Kevin Kilty
The most commented threads on WUWT involve anything having to do with thermal radiation and their almost intolerable length results from what I can only classify as bickering. The thread from this effort from several days ago contained a few worthy nuggets, though, which form the basis of this essay. Hopefully we aren’t going to generate a lengthy thread today.
One point of contention that always arises is the dogma that all objects and substances at any finite temperature will emit Infrared radiation. In addressing this persistent dogma, one commenter said this
“I suspect that, like me, he was told somewhere in his education that ALL things radiate thermal IR. I suspect that goes for all of us. I also suspect that it would be interesting to peruse some older texts to ascertain exactly when the blurring of radiative behaviors between ‘things’ (objects) and atmospheric gases began.”
Indeed, it probably does go for us all. No doubt I believed this at one time. Even after modifying my view decades ago I have made careless statements that haven’t always helped clarify matters. But as consolation, I can point to erroneous statements at NASA like this one:
How Atmospheric Sounders Work
Everything that has a temperature radiates. We radiate. Snow cones and swimming pools and pine trees radiate. So do the molecules of all the gases that make up the atmosphere.” [1]
Here is an even trickier statement that is half correct and half completely wrong. It is in a textbook I have taught from, Fundamentals of Heat and Mass Transfer, Incropera and Dewitt.
“For non-polar gases, such as O2 or N2, such neglect is justified, since the gases do not emit radiation and are essentially transparent to incident thermal radiation….” (p. 896 in the 7th edition).
It is absolutely correct that nitrogen does not emit IR radiation. Why is it half completely wrong? Let’s find out…
With inspiration from the discussion above, let’s tackle the subject of radiation interacting with matter. This is an exceptionally complicated topic. However, quite a lot can be understood with a couple of simple models and applying these to a number of materials.
The Electric Dipole
Figure 1 shows an EM wave, with S indicating its Poynting, or propagation, vector. From the standpoint of a stationary dipole, which consists of a mass carrying positive charge separated from another mass carrying negative charge attached through a spring, the EM wave is an alternating electric field. The dipole is a model of a gaseous molecular substance, with the masses standing in for atoms and the spring representing the molecular binding. We can treat this model purely with classical dynamics, or as semiclassical by quantizing the dynamics of the masses and spring, or as entirely quantum mechanical; and arrive at the same conclusions.
Figure 1.
As the EM wave (A) passes by, what our atom sees is first a downward electric field vector that squeezes the two masses together (B), then a passing upward electric field vector that stretches them apart (C), and so on. The passage of the EM wave is setting the dipole into vibration, which also means putting energy into it.
The most effective input of energy would occur when the frequency of passing upward-downward motion occurs at the resonant frequency of the masses and spring. It should go without saying that a dipole already in motion could release its energy back to the EM field, if the phase difference between a passing wave and the dipole were such that the wave could absorb the energy involved.
Quantum mechanics modifies this picture only a little. Instead of an oscillator that can absorb an arbitrary amount of energy, a molecule has allowed energy levels of vibration and the quantized EM field has to provide the correct amount of energy to move the oscillator from one allowed energy level to another. (We will ignore the complication of linewidth.)
However, the essential features of interaction are on display. The passing electric field (EM wave) has an electric field vector (E) and the matter contributes a dipole able to couple to the EM field. Not possessing a dipole handicaps a molecule from this very important interaction.
Which molecules in the atmosphere does this exclude?
Argon for certain because it is a monatomic noble gas. There is only one mass and no spring at all. No matter how a person dices this situation the noble gasses have no way of coupling to an EM field. Not in the infrared anyway. One has to await very energetic radiation that is capable of coupling to electronic orbits in the visible and ultraviolet. As Willis explained, the purge of an IR spectrometer with argon to prevent spurious signals is an experiment demonstrating conclusively that some materials do not radiate IR.
Forsaking isotopes for the moment, nitrogen has two identical molecules covalently bonded.There is no ionic character to this bond and so nitrogen is excluded from interaction with an EM field at IR wavelengths. Oxygen too is a covalently bonded molecule and does not participate in interactions with an IR EM field.
Now CO2 is interesting. Figure 2 shows the CO2 molecule. It is linear and symmetric. Each oxygen is covalently bonded to the carbon. Thus it seems that this molecule might not couple to an EM field. However, the covalent bonds are not perfectly so, but because of the difference in electronegativity of carbon (2.55) compared to oxygen (3.44), there is a small ionic character to the bond. With a difference in electronegativity of only 0.89 the bond is known as polar-covalent. Thus, in the Incropera quote above, the overall molecule being non-polar, may still couple, and strongly so in the case of CO2 to an EM field because the details of the molecule contains effective dipoles.
Figure 2.
The magnetic dipole
Molecular oxygen, though covalently bonded, is a bit of an oddity. Despite its similarity to nitrogen, each oxygen in the molecule ends up with an unpaired electron in its ground state – a state known as a triplet (two paired, and one unpaired electron). Molecular oxygen is a diradical. The electron spin plus orbital angular momentum of the unpaired electron gives ground state oxygen a magnetic dipole – something like a magnetized, spinning top. This, in turn, allows oxygen to align itself with a local magnetic field. It is the reason for oxygen being notably paramagnetic.
Oxygen molecules, just like any other molecules in the atmosphere, are constantly subject to collisions, a process that disturbs the equilibrium orientation of the magnetic dipole. Once disturbed the magnetic dipole will seek to re-establish its equilibrium and will do so through precession around an orienting field at a fixed frequency in the microwave region (around 60GHz). In other words, the signal that the magnetic dipole emits depends on there being a small amount of disequilibrium in the complex state of the atmosphere.
Greater temperature leads to greater rate of collisions, greater disequilibrium, and greater signal magnitude or what people sometimes refer to as brightness. By measuring the brightness of particular microwave frequencies, satellites are able to determine atmospheric temperature. However, the brightness of these signals is measured in microwatts per meter squared per steradian of solid angle – useful for measurement but negligible for transport of energy.
Condensed matter
The biggest difference between gaseous material and condensed material is that in condensed materials each molecule is influenced greatly by all the other molecules in close proximity. This leads to absorption and emission of radiation taking place in broad bands or continuums rather than in discrete frequencies. People often assume that this means that condensed matter will radiate like a blackbody according to the Stefan-Boltzmann (SB) law. Some background on SB is in order.
Two laws of radiation are of importance here. Conservation of energy demands that the transmissivity (t), absorptivity (a), and reflectivity (г) must satisfy the relation t+a+r = 1. Kirchhoff’s rule is that emissivity and absorptivity at any frequency are equal (a=ε). Thus ε+r+t=1 at a given frequency.
What we often call blackbody radiation was called cavity radiation initially. Within a cavity with a very small exit, t is zero because the cavity is opaque; r is zero because the exit is so tiny that entering radiation only escapes after very many reflections, and thus a=ε=1. From Planck’s efforts, but also as a result available from classical thermodynamics, the emitted power in this cavity radiation is W=σT4, the SB law.
Now if we dispense with the cavity by opening the condensed matter up to become a surface, does the SB law still hold? Obviously SB depends upon so many reflections in a cavity that in the limit all radiation is absorbed (ε=1). In contrast, only one reflection occurs on an open surface which means the actual reflectivity of the material is now important.
Thus, metals, which are often reflective, will have an effective ε less than 1. Gold, for example, when well polished, has an emissivity of 0.02 to 0.05. Gold can be made very emissive, however, if its surface is rendered porous by processing to a condition known as black-gold. It now presents a surface in which EM energy can fall into “canyons”, to make an analogy, and be absorbed effectively. So, the surface state of any material can produce a designed emissivity.
Just as is the case for molecules with no dipole to couple to an EM field, condensed materials may lack a dipole as well. Diamond, for example, has a highly symmetric structure with all covalent bonds. Its IR emissivity is very low; one table lists it as 0.02. It is not reflective but highly transparent with only two instances of two-photon absorption at 2.5 and 6.5 micrometers. Quartz, on the other hand, which looks superficially like diamond, is built from mainly covalent silicon oxygen bonds. However, just as is the case with CO2, a difference in electronegativity between silicon (1.90) and oxygen (3.44) makes the bonds effectively polar. Quartz has an infrared emissivity above 0.9.
