By Jim Steele
Focusing only on CO2’s warming effect distorts our understanding of how the atmosphere warms and cools. More importantly it ignores how CO2 is critical for cooling our atmosphere. Oxygen and nitrogen comprise 99% of our atmosphere. Those molecules do not absorb or emit radiation. Our air is heated by collisions with solar heated surfaces and warmer atmospheric molecules. So, we must understand how oxygen and nitrogen shed their absorbed energy, so it can exit back to space.
Only the sun’s radiation adds heat to our world. Of the arriving solar heat 30% is reflected without warming the earth. 20% warms the atmosphere directly and only 50% warms the earth’s surface.
About 25% of that surface heat (12 units) gets radiated back to space immediately at wavelengths that are not affected by greenhouse gases.
About 14% (7 units) of the surface heat warms the air above via surface collisions with that heated air carried away from the surface by rising convection.
Slightly less than 50% (23 units) of the surface heat causes evaporation with moist convection also carrying heat away from the surface.
The surface only radiates 16% (8 units) of absorbed heat away, which is however also intercepted by greenhouse gases, primarily water and CO2. But greenhouse gases redirect some of that heat back to the surface and slow the earth’s cooling rate.
Eventually at the top of our atmosphere (the troposphere) greenhouse gases must radiate about 58% of all incoming solar heat back to space.
This diagram illustrates the physics showing how the earth cools by radiating infrared heat back to space according to Planck’s Law.
The 320k dashed line represents how much heat via each infrared wavelength should theoretically radiate back to space if there is no interference from greenhouse gases. Here, its from the Sahara desert’s surface when the surface is heated to 47 Celsius. This theoretical line is based on solid reproducible & trustworthy science.
The other dashed lines represent how much heat each wavelength should radiate from cooler temperatures. These cooler temperatures happen at higher altitudes as seen in the table on the right.
The solid black line represents what wavelengths of heat that satellites actually detected leaving earth. The thin red lines are model predictions. Only infrared heat with wavelengths between 10 and 13 microns as well as between 8 and 9.5 match theory. That’s because greenhouse gases do not interfere with those wavelengths, allowing 25% of the surface heat to escape immediately.
Water vapor absorbs infrared wavelengths between 14 & 25 microns as well and between 6 & 8 microns. Satellites detect those wavelengths but with intensities suggesting the heat has radiated from cooler temperatures. Satellites can’t detect the latent heat absorbed during water’s evaporation at the surface, but satellites do see the heat once released when water vapor cools and condenses to form clouds and rain at 1 to 4 kilometers of altitude. Above water vapor’s condensation altitudes, water vapor provides an insignificant greenhouse effect as the air is very dry.
CO2 primarily absorbs infrared centered around 15 micron wavelengths but satellites do not detect any heat from 15 micron wavelengths being emitted from the surface or lower atmosphere. So, a second critical question is why don’t satellites detect 15 micron wavelengths until the top of the troposphere and stratosphere? Radiant heat proven to be absorbed and emitted by CO2.
Greenhouse theory argues infrared absorbed by CO2 is trapped near the surface according to the following model.
Surface heat radiates upward but some is intercepted by greenhouse gases.
Those gases absorb the infrared heat but then quickly emit it. However, about half is redirected back towards the surface, slowing the surface’s cooling rate.
The infrared that continues upwards is again intercepted sending half back towards the surface
The same dynamic continues so by 2 to 4 kilometers altitude, virtually no surface radiation absorbed and emitted by CO2 escapes. So, satellites simply don’t detect it.
The same diminishing dynamic reduces the amount of infrared emitted from higher altitudes that can reach the surface. So, most scientists believe warming via a greenhouse effect happens mostly in the lowest atmosphere.
But this dynamic only partially explains atmospheric warming.
That greenhouse theory does not account for how collisions, as illustrated by Newton’s Cradle, that transfer energy from greenhouse gases to non-greenhouse gases and vice versa.
Water and CO2 do not trap absorbed infrared heat, but quickly emit it in just milliseconds after it is absorbed. This is called the relaxation time.
Now compare well studied collision frequencies to relaxation times.
Near the surface, nitrogen molecules can collide with CO2 6 billion times in just one second. So, before a CO2 molecule can relax and emit any radiation, nitrogen will collide 600,000 times with CO2 and steal it’s absorbed energy and transfer it to other atmospheric molecules via more collisions.
Collisions with nitrogen and oxygen constantly steal heat from CO2 and because nitrogen and oxygen don’t radiate that heat, satellites don’t detect any wavelengths centered around 15 microns until CO2 emits that heat in the stratosphere.
Heat trapped by oxygen and nitrogen can only lose that collisionally absorbed heat by transferring energy back to a greenhouse gas via more collisions. Only then can greenhouse gases radiate that energy back to space. However, CO2 will only radiate heat back to space where heat stealing collisions are greatly reduced, and that only happens in the less dense upper troposphere and stratosphere, exactly where satellites detect 15 micron wavelength emissions back to space.
Heated air constantly convects to the upper atmosphere, but without CO2, there would be no shedding of the air’s absorbed energy. There would only be runaway warming.
Greenhouse warming in the lower atmosphere is balanced by CO2 cooling in the stratosphere. This explains why CO2 concentrations do not correlate with earth’s temperatures over millions of years!
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I said that here more than a decade ago.
And it is still wrong..
You get my negative vote as you made an assertion without providing any objective scientific evidence. Back up you bold statements.
and E.Schaffer is wrong – see how easy that is?
So perhaps you can explain why we still cannot extract a human CO2 heating signal, and why the ”Global average temperature” is still within natural variation range.
So why do refrigerators use R-744?
They don’t. R-744 is CO2, and it’s use as a refrigerant requires specialized high pressure systems, so is only used in large scale, commercial freezer set ups. Consumer fridges still use R134a for the most part, as do most automotive systems. Eco nuts have banned most good refrigerants and replaced them with poorer efficiency ones, and in some cases like YF1234 are now flammable and extremely toxic in leaks and fires, whereas all the old, better performing refrigerants were inert and non flammable.
You did.
That is true.
Sorry I missed it Stephen
Outstanding article, Jim. Thanks!
This article and many others clearly show that Global Warming aka Climate Change aka Boiling et al was never about the science. Once the left realized that courts stated they cannot judge scientific papers and results, but had to weigh them all equally then it opened the doors for idealogues to use science toward their political and financial ends. Hence the mess we are in.
DEI in science is the woke IPCC ideal
and any model that assumes CO2 is a well mixed gas is flawed since the satellites have shown it is NOT well mixed globally …
How could it be well mixed? It is almost all emitted at the surface, and it is almost all absorbed by the surface. The air is densest at the surface, so the concentration would have to be highest per volume unit at the surface. I have never understood that claim. Maybe in a closed box it will mix evenly, but not on earth.
Zzzz
Every greenhouse proves that wrong.
Otherwise gardeners could dose up with CO2 in March and it would still be high in September.
That’s September in the same year, not 100 years later.
Greenhouses do not prove anything about the carbon cycle across the entire planet.
The attached chart shows the seasonal photosynthesis CO2 swings for Barrow Island. Every single in situ hourly CO2 temp and CO2 monitoring shows this pattern, except for a very flat pattern at SPO
It is well mixed..
Substantiation?
Check the Scripps Institute site. https://scrippsco2.ucsd.edu/ The monitoring network that Charles Keeling set up wasn’t just the “continuous” monitoring at Moana Loa. It also included weekly(?) flask measurements at a number of sites strung out from the Arctic Coast of Alaska to the South Pole. There is a variation in CO2 between the tropics and the polar areas. But it’s slight.
The seasonal variations for the South Pole and Pt. Barrow are very different. See the graph supplied by Banton, below.
Here:
The satellite data.
Here’s a paper from 2017, fig 4, showing a range of 10 ppm. out of 370.
https://acp.copernicus.org/articles/17/3861/2017/acp-17-3861-2017.pdf
The observations from OCO satellites say otherwise. The concentrations var between 390 and 408 ppm in this animation from 2015 prepared by the OCO-3 project of NASA/JPL
Thanks.
I couldn’t remember the scale bar range on those animations when I watched them when they were first released.
To be fair, this is a variation of little over 2% either way relative to the mean of the scale, so it depends on what you consider well-mixed. That’s the issue with scales that do not start at 0, they are prone to visually exaggerate differences.
As Willis has shown, over the ages the earth’s temperature has been stable within 0.25% if you look at the temperature in Kelvin.
399+/- 8 seems pretty well mixed to me. Just saying.
The substantiation for CO2 being “well-mixed” depends entirely on the definition one is using for “well-mixed.”
CO2 levels are at a bit over 420ppm.
CO2 levels vary from place to place by 3 to 5 ppm.
That is the definition of a well mixed gas.
That is well-mixed horizontally.
The original commenter was talking about vertical mixing; he referred to the surface, not the latitude.
CO2 is a heavy molecule.
There would be even less of it at higher elevations, where there is little mixing, say 3 km.
Also, H2O freezes on molecules, or what ever, at about 2 km.
Low ppm above the clouds
Most of the outward low frequency radiation photons leave at the poles, thereby cooling the earth
“CO2 is a heavy molecule.
There would be even less of it at higher elevations, where there is little mixing, say 3 km.”
No, the lower atmosphere is what’s known as a homosphere, where atmospheric turbulence maintains relative concentrations approximately constant.
This article promulgates the idea that cold molecules can radiate energy to hotter ones. This is contrary to the 2nd law of thermodynamics, which has never been disproven! The rest is fine, but no energy can be back radiated from the higher parts of the atmosphere. It must only go out to space, which of course is much colder. Physics anyone?
It is not contrary to the second law of thermodynamics, which concerns net energy exchanges. All layers of the atmosphere are radiatively exchanging energy with the layers above and below.
Incorrect to say all layers radiatively exchanging energy. Air parcels expand and carry heat w/o exchanging heat outside the parcel. No heat enters or leaves a air parcel of air.
‘All layers of the atmosphere are radiatively exchanging energy with the layers above and below.’
The word ‘radiatively’ invalidates the above sentence. Kirchoff’s Law, as well as the other properties of thermal radiation, only applies to condensed matter, It most decidedly does not apply to gases, at least under conditions that prevail anywhere outside of plasma physics laboratories on this planet.
Presuming that you will continue to insist that radiant transfer models accurately describe the physics of energy transport in the troposphere, perhaps you’d be willing to show us how photons, rather than energy, are conserved in the process. Your Nobel Prize awaits you!
https://physics.stackexchange.com/questions/290528/difference-in-thermal-radiation-between-condensed-matter-and-gases
Frank, you obviously didn’t bother reading your reference. In part –
Even your own reference doesn’t support you. The comment you probably accepted is complete nonsense – probably a physics lecturer who doesn’t understand what he teaches.
‘Frank, you obviously didn’t bother reading your reference.’
Actually, Michael, you’re the one who didn’t bother reading the reference – take another look at the ‘JPL table’ and let me know if any of the’Typical Sources’ pertain to gases in Earth’s atmosphere.
Now read the source the ‘JPL response’ actually links to:
“Continuum Emissions from Ionized Gas
Thermal blackbody radiation is also emitted by gases. Plasmas are ionized gases and are considered to be a fourth state of matter, after the solid, liquid, and gaseous states. As a matter of fact, plasmas are the most common form of matter in the known universe (constituting up to 99% of it!) since they occur inside stars and in the interstellar gas. However, naturally occurring plasmas
are relatively rare on Earth primarily because temperatures are seldom high enough to produce the necessary degree of ionization.”
Notice that a ‘continuum of emissions’ only arises from ionized gases, i.e., plasmas, and that these are rare on Earth.
You are a tiresome troll who aids the cause of climate alarmists by spewing easily refuted nonsense.
Frank, your reference also states –
I suppose you are now going to say that the author really meant something else?
I’d suggest some fairly basic physics textbooks, but they might be written by someone like Raymond Pierrehumbert – or even Michael Mann PhD, or Gavin Schmidt PhD.
Believe as you wish, but the Earth has cooled over the past four and a half billion years, and continues to do so (being hotter than its surroundings.)
Still quoting from JPL’s lame response, where they conflate atmospheric gases and plasmas. Again, with respect to differentiating between the radiative spectra of condensed matter and gases, the key word is ‘continuum’.
‘Believe as you wish, but the Earth has cooled over the past four and a half billion years…’
I think most folks would agree that the Earth is cooler now than it was during the Hadean Eon, so take a bow. In fact, I’m going to nominate you for a ‘Lord Kelvin’ award – if you’ll recall, he was the 19th century authority figure who locked the scientific consensus of his day into a low-ball estimate of the Earth’s age, based on his estimate for how long it would take molten rock to solidify and cool. Of course, he had no knowledge that heat from the decay of radioactive minerals in rock made his estimates, shall we say, a bit light, but apparently that’s a flaw the two of you share.
Which of course sounds scientific, but you can’t explain what it means. That’s because you don’t know what you are talking about. All matter above absolute zero emits IR radiation, with frequencies proportional to absolute temperature. Temperatures do not move in discrete units. Bad luck for you.
Really? What flaw do I share with Kelvin? The knowledge that the Earth has cooled over the past four and half billion years?
The mistake here is thinking Kirchhoff’s Law only works for continuous spectra like solids. In gases, the spectrum isn’t smooth, it’s line by line. But the law still applies. At every absorption line, the emissivity equals the absorptivity. That’s why CO2 and H2O both absorb and emit at the same infrared bands. The physics doesn’t break down for gases, it just has to be applied discretely rather than as a continuum.
‘At every absorption line, the emissivity equals the absorptivity.’
The mistake here is thinking Kirchhoff’s Law works for mixtures of IR-active and non-IR active gases. The reality is that thermal radiation absorbed by GHGs is converted to sensible heat via collision with non-IR active gas species within meters of the Earth’s surface. In other words, since the rates of spontaneous emission lag those of absorbtion by several orders of magnitude in the lower troposphere, Kirchhoff’s Law, and therefore radiative transfer models, are not applicable.
That’s not how it works. Collisions between IR-active and non-IR-active gases are exactly why we treat the troposphere as being in local thermodynamic equilibrium. In LTE, the populations of molecular energy states are governed by the local temperature, which means emission and absorption coefficients remain tied by Kirchhoff’s Law. Yes, absorbed radiation can thermalize quickly into translational energy of the bulk gas, but because collisions also repopulate excited states, those same molecules emit according to the local temperature. Radiative transfer models explicitly use this: absorption → collisional redistribution → emission, all consistent with energy conservation and Kirchhoff’s Law. Far from being “not applicable,” LTE is precisely what justifies their use in the troposphere.
Try again. Collisions occur faster than relaxation regardless of which way the energy transfer occurs. No emissions have time to happen.
Collisions do happen faster than spontaneous emission in the dense lower atmosphere, that’s why we assume LTE. But LTE doesn’t mean “no emission”; it means collisions continually repopulate excited states according to the Boltzmann distribution. Those states then emit at rates set by temperature, giving the observed thermal spectrum. If collisions really quenched emission entirely, the atmosphere would be dark in the infrared, which is flatly contradicted by satellite spectra.
‘If collisions really quenched emission entirely, the atmosphere would be dark in the infrared, which is flatly contradicted by satellite spectra.’
Satellites measure the Earth’s IR emissions as seen from their positions beyond the atmosphere and the spectrum of H2O largely overlaps that of CO2, hence the above statement is deceptive.
No one is contesting the facts that spontaneous emission of photons and radiation to space by GHGs takes place in the upper troposphere. What’s finally beginning to come under scrutiny is the predominant process by which GHGs in the lower troposphere facilitate the transport of heat from the surface to the upper troposphere, i.e., enabler of convection or radiative pinball machine.
Well said Frank!
My article simply acknowledges there is a certain degree of radiative pinball happening. What I hoped to show was convection and collisions that are the predominant processes transporting heat away from the surface and out to space
Glad you liked the metaphor, Jim. An equally good one is likening the alarmist model of radiative-centric energy transfer to the apparatus used for my state’s weekly lottery drawings – lot’s of vertical movement and ‘conservation’ of
ping-pong balls, er photons, as opposed to energy.That’s a misrepresentation. Yes, water vapor and CO2 absorption bands overlap, but not completely. CO2 dominates in the 15 µm region where H2O absorption drops off, and satellites measure this directly. More importantly, radiative transfer in the lower troposphere doesn’t require every absorbed photon to be re-emitted before a collision. Collisions maintain local thermodynamic equilibrium, ensuring the gas emits according to temperature. That’s why the atmosphere radiates upward at every layer, not just at the top. Convection is certainly important for moving energy, but it complements radiation rather than replacing it. Radiative transfer models already couple both. The “radiative pinball” idea is just a colorful restatement of the well established fact that photons are repeatedly absorbed, re-emitted, and scattered until escaping to space.
‘CO2 dominates in the 15 µm region where H2O absorption drops off, and satellites measure this directly.’
Looking at the so-called Planck curve, that emission peak is miniscule. Another problem for CO2 is that everyone who has run the radiative transfer codes agrees that it only emits to space above 80km.
As for what’s really going on in the region of the so-called notch, the satellites can’t tell us anything definitive because of the overlap of the H2O and CO2 emission spectra. Rather, we have two conjectures for heat transfer in the troposphere, one radiative and one convective.
Of these, only the latter is consistent with the fact that we’re all here today to opine about this, notwithstanding that CO2 concentrations have varied significantly over time and that 70% of the Earth’s surface is covered by water, which in its vapor form is the most potent GHG.
Van Wijngaarden and Happer have generated spectra with zero CO2, and Harde (2013) showed the difference with and without CO2 explicitly in his figures 17 and 18.
The spectrum tells us exactly what is happening. In the mid and upper troposphere where H2O radiates to space, CO2 partially absorbs the H2O emissions that span the Q band of CO2. Because H2O emits almost in a continuum of wave numbers in this region, (~14-16 μm), it excites a broad band of the A branch rotovibrational states. In the lower atmosphere the still high collision rates deactivate the excited CO2 molecules across the band. Hence, the “notch.” The tiny emission band in the middle is from CO2 emitting near the mesopause, where there is very little energy to emit but approximately 1.5% of the molecules have enough kinetic energy to excite the mode and the collision rate is low enough that most of the excited CO2 can radiate to space. It’s not rocket science, it’s just spectroscopy.
In the troposphere, collisionally activated CO2 overwhelmingly decays collisionally.
Photons are emitted by electrons. They just appear, travelling at the speed of light. Instantaneously, no acceleration involved.
This takes no time that you can measure, which you couldn’t establish precisely anyway, because you cannot precisely establish the position and momentum of either the electron or the emitted photon.
Still no GHE – it’s all nonsense.
In the case of CO2 the photons are emitted as the result of vibration of the C=O bonds, spectra of molecules are electronic, vibrational and rotational in origin. In the case of the bending mode of CO2 the mean time between emissions is much longer than the time between collisions in the lower atmosphere which is why most of the excitation energy is transferred to other molecules by collisions there.
Don’t be silly. CO2 cannot be distinguished from any other gas at the same temperature merely by measuring the frequency of emitted IR.
Somebody is obviously treating you as ignorant and gullible if they told you otherwise.
Certainly can, that’s one of the uses we put the spectrometer to in my research lab!
The issue is how long after absorption does it take for the photon to “appear”, not how long it takes for any acceleration to reach the speed of light. If the absorbed energy is lost by collision before it can radiate, then it never radiates that energy.
