. The part that said “power is work per unit time” and “work is what happens when a force moves an object”. But the part of your brain that knows that doesn’t seem to be connected to the part that’s typing on the keyboard here.
And this claim is not what you said earlier, either, is it? You said “radiation [energy] is emitted”. Now you say “radiant flux (power) emitted”. Yet energy is not the same as power, is it? So, were you being sloppy then, or are you being sloppy now?
“But hey, why let facts get in your way”
What facts? All you have are contradictions and hallucinations. And quotes from people who are also suffering from a variety of contradictions and hallucinations. Remember when you asked the AI to stick to directly observed facts? Can you do that too, please?
Nothing else to add, Willis? That’s your final word? You found something “from the web”, you fell for it, despite knowing better (!), and now you think the rest of us should too? But what if the rest of us didn’t just fall off the turnip truck yesterday? What if we are less gullible than you are? Then what?
While you are at it, can you explain precisely why you objected to me adding “energy” to your word “radiation” after you explicitly told us that “radiation is energy”?
Steve, look “radiant exitance” up yourself. There are literally dozens and dozens of definitions of radiant exitance out there, all saying the same thing. The graphic below is only the tip of the iceberg.
Now, you’re free to say that the entire scientific community and all the dictionaries and Grokipedia and everyone else put up those dozens and dozens of definitions just to fool us, and you’re the true genius who sees through it all … sane people, not so much.
w.
There is no such thing as “objects emitting power”, Willis, regardless of how many uninformed people claim that there is. And you know that very well, since you told us the following definitions:
1) Radiation is energy
2) Energy is the capacity to do work
3) Work is what happens when a force is applied to an[other] object
4) Power is work per unit time
Where is the “other object” that a force is being applied to, in your definition of “radiant exitance”, please? And what is its temperature?
“the entire scientific community […] just to fool us”
The Argument from Authority again, Willis? I told you before, that is a classic Logical Fallacy. You should try to avoid it if you want to pretend to be an amateur scientist. Note that none of the people you have quoted are physicists, oddly enough. Why don’t you find one and ask him? No need to take my word for any of this.
“you’re the genius”
No, just a guy on the Internet who knows his physics from his basket-weaving. You know yours too, or at least part of your brain does, but you keep forgetting anyway, don’t you?
Oh and in case you don’t get around to replying before this comment thread is closed (probably tomorrow), one more relevant definition you gave us is this one:
5) Anything that cannot be measured is imaginary
Since “radiant exitance” cannot be measured, your definition tells us that it is imaginary. Right? Of course it is. Glad we got that cleared up!
Again, “I know everything!”
Do you know what a field strength meter is?
How about this reference (just so you don’t think I am offering my opinion).
https://ntrs.nasa.gov/api/citations/20150021314/downloads/20150021314.pdf
And, as usual, you give no resource that refutes what I posted other than “I know everything!”
“you give no resource that refutes what I posted”
“Resources”, Jim? How about a physics textbook? Or failing that, Willis himself? Here is what he has told us:
1) Radiation is energy
2) Energy is the capacity to do work
3) Work is what happens when a force moves an object
4) Power is work per unit time
These are all correct physics definitions. But do we measure “energy” in Watts, Jim? What is a Watt, anyway? Can you tell us? (the clues are above)
““I know everything!””
I never said that, of course. Now you just sound like a six-year-old. I do know my physics better than engineers and fishermen, though, since I’ve actually formally studied the subject. All my life.
There you go again! “I know everything”. And even better, I know more that engineers and fisherman. Show us a textbook you have written that is used in colleges and universities. You don’t even post with your real name as Willis, I, and others do. If you are so confident in you knowledge, then you should have no problem posting your real name.
And, radiation is energy in transit. Radiation is an EM wave with an intensity that describes how much energy is in the wave.
(editing this comment because I didn’t see the one you posted five minutes previously)
“I know everything”
No, that is a lie. It is not what I said.
“You don’t even post with your real name”
That is not my fault. I tried to get WordPress to show it, but it doesn’t. Other people have the same problem too. My full name is Steve Keppel-Jones.
“radiation is energy”
Correct. Objects emit energy. Not power. Try to keep that in mind. There is no such thing as “radiant exitance” in Watts, independent of the conditions of the surroundings.
A watt is a Joule/sec. It *is* a measure of power. It is the rate of energy transfer. W/m^2 is the rate of energy transfer, i.e. the power, transferred per square meter. (Joule/sec)/m^2.
Multiply (Joule/sec)/m^2 by the area and you get Joule/sec.
Power is either how quickly work is done or how quickly energy is transferred.
Radiation is associated with the second. How quickly energy is transferred.
For an EM wave of a given intensity in W/m^2 the larger the area illuminated the more energy is transferred per unit time. If you have one square meter illuminated by an EM wave of 1 W/m^2 then you transfer one joule in 1 sec, i.e. 1 watt. If you have two square meters illuminated by that same intensity of 1 W/m^2 you get 2 joules transferred in one second or 2 watts. Again, watts is power, not energy.
Radiation *is* power transferred over time, not energy. Energy is measured in joules. Radiation is measured in energy/time, i.e. power, not in joules, i.e. energy.
It continues to baffle me that we don’t use Joules when discussing the effect of solar radiation onour Earth.
A full day of sunshine delivers anything from 5-30 MJ/m^2
https://www.pveducation.org/pvcdrom/properties-of-sunlight/isoflux-contour-plots
This is just enough to only INCREASE the temperature of 1 m^3 of ocean water 1-7 K.
Nothing like the temperatures we see on our moon.
You have to know joules/sec/area before you can calculate total joules delivered over a time period to an area. You have to know the power being delivered before you can know the energy delivered.
The real issue is the path loss between the source and receiver. The power at the receiver doesn’t tell you the source power. That path loss may be from the square-law and/or from absorbers/diffusers. You can’t really use the S-B law based on what is received unless you assume 0 (zero) for the path loss between the source and receiver.
“Radiation is associated with the second. How quickly energy is transferred.”
No it isn’t. Radiation is energy, not power. Even Willis knows that. And “energy” is not the same as “how quickly energy is transferred”, is it?
“For an EM wave of a given intensity in W/m^2”
That’s not how electromagnetic fields work. Electric and magnetic field intensities are measured in Newtons/coulomb and Gauss, respectively, not Watts. Electromagnetism is a force. Where did you study your physics, again?
Do you think no one notices your Equivocation argument fallacy?
Radiation is *NOT* radiant flux. Radiant flux is W/m^2. Watts is joules/sec.
Joules per second is a RATE of energy transfer.
You keep hoping people will miss what you are doing! Radiant flux is not energy, it is the RATE OF ENERGY TRANSFER. Get your definitions correct and address the actual definition under discussion and not something else.
“Electric and magnetic field intensities are measured in Newtons/coulomb and Gauss, respectively, not Watts.”
Why do you think they measure the output of a radio transmitter in WATTS? Watts is power. Power is the RATE of energy transfer. You don’t describe radiation in joules, you describe it in POWER, joules/sec. Nor do you describe radiation in terms of field intensity. Field intensity is more commonly specified in V/m, volts per meter, not the rate of energy transfer.
You are a troll, pure and simple. No one is speaking to field intensity. For heat the important value is the rate of energy transfer in joules/sec. not in volts/meter. The field intensity will be dependent on the charge value of each component, e.g. the earth and sun. Since the charge on each is in the millionth of a V/m the electric field between the two is very, VERY small.
Radiant flux is POWER, not energy. Again, get your definitions correct if you are going to address the issue. Stop telling us that radiation is energy when what is being discussed is POWER, the rate of energy transfer. It is the POWER in the radiant wave that determines how much energy will be transferred per unit time, not joules. To get to joules you need to specify a time interval. Watts doesn’t specify a time interval, its joules/sec. More joules/sec means more energy transferred in the same amount of time.
It’s why the power company charges you based on Watt-hours. See that term “hours”? They charge you based on the energy used. But it is the Watt component that tells you how quickly that energy accumulates.
Why do you think heaters are rated in watts and not watt-hours?
You are a troll, nothing more. Using an argumentative fallacy to try and tell people why they are wrong. *YOU* are the one that is wrong. Admit it to yourself and move on.
“Radiation is *NOT* radiant flux.”
Correct. So what do you think radiation actually is? Both Jim (also a Gorman – a relative?) and Willis have told us that radiation is energy. They are correct.
“You don’t describe radiation in joules”
Physicists do. Because, as we have all established by now, it is energy. And energy is measured in Joules, isn’t it? Of course it is.
“you describe it [radiation, i.e. energy] in POWER,”
That would certainly be a dumb thing to do.
I’m not the one who is “equivocating” here, am I? Of course not. Sit down and stay in your lane, Tim. It obviously isn’t physics. Can you answer my question and tell us who taught you your physics?
Radiation is an ENERGY FLOW, it is a FLUX. Flows are described by rate! E.g. Watts/m^2.
No, they do *NOT* describe radiation by joules. Joules is meaningless in terms of radiation. You do *NOT* measure radiation at a point by measuring how many Joules exist at the measurement point. You measure radiation by energy FLOW. Joules per second, NOT JOULES.
You seem to want to keep mixing up electrostatic fields, i.e. V/m, with radiation that is measured as an energy FLOW. A laser is radiation! It is measured in POWER (i.e. WATTS) using a power meter, a thermal sensor that measures the heat generated by the laser hitting a surface, a calorimeter, etc. None of these measure JOULES. They measure Joules/sec!
How many JOULES does the signal from a 100watt transmitter have? How many JOULES does the signal from a 1000watt transmitter have?
My bet is that you will ignore the questions.
Then why do they characterize the output of a transmitter in WATTS and not in JOULES?
“Can you answer my question and tell us who taught you your physics?”
Francis Sears and Mark Zemansky from their university level series of physics. From thier Volume 1, Page 379:
You are beating a dead horse and just demonstrating your total lack of experience in the real world. Radiant energy is *NOT* defined as “joules”, it is defined as “joules/sec”. It is a FLOW, a rate of energy transport. It just happens to be carried by radiation instead of by conduction or convection.
All you are doing here is wasting everyone’s bandwidth trying to convince you know more than EVERYONE, and you are just winding up looking like a fool.
WE, the ‘money chart’ is your figure three—something I have not seen elsewhere. But I think you need to redo your interesting analysis using different values for albedo and emissivity based on a quick fact check.
As mapped by NASA, the average albedo of Earth is ~31%. The data is at earthdata.nasa.gov. This includes clouds, so is the ‘with atmosphere’ case. The Earth’s average cloudless albedo was estimated by Otterman (1977) to be 0.154, so the ‘without atmosphere’ case.
As mapped by NASA, the average emissivity of Earth is 0.94.The data (in the NASA/Japan ASTER database) is at terra.nasa.gov.
I haven’t crunched the SB numbers, but directionally you should find a significantly higher ECS than the ~0.44-0.47C in this otherwise thought provoking post.
Thanks, Rud, always good to hear from you.
I’m using the gridded MODIS emissivity figures described below, available at ftp://ftp.ssec.wisc.edu/pub/g_emis/. The NASA figures at the Terra site say 0.96 average. The MODIS figures are slightly higher at 0.976.
As to albedo, I’m using the CERES figures (surface reflected solar divided by surface downwelling solar) to maintain consistency with the other data.
However, this makes very little difference. The S/B calculated change in temperature from the emissivity figures I used in the head post is 20.43°C. Using your emissivity of 0.94 gives us a temperature change of 20.62°C.
And if I use your surface albedo figure of 0.15 instead of the 0.125 CERES data I used in the head post, it only changes the sensitivity from 0.47 to 0.50 °C/2xCO2 … as I said, I like this method because changes in the values don’t much change the answer.
It was to allow for these minor variabilities and uncertainties that I describe it as “~ a half-degree” per doubling of CO2.
Let me know what you find when you do run the numbers.
Best to you, my friend,
w.
===
###########################################################################
README file for the global IR land surface emissivity database
Version: 3.0
Created by Suzanne Seemann and Eva Borbas CIMSS/SSEC University of Wisconsin
Contact: Eva.Borbas@ssec.wisc.edu
Date of release: August 2009
Note: This dataset was released in the hope that it will be useful,
but without any warranty.
!!!!! CHANGE IN DATA VERSION !!!!!
###########################################################################
OVERVIEW AND REFERENCE:
The UW-Madison baseline fit (BF) global emissivity data available here were derived using a
procedure that fits monthly averaged MODIS MYD11 (MODIS level 3 operational land surface
emissivity product) values to a baseline emissivity spectra determined using laboratory
measurements of land surface materials.