Natural Earth materials
Materials like water, ice, and snow contain dipoles from water itself. They are highly absorptive and emissive. Rocks and bare soil are composed generally of silicate minerals and have a rough surface of loose particles where multiple reflections aid absorption and emission; Emissivity is generally above 0.90. Vegetated soils are occasionally damp, and their reflectivity is reduced by scattering within the vegetation. They are highly emissive.
The result is that quite a lot of the Earth’s surface by virtue of composition, roughness, water or vegetation is very nearly a black body at IR wavelengths. MODTRAN, for example, uses a surface emissivity of 0.971 for all its default calculations.
The Model Atmosphere
A number of people who comment on these threads speak of thermalizing the blackbody radiation. In typical usage of this word it would mean the radiation coming into equilibrium in some way with its environment. What I suspect people mean by their usage, is that black body radiation is converted into a distribution of motion of the material constituents of the atmosphere appropriate to the local temperature; i.e. radiation disappears into kinetic energy quickly.
Consider an atmosphere composed of nitrogen only. Place it in a cavity if one wishes, to act as a heat bath to maintain it at a temperature of, say, 300K. Even though the cavity is filled with blackbody radiation there is no interaction of the radiation with the nitrogen because the nitrogen does not possess a dipole and has no means to couple to the EM field. Figure 3 shows what an IR spectrometer would detect at satellite levels looking down at the Earth’s surface in this instance – a pure blackbody curve at a temperature of 294.2K.
Figure 3.
Figure 4 shows the heat capacity at constant volume for nitrogen and carbon dioxide between 200K and 1800K. From 200K to around 400K or so, the curve for nitrogen is very flat and equal to 5/2 R, where R is the gas constant. By kinetic theory of gasses this suggests at typical Earth atmosphere temperatures 200-300K, a nitrogen atmosphere has five degrees of freedom – three translations in x, y , and z, plus two rotations are all fully participating in energy storage, and the average energy per nitrogen molecule is 5kT, where k is Boltzmann’s constant. At 300K this is 2×10-20 J or about 0.13 eV per atom. Vibrational states for nitrogen are not involved until much higher temperatures.
Carbon dioxide, on the other hand, has a Cv at 300K indicative of the participation of about six degrees of freedom, or a little more, thus the activity of three translational modes, two rotations, and one or more vibrations. The slope of its curve suggests more vibrational degrees of freedom are becoming effective quickly.
Figure 4.
Adding CO2 to our model nitrogen atmosphere
CO2 has three vibrational modes in the IR portion of the spectrum. The symmetric stretching mode (n1), degenerate bending mode of CO2 (n2), and the asymmetric stretching mode (n3) can all be excited by collisions with N2 gas. Average kinetic energy per degree of freedom at 300K is roughly 2×10-21 J or 0.013 eV and the least energetic mode of vibration of CO2 (n2) requires 1.3×10-20 J or 0.08 eV. Another way to look at this is that the equivalent temperature of the n2 mode is about 960K, so even at a gas temperature of 300K the more energetic N2 molecules in the (Maxwell) speed distribution are capable of raising a ground state CO2 molecule into an excited n2 state during a collision. If we consider CO2 as a two-level atom, the Boltzmann factor e-∆E/kT indicates that around 4% of nitrogen molecules would reside in the n2 state at 300K equilibrium. That the n2 mode is more rapidly de-excited by a collision than by emission matters not because detailed balance demands that the reverse occurs just as often.
By adding CO2 we now have an atmosphere that can radiate to other parts of the gas, to the enclosing surface, or through a window . A spectrometer placed outside the body of such gas can detect IR radiation coming from the gas. It is an experiment exactly like this that produces Figure 5, which is a graph helpful in calculating how CO2 in combustion gas transfers heat to the enclosing vessel in an oven, furnace, boiler, jet engine, and so forth.
Figure 5. The effective emissivity of CO2 mixed with non-participating gasses depends on pressure, temperature, and path length. From Incropera and Dewitt, Fundamentals of heat and Mass Transfer.
An Aside on the IR Spectrum of CO2
While we often refer to a powerfully absorbing CO2 transition at about 15um wavelength (the n2 mode) as a single feature, the reality of this feature is quite complex.The reason is that rotation of an atom in mode n2 stretches the bond of O to C and this affects the bending mode vibration (n2) energy. It produces a series of many closely spaced lines each one differing in total angular momentum. There are three branches to this line known as P,Q, and R. The difference between one and another is that on the P-branch the total angular momentum (J in quantum mechanical parlance) changes during a transition by -1. On the Q branch J changes by zero. On the R branch J changes by +1.
The lines are part of a very fine structure where the wavenumbers of individual lines differ from adjacent ones by only 1 to 10 cm-1. A spectrometer with resolution of several cm-1 or worse cannot see them individually but only as a sort of continuum. One cm-1 difference at this wavenumber is about 30GHz difference in frequency.
A Radiating Surface in Contact with the atmosphere
Adding CO2 allows our atmosphere to interact with a passing EM field. Considering an isolated spectral component of the blackbody radiation emanating from the surface is analogous to an EM field excitation of gas – as in a maser for example. Some of the EM field is absorbed, then de-excited by collisions, excited by other collisions, and so forth. This passing EM field raises the population in the excited state slightly above what equilibrium at the gas temperature would suggest.
For segments of the IR spectrum where atmospheric components are highly absorbing, like the 15 um CO2 n2 complex, a spectrometer will detect only emissions from very near because those coming from far away have been absorbed. Figure 6 shows this in the segment indicated by blue arrows.
Non-equilibrium effects of gradients
The radiation emitted by a CO2 atom making transition from the vibrational state to ground state is not isotropic, but there are so many CO2 molecules per unit volume of atmosphere, all of which are randomly oriented, that the resulting radiation from bulk material is isotropic.
Considering the isotropic nature of this radiation, it is quite apparent that radiation emitted from bulk materials from a cooler place can land upon warmer materials, such as the ground surface and be absorbed there. No physical principle suggests otherwise, and without considering all such radiation it is not possible to determine net heat transfer correctly nor achieve conservation of energy. Net transfer of heat occurs spontaneously from warmer environments to colder ones, even considering that individual packets of radiation are not constrained by temperature and may go the other way.
Energy to maintain a populated n2 state is dependent on local temperature plus the brightness of any passing EM field. The brightness of spectral lines generated also depend on the concentration of radiating species. What a spectrometer looking into an atmosphere of such will detect is no longer a continuous blackbody curve, but rather segments, or even non-recognizable individual broadened lines of brightness dependent on what part of the atmosphere supports the observed spectral feature.
For example, Figure 6 shows a spectrographic measurement of Earth’s surface measured from a satellite. In the broad section from wavenumbers 800 to 1200 cm-1 demarcated by red arrows, apart from a dip in the spectrum near 1050 cm-1, the curve looks almost like a blackbody spectrum from a source at 290K – the Earth’s surface, in other words. This is a clear window in the atmosphere where the spectrometer sees the surface unimpeded. Meanwhile the blue arrows denote sections of the spectrum where CO2 and H2O block direct view of the surface and the spectral brightness comes from the last few km of CO2-rich atmosphere near the tropopause (220K) or top of water vapor-rich atmosphere near 2 km above the surface.
Figure 6. From Harde, International Journal of Atmospheric Sciences, Volume 2013, Article ID 503727, 26 pages, http://dx.doi.org/10.1155/2013/503727
The Impact of Clouds
Clouds present a complex picture. The minute particles that make up clouds are undoubtedly large enough at times to behave much like an ensemble of small blackbodies, which if dense enough, would, as a thick enough layer, act as black bodies themselves at the local cloud temperature.
The effect of smaller particles in clouds redirect visible radiation. In the summer a person can stand outside on the south flank of a tall cumulonimbus tower and notice, not only the very bright sky to one’s north, but also increased warmth on exposed skin and on the ground surface that does not much affect air temperature. See for example Figure 5 in reference [2].