‘Yes, absorbed radiation can thermalize quickly into translational energy of the bulk gas, but because collisions also repopulate excited states, those same molecules emit according to the local temperature.’
I’m glad we agree that thermalization heats the pool of gases, but then, why wouldn’t the reverse thermalization that ‘repopulates the excited states’ cool the same pool of gases? In other words, the same air parcel simultaneously absorbs, heats, cools and radiates, all without any work, i.e., convective lifting, being done on said parcel.
I think the obvious answer that requires the least amount of hand waving is that absorbtion and heating predominate in the lower troposphere, while cooling and emission predominate in the upper troposphere, and convection, not radiation, explains energy transport between the two regions.
It’s not “hand waving,” it’s physics. In LTE, collisions both thermalize absorbed radiation and repopulate excited states. That does mean emission and absorption are constantly occurring in the same air parcel, but with net rates that balance so the radiation field matches the local temperature. That’s why the lower troposphere doesn’t just absorb, it also emits, which is exactly what satellites detect looking down. Convection is certainly a major transport mechanism, but it doesn’t replace radiation; the two operate together. In fact, the lapse rate itself arises from the interplay of convection and radiative cooling. If radiation in the lower troposphere were negligible, the whole greenhouse effect would vanish, which is flatly contradicted by both measurements and models.
‘That’s why the lower troposphere doesn’t just absorb, it also emits, which is exactly what satellites detect looking down.’
Not possible.
‘In fact, the lapse rate itself arises from the interplay of convection and radiative cooling.’
Absorbtion and thermalization heat the lower troposphere. Reverse thermalization and emission cool the upper troposphere. Convection is how heat from the former is transported to the latter. The lapse rate is simply a manifestation of the resulting temperature gradient.
‘If radiation in the lower troposphere were negligible, the whole greenhouse effect would vanish, which is flatly contradicted by both measurements and models.’
You need to clarify what you mean by GHE. If you simply mean warmer air below and colder air aloft, we obviously don’t need spontaneous radiation in the lower troposphere to maintain convection or the lapse rate, all of which is observable.
You’re trying to separate radiation and convection as if they were mutually exclusive, but the atmosphere doesn’t work that way. Convection moves heat upward, but radiation is still occurring at every level, even in the dense lower troposphere. That’s not speculation, it is measured directly with downward-looking instruments (e.g. radiosonde IR sensors and aircraft), which detect emission from H2O and CO2 in the lower atmosphere.
The lapse rate isn’t “just convection”: it’s set by convection and radiative cooling. If convection alone governed it, the troposphere would adjust to the dry adiabatic lapse rate everywhere, yet observations show deviations where radiation matters. And the greenhouse effect isn’t merely “warmer below, colder aloft”; it’s the fact that gases radiating to space do so from colder, higher layers, which reduces the efficiency of cooling and raises the surface temperature required for balance. That effect requires emission throughout the troposphere, not only at the top, which is exactly what Kirchhoff’s Law and radiative transfer predict and what observations confirm.
‘Convection moves heat upward, but radiation is still occurring at every level, even in the dense lower troposphere.’
I believe that’s the crux of our debate. Not just whether it’s occurring at ‘every level’ of the troposphere, but what exactly it’s doing in terms of transporting energy from the surface, and by that, I mean independent of any energy (sensible or latent heat) conveyed by convection? And just to be clear, I’m only referring to the thermal energy from the surface that is absorbed by IR-active gases.
‘That’s not speculation, it is measured directly with downward-looking instruments (e.g. radiosonde IR sensors and aircraft), which detect emission from H2O and CO2 in the lower atmosphere.’
Again, I’d like to know what they’re actually measuring. If they’re doing spectroscopy, are the sensors of their instruments cryogenically cooled? If so, that would certainly bias which direction energy might be flowing, if at all.
‘That effect [GHE] requires emission throughout the troposphere, not only at the top, which is exactly what Kirchhoff’s Law and radiative transfer predict and what observations confirm.’
Thanks for clarifying what you mean by the GHE. If I may, you’re saying that it’s strictly a radiative phenomenon that requires Kirchhoff’s Law to hold throughout the troposphere, the result of which is that any increase in atmospheric CO2 concentration must result in warming, not only from CO2, but also from H2O (water vapor) amplification.
On this point, I’m highly confident that there is no evidence from ice or deep-sea cores to support this premise. Specifically, ice cores from both hemispheres demonstrate that changes in CO2 concentration lag changes in temperature on the order of 1,000 years, or so. Similarly, temperature changes over the past 65my derived from carbonate cores do not show any association with CO2 concentrations, and, in fact correlate well with changes in land movements caused by plate tectonics.
Aside from the last item, what I’m inquiring of you requires a background in physics that is beyond mine. However, that doesn’t mean I can’t follow ‘good’ science, hence the only frustration I really have is that so-called informed ‘alarmists’ (and ‘luke warmers’ for that matter) refuse to engage in debate with those who believe molecular collisions radically change the radiative properties of GHGs to the extent that radiative transfer models don’t accurately portray the physics of energy transport through the troposphere.
Having said that, I’d be interested in your opinion of the following:
https://wattsupwiththat.com/2025/08/21/how-co2-both-warms-and-cools-our-atmosphere/#comment-4109702
Collisions in the lower troposphere don’t suppress emission; they shape it. Absorbed radiation quickly thermalizes into the bulk gas, and collisions continually repopulate excited states so that, under local thermodynamic equilibrium, the gas emits according to its temperature. That’s exactly what Kirchhoff’s Law guarantees. Radiation and convection are not competitors; they act together, with convection moving heat upward while radiation allows every layer to exchange energy with its surroundings.
The measurements confirm this directly. Radiometers and spectrometers at the surface, on aircraft, and on satellites all detect both upwelling and downwelling infrared radiation in CO2 and H2O bands. These signals are the fingerprints of emission from the lower atmosphere. If radiation were absent there, those observations simply wouldn’t exist.
The greenhouse effect itself is not just “warmer below, colder above.” What matters is that emission to space comes from higher, colder layers. Raising CO2 concentrations lifts the effective emission height into colder air, reducing the outgoing flux at a given temperature and forcing the surface to warm until energy balance is restored. Water vapor then amplifies this process.
The familiar ice-core lag is no contradiction. Orbital changes start the cycle, oceans then release CO2 as they warm, and that added CO2 amplifies the change. On geologic timescales the pattern is the same: high CO2 aligns with warm climates, while low CO2 aligns with glaciations. These correlations are well established in the paleoclimate record.
The post you cite rests on the misconception that collisions prevent re-emission. In fact, collisions enforce LTE, which is precisely why Kirchhoff’s Law holds in the troposphere. Radiative transfer depends on that principle, and its validity is confirmed across spectroscopy, astrophysics, and atmospheric science. If collisions really quenched radiation, none of these fields would work.
“emission height into colder air, reducing the outgoing flux at a given temperature and forcing the surface to warm”
Do you not see the contradiction here? If concentration goes up you have mire molecules radiating. Does that increase the flux or decrease it?
More molecules means more absorption and more emission, but what matters is where the last emission to space happens. When CO2 increases, the “optical depth” thickens, so photons at 15 µm escape only from higher, colder layers. Colder emitters radiate less (Stefan–Boltzmann applies locally), so the outgoing flux at that band decreases. The surface and lower troposphere then warm until the total outgoing IR once again balances incoming solar.
So yes, there are more emitters, but because emission to space comes from colder levels, the net flux to space drops. That’s the greenhouse effect in a nutshell.
“but what matters is where the last emission to space happens.”
What matters is how much HEAT is sent into space in the form of an EM wave.
More emission from a colder temperature can easily be the same as less emission from a warmer temperature – it depends totally on the radiating masses of each.
“Colder emitters radiate less (Stefan–Boltzmann applies locally), so the outgoing flux at that band decreases.”
Again, more MALARKY! You admit in the very first sentence that more molecules emitting means more emission. And then you turn around and say that more molecuoes emitting means less outgoing flux.
PICK ONE AND STICK WITH IT!
“So yes, there are more emitters, but because emission to space comes from colder levels, the net flux to space drops.”
The very same malarky in the same sentence! More emitters at less flux per emitter can easily equal fewer emitters at more flux per emitter. YOU NEVER give any figures for either the number of emitters or the flux per emitter. You just keep on saying colder temps mean lower flux and ignore the number of emitters involved!
FM, pure and plain.
I’ll respond to both of your comments together, since they’re really making the same set of points.
There’s no contradiction in saying “more molecules” and “less flux.” Adding CO2 increases the number of emitters, but what matters isn’t raw count, it’s optical depth, i.e. how far radiation at 15 µm can travel before being absorbed. With higher CO2, photons escape to space only from higher, colder layers. Colder layers emit less per unit area (per Stefan–Boltzmann and Planck’s law), so the outgoing flux drops until the surface warms enough to restore balance. That’s why increasing CO2 reduces flux at those bands even though there are more emitters.
On the “temperature vs. heat” objection: no one is confusing the two. Radiative flux (power per area) is the relevant quantity for energy balance at the top of the atmosphere. Satellites measure radiance spectra directly, integrate them over wavelength, time, and latitude, and produce global energy budgets. Surface and radiosonde measurements anchor those fluxes to actual temperatures. This isn’t hand-waving, it’s the basis of modern radiation climatology, validated by multiple independent measurement systems.
As for emission: CO2 doesn’t radiate at 15 µm because it has the same temperature as the surface; it radiates there because that’s its vibrational transition frequency. The intensity at that frequency depends on the local temperature through the Boltzmann distribution. Planck’s law and Kirchhoff’s law both apply to gases in local thermodynamic equilibrium, not just solids. That’s why line-by-line radiative transfer models built on measured molecular absorption coefficients reproduce both laboratory spectra and satellite spectra with high accuracy.
In short: more CO2 shifts the effective emission level higher and colder, reducing flux until the surface warms; radiative flux, not “heat,” is the right measure of energy loss; and CO2 emission lines arise from molecular physics, not blackbody assumptions. These aren’t guesses and they are confirmed every day by spectroscopy and satellite observation.
In the troposphere, collisionally repopulated excited states decay collisionally.
Collisions certainly dominate de-excitation in the dense troposphere, but that doesn’t mean radiative decay doesn’t occur. What collisions actually do is enforce LTE, so the distribution of molecular states follows the local temperature. That equilibrium guarantees that emission and absorption coefficients remain linked by Kirchhoff’s Law.
If collisions truly quenched emission altogether, the lower troposphere would be dark in the infrared. Yet we directly measure strong downwelling IR at the surface and from aircraft, and satellites looking down see spectral emission features from lower layers as well. Those observations are exactly what you’d expect if collisional excitation and radiative decay coexist in LTE, and exactly what you wouldn’t see if collisions eliminated emission.
Lot’s of confusion here. Absorption of IR from the surface by CO2 is supposed to be the source of GHE. Yet if the satellites can actually measure the IR radiation from the troposphere, especially the lower troposphere, just how much IR from the surface *is* being absorbed by CO2? And that doesn’t even speak to what H2O absorbs. If there is absorption then how do the satellites know what the total IR radiation from the surface is? There just seems to be a lot of “assumptions” in all of this and assumptions are guesses. Guesses obey the “a$$ u me” rule.
There’s no contradiction here. Satellites measure upwelling radiance at different wavelengths, not just “the total IR from the surface.” In the spectral regions where CO2 or H2O absorb strongly, satellites see reduced surface emission and enhanced emission from higher, colder layers of the atmosphere. In the “atmospheric window” regions where absorption is weak, they see surface radiation directly. That’s exactly how we know which gases are absorbing and emitting: the spectrum tells you.
So yes, CO2 and H2O absorb surface IR, and satellites detect that absorption because the spectrum shows the imprint of those gases. The greenhouse effect isn’t an “assumption,” it’s a measurement: the dips in the outgoing spectrum at 15 μm (CO2) and in water vapor bands are the direct fingerprints of absorption and re-emission in the troposphere.
More malarky. It isn’t total radiation across all frequencies, it’t the total at CO2 frequencies. If you don’t know the total then you can’t know how the radiation measurement at that frequency relates to the total at that frequency.
When you say “reduced surface emission” how do you know its reduced if you don’t know where it started? The earth is *not* a black body. So again, how do you *know* the total?
We don’t need the Earth to be a perfect blackbody to know what’s leaving the surface. The surface behaves as a near-blackbody in the thermal infrared with an emissivity close to 1, and its emission spectrum is measured directly by ground-based radiometers. When satellites see less radiance at 15 µm than a warm surface at that temperature would produce, and instead see a spectrum corresponding to colder air aloft, that is “reduced surface emission.” The reference is not a guess but is the Planck curve at the measured surface temperature, adjusted for known surface emissivity.
That’s exactly why line-by-line radiative transfer models match satellite spectra so well: we know the surface emission, we know the absorption coefficients, and the result matches observation. If CO2 weren’t absorbing and re-emitting, those 15 µm dips in the spectrum wouldn’t exist.
The surface emissivity varies considerably. In addition the surface of the earth is a heat sink. Heat from the sun is diffused into the soil to a considerable depth for later dissipation. Think seasonal. Anything diffused into the soil is not immediately emitted. That is one reason the earth is not a black body.
Even more malarky! 1. Temperature is not heat. 2. Flux is not heat.
At best, heat loss is related to the area under the temperature curve, i.e. it is a time function. The surface temperature varies a LOT during the daily profile, it varies even more during the month due to seasonal variation. That means that 1. the emission measurements have to be integrated over time as well as the surface temperature in order to determine what *HEAT* is being lost/gained.
In addition, the surface temperature varies hugely from 0° latitude to 90° latitude. Unless the emission data is somehow related to the surface temperature at the same exact time the emission measurement is made you *still* can’t figure out what the “reduced” emission level is. Because of the paucity of surface temperature measuring locations how is the satellite emission measurements physically related to the surface temperature at the time?
This just all appears to be hand waving FM.
“f CO2 weren’t absorbing and re-emitting, those 15 µm dips in the spectrum wouldn’t exist.”
More FM. For CO2 to emit at the same frequency as the surface, i.e. 15μm, they two would have to be the same temperature in order to use S-B. Without S-B you can’t calculate temperature from frequency. And this is assuming that CO2 can be treated as a black body under the requirements of Planck, namely a solid, homogeneous object in equilibrium where dτ is large enough to be treated as emitting equally in all directions with no quantum effects at play.
Hand waving FM from start to finish.
“If radiation in the lower troposphere were negligible, the whole [radiant] greenhouse effect would vanish,”
It doesn’t exist in the first place, so it doesn’t need to vanish.
“which is flatly contradicted by both measurements and models.”
What measurements? (models are just fantasies)
The GHE necessarily exists, else the observed surface temperature of earth would be well below freezing.
Fantastic – simultaneous heating and cooling! A “climate science” staple. Cooling is heating, heating is cooling – all good “climate science” religious precepts.
Your religious beliefs are your affair. Not science, in my view.
That the same air parcel could simultaneously do all those things was my retort to AJ. Do pay attention, Michael.
Wouldn’t normally up-vote Alan J but the blind squirrel and all that.
No, your first instinct was the correct one. He still has no clue what he’s talking about. There is no such thing as “net energy exchange”, i.e. “net work”, because there is no such thing as “gross work” either.
Not true. Radiation strikes everything. Earth’s radiation hits the sun.
The ‘cold to hot’ argument applies only to conduction. Everything above absolute zero radiates and that radiation goes until it is absorbed or deflected by something.
“The ‘cold to hot’ argument applies only to conduction”
Who told you that? The 2nd Law doesn’t say anything about the mechanism of energy transfer… are you hallucinating?
Try reading the science, not just the summations.
I did read the science. I’ve been studying it all my life. What do you think the 2nd Law says, MarkW?
What’s the matter, Professor? Are you taking time to compose your erudite lecture on how the Second Law of Thermodynamics somehow exempts electromagnetic radiation? Complete with experimental evidence? I was really looking forward to learning how that works. Being able to evade that law sounds quite exciting. Or are you just pulling baloney out of your fundament, like so many others?
You know, I used to have a lot of respect for your viewpoint, because on non-physics-related topics, it’s generally bang-on. Now, though, in one fell swoop, you have revealed your true self as just another arrogant ignorant egotistical charlatan, a Dunning-Kruger basket case far outside his field. Too bad, really.
Conduction means when a air parcel expands due to increased heat, the air parcel rises and is replaced by sinking colder air parcel, which eventually gets heated to begin to rise. Conduction is a cycle of heat loss.
You’re confusing conduction with convection.
Exactly. Unless quantum mechanics is even weirder than we think, Photons and photon emitters are not clairvoyant. There’s no way for photons to know if they are heading for something hotter than the source that emitted them.
There’s no way for individual water molecules to know whether they’re heading upriver or downriver at any given nanosecond, either, yet the river always flows downhill. Why?
Yes, quantum mechanics is weirder than you think.
A photon is not a particle (ie- localized) until it is detected. Before that it’s a wave, ie- a probability function.
And does not interact with said Sun. Just like photons from ice hit water, but don’t affect it at all.
They certainly do, they’re absorbed and transfer their energy to the water, of course the water also emits photons which cannot the ice and transfer their energy to the ice!
Should be: They certainly do, they’re absorbed and transfer their energy to the water, of course the water also emits photons which can hit the ice and transfer their energy to the ice!
What you claim is a misapplication of the second law. The second law deals with “temperature”, but molecules do not have a temperature. They have some level of energy, which can be transferred even contrary to the energy gradient. Temperature can only be assigned for an ensemble of molecules. Even when the ensemble contains a small number of molecules there are fluctuations in which the second law, as you state it, is violated.
“They have some level of energy, which can be transferred even contrary to the energy gradient.”
At a quantum scale that is true. At a classical scale, which is what we are discussing here, it is not. Be careful not to mix up your scales!
What The Real Engineer should have said, to be more precise, is that Jim’s article claims that classical work is done from colder objects to hotter ones. That never happens. The Second Law says so.
(Jim’s claim was “Those gases absorb the infrared heat but then quickly emit it. However, about half is redirected back towards the surface”, which is false as written, since “heat” and “energy” are not the same thing in physics)
All atoms operate on the quantum scale. Just because you lump a bunch of atoms together does not change this fact.
Lumping a bunch of atoms together and trying to prevent their aggregate behaviour from decohering is very difficult indeed, so yes, it does change this fact, as a matter of fact. Who taught you your physics, anyway?
It turns out that the quantum scale and the classical scale do not, in fact, follow the same rules. For example, entropy (the foundation of the Second Law) does not exist at the quantum wavefunction scale, but it does at the classical level. Why?
Energy transfer does occur at the atomic level. However, “heat” requires a large enough volume to insure sufficient molecules exist to make macro analysis possible.
Incoming radiation has three characteristics, absorbtion, reflection, and transmission. The analysis of these don’t work at the atomic level.
Planck says this.
From article:”But greenhouse gases redirect some of that heat back to the surface and slow the earth’s cooling rate.”
If we are using the term “heat” to describe what is going on then saying heat can’t go from cold to hot seems reasonable.
That depends on how the heat is being transferred. Radiation does not care what the temperature of the radiating molecule is , nor what the temperature of the receiving molecule is.
On the transmitting molecule, when an electron drops from one shell to a lower one, a photon is released. The energy of the photon is determined by the energy difference between the two electron orbits. The temperature of the molecule is irrelevant.
A molecule hit by a photon will absorb that photon if there is an electron on that molecule that can jump to a higher shell by absorbing exactly the amount of energy that the photon contains. Once again, the “temperature” of the receiving molecule is irrelevant.