The JAMC paper that details the baseline fit procedure for deriving the database
and its application to atmospheric retrievals is available at the website
http://cimss.ssec.wisc.edu/iremis/
Please reference this paper for any use of the database:
Seemann, S.W., E. E. Borbas, R. O. Knuteson, G. R. Stephenson, H.-L. Huang, 2007:
Development of a Global Infrared Land Surface Emissivity Database for Application to
Clear Sky Sounding Retrievals from Multi-spectral Satellite Radiance Measurements.
J. of Appl. Meteor. and Climatol., Vol. 47, 108-123.
isn’t the real difference between a planet with oceans and one without?
Absolutely essential. Haven’t seen any sort of explanation how our ice cold atmosphere can heat our almost 4km deep oceans to ~270K and above.
Ah, CO2 magic – the wizardry and arcane knowledge of “climate scientists”. Ask Willis – he thinks he’s a proud possessor of these secrets.
The oceans are the dog, the atmosphere is its tail. The enthalphy of the oceans is about 1000x the enthalpy of the atmosphere because the mass of the oceans is about 250x the mass of the atmosphere and the specific heat of water is 4.2x dry air.
The atmosphere does not warm the oceans. if air is warmer than water, the water evaporates and cools the air. If air is colder than water, the water warms the air. What keeps the oceans warm is the rays of the sun which it absorbs quite well.
Only the upper 10-20m are directly heated by solar. Mixing and conduction take care of another few hundred meters. In exceptional cases like the Mediterranean Outflow water you’ll see warm, very salty water sink deeper.
For the bulk of the oceans solar has no influence.
Any warming during spring- summer will leave at the surface again during autumn-winter.
….for the bulk of oceans solar has no influence….
…c’mon now…. it has several dozen watts per sq.M acting over about 4 billion years…
You’ll be hard pressed to find any solar influence below ~500m, since warm water does NOT sink towards the ocean bottom.
So for the last 4 billion years the sun has been increasing the temperature of only a small upper layer of the oceans and that added energy has been lost to space again during diurnal and seasonal cooling. The reason the deep oceans are so warm (~270K) is NOT the sun.
Think about the >99% of Earth that consists of molten stone and molten metal.
It is true warm water (given constant salinity) does not sink.
However, heat (flow of thermal energy from higher to lower temperature) does flow from the warmer surface to the colder depths.
The reason the deep oceans are “so warm” is due to a number of factors. Core thermal energy, solar EM energy, and pressure/density.
Specific heat capacity is part of the equation.
It does indeed, until the surface begins to cool in autumn and becomes colder than the warmed water at depth. This water will now rise back to the surface and cool as well.
Typical profile:
Not for long, it doesn’t! Any water heated by this “flow of thermal energy” expands, and rises.
if you step into a shallow tidal pool, or even a deep cold lake heated by the Sun, you will quickly discover that the surface layer may be quite hot, while the water underneath is not.
Solar ponds can heat water to maybe 90 C at a depth of 1.5 metres or so by artificially influencing salinity and density. However, they are not a general economic source of energy, as Nature can’t be fooled for long, and buggers all your cunning thinking with physics. Diffusion, Brownian motion, energy density, no free lunch, and so on.
The vast bulk of the oceans are heated by their proximity to the glowing hot interior of the Eartb just a few km below.
The core of the earth is hot, something around 5000C.
The earth is not a perfect thermal insulator.
Estimates are, the thermal energy flowing from the core to the surface is ~ 0.57 W/m^2.
Over time, the ocean warms even with no atmosphere.
In point of discussion, with no atmosphere, the ocean warms faster given no convection or latent heat losses.
Geothermal flux for oceanic crust is supposed to be ~100 mW/m^2.
This is enough energy to warm ALL ocean water 1K every 5000 years, assuming the sun provides the isolating warm surface layer.
So without the cooling of sinking brine (AABW = AntArctic Bottom Water) the oceans would be warming pretty “fast” indeed.
Which occurs everywhere. Don’t even need brine, the same process occurs in freshwater lakes – Lake Baikal, for example. That’s why lakes don’t freeze right through, and there are freshwater lakes under kilometers of Antarctic ice.
Just sayin’.
Lake overturning does not happen in the oceans, they are too deep.
Lakes under Antarctic ice like Lake Vostok are there because of geothermal warming under the ice.
…and clouds with a local albedo of up to 0.8 above an ocean with an albedo of about 0.1…..With the worlds 70% ocean and 65% cloud cover (mostly) averaging out to an albedo of about 0.3 depending on how much afternoon cloudiness forms…itself a function of how much sunlight the ocean absorbs….
Plus all of the life, without which this whole thing disappears. There is at least some merit to Lovelock’s gaia hypothesis.
Totally, ocean surfaces barely cool at night, making the global average surface temperature much higher.
From the above article (my bold emphasis added):
“Multiplying the surface sensitivity by 1.26 to convert to sensitivity to greenhouse gases increases the equilibrium climate sensitivity from the surface value of 0.35°C per doubling of CO2 calculated above, to a final figure of 0.44°C per CO2 doubling.”
That would be true only if the second half of a doubling of the greenhouse gases concentration (take the case of CO2 going from 400 to 800 ppmv atmospheric concentration) had the same effect as the initial change (going 0 to 400 ppmv) . . . ie., if there was a linear relationship. However, that is not the way gas absorption of IR radiation works . . . the Beer-Lambert Law states that the absorption of radiation is a logarithmic function of the concentration of the absorbing gas and the path length of the radiation.
If William Happer, Richard Lindzen and other atmospheric scientists are correct in asserting that CO2 has basically reached its logarithmic asymptote with respect to any further concentration increase causing any significant additional absorption of LWIR radiation energy, then ECS could be expected to be much less than 0.44°C per CO2 doubling . . . perhaps as low as 0.04 °C per CO2 doubling.
Overall, I believe Willis has done a very fine job in his above article in mathematically establishing an upper limit of 0.5°C per CO2 doubling, but it could be—probably is, actually—much less.
There is a whole other aspect to this of which most people are oblivious. As Willis mentioned the “canonical” CO2 forcing is 3.7 watts per doubling of CO2….with a lot of discrepancy between how much temperature increase at surface that 3.7 watts/sq.M is going to cause.
Well the answer is precious little. The atmosphere is in a constant state of convection, and advection due to thermal and water vapor buoyancy and coriolis forces, causing weather fronts, thunder storms and so on. Kinetic energy that has resulted from heat energy. Over any given square meter of the planet’s surface the kinetic energy of the moving parcels of atmosphere sum to something like the following (courtesy David Dibbell and ERA5) tap the arrow to make it run :
https://www.youtube.com/watch?v=hDurP-4gVrY
As you can see, the effect of 3.7 watts in these existing few thousands of watts per sq.M. is probably irrelevant…certainly not going to all show up as “surface temperature” as many thermodynamically challenged assume.
Thanks for the honorable mention here. “There is a whole other aspect to this…” – exactly so.
This is a also a good time to mention the comments from meteorological experts on Callendar’s 1938 paper attributing a reported warming trend to rising concentration of CO2.
In short, it was pointed out that the motion changes everything about what to expect as an end result from the incremental IR absorbing power. More here.
https://wattsupwiththat.com/2025/04/06/open-thread-138/#comment-4058322
Bottom line – energy conversion within the general circulation is the proper context, in my view, to judge the order of magnitude of the incremental radiative effect.
“Overall, I believe Willis has done a very fine job in his above article in mathematically establishing an upper limit of 0.5°C per CO2 doubling, but it could be—probably is, actually—much less.”
This is a good point. Starting with the “forcing” + “feedback” framing of the climate issue, Willis uses a logical approach to arrive at a non-crisis answer. So even though I consider the framing itself as physically unsound, for reasons I often post about, I recognize that the exercise is meaningful to many.
Thank you.
No refund?
What a scam.
Willis,
Two significant “missing from actions” are –
Somewhat irrationally on my part, I find it intuitively a difficult concept that the addition of a GHG can increase temperature at the surface, as if energy was being created. It leads to thoughts of perpetual motion machines using sunlight and pumps to move gas. I do not exclude a climate sensitivity value of zero, but it is hard to explain why beyond “Does it fail sniff test.”
Practical experiments on these topics have forever been in short supply, sometimes with the excuse that there is no parallel Earth B.
Is it time to look for differences between temperatures measured historically, as in air just above land or sea, as opposed to the temperature of said land or sea, in the first short distance from the surface. Can they be taken as equal? Seeking an answer in 1974, I asked a friendly geologist to measure the land temperature profile near his home every hour, from surface down to 1.5 metres, for 2.5 days. Interpretation used traditional heat flow equations, with lags. Found that more sites needed study, so no firm conclusion worth reporting is offered. (Some studies of near surface sea are in the literature.)
My tiny experiment was also hard to reproduce. The observer was chosen because his wife had produced a child a week before, so he would be up and about a lot during the night.
Geoff S
See my posts “The Steel Greenhouse“, “People Living In Glass Planets“, and “The R.W. Wood Experiment“. There’s no violation of the 2nd Law of Thermodynamics.
From the steel greenhouse link –
Physically impossible. Sorry about that – no matter how close your steel shell is, it has a greater area than your spherical radiator. Therefore, it may (if a perfect mirror) reflect 100% of the radiation – something less than 235 W/m2.
Trying to weasel out by saying it will be so close as to make no difference is just stupid. The shell will be slightly cooler than sphere, radiating slightly less than 235 W/m2. And, of course, a hotter body does not absorb radiation from a colder – otherwise you could heat your coffee by adding energy in the form of ice cubes.
Again, Willis, learn that you can ignore physical and mathematical rules all you like, but it still won’t allow you to make thermometers hotter by adding CO2 to air.
Appealing to your own authority won’t help, will it?
And, of course, a hotter body does not absorb
radiationthermal energy from a colderBetter yet: “NET THERMAL ENERGY”!
Here is part of the work re temperature of land just below surface. (I thought I had lost it.)
Without going into detail, a broad question is: how/where does one specify the land temperature of the Earth affected by the Sun?
Geoff S
Blimey, that’s a lot of solar energy being absorbed and re-emitted every day! Does WE’s calculations take this into account?
Another source for ground temperatures.
It is simply IMPOSSIBLE for “lab experiments” to objectively and scientifically represent the complex details of Earth’s atmosphere absorbing surface-emitted LWIR radiation, given the range of variables such as representative long path-lengths (on the orders of kilometers), influencing parameter continuous variations with altitude/atmospheric density and temperature and pressure, convection/wind/turbulence effects, and the discrepancies introduced by an experiment’s physical boundary conditions (i.e., physical container size, materials, and unavoidable heat transfer).
Agreed. As Feynman said –
You can’t even devise a testable hypothesis, let alone an experiment to test it, can you? Just spout more ignorant and gullible nonsense. You believe that adding CO2 to air makes thermometers hotter. How stupid is that? John Tyndall’s experiments show the exact opposite, but you refuse to accept fact, don’t you?
Well, I simply never claimed that anywhere, any time.
However, I do believe that LWIR-absorbing gases in Earth’s atmosphere (including CO2) do absorb radiation energy emitted from Earth’s surfaces up to a certain concentration level and subsequently distribute that energy throughout the atmosphere via molecular collisions with other non-LWIR-absorbing molecules, mainly N2 and O2.
That doesn’t necessarily mean that a thermometer in your refrigerator or in your oven gets hotter with each incremental increase in atmospheric CO2 concentration. /sarc
BTW, in context, Richard Feynman was referring to “agreeing with” good scientific experiments and observations, NOT improper or bad ones.
In that regard, nature has already conveniently “run” the full scale, realistic “experiment” for us . . . paleoclimatology proxies over the last 500 million years or so clearly reveal the ridiculously poor correlation of global lower atmospheric temperature versus atmospheric CO2 concentration levels. As one example, see the attached graph.
Now, you were saying something about “ignorant and gullible nonsense” and “stupid” . . .
Bottom line, Michael, better (1) check on you med dosages, or (2) check in with a good psychiatrist.
Ooops . . . here’s the graph that I meant to attach
I didn’t say you did. I said you believe that adding CO2 to air makes thermometers hotter. Wriggle and squirm as much as you like.
So, adding CO2 to air makes thermometers hotter, is that it? A thermometer in this case being an instrument designed to measure the degree of hotness of matter. Play some silly semantic game at this point, if you wish.
I agree. For example, the amateurish “experiments” performed by people like Roy Spencer PhD – trying to demonstrate that the radiation from a hotter object can be absorbed by a colder, raising the temperature of the hotter. Magic, of course, because the colder object doesn’t lose any energy at all! If it did, it would become colder. But the ignorant and gullible only rattle on about “the conservation of energy” when it suits their fantasy.