Many cloud particles are of the order of the wavelengths of light and thermal IR, or slightly larger, and scatter radiation via Mie Scattering, but in general the true phase function for cloud scattering is poorly known. Wijngaarden and Happer (2005) [2] have formulated a 2n algorithm that replaces the poorly known phase function with a scattering matrix that could be stated accurately with measurements. Thick low level clouds redirect thermal IR originating at the ground surface.
Undoubtedly clouds act to lower Earth temperature from what it would be under a cloudless planet – an obvious negative feedback in this sense. However, from the standpoint of orthodox climate science this large effect is what maintains the current mean climate, and by cloud feedback they intend the impact that a slight increase in cloudiness would have on the mean climate state. This divergence in definitions fuels much debate.
Considering IR radiation alone, how does the Earth’s surface cool?
The SB feedback factor is the increase in emitted thermal radiation from a surface by virtue of increasing surface temperature. To calculate this factor one would simply differentiate the SB law with respect to temperature, then invert the result.
W = σT4; where σ, the Stefan constant, is 5.67×10-8 in SI units. By the power rule of differentiation dW/dT is then 4σT3, and its inverse, dT/dW is, at 300K, 0.16 K per W/m2. In other words, for the SB law to radiate an additional watt per square meter at 300K demands only an extra 0.16K of surface temperature.
However, the extra watt of blackbody radiation will be partially absorbed by water vapor and carbon dioxide and some of this returns to ground to frustrate cooling. In this instance of the present greenhouse effect, a greater temperature increase is needed to fully radiate away one watt per meter squared to space. In an essay posted six years ago this was explained by treating the entire atmosphere as a surface coating to give the Earth a better “figure of merit” as a solar collector. The atmosphere acts to reduce the effective surface emissivity from near 1.0 to about 0.62. Thus surface radiation is actually W=εσT4; and dT/dW is 1/(4εσT3), which at 300K is 0.26.
Now, one can argue that not every place on the planet has this same “coating”, but the point is that unless there is no local greenhouse effect at all, it takes more than the SB derived sensitivity of 0.16K and perhaps as much as 0.26K increase of surface temperature to cool the surface by one watt per meter squared.
References
- https://www.nesdis.noaa.gov/news/how-atmospheric-sounding-transformed-weather-prediction
- W. A. van Wijngaarden and W. Happer, 2025, Radiation Transport in Clouds, Climate Science, Vol. 5.1 (2025) pp. 1-12.
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Without any comment, there’s this web page:
https://climatepuzzles.org/
Nice set of pages. Thanks for the URL.
Great article, Kevin, thanks.
Thanks!
In figure 5, pc,L = 4.0 ft.atm.
what is the meaning of that term?
Heat is transferred to jet engine surfaces, but almost all energy is due to expanded, very hot, combustion gases entering, at high velocity, the very short blades of the compressor, and then the turbine
Partial pressure of carbon dioxide and L is path length.
What mkelly says in customary units — you know, U.S. engineers still use lots of non-SI units.
the charts produced in the 1930s for calculating radiation from non luminous gases at high temperatures and used in Macadams show emissivity as a function of gas partial pressure (ats) and mean beam length (ft). T hey are for the principle radiant constituents Co2 and H20 in flue gases. I use them all the time in furnace calculations. They were developed long before SI units were imposed on us. Great units – BTU/,ft^2/F/in etc
You chose the best screen name. Great job.
The Fig 5 chart used for furnace calculations can also be applied to IR absorption by CO2 bands in the atmosphere. The small proportion of IR concerned is absorbed to extinction in quite a short traverse the the atmosphere. In a dry atmosphere m0st of the IR escapes unabsorbed as shown in arid desert conditions leading to huge day night temperature swings.Co2 at 380 – 400ppm does not of itself absorb sufficient IR to cause runaway global warming; it is simply quite impossible
Pc,L
Partial pressure times length of the gas medium between the emitting and receiving surfaces. Hottel charts were made for furnaces at a time when many engineering calcs used nomograms and charts (before electronic calculators). This particular usage is a shorthand method of “calculating” the number of CO2 molecules in the IR path… and coming up with a broadband emissivity. You wouldn’t use it today except as a quick backcheck on your computer output….
Here’s a deeper dive with the usual lack of description of how to guesstimate “L”
but includes “from-the-horses-mouth” references.
https://www.thermalradiation.net/On-line%20appxs%20text%2012-11-22.pdf
The charts work remarkably well . Setting up the furnace heat transfer model is key. Satisfying heat balances and furnace gas and refractory surface temps eliminating seat of the pants guesswork – great. Chemical engineers of the 1930s facilitating design of heat transfer equipment before spreadsheets were a godsend to the slide rule generation.
Thank you.
This is the essence of why Shula & Ott are incorrect:
The population of molecules energetically capable of emission is constantly maintained by collisions. It is not diminished by them. It’s a fundamental feature of the Kinetic Theory of Gases.
The corollary is that Wijngaarden and Happer have it right, along with Hermann Harde.
‘The population of molecules energetically capable of emission is constantly maintained by collisions.’
Yes, but their point is that these molecules don’t spontaneously emit photons in the lower troposphere, hence making radiative transfer models inappropriate in this layer. The upshot is that the thermalization of excited GHG molecules by collision with no-GHG species produces sensible heat than is then convected aloft to where the reverse process will excite GHGs in layers where spontaneous emission can occur.
Note, if radiant transfer does apply to the lower troposphere, then you’ll need to explain why there isn’t any evidence that CO2 is ‘the control knob’ of the Earth’s climate any where in the geological record.
That point is wrong (though I don’t think they really say that). There is a proportion of the population of molecules that depends on local temperature and density that is emitting constantly with probabilities in accordance with Einstein coefficients, which are quite independent of residence times in any particular energy state. The number of emissions per second per unit volume is higher for higher temperatures – I.e. at ground level, because the proportion of energetically competent molecules is higher.
Here is Wehrbein & Leovy (1981) “A RADIATIVE HEATING AND COOLING ALGORITHM FOR A NUMERICAL MODEL OF THE LARGE SCALE STRATOSPHERIC CIRCULATION (pdf)” A NASA report.
p. 14 provides the Einstein coefficient for CO₂ emission to be 0.74 sec⁻¹. They also provide the rate of collisional decay at 180 K, 1 atm = 0.67E5 sec ⁻¹.
Given the rate disparity of 9.05E4 favoring collisional decay, there’s no way radiative decay can contribute to sensible heat.
The constant Boltzmann concentration of excited molecules overwhelmingly decays by collision.
These numbers corroborate Curtis & Goody (1956).
‘There is a proportion of the population of molecules that depends on local temperature and density that is emitting constantly with probabilities in accordance with Einstein coefficients, which are quite independent of residence times in any particular energy state.’
Discussions about Einstein coefficients are beyond my ken, but you shouldn’t presume Shula & Ott haven’t considered this aspect of radiative physics:
‘The Einstein coefficients are valid for atoms in a radiation field where both excitation and de-excitation are driven by interactions with photons. After thermalization near the surface the excitation and de-excitation in the troposphere are dominated by collisions with other gas molecules, thermalization and thermal excitation. There no longer exists a radiation field for radiative transfer to occur.’
As for the idea expressed elsewhere in the comments here that radiant transfer models can be applied to the lower troposphere on the assumed existence of LTE, they say this:
‘Some also invoke local thermal equilibrium (LTE) to make some of these assumptions. This is not valid. LTE implies that the Maxwellian (kinetic) temperature is the same as the Planck (radiation) temperature and the Boltzmann (energy state distribution) temperature. LTE cannot exist in the lower atmosphere which is never in equilibrium.’
Yes,
but, the reverse process, occuring just as frequently un-produces that sensible heat. The idea of local thermodynamic equilibrium, which is pretty central to calculating brightness of lines, demands something extremely close to detailed balance.
exactly
‘but, the reverse process, occuring just as frequently un-produces that sensible heat.’
That would be true, except that in a gravitational field the sensible heat is convected upwards, which makes the process irreversible since work is done.
Upon review, I am left with the impression that the purpose here was to paper over any science that would suggest that IR-active gases can’t be treated as condensed matter for purposes of radiant energy transfer in the lower troposphere.