No, that’s a special case – an excited atom within a molecule.
Matter at say, 2 K, is radiating photons. Matter at 1 K absorbing enough of these photons will be heated to 2K, and radiate photons with energies proportional to the temperature.
All matter. Gasses can both absorb and radiate IR, as Tyndall demonstrated.
Without intending any offense, you simply do not know what you are talking about, or you are intentionally trying to mislead.
That’s what’s being discussed, a CO2 molecule at its ground vibrational state absorbs an IR photon of the appropriate wavelength which excites it to the first vibrational state, The excited molecule will then need to lose that excess energy either by emission or by collision with other molecules.
Phil, don’t anppear even sillier than you already seem.
CO2 is heated by absorbing energy. No “appropriate” wavelengths necessary. Just photons emitted by anything hotter than the CO2.
By the way, molecules don’t actually collide – that’s a hangover from the 18th and 19th century. Physics has moved on since then.
“CO2 is heated by absorbing energy. No “appropriate” wavelengths necessary. Just photons emitted by anything hotter than the CO2.”
Not true, CO2 will absorb photons of the wavelength necessary to excite the molecule from its current energy level to another, higher energy level. The temperature of the source is irrelevant, a blackbody at any temperature is capable of exciting the CO2 bending mode by emitting a 15μm photon.
I’m not the silly one here, I suggest you read a freshman text on molecular spectra!
Dimwit! CO2 can be raised from absolute zero to 1K (although you probably refuse to believe it). The energy of photons emitted by a body ar 1 K are incapable of exciting anything – except your imaginary fantasies.
“That depends on how the heat is being transferred.”
No it doesn’t. Sit down and stay in your lane, MarkW. It isn’t physics.
The second law deals with conductance of heat. Not radiation.
A molecule of any temperature above absolute zero will radiate.
When that photon hits another molecule that is capable of absorbing at that frequency, it will be absorbed. Period. The photon does not carry any information regarding the temperature of the emitting molecule or atom.
“The second law deals with conductance of heat.”
No it doesn’t. You’re hallucinating.
In fact, the Stefan-Boltzmann Law says that all matter with a temperature above absolute zero radiates thermal energy. That law states NOTHING about the presence or absence of one or more bodies being external to the radiating body. Therefore, in fact a colder body can radiate energy onto a warmer body.
There is nothing there that violates the 2nd Law of thermodynamics.
If the issue is the net energy exchange between TWO radiating bodies with view factors of each other and at different temperatures, that’s another matter altogether!
Yeah, physics.
“Therefore, in fact a colder body can radiate energy onto a warmer body.”
That is correct. Note, however, that energy is not measured in Watts, is it?
Energy transferred per unit time can be measured in watts . . . we live in a universe where time marches on as physical process occur.
For the linguists/morphologists out there, “radiate” is an active verb.
I’m not sure what you’re trying to say, because you seem to be deliberately vague, but if you are trying to say that “radiating energy” is identical to “doing work”, no it isn’t.
LOL ENGINEER,
Please explain how an excited greenhouse molecule knows to only emit radiation towards colder regions????? You abuse the 2nd law of thermodynamics. Emissions are mostly isotropic so that the photon is emitted in any direction with equal probability. The emitting molecule has no preferred orientation for emission.
Jim, you obviously don’t understand that matter does not necessarily interact with radiation. A simple example is an ice cube submerged in water.
You have no idea what happens to the photons emitted by the ice, do you? Hence your demand to “Please explain . . . ” is your attempt to disguise your own lack of knowledge.
You really have no clue, do you?
THAT’S A REALLY STUPID REPLY MICHAEL
Jim, SHOUTING an opinion based on an appeal to your own authority, might make you look more stupid than I.
You really have no idea what happens to the IR photons emitted by an ice-cube totally submerged in water, have you?
Just like most of the fanatical commenters here, you refuse to accept that you can’t even describe this mythical greenhouse effect in any consistent and unambiguous way!
You might as well just keep SHOUTING. That is often the tactic used by the ignorant and gullible to disguise their ignorance and gullibility.
“Please explain how an excited greenhouse molecule knows to only emit radiation towards colder regions?????”
Nobody said that. Sit down and stay in your lane, Jim. That’s landscapes and cycles, remember? Not physics.
Lots of folks do say this when they try to discuss the emission of photons on an atomic basis. Dipole molecules radiate in randon directions just like a dipole antenna that is randomly changing its orientation.
This requires probability treatment to analyze heat.
Planck recognized this early on and made his analysis based upon an homogeneous volume of size dτ.
JM, +100
An homogeneous volume of size dτ, is needed to analyze heat. dτ must be of a size such that the random emissions of a large number of atomic structures radiate in all directions, i.e., is tropic.
Without this probability must enter the equations for heat radiation transfer.
Actually, it’s more pernicious. There is an implication that matter must absorb all radiation which impinges upon it, which is, fortunately, nonsense. Otherwise, all matter in the universe would be infinitely hot, continuously absorbing radiation.
About as silly as expecting ice cubes to heat your soup. CO2 warming believers are both ignorant and gullible – not necessarily stupid or ill-educated. Just deluded.
According to you:”All matter above absolute zero emits IR radiation, with frequencies proportional to absolute temperature.” Also according to Kirchoff’s Law a good emitter is also a good absorber. So the “implication that matter must absorb all radiation which impinges upon it,” is your theory!
Keep dreaming.
This article promulgates the idea that cold molecules can radiate energy to hotter ones. This is contrary to the 2nd law of thermodynamics, which has never been disproven! The rest is fine, but no energy can be back radiated from the higher parts of the atmosphere. It must only go out to space, which of course is much colder. Physics anyone?
Agree, so this seems to discredit the entire article.
While Jim’s grasp of physics is, shall we say, “loose” at best, I don’t think we need to throw out the entire article. The graph of radiant energy measured from space correctly shows that some wavelengths of energy are being emitted at different temperatures (thus altitudes) than others. That is valid science, and worth keeping in mind, I would say.
Very curious stevekj, what exactly how do you fine my “loose” physics” ? Clearly your comment suggests that by demeaning my proven physics, you reveal your total lack of understanding of physics. Prove me wrong by specifically critiquing my physics. Do you disagree about the collision dynamics???
No, Jim, the collision dynamics isn’t the issue with your “loose” grasp of physics. That part is accurate, as far as I can tell. This, however, is not: “Those gases absorb the infrared heat but then quickly emit it. However, about half is redirected back towards the surface”
That’s not how heat and radiation work.
Well stevekj all I can surmise is you are totally ignorant of the physics!
Surmise away, and then try to explain what you think “heat” means, and what exactly the Second Law tells us about it. Go for it! I’ll wait here.
Since the claim is 100% false, the article stands.
Claim isn’t false, the graph is in inverse mode. (10 microns should by peak not 15 microns). It’s like splitting 340 watts into 2.63 10^25 (99% N2 & O2) pieces with an energy of 1.3 x 10^-21(15 microns -80°C heat) joules per piece. Carbon dioxide has much less mass (ignored in the graph) so requires much higher energy (visible light energy) to have same energy being claimed. The article is based on model theory, not reality.
10 microns at same mass equals total energy of sunlight IR 528 watts. Heat found at the tropics.
LOL Danley,
Please explain how an excited greenhouse molecule knows to only emit radiation towards colder regions????? You abuse the 2nd law of thermodynamics. Emissions are mostly isotropic so that the photon is emitted in any direction with equal probability. The emitting molecule has no preferred orientation for emission.
“Please explain how an excited greenhouse molecule knows to only emit radiation towards colder regions?????”
Literally no one said that. You’re hallucinating.
“You abuse the 2nd law of thermodynamics.”
No he doesn’t.
“Emissions [of energy] are mostly isotropic so that the photon is emitted in any direction with equal probability.”
That is correct. But says nothing about doing work, i.e. heat. What do you think “heat” means, Jim? How about “work”? And how do these concepts relate to the 2nd Law?
So an ice crystal in the atmosphere at a temperature of 273K emits a 15μm photon which hits a CO2 molecule in its ground state at 274K. The CO2 molecule absorbs that photon starting the bending mode vibration, i.e. ‘work’ has been done.
No, individual photons do not do “work”. That phenomenon requires entropy, which doesn’t exist at the quantum scale. You are confusing totally separate domains of physics. So no, you can’t buttress your ignorance of classical thermodynamics with your ignorance of quantum physics. Science doesn’t work that way.
Then take the time to do your own write up and educate the willing.
This claim is 100% false. All atoms above absolute zero radiate. They radiate based on the energy levels of their electrons. The energy of the photon released is based on the potential energy levels of the electrons in the atom. The temperature/energy of the atom itself is irrelevant.
Once a photon is radiated, it does not carry any information regarding the temperature of the emitting atom. When it strikes another atom, it will be absorbed by that atom based on the potential energy levels of the electrons of that atom. Once again, the temperature of the atom is irrelevant.
While that applies to atoms, atmospheric physics is mostly concerned with relatively low temperature molecules. When considering LWIR photon absorption, molecule-to-molecule collisional energy exchanges, and photon re-emission processes happening in Earth’s atmosphere, at temperatures from, oh, about 330 K down to about 140 K, the energy changes are tied up in mechanical modes of energy (translational velocity and the oscillations of molecular bonds in stretching, rocking/bending, and rotating modes) and NOT in energy changes in electron shells, that is electron excitation.
“All atoms above absolute zero radiate”
energy
What units do we measure energy in, MarkW? And what do you think “energy” means?
Your quantum-scale description of photon movement is irrelevant at the classical scale, which is where we carry out all our measurements. No one has ever measured energy being transferred via radiation from a colder object to a hotter one. It’s a fiction. The quantum energy foam is like Brownian motion in a hot cup of tea. Nothing more.
There are three things that can happen to a photon: It can be absorbed, upon which it is no longer a photon. It can be deflected, think pool balls. It can be reflected, meaning it returns to the direction it came from. The reason telescopes can see light from million to billions of years ago is because those photons that were emitted at that time are just arriving at the telescope.
nitpick. There really isn’t any “deflection”, only absorption or reflection. Energy is an electromagnetic field, i.e. a vector with a normal component and a non-normal component. Only the energy in the normal component is available for absorption. The non-normal component and the non-absorbed normal component just continue on in the original directions thus the direction of the EM field remains the same. No deflection. The magnitude of the EM field will change but not its direction.
Let us test you’re hypothesis with a thought experiment.
Imagine excited molecule A.
It’s spinning around furiously but all around it are hotter molecules. Excited molecule A wants to relax a bit, But it can only emit energy in certain quantised packets (that sounds silly but trust me on that). Now, the surrounding hotter molecules have electrons that can jump up to an excited state themselves. And those jumps – they match the quantised packet of excited molecule A.
What happens?
Option 1) Excited molecule A sees the surrounding hotter molecules and decide that now is not the time to release a photon. This is a bold re-interpretation of molecular physics and a big win for the Pantheists. I’m sceptical.
Option 2) Excited molecule A releases the energy but the hotter molecules around see where it’s come from and snub it. They duck or shimmy or just let it pass through. Snobby molecules will wait for a photon from a hotter molecule, thank you very much. This has the same challenge as Option 1.. I’m still sceptical.
Option 3) Excited molecule A releases the energy and the hotter molecules around interact with it. They get excited. Cold to Hot. Of course, with more high energy (hot) molecules than low energy (cold) molecules, on average this doesn’t happen as often as the other way round. That’s just statistics (or thermodynamics). But it must happen. It must happen in the middle of the atmosphere where it’s swamped by the reverse operation. And it must happen at the top of the atmosphere where that energy doesn’t meet a molecule above it very often and soi the energy escapes to space.
What other scenario can you see?
Agree. All molecules 2.63 x 10^25 (at certain levels) have the same energy 1.3 x 10^-21 joule. 99% N2 & O2 have the right amount to equal 340 watts, 1/4 of the solar constant 1360 watts. Hotter molecules (either heated by the sun, or colliding w/ hot surface) ceases cooling (after rising) at a certain level (usually 250hpa). Above that energy transfer occurs with incoming sunlight which increases radiative active molecules energy.
No 2nd law doesn’t apply for an emitting gas molecule, the excited molecule has no ‘knowledge’ of the state of it’s surroundings and is equally likely to emit a photon in any direction regardless of the state of the ultimate destination of that photon.
Question for the discussion:
I sometimes pull up the infrared satellite view when supercell thunderstorms are approaching. The cloud tops are bitterly cold in that data presentation. There are sometimes (presumably) cirrus clouds also within the satellite field of view. These clouds do not register as bitterly cold, and in fact do not register at all on the color scale. (Perhaps they are too thin to be measured by the satellite sensor?)
I would think the tops of the thunderstorms would be very hot compared to the surrounding atmosphere since they are usually packets of moist air that rose from the ground (before they expanded and cooled), but more importantly there should be a massive amount of heat of condensation occurring near the tops of these storms.
If the sensor could measure the air at the same altitude adjacent to the tops of the thunderstorms, would we observe that the thunderstorm was in fact much warmer air?
Finally, I added this question, because it does seem to be a way to transfer a lot of atmospheric heat from the surface to the top of the troposphere where it can be readily radiated out to space. However, it also could be insignificant if it is a tiny number like only 0.1% of the atmospheric processes.
Thanks for any teaching responses.
I don’t know what band you are looking at when you say the cloud tops are bitterly cold, but take Band 13 as an example. Band 13 allows cloud top temperatures day or night, and from cloud top temperatures one can get an idea of things like convective intensity. The colder the cloud top most likely the greater the convective intensity. Thunderstorm tops are never hot. One common misconception is to overlook that temperature is determined from a local energy balance and that depends on more than just heat transfer. It involves the first law of thermodynamics.
the left hand side is change in internal energy which translates to changes in temperature, but note that it involves not only contributions from heat (Q), but also contributions from work processes (W). The rise of heated and moisture laden air has to perform work against its surroundings because it expands as it rises. This work comes from its internal energy because the rise is too fast for heat transfer to be significant. This rising air cools as it does work.
As to why the cirrus clouds don’t register…
Band 13 is in the clear window, so it provides an intensity dependent on what materials provide most of the emissivity in the view. The atmosphere proves nothing. Cirrus clouds provide very little emissivity and the ground surface then provides almost all the emitted power, as long as lower clouds aren’t in view beneath them. Thus, the cirrus regions might appear hotter, but this is because the ground surface is hotter.
I hope this helps.
High convective clouds look cold because they are at an altitude where it is cold. Other than the surrounding atmosphere, they are optically thick. So you see their cold IR signature, while around them you see the IR signature of lower atmospheric levels.
Barely! It is their excessive convective heat that drives them up (also consider condensation), but once they hit the ceiling (the tropopause), they can not go further and thus produce an anvil kind of shape, as below. If they were still warmer than the surrounding air, they would still be going up..
“If they were still warmer than the surrounding air, they would still be going up.”
Keep in mind that buoyancy is a function of differences in density, not temperature.
Cold dry air is more dense than moist or saturated air (i.e., clouds) at the same pressure level (i.e., altitude).
Therefore, even at the same temperature and pressure moist air/clouds will rise above surrounding dry air.
I don’t know why you got a down vote, virtual temperature involves humidity for exactly the reason you describe. I also thought of speaking to the fact that as convection is busy converting buoyancy to work and thence to kinetic energy, very active thunderstorms will overshoot their equilibrium level — very common. I suspect these storms will be even colder than their environment.
Thanks for the informative replies!
Working in geostationary satellite meteorology for the past 40 odd years, I’ll see if I can some clarity to the other replies…
>I would think the tops of the thunderstorms would be very hot compared to the surrounding atmosphere since they are usually packets of moist air that rose from the ground (before they expanded and cooled), but more importantly there should be a massive amount of heat of condensation occurring near the tops of these storms.
The thing that has been left out is the rising adiabatic lapse rate (dry = 5.4 °F cooling per 1,000 feet, moist = 2.7 °F cooling per 1,000 feet). Notice – COOLING. The air parcels do not warm while they rise, they cool. In fact, the coldest portion of the anvil is at the center of the updraft and it warms toward the edge of the anvil/cirrus because of adiabatic warming from the sinking air (opposite effect from rising/cooling air).
For cirrus clouds, as with all other clouds, they are basically the same temperature as the air around them (quite cold) but if they are thin cirrus, the overwhelming IR energy from the warmer lower levels may overwhelm the lesser IR energy they are radiating so they will not be as obvious.
>If the sensor could measure the air at the same altitude adjacent to the tops of the thunderstorms, would we observe that the thunderstorm was in fact much warmer air?
No because, again, the clouds are basically the same temperature as their surrounding air (minus the temperature of the rising parcel, of course) for convective parcels. For stable parcels being forced to rise (isotropic lift, orographic lift, etc.) the temperature difference is much less.
>Finally, I added this question, because it does seem to be a way to transfer a lot of atmospheric heat from the surface to the top of the troposphere where it can be readily radiated out to space.
That’s right, because it is not! In reference to the term ‘Outgoing Longwave Radiation (OLR)’, it is a negative anomaly in areas of deep convection where the cloud tops are COLD and it is a positive anomaly in areas of clear WARM sky’s or shallow (WARM) cloud tops are at.
Ref: Weldon, R.B. and S.J. Holmes (1991). Water Vapor Imagery: Interpretation and Applications to Weather Analysis and Forecasting, NOAA Technical Report NESDIS 57, Washington, DC. (the quintessential water vapor imagery reference)
Jim, a couple of points
1) Ocean heat content also comes from venting at the ocean floor. I have read that up to 20% of the ocean heat may come from venting.
2) On the one hand, you say that once CO2 absorbs radiation, it reemits half back to the surface and half toward space. Yet you also say that CO2 collides with N2 and passes on the energy to N2. I believe that 99%+ of excited CO2 in the lower atmosphere returns to its ground state through collision.
When a photon is absorbed, it ceases to exist. It is not ‘reradiated. When a photon is radiated, the direction it takes is not definable, but from the molecule’s point of view it can only strike earth or go toward space. Because the earth is round more is sent toward space than toward earth, but for all practical purposes, we can say that half is space-directed and the other half is earth-directed.
You bring up a good point even though it is a minor consideration. MODTRAN is a very good tool to learn about IR radiant transfer in the atmosphere, but it has very substantial limitations. Something erroneous within MODTRAN is that the UofChicago version does a radiant intensity calculation on a vertical angle only (units of
) then produces an irradiance value (units of 
) by assuming isotropic radiation — I.e. it only takes the radiant intensity and multiplies by 
. In fact it multiplies by 3.14 and not actually 3.14159265…
Obviously as one rises away from the Earth’s surface the lower hemisphere view contains some of cold space in the periphery and this violates the isotropic assumption.
No, it does not violate the assumption of isotropic radiation, which remains true.
Instead, as noted, the situation fails to consider the change in view factor as a function of distance from an essentially-spherical object. Very near the surface of the sphere, the view factor of the surface is essentially equal to the view factor away from the surface. However, at an altitude equivalent to one radius of the object, the view factor of the sphere is approximately only 7% that of all surrounding space.
The falloff in view factor of Earth with altitude is much more rapid than most people imagine!
The formula for calculating the solid angle subtended by a sphere located a certain distance away from an observer is given by:
Ω = 2π (1 – (√(d² – R²) / d))
where:
Ω is the solid angle in steradians (sr),
R is the radius of the sphere,
d is the distance from the center of the sphere to the external point, and
a full sphere subtends 4*π steradians at its center.