Really? That’s not an “experiment”. Four and a half billion years of history merely shows that the surface is no longer molten, liquid water exists, and so on. In other words, the Earth has cooled.
A common response by an ignorant and gullible cultist, who has no facts to back up his delusional thinking that adding CO2 to air makes it hotter.
Your humor is palpable . . . please do carry on.
Would that be patronising sarcasm from an ignorant and gullible cultist who believes that adding CO2 to air makes thermometers hotter?
Or is it a true encomium, praising my obvious wit and superior intelligence?
I accept your flattery. Thank you.
You are welcome. Take it for what it’s worth.
Literally hundreds of experiments performed in undergrad lab projects, shining light through a long tube of air with transparent ends…then adding CO2 show the heat absorption, thus calculable temperature increase, using an infrared camera. Thus making you undeniably a “denier” in their view.
It is actually one of the experiments (or videos of it) that cements the concept of CO2 atmospheric warming in college educated minds. Unfortunately little time is spent measuring how very little the 3.7 watts per CO2 doubling is actually warming the tube, since convection and random air currents in the lab make tube surface temperature reading inconsistent.
https://youtu.be/kGaV3PiobYk?si=41Gn-isMuSa6wq4-
Go back and check: I clearly posted “It is simply IMPOSSIBLE for ‘lab experiments’ to objectively and scientifically represent the complex details of Earth’s atmosphere absorbing surface-emitted LWIR radiation, given . . .”
Based on your comment, it appears YOU did not comprehend the meaning of “objectively and scientifically represent the complex details of Earth’s atmosphere absorbing surface-emitted LWIR radiation”.
What? . . . you really think “shining light through a long tube of air with transparent ends” fairly represents Earth-surface emitted LWIR (“light” that is not visible to the human eye) and the Earth’s atmospheric variations of temperature, pressure and humidity over a path length in the range of 3-5 km???
And should I be surprised that adding “light” energy to a closed tube of air, with or without adding in CO2, causes a temperature increase.
And doing hundreds of bad undergrad lab experiments is supposed to give one confidence in the cumulative results???
That you admit “measuring how very little the 3.7 watts per CO2 doubling is actually warming the tube, since convection and random air currents in the lab make tube surface temperature reading inconsistent” speaks volumes about the humorous points that exist in your overall reply . . . among them, don’t you think the tube surface temperatures have a direct influence on the temperature of the gas contained within?
Correct.
Cp versus Cv plays into this as well.
The properties of glass, EM and thermal are not properly adjudicated in the experiment.
TYS,
I agree that the complexity of the whole earth system makes lab modelling of the whole system practically impossible.
But I am seeking lab experiments that show if parts of the system can be measured by lab work. For example, it should be within the wit of science to see if a synthetic gaseous atmosphere can be heated differentially when various levels of air +/- CO2 are irradiated by the same sunlight. The experiments I have seen so far have been amateurish. How about a proper effort?
Geoff S
Geoff, I certainly welcome any good experimental data that might shed light on this complex subject. The main problem that I see is as follows:
The characteristic path length for an e^-1 “folding” of LWIR absorption of surface-emitted LWIR is stated to range from meters to kilometers in Earth’s lower atmosphere (I’m guessing that range of variation is due to the specific wavelength being considered and complex variables such as percent of water vapor actually present in the path-length column and the rate of molecular collisions, itself a function of temperature and pressure). I believe that most atmospheric physicists agree that the “extinction” path length for absorption of surface-emitted LWIR is about 5 km above Earth’s sea level surface, admitting there is a wavelength-dependent “atmospheric window” that permits some surface-emitted LWIR to escape directly to space.
Now, LWIR-radiation off Earth’s surfaces (generally considered to cover a wavelength range of 3–70 microns for the range of Earth’s surface temperatures) is emitted hemispherically, with most of the energy leaving in a direction close to normal to the surface . . . that is, generally vertically. So, for a realistic “lab” experiment one would need to mimic the change in atmospheric temperature and pressure and humidity, thus density, as a function of altitude along an experimental column on the order of 5 km long. I don’t see any means to produce these variations using a horizontal test pipe, and I don’t think it can be done in a vertical tube of any diameter that would need to reach an altitude of 5 km (well, to even reach an altitude of at least 3 km to characterize the overall trend).
Your thoughts on this are always welcome.
Although the EM wave front does have a vector orthogonal to the surface, the EM was is not planar, it is spherical (1/r^2). A basic assumption is all EM emitted from the surface goes straight up is flawed and is rarely challenged.
Given the temperature, the preponderance of EM emitted by the surface is in the LW frequencies, including LWIR through uWaves.
The SB T^4 calculations demonstrate a full spectrum of emissions. The total energy and peak energy frequency shift with temperature as we all know, but there are low levels of light and UV and near IR emitted in addition to LWIR.
Part of my objections to the whole “climate science” debacle is the number of items omitted or hand waved off with the results published with preposterous resolution. CERES bandwidth covers 99.95% of the spectrum with a published measurement accuracy of +/- 0.5% to 1.0% and with that we can reach a conclusion of an energy imbalance of 0.6% and a delta temperature of 0.01 C?
This is so much gobbledygook composed with scientific terminology.
A wavefront that is not planar does not have of vector that is orthogonal to any specific plane, and in many common cases it is called “radial”.
The term “1/r^2” does not define a spherical wavefront; instead it characterizes the dependence of many different physical parameters on variation with distance, such as gravitational attraction, electrostatic field strength, EM radiated energy, acoustic energy, and intensity of ionizing radiation.
Other physical parameters vary as 1/r^3, such as the electric field of an electric dipole, the magnetic field of an electric dipole, and gravitational tidal forces.
I specifically pointed out “LWIR-radiation off Earth’s surfaces . . . is emitted hemispherically”, so question your need to reply with “A basic assumption is all EM emitted from the surface goes straight up is flawed and is rarely challenged.”
My question is if the returned energy is originally from the surface and has been absorbed by CO2 in the atmosphere, then how much of that re-emitted energy from CO2 actually reaches the surface? Why isn’t it absorbed by the CO2 in the atmosphere and re-emitted toward space? I have a hard time picturing how that “returned energy” gets to the surface without being absorbed first. I can believe that the atmosphere might warm from the absorbed energy but that isn’t the same thing as warming the surface.
‘I have a hard time picturing how that “returned energy” gets to the surface without being absorbed first.’
You’re having a hard time picturing that “returned energy” because it’s not happening.
https://issuu.com/johna.shanahan/docs/241128_thermaliztion_of_sunlight
Interesting read. I mostly agree but there are points that are incongruent with my understanding of physics.
Thanks for the feedback. I’m certain my understanding of / training in physics is inferior to yours, but after a good amount of slogging through the concepts, I’m comfortable that Shula et al are on to something.
Oh, but it IS happening. It is called the reflection of EM energy and does NOT involve the absorption of incoming energy followed by radiation of that energy.
Theoretically, a 100% reflective surface would never get hot no matter how many terrawatts of incoming energy impinged upon its surface.
In common discussions of Earth’s climate and its net power flux balance, reflected EM power/energy is commonly referred to as Earth’s albedo, which is a composite of power/energy reflected from Earth’s surface AND from its atmosphere/clouds.
You need to read Planck’s discussion of electrodynamics using Maxwell’s equations. See Part II to start.
The upshot is that EM waves are not reflected (reversed in direction) like a ball against a wall. They must be absorbed and remitted.
I am familiar with both Planck’s theories and laws, as well as Maxwell’s equations of classical EM theory.
So,
1) Specular reflection—as opposed to diffuse reflection—is the mirror-like reflection of light from a smooth surface, where incoming parallel light rays bounce off the surface in a single, continuing direction . . . analogous to a ball bouncing off a wall. The most important physical law for specular reflection is “the angle of incidence equals the angle of reflection”. And this law is successfully applied to the theory, design and manufacturing of many optical surfaces, some to a precision of one-tenth the wavelength(s) of incoming light (EM energy). For example, each of the James Webb Space Telescope’s (JWST) primary mirror segments was polished to an average surface figure precision of better than 25 nanometers RMS . . . and is then mechanically deformed to achieve a much better operating surface figure.
2) If one really believes that incoming light energy (considered as either photons or EM waves) is first absorbed and then subsequently radiated (again, as either photons and waves), then there is the apparently intractable issue of explaining under quantum theory how the involved atoms/molecules/electron clouds of the mirror’s surface material all “know”/”calculate” the exact same direction in which to radiate that energy after some time constant. Quantum behavior generally involves random in time and random in space (omnidirectional) emissions of radiation (excluding special cases such as lasers stimulating coherent, planar emission as light passes through a medium with an special “inverted electronic state” and in a device custom-designed to produce such). But specular mirrors are not such devices.
In conclusion, please explain how the hypothesis of radiation-absorption-followed-by-radiation is consistent with the observed angle-of-incidence-equals-angle-of-reflection characteristic of specular mirror surfaces.
“then there is the apparently intractable issue of explaining under quantum theory how the involved atoms/molecules/electron clouds of the mirror’s surface material all “know”/”calculate” the exact same direction in which to radiate that energy after some time constant. “
Red herring. If you are familiar with Planck then you know that his treatise only addresses “oscillators” and he doesn’t try to explain any quantum effects of those “oscillators” .
Planck states: “ For in this case, as we know, the familiar formulæ of the
classical dynamics hold for any period of the oscillator whatever, since
the quantity element of action h may then, without any appreciable
error, be regarded as infinitely small.”
His application of “classical dynamics” is a major clue.
“In conclusion, please explain how the hypothesis of radiation-absorption-followed-by-radiation is consistent with the observed angle-of-incidence-equals-angle-of-reflection characteristic of specular mirror surfaces.”
Heat transfer is *ONLY* accomplished by the amount of radiation perpendicular to the surface. As the angle of incidence changes the area perpendicular to the radiation vector changes. A_eff = Acosθ.
This does *not* mean that heat can’t be transferred to an oblique surface. It means that the amount transferred is less than the total available. The rest is reflected. Jeesh, just think about it. When do you get the most warmth from the sun when you stand outside? Just after sunrise? Just before sunset? At noon?
Planck’s entire work is based on black bodies. Black bodies by definition absorb *all* impinging radiation. That is not the Earth. Panck: “Generally we may say: Emission without simultaneous absorption is irreversible, while the opposite process, absorption without emission, is impossible in nature.”
For a non-black body, absorption only occurs for the radiation perpendicular to the surface. What do *YOU* think happens to the rest of the energy if it isn’t reflected?
Remember, Planck doesn’t say *when* the emission occurs, just that it does.
Are you kidding me???
Do you really assert that if I shine an infrared heat lamp at a flat piece of iron or steel having a diffuse finish at, say, a 30 degree angle-of-incidence to its surface it WILL NOT heat up as a result of that intercepted energy flux.
Then too, does your comment mean that you believe Earth’s surfaces don’t receive heating from the Sun except around the time of local noon, and even then only if the Sun’s elevation in the sky is 90°?
Not true. According to https://en.wikipedia.org/wiki/Emissivity ,
“By 1900 Max Planck empirically derived a generalized law of blackbody radiation, thus clarifying the emissivity and absorptivity concepts at individual wavelengths.”
with a specific bibliographical reference to
Planck, Max (1901). “Über das Gesetz der Energieverteilung im Normalspektrum”. Annalen der Physik. 4 (3): 553–563.
The terms “emissivity” and “absorptivity” were introduced to characterize radiation variations of REAL WORLD OBJECTS compared to theoretical blackbodies.
No need for me to comment further.
Why do you keep throwing up red herrings? Do you not understand the difference between a spherical wave from a near point source and a plane wave from a distant source?
Attached is a page from a Thermodynamics textbook published clear back in 1942.
cosθ_1 and cosθ_2 are the angles the surfaces make with regard to the perpendicular.
With a plane wave, the energy hitting dA2 from dA1 and being absorbed is maximum when cosθ_2 is perpendicular to the plane wave.
Please note that the equations here is for black bodies. For an object that is not a black body then the reflectivity factor must be added to the equations. The amount of energy reflected based on the reflection factor is *also* based on the angle of incidence.
Where in Pete’s name are you coming up with this? Do you *really* think that cosθ is 1 at 90deg and 0 for all other angles?
There *is* a reason that the daytime temperature is very much like a sine wave. You seem to believe that a point on the Earth’s surface is heated at the same rate from sunrise to sunset. That would *NOT* generate a daytime temperature curve that is close to a sine wave. It’s not a pure sine wave because the actual temperature curve is a mixing of the sine wave from the sun and the exponential decay from the Earth radiating away the heat it gets from the sun.
Planck’s work is based on black bodies. He recognized that a given homogeneous and isotropic could absorb and emit differently as the wavelength of radiation changed. This doesn’t mean he didn’t evaluate the process as with a black body.