I am not a physicist, but would suggest that ‘detailed balance’ doesn’t mean that photons have to be conserved when excited CO2 molecules undergo collisions, just energy.
You are making the mistake of conflating Kichoff’s law for material bodies to apply to individual GHG molecules.
Your statement, “…but, the reverse process, occuring just as frequently un-produces that sensible heat.”, is incorrect. At STP, about 4% of the gas molecules have sufficient energy to excite the vibrational state of CO2, but 100% of the molecules can de-excite it. It is not symmetric. Of the 4% that are excited by collisions, only one in 100,000 escape de-excitation an emit IR, and that IR is immediately absorbed by another CO2 (or more likely, H2O) molecule.
That being said, the population of CO2 and H2O is sufficiently large that a detector will “see” this radiation even though it is not transporting any energy.
Schwarzschild stated in his paper: ““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.”
Schwarzschild was attempting to describe the Sun’s photosphere. It should be obvious from his statement that his radiative equilibrium model would not be applicable to the Earth’s atmosphere. It was derived from Shuster’s one dimensional two-stream model of radiative transfer.
These models phenomenological and based on radiance, not radiative flux which are very different.
It’s nice to see someone else invoking Harde, whose treatment is much more analytical than Happer and van Wijngaarden (and predates them by ~7 years), but you have cherry picked a few points to continue to support the “radiative transfer” narrative. Both are deeply flawed and ignore the true nature of energy transport in the atmosphere.
That the one-dimensional radiative equilibrium model can reproduce a satellite spectrum is not a sufficient condition to conclude the atmosphere is driven by radiative transfer.
Energy transport in the atmosphere is via convection, which involves ALL of the molecular species in the atmosphere. They obtain this energy via absorption of radiation from the surface by IR active gases (mostly H2O) which then lose their vibrational energy via collisions converting it to sensible heat. This is not controversial, but it is an irreversible process, reference here:
https://repository.up.ac.za/server/api/core/bitstreams/4873bb3c-2b97-4530-b113-4592807780b3/content
More importantly, sensible heat is added via conduction which is ignored in the popular equilibrium models because in the models insolation is constant and the model represents “equilibrium”.
It is the sensible heat in the atmosphere that drives the collisional processes which create the radiation field that exists from the surface to the mesopause. The field is explained in Einstein’s “The Quantum Theory of Radiation” (1917).
https://s3.cern.ch/inspire-prod-files-9/9e9ac9d1e25878322fe8876fdc8aa08d
Einstein recognized the relationship between the Stefan-Boltzmann distribution and the Planck distribution. This explains why atmospheric collisions generate a radiance spectrum with the shape of a Planck spectrum.
In a nutshell, radiation can escape to space when the rate of spontaneous emission, as measured by the Einstein A coefficient, exceeds the rates of absorption and collisional de-excitation.
For water vapor, this occurs in the mid to upper troposphere in clear skies. It also occurs at the tops of clouds where there is always a population of free water vapor molecules in dynamic equilibrium with the condensed water that can radiate fiercely to space. This can be observed in BAND 16 of GOES (thank you David Dibbell).
Water vapor is responsible for almost all of the heat loss to space. This is obvious from the spectra if you read Harde’s description of his figures 17 and 18, though the spectra are commonly misinterpreted. His figure 17 also shows that water vapor is quite active in the so-called “atmospheric window”.
Ozone is in emission in the stratosphere, and CO2 isn’t in emission until we reach the mesopause, the tiny peak at the bottom of the spectrum “divot” between 14-16 μm.
CO2’s role in atmospheric heat transport is close to zero. Water is the driver.
“The population of molecules energetically capable of emission is constantly maintained by collisions.”
And 99.997% of collisionally excited molecules decay back down by collision. The intensity of decay-radiated IR is always tiny.
You have not understood the fundamental point that regardless of the frequency of transfers between energy states, the population of molecules capable of emission remains constant for given local temperature and density, and therefore so does the rate of emissions. The rate of emission depends on that population, not the fact that the members of the population are in constant flux.
The rate of emission depends upon the partitioning of decay modes. Collisionally activated CO₂* decays 99.997% by collision.
For any constant population of collisionally activated CO₂*, the intensity of emitted radiation will always derive from 0.003% of that population. Effectively zero contribution to surface warming.
Pat,
I’m going to suggest that you research the concept of “detailed balance” as it applies to statistical physics. Unfortunately you can get into the weeds pretty fast regarding detailed balance on the internet, much of which is about chemical reactions and not statistical physics.
How would doing so change my view that collisional deactivation always dominates?
Because with detailed balance, Pat, the activation is equal. These microscopic processes are time invariant. That’s the basis of Onsager’s irreversible thermodynamics.
not time invariant, darn it, but time reversal invariant…
Detailed balance just refers to the stationary state. Here, that state is the constant Boltzmann fraction of vibrationally excited CO₂* at a given temperature.
About 20m or so above the surface, that state is maintained by collisional activation and collisional decay.
Not correct. Try this analogy. Imagine at the end of an athletics meeting the 4,000 spectators spill onto the 400m running track, 10 across at 1m intervals. Consider the population divided into 100m segments, representing 4 different populations of “molecules”. The high energy ones, waving their arms about, are in the finishing straight, and the lowest energy, arms to their sides, in the first bend. While they stand there, there is no change in which people are part of which population, with their “energy” represented by how vigorously they wave their arms (dictated by which section of track they are in). Now have them walk around the track slowly at 1 metre per second – the people in each segment will change slowly, but the total number in each segment will remain at 1,000. Here, the speed of walking is the rate of change of population of each energy level. Speed them up to marathon pace of about 5m/sec, and more of them change from one population to another, but still the number in each section remains 1,000 all the time. Speed them again to 10m/sec for the 100m dash pace – same thing. Turbocharge them to 29,000m/sec – still the same thing. The rate of change of the populations doesn’t alter how many are in each segment. There are always 1,000 “high energy” ones in the finishing straight.
Now, for any given excited energy state there is an instantaneous probability of decay to a lower energy state with the spontaneous emission of a photon. We can imagine a marshal in the stands manning a paintball gun aimed in the general direction of the finishing straight with his daughter wiggling the direction of the barrel, firing single shots at the passing crowd. Anyone hit has to drop out and cross over to the lower energy back strait and hand out their “paintball” photon on the way. The rate of hits doesn’t depend on how fast the crowd is moving – it just depends on the number in the firing line, which is constant.
There is a good exposition of Maxwell’s derivation of his eponymous distribution of gas molecule speeds and energies here:
https://phys.libretexts.org/Bookshelves/Thermodynamics_and_Statistical_Mechanics/Supplemental_Modules_(Thermodynamics_and_Statistical_Mechanics)/Thermodynamics/1.5%3A_The_Kinetic_Theory_of_Gases
Some calculus required.
“Now, for any given excited energy state there is an instantaneous probability of decay to a lower energy state with the spontaneous emission of a photon.”
Which, at 1 atm, is ~(1/29000) of the probability of decay by collision.
The ~5% Boltzmann population is dynamically maintained as constant, by collisional promotion and collisional decay. The concentration of excited molecules is constant.
The back-to-ground-state decay channel of the constant Boltzmann population of collisionally excited CO₂ molecules at 288K (~5%), will necessarily be either by collision or by radiation.
The ratio of these energetic decay processes is governed by the relative rates of the two modes of decay. These are ~2.3/sec for radiation and ~6.7E4/sec for collision.
The dynamics of energetic decay is almost entirely governed by collision.
Nothing in your explanation gainsays that.
Let me add that the photons released by the rare radiative decay of CO₂* will in turn be rapidly absorbed by ground-state CO₂. Photonic absorption produces CO₂*. This CO₂* in turn will overwhelmingly decay by collision (at 1 atm).
Radiative decay is invariably frustrated by collisional decay at every step of the process. The intensity of radiation produced by radiative decay will always be extremely low. Back radiation will always be negligible.