For d=2R, Ω = 2π (1 – (√(3) / 2)) = 0.27π steradians, equivalent to the sphere occupying only about 0.067 of all surrounding space.
Why in the world do you argue with me when you are saying the same thing? View factor and hemispheric view derive from exactly, and I mean exactly, the same considerations. If the proportionality between a differential vertical view and a hemispheric view is the value pi, then a view high above the Earth where some background is cold space with no temperature (MODTRAN sets the temperature of background space to zero K) and some is warm Earth surface will have a proportionality less than pi.
A distance of one Earth radius away from the surface is very far away compared to the sorts of considerations here.
Sorry, I didn’t mean to upset you. I was clarifying that isotropic radiation is a physical characteristic of the individual radiating atom or molecule, reflecting the randomness of emissions as governed by quantum mechanics. Isotropic radiation, per se, is not dependent on a view factor to external objects.
BTW, at an altitude of only 32 km (0.005 Earth radius), Earth would subtend a solid angle encompasses only 45% of all surrounding space, a significant change from the 50% coverage at the surface.
You can’t analyze “heat” molecule by molecule. Isotropic radiation from a single atomic structure is a time function of probility. Think of a die with 360 sides. How many throws and how long does it take to land on every side?
Now think of how many CO2 molecules are in one cubic meter. How long does it take for all points on that volume to receive a heat ray. Is it a short enough period that it is essentially isotropic at any given instant?
Good point on 2). Collisions does take the energy absorbed from the surface and spreads it around to other molecules. However, that process is symmetric. Collisions also provide energy to CO2 molecules which then can emit that energy either upward or downward.
This is a key point which I think climate science had wrong for many years. The energy now emitted has nothing to do with surface energy initially absorbed by a CO2 molecule. It could be energy which entered the atmosphere from evaporation, conduction, water vapor absorption, etc. It no longer has anything to do with surface radiation.
This is now atmospheric energy which eventually might get radiated to space by a CO2 molecule but there is no connection to the original 15 µm surface radiation. Hence, graphs, such as presented here showing changes in that spectral band, really tell us very little when they are based on satellite observations.
What actually happens is energy is constantly being reabsorbed and reemitted low in the atmosphere and passed back and forth from O2 and N2 to CO2 and H2O. A little energy makes it back to the surface, but most of it works it way upward over time due density changes.
Because there is so much water vapor overlapped with CO2, almost no energy makes it out until it gets high enough and the air dries outs. This is where CO2 takes over and moves energy at a constant rate to space. The constant rate is independent of CO2 concentration.
For the same reason why CO2 wasn’t able to re-radiate the energy it absorbed from a photon, but instead lost it to a collision. That same CO2 molecule is much more likely to lose the energy it gained from a collision to another collision before it would have a chance to radiate it.
Beyond that, all the molecules in the area around the CO2 molecule have the same spectrum of energies. Because of this, there isn’t much energy being passed from one molecule to another, because they are all close in energy level to begin with.
When the CO2 atom absorbed the photon, its energy level was temporarily raised above the level of the other atoms in its local group. Once this energy was transferred via collisions to it’s neighborhood, the normal distribution was restored.
I have a different interpretation of decay time due to photon interaction which leads to almost instantaneous emission after a collision. This view supports Kirchhoff’s Law of Radiation and lots of experiments. See Feldman et al 2015 among others.
The decay time is due to the molecule itself being unable to interact with another photon, however, the photon causing the decay is gone immediately.
This view makes more sense when you consider photon absorption. If the decay was due to a delay in energy movement into the CO2 molecular bonds, it wouldn’t be available until the decay time ended. That would destroy the thermalization view. My view has the energy immediately moved into the bonds which also means the photon is long gone. Thermalization is then available. The decay time caused by the photon interaction is internal to the CO2 molecule.
I realize this completely invalidates the Marcus/Ott theory. Turns out it is unnecessary. The reason CO2 does not warm the surface is due to conduction and the 2nd Law.
Looking at it another way.
How far a photon of a particular energy is likely to travel before it hits a molecule of CO2 depends on the density per unit area of CO2 molecules.
Whether the CO2 molecule is more likely to lose that extra energy via radiation or collision depends on the density of other molecules in that same unit of area.
In the lower atmosphere, the distance that this photon can travel is measured in just a few meters. Doubling the CO2 concentration just shortens this distance by a bit. Since a few meters is just a tiny fraction of the many kilometers that the photon would have to travel to reach space, the exact concentration of CO2 here in the lower atmosphere is almost inconsequential.
Where CO2 makes a difference is in that region where air density is starting to drop low enough that a CO2 molecule that gains extra energy (whether by collision or absorption doesn’t matter), has a non-trivial chance of losing that extra energy by emission rather than collision. In this region, the question of whether that photon is emitted towards space or towards the planet becomes relevant.
A photon emitted towards the planet will get reabsorbed.
A photon emitted towards space has a chance of escaping.
In this region, if you increase the concentration of CO2 molecules, then you decrease the odds of a photon emitted towards space, making it to space without being absorbed again.
15 µm surface radiation only applies to particles travelling at the speed of light. Bonded molecules travel much slower, where heat is divided by kelvin temperatures.
198597/713(constant pressure heat capacity) = 278.5K
You’re right. At 1 atm, it’s 99.997% return to ground state by collision.
In the lower atmosphere.
That percentage goes down as the air pressure lowers.
True. But collisional decay still vastly dominates throughout the troposphere.
Nelsen did you read the article. My main point precisely revolved around the fact that ” CO2 collides with N2 and passes on the energy to N2.”
I stated, “NEAR THE SURFACE, NITROGEN MOLECULES CAN COLLIDE WITH CO2 6 BILLION TIMES IN JUST ONE SECOND. SO, BEFORE A CO2 MOLECULE CAN RELAX AND EMIT ANY RADIATION, NITROGEN WILL COLLIDE 600,000 TIMES WITH CO2 AND STEAL IT’S ABSORBED ENERGY AND TRANSFER IT TO OTHER ATMOSPHERIC MOLECULES VIA MORE COLLISIONS”
Jim, I read the article. I totally agree with the collisional analysis. My point is that if you believe in the collisional story, the greenhouse story falls apart.
I said that at least 25 years ago. As a laser engineer, and weather instrument engineer, I had to learn about these things.
The one thing I didn’t see was the cooling of the atmosphere by the radiation from the surface, especially at night. Surface radiates and cools. The atmosphere at the surface cools with it. The longer that goes on, the deeper the cooler air pool becomes. This is cooling via conduction with the surface. It is simply the opposite of that atmosphere being heated by the surface and convecting aloft. That happens fast, with the sun providing the energy, but the cooling is slower, since it is by conduction only. There is no opposite of convection at night, so it cools until the sun returns. Seen over a day, it looks like a sawtooth – fast rise, slow fall.
I agree entirely except that there is a sort of convection at night when radiation cooling at the surface produces a katabatic wind which stirs the atmosphere and produces deep cool layers especially down the surface gradient into river valleys.
Dew point is a large factor here too. As the H2O vapor condenses, it releases heat to the surrounding air.
From Curtis & Goody, 1956, p. 196: “The collisional relaxation lifetime for CO₂ in atmosphere of nitrogen at 220 K is about 15 microseconds. The radiative decay lifetime is 0.43 sec. For CO₂, the ratio of (radiative lifetime)/(collisional lifetime) = λ/φ = 3.4×10⁻⁵ at 220 K and 1 atm pressure.”
Page 199: “The height at which λ = φ is about 74 km.”
The H₂O vibrational lifetime is about 0.06 sec., making the emission/collision ratio about 2.4×10⁻⁴.
You’ve got it right, Jim, collisions dominate energy transfer in the troposphere, not radiation. Radiative transfer is negligible.
I’ve done a first-pass calculation of the Boltzmann concentration of collisionally excited CO₂ in a 1 sq.m column of the atm to 11 km. and the corresponding intensity of spontaneously emitted 15μ IR. The collision/emission decay ratio is ~29000 (1 atm) is scaled by the number ratio at altitude.
Total IR back-radiation sums to about 0.25 W/m².
The radiative transfer model is wrong.
Pat, what you say would be so if the transfer from CO2 to the diatomic molecule could take place only on a one-way street. The totality of all transfers through a plethora of interactions, however, maintains a Boltzmann distribution of CO2 levels, along with all others under the principle of detailed balance.
I see Richard M, nearby, says the same thing in a different way.
I calculated the Boltzmann populations. A collisionally excited CO₂ molecule will decay collisionally 99.997% of the time (at 1 atm). Radiative decay is invariably negligible.
Pat, I believe you are using an incorrect view of decay times. I believe the actual photon emission occurs nearly instantaneously during the collision itself. The decay just prevents another photon interaction.
‘I believe the actual photon emission occurs nearly instantaneously during the collision itself.’
????
You have a collision of CO2 with another molecule.
The energy gained by the CO2 molecule is enough to cause the emission of a 15 µm photon.
Pat (and many others) believe the molecule must wait for .43 sec before the photon is emitted and hence the energy can be (and usually is) lost in another collision. No actual emission occurs.
I believe a photon emission occurs immediately and the .43 sec is the time it takes before the molecule can interact with another photon.
Curtis and Goody are unambiguous, that 0.43 sec is the lifetime of the excited state, before the spontaneous emission of a 15μ photon occurs.
For example, p. 196: “The important factor in the equation of transfer is the ratio λ/φ which for both water vapour and carbon dioxide appears to be about 3.4 x 10⁻⁵ at 220° K and 1 atm pressure. Deviations from the Boltzmann distribution will start to be important when this ratio is unity, i.e. at a pressure of about 34 dyn cm⁻² or a height of about 74 km, the uncertainty in this last figure being of the order of (+/-) 10 km.”
λ is radiative lifetime; φ is collisional lifetime.
Other sources, which I don’t have handy just now, have given alternative CO₂ radiative lifetimes of 0.74 sec or 1.1 sec.
Emission decay time is a property of vibrationally excited CO₂, and is a constant; independent of pressure.
Collision doesn’t induce a simultaneous emission of a photon. The CO₂ vibrational energy is converted to kinetic energy, which is likely partitioned between the two colliding molecules.
Pat, you should write a WUWT article on this subject. Or are you already working on another paper?
Thanks, Thomas. I’m working on another paper. I hope my last concerning climate.
Well, I for one, hope not. 😆
You’re too kind, Tom. But lots of other stupid ideas call for attention.
The relaxation time constant comes about from photons being a quantized emission of energy. A vibrating CO2 molecule acts as an antenna and 0.43 seconds is the amount of time needed to emit the energy in a photon from Maxwell’s equations. Where the weirdness in quantum mechanics comes in, is that the probability of an excited CO2 molecule emitting a photon in a 1 millisecond interval is 1/430 (0.43 seconds is equal to 430 milliseconds). This exactly the same math as radioactive decay.
Try this:
1) CO2* + N2 ==> CO2 + N2(+V)
[aka non-radiative deactivation] => lower troposphere
or
2) CO2 + N2 ==> CO2* + N2(-V)
[aka non-radiative excitation], => upper troposphere
Where:
+/-V means higher / lower velocity (kinetic energy)
Most radiative lifetimes are much shorter. The 15 μm CO2 ban has an unusually long radiative lifetime that has been measured by many different groups. This gives it some unusual properties that are interpreted by many as magical powers (sarcasm intended.).
And the long “radiative lifetime” is why CO2 makes for great lasers. It wasn’t until I started pondering about the radiative lifetime that I understood what “stimulated emission of radiation” was all about.
Trapping heat maybe?
I checked Feldman et al 2015. Here’s what they claimed:
“The time series both show statistically significant trends of 0.2 W m−2 per decade”
This was for a CO2 concentration increase 22 ppm/decade. That is close to 5% of the total CO2 in atmosphere. Even a linear trend would lead to just under 4 watts/m2. Isn’t this experimental falsification of your number?
How is it possible to measure a trend of 0.2 W/m^2/decade = 0.02
W/m^2 annually, when the outgoing long-wave at the TOA isn’t known to better than (+/-)3.3 W/m^2?
In addition, assuming CO2 forcing directly causes warming assumes all other parts of the climate remain unchanged. Not believable.
This is “back-radiation” measured at the surface. They are just reporting what the device has measured.
‘They are just reporting what the device has measured.’
What device(s)? If I recall correctly, many IR instruments, e.g., cameras operate within the so-called atmospheric window and detect IR wavelengths between 8-14 microns. Obviously, this isn’t scary ‘back-radiation’ from CO2 (15 microns), but rather IR absorbed and emitted by dust, ice, water droplets and other forms of condensed matter in the atmosphere.
Unfortunately, in order to ‘see’ the scary back radiation from CO2, one needs a detector that that is cooled below any temperature that exists on earth, which, at least to some folks, seems a bit disingenuous given that heat flows from hotter objects to cooler objects. In other words, does back radiation from CO2 actually exist, or is it a manifestation of the instrumentation?
“Atmospheric Emitted Radiance Interferometer (AERI) instrument and atmospheric state data at these two sites.”
https://escholarship.org/content/qt3428v1r6/qt3428v1r6.pdf
I believe this device meets your specifications.
‘I believe this device meets your specifications.’
If, by ‘my specifications’, you mean it’s crap science, then yeah, it does.
Feldman et al is alarmist hand waving at its worst. First of all, they invoke radiant transfer models to validate the use of radiant transfer models, which is obviously circular reasoning.
Second, as Pat Frank correctly stated, above, there is absolutely no way they could have obtained meaningful results given that the uncertainty of the their data vastly exceeded their computed trend.
And finally, as I alluded to above, using a cryogenically cooled IR ground sensor means that the so-called back radiation from CO2 most likely originated within several feet of the sensor, i.e., not from the upper reaches of the troposphere. In other words, they’re effectively just measuring air temperatures in the immediate proximity of their instruments.
Note:
‘Two detectors are used, an HgCdTe and an InSb, cooled to cryogenic temperatures, to cover these spectral ranges.’
https://www.ssec.wisc.edu/aeri/
“In other words, they’re effectively just measuring air temperatures in the immediate proximity of their instruments.”
Took the words right out of my mouth, Frank. 🙂
I believe the models are only used to remove the energy coming from water vapor. There’s no way to identify the photons arriving at the sensor exactly. They could be coming from either CO2 or H2O. It is not circular logic. As long as they are doing it correctly, I have no problem with this technique.
The uncertainty mentioned by Dr. Frank does not apply to these measurements.
Don’t get me wrong. This doesn’t do anything to support AGW as they have claimed, however, I believe it disproves the claims that CO2 molecules will not emit IR radiation near the surface.
Knowing the “back radiation” value IS measuring the temperature of the emitter. If it can be isolated to one species of molecule, i.e., CO2, then you can know the temperature of that group of molecules. Otherwise, it is only a compilation of various molecules.
The next question that needs to be known is what the temperature of the surface is. If it is warmer than the “back radiation”, the “back radiation” cannot heat the surface. It may increase the time portion (1°/sec to 0.9°/sec) of the cooling gradient, but it cannot change the sign of the gradient.
There is already energy transfers going on at the surface – atmosphere boundary working to remove any temperature difference. Conduction never stops and is driven by surface collisions by all atmospheric molecules. All that happens with this radiation is a slight change in the amount of conducted energy moving one way or the other.
From the Methods section: “Spectra observed by the AERI instrument channel 1 were analysed over the period from 2000 through the end of 2010 at the ARM SGP and NSA sites.”
SGP (Southern Great Plains) and NSA (North Slope Alaska) are both ground-level sites. Feldman, et al., were measuring surface radiation intensity.
Yes, that’s how you measure back-radiation. Almost all the IR which reaches the surface in the near 15 µm bands originates from a few meters above the surface.
The IR is coming from the same part of the atmosphere you stated would be unable to emit more than insignificant IR. Is experimental proof your claim is wrong? Appears that way unless the methodology is wrong.
“The IR is coming from the same part of the atmosphere you stated would be unable to emit more than insignificant IR.”
What does that sentence mean?
The warm surface emits IR with the ~288 K Planck intensity distribution. That has intensity at 15µ. If one points a ground-level AERI spectrometer at the sky, the measured radiation is the downward ~50% of the local isotropic IR radiation.
Jim, you say oxygen and nitrogen do not absorb or emit radiation. This is wrong. They do, maybe “weakly,” and maybe not at wavelengths relevant to atmospheric warming, but making an obviously wrong statement like this about basic physics weakens the credibility of whatever else you say. Please me correct!
I think the only clarification needed is to say that oxygen and nitrogen don’t absorb or emit in thermal IR bands. This eliminates the microwave emissions of oxygen from consideration, which then leaves only Raman emissions to consider which I don’t think are of any significance.
Oxygen and nitrogen are pressurised gases, They have a temperature in the thermal IR (15 microns), Mass size is what determines total heat or emission at higher thermal energy band. Energy per particle 1.3 x 10^-21 (higher than 1.38^-23 ideal gases). At surface what does these particles collide with most. The ground. The ground is heated by the sun. Pretending a gas (CO2) with less mass does all the work is stupid.
Where’s Michael Flynn when you need him?
/sarc
Frank – right here.
Yes they can. You are making stuff up, and cannot produce a single reproducible experiment to back up your statement.
Sorry about that, but I’m right, aren’t I?
You Michael are making stuff up! Again just show the emission and absorption spectra of O2 and N2 to prove your not retarded
I’ll repeat – You are making stuff up, and cannot produce a single reproducible experiment to back up your statement.
Sorry about that, but I’m right, aren’t I?
Duck and weave all you like. Just read Tyndall’s experimental results. He compared the IR absorption of gases like CO2 and water vapour, to the absorption of a mixture of O2, N2 etc (dry air purged of CO2 and particulate matter. Yes, O2 and N2 absorb and emit IR.
You don’t have to believe reproducible experimental support if you don’t want to. Believe in fairytales if you wish – I don’t mind at all.
Perhaps I should have been clearer. Oxygen and nitrogen do not absorb or emit radiation at the typical temperatures of earth’s climate system normally >3-4 microns. Oxygen does emit nanometer wavelengths but that is irrelevant to this article.
Jim, that’s just nonsense, unless you have reproducible experimental support for your breathtaking assumption.
All matter above absolute zero emits radiation at frequencies dependent on its absolute temperature. No exceptions.
Again just show the emission and absorption spectra of O2 and N2 to prove your not retarded
Jim, go on then – show your emission spectra due to temperature. Show a temperature at which O2 and N2 do not emit IR, and explain why they remain at absolute zero (emitting no IR at all – the definition of 0 K) whilst surrounded by hotter gases.
Feel free to have a tantrum if you can’t answer.
From Google AI: For a molecule to absorb or emit IR radiation effectively, there needs to be a change in its dipole moment during vibration. A dipole moment arises from an uneven distribution of charge within a molecule. Because O2 and N2 are symmetrical, stretching or bending their bonds does not create a change in their dipole moment. This prevents them from interacting with the electric field of IR radiation.
There is an exception, they are called gas molecules!
All matter above absolute zero emits IR. Show me a reproducible experiment which demonstrates otherwise, and I’ll gladly change my view.
But you can’t, can you?
I thought so.
What you posted was:
“All matter above absolute zero emits radiation at frequencies dependent on its absolute temperature. No exceptions.”
Not true in the case of gas molecules as I pointed out!