Where do you think the term and resulting curves known as Planck curves originated. These were not determined from grey body evaluation.
If you believe everything you read at Wikipedia, then you are missing the boat.
Planck’s work is available on the internet for a small cost. Obviously you have not studied it for understanding if you must use Wikipedia as a primary source.
Go thru this exercise.
Look carefully at what K↓ becomes. It is the perpendicular component of the impinging ray.
Have you never separated a vector into its vertical and horizontal vectors in a physics lab so you can add forces?
Hah! Boldly stated but misleading and fundamentally false. “Classical dynamics” DID NOT consider electrically-charged oscillators having only certain allowable step changes (i.e., “minimal increments”) in energy as part of a physical material’s composition.
From https://en.wikipedia.org/wiki/Planck%27s_law (my bold emphasis added):
“In 1900, German physicist Max Planck heuristically derived a formula for the observed spectrum by assuming that a hypothetical electrically charged oscillator in a cavity that contained black-body radiation could only change its energy in a minimal increment, E, that was proportional to the frequency of its associated electromagnetic wave. While Planck originally regarded the hypothesis of dividing energy into increments as a mathematical artifice, introduced merely to get the correct answer, other physicists including Albert Einstein built on his work, and Planck’s insight is now recognized to be of fundamental importance to quantum theory.“
“Boldly stated but misleading and fundamentally false”
Malarky. If Planck had considered the quantum effects of individual ATOMS then he would have also had to characterize all the quantum effects from different atoms. He didn’t since a blackbody made up of iron atoms would have a different characteristic than a blackbody made up of sodium.
wiki: “hypothetical electrically charged oscillator”
Planck analyzed hypothetical electrically charged oscillators. He *also* assumed that enough of those hypothetical electrically charged oscillators would exist in the blackbody that they would generate isotropic emission since he also assumed that the energy from each individual hypothetical electrically charged oscillator would be emitted in a specific direction. Thus there had to be enough oscillators to cover the entire sphere.
You claim you have studied Planck but it’s sure not obvious from what you are asserting!
This is all covered at the very start of his thesis. It is fundamental to all of his calculations. Knowing quantum level motion is not necessary for determining the macro amount of radiation emitted and absorbed from a homogeneous and isotropic body. Just as in a chemical reaction, one doesn’t need to evaluate each and every atom in order to understand the macro behavior.
Wikipedia again? Here is the real issue. Is Planck’s work incorrect? Has it been refuted or just refined and built upon?
If you can’t refute his work or find a refutation that has been published, then don’t claim references to it are incorrect.
Here is a graphical depiction of what occurs.
Here is a section from a class at the University of Wisconsin. Maybe it will help explain to you how reflection, absorption, transmission and emission occurs.
If you have a problem with these, please explain in detail with the appropriate math how these resources have managed to be wrong. Maybe you can also contact them to advise them about changing their course materials.
Please note that material that only absorb partially, do radiate that absorbed energy in all directions. However, the remaining energy is either reflected or transmitted.
Your “graphical depiction” does not meet my request for a quantum theory-based explanation of light reflection with the understanding that quantum processes are characterized by random times of radiation emission (aka statistical “relaxation” time constants) as well as by isotropic (i.e., random direction) radiation.
You did not provide any link to a University of Wisconsin class “section” . . . but even if you did, it is quite common for class textbooks or even a given professor’s notes to contain factual errors.
Answering my request for a scientific hypothesis that YOU propose with a request that I provide evidence that you or others are wrong is known as a logical fallacy.
Finally, in your last sentence in you post immediately above you state:
“. . . the remaining energy is either reflected or transmitted.”
which stands in sharp contrast to your prior declaration that:
“The upshot is that EM waves are not reflected . . . They must be absorbed and remitted.”
Hedging your bets, huh?
“Your “graphical depiction” does not meet my request for a quantum theory-based explanation of light reflection with the understanding that quantum processes are characterized by random times of radiation emission (aka statistical “relaxation” time constants) as well as by isotropic (i.e., random direction) radiation.”
Go *actually* study Planck. Heat transfer is *NOT* analyzed at the atomic level but at the macro level. There have to be enough radiators in the object being studied to generate isotropic radiation. That’s what Planck refers to as “macroscopic”.
Sec 121: ” If we now take into account the considerations stated above
in Sec. 113, it is evident that we must distinguish in the theoretical
treatment two entirely different kinds of states, which we may denote
as “microscopic” and “macroscopic” states. The microscopic state is
the state as described by a mechanical or electrodynamical observer;
it contains the separate values of all coordinates, velocities, and field-
strengths. The microscopic processes, according to the laws of mechan-
ics and electrodynamics, take place in a perfectly unambiguous way;
for them entropy and the second principle of thermodynamics have no
significance. The macroscopic state, however, is the state as observed
by a thermodynamic observer; any macroscopic state contains a large
number of microscopic ones, which it unites in a mean value. Macro-
scopic processes take place in an unambiguous way in the sense of the
second principle, when, and only when, the hypothesis of the elemental
chaos (Sec. 117) is satisfied.”
Sec 135: “Since, according to this law, we are free to choose any system whatever, we now select from all possible emitting and absorbing systems the simplest conceivable one, namely, one consisting of a large number N of similar stationary oscillators, each consisting of two poles, charged with equal quantities of electricity of opposite sign, which may move relatively to each other on a fixed straight line, the axis of the
oscillator.” (bolding mine, tpg)
BTW, Planck also understood reflectivity vs absorption. See his Section 36 and 59.
Again, you claim to have studied Planck but it’s simply not obvious from your postings.
Your need to know the micro details of EM waves and their absorption and emission is misplaced for determining the heat being transferred in a macro assessment.
Here is what Planck says about the issue. If you have a problem with his description, then I suggest you write a refutation and include it in your response.
LOL! So university textbooks have factual errors. Is that all you have?
Why don’t you describe your refutation of the graphic? Is it a factual error?
Maybe you have some textbook references that directly contradict what I have shown from the U of Wisconsin. I would love to see them!
The answer is as clear as night and day! Sorry for the play on words, but the surface heats during the day, and cools at night. The phenomenon is explicable using known physical laws, and no “CO2 magic” of the “climate wizard” variety is needed.
Actually, people like Baron Fourier and Professor John Tyndall (amongst many others), wrote fairly lengthy explanations – not involving quantum physics, photons, or electron orbitals at all.
I accept that many people who are afflicted with ignorance, gullibility, or both won’t be happy, but I really don’t care.<g>
Tim,
Please critique.
When a bush fire heats the air abnormally, the conditions of atmospheric circulation, emission to space and so on change briefly to restore the overall temperature to its long-term narrow bracket.
When a volcano erupts and heats the air abnormally, the conditions of atmospheric circulation, emission to space and so on change briefly to restore the overall temperature to its long-term narrow bracket.
When long wavelength radiation from greenhouse gases heats the air abnormally, the conditions of atmospheric circulation, emission to space and so on change briefly to restore the overall temperature to its long-term narrow bracket ……
Geoff S
You are correct.
Le Châtelier’s principle applied to climate. Valid for a climate in an quasi-equilibrium condition and having multiple negative feedback loops, such as Earth now and as has existed for
millionsbillions of years.It is one of the primary reasons that AGW/CAGW climate alarmists were forced to come up with imaginary “tipping points”.
Bravo.
The other part is that “if absorbed by the surface again”, then the surface should radiate it across the whole spectrum which means most goes directly to space through the atmospheric window.
How long does it take?
(from when the CO2 is added to when the entire planet settles at its new temperatures)
I don’t think that this can be answered as the (man made)co2 is added only in very specific places on earth and then needs some time to disperse and it happens at a very slow pace.
But if we’d able to add instantly 10% to a system,
the new temperature level(not the climate)on a continental bases should be reached very fast as those electromagnetic waves travel at the speed of light.
Go on then. Replace 10% of the air in a container with CO2. Not enough? Replace it all, 100%!
Gee, the temperature didn’t change at all. Maybe you’re dreaming?
1) I suggest you use Grok with what you want to post and take its suggestions. Otherwise, you come across like a pompous ass. The way you respond is horrific and destroys anything you wish to convey.
2) You would get a dramatic temperature change with that. That is if you were Bill Nye and Al Gore (and their favorite NLE). There are links that have not withstood the test of time, but Anthony nails it here.
Video analysis and scene replication suggests that Al Gore’s Climate Reality Project fabricated their Climate 101 video “Simple Experiment” – Watts Up With That?
Oh dear, taking offense, are you? What mental deficit leads you to think that I am liable to accept advice from someone who uses phrases like “pompous ass” to describe me? Flattery will get you everywhere, you know.
I respond as I wish.The moderators peruse and approve every comment I make before it appears, as far as I know.That’s why it sometimes seems I don’t respond quickly.
Grok? You jest, surely. What’s wrong with using your own intelligence? Dig up a few facts, show that I’m wrong, and I will change my views.
Wouldn’t you?
Our moon has an albedo of ~11%. So an easy guess would be:
Slightly colder then our moon.
Average surface temperature of our moon ~197K as measured by the Diviner project.
Taking Earths higher rotation rate into account might even out the difference in albedo more or less.
Great answer! Assuming the hypothetical Earth-with-no-atmosphere is consistent with a planetary body-without-oceans.
Sorry, Ben, but you can’t properly calculate average temperature that way. Temperature is an intensive property. To get the true average temperature you need to average the surface radiation, then convert it to temperature.
w.
“To get the true average temperature you need to average the surface radiation, then convert it to temperature.”
I don’t understand that comment. Per the Stefan-Boltzmann equation (P = eσAT^4), one can calculate radiation power from a body’s average surface temperature just as easily as one can calculate a body’s average surface temperature from its net radiation power.
The S-B equation is based on Planck’s work and thus has some restrictive assumptions. E.g. homogenous material in thermal equilibrium. If there are thermal gradients, materials with varying emissivity (think wavelength dependent emissivity) then you need to be able to describe them and integrate them.
Just measuring radiation at “space” and averaging it winds up with a measurement uncertainty for the surface temperature that is probably greater than the 0.5C per doubling of CO2 calculation.
Tim, thanks for your reply, but the gist of the above article was an essentially mathematically treatment to develop a surface temperature from a radiation balance with an appeal to using the Stephan-Boltzmann equation.
Up until introducing the scatterplot of Figure 3, the article made no reference to measurement uncertainty (appropriately) since the discussion was framed in theory, not practice.
At some point you have to leave the blackboard and actual interface with reality. All the blackboard stuff doesn’t “prove” anything, observations in reality does. If your calculation depends on satellite measurements then you’ve already left the world of the the blackboard and theory. Thus measurement uncertainty *must* be considered.
Uhhhh . . . did you mean to direct that comment to Willis Eschenbach, author of the above article, and NOT to me?
Radiation varies as the fourth power of the temperature (K), which is wildly non-linear. If you double the temperature, the radiation increases sixteenfold.
And temperature, unlike radiation, is not conserved. It’s an intensive value, like say density, and not an extensive value, like say mass or W/m2.
Both of these mean you can’t just average temperature. Doesn’t work.
w.
Hang on thar jest a darn minute, pilgrim!
So the Earth’s “global average surface temperature” is worthless? Come on, Willis, tell me you really meant something else, but you’ve forgotten what it was.
Temperature is the degree of hotness, and temperature measuring instruments are calibrated in these “degrees” – of varying sizes, Celsius, Fahrenheit, Reaumur, for example.
Now, if your “calculations” give you a different answer than a properly calibrated “degree of hotness” measuring instrument, then your “calculation” is worthless. “Should be”, “could be”, “would be”, fail in the face of “is”.
No GHE.The Earth has cooled, and continues to do so. Accept reality.
I’m afraid the current stock of climate “scientists” has to die out before this reality will become accepted. Hope to be still around when that happens.
Ben, you can but hope. Fully entrenched popular madness can last a long, long, time.
If I live that long, I’m worried that I might not want to! <g>
“So the Earth’s “global average surface temperature” is worthless? “
Yes, it is. Readings at individual stations aren’t, for the most part, but averaging them all together is utterly meaningless.
Those statements are contradictory . . . as well the fist sentence being meaningless without being referenced to a time frame.
Earth has definitely cooled since its formation. So what?
Earth definitely cools as it enters an Ice Age (there have been at least five so far in Earth’s climate history). However, Earth definitely warms as it exits a major Ice Age . . . how could it not?
Earth definitely cools in transitioning from an interglacial (aka interstadial) period to a glacial (stadial) period. However, Earth definitely warms in transitioning from a glacial period to an interglacial period . . . how could it not?