Read the example again. The rate of collisional decay is matched by the rate of collisional promotion, so there is no net change in the high energy population, and it has zero effect on the rate of emissions. The rate of emissions only depends on the number of energetically capable molecules which is quite sensitive to temperature.
There are two consequences of photon emission compared with collision. Photons move at the speed of light, whereas molecules are moving at speeds of a few hundred m/sec on average. That means that the effective cross section of a molecule is much smaller for a photon than another molecule, because the molecule can be regarded as a static target for a photon, whereas molecules are rotating and maybe vibrating at high frequencies, so presenting a much larger effective target to each other. That means that the mean free path of photons is longer before their next collision (also because they will only interact with a molecule that has a corresponding quantum energy level free), although their relative speed means that it occurs much sooner. Convection as a process is slower still, relying on the bulk movement of air parcels at a few m/sec aside from hurricanes and tornados.
“The rate of collisional decay is matched by the rate of collisional promotion, so there is no net change in the high energy population, and it has zero effect on the rate of emissions.”
The rate of emission is thousands of times slower than the rate of collision throughout the troposphere.
And? All that means is that any deviation from the Maxwell distribution caused by either the emission or absorption of a photon is repaired extremely rapidly.
You appear to be equating to a photon to a bullet. A photon is a quantified packet of energy inherent to an ever expanding EM wave. Analogizing a molecule as a “static target” isn’t exactly correct.
It’s a good enough model for estimating comparative mean free paths. If you really want to get precise you can always try using the most detailed parameterisation of behaviour.
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.133.203604
To help you understand the physics, see my comment to Kevin here:
https://wattsupwiththat.com/2025/06/07/radiation-and-matter/#comment-4081596
I don’t understand the analogy – energy transfer from the surface to space is not a closed loop. Are you suggesting that photons, as well as energy, are conserved?
The closed loop idea was a simplified attempt to reflect the idea that the distribution of energies remains unaltered by the rate at which molecules change from one energy level to another via collisions, and therefore there is an essentially constant population of enmission capable molecules at any given temperature – it does not decline because some molecules lose energy since they are replaced by others gaining it. A more technical approach is via Boltzmann’s H theorem but that requires much more mental gymnastics to follow.
The other reason was to separate out the probability of emission from the time spent in a high energy state. I think there is a lot of confusion caused by the idea that collisions are frequent.
Would the result be the same if the average time for an excited molecule to radiate was a billion years?
I would never argue too much with Wijngaarden or Happer or Harde for that matter. Very capable guys. The trouble though in this topic in particular is that the physics education is distressingly different from engineering and chemistry as well. Sometimes Happer discusses things in a way not especially accessible to non-physicists. At other times he’s really good at it.
In physics education “thermal physics” looks different from kinetic theory which looks different from engineering thermodynamics. It’s the reason that I put in Figure 4, Cv versus Temperature for CO2 and N2, and labelled Cv in units of R. I wanted to show that macroscopic measurements of Cv (engineering) conform nicely to ideas about molecular degrees of freedom in kinetic theory.
‘I would never argue too much with Wijngaarden or Happer or Harde for that matter. Very capable guys.’
They are capable, and I wouldn’t argue with them, either. On the latter point, I recall from one of vW&H’s papers that our predominantly O2 and N2 atmosphere would be largely isothermal but for the presence of GHGs, which if memory serves, impart sensible heat near the surface and extract same at altitude, thereby resulting in a convective troposphere.
My difficulty, then, is with understanding the ‘consensus’ climate physics (detail balance?) that collisions of excited IR-active gases with non-IR-active species at the surface (or the inverse process aloft) has no impact on the quantity of spontaneously emitted photons at either level.
If true, and given that basic energy conservation means that there has to be a trade-off between the quantities of kinetic and photonic energy, then the inescapable conclusion for this non-physicist is that vW&H are incorrect with regard to the process of convection.
No, molecules are constantly colliding and being deflected up or down. The force of gravity causes them to slow down a little as they go upwards, and speed up as they go downwards. The result is the “lapse rate” of the atmosphere…colder with increasing elevation. There are those who attribute the lapse rate to convection and greenhouse gases, and say that if the atmosphere was entirely non-GH gases, it would be isothermal. Mere mention of this topic at WUWT generally causes vitriolic comments…so have at me…
Not in the mood for vitriol – enjoy the day, sir.
What does your intuition tell you about the distance they travel before colliding?
This was an interesting and extremely well-written article to be sure, but the detailed presentation on IR seems entirely irrelevant to the conclusion. The real question is, where does the “extra watt” come from in the first place?
That concluding paragraph is not meant to summarize the whole effort, but rather to address some contention over Stefan-Boltzmann feedback.
Thought experiment: If adding a bit of CO2 to the atmosphere were to lower outgoing IR by
, then what temperature increase at the surface would be required to bring the outgoing IR back up by that amount and restore balance?
I was just pointing out that the reality of greenhouse gasses above the surface means that the derivative of the SB law w/resp to T, inverted, does not present the entire answer to this.
Adding CO2 to the atmosphere does not stop the surface cooling. Step outside at night, and you will see a cooling surface.
Your waffling about “balance” is delusional. A cooling surface is cooling. Losing energy. It doesn’t magically heat up to “restore balance”. It gets colder!
After four and a half billion years of continuous sunlight, the Earth’s surface has cooled. Lost more energy than it received, both externally and internally. No “balance” at all. The energy has gone. Vanished – pffft! Never to be seen again. Conserved, but not around here, that’s for sure.
Bickering means arguing about petty matters and the debate over radiation physics is not petty. It is fundamental. You need to recognize it.
Quite. I think Kevin has made an excellent stab at laying out the correct physics.
David, radiation physics is important. Yet, please have a look at the length of some subthreads on the article that prompted my effort here, and I think that bickering classifies it accurately.
My point is not that radiation physics is important. It is that there is deep debate about it that often plays out here at great length. I see a little bickering but mostly it is serious debate. It is one thing to present a position but claiming that opposition is bickering is just name calling. There is no agreement ate fundamental level.
This is where I always run into trouble: “However, the extra watt of blackbody radiation will be partially absorbed by water vapor and carbon dioxide and some of this returns to ground to frustrate cooling.”
Curtis & Goody (1956) calculate, and present, the rates of collisional and radiative decay of both H₂O* and CO₂* = the vibrationally excited molecules.
At 1 atm. pressure, the collisional decay of each molecular type is ~2,900× and ~29,000×, respectively, faster than its radiative decay. Energy transfer is >99.99% collisional.
“Back-radiation” from the decay of CO₂* or of H₂O* is of extremely low intensity. Negligibly low.
Either I’m overlooking something critical, or back-radiation can play no important role in further warming the surface,
Molecules in a gas have a distribution of kinetic energy or equivalently of velocity that is a function of temperature. See Maxwell-Boltzmann distribution. Via collisions, they are constantly exchanging energy and momentum with each other. There is no change in the distribution of molecular energies as a result of collisions. There are always the same proportions of faster and slower moving molecules. Energy and momentum are conserved unless there are other means to add it or remove it. For every molecule that loses energy in collision another gains it. There is no decay in the proportion of high energy molecules in the gas unless energy is removed from it, e.g. by radiation.
The rate of molecular collisions for a given temperature fundamentally depends on the density of the gas.
Nevertheless, the intensity of radiation emitted from any constant (small) Boltzmann population of collisionally excited CO₂* will always be derived from 0.003% of that population (at 1 atm).
The tiny intensity of emitted radiation contributes nothing to warming.
Yes, that excited state maintains a steady population (fraction) of about 1/ 30,000, at standard temperature, but …
First, the emission rate (lifetime) may be about 300,000 per second, so that each molecule on average ‘turns over’ ~ 10 infrared photons per second!
Second, that population increases steeply with every (slight) change in temperature
From that point, one complete the calculation (based on CO2 density, i.e. number of molecules per unit volume) to turn this into emitted Watts per cubic meter, or fluxes of Watts per m2 (in a particular direction, e.g. ‘downward’ / downwelling).
Pat, Robert Whetten, below, addresses part of this in the italicized quote. I keep mentioning “detail balance” which is what that quote is about. Unfortunately detailed balance is a concept not encountered in undergraduate physics, which is an pedagogical deficiency in my view.