The figure that Jim supplied, 25% of surface heat is carried away in frequency bands not absorbed by greenhouse gasses, varies quite a lot depending on surface elevation and local humidity profile. In the tropics near ground surface the atmosphere is about 10-12% transparent. Where I live, at 2200m AMSL, the humidity is low and space not far away and 30% or more of that radiation passes directly out the transparent bands. Elevated and dry portions of the Earth play an important role in cooling the surface, just as do regions with descending dry air above such as deserts and even ocean surface in the subtropics.
Jim’s description of the process shown in the graphic from Stephens et al, explains the process for one iteration. Left unexplained is that this process repeats endlessly for all of the radiation sent downward and absorbed again by the surface. When absorbed by the surface it is “thermalized” to use a common term, losses its identity as to frequency, and some fraction, 25% in this case, travels transparently to space.
This is even true for the energy stored temporarily in nitrogen and oxygen molecules.
oxygen and nitrogen can lose energy also by collisions with the ground surface. Such energy transferred is thermalized again and some of it passes directly to space in whatever bands are transparent. It is an important process explaining how air near a ground surface cools rapidly after sunset and produces such things as a katabatic wind. The bigger point is that much of the Earth’s surface is warmed and cooled over different portions of the day/night cycle by whatever process dominates during each part of the cycle.
The topic of heat transfer in the atmosphere is so complex that no matter how carefully one tries to explain it, and this essay is very good, a person always falls a bit short of a full explanation.
Indeed Kevin, Climate complexity requires lengthy discussions.
My intent was to explain in about 6 minutes a critical dynamic showing how CO2 cools the atmosphere so that the average Joe appreciates CO2’s cooling effect. 99% of the pubic has not heard of its cooling affect because the narrative pushes a warming crisis.
To keep it short and concise, I cut out the fact that any warming of the surface by a singular wavelength gets radiated away over a full spectrum of wavelengths by the surface, meaning nearly 25% escapes via the atmospheric window.
I asked two simple questions to limit the frame this discussion to help explain CO2’s cooling effect: 1) How does N2 and O2 cool without radiating the absorbed heat away & 2) why don’t satellites detect wavelengths around 15 microns until the tropopause and stratosphere? Yet I haven’t seen any criticisms dealing with those questions. Instead all the comments are people pushing their own pet peeves and fail to address th main points.
Jim, the article accomplishes exactly what you intended it to accomplish. The difference between collision time and relaxation time is a key point that I had not read about before. I always enjoy your posts, and I always come away better informed. Well done. Keep it up.
In the tropics near ground surface the atmosphere is about 10-12% transparent (wrong). Tropics gets 120°C per second coming in. At 70hpa the temperature is -73°C. Dry air ground would heat to 47°C reduces to 37°C. This means the solar 120°C escapes, 90% atmospheric and ground heat remains. Ocean at 23-25°C + -73°C means 82% heat already present. Solar adds again and goes out. You ignore 10 x ^25 particles keep motion (emitting heat) when solar isn’t present. And think radiative process (greenhouse effect, otherwise absolute zero at night) only is how the atmosphere works.
Wow, that makes no sense at all…..
This article reminds me of a Star Trek scene..
https://www.youtube.com/watch?v=v4fU0Ajo4RM
Jim, you are riding a dead horse! Whatever you think those exchanges on a molecular level, be it collisions or radiation, would mean, whatever you interpret into it – in reality it does not matter.
The article is just full of mistakes. Most significantly you do not understand the GHE. It is a very simple concept that requires
a) an optical thickness of the atmosphere and
b) a (adiabatic) lapse rate
And that gives you this:
That is all. Thinking about how molecules bump into each other, how they isothermically exchange radiation is great to occupy your mind, but it leads nowhere. You might just as well deal with the hen-egg problem. For the same reason “back radiation” does not do a thing and is irrelevant.
What CO2 does after all, is to elevate the emission altitude, which with a given lapse rate, will yield higher surface temperatures. More CO2 thus WILL cause warming. That is something to be understood in the first place, and only then one could start discussing the issues with the theory, of which there are plenty and serious ones.
“What CO2 does after all, is to elevate the emission altitude, which with a given lapse rate, will yield higher surface temperatures.”
That conclusion assumes the radiative transfer model.
Collisional transfer dominates in the troposphere. It’s not clear at all (to me, anyway) that the emission altitude increases under a collisional transfer mechanism.
Dominates WHAT? The transfer of energy? In an isothermic environment, which the atmosphere basically is on a micro level, there is no transfer of energy. Whether molecules pump into each other, or exchange radiation, it does not do anything!
I know there is all kind of little cognitive challenges in (climate) physics, but they are really not hard. One should be able to master dozens of them flawlessly. And then you can put the pieces together to see a bigger picture.
But damn it, how do you manange to fail on every single one..
“Dominates WHAT?”
The transfer of energy initially absorbed from the IR radiation emitted by the warm surface.
Obviously, a warm radiant surface is inconsistent with your offering of an isothermal system. Surface IR, after all, is the context of the discussion here.
Thank-you for expressing immediate disdain — a dead give-away of someone incapable of civil debate.
???
This is super simplistic. You can already see it in the “Earth energy budget”.
There are 356W/m2 going from the surface into the atmosphere, and 333W/m2 going from the atmosphere onto the surface. The figures not quite correct and actually they would be even closer by. Yet it is evident enough.
It is just a (largely) isothermic exchange of radiation. It is not doing anything. The same thing happens in your body. The molecules bump into each other and exchange radiation, about 1-3MW if the AI guesses right. If that would do anything, you’d either burst into flames, or turn into a popsicle.
Whether it is collisions or radiation “dominating” the non-transfer of energy, will not matter. It is still irrelevant.
Any rational human being would see this, check the box and move on. You however get totally lost in confusion over it..
Schaffer must be getting paid to push this misinformation. No one fails physics this badly.
Schaffer say “there is no transfer of energy. Whether molecules pump into each other, or exchange radiation, it does not do anything!”
Oh my!
Are you getting paid to play stupid?
“The molecules bump into each other and exchange radiation”
No, they don’t. They exchange energy. Radiation is the propagation of energy through space.
Your graphic doesn’t gainsay that collisional energy transfer dominates in the troposphere.
LOL Schaffer assumes the earth is in isothermic exchange when such thermodynamic equilibrium requires uniform temperature and pressure everywhere, no net flows of energy or matter, which are conditions impossible on a dynamic Earth.
Schaffer is so ignorant!!!
The altitude of the emission layer is determined by the collisional deactivation frequency. Emission to space can occur when the frequency of collisional deactivation is less than the rate of spontaneous emission, which one can calculate based on the Einstein A coefficient.
Right. I got that from Curtis & Goody, 1956. But a question to you, Tom: Is the total integrated power intensity emitted from earth a constant, whether CO2 is present, or not?
If yes, when CO2 is present, does the Planck distribution of spectral frequency shift to longer wavelengths?
A great question. Since there does not exist an instrument that can measure the total integrated power intensity emitted from the Earth, the answer is speculative. To clarify, I assume your question refers to a scenario where all other conditions are fixed, and the only thing that changes is whether or not CO2 is present.
The IR-active gases are simply energy converters. They convert radiation into sensible heat near the surface, and they convert sensible heat into radiation where it can escape to space. In the space(s) in between, there is a flurry of collisional excitation and de-excitation, and (relatively) occasional spontaneous emission which is quickly absorbed. There is no energy transport via radiation as a result of this process. The energy transport is via bulk transport of convection.
There is also no mechanism for them to “store” or “trap” heat as is often claimed. The CO2 “notch” in the spectrum leads to this interpretation, but it is not “trapped” heat. It is simply a fraction of the radiation from water vapor being absorbed by CO2 and immediately converted back into sensible heat which is equipartitioned quickly to all molecules. That will continue to drive convection and eventually collisional excitation that will enable radiation to space.
Some might argue that this “must” change the radiative energy transport in some way. Perhaps it does in some subtle way that we do not understand.
Given the myriad of energy conversion processes that take place in the atmosphere, however, I don’t see how it can make a significant difference.
As I alluded to in another comment, there is far too much inferred from the one-dimensional radiative transport models that are based on something with no resemblance to Earth excepting the atmospheric temperature and composition profiles. These are quite complex boundary conditions and can be adjusted to produce any spectrum you want.
All wrong. The lapse rate (whether the “dry” one or the “wet” one) is a metric based on measured values of temperature versus altitude. The lapse rate, being a metric, is incapable of transferring energy/heat. Therefore, to the extent the “emission altitude” (whatever that is supposed to mean) is raised, if a temperature is associated with its definition, it will likewise raise the altitude over which the lapse rate is calculated . . . leading to an effective decrease in deg-F change per unit altitude increase . . . meaning actually resulting in a lower lapse rate.
The lapse rate also depends on the Cp of air. The Cp changes as the WV or CO2 increases. All in all a good post.
No, as above, elevating the emission altitude per se will increase the GHE.
Sorry, this is not true. The emissions altitude of CO2 is constant and just like the lapse rate, set mainly by gravity. This error is based on assuming a constant upward flux of surface energy by CO2. It is derived from assuming surface energy emissions are mostly constant (based on temperature) and remain constant as they move through the atmosphere.
Thermalization of the surface emissions destroys this view. The upward flux of energy is no longer tied to surface emissions. If you increase CO2 you do get increased absorption, but you also get increased emissions. All atmospheric energy is now available. Turns out Kirchhoff’s Law keeps two processes in sync. The energy flux moves slower due to increased absorption, but there’s more energy in the flux due to increased emission. They cancel out keeping the net flux constant.
..and by the concentration of CO2 of course
Again, seems hard to figure out what you meant when writing these words..
No, it is NOT set by the concentration of CO2. That is my point. The assumption that dZ is not zero, which drives the change in the emissions height used in all your diagrams, is not true.
I already explained why which you appear to not understand. Until you figure it out you will remain confused.
Looks more like the “lapse rate effect”. Is it the intercept that changes or the slope? And why?
What you mean with “lapse rate effect”?
Schaffer for years you have ridden ypur dead horse from which you try troll every discussion. At first I thought you were totally ignorant but I come to believe you are a malicious troll spreading misinformation. Do you get paid?
Yes, the oil industry is paying me millions to teach fundamental climate physics, so that the “critical side” might eventually find a leverage to falsify the science.. *facepalm*
If you were somewhat familiar with the literature, let alone if you had reflected on it, I would not be telling you news. You can have the same story from many reasonable sources, like..
or this..
or this..
And btw., whenever I put up something reasonable and thought provoking, you guys are like naaah, you will not provoke us to think! Well, ok..
Again assuming the radiative transfer model, when the troposphere is dominated by collisional transfer.
You are simply proving that the so-called GHE is a mathematical artifact of the claim that radiative transfer theory can describe the dynamics of the Earth’s atmosphere. There is nothing wrong with the mathematics, and the RTE can provide a reasonable facsimile spectrum as shown by Harde (2013).
The most egregious example of this can be found in van Wijngaarden and Happer (2023). Their hypothetical “Gray Atmosphere” according to the mathematics of RTE produces a troposphere height of 16.6 km and a surface temperature increase of 162 K! According to that the surface temperature for “radiative balance” would be 440K.
They comment, “The simple “one-dimensional” model of a gray atmosphere is only semi-quantitative, but it leads us to a number of important insights on how greenhouse gases work.”
Page 20 of https://wvanwijngaarden.info.yorku.ca/files/2023/03/GreenhousePrimerArxiv.pdf?x45936
I do not quite know why they would assume an average emission altitude of 16.6km. It is just about 5km. And then of course we have a moist-, not dry adiabat.
It is explained in their section the Gray Atmosphere, so I’m inclined to assume either you didn’t read the whole section or you didn’t understand it.
perhaps you need to read what precedes the single page that I referenced.
CO2 has no role in the troposphere. Nitrogen and oxygen have enough mass (43 moles) to equal solar input with 15 micron energy per particle at speed of light.
Carbon dioxide itself does not. Only fantasy models think up magic for CO2 to do.
Air molecules power 1 bar pressure.
28.96g x 43.8 moles = 364 J x 278.5K = 1 bar pressure
278.5K x heat capacity 713 = 198597 (1.4 x 1.268 x 334.4 m/s).
Dry air lapse rate 9.8°C per km is the difference between hot desert and cool ocean.
Wet air lapse rate 6.5°C per km means heat at hot desert is hidden in the moist air from the ocean (only living things feel, not heating the air to desert temperature).
Don’t know what you want to say. But yes, CO2 has a role..
Wow, that makes no sense either.
Johnny! We have a winner! . . . for the most absurd mashup of science ever posted on WUWT.
You’ve accurately summarized the ‘canonical narrative’ of climate alarmism. Unfortunately for the alarmists, the satellite data of the past 20+ years has indicated that the slight increase in GAST (global average surface temperature, whatever that is) has been accompanied by commensurate increases in OLR (outgoing long-wave radiation) and ASR (absorbed solar radiation), a ‘surprising paradox’, to be sure. Not to worry, though, climate alarmists never let the facts get in the way of a ‘good’ narrative.
If you want to criticize a book, or a film, you should read it, or watch it respectively, first. I am one of the crazy, stupid people believing that is how it works.
What you people do not like is taking that one first necessary step. You want to falsify “the science” without knowing about it. It is NEVER going to work, just useless..
Only where “all other things” are “held equal.”
Which they have never been, are not, and will never be.
From the first paragraph of the above article:
“Oxygen and nitrogen comprise 99% of our atmosphere. Those molecules do not absorb or emit radiation.”
Sorry, Jim, that statement is incorrect based on fundamental physics. The Stefan-Boltzmann Law states, and gives equations governing, the fact that all matter above absolute zero temperature emits thermal radiation. The characteristic spectrum of that radiation can range from (theoretical) blackbody, to greybody (emissivity <1.0), to relatively widely-spaced spectral bands depending on the specific substance involved. Metals behave differently from solid non-conductors which behave differently from liquids which behave differently from gases, but all with a temperature above 0 K (and, importantly, when considered as an grouping of atom and molecules, not as an single isolated atom or molecule) will emit thermal radiation.
In addition, Kirchhoff’s Law of Thermal Radiation states that a good emitter of radiation at a given wavelength is also a good absorber of radiation at that same wavelength.
There is a common misconception that non-LWIR-active gases (often referred to as non-greenhouse gases) cannot and do not absorb or emit IR radiation (which is a part of the EM spectrum encompassed by the term “thermal radiation”) because they don’t have a permanent or vibrationally-inducible electric dipole moment (such the situation with CO2). And that, of course, sounds reasonable as presented.
While this is “somewhat true” as the first step is evaluating energy exchange in Earth’s atmosphere, it neglects two very important—but generally less recognized—physically processes:
(a) molecule-molecule collisions are occurring in Earth’s low atmosphere of a rate in the range 10^6 to 10^9 per second (as a figure your article presents indicating “CO2 Collision Frequency” summarizes), and since 99% of those colliding molecules are either N2 or O2 (both non-LWIR-active) those collisions extremely rapidly equilibrate all LWIR energy absorbed by GHGs (predominately water vapor, followed by CO2) throughout the entire atmosphere, and
(2) when any two molecules (polar or non-polar) undergo a collision, that collision induces a momentary dipole moment in both colliding molecules due to the momentary acceleration in the electron clouds surrounding each atom. It is the acceleration of electric charge that produces EM radiation, in accordance with Maxwell’s equations.
And note that even N2 and O2 should theoretically be able to absorb some LWIR directly from Earth’s surface at times of molecule-molecule collisions, but I expect that the time of the resulting induced dipole moment is so short during each collision that this potential is actually realized as insignificant, despite the very high rates of collisions . . . at least I believe that is confirmed by MODTRAN/HITRAN.
So, bottom line, by the time one reaches the top of the atmosphere, all of the atmosphere (predominately N2 and O2, of course) is radiating the amount of LWIR energy that was originally absorbed only by greenhouse gases, with each species having its own spectrum/bands of emissions, generally covering the near-IR down to radio frequencies.
And as a final note, since each atmospheric molecule radiates thermal energy isotropically, this also clarifies that the “greenhouse effect” of thermal radiation of energy back toward Earth’s surface also results from the entire atmosphere (being thermally-equilibrated, recognizing bulk temperature varying as a function of altitude), not just from those molecules of greenhouse gases.
Thank you. Thank you. A million times, thank you. I am so sick of hearing the moronic canard that non-greenhouse gases do not radiate any energy. It is a physically ridiculous statement.
Your point #2 above is especially well taken. Momentary dipoles are critical for understanding how hot bodies radiate away energy, and this obviously applies to all matter – non-greenhouse gases included.
The amount of IR radiation from N2 and O2 is small enough that it can be approximated as zero for IR as single element diatomic molecules usually don’t have a significant electric dipole moment associated with energy levels in the IR region. Water has a larger electric dipole moment than CO2, hence the shorter relaxation time.
It appears that your missed the point upthread that “when any two molecules (polar or non-polar) undergo a collision, that collision induces a momentary dipole moment in both colliding molecules due to the momentary acceleration in the electron clouds surrounding each atom.”
In Earth’s atmosphere, those “single element diatomic molecules” (more simply, N2 and O2) isotropically radiate a MASSIVE amount of thermal energy due to their high collision rates and their temperatures. That’s a major contributor to explaining why Earth’s surfaces in equatorial and temperate latitudes don’t freeze over during the night.
LOL TYS, You demonstrate why a little bit of knowledge is a dangerous thing!
The collisions DO NOT affect radiative heating. Equatorial surfaces remain wam due to water vapor’s greenhouse effect in the lower atmosphere. The tropical heat that is transferred to N2 and O2 can only be shed by colliding with greenhouse gases that emit the heat back to space and that depends on CO2 in the upper atmosphere where water vapor is greatly reduced. Because the balance between non-greenhouse gases to greenhouse gases than can radiate heat back to space is so lopsided, cooling of the tropical atmosphere is slowed by the cooling chokepoint created by the relatively small concentrations of CO2, NOT DUE to RADIATION RELEASED by COLLISIONS
Jim, please tell me that, by that statement, you are NOT asserting that N2 and O2 at the temperatures they are in Earth’s atmosphere (which are, of course, are above absolute zero) do not directly emit thermal radiation.
Again TYS, just show the emission and absorption spectra of O2 and N2 to prove you too are not retarded
Jim,
WOW! . . . you should know that in the past I honestly had a higher regard for you than that statement engenders.
So, to evidence to you—realizing of course that you seem ill prepared to accept a “proof” based on the totality of your posts under your article—that I am not retarded to the best of my knowledge, I offer these items is response:
1) You yourself have now provided the statement to show all WUWT readers the insightful wisdom of Socrates who is reported to have said “When the debate is lost, slander becomes the tool of the loser.”
2) I posted here on WUWT (see below) at a posting time of 5:22 pm August 21, 2025 a reply to your request for emission and absorption spectra information over the range of 3 to 200 microns wavelength, including reference to a spectrum plot showing an absorption peak for O2 within that range. The other references I provided have text that provides absorption peaks for N2 within the same range of wavelengths.
It is revealing that more than 3 hours later, with your post above at 8:36 pm, you are repeating that same request for emission and absorption spectra. As a lawyer might say, “ASKED AND ANSWERED!”
Still no relevant absorption or emission spectra, but TYS insists on repeating blah bah blah
You can continue making that statement every 23 minutes or so for a long as you wish . . . but the fact, for all WUWT comment readers to clearly see, below is that I provided the equivalent of the information that you requested, what? . . . now more than 4 hours ago in a reply that I made directly back to you.
Actually, the Earth’s atmosphere stays warm at night everywhere because the rate of energy transport is governed by convection, with a speed about 9 orders of magnitude less than the speed of light.