Paleoclimatology science has documented tens of glacial/interglacial cycles of cooling/warming over the 2.6 million years since Earth entered the current Quaternary Ice Age. See the attached graph for a graph of the last 10 or so such cycles . . . and note the indication of warming.
And, “recently” in geological terms, the Earth warmed up as it exited the Last Glacial Period (LGP) and entered the current Holocene that started about 11,700 years ago.
These facts are reality . . . accept them.
That particular piece of word salad is pretty stupid, isn’t it?
The Earth is a large glowing blob of matter, with a very thin congealed crust. If you imagine the Earth as the size of an apple, the non-glowing hot crust is about as thick as the apple’s skin.
About 300,000,000 km or so from the Sun. Not much warming there. Even a bowl of boiling hot water left in full sunlight – say 1,000 W/m2 – will cool, and it’s less than 100 C!
In my worthless opinion, you are ignorant and gullible. The Earth doesn’t “warm up”. You are confused about the effect of thermometers responding to increased heat by indicating higher temperatures.
Or maybe you share the delusion that adding CO2 to air makes thermometers hotter?
Now that’s funny! Did you forget your earlier post (above, posted November 24, 2025 6:36 pm) wherein you declared in reply to to me:
Are you now doubting your previous hardline declaration?
I’ve already provided some corrective actions for you.
Of course not. You share the delusional belief that adding C2 to air makes thermometers hotter with a host of other ignorant and gullible people. Anyone who believes in a “greenhouse effect” (which of course cannot even be consistently described!).
Ah, the unstated “corrective action”. What might that be? Should I be quaking in fear at the prospect of being “corrected”, or merely be very afraid?
No, I also consider the melting of glacial ice on land to be an indication of temperatures WARMING over time . . . no need for a thermometer to prove that!
Here is what Wikipedia (https://en.wikipedia.org/wiki/Laurentide_ice_sheet )
has to say about the Laurentide ice sheet that covered millions of square miles, including most of Canada and a large portion of the Northern United States, multiple times during the Quaternary glaciation epochs, starting about 2.58 million years ago:
“The last advance covered most of northern North America between c. 95,000 and c. 20,000 years before the present day and, among other geomorphological effects, gouged out the five Great Lakes and the hosts of smaller lakes of the Canadian Shield.
“In New York, the ice that covered Manhattan was about 2,000 feet high before it began to melt in about 16,000 BC. The ice in the area disappeared around 10,000 BC. The ground in the New York area has since risen by more than 150 ft because of the removal of the enormous weight of the melted ice.”
Thank you for providing the opportunity to present this scientific data to other WUWT readers, as I a confident it will have no impact whatsoever on your stated beliefs.
So you now think that thermometers respond to ground level changes? Or are ground level changes due to the mythical GHE?
You are obviously ignorant and gullible. You bang on about “sea level” changes smaller than the thickness of a human hair, then blithely ignore the fact that a rise of land abutting the sea of 150 ft results in a drop of sea level of 150 ft!
So do you think increases of CO2 cause sea levels around Manhattan to drop by 150 ft, or rise by 0.1 mm?
Sorry, adding CO2 to air doesn’t make thermometers hotter, and marine fossils found at the tip of Mt Everest do not mean that sea levels dropped by more than 8848 m due to CO2 levels either.
Accept reality. You’re ignorant and gullible, as well as confused.
As I predicted: “it will have no impact whatsoever on your stated beliefs”.
Yes! The GAST is a “worthless” calculation, if the purpose of the calculation is to measure any aspect of “climate” experienced by any human at any location on Earth. It is even of negative value if the intent is to infer that a minuscule change in “average” “degree of hotness” over the course of decades is sufficient to change all, or any, aspects of the various local climates that humans actually care about!
Your last sentence is what I’ve been screaming for the last three decades! Unfortunately, though true, it seems to make no difference to the Global Warming activists, who keep fretting about supposed “changes” of 0.05K in the global average of the seasonal average of the daily average of the range of thermometer readings of thermometers scattered somewhat willy-nilly about the Earth’s surface over the course of decades! “Bless their hearts!”
Radiation depends on temperature if you are using S-B. Temperature at night is basically a decaying exponential. Thus the radiation from the surface is also an exponential decay at night regardless of how you calculate the value of the radiation in W/m^2.
The actual heat loss then becomes the integral of the nighttime temperature curve (i.e. the area under the curve) or the integral of the nighttime radiation curve.
The radiation returned toward Earth by CO2 is “reflected” radiation. According to Planck, reflected radiation back toward the source is “compensated” for by future radiation from the source.
Planck: “For example, if we let the rays emitted by the body fall
back on it, say by suitable reflection, the body, while again absorbing
these rays, will necessarily be at the same time emitting new rays, and
this is the compensation required by the second principle.”
What this means is that the integral contains a factor of 1/λ. Reflected radiation makes λ smaller, i.e. a slower decay. Thus the integral gets larger for a slower decay. This is the compensation Planck speaks of. Higher temperature lasting for a longer time radiates away more total heat, the original heat plus the reflected heat.
ROFL! A downcheck but absolutely *NO* refutation of Planck. Why am I not surprised?
Willis, thank you for the reply, but that leaves me even further at a loss. In your third paragraph of the above article you state:
“So we can calculate the temperature directly from the absorbed radiation using the Stefan-Boltzmann equation.”
If that calculated temperature (a single value) is not the “average” temperature of the hypothetical body in the gedanken experiment, then what is it???
TYS, the CERES dataset contains 1° latitude x 1° longitude gridded datasets of radiative flows. Among these are the upwelling longwave (thermal) radiation from the surface.
Using the monthly average emissivity figures, using the S/B equation we can calculate the gridded surface temperatures from the upwelling LW radiation.
HOWEVER, if we want the average temperature of the planet, we can’t just average the gridcell temperatures.
Instead, we have to average the radiation, then convert that to temperature using the S/B equation and the average global emissivity.
The difference is not trivial. For the earth, averaging the temperatures directly gives us 15.4°C. But averaging the radiation, then converting to temperature, gives us 18.1°C.
Why? Radiation varies as T^4, and radiative flux is conserved … but temperature isn’t.
Here’s another way to see it. Average emission of upwelling surface LW is ~ 400 W/m2. S/B says that equates to 18°C.
Now let’s divide the earth in half, and have one side emitting 600 W/m2 and the other side emitting 200 W/m2. We still have the same average radiation in W/m2, so we haven’t added or subtracted any radiation.
But 600 W/m2 is ~50°C and 200 W/m2 is -28°C, so the (incorrect) temperature average is now only 10°C.
Exact same amount of energy, but different temperature average … which is why we have to average radiation first, and only convert it to temperature at the final step.
Hope this helps,
w.
Rearranging the S-B equation (P = eσAT^4), we can get a constant of proportionality between radiated power, P, and the term (A*T^4):
P/(A*T^4) = eσ = constant for your gedanken experiment.
Applied to the stated values in your above reply (converting °C to K) and keeping things equally dimensionless:
Case 1: 400/(1*(18+273)^4) = 5.58E-8
Case 2a: 600/(0.5*(50+273)^4) = 1.10E-7
Case 2b: 200/(0.5*(-28+273)^4) = 1.11E-7
I think this shows there must be an error (appears to be a factor of two) between your calculations for the uniform spherical, +18°C temperature case versus the values you give for the two-hemispheres in the +50°C/-28°C case.
Apologies, Willis . . . my bad, actually my careless BAD!
Since your radiation values in all cases were expressed as a power flux (W/m^2), and not as absolute power (W), I should have more properly rearranged the S-B equation as (P/A)/(T^4) = eσ = constant.
I mistakenly used 400, 600 and 200 watts instead of watts/m^2 in my dimensionless calculations. Your mathematics and comparisons of the two cases (including the derivation of the temperature difference between Case 1 and Case 2) are correct, and indeed all result in essentially the same constant of proportionality (about 5.5E^-8, without regard to proper dimensions), recognizing slight variations of such are due to rounding off temperatures to nearest integer values.
Again, mea culpa.
And, so, thank you for clarifying my misunderstanding of using total radiation flux to back out an “average” temperature value whenever a surface is not isothermal.
Having created a spreadsheet that revealed my bonehead mathematical error, as noted in my above reply to Wills, I decided to exercise it to discover the following:
For Willis’ limit case comparison of a spherical body with two hemispheres at different temperatures compared to a full spherical body at a single temperature (assumed to be +18°C, the same as Willis considered), with all other things being qual, it turns out that the case of for one hemisphere being at +30°C (12°C hotter than +18°C) and the other hemisphere being at +6°C (12°C cooler than +18°C)—a total swing of 24°C—the computed net radiated power flux difference (aka “error”) compared to an isothermal sphere at +18°C) is 1%.
To get up to a 3% difference in computed net radiated power flux between for the two-hemisphere case with temperatures “averaged” to +18°C versus the isothermal sphere at the same temperature, one would need one hemisphere be at +38.5°C and the other hemisphere to be at -2.5°C, a 41°C temperature swing centered about the reference +18°C of the isothermal sphere.
Bottom line: true, one cannot use an averaged surface temperature of a non-isothermal, airless body in combination with the S-B law to accurately calculate its surface-emitted radiation flux . . . but doing so for up to a 24°C equally-centered difference-about-the-mean will only introduce an emitted power error of 1% or less.
For reference, the Earth with its current atmosphere has a typical swing of magnitude up to 15°C between daytime and nighttime average temperatures at temperate latitudes, but this swing is near the surface . . . I don’t have a good number for the corresponding day-night swing at TOA.
1% of 18C is about +/- .2C. With that kind of uncertainty how do you identify temperature differences in the hundredths digit?
Sorry, you can’t use percentages with Celsius, or with Fahrenheit for that matter. You have to use Kelvin. Basic thermodynamics.
1% of 18°C is about 2.9°C.
w.
That is not something that I do.
Having said that, please note that I specified there was a 1% error in the computed power flux (W/m^2) introduced by assuming a +18°C isothermal spherical case versus the case of the +30°C/+6°C two-hemisphere scenario where the two different hemispherical surface temperatures average to the same +18°C. That 1% error in radiated power flux does not transfer to a 1% error in surface temperature.
Here’s a real-world example, guys.
If I take an area-weighted average of the 64,800 individual gridcell temperatures of the earth, I get 17.1°C.
But if I do it properly, by first averaging the upwelling surface LW radiation and then converting to temperature, I get 18.1°C.
w.
Is W/m^2 an intensive property like temperature? It appears to be a measurement similar to density (mass/volume) even if it is called intensity instead of density.
If you can’t average temperature because it is an intensive property then can radiation intensity be averaged?
I believe it is an extensive property. However, that doesn’t mean you can deal with this property with simple arithmetic calculations. The conditions of measurement must be repeatable in order to do so. It’s like adding the lengths of two compressible materials. Unless one also includes any tension or compression, one has no idea if the mathematical calculations using the lengths make any sense.
Tim, suppose I have a 10 m^2 plate absorbing 10 W/m2 of radiation. That’s a total flux of 100 watts.
Next, suppose I have a 20 m^2 plate absorbing 20 W/m2 of radiation. That’s 400 watts
What is the average flux in W/m2?
Well, between them, they are absorbing 500 watts over an area of 30 m^2. So the average flux is 500 watts / 30 m^2, which is 16.67 W/m2.
It’s called “area-averageing”.
w.
It’s called “area-averageing”.
Let’s examine some of the problems with what you have done.
In order to get the wattage values you have calculated, you must have assumed that each m² absorbs either 10 W/m² or 20 W/m².
Let’s calculate the temperatures at that intensity using the simple SB equation of I=σT⁴.
@ur momisugly 10 W/m²-> (10/5.67×10⁻⁸)¹/⁴ = 115K
@ur momisugly 20 W/m²-> (20/5.67×10⁻⁸)¹/⁴ = 137K
Temperature Average at equal areas = (137 – 115)/2 = 126K
Now, let’s use the average of 16.67 W/m²
@ur momisugly 16.67 W/m²-> (16.67/5.67×10⁻⁸)¹/⁴ = 131K
Hmmm! Some defugalty somewhere.
[|126 – 131|/126]*100 = 4%
Let’s look at standard deviation which is a measure of uncertainty.
s = √(242/1) = ±15.6K
SDOM = ±15.6K/√2 = ±11K
Let’s see what the percent difference between the W/m² work out to be.
Yours -> 16.67 W/m²
I = (5.67×10⁻⁸)(126)⁴ = 14.29 W/m²
[(16.67 – 14.29)/16.67]100 = 14.3%
The end result seems to be inaccurate in one way or another. To be consistent the math should be reversable and this doesn’t seem to be the case.