However, when you say the tiny intensity contributes nothing, I would point to Figure 5 in my text which shows that some CO2 added to a nitrogen atmosphere will radiate like a black body with an emissivity less than one, and which depends on temperature. Note that the family of curves shows that the effective emissivity is greatest at lower temperatures. That observation shows a relation between the Maxwell speed distribution, the n2 vibration of CO2, and the SB curve. Transport problems in many engineered systems can’t be done with sufficient accuracy without graphs like Figure 5.
Kevin, your Figure 5 is Emissivity. It’s not the emission intensity when collisional decay is the overwhelming route back to the ground state.
This is why the warming crowd left the back radiation argument years ago. Now they’re beginning to realize that collision transfer is a 2 way street and also can only have a minimum impact on temperatures. So what’s the next play?
‘So what’s the next play?’
By my reckoning, there have been at least two changes in the narrative. Since an increase in radiative forcing will always result in warming when run through a radiant transfer model, the current / next play will be to defend the use of these models via the continued conflation of the radiative properties of GHGs with those of condensed matter.
Wrong. The proportion of molecules energetic enough to emit spontaneously is much higher: in the example Kevin gave it is ~4%. Do the math.
Energy of the 667cm-1 line or 20THz is hv =~6.3e-34 x 2e13 or ~1.3e-20J as he stated: that is ΔE. k is 1.38e-23 and T=300K so kT is ~4.1e-21
-ΔE/kt is therefore -3.2, and exp(-3.2) is ~0.042, or just over 4%.
The fact that the energies of individual molecules are constantly changing due to collisions, and that therefore the great majority that had substantial energy have given it up in collisions by the end of a second is irrelevant: the population of molecules with the requisite energy is sustained by the ongoing process of collisions.
I’ve previously done the equivalent calculation. The fact that there’s a constant fraction of vibrationally excited CO2 does not gainsay the fact that the overwhelming majority of them always decays collisionally.
There are ~6.9E21 CO2 molecules/m^3 at 400 ppm and 1 atm. Let 4.2% be CO2* => 3E20 (molecules/m^3)*0.74/0.67E5 = 3.3E15 photons/sec/m^3*1.3E-20J/photon =>1.0E-4J/sec/m^3 = 1E-4 W/m^3.
If half go up and half down, the downwelling flux is 5E-5 W/m^2. Negligible.
Decay by collision is not decay: it is merely the rate at which more and less energetic molecules move between energy states due to collision, and is equal and opposite for the collision process. Decay by emission is an entirely independent process that does not depend on residence time in the higher energy state: the rate depends solely on the number in that state.
Why are you dividing by 0.67E5? You are also ignoring the inward flux of photons from surface emission. If the energy they represent isn’t dissipated then it must contribute to ongoing heating. They do, but only to the extent that their average upward path is delayed by collisions, which establishes an equilibrium temperature.
“Why are you dividing by 0.67E5?”
0.67E5 s⁻¹ is the collisional relaxation rate for CO₂* at 1 atm and 180 K, from Wehrbein 1981 (pdf) previously mentioned. 0.74 s⁻¹ is the Einstein coefficient for spontaneous photon emission.
The ratio 0.74/0.67E5 provides the ratio of radiative decay/collisional decay in any given molecular assembly of CO₂* = 1.1E-5.
For every 3E20 CO₂* molecules per m³, 3.3E15 will decay by emission and ~3E20 will decay by collision.
A constant flux of 3.3E15 photons/sec/m³ will be maintained by the Boltzmann population of CO₂*.
(3.3E15 photons/sec/m³)*(1.3E-20J/photon) =>1.0E-4J/sec/m³ = 1E-4 W/m³.
If half go up and half down, the downwelling flux is 5E-5 W/m². Heating of the surface by back-radiation is negligible.
That’s only true if one is considering a Maxwell-Boltzmann distribution of mixed gases at asymptotic equilibrium and without input of energy. If one wants to consider what is happening in reality to Earth’s atmosphere with the continuous input of LWIR energy (as radiation off Earth’s surface) and with the continuous outflow of thermal energy off the ensemble of mixed gases—some “LWIR-active”, some not so much— back to the surface and to deep space, then the distribution of molecular energies is necessarily not in equilibrium (i.e., cannot be unchanging as a result of collisions).
If there is a disequilibrium in energy input and output then there will be heating or cooling. That will result in changes in pressure and volume to match. A new equilibrium is established.
E = PV = nRT
The ideal gas law (PV = nRT) is an equation of state in the absence of energy transfer across the control volume under consideration. It simply relates pressure, volume and temperature to each other for a fixed mass (number of moles, n) of gas in the absence of any energy crossing the control volume of that gas mass.
This equation, per se, is not useful for calculating the energy contained a control volume because, by definition, ii is just a constant value for a fixed (i.e., unchanging) temperature, although it does have the units of joules (of energy) in the MKS system.
So when you add or subtract energy from the gas a new state results. It does not explain the rate of energy input or output, but it does explain the consequences of change, athough factoring between pressure and volume gets more complex in a atmospheric column rather than a sealed lab flask where all the change is to pressure because the volume is held constant.
But the fact is, it (the ideal goes law) doesn’t even do that. For example, from state 1 for a given control volume, I added thermal energy to raise the control volume’s uniform absolute temperature at state 2 by 5% (say from 300 K to 315 K) . . . please tell me what then are the resulting changes in the pressure and volume within the control volume.
You simply can’t do that for each separate parameter (P and V) without introducing a totally separate thermodynamic parameter (specific heat of a substance) and a related thermodynamic equation, such as that for isentropic expansion (P*V^k = constant, where k is the ratio of specific heats, c-subP/c-subT). Beyond isentropic expansion of a control volume, there are also posslble adiabatic and polytropic expansions of such.
Your example has an inherent constant volume, i.e., “control volume”. That means any change will be evident only in a change in pressure.
One doesn’t even need the ideal gas law. The combined gas law suffices.
PV/T = k
Errrrr . . . a “control volume” is NOT defined as being constant volume, instead only defining a constant mass and a “surface” over which the transfer of energy/work can occur.
For example, consider the numerous examples of the thermodynamic “control volume” being defined as the volume confined by a movable piston in a otherwise closed container.
BTW, there is no scientifically accepted equation that PV/T = k for all gases . . such would imply that nR = constant for all gases, which is patently absurd.
ROTFL.
https://www.mechstudies.com/control-volume/
Control Volume is defined as the fixed volume over which mass can transfer in and out across its boundary.
https://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node19.html
https://ch301.cm.utexas.edu/gases/#gas-laws/combined-law.html
I’m not sure what you think the term “Law” means, but to me it means generally applicable. There are instances where the Gas Laws do not apply for sure, but for general application in the atmosphere, they are “Laws”.
As for your references to “control volume” necessarily being a fixed volume, one needs to exercise care as to what is being discussed/analyzed vis-à-vis an “open” versus a “closed” system:
— In an inertial frame of reference, a “control volume” can be a fictitious region of a given volume fixed in space through which the continuuum (a continuous medium such as gas, liquid or solid) flows; this appears to be the basis for your assertions as to control volumes always being fixed because one references mass transfer into and out of the boundaries of the control volume and the other refers to a conservation law. This is consideration of a control volume when evaluating an “open” system.
— However, in considering the thermodynamics of heat engines (of the simple piston-in-cylinder type that I specifically referenced) the control volume is defined as the arbitrary space within the cylinder that contains the fixed mass of fluid (typically an ideal gas). The boundaries of this control volume are the cylinder walls and the movable piston itself. Therefore, this “control volume” is NOT a fixed volume. And obviously, there is no continuum flow through a piston-in-cylinder arrangement. The piston-in-cylinder can hypothetically be considered to be adiabatic or not, and the piston can also hypothetically be considered or not as enabling work to be performed on or by the gas. This is consideration of a control volume when evaluating a “closed” system.
Next, as for your comment,
That may be so, but that law is of limited use. The equation PV/T = k applies only to a specific mass of a specific ideal gas. The value of “k” varies from gas to gas . . . and is obviously equal to the more scientifically appropriate quantity “n*R”.