Convection is ignored in the radiative transfer models because, as Schwarzschild himself said, “Radiation equilibrium will occur in a strongly radiating and absorbing atmosphere, in which the mixing effect of ascending and descending currents [convection] is insignificant compared to heat exchange by radiation.”
Unfortunately, his admonition continues to be ignored.
Collisional contact is picoseconds. If the instantaneous collisional dipole permitted radiation, the photon could carry off only the energy of collision. That would leave each molecule bereft of K.E.
The mechanism doesn’t seem plausible to me.
Please keep in mind that the a priori assumption is that one or both of those colliding molecules had been previously “energized” above the average Boltzman energy distribution curve by virtue of have collided with, and received energy, from LWIR-energized greenhouse gas molecules. LWIR photons transfer a “punch” compared to the normal molecular energies associated with the mechanical modes of diatomic molecular bond stretching, bending and rotation for gases over the range of atmospheric temperatures.
At a surface air temperature of 288 K, about 5% of CO₂ molecules will contain vibrational energy equivalent to having absorbed a 15μ photon.
15μ IR from the surface is absorbed within about 90 m. After that, it’s virtually all collisional transfer.
(I am trying to follow your comment but have a large gap in my limited study of physics.)
Question: If we compared a cylinder of 100% N2 versus a cylinder of 90% N2 and 10% CO2, both at the same temperature and pressure, would we observe a difference in the EM emissions of the two cylinders? (Assuming the glass was not opaque in some band that we really wanted to observe.)
Answer: Yes, as you’ve defined the comparison and assuming both cylinders are exposed to identical amounts of incoming LWIR radiation.
By itself (the 100% case), nitrogen does not significantly absorb LWIR, and so would not gain additional energy to re-emit (above the base thermal radiation associated with its given temperature and density, independent of incoming LWIR).
In comparison a 90% N2–10% O2 mixture would have the CO2 absorbing some of that incoming LWIR radiation, which would then be equilibrated throughout the mix due to collisional energy exchanges while at the same time those molecule-molecule collisions would enable both N2 and CO2 molecules to radiate away as thermal emissions some of their received additional energy (that is, both the CO2 and the N2 would consequently be radiating more energy than the base thermal radiation associated with just their temperature and density in the absence of any incoming LWIR).
Keep in mind that in terms of “observing the EM emissions”, the thermal radiation from all matter having temperatures above absolute zero is governed by the laws of quantum mechanics (i.e., is statistical) such that even for room temperature gases the spectrum for “thermal” emissions can range basically from near-IR all the way to radio frequencies and lower. That is, the thermal emissions for the 90% N2–10% O2 mixture case will be spread out, not necessarily concentrated only in the frequency range of the incoming LWIR.
Actually, a mixture of N2 and CO2 will create its own radiation field as a result of collisional excitation. There will be no net energy flow in or out of the system, but a self-generated isotropic radiation field will exist. The same is true in the Earth’s atmosphere from the surface to the mesopause. This is explained in Einstein’s “Quantum Theory of Radiation” (1917), and discussed in detail in Harde (2013).
LOL, To all the idiots that object to the fact that O2 and N2 do not absorb or emit radiation between 3 and 200 microns which are the critical wavelengths for the earth’s cooling. Ozone does interact around 10 microns. Oxygen does in the microwave and ultraviolet wavelengths but these are insignificant and irrelevant for earth’s cooling radiation.
So will the stupid trolls please show me the emission and absorption spectra for O2 and N2 and avoid all the irrelevant theoretical BS.
Sorry Jim, you are delusional. You can’t back up your assertion with experiment, can you? All matter above absolute zero emits infrared radiation.
There is no greenhouse effect, and as a matter of fact, no consistent and unambiguous description of this mythical creature can even be found.
As proved multiple times you are the delusional one! Here’s a line spectra comparing O2 and CO2, note the intensity is plotted as log10.
Phil, that’s meaningless, isn’t it?
I suppose you are implying the Earth hasn’t cooled to its present temperature, and that the surface doesn’t cooled at night, are you?
Accept reality, laddie.
“Phil, that’s meaningless, isn’t it?”
No it’s an actual comparison of the IR spectra of O2 and CO2!
“I suppose you are implying the Earth hasn’t cooled to its present temperature, and that the surface doesn’t cooled at night, are you?”
Why on Earth would I be implying that irrelevancy!
“Accept reality, laddie.”
The one is unable to accept reality is you.
Phil, the absorption spectra of gases obtained by shining a 5500 K light source through them is irrelevant to the fact that the Earth has cooled over the past four and a half billion years.
Or to the fact that the surface cools every night, and most of the day.
Maybe you are ignorant and gullible enough to believe in a GHE – which you can’t even describe in any consistent and unambiguous manner! Who are your heroes? Fools or frauds like Gavin Schmidt and Michael Mann?
Sucker!
“. . . the fact that O2 and N2 do not absorb or emit radiation between 3 and 200 microns which are the critical wavelengths for the earth’s cooling.
Well, this reference presents a graph of an O2 absorption peak at 120 microns (83.47 cm-1):
“The O2 far-infrared absorption spectrum between 50 and 170 cm-1”, M. Toureille, et.al, Journal of Quantitative Spectroscopy and Radiative Transfer, February 2020
(abstract and graph available at https://www.sciencedirect.com/science/article/abs/pii/S002240 )
I don’t know if that absorption peak is particularly strong or weak relative to water vapor or CO2 at other frequencies, but you didn’t ask about that.
As for N2 absorption bands between 3 and 200 microns, I couldn’t quickly locate a similar reference nor spectrum, but note that Google’s “AI overview” makes these statements:
“Collision-Induced Absorption (CIA): When N2 molecules collide, a transient dipole moment can be induced, leading to temporary absorption of infrared radiation. This phenomenon is observed in dense nitrogen gas and in the atmospheres of Earth, Titan, and Triton.
— Fundamental band near 4.3 μm (2325 cm⁻¹): This is a significant collision-induced rotovibrational absorption band of N2.
—Rototranslational band near 100 cm⁻¹ (100 μm): This is another collision-induced absorption band of N2.”
You may also find this interesting:
“Collision-Induced Absorption (CIA) of infrared radiation contributes appreciably to the total absorption of radiation in planetary atmospheres. Indeed, even gases that consist of molecules that have no intrinsic electric dipole moment (including molecular hydrogen, oxygen and nitrogen) absorb radiation if densities are sufficiently high. In the terrestrial atmosphere N2single bond-N2, N2single bond-O2, and O2single bond-O2 collision complexes are important absorbers at various wavelengths and are routinely monitored by different remote sensing (Sioris et al., 2014 Chimot et al., 2017 Kataoka et al., 2017) and ground-based missions (Hartmann et al., 2017 Ortega et al., 2016 Spinei et al., 2014 Gordon et al., 2010).”
from “Update of the HITRAN collision-induced absorption section”, Tijs Karman, et.al., Icarus, August 2019
(abstract, introduction and highlights available at https://www.sciencedirect.com/science/article/abs/pii/S0019103518306997
And finally this, FWIW from Wikipedia:
“Collision-induced absorption (CIA) and emission (CIE) spectra are well known in the microwave and infrared regions of the electromagnetic spectrum, but they occur in special cases also in the visible and near ultraviolet regions. Collision-induced spectra have been observed in nearly all dense gases, and also in many liquids and solids. CIA and CIE are due to the intermolecular interactions, which generate electric dipole moments.”
(Ref: https://en.wikipedia.org/wiki/Collision-induced_absorption_and_emission , with my bold emphasis added)
The above is not just theoretical nor is it BS, it reflects scientific measurements.
And I’ll leave it to others to judge if this comment has been posted by, in your words, “a stupid troll”.
LOL “I don’t know if that absorption peak is particularly strong or weak ” of course you dont!
Errrr . . . your specific ask was:
You DID NOT ask for any comparison of relative strengths of emissions/absorptions across the defined spectral range . . . therefore, your bad, not mine.
And you are too stupid to think “relative strengths of emissions/absorptions across the defined spectral range” is not key to the discussions here? Your stupid not mine
What you idiots keep arguing are irrelevant factoids like “everything radiates” and mentioning “peaks” without ever showing the radiance at those peaks. Your mentioned absorption and emission wavelengths are totally irrelevant and insignificant in the context of the LWIR that the earth emits to cool.
Further TYS, Your link doesnt work to support your statement O2 has an absorption peak at 120 microns. It doesn’t matter. Attached is the spectrum for earth’s LWIR showing NO SIGNIFICANT radiance at 120 microns.
No.
P.S. While the URL I provided to you may not have worked on your cell phone, computer or iPad/tablet, I also provided in parallel the text citation for the same reference paper. I am neither surprised nor saddened that you are unable to use that information to thus locate the paper, INDEPENDENT of relying on just the URL. Perhaps you have a friend that can help you with that.
The radiative power density of a black body at 300K is much less at 100um than it is at 10um. While O2 may be a great absorber and emitter at that wavelength there isn’t going to much in the way of watts per meter squared coming from O2.
It is not about absorbing or emitting that kind of radiation at all. It is about gases radiatively cooling down just like any other kind of matter. Are you honestly implying that a sample of N2 or O2 simply doesn’t lose heat to the environment, that it would never cool down? If you are, please rethink your position.
Dasein Enlighten us with your cooling mechanism
Jim,
It has not gone unnoticed by me that you have not commented on any of my posts, including those directly to you. Of course, that is your prerogative.
Perhaps I should be grateful that you have not “LOLed” my posts, or referred to me as “stupid”, “an idiot”, “ignorant”, or “retarded”, as you have some others.
There is much wrong with your post, and you clearly have gaps in understanding.
To begin, in reference to the comment by Dasein, Here is how the gases can cool without radiation:
At the surface, the energy density of the atmosphere including latent heat assuming about 1% water vapor is about 300,000 joules/m^3.
At the tropopause, it’s about 60,000 joules/m^3. (I am doing some rounding here.)
This means the between the surface and the tropopause, the energy loss is about 240,000 joules/m^3.
About 10%, 24,000 joules/m^3 is from latent heat released in the condensation of water vapor.
About 16%, 40,000 joules/m^3 goes into increased gravitational potential energy of the elevated molecules.
These mechanisms will result in cooling the atmosphere independent of radiation to space.
Interestingly, because of the low density of the atmosphere, the total gravitational potential energy actually decreases with altitude from the tropopause to the mesopause by an almost identical amount.
It does answer the question of how the atmosphere can cool without radiative loss, however.
Now I’ll discuss some of the shortcomings/errors in the article.
Conflating the “work” done by CO2 with the “work” done by water vapor. This is a universal problem in the climate narrative. At the surface, water vapor is typically 20-50 times more abundant than CO2. It also has a broader and denser range of frequencies the effectively absorb/emit CO2. Generate a spectrum with CO2 only and set a lower wave number limit of 200, or 100 if you can, and you will see this. You can also look at figure 17 of Harde (2013) https://onlinelibrary.wiley.com/doi/10.1155/2013/503727
In absorbing radiation from the surface, H2O competes with CO2 in the 14-16 μm Q band and also absorbs across a much broader band, often referred to as a “continuum” up to about 13 μm. It also overlaps the “atmospheric window” at some frequencies as you will see if you take the time to look at Harde’s work.
Tyndall estimated that the absorptive/emissive capacity was 16,000 times that of CO2. While that might be exaggerated, Tyndall was quite meticulous and is it likely good within an order of magnitude.
Water also has a permanent dipole moment which facilitates both absorption and emission relative to CO2.
This means that in the competition for absorption of surface radiation, water vapor is overwhelmingly dominant.
I do not know where you came up with 10-100 msec for the “relaxation time of CO2. Perhaps an average of the entire HITRANS database? Because of the dominance of water vapor, the only CO2 transitions that
matter are the Q band, and perhaps some P and R band transitions. The relaxation times for those are ~ 1 second, an eternity on the timescales we are concerned with. That is why they are the only prominent peaks in the spectrum.
A deeper explanation can be found in the following linked comments, elsewhere in the thread. No need for me to retype everything.
https://wattsupwiththat.com/2025/08/21/how-co2-both-warms-and-cools-our-atmosphere/#comment-4108954
https://wattsupwiththat.com/2025/08/21/how-co2-both-warms-and-cools-our-atmosphere/#comment-4109713
https://wattsupwiththat.com/2025/08/21/how-co2-both-warms-and-cools-our-atmosphere/comment-page-2/#comment-4109865
With regard to what you call “stealing” of energy through collisions, this is not limited to N2-CO2. The overall collision rate is about 7,000,000,000/second. Only a small fraction of these, about 100,000/second, result in non-radiative de-excitation ( your “stealing”) in the case of CO2. In the case of water vapor, the rates are much higher.
You said, “Satellites can’t detect the latent heat absorbed during water’s evaporation at the surface, but satellites do see the heat once released when water vapor cools and condenses to form clouds and rain at 1 to 4 kilometers of altitude. Above water vapor’s condensation altitudes, water vapor provides an insignificant greenhouse effect as the air is very dry.”
continued in following post…
There is no radiation released directly as the result of condensation of water vapor. The latent heat manifests as sensible heat. That is why the moist lap rate is lower than the dry lapse rate, and why there is a tropoPAUSE when there is no more H2O to condense.
Water vapor, and all “GHGs”, cannot emit to space until two conditions are met. First, the density must be low enough so that the radiation is not re-absorbed. Second, the rate of non-radiative de-activation (“stealing”) must be lower than the spontaneous emission rate, the inverse of your “relaxation time“.
Given the length limit of posts, I’ll leave it at that. One must read the linked posts as well.
The premise of your article is misguided. While CO2 does absorb a small amount surface radiation, and emits a bit to space, it is not capable of driving energy transport at a significant level is in the atmosphere.
The heavy lifting of absorption/emission is done by water vapor. Energy transport is driven by convection. Release of radiation to space is driven by sensible heat through collisions, where the conditions to allow radiation to space are met.
There is nothing wrong with the radiative transfer models per se. They can reproduce the satellite spectrum for reasons that are well understood.
That does not imply that radiative transport is the actual mechanism of energy transport in the atmosphere. Explaining that would require a detailed discussion of radiative transfer theory and the radiative transfer equation.
Well Tom Shula, since you are begging for engagement with your comments I will oblige in a small way.
First I agree in large part with all you have said with only minor disagreements. You definitely have studied the issues well, have a solid understanding of the physics, and you like to showcase your knowledge.
So mostly I have felt no need to interject into your pontifications that mostly repeated my collision based arguments. I was writing simply to convince the lay person that CO2 has a very important cooling effect.
I shied away from the detailed physics you push because the public gets overwhelmed and shuts down due to scientific overload for their limited understanding. Parading those details are best reserved for “nerd battles” between us. So when “alarmed” lay people tell me rising CO2 is causing runaway warming, I ask them about its cooling effect, which they never consider or heard of. Then I ask them, as I did in the beginning of this article, how does oxygen and nitrogen shed the energy they absorbed via collisions.
Yet you say (perhaps just to goad me and not ignore you) “The premise of your article is misguided.” CO2 “is not capable of driving energy transport at a significant level is in the atmosphere” But that was never my premise, but does appear to be your straw man to continue pontificating. Oddly you then reiterate exactly what was the premise of my article that “Release of radiation to space is driven by sensible heat through collisions, where the conditions to allow radiation to space are met.”
I agree that water vapor does the heavy lifting as it accounts for 70-80% of the greenhouse effect. But because the air becomes extremely dry in the upper troposphere and above to mesosphere “the release of radiation to space by sensible heat through collisions” is virtually carried out by collisions with CO2.
You argue “With regard to what you call “stealing” of energy through collisions, this is not limited to N2-CO2.” But again I never claimed that it was limited. Just another straw man. I provided a chart showing different collision frequencies. N2-CO2 was just an example to illustrate the collision dynamics.
And my last comment in reference to your “Here is how the gases can cool without radiation: …..About 16%, 40,000 joules/m^3 goes into increased gravitational potential energy of the elevated molecules. These mechanisms will result in cooling the atmosphere independent of radiation to space. ”
Such a reply seems disingenuous. Adiabatic cooling due to the drop in pressure cools that atmosphere as altitude increases. But if that air sinks to lower altitude, the temperature will rise back up. Perhaps you need to discuss enthalpy.
btw I would not ever call you an idiot or retarded. I reserve that for people who claim oxygen and nitrogen are greenhouse gases too, and those who claim radiation can’t travel from a cold object to a warm object because it violates the 2nd Law.
Thanks for responding Jim. While you may view it as “begging”, non-response is a tactic I’ve become accustomed to, and sometimes it requires being overly provocative (call it “goading” if you like) to elicit a response. My hat’s off to you for having a sufficiently thick skin to engage.
While I understand the intention of your post, consistent with the “CO2 is good” stance of the “Coalition”, it is misleading nonetheless.
The relative abundance of water vapor at the surface, typically 20-50 times the CO2 concentration, coupled with it’s ability to absorb IR over a broad range of the spectrum as compared to CO2 tells us that CO2 is a minor player insofar as absorption of surface radiation is concerned.
It’s the “notch” in the spectrum that leads to the belief that CO2 has these magical powers and that a “greenhouse effect” exists. It’s simply a region in the atmosphere where H2O is emitting and CO2 is still absorbing. Its long radiative lifetime means that at altitudes in the mid to upper tropopause it is deactivated by collisions before it can emit. It is a characteristic of the radiation field produced by collisions in the atmosphere.
The radiation is not driving the atmosphere, the atmosphere is driving the radiation.
The tiny peak for CO2 at the bottom of the notch is the only contribution CO2 makes to cooling the atmosphere. It’s already quite cold at the mesopause, and the paucity of molecules means there is very little energy left. The peak would appear to represent < 1 W/m^2, probably much less.
That is why the premise of the article is misleading. CO2 plays almost no role in energy transport. Your article title implies to the “lay person” that the role of CO2 is significant, otherwise, why would you write about it?
A more interesting article would be how water vapor, which exhibits the same behaviors as CO2 except at a rate more than two orders of magnitude higher, accounts for almost all of the energy radiated to space. In addition to that, water vapor which is condensible and has a remarkable capacity for latent heat, is responsible for the energy transport and other phenomena that create our weather.
FYI,you made a comment about the “extremely dry” upper atmosphere in reference to water vapor emissions. What do you think the water vapor concentration is when it can radiate to space?
The volumetric concentration of CO2 at in the upper atmosphere where its radiation can escape to space is 1/50,000 what it is at the surface.
The radiative lifetimes of H2O are considerably shorter, but most H2O emission will occur when the concentration is ~ 10-100 ppm. It has to be low, or it would be self absorbing.
You said the I reiterated your premise when I said “Release of radiation to space is driven by sensible heat through collisions, where the conditions to allow radiation to space are met.” Revisiting the article, I guess I can see that, but it’s a process that can be described very precisely, which you did not.
Regarding your “I shied away from the detailed physics you push…” comment: Really? The let’s “dumb it down so that people can understand it” approach? How has that worked out?
If we could have the “nerd battles” you propose, that would be fine. The CO2 Coalition, Heartland Institute, and the editorial staff of this site are very clear that they don’t want radiative transfer or the “greenhouse effect” challenged. I have “the receipts” as they say. A great example is the post I co-authored with Andy May earlier this year. In addition to having to “dumb down” the technical content, remove “undesirable” concepts, and water it down with a section on paleoclimatogy, the editorial staff felt it necessary to include a “disclaimer” as a preamble to the article. Censorship is alive and well.