Exponential variables are difficult to deal with and aren’t amenable to simple arithmetic averaging. I learned this when dealing with characteristic curves of active devices in circuits. Vacuum tube and semiconductors do not have linear responses, they are exponential and you are only asking for trouble if you forget it.
Thanks, Jim. You’ve just explained why you cannot simply arithmetically average temperatures.
You have to area-average the watts involved and convert back to temperature.
Watts are conserved. Temperature isn’t. Which is why you need to average the watts and not the temperature.
w.
You seem to have missed the point, Willis. Watts may be conserved but the conversion to temperature is not arithmetically linear.
T = (P/σεA)^(1/4)
This is not an arithmetic linear relationship. The average watts won’t tell you the average temperature nor will the W/m^2 average.
In addition, the S-B equation assumes that the emission curve is a Planck black-body curve. The earth does not have a Planck-shaped emission curve and an emissivity constant won’t fix that. The shape of the emission curve also has an impact on the measurement uncertainty reached by using an “average” radiation intensity to calculate an “average” temperature.
Here are just some of the items that need to be included in a measurement uncertainty budget for calculating the Earth average temperature from the radiation intensity measured by satellite.
This is just a start. I could add more. Things like skewness of the measurement data, the difference between the mean of the measurements and the mode of the measurements, etc. It all *adds* up. You just can’t assume that it is all random, Gaussian, and cancels.
I’ll repeat what I’ve said about so much in climate science: the measurement uncertainty is so large that it totally subsumes any differences that might be generated by the stated values of the measurements. You don’t actually know what the differences are!
I’ll politely disagree. If the conversion doesn’t doesn’t work with average values, then the averaging is the problem. It is not a reversible process.
Why do I say that? What temperatures give the correct values of absorbed radiation? What function gives correct and reversible values.
Area averaging is not the problem. Using an arithmetic average on an exponential function is the problem.
Using the SB, these are the individual temperatures to be expected of each body.
10 W/m² absorbed by 1 m² -> 115 K
20 W/m² absorbed by 1 m² -> 137 K
(115 + 137) / 2 = 126 K
126 K -> 14.5 W/m²
Your average intensity comes out 16.67 W/m²
14.5 ≠ 16.67 Why is that.
The proper way to evaluate an exponential is to integrate over an interval of “a” to “b” and multiply by (1 / \a – b|).
If Ta and Tb are the correct temperatures, then one should be able to integrate the SB equation over the expected temperature interval and obtain the intensity needed to realize the average temperature.
Here is an integral of the linear interval between 115 K and 137 K.
As you can see, the average of the temperatures is 126 K.
Here is an integral of the exponential side of the SB equation with an interval using the same temperatures, 115K and 137 K.
Funny how the exponential integral gives the correct value (14.5 K rather than 16.67 K) that allows SB to achieve the 126 K average value.
Doing this makes the results reversable which is what is needed to make the “=” operator meaningful.
Excellent! . . . I dared not try to back out the “average” temperature of, nor the averaged surface LW radiation from, 64,800 individual gridcells covering Earth’s surface.
A 1°C difference between the two methods, less than 0.4% difference based on degrees K, is completely believable!
Thank you.
Again, the S-B formula is developed from Planck’s work. Planck’s work is based on some restrictive assumptions. Two are a homogenous body and no internal temperature gradients within the material. The Earth’s biosphere does not meet these requirements. The use of a parameterized constant to correct for a particular this, known as “emissivity”, introduces an uncertainty in the result similar to the uncertainty in climate models from parameterizing clouds. My guess is that due to the highly variable surfaces involved that the temperatures calculated are larger than any differences trying to be identified.
You do realize that there are “standoff” infrared thermometers, I assume. These calculate the “spot temperature” of a spot on an object by using the S/B equation. Cheap ones use an assumed emissivity of 0.95. However, more expensive models allow you to set the emissivity. Using those, the conversion is accurate to within a couple of degrees K.
So no, introducing the emissivity doesn’t INTRODUCE an uncertainty, it REDUCES the uncertainty.
Next, you say:
While this is true, it’s incomplete.
Regards,
w.
“These calculate the “spot temperature” of a spot on an object by using the S/B equation.”
The operative words here are “spot temperature”, which is different than “object” temperature. You are trying to impute the temperature of an OBJECT, namely the Earth, from a collection of SPOT temperatures themselves imputed from spot radiation measurements.
”This means you can invert the relation locally to infer a “radiative temperature” (often called equivalent blackbody temperature or brightness temperature) of a spot”
same problem. If the object is not in thermal equilibrium then you can’t use a spot temperature for the object temperature. You wind up “averaging” temperatures, intensive properties. Which you said yourself doesn’t work. It doesn’t matter if you first measure radiation if you are doing so as a “spot measurement”.
“So no, introducing the emissivity doesn’t INTRODUCEan uncertainty, it REDUCES the uncertainty.”
If you are using an “average” value with a spot measurement then you *are* introducing more uncertainty since the radiation/temperature is a power function that is location specific. You can’t mix “spot” with “global” without increasing uncertainty.
Planck used the thermal equililibrium assumption in order to treat the black BODY, not spots on the body.
Thanks, Tim. You say:
I would not dream of imputing the average surface radiation of the earth from a few, a dozen, or even a hundred spot radiation measurements.
However, the CERES data I’m using have spot measurements for 64,800 different spots covering the entire planet, both clear-sky and all-sky … at which point yes, we can indeed impute the global average surface radiation and thus the global average surface temperature.
w.
The issue is that you don’t know an accurate emissivity value for any of those spots let alone an accurate average value for them. That lack of accuracy does have a measurement uncertainty value.
Since T^4 = M/εσ, if both M and ε are variables then the measurement uncertainty for T^4 is u(M) + u(ε) assuming direct addition. Assuming an average value for M implies that the variance of M is zero or that the distribution of M across the measurement points is random and Gaussian, resulting in cancelation thus leaving the average as an accurate “best estimate”. Same for ε.
The distribution of M across the globe is probably not Gaussian. Same for ε.
The typical estimate for the uncertainty would be the standard deviation of measured values. With 68K measurements you should be able to get a pretty good estimate of the variance of the data and thus the standard deviation. That would give an indication of whether the differences being found are greater than the uncertainty or not. If not, then you don’t really know if the differences are real or not.
BTW, if you can’t average temperature because it is an intensive property exactly what is W/m^2? W/m^2 appears to be the same type of measurement as density (mass/volume) even though it is called intensity. Density is an intensive property. Can you actually average W/m^2 ?
Yes, you can average it, but only if you area-average it.
w.
Sounds a bit like the epicycle believers, who just came up,with ever more complicated explanations when faced with logical or factual difficulties.
An area weighted arithmetic average of a variable such as insolation whose absorption rate varies by latitude (i.e., by cosine) is questionable.
Each area can have numerous variables that make the absorbed flux vary by a substantial amount. Dust, clouds, fog, etc. are not an area function.
That would be the same thing as adding up all temperature data taken at a specific time from around the earth, e.g. 0000GMT, and dividing by the surface area of the earth, would it not? Or maybe adding up all Tmax values each day around the earth and dividing by the surface area of the Earth.
Would that give you an acceptable global average Tmax temperature?
I think the issue is still that radiation intensity from the surface is a function like f(x,t). The problem is that the satellites don’t provide values for all x at a specific t. So you don’t really know what the total radiation intensity for the earth is at any given time. Thus you are going to have a pretty large standard deviation in your data – meaning a pretty large uncertainty.
Willis, please read JCGM 100:2008, Section F.1.1.2.
What you are determining by averaging those 64,800 samples is the PROPERTY OF A MATERIAL by sampling. Those measurements are not made on a single sample. Therefore, you must add the variance of possible differences among samples, to the repeated observations of a single sample.
Beware the lumber salesman who offers a train load of lumber that has an average of 8′ ±0.03″. You need to ask what the variance of that lumber truly is instead of not recognizing that you are being quoted the standard uncertainty of the mean.
The variance/standard deviation and the standard uncertainty of the mean are two different things with two different connotations.
If conditions change between measurements (drift, different setups, different operators, different environments), the observed spread will include systematic changes and correlated errors, not just random variation. That violates the independence and identical-distribution assumptions and makes s/√n misleading.
In metrology you separate repeatability (same conditions) from reproducibility (different conditions) and treat systematic components separately, then combine them (root-sum-square) to form the full measurement uncertainty.
And of course, you guess the emissivity. Even this doesn’t work when you are measuring temperatures of different objects with varying properties at different times.
Sorry Willis, pretending to measure air temperatures, or surface temperatures under kilometers of water or ice is just silly. Good for ignorant and gullible pseudoscientists, and their ignorant and gullible followers.
In my worthless opinion, of course. Next thing you’ll be claiming that adding CO2 to air increases these thermometric readings!
You can’t calculate the effect of incoming solar radiation on the surface temperature without taking the heat capacity of that surface and the volume that is warmed into consideration.
Oceans warm up very slowly, while deserts heat up fast for the exact same amount of radiation.
“You can’t calculate the effect of incoming solar radiation on the surface temperature without taking the heat capacity of that surface and the volume that is warmed into consideration.”
I do think the whole argument and conclusions of the above article are based on an assumed equilibrium condition of incoming radiation power = outgoing radiation power. In that case, the mass, effective specific heat and effective thermal conductivity of the hypothetical planet don’t matter.
Correct. Thanks, TYS.
w.
On the moon equatorial temps go up to ~400K, which is about the radiative balance temperature.
Where do we see such temperatures on our oceans?
I do not see any radiative balance temperatures over our oceans.
That’s because Earth, unlike the Moon, has a relatively dense, convective atmosphere which serves to both restrict surface radiation to space and to “smooth” atmospheric temperature variations over different latitudes and longitudes.
Earth’s hydrological cycle, which is largely independent of GHGs, plays a major part in this difference, particularly the heat transferred to the atmosphere as a result of ocean water evaporation rate as a function of surface temperature. 400 K is ABOVE the boiling point of ocean water, which is 100.5°C, equivalent to 373.7 K.
The fact that the Earth rotates once-per day relative to the Sun, compared to the Moon rotating once every 29.5 days relative to the Sun is also a major factor affecting the degree of relative radiative “balance” at the surfaces of these two bodies.
Clarification on my second paragraph: Earth’s hydrological cycle obviously involves water vapor, which of course, is actually the strongest of all “greenhouse gases” in the atmosphere. To be completely correct I should have stated “. . . largely independent of GHGs other than water vapor, plays . ..”
TYS:
“The fact that the Earth rotates once-per day relative to the Sun, compared to the Moon rotating once every 29.5 days relative to the Sun is also a major factor affecting the degree of relative radiative “balance” at the surfaces of these two bodies.”
If the Moon rotated every 24 hours, the day side surface temperatures would still be close to equilibrium with the solar irradiance, there is very little thermal lag in the lunar surface. The night side average temperature would be about the same. The main difference would be that the dawn and dusk terminator temperatures would be closer with a faster rotation.
Ben, those are instantaneous temps, and we’re discussing average temps.
w.
To be more specific:
solar radiation at the surface in the tropics goes from 0 (zero) at night to ~1000 W/m^2 at noon. Corresponding temps would 0 – ~365K.
In reality ocean surface temps vary 1-2 K over a day.
Where is the radiative balance?
Simple . . . the radiative balance of incoming solar power versus Earth’s outgoing, radiated power is established by temperature/power emitted at the top of the atmosphere (TOA), not at Earth’s surface.
The energy balance is at TOA, but all SB calculations are for SURFACE temperatures, so radiative balance means that the incoming solar has heated the surface to the temperature where that surface emits the same amount of radiation as it receives. Simple.
“. . .so radiative balance means that the incoming solar has heated the surface to the temperature where that surface emits the same amount of radiation as it receives. Simple.”
Maybe simple for you, but your prior sentence is questionable. Earth’s average albedo, most of which arises from cloud coverage, means that about 30% of total incoming solar radiation is either reflected off clouds or is reflected off Earth’s surface, thereby never being absorbed to create thermal energy. This reflected energy can be/is measured at TOA.
Then too, cloud coverage as well as absorption of surface-emitted LWIR by greenhouse gases and their subsequent themalization of that energy throughout the full atmosphere (about 99% being N2 and O2) via molecular collisions means that Earth’s atmosphere radiates a large amount of energy to space . . . it is not just Earth’s surface that radiates.
It is true that in the short-term—since Earth’s surface temperature (including oceans) is relatively constant over thousands of years— Earth’s surface radiates as much energy as it receives, but this is only part of the complex, multiple processes that are occurring.
A true “radiative balance” (à la Kiehl and Trenberth, 1997, et seq.) for Earth considers both Earth’s surface and its atmosphere, clouds and overall albedo.