Furthermore, the ideal gas law was derived from observation of Boyle’s Law and Charles’s Law with the inclusion of Avogadro’s Law so as to reference mass (as the number of moles, n) being necessary to derive the equation: PV = nRT.
Pat Frank writes “Either I’m overlooking something critical,…”
I believe that you are correct — the infrared-radiation energy that is absorbed by molecules in the upper atmosphere is efficiently ‘thermalized’ by collisions — so that there is indeed a missing step in the argument as it is usually presented.
But it seems to me that one can recover the essential effect by inserting an additional statement (marked in italics below):
In the upper atmosphere (as in any dense gas), the molecules excited vibrationally by absorption of infrared radiation, efficiently transfer that energy (via collisions) to nearby gas molecules (predominantly di-nitrogen N2, di-oxygen O2), degrading it into the usual (Maxwell-)Boltzmann distributions for molecular vibrations, rotations, translations.
[‘Thermalization‘ is a handy generic term given to this such a cascade of processes.]
But that degradation doesn’t mean the energy is lost;
rather it shows up as an increase in the temperature for that parcel of gas …
and thereby in an increase in the (minute) portion of the infrared-radiating molecules
that are always being returned to the excited (radiating) state,
in accordance with the Boltzmann distribution (at that temperature).
Once you’ve accounted for this —
a (slight) increase in the rate of the usual spontaneous emission process (‘isotropic’ in direction), which should probably no longer be termed ‘re-radiation’ with all that could imply,
— the rest of the calculation proceeds as usual, to arrive at the correct formula for the (ugh) ‘radiative GHE’.
Nevertheless, the emission intensity (0.00003 of the decays (at 1 atm)) is extremely weak.
Curtis and Goody note that radiative and collisional decay rates become equal at about 74 km.
This topic came up before and I still think some are reading too much into the term “radiation decay”. Consider what happens during absorption.
The molecule interacts with the EM field and collects the energy almost immediately. If it wasn’t immediate, it could not transfer than energy to another molecule during a collision. That means the energy already is internal to the molecule and the connection to the EM field is irrelevant but could have ended immediately after the energy becomes available for kinetic transfers.
The molecule still has the long radiation decay time even though the energy is long gone. This tells me the decay time is internal to the molecule but not directly tied to the interaction with the EM field.
When a kinetic collision occurs the energy is either going into the molecule or at least some of it will be emitted. This must also happen quickly if collisions can once again share energy with another molecule. There’s no place for that extra energy to exist. As a result I think the energy is emitted during the initial collision. Once again the radiation decay time is internal to the molecule, as in the absorption case, even though the actual energy is gone.
With this view, the interaction with the EM field occurs immediately for bother absorption and emission. The long radiation delay only affects a future interaction with the EM field.
Now everything works symmetrically and Kirchhoff’s Law is maintained for these gases.
I think that is to misunderstand the quantum process. There is no requirement for a molecule to remain in an excited state for any time at all. The rate of decay is simply in proportion to the number of excited molecules, regardless of how long they have been excited, at least at the timescales relevant here.
Please see the calculation here.
Well done. BZ.
Nice article!
There is a bit of room for improvement in discussing how the quantum nature of photons interacts with Maxwell’s equation, which explains why an excited molecule doesn’t radiate instantaneously. As you noted, the CO2 molecule is a relatively poor antenna requiring a longer average time (time constant as it is an exponential decay) to radiate when compared to water with its electric dipole moment.
You are correct, of course. It’s difficult to explain why that darned dipole, once vibrating, doesn’t emit IR radiation immediately without using QM. I decided that going off in that direction would be not productive because I wanted mainly to show that a dipole of some form is need to couple to IR fields, which explains a lot of observations and puts an end to the “all materials radiate IR” dogma — or I hoped it would.
Understood with respect to not wanting to go down the rabbit hole of explaining delayed emissions. The emphasis on electric or magnetic dipole was exactly right.
One of my learning experiences came from reading about how CO2 molecules once excited are much more likely to de-excite as a result of a collision than by emitting a photon (this was from Roy Spencer’s article on what skeptics get wrong). This led to me to consider what was going on with the process of radiation, thinking about gamma rays where the excited nucleus acts as antenna (emission of gamma rays are not isotropic with respect to the axis of the nucleus), then thinking the spectral lines should be Lorentzian and finding out they are. This line of thinking also clued me into what was going on “stimulated emission of radiation”.
A very solid effort, Kevin, and a good read over morning coffee. It’s hard to explain these nuances without everyone involved having a semester or so of quite-a-bit-more-advanced-than-high-school physics and thermo….and even then, the somewhat off-base understanding of people who passed those more advanced courses and still believe N2 and O2 emit IR at ambient Earthly temperatures instead of being transparent, because their textbook “said so” in relation to black bodies is difficult to overcome….Someone is always going to hold up their old textbook and say “Why should we believe you compared to this ?” I think it’s a lack of emphasis in those courses of the different nature of gases and solid surfaces as far as IR emissions…and maybe poor wording in textbooks…
Thanks, As I pointed out by way of my two examples, one of which is from NASA, there is no shortage of “poor wording”. As far as your “more advance than…” comment goes, one missing element in this whole topic is “detailed balance” a concept which I didn’t encounter until my statistical physics course (textbook by P. Morse) in graduate school — so its much more than high school physics, but absolutely essential to understanding energy partitioning in equilibrium.
One point the n2 mode of CO2 has two orthogonal vibrations and the symmetric stretching mode is IR inactive.
Yes, correct. I think that the symmetric mode, however, can participate in a combined mode with n2.
MODTRAN tells you everything you need to know:
1) CO2 backradiation of 15 micron shows a log decay
2) Chaning CO2 levels from the current level mean almost nothing
3) The CO2 signature can’t even be seen until you are at an altitude where H2O doesn’t exist
4) H2O absorbs the same wavelengths and many more than CO2, making CO2 irrelevant at lower altitudes
5) CO2 doesn’t warm the atmosphere, it prevents the temperature from falling below -80C, CO2 puts in a temperature floor at the same temperature of dry ice or consistent with 15 microns
6) Simply add a cloud layer and CO2 becomes irrelevant
7) If 100 ppm CO2 could warm the atmosphere by 1 degree C we could warm our homes by bubble wrapping them in CO2
This video shows that the numbers don’t add up.
https://app.screencast.com/neFlZzIsVVHOG
I would use Modtran to reply to your point number 6, but for some reason I am getting a 403 error that I have no permission to use the resource. They may be doing maintenance right now. Later, though.
OK, I ran a tropical model, from the ground looking upward, took out all GHGs except CO2. The result is a downward flux of
. Once this reaches the ground surface, information that it came from any temperature at all in the atmosphere is gone, and it is just energy reaching the ground surface. Certainly 81 is big enough to supply warmth to the surface.
This is where I always have a problem. That 81 W/m^2 is merely returning what the earth has already sent skyward. It’s just replacing what has already been lost, admittedly time shifted. CO2 is *not* a standalone heat source like the sun. Therefore it can’t warm the earth, at best all it can do is slow cooling. And slowing cooling just means more heat is lost per unit time since the earth remains at a higher temp for a longer time.
I think your result should put an end to the Marcus/Ott claims that CO2 emits almost no radiation low in the atmosphere. We know from other work that when CO2 does emit IR it travels only a short distance at low atmospheric concentrations. This means almost all of the 81 watts/m2 must have originated within just a few meters of the surface.
I agree this must provide some warmth to the surface, but that doesn’t mean adding more CO2 will provide more warmth. In fact, it doesn’t due again to the emission occurring so close to the surface.
What happens is the emission cools the low atmosphere in addition to warming the surface (both temporary). This energy imbalance is corrected by simple conduction increasing energy transfers from the surface back into the low atmosphere.