I did not engage seriously here at WUWT until I a few other members expressed interest in my work with Markus Ott. Prior to that there was little discussion outside the mainstream narrative. At least now there is a bit of discussion here and I have private communications with other interested scientists.
Thanks for engaging. My only interest is getting the physics right. The RTE can produce a proper spectrum, but it does not follow that radiative transfer is the mechanism at work in the atmosphere.
The atmospheric physics are not difficult to understand, they are just impossible to model.
The physics seems difficult to understand because they are using a model that doesn’t apply to the Earth’s atmosphere.
At the tropopause CO2 concentration is 420-430 vs H2O which 1-10 ppmv . Radiation to space at the tropopause would be mostly due to CO2.
I also question the premise of restricting radiative cooling to the mesosphere. There is very insignificant convection above the tropopause so how does warm air from the troposphere reach the mesosphere.
Jim, I do not know whether or not you read my other comments in this thread which I linked in my comment that initiated our present exchange. I have gone back and reviewed them again. I would be curious how from my comments you concluded that I was restricting radiative cooling to the mesopause.
Regarding the comment related to concentrations, the ability of to radiate to space is not related to the relative (ppmv) concentration. It depends on the volume density of molecules, among other factors.
The “big three” active gases, H2O, O3, and CO2 each behave differently in the atmosphere. I will try to explain in more detail below.
H2O is distinguished because it is a condensible gas at atmospheric temperatures. At increasing altitudes the concentration of water vapor drops very quickly. The condensation reduces the lapse rate by releasing sensible heat into the atmosphere.
In order for H2O emissions to escape to space, on a line by line basis, the following conditions must apply:
First, the absolute concentration (which determines the spacing between H2O molecules independent of the concentration of other molecules) must be sufficiently low that the emission by one H2O molecule is not absorbed by another H2O molecule.
Second, the collision rate (which determines the deactivation rate [your “stealing”]) must be lower than the “relaxation time” for that particular spectral line.
These same conditions will apply to O3 and CO2 as well.
Here is a link to Harde (2013):
https://onlinelibrary.wiley.com/doi/epdf/10.1155/2013/503727
Hopefully the “upgrade” to the site will provide the ability to post pictures in comments.
Please open it and look at Figure 17 on page 21. This is the emission spectrum with H2O only in the atmosphere. It is “noisy” because the water vapor lines have a broad range of relaxation times and each one emits at a slightly different altitude. Also note that H2O emits across the entire spectrum, even in the so-called “atmospheric window”, and that the emission spectrum goes all the way down to wave number 10/cm, a wavelength of 100 μm. The emissions of H2O observed in a clear sky spectrum from space occur from the mid-troposphere to the tropopause, and account for the vast majority of radiation to space.
Generally, in the spectrum, “peaks” are indicative of emission and “troughs” are indicative of absorption.
Now look at figure 18, which shows that the addition of CO2 results in some of the H2O emissions being absorbed by CO2. This does not mean that the H2O is no longer emitting according to figure 17. The emission is there but some is absorbed by CO2. You will also note that there is no CO2 peak at the bottom of the “notch.” That is because these spectra were modeled at an observation altitude of 12 km, and it is impossible for CO2 to emit to space at that altitude. The absorbed water vapor emissions are still being “stolen” and converted back into sensible heat in the atmospheric pool.
Getting back to the spectrum you posted, O3 is not uniformly distributed. The shape of the ozone peak is the result of absorption beginning in the mid-troposphere creating an “trough” and emission in the stratosphere, the small “peak” in the center of the “trough”. the stratosphere is where the conditions for emission to space hold for O3.
CO2 is unique in that its relative concentration is approximately continuous in the atmosphere, and it does not condense as H2O does. The conditions for CO2 to emit to space are not satisfied until the altitude is near the mesopause where the collision rate is ~ 1/sec or less. The energy density in the mesosphere is very low, so there is only a minuscule amount of energy it can emit.
In summary, to be clear, H2O emits from the mid troposphere to the neighborhood of the tropopause, and accounts for almost all of the radiation to space. That is why we have a tropoPAUSE.
O3 emits in the stratosphere.
CO2 emits around the mesopause.
All of the radiation that is radiated to space by these gases is produced as the result of collisional excitation. Outside of the atmospheric window frequencies, there is effectively no IR radiation emitted by the surface that escapes to space.
To be continued….
The radiative transfer equation (RTE) is a correct and proper equation, and it correctly models the spectrum of the atmosphere. The methodology of the RTE however, is purely phenomenological and has nothing to do with real atmospheric dynamics.
In the RTE, the sources of the radiation are “volume elements” of irradiance that obey Kirchhoff’s law and emit radiation according to Planck’s radiation law. In the RTE, radiation migrates from the surface to space in a series of absorption and emissions by these “volume elements”. The emissive characteristics of these “volume elements” are determined by temperature and concentration of the individual active species based on radiosonde data. The precision of these characteristics is quite good because the HITRAN database is used to determine the spectral characteristics of emission.
That being said, the RTE does not reflect the actual processes in the atmosphere. It is the “absorption/emission ladder” in the RTE that creates the illusion of the “greenhouse effect.” That is, that the RTE is a literal representation of atmospheric processes when it is not.
The RTE at its core is a purely radiative, classical model. There are no particles, there is no convection, there is only radiation modeled as “light rays” propagating through an absorbing atmosphere. This may be a surprise to you and many others, but it is correct.
When one understands the true mechanisms at play in the atmosphere, it becomes obvious that there is no “radiative barrier” between the Earth’s surface and space.
The “barrier” of energy that keeps the Earth “warm” is the atmospheric speed limit of convection which along with water vapor and other global forces drives atmospheric circulation.
Energy arrives from the Sun at the speed of light. It is only allowed to leave at a speed that is about 9 orders of magnitude less.
We do not have adequate tools to model convection on an atmospheric level, and likely never will. Trying to model it on the basis of radiative processes is a fool’s errand.
Weather and atmospheric circulation are driven by sensible heat, not radiation.
Ocean currents are driven by sensible heat, not radiation.
Even the radiation is Ithe atmosphere itself is driven by sensible heat.
Tom, a quick question if I may re. your reference to the HITRAN database. Does anything ‘there’ take into account the ‘spectral characteristics of emission’ for mixtures of GHGs and non-IR active gas species under conditions that exist at various levels within the troposphere? I ask only because as often as HITRAN is cited to lend credence to GCMs and other models incorporating radiative transfer codes, it seems odd that investigators such as Heinz Hug found it necessary to collect their own data on such mixtures.
Great question, Frank. The HITRAN database includes a lot of parameters for each “line”, including the effects of self broadening, pressure broadening, and temperature sensitivity. You can see a table of the database format here:
https://lweb.cfa.harvard.edu/hitran/formats.html
The database was established to provide the necessary data to model radiative transfer in gaseous environments. The presence of non IR active gases is also accounted for by database entries that are available for collision induced absorption (CIA), where normally inactive species can absorb/emit as the result of a temporary dipole moment induced by a collision. These insignificant in our atmosphere but can be important in other situations.
NASA’s Planetary Spectrum Generator is an online tool that I found and like to use. Here’s a link to the atmosphere setup page to give you a sense of how many boundary conditions there are.
https://psg.gsfc.nasa.gov/atmosphere.php
Van Wijngaarden and Happer, for example, divide the atmosphere into 500 layers.
The models obviously do a good job of reproducing the spectrum. In principle, with the number of boundary conditions available one should be able to get quite close to any real spectrum providing you choose the proper boundary conditions.
In Heinz Hug’s case, I think he was old school and had the equipment at his disposal so he did a series of experiments. Old school science. We could use a bit more of that, but it requires skepticism, curiosity, and a willingness to challenge the status quo.
Tom: “At the surface, the energy density of the atmosphere including latent heat assuming about 1% water vapor is about 300,000 joules/m^3.”
How do you calculate that? A citation would be fine.
Hi Pat,
I used Perplexity AI to do the calculations for me.
Keep in mind that there are multiple iterations/ levels in the conversation. There is the challenge of communicating to the AI exactly what you want which didn’t happen on the first try, and then there were follow ups which also had to be respecified in some cases.
id suggest you peruse the entire conversation to see what is there. Once you have the context I think it will all make sense.
A link to the conversation is below.
https://www.perplexity.ai/search/9d113fad-28c6-45be-8d20-bee7fcb2683b
Thanks, Tom. It made sense immediately.
The first equation pretty much answered my initial question.
I’ve downloaded the whole conversation, for study at leisure.
Thanks again for making it available.
The Stefan-Boltzmann law applies to cavity radiation, which can be a useful approximation for large enough samples of condensed matter, but gaseous matter is about as far away from being approximated by a cavity and/or condensed matter as can be.
Yes, the S-B law applies to “condensed matter”, matter which has a well defined surface whether interior or exterior. This doesnot apply to gases.
You can provide experimental support for this statement, I presume?
Only joking – of course you can’t. Tyndall’s experiments showed otherwise, I believe.
I’m not saying that gases cannot emit radiation, but the mechanisms are different.
Consider a volume of Argon enclosed in a container internal walls that have an emissivity of zero. What is the mechanism by which the Argon will generate a radiation field? There isn’t one.
If we place a sample of air in that same container, it will generate its own infrared radiation field as a result of collisional excitation of IR active molecules. The kinetic energy of the molecules will equilbrate with an isotropic, self generated radiation field. This is explained in Einstein’s “Quantum Theory of Radiation” (1917).
The S-B law, as originally stated, applies to “condensed matter.” Liquids and solids, but not gases.
Good. So you accept that gases do emit radiation. You, like most people, have no clue about the precise mechanics of how this radiation occurs.
It’s pretty simple – an electron emits a photon. Why? Much is known, but as Feynman said, nobody really understands quantum mechanics.
Who cares, anyway? All matter (including gases) above absolute zero emits infrared. End of story.
Well the emission of IR has nothing to do with electrons it’s actually due to rotational/vibrational transitions!
Geee . . . I could swear that many professional astronomers and theoretical physicists apply the Stefan-Boltzmann law to relate a star’s photosphere temperature to its radiation power (i.e, absolute luminosity) . . . stars . . . you know, those great big masses of gravitationally-bound gaseous matter (mostly H2 and He) that are generally spherical in shape and without any evidence of concavity.
Perhaps I’ve been misinformed . . . in which case all astronomers will need to throw out their Hertzsprung-Russell diagrams.
ROTFL.
Actually the stars consist of plasmas, the fourth state of matter.
Actually, a plasma is defined as ionized GAS that consists of a collection of positively charged particles and free electrons that collectively have little or no net charge. “Plasma” may be considered to be a fourth STATE of matter but it is nonetheless just a collection of atoms and/or molecules, all of which have been stripped of one or more of their normally-bound electrons, which are then moving freely about in the essentially electrically-neutral collective, which is not dense enough to be considered as a liquid or solid.
TYS. Your statement ” by the time one reaches the top of the atmosphere, all of the atmosphere (predominately N2 and O2, of course) is radiating the amount of LWIR energy that was originally absorbed only by greenhouse gases” WRONG WRONG WRONG . The data does not support your wrong-headed narrative.
Please provide the absorption and emission spectrum for O2 and N2 at the top of the atmosphere’s temperatures.
Why? What “emission spectrum” are you on about? Maybe you are confusing spectrometry and spectroscopy with emitted wavelengths of IR due to temperature.
Or maybe, like other ignorant and gullible CO2 believers, you don’t know what you are talking about. Children often accept fairytales as fact, grown-ups not so much – apart from people who believe in a “greenhouse effect” they can’t describe in any consistent and unambiguous way.
See the comparison, Jim?
Clearly Flynn you cant provide because clearly you don’t understand physics.
Jim, Papers at this link disagree with your assertion that N2 and O2 do not radiate
Scientists: Oxygen & Nitrogen ‘Radiatively Important’ Greenhouse Gases With IR Absorption Temps Similar To CO2
A 3-body collision involving the dissociation of N≡N triple bond – the strongest bond in all of Chemistry, is bound to be a rare process. I could be wrong, but it seems to me that the radiative intensity of that process is extremely low.
Requires 950 kJ/mol for dissociation which corresponds to vacuum-UV radiation leads to formation of NO in the high atmosphere
That, I’ll believe. 🙂
Well Nelson I now see how easily deluded you are. I confess you gave me a great belly laugh!Clearly you are not a scientist. From your link too the paper Experiment: nitrogen, oxygen absorb IR to about the same limiting temperature as CO2:
“Twin styrofoam Saran-wrap-sealed tubes exposed to sunlight were used, one with pure (1,000,000 ppm) CO2 and the other with air (N2, O2) and/or Ar.
The results were admittedly “surprising” given expectations CO2 would operate as a radiatively distinct GHG.
The tube absorbing IR with N2 and O2 (air) and Ar warmed to a temperature limit quite similar to (55°C to 58°C) the temperature limit in the 100% CO2 tube (58°C).”
While their result tell me they more likely proved CO2 caused no significant warming, Nelsen sees that woeful experiment as proof that N2 and O2 are greenhouse gases!
Damn. Doesn’t anyone engage in critical thinking anymore!
You can’t even describe the “greenhouse effect” in any consistent way, so talking about “greenhouse gases” is just irrelevant pseudoscience, isn’t it?
Jim, there were a number of papers at the link I supplied. If you didn’t like one of the experiments that’s fine. Why not respond to the GRL paper. I’m not sure why providing a link to articles discussing the issue makes me deluded. I never said N2 and O2 were greenhouse gases. O2 and N2 are homonuclear diatomics. You should stop the name calling as it makes you look petty.
_________________________________________
It’s a time thing.
From 1945 to 1979, CO2 cooled our atmosphere.
Since 1979 to date, CO2 warmed our atmosphere.
Is that too difficult to understand?
CO2 heat is tricky. Sometimes it even hides in the oceans.
😆😅🤣😂
15 microns is -80°C that is why satellites only detect it from the stratosphere. Energy transfer occurs above the tropopause. Below the tropopause is expanded air, surface has compressed rising air parcels that do not gain or lose heat. No energy transfer while expanding.
CO2 does not absorb sunlight. Surface has window to allow IR to escape. CO2 has less than 1% of the mass 1.05 x 10 ^22 particles and same energy per particle at 15 microns but magically can heat of 99% of the other particles 2.61 x 10 ^25 particles.
Therefore only in models can CO2 heat to heat 99% of the atmosphere.
99% of the atmosphere 2.61 10 x ^25 particles.
Atmospheric pressure gives each of the 2.61 10 x ^25 particles, 7.6 x 10 ^-21 joules.
198597 Joules /713 Joules (Constant pressure heat capacity) =278.5K (mass (1.4×1.27kg) x motion (334 meters per second)).
Solar input (1/4 of solar constant) gives all particles in the atmosphere 1.3 x 10 ^-23 joules w/ 15 micron wavelengths. Providing the 340 watts of heat to the surface due to 2.61 x 10 ^25 particles.
CO2 1.05 x 10 ^22 particles gets heat through collisions of 25% of the 99% N2 & O2 not the other way round.
Our blue sky is due to 0.483 microns heating 0.43 moles at 4.11 x 10 ^-22 joules per particle.
Equivalent to 104 watts allowing 236 watts to escape to space through the atmospheric window. (claim to be reflected, its true that the illuminated side reflects for the dark side of the earth). Remember models assume flat earth w/o any dark side.
The 50% absorbed at the surface claim comes from 26.9°C (tropics) to -5.9°C (zero entropy in water vapor) as the heat loss of 170 watts.
Reality: 278.5K (5.3°C) to 207K (-66°C), 240 watts, is what satellites detect.
278.5K earth’s mean temperature proportional to 340 watts (1/2 of total earth gets, NH 380w, SH 300w Oct & May). 680 + 240(OLR) + 440 watts (visible light) = 1360 total.
This article is articulating model theory of the greenhouse effect, which flaws I have pointed out in this post.
“…and same energy per particle at 15 microns …”
__________________________________________
Microwave ovens, you know, RADAR ranges,
use wavelengths longer than 15 microns and
you can boil water in one of those things.
Yes, a black body that has a Plank curve that
radiates predominately at 15 microns would
be a brick of dry ice. Great argument, but as
far as I know that isn’t the way it works.
There are lots of arguments why “The Climate
Crisis” is a load of steaming propaganda, but
the 15 micron argument isn’t one of them.
When you’re in a war of words, don’t use an
argument that is false as that will cost you
your credibility.
You’re welcome (-:
Microwave ovens heat by exciting rotational motion of the water molecule, where the resonance occurs at around 21 GHz with what’s probably the broadest line in physics.
This is a bit like DC used to heat tungsten to 2800 – 3400K, nothing to do with energy of a photon.
And my microwave oven is cold ! Yet keeps transferring heat to the hot object inside until turned off. My goodness how can that be ? Transfer from cold to hot !
ans…a photon and its energy, e=hf, is a good start…
Micro waves aren’t heat. The air inside the micro wave isn’t heated. Ceramic cups aren’t heated. Yes you can boil water but that is because it’s water.
Thanks for the replies. A while back I posted the 15 micron argument here at WUWT and one of the regulars pointed out why it was a poor argument. LINK
I didn’t understand the reasoning then and still don’t but I’ll accept that it’s not a valid argument and ought not be repeated.
Steve, your LINK took me to a photo of Bernie Sanders – yikes! Apologies again for the unclear explanation, but otherwise stand by the conclusion.
A photon isn’t “heat” until it is absorbed by something….is a good concept to keep in mind.
Sounds fair to me, in general. The photons emitted by ice stubbornly refuse to be absorbed by water, it seems.
Oh dear, that makes a mockery of the greenhouse effect, doesn’t it? I’ll probably be told to stand in the corner again for challenging the teacher.
Yes, the photons emitted from the surface of water ice (at a temperature of 273 K) will have a different EM spectrum distribution than will the photons emitted from a land, foliage or water surface at 273+16=289 K. But NOTHING prevents those photons from being absorbed by surrounding objects having temperatures either higher or lower than 273 K.
Surprised by that?
Indeed I am. However, if you can heat water with the energy emitted by ice, I would be even more surprised.
You’re dreaming.
Except in the case of microwave ovens no heat is transferred via the microwaves. Thanks for “good concept” but I will stick with my thermodynamics book.
Hmmm . . . who knew?? And all along I was taught —and therefore thought—that the average translational kinetic energy of gases was based on and calculated as equal to 3/2 kT for each atomic species and as 5/2 kT for each ideal diatomic species at moderate temperatures, where k is the Boltzmann constant and T is the absolute temperature. No indication there of a pressure dependence.
So, please explain how pressure conveys energy to particles . . . and note the this is distinctly different from atoms and molecules in collective ensembles creating pressure by dint of being confined in a given volume and having a temperature above absolute zero.
So, please explain how pressure conveys energy to particles . . .
Of course, he can’t. A SCUBA tank at 2500 psi cannot be distinguished from an empty tank by temperature alone.
The translational kinetic energy of an ideal gas molecule is 3/2kT, 5/2kT includes the rotational energy for a mononuclear diatomic.
5/2 kT is the mechanical energy for all diatomic molecules, as I previously posted. This applies to molecules such as CO and NO.
You said: “the average translational kinetic energy of gases was based on and calculated as equal to 3/2 kT for each atomic species and as 5/2 kT for each ideal diatomic species at moderate temperatures,” which is what I was replying to. 5/2kT includes the rotational energy.
Since one mole of a gas contains 6.02×10^23 particles—and assuming I’m deciphering your mashup of math symbols correctly—you appear to be asserting that there are only something like 43 moles of gas in Earth’s atmosphere!