Actually, as Baron Fourier and others have pointed out, it radiates a little more than it receives from the Sun. That is why the surface has cooled over the past four and a half billion years.
I originally had to think gently for about 15 seconds to come to that conclusion. I reserve the display of my ignorance and gullibility for other areas.
Luckily for me, you probably don’t know enough to know the areas in which I am likely to be ignorant and gullible. <g>
My exact sentence to which you refer was:
“It is true that in the short-term—since Earth’s surface temperature (including oceans) is relatively constant over thousands of years— Earth’s surface radiates as much energy as it receives . . .”
That you chose to leave out the qualifying words “in the short-term—since Earth’s surface temperature (including oceans) is relatively constant over thousands of years—” speaks volumes about the areas in which you are “ignorant and gullible” (you own admission).
I don’t need to see/know any more.
I’m not sure about that. The satellite may have a designed “spot” beam width but I doubt there is any way to adequately address the issue of radiation from a surrounding area from leaking into the device. In other words, it is not measuring only the vertical component. This would be a systematic offset that would affect the uncertainty.
The fact that orbiting satellites measure TOA radiation in sequential, near-nadir “footprints” having very small FOV angle that cumulatively more-or-less cover Earth’s entire surface in about one day pretty much rules out “leakage” from surrounding areas degrading the overall measurements.
From Google’s AI:
“A single CERES satellite doesn’t fully cover the Earth’s surface in one pass, but a full mapping takes a little over a day because it takes approximately 1.5 hours to orbit Earth and views most places on Earth at least twice a day. The instruments are able to collect a large number of independent observations in an hour, covering about 100,000 footprints.”
Most CERES-instrumented LEO satellites are in polar orbits.
How does NOAA account for the path loss encountered by the “radiation” it measures? What it actually sees is *NOT* what is sent by he ground. There are lots of things in the atmosphere that absorb radiation of different frequencies. If you don’t know the path loss then you do *not* know what was sent from the ground. That path loss varies tremendously from location to location as well as over time. The variation in the path loss contributes even more to the measurement uncertainty and it cannot be “averaged” away. Nor does it get divided by √n. The measurement uncertainty is the variance (square of the standard deviation) of the measurements, not how precisely you can locate the mean of the data. How precisely you locate the mean doesn’t tell you the variance of the data which is the measurement uncertainty of the mean.
BTW, I believe that the average of a function M = εσT^4 is related to T^5. It is not the arithmetic mean of (M_max + M_min)/2. It would be something like (T_max^5 – Tmin^5)/5(Tmax-Tmin). Is this what NOAA uses to find the mean global radiation value?
Using the AI data, show us how wide a view in degrees is needed to “see” the entire earth in about 12 hours to get twice a day coverage.
I show 360° / 12 ≈ 30°.
Do the passes follow the sun at the same lag? If not, it is not measuring comparable values from repeatable conditions.
Jim, the “total radiation measuring device” simply does not exist. All this TOA stuff is complete nonsense.
The Earth has cooled, continues to cool, and anyone who still believes that adding CO2 to air makes thermometers hotter is suffering from delusional psychosis, for which no cure is currently possible.
Radiative balance does not exist on Earth. At TOA we will be close to an ENERGY balance.
Since power is energy per unit time, for a specified interval of duration, power balance is equivalent to energy balance.
To normalize Earth’s power balance for Earth being (a) spherical compared to solar energy arriving in essentially a unidirectional planar fashion, (b) Earth rotating once per every ~24 hours with respect to the Sun, and (c) solar insolation varying during a year due to Earth’s elliptical orbit around the Sun, climate scientists have found it most convenient to use geographically- and annually-averaged power flux contributions and net balance (units of W/m^2, or equivalent). See Kiehl and Trenberth, 1977, et seq.
BTW, the above use of “balance” is solely for general convenience in related discussions. In reality Earth is never in a true balance of incoming power flux versus outgoing power flux beyond, perhaps, a day. This is primarily due to its continuously varying albedo and the time constants associated with heat energy going into or coming out of the oceans.
What is it: radiative balance or energy balance?
Can’t have it both ways.
None of ithe analysis relates to physical processes occurring on Earth.
Earth without an atmosphere would be ice all over the regions now oceans. It would likely be highly reflective. We know water based objects remain ice in close proximity to the Sun.
If you go to no water planet then you need to do the complete analysis over the entire area allowing for day and night because you are not dealing with a linear relationship between temperature and long wave emission. You will need to make some estimates of the thermal inertia of the ground to allow for the daily cycling. I suggest a minimum of 1 hour time interval over a 1 degree grid for 21 years using JPL Sun-Earth declination and distance data. Run a few times using the end conditions as the next run start conditions until the annual cycle is stable.
There is no such thing as downwelling long wave radiation. It is an inferred number for the atmospheric absorption based on applying the S-B relationship.
This sort of analysis is fodder for the incompetents who believe CO2 can have direct influence over Earth’s radiation balance. We know through the CERES period this century that all additional energy retained is a result of lower reflectivity and and slightly higher emission temperature:
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This is generally due to less cloud but not exclusively. Some is due to less land and sea ice. And the situation is reversed just north of the equator where there has been an increase in cloud on average.
So the only way CO2 can have an influence on the energy balance is if it is reducing cloud forming. And what ever that process is, it is doing the reverse just north of the equator.
Surely not when the surface was molten? That’s just my assumption, but evidence seems to support it.
There is conjecture that an impact formed the moon and that resulted in the loss of the atmosphere that existed then causing the Earth to be a more watery planet. There should be a higher concentration of CO2 based on the space rocks that are though to have bound together to form Earth.
At time when Earth has lost atmospheric mass, it has become an icefall. At times when the atmospheric mass was greater, it was warmer than now.
I determine that 15C is the minimum temperature required to support convective instability with the present mass. Once convective instability shuts down, the Earth becomes an ice ball. Convective instability is by far the most important process in Earth’s atmosphere.
Hi Willis, I calculated the localised climate sensitivity, depending on temperature https://klimaathype.wordpress.com/2018/03/24/localised-instant-climate-sensitivity-for-2xco2/
local climate sensitivity for 3.7 W/m2, based on Stefan-Boltzmann
If I recall it correctly, Happer and vanWijngaarden calculated about 0.6C as the maximum increase from doubling CO2 from its current value less any countervailing phenomena that Nature likes to throw at changes (Nature abhors a vacuum and such.) But our climate “experts” insist that the change will be 3, 4, or is it 5C but somehow never tell us why H/vW are wrong. Now they have one more guy to ignore. Well done Willis.
Denis, here are the authors, dates, and values of the best estimates shown in Figure 4 that are under 1 W/m2.
Author Year bestest Class
1 Specht et al. 2016 0.37 Theory & Reviews
2 Idso 1998 0.42 Theory & Reviews
3 Lindzen and Choi 2009 0.47 Observations
4 Harde 2017 0.65 Theory & Reviews
5 Lindzen and Choi 2011 0.72 Observations
6 Bates 2016 0.92 Observations
References are in the paper here.
w.
Many thanks.
It’s not just photons that warm things up. CME’s substantially warm the thermosphere and some of it does find its way down to where we live. According to GROK (at least half as smart as Willis), a moderate one can warm it 100K or more (thermosphere). The SSW currently forming over the Arctic is a relic of the solar storm that just occurred. SSW’s cause changes in the polar vortex which changes the jet stream which influences storm tracks.
This may not seem like a big deal but though geothermal energy is relatively insignificant, Krakatoa influenced temps for a couple of years. Same with solar storms.
Thanks, RB. I fear you’ve neglected the density of the air at the surface (1.2 kg/m^3) and in the mid-thermosphere (1E-7 kg/m^3),
This means that the amount of energy required to raise a cubic meter of thermosphere air by 100°C would raise a cubic meter of surface air by 0.00001°C …
Not a significant player in terms of temperature, although it does allow for long-distance radio propagation.
w.
Willis,
You convinced me several years ago that the real ECS from doubled CO2 is effectively zero:
https://wattsupwiththat.com/2020/09/15/cooling-the-hothouse/
While the above estimates (0.44 / 0.47) are in that neighborhood, and are certainly much better than the garbage emanating from the alarmists’ GCMs, I’m not comfortable with any calculation based on radiant transfer theory or any other phenomenological explanation that doesn’t take into account that thermal energy absorbed by IR-active GHGs is overwhelmingly converted to sensible heat by collisions with non-IR active gas species within meters of the Earth’s surface.
Some daytime solar energy is convected aloft into the atmosphere rather than radiating to space. This energy is then returned to the surface during the night to limit cooling. Functionally this surface to atmosphere exchange delays outgoing radiation. Why do radiative delays warm while convective delays do not. Joules are joules regardless of source.
A general comment for all, to clarify my position.
I do think that radiatively active gases affect the temperature. Just nowhere near as much as the conventional wisdom would have you alarmed to believe.
In addition to theoretical arguments and other evidence, my own climate model is able to do a very good job of emulating the real-world temperature using only albedo and the Ramanathan “greenhouse factor”, the percentage of upwelling surface radiation that is absorbed in the atmosphere.
Here is a more recent graphic from my ever-evolving Constructal climate model, showing how those two measurable environmental variables, albedo and greenhouse factor, are all you need to emulate the absolute temperature.
This also makes physical sense. Albedo controls how much energy is entering the system at any time. The greenhouse factor controls how much energy is leaving the system at any time. Any imbalance between those two will be reflected in a temperature change.
The excellent goodness of fit of the Constructal climate model is further evidence that the greenhouse factor, the percentage of upwelling radiation absorbed by radiatively active gases in the atmosphere, does in fact affect the surface temperature … just not as much as people think.
Regards to everyone,
w.
‘I do think that radiatively active gases affect the temperature. Just nowhere near as much as the conventional wisdom would have you alarmed to believe.’
346 W/m^2 of ‘downwelling’ radiation seems inconsistent with ‘nowhere near as much’, particularly in comparison to 187 W/m^2 of downwelling solar radiation.
Thanks, Frank. For clarity, I’ve changed it in the head post to say:
Best to you,
w.
I think that we need your definition of “a very good job.” A correlation coefficient of 0.71 means that only about 50% of the variance is explained and only about 31% is explained by a correlation coefficient of 0.56. That puts us down in the territory of coin flipping.
Basically the earth would be as warm as the other stony body orbiting the sun in close proximity… The Moon. Though the night side wouldn’t get as cold because of the 24 hour day period where the moon has a 14.5 x 24 HR 348 hour heating cycle followed by a 348 hour cooling cycle.
The Moons surface does not “accumulate heat”. It reaches its maximum temperature when receiving maximum sunlight, but its slower “rotation” compared to the Earth means it takes a little longer to reach maximum.
Here’s what NASA says –
No atmosphere, you see. Both hotter and colder than anything on Earth. Willis probably doesn’t believe NASA, either.
Willis was positing “A World without Air” much like the Moon. The Earth would basically have the same atmosphere as the Moon…No Atmosphere and such should be subjected to the same initial heating at sunrise as the moon (and the same initial cooling at sunset)
In the first 12 hours after sunset the lunar surface temperature drops to about -133°C (-208°F) and would also be the temperature on an atmospherless stony body with a 24 hour rotational period. Conversely in the first 12 hours after sunrise the lunar surface reaches a temperature of 127°C (260°F) and should also be the maximum temp (at the equator) for an atmosphereless stony body with a 24 hour rotational period.
Now because of the axial tilt there would still be apparent seasonal differences as well as increased heating at higher latitudes due to increased solar exposure during local summer as well as increased night time cooling during local winter.
So remove the atmosphere and the oceans boil away, the poles sublimate and the Earth assumes a similar heat/cold system as the Moon with daytime highs of 260°F just prior to Sunset and nighttime lows of -208°F just before Sunrise with warmer high temperatures at the poles in local summer and colder lows in local winter.
Your explanation correctly addresses the need to know the variance of temperatures and not just an average of temperatures.
A separate comment about very interesting and important figure 3 and the resulting net positive feedback 1.26 based purely on observational data. There is a round about way to reach the same result ‘theoretically’.
In 2011, Lindzen estimated the no feedbacks ECS as 1.2. Some years later, the data from Monckton’s ‘Irreducible Minimum’ paper gave 1.16. Close enough. We simplify to 1.2
AR4 gave the ‘best’ ECS estimate as 3. (AR5 explicitly did not, as EBM observational estimates were about half. AR6 dropped the discrepancy entirely.)
Lindzen’s 2011 paper explicitly translated IPCC ECS into Bode Curve (from electronic engineering) feedback—ECS 3 is Bode 0.65. IPCC AR4 explicitly said that water vapor feedback alone doubled the no feedbacks case; so water vapor alone ECS (1.2*2) 2.4, or Bode 0.5 per IPCC. Both AR4 and AR5 said net feedbacks excluding water vapor and clouds were ~0. So assumedIPCC cloud feedback must be (0.65-0.5) 0.15.