Who knew it was so thin!!!
ROTFL.
At 15 °C 7.2% of the CO₂ bending mode is excited by thermal collisions. So, even without incoming radiation, thermal energy of the atmosphere is radiated away.
At equilibrium (the temperature of the atmosphere is the same as the temperature of the ambient Planck radiation) the same amount of radiation that is absorbed by the greenhouse gases is also emitted.
“Oxygen and nitrogen comprise 99% of our atmosphere. Those molecules do not absorb or emit radiation. Our air is heated by collisions with solar heated surfaces and warmer atmospheric molecules. So, we must understand how oxygen and nitrogen shed their absorbed energy, so it can exit back to space.”
I’m confused.
Could you be more specific about your confusion Jeff
I know squat about physics, except that it works 🙂 So the bolded sections. “do not absorb or emit radiation” then “shed their absorbed energy”. Those two things don’t seem to agree with each other.
Ah, it’s magic of the “climate science” variety, Jeff. Pure pseudo-scientific gibberish. Don’t question the author’s fantasies, or he will have a breakdown and shout “YOU ARE STUPID FOR NOT MINDLESSLY ACCEPTING WHAT I WRITE!”.
Jeff they bump into a CO2 and transfer energy to it and it can radiate it out. I believe that isvwhat Mr. Steele is getting at.
Same old nonsense about back radiation from cold to hot.
Radiation is Shortt for electro-magnetic radiation. It exists in the electro-magnetic field. The field equilibrates at the speed of light. Energy does not flow up field.
All objects communicate with other objects at the speed of light in both the gravity field and the E-M field. If the Sun disappeared now, we would sense the loss on Earth in both gravity and E-M fields after 8 minutes.
Rick, you have stated this many times before. Let me try to explain again. The Stephan Boltzmann equation between two parallel planes is
q/a= [k/(1/ehot+1/ecold-1] x (Thot^4-Tcold^4)
or basically q/a= K x (Thot^4-Tcold^4) to overly simplify
The -Tcold^4 part is the “back” radiation.
You can consider both the “fore” and “back” radiation to be photons. They aren’t “heat” until they have been absorbed. If you look at a Planck curve you can see that the hot body is also emitting photons at the same frequencies and amounts as the cold body but lower on the curve…and those exactly cancel each other out…so photons can flow either way but “heat” is a net conversion of those photon’s energies and is from hot to cold….at least for naturally emitting human scale emitting objects…but doesn’t apply to microwave ovens, lasers, gases, many things.
And water doesn’t absorb the photons from ice and get hotter, does it?
Back radiation is just meaningless “climate science” jargon, aimed at the ignorant and gullible, trying to assure them that a cold atmosphere can raise the temperature of a hotter surface, by the miraculous powers of CO2 or water vapour, or something!
No, the laws of physics can’t be discarded at the whim of fantasists who cry “Slow cooling results in heating!”.
No, in your scenario, the water isn’t going to get hotter because the water is sending more photons of equal and higher energies the other direction. Look at the graph above.
BTW, I agree with you that calling it “back radiation” lends itself to thinking it can “heat” something, when in fact it only cancels photons going the other direction(wow that’s an unscientific description)
You can do an experiment. Get a laser that is high-powered enough to burn paper (say 200 milliwatt) and shine it at a mirror that you have soaked in ice water. I think you will find that the reflected beam will burn paper, or at least get it much hotter than the cold mirror.
Don’t be stupid. Radiation from ice is not a laser. A mirror reflects – it doesn’t matter what temperature it is.
You cannot do any sort of experiment that shows radiation from a colder object is absorbed by a hotter – making the hotter increase in temperature.
From post:”…doesn’t apply to microwave ovens, lasers, gases,”.
If it doesn’t apply to gases and we are talking about CO2 (a gas) then why is Rick wrong?
‘Cuz he keeps saying there is no such thing as “back radiation” from cold to hot. This is a disservice to those who wish to understand radiation and/or the second law of thermodynamics. It is clear that there is a [Thot^4-Tcold^4] term in radiative transfer and that the negative term is the “back radiation”, and the formula is not at odds with the second law….only our extraction and consideration of the negative term results in misunderstanding. I have to say that I have taken graduate level heat transfer courses and didn’t understand the concept until I considered (now years ago) the graph above https://wattsupwiththat.com/2025/08/21/how-co2-both-warms-and-cools-our-atmosphere/#comment-4109085
And that objects that are emitting on a “cooler” curve emit photons that exactly counterbalance the equivalent photons of the hotter body. If you heat the cooler body, eventually the photons transferred will offset each other (obviouslywhen the bodies are at the same temperature), and there will be no heat transfer.
This article and many of the related comments comprise a wonderful example of the many convoluted misunderstandings and misinterpretations of energy transport in the atmosphere. That being said it is consistent with the “narrative” and so not subject to any editorial disclaimers, as was:
https://wattsupwiththat.com/2025/03/09/beyond-co%E2%82%82-unraveling-the-roles-of-energy-water-vapor-and-convection-in-earths-atmosphere/
I will focus on one item, the interpretation of the spectrum particularly as it pertains to CO2. In the article linked above you will find a spectrum of emission of water vapor only, published by Herman Harde in his 2013 paper. He also used a line-by-line radiative transfer model that ran on a personal computer, preceding the work of van Wijngaarden and Happer by almost 8 years. Why it never received recognition I do not understsnd.
Harde understood and asserted that almost all excitation and de-excitation of IR-active molecules from the surface to the mesopause are due to collisions. Excitation via absorption and de-excitation via emission are orders of magnitude less frequent. Outside of the atmospheric window frequencies, all of the radiation that escapes to space is generated via collisions with the sensible heat of the atmosphere as its source function.
Water vapor emits in a continuum across almost the entire infrared spectrum from the Earth’s surface. It even overlaps with some parts of the “atmospheric window”, which is why the band from 8-14 microns is jagged, not smooth.
More importantly, the peak of the water vapor emission band overlaps with the Q-branch of CO2 at 15 microns.
The TOA spectrum is an emission spectrum. If the CO2 was zero, you would see the water vapor emission band as continuous. Van Wijngaarden and Happer show this in some of their modeled spectra.
The author of the article, as many do, seems to interpret the “notch” around the Q-branch CO2 peak as both absorption and emission of CO2. This is not correct.
The area under the notch (the”emission” part) is from water vapor emission. The “notch” itself represents suppression of water vapor emission due to partial absorption of the water vapor emissions in this band by CO2.
Because the water vapor emissions are abundant and occupy a continuum of wave numbers, the absorption occurs not only at the 15 μm peak, but also at many of the rotational sidebands.
The water vapor emissions absorbed by CO2 are rapidly converted back to sensible heat via collisional de-excitation. The energy is not trapped, It is in the heat pool of the atmosphere.
The emission of CO2 in the spectrum is represented by the tiny peak at the bottom of the divot. This is CO2 emission near the mesopause, most estimate at ~ 83 km altitude. This is where CO2 can still be excited by collisions, and the collisional de-excitation rate is longer than its radiative lifetime, 0.5-1.0 sec.
The energy density of the atmosphere at sea level is about 280,000 joules/cubic meter. At the stratopause it is reduced to about 280 joules/cubic meter, and at the mesopause it is about 14 joules/cubic meter.
There is precious little energy available to excite CO2 at the mesopause, which is why the peak is so small and so narrow.
Tyndall estimated in his experiments that at typical relative concentrations the radiative power of water vapor is about 16,000 times that of CO2. Water vapor is responsible for most of the energy radiated to space. Water vapor is also responsible for most of the absorption of surface radiation, which is converted into sensible heat which drives convection.
It is water vapor that drives weather and the climate. CO2 is insignificant insofar as energy transport is concerned.
David Dibbell repeatedly brings up the importance of energy conversion in the atmosphere. This is extremely complex, largely ignored, and cannot be accounted for in one dimensional models that generate a spectrum.
There is work being done in the atmosphere continuously, and it is silly to conclude we can draw any global conclusions from single column atmospheric models.
No, it’s heat. No heat, no climate. Just a lot of frozen gases.
No it’s not. The surface radiates directly to space. That’s why images (IR, visible light, etc) can be taken from satellites. As a matter of fact, deserts radiate extreme amounts of energy directly to space, resulting in fast cooling in the absence of sunlight and water vapour. That’s how ice has been made in blisteringly hot desert regions for millennia.
Practical physics, you could call it.
The real problem with radiative solutions is that the earth itself is a big heat sink. As the surface warms, heat is moved in two directions. Upward via radiation and conduction/convection. Downward by conduction, i.e. diffusion of heat through a solid. That stored heat is held for a TIME before it is released.
What is the consequence? Temperatures in the atmosphere are not raised as much as expected. I’ve never sat down and computed how much this effects radiative flux upward from the surface. It may be small enough to be ignored. However, it does exist and I never see anyone motion it.
I guess my pet peeve is that climate science never deals in gradients and time. It is all concened with what happens at an infinitely small instant of time. As an EE, we had to take ME thermodynamics to learn how to deal with heat sinks. This was calculus based with time as variable. It dealt with conductivities, diffusion, radiation efficiency, convection from fans, etc. I never see climate science papers deal with this, ever. It is all about temperature time series that are never treated as time series by dealing with auto-correlation and seasonality that averages just cover up and hide away from sight.
Not only an instant in time (I.e. a steady state solution), but a pencil thin atmospheric column.
I share your frustration. I entered graduate school to earn a PhD in mathematical physics and discovered that it was far too removed from reality. Radiative Transfer Theory is considered a branch of mathematical physics, and in studying the history of it I learned that within the RTT community at large the phenomological, heuristic RTE as used in “climate science” was always considered problematic because it is NOT based on fundamental physical principles. This was largely resolved by the work of the late Michael Mishchenko, but the old ways persist.
I worked as an engineering physicist in the development of analytical instruments and thin film processing equipment, primarily in high vacuum systems where an understanding of heat transfer via radiation, conduction, and convection were paramount. Thermally isolating a substrate at 1000 C from its support structure, or cooling a slab of conductive ceramic composite with an incoming heat load ~ 100 W/sq-cm is no small feat. You also learn that convection kicks in at 2-5 Pa of nitrogen as you see red-hot disks of metal immediately cool in an atmosphere equivalent to the mesosphere.
The only places in the atmosphere where “radiative transport” actually occurs are at the respective emission heights for H2O, O3, and CO2. The atmosphere is not an abstract, mathematical construct.
Volcanoes, magma rifts and radioactive elements might beg to differ that the sun is the only source of heat on the earth.
After writing this paper, I’ve come to wonder whether geothermal heat from the oceanic crust is what keeps the oceans liquid.
Every 1 M km³ of basalt emerging in a submerged Large Igneous Province can raise the temperature of the world ocean by ~1 C.
Geothermal heat flux is presently about 50 mWm⁻². Over the 3.6E14 m² of ocean crust that’s 18 TJoules/sec every second. Unidirectional from the bottom, the energy is retained until sea surface radiation or evaporation.
In the deep past, geothermal heat was greater. It would take some complex ocean modeling to answer the question, whether geothermal flux is responsible for keeping the ocean liquid. But it seems a reasonable idea.
Wonder no more. It does. Hence convection, causing deep ocean currents. Chaotic convection even, causing some deep currents at different levels to flow at 180° to each other.
Or you can believe in the miraculous powers of CO2, I suppose.
Thanks, Mike. If you have a citation for that, I’d be grateful if you’d post the link.
I’ve written to 2 ocean physicists with that question, but received no answer.
If it’s true, then most of the energy of the ocean is geothermal. The warm sea surface is just the result of superficial solar heating of the top of an already warm ocean.
This, in turn, is suggestive that the ~390 W/m^2 of SB power emitted by the ocean is mostly the product of geothermal energy. Not from solar heating or the GHE.
I don’t do citations, generally. I let people make up their own minds, as I find appeals to authority (citations) lead to people providing opposing citations.
However, asking Google produced the following –
Google can hopefully provide a citation list
I’ve done many keyword searches in Google Scholar.
Nothing has come up about the consequences of steady geothermal heating on the retention of the liquid ocean state.
Frank, I can’t remember my previous query, but here’s one just now –
“sources of abyssal heat”. Response –
A bit different from the one before, but should lead you to the source of the AI comment.
‘I don’t do citations…’
Because, generally speaking, a lot of your comments are unsupported, hence citations don’t exist.
And, generally speaking, none of your criticisms are unsupported by fact, and hence opinions of no value whatsoever. <g>
So, they are supported by fact?? 😉
Geothermal heat is about 40TW, about half of which is due to radioactive decay, in contrast incoming solar is about 170,000TW.
Right. But it’s up from the bottom, and it’s been there for the last ~3.5 billion years.
Pat (sorry, think I called you Frank, earlier) –
Yes, and the heated water expands, becoming less dense, and rises. Due to the somewhat peculiar properties of water, the result is convection, as the densest water is just above freezing.
Hence, in brief, the phenomenon where the liquid temperature profile shows falling temperature with depth, unlike the solid form (ice caps, say), where temperature rises with depth.
Given a geothermal profile of 25 C per km, at 10 km the crust temperature would be 250 C, but the water temperature at the same depth is around 2 or 3 C. So places like the Mariana Trench are essentially 250 C rocky buckets of very cold water.
So much for sunlight warming the deep oceans, or deep ocean currents being caused by the wind – as some ignorant and gullible “experts” at NOAA proclaim. A good example of Richard Feynman’s view that “Science is belief in the ignorance of experts”.
As has the incoming solar!
And the 40TW is being measured? Or is it “estimated?” Call me skeptical out precise numbers about ocean anything.
It’s measured.
Sorry, but that’s just nonsensical. There is no “greenhouse warming”. That’s just pseudo-scientific jargon without any physical basis at all, promoted by the ignorant and gullible.
As Fourier correctly observed, the Earth loses all the heat it receives from the sun, plus a little internal heat (the Earth being more than 99% glowing hot matter, slowly cooling).
Thus, after four and a half billion years, the Earth’s surface is no longer molten.
And at night, of course (and most of the day), the surface is radiating away more heat than it receives from the Sun. Maybe “climate scientists” don’t accept nighttime, because their brightly coloured cartoons about the Sun’s influence at night would make them look quite mad.
Which they are, obviously.
I have so many questions I don’t know where to start. Short wave radiation enters the atmosphere from the sun. Some of that radiation (heat?energy?) is reflected. Some is absorbed in the atmosphere. Absorbed by what? Is it bounced around the way CO2 reacts with long wave radiation? Some short wave radiation reaches the earth and is reflected back. Finally a good amount warms the earth. Once the earth warms it emits long wave radiation to space. My understanding is that it is this long wave radiation that the CAGW guys are concerned about. In any case before the long wave radiation makes it to space it is absorbed(?) by water vapor and or CO2. Water vapor and CO2 do not hold the long wave radiation rather they emit (reradiate?) it in all directions. A little less than half is emitted towards earth and the rest away from earth. And this goes on until the energy escapes to space but only after many absorptions and is re emitted. Now you are saying that oxygen and nitrogen can receive some of this energy from a collision with CO2 and releases that energy after another collision with CO2. Is that right? If we don’t have this interaction between CO2, oxygen and nitrogen we would have runaway warming. Is that right? I am having a hard time wrapping my head around this.
The surface heats during the day, and cools at night. The Earth continuously loses about 44 Tw at present. Losing more energy than it gains is called cooling.
Thermometers respond to the temperature of their surroundings – not CO2 or H2O.
Just reality.
nitpick- the surface actually cools during the day as well as at night. It’s just during part of the daylight hours (less than 12 hours for most of the planet), the heat in (from the sun) exceeds the heat out.
Fair enough. Most CO2 worshippers just refuse to accept reality.
“ The Earth continuously loses about 44 Tw at present.”
That’s the Earth’s core heat transfer to the surface, and about half of it is due to radioactive decay, solar irradiation is about 170,000TW. Geothermal heat is a trivial part of the Earth’s energy balance.
I suspect steady geothermal heat is what keeps the oceans liquid.
Phil, and yet the surface cools every night, and has cooled over the past four and a half billion years – which makes a mockery of your ignorant and gullible beliefs.
The Earth continuously loses about 44 TW at present. That’s called “cooling”.
The Earth’s core loses about 22TW at present and generates about 22TW by radioactive decay which heat the surface as a tiny fraction of the 170,000TW from the sun which also heats the surface. To maintain the Earth’s temperature radiational cooling would need to be about 170,000TW.
Science was never my strong suit and it shows.
Infrared (heat) is one form of radiation. But you say that isn’t absorbed. But then you say oxygen and nitrogen absorbs heat from other molecules and then sheds it. Totally lost.
Unfortunately, the author doesn’t seem to understand.
Sensible heat (temperature) is molecular kinetic energy, Ben. Faster molecules = more energy = more heat transfer.
Ben, One form of heat transfer is via radiation. Oxygen and nitrogen do not absorb or emit radiation in the infrared wavelengths involved in cooling the earth. Another form of heat transfer is conduction. Oxygen and nitrogen can gain energy via conduction when colliding with solar heated surfaces or heated molecules, and at earth’s temperatures the only way they can shed that heat is via more collisions. There is nothing contradictory with those dynamics.
Only in your imagination, of course.
After all these years I still see the same fundamental misunderstandings.
The higher than expected temperature at an irradiated surface beneath an atmosphere (the greenhouse effect) is a consequence of convective overturning and not caused by the presence of radiating material within the atmosphere.
Radiative gases have a thermal effect but that is always neutralised by a change in the speed of convective overturning.
That change in speed from our emissions would be too small to notice.
Convection converts surface heat to potential energy higher up and PE does not radiate so energy accumulates in the system until it is reconverted back to heat in the downward leg of overturning.
That conversion and reconversion takes time and the length of time determines the size of the surface temperature rise.
More atmospheric mass gives a larger greenhouse effect at any given level of irradiation and strength of gravitational field.
That is all there is to it.
We have all been lied to and thoroughly confused by bad science based on radiative physics.
No, if the “expected” temperature does not agree with the actual measured temperature, then your “expectations” were wrong.
Don’t you agree?
The expected temperature for a planet with no atmosphere is calculated by the Stephan-Boltzmann equation.
When one adds a convecting atmosphere the measured temperature is always higher.
Which is completely stupid. You cannot calculate the temperature of a body merely by knowing how much radiation falls upon it. How hot is your soup? In the restaurant? Outside in the sun?
The Earth’s surface temperature has varied between glowing hot, to its present range, including an infinite number of temperatures between the start and final figures.
Go on, tell me what you expect the Earth’s temperature to be, misusing a calculation which you obviously misunderstand!
Temperatures are what they are measured to be – no more, no less.
I think the Stephan-Boltzmann equation tells us the temperature of an object based on the frequency of light that is emitted by that object, not based on how much radiation is falling on the object. Radiation absorbed will heat the object, but radiated energy is what tells us what its temperature is.
Thomas, correct. You cannot determine the temperature of an object in sunlight using any “calculation” relating to incoming solar radiation
Anybody who believes otherwise is quite mad.
Assuming you’re talking about the AVERAGE temperature. Best be clear on that or the Moon will soon be rolled out as a daytime example…
Yes, and there is no gravity in the radiative transfer equation. No gravity, no convection, and what few understand no molecules.
Only heuristic, phenomenological packets of irradiance that emit according to the S-B Law and follow Kirchhoff’s law. It is a fantasy atmosphere, a purely mathematical construct.