In 2010, Dessler published a paper showing (despite his erroneous claims) that statistically cloud feedback alone was ~0. McIntyre reworked the paper in 2011, fixing some Dessler errors, but reached the same conclusion—about zero.
The INM CM5 CMIP6 climate model produces no spurious tropical troposphere hotspot. They published a paper on why—simply put, they parametrized tropical ocean rainfall using ARGO data. Twice the rainfall, half the water vapor feedback, no hotspot. This directly implies that water vapor feedback is half of the AR4 explicit statement, so Bode 0.25 rather than 0.5.
Now, (0.65-0.15-0.25) is Bode 0.25, which is a ‘theoretical’ positive feedback of 1.25x, corresponding almost exactly with WE Figure 3 purely observational 1.26!
Good comments, Rud! And Willis!
CO2 is a benign gas, essential for life on Earth.
There is no evidence CO2 has any discernible effect on Earth’s weather or climate.
And now we see why.
Any CO2 effects on Earth’s weather or climate are “in the noise”.
Thanks, Rud. From my post above:
Here are the authors, dates, and values of the best estimates shown in Figure 4 that are under 1 W/m2.
Author Year ECS Class
1 Specht et al. 2016 0.37 Theory & Reviews
2 Idso 1998 0.42 Theory & Reviews
3 Lindzen and Choi 2009 0.47 Observations
4 Harde 2017 0.65 Theory & Reviews
5 Lindzen and Choi 2011 0.72 Observations
6 Bates 2016 0.92 Observations
References are in the paper here.
w.
Willis, your estimate seems to agree with Idso’s 8 natural experiments.
Nice work Willis. While I don’t always understand the math and science one thing that always concerned me was people estimating the sensitivity of a doubling of CO2 without mentioning any concentrations. Are we saying it’s the same whether we are talking 50 ppm to 100 ppm or 30 to 60 or 150 to 300?
Don’t expect a rational answer – they just make stuff up as they go along. It’s all nonsense – adding CO2 to air doesn’t make thermometers hotter. CO2 at 20 C is exactly the same temperature as O2, N2, NH4 or any other gas at 20 C. The ignorant and gullible simply refuse to accept facts.
I suspect that the climate sensitivity for a doubling of CO2 varies with the temperature, but I can’t give you a proof of that.
Complete nonsense. Assess all you like, believe as many Constructal climate model fantasies as you like – adding CO2 to air doesn’t make thermometers hotter. You wrote, possibly in a moment of madness –
Well, I have to point out that when the Earth’s surface was molten, it was above freezing – atmosphere or no atmosphere. When it cooled to the point that the first liquid water appeared, it was still above freezing.
Your faux “thought experiment” shows that you are no Albert Einstein, or Nikola Tesla. Now you may choose to believe that Earth is not more than 99% glowing hot matter if you wish. Unfortunately, this would put into the class of the ignorant and gullible self promoters, trying to appear intelligent.
Sorry Willis, but the airless Moon shows what happens to a molten body in space which cools far faster than the Earth due to its greater surface to volume ratio (and a couple of other physical considerations).
Talking about an Earth without an atmosphere is about as pointless as talking about water which is not wet. As Feynman pointed out, certain calculations involving water flow and pressure are wonderfully easy to – until you use water which is wet.
By the way, trying to calculate the surface temperature of a mostly molten sphere, in direct sunlight, using the Stefan Boltzmann equation is pretty stupid. As Baron Fourier pointed out quite some time ago, the Earth loses to space all the energy it receives from the Sun, plus a little remnant energy. Hence the Earth has cooled since its surface was molten.
Before you launch into a torrent of mindless invective, and threaten to “ban” me, I ask that when you comment, you QUOTE THE EXACT WORDS you are discussing. This is essential to prevent misunderstandings.
Maybe that sounds familiar? I have copied your words.
“Here’s a thought experiment. Consider the Earth with no atmosphere and with the same surface albedo of 12.5% that it has now. How warm would it be?”
About the same as the Moon if it also had no oceans.To calculate that, you have to apply S/B to the illuminated hemisphere only, which is two times and not four times the disk area, and average that with the dark side mean temperature.
394K * 0.5^0.25 = 331.3K
minus 12% albedo
331.3K * 0.88^025 = 320.89K
and averaged with a dark side mean temperature of 80K
(320.98 + 88) / 2 = 200.4K
At any given time, Earth’s sunlit side is cooler than the sunlit side of the Moon, but its global average temperature is far higher than the Lunar global average because of all the thermal reservoirs keeping Earth’s dark side so warm.
A value of the climate sensitivity that is that low would appear to be directly contradicted by measurements. The global temperature has risen by over 1 degree in less than a century and the trend from UAH satellite measurements is 0.15 degree per decade. So if the climate sensitivity is as low as you suggest then the current rate of temperature increase would have to be almost entirely due to something other than rising CO2 levels.
Furthermore if the current temperature increase is unexplained by CO2 levels then it is impossible to say what will happen in the future nor to say how high temperatures might go.
Lol.
The arithmetic global average of the local average of the seasonal average of the daily average of the range of various (real or imaginary, recorded or “adjusted”) thermometers scattered around the world has gone up who knows how much in less than a century for who knows what reason(s).
According to the written temperature record, the temperatures haven’t “gone up” since the 1880’s. It is no warmer today than it was in the recent past, bogus Hockey Stick chart, or no bogus Hockey Stick chart.
Global temperature is an unphysical nonsense.
Does that apply to Willis entire article as well? If the global temperature doesn’t exist then Willis’ entire article would be nonsense since clearly the ECS wouldn’t exist as well.
Maybe a global temperature does exist, but it is NOT represented by the bogus, bastardized Hockey Stick global temperature chart you worship.
You are a meteorologist, are you not? So surely you have seen regional temperature charts that look nothing like the bogus Hockey Stick chart temperature profile. Doesn’t that raise questions in your mind as to the legitimacy of the bogus Hockey Stick chart? It should. One or the other, the regional chart, or the Hockey Stick chart, is wrong, and does not represent reality.
Since there are numerous regional temperature charts that look nothing like the Hockey Stick chart profile, the logical conclusion is that the bogus Hockey Stick chart is indeed bogus.
But you don’t see it that way? I wonder why. It’s plain as day.
The bogus Hockey Stick chart is a misrepresentation of reality. A deliberate misrepresentation of reality. And it has fooled millions of people because most of them have no clue about weather and temperature history.
It should not fool people who are familiar with regional temperature charts, like meteorologists.
Here are 600 regional charts for you from around the world. See if you see any of them that have a Hockey Stick chart “hotter and hotter and hotter” temperature profile. What you will find if you look is that NONE of them have a “hotter and hotter and hotter” temperature profile. They all show it was just as warm in the recent past as it is today.
REAL temperature records:
https://notrickszone.com/600-non-warming-graphs-1/
Indeed, I have always been puzzled that Global Warming only ever appears in global temperature calculations, and not in real, measured temperature records from discrete locations.
We know why, too: Because the temperature data has been mannipulated for political/personal purposes.
And this mannipulation has devastated the Western world by causing their politicians to do very stupid things in an impossible effort to reduce CO2.
All because of a bogus Global Temperature record. If the politicians were looking at a real representation of the temperatures they would see that it is no warmer today than it was in the past, and they would have to say to themselves: What’s the problem? It seems CO2 has no discernible effect on Earth’s temperatures because it is no warmer today than in the past even though there is more CO2 in the atmosphere than there was in the past.
But they can’t say that looking at a bogus Hockey Stick Global chart. That’s the problem.
As far as I’m concerned, the creators of the bogus, bastardized Hockey Stick chart are criminals of the worst kind. The destruction they have created with their lies is unbelievable, and ongoing.
As Willis wrote –
He’ll probably say “But this is different. I really meant . . . “
Or your “biased and unquestioned” intuition that ECS > TCS is wrong.
“The global temperature has risen by over 1 degree in less than a century”
Not in the United States.
In the United States it is no warmer today than it was in the 1930’s.
That also applies to all the land surfaces on Earth according to the written, historic temperature records.
Your “global temperature” is a bogus figure made up out of whole cloth by Temperature Data Mannipulaors. The written, historic temperature records refute the global temperature estimates made before 1979.
And yet, you use this bogus concept to form your opinions of the Earth’s climate. I guess you don’t see the error of your ways.
I can give you a link to 600 regional charts from around the world that show it was just as warm in the recent past as it is today. Your 1.0C figure is bogus. It’s a computer game that you have bought into. Thus, your confusion about reality.
Tom, it’s interesting that the 70% of the surface covered by water doesn’t seem to count when trying to insist that a few thermometers scattered over the remaining 30% can be used to magically infer the average temperature of a few squillion tonnes of glowing hot planet.
The ignorant and gullible have convinced themselves that thermometers get hotter due to the magic of CO2, rather than heat. Unfortunately, they have to resort to abandoning physical concepts like temperature being actually measured in “degrees”, and instead using things like W/m2 as a proxy, allowing them to gaily add and subtract temperatures, whilst sounding as though they know what they’re doing.
A pack of delusional cultists, sharing a common fantasy – luckily becoming more irrelevant with each passing day. However, the damage has been done. Tens of billions of dollars down the drain, millions of lives blighted, and a generation or two believing that they can “Stop Climate Change”, or similar claptrap.
I’m not a fan of “pollution”, but my desire for adequate potable water, sewerage, a weatherproof dwelling, and so on, means that both I and others have to forego something else, somewhere, some time. That’s life, and then you die anyway, I suppose.<g>
Here is a link to Australian data showing that the claimed 1.51 +/- 0.32 deg C of national warming 1910 to 2024 is WRONG.
Closer to 0.95 deg C is more likely.
The difference comes from ADJUSTED official data.
Geoff S
https://www.geoffstuff.com/halfwarm.docx
According to this chart below it is no warmer today in Austrailia than it was in the 1880’s or the Early Twentieth Century.
In reality, there has been no increase in the temperatures in Australia since the 1880’s.
From post:”…temperature increase would have to be almost entirely due to something other than rising CO2 levels.”
Yes, yes, yes. CO2 can’t cause a temperature increase. You finally got it.
” it is impossible to say what will happen in the future nor to say how high temperatures might go.”
Correct if you add “or how low.”
It would not take much variation in the solar surface temperature and variations in orbit, etc., to account for the temperature variations observed.
The unexamined assumption is that all the warming is a result of CO2. As you point out, if that isn’t the case, then the climate sensitivity from CO2 doesn’t have to be as high as is commonly assumed. If the measured warming isn’t from a delay in IR emissions from CO2 increases, but is as other have suggested, a decrease in cloudiness resulting in an increased effective insolation, then there is an alternative explanation.
First, I’m always impressed w/ Mr. Eschenbach’s work. The same with this one. We all know the IPCC equation, which uses the constant of 5.35, is just something made-up, as far as I’m concerned.
Second, I’ve long thought any ‘doubling’ effect is close to zero (for carbon dioxide); anything under unity is no cause for concern. The discussion of Mr. E here reinforces that belief.
Third, I may be mis-reading something, but in my Astrophysics classes, the discussion of planetary atmospheres and such, we had to analyze the case for the Earth, two ways: 1) without any atmosphere; and 2) the case of atmosphere, but with no (so-called) ‘greenhouse gasses’. The SECOND case would require that the Earth, as we know it, would have NO water, in any form (solid, liquid, or gas).
Most of us should know that the Earth has two primary rock types: continental crust (generally about 2.7 g cm^-3), standing above the mafic crust (generally 3.3 g cm^-3) or oceanic crust.
Roughly speaking, Venus displays something similar to this, with the small ‘continents’ standing above the lower peneplains.
The upshot of all of this discussion is that without water, the Earth’s albedo is different from what we use on the REAL Earth, and, the atmospheric depth (air replacing ocean) would be some 3,500 metres (give-or-take) deeper than at present.
Throwing all of that into the mix, we were able to calculate that Earth’s “average” temperature in an N2/O2 atmosphere would be about 277 – 280 Kelvins. Variance came from having to guess at continental crust albedo, since there would also be no land plants, and erosion/deposition of weathered detritus would be far different from what we see today.
And, in case any one is wondering, there was also a noted dependence upon latitude in the ‘model’ Earth with no water, but an atmosphere. There was approximately 5 – 6 Kelvins difference between equatorial “Earth” and polar “Earth”.
Yeah, we did this with slide rules, back in the B. C. days (before computers … ).
Mark H
That’s not a lot different than what I get with using trig functions to define the radiation at points on the earth.