I’ll agree that this is a much better effort than Willis’s article which is linked to. I posted a link to a video in the comment section of the article which Willis ignored because he “doesn’t do science videos”. I would judge the video as at a more basic level than this article so would be more of an aid to Willis, but it covers many of the same points. For ease of reference I re-post it here: https://www.youtube.com/watch?v=oqu5DjzOBF8 Instead of constant back and forth maybe you could contact Willis directly and get him to watch the video or read this article (he shouldn’t have any issues with reading an article) and then discuss with him any questions raised. You never know, in that way people may become more educated and comfortable with the way global heating works (heating above what a black body temperature with no GHGs is calculated to be) and the discussion can move on to more relevant and important issues. You never know, progress may be made.
Willis, though, had a great insight in the article from the other day by way of argon purging of spectrometers.
Thanks for repeating that URL I’ll have a look at your video after I dig weeds, mow the lawn, etc…
BTW, there isn’t much back and forth between Willis and I , but others….
I’ve left a couple of critical comments under Sabine’s video. One two years ago, another now, following your reminder. Her explanation is deficient.
I suspect Sabine Hossenfelder’s understanding of a lot of things may be deficient.
Thank you Kevin. Interesting. About MODTRAN, does the Grok AI answer below make sense to you?
I tried to go to the MODTRAN documentation just now, but it tells me I do not have “permission.” Weird. And the model page (U of Chicago) itself says the same thing. Hopefully it is a temporary glitch.
Be that as it may, I asked Grok:
“Does the MODTRAN model assume that absorption and emission of IR by H2O and CO2 occur essentially instantaneously, or does it instead go through a computation from statistical physics of collisional excitation and de-excitation involving non-IR molecules?”
Grok AI answer:
“The MODTRAN (Moderate Resolution Atmospheric Transmission) model primarily focuses on calculating the absorption and emission of infrared (IR) radiation by atmospheric constituents like water vapor (H₂O) and carbon dioxide (CO₂). It does not explicitly compute the microscopic processes of collisional excitation and de-excitation involving non-IR molecules using statistical physics. Instead, MODTRAN uses a band model approach to approximate the radiative transfer processes, including absorption and emission, with high computational efficiency.
In MODTRAN, the absorption and emission of IR radiation by molecules like H₂O and CO₂ are modeled using precomputed spectral line parameters from databases such as HITRAN (High-Resolution Transmission). These parameters account for the line-by-line absorption coefficients, which implicitly include the effects of molecular interactions, but the model does not dynamically simulate the collisional processes (e.g., excitation and de-excitation due to collisions with non-IR molecules like N₂ or O₂). The absorption and emission are treated as effectively instantaneous for the purposes of radiative transfer calculations, assuming local thermodynamic equilibrium (LTE) in most atmospheric layers.
In LTE, the energy level populations of molecules like H₂O and CO₂ are governed by the Boltzmann distribution at the local temperature, and the model uses this to compute emission and absorption without explicitly resolving the time-dependent collisional dynamics. For non-LTE conditions (e.g., in the upper atmosphere), MODTRAN can incorporate simplified non-LTE effects, but these are parameterized rather than computed from first principles of statistical physics.
In summary, MODTRAN assumes absorption and emission occur effectively instantaneously based on precomputed spectral data and LTE assumptions, rather than explicitly computing collisional excitation and de-excitation processes.”
Yes. MODTRAN has gone through many versions. It think MODTRAN at UofChicago is a version built upon LOWTRAN6. At least that is what their documentation says. LOWTRAN 6 had 20 cm-1 resolution, and the algorithm in MODTRAN bettered this to 5 cm-1. None of these available codes use line by line computations from HITRAN as far as I know, but rather they gather the significant lines into bands, and use them over and over with an approximate method of handling temperature changes (that surface offset). This is much faster than line by line computations. Will Happer would know the sorts of computing times needed for line by line efforts.
With regard to the instantaneous assumption in the calculations, nothing in the atmosphere could change significantly in the time it takes to complete transitions, and if one is worried about delays, then one would have to worry about the delays involved in propagating highly oblique as opposed to verticle traveling rays. And worry that there might not be LTE (LTE breaks down maybe above 70km, maybe higher than that) MODTRAN just takes the vertical results and multiplies by 3.14 (not even pi) to calculate an irradiance on a surface from a vertical radiant intensity. Lots of things about MODTRAN are quick and dirty so its absolute accuracy is probably not much better than, say 5 W/m2. But its relative accuracy in comparing two cases is probably better.
Did you have a specific question?
Thank you. “Did you have a specific question?” No, your response confirms what I have supposed all along about how the model works. By the way, I recall a few years back we had each discovered that 3.14 factor instead of a more precise pi. I thought that was interesting.
Anyway, I have taken MODTRAN as a valid approximation all along. In any case, you know from my posts about atmospheric dynamics and from the Band 16 visualizations, that I don’t consider the radiative effect of incremental CO2 to be capable of any more than a negligible influence on the end result in the real atmosphere with clouds.
“Did you have a specific question?” No, your response confirms what I have supposed all along about how the model works. By the way, I recall a few years back we had each discovered that 3.14 factor instead of a more precise pi. I thought that was interesting.
Anyway, I have taken MODTRAN as a valid approximation all along. In any case, you know from my posts about atmospheric dynamics and from the Band 16 visualizations, that I don’t consider the radiative effect of incremental CO2 to be capable of any more than a negligible influence on the end result in the real atmosphere with clouds.
‘In summary, MODTRAN assumes absorption and emission occur effectively instantaneously based on precomputed spectral data and LTE assumptions, rather than explicitly computing collisional excitation and de-excitation processes.”’
Thanks, David.
My takeaway from this is that since MODTRAN, and I presume other radiative transfer models (RTMs), don’t account for collisional processes, I don’t understand why they are considered applicable to the lower troposphere where we know that thermalization is ‘favored’ over spontaneous emission by many orders of magnitude.
Further to the above, I also wonder if HITRAN, which is the database / source for spectral line parameters that feed the RTMs, actually includes spectra of GHGs in mixtures with non-IR active gas species, and also how good is the assumption of LTE in the lower troposphere given the turbulent conditions evident in the GOES Band 16 visualizations you frequently provide?
Thanks for your thoughts. I have a lot more to think about on this topic. About MODTRAN, it seems logical to accept its modeling of transmittance values, e.g. the incremental loss of overall transmittance to space with increasing CO2 in the “wings” like for Band 16. This just means there is absorbing power being added. But about the “forcing” aspects, I too wonder how LTE could be a good assumption. Consider how much time a column of the atmosphere spends away from the 0 index mark in these histograms of the “vertical integral of energy conversion” and in the video I often post about. And the W/m^2 values are huge compared to the incremental “forcings” – and they change rapidly in both directions. https://drive.google.com/drive/folders/1PDJP3F3rteoP99lR53YKp2fzuaza7Niz?usp=drive_link
Another aspect is that surely a unit volume of the atmosphere, spiked with water vapor and CO2, is a good IR emitter from its own internal energy. The IR from the surface of the earth is not the only source to replenish the internal energy that was converted to outgoing IR.
I may be rambling a bit here. Sorry about that. Be well.
Hmm. Complex and confusing essay, at least to me. Does it not remain true that any matter, gas, liquid or solid, that is above a temperature of absolute zero radiates in the infrared until it reaches absolute zero at which point it stops radiating energy? How else can there even be absolute zero? Whether any particular gas, liquid or solid intercepts incoming radiation and either increases its internal temperature or re-radiates some energy immediately or later as a consequence is a very different and very complex question. No?
What I hoped to accomplish here was to show that a dipole is needed to couple to thermal radiation, and some materials do not have a dipole available. Noble gasses for instance do not. So, you will never see an IR spectrum of a noble gas. Noble gasses have visible and UV electronic transitions, but nothing in the band we’d call IR.
I hope this helps.
If they radiate that represents energy liss, correct? Energy loss means heat loss, correct? Why is heat loss in the boisphere always restricted to IR ONLY?
Complete nonsense. All matter above absolute zero radiates infrared
Nonsense. All matter above absolute zero emits infrared. You are totally confused. Noble gases like neon, argon, etc,, only emit visible light or UV when excited by high voltage electricity. The same for sodium vapour lamps (mostly visible light), and mercury vapour (UV), which stimulates a phosphor coating to emit visible light in a fluorescent fitting.
You appear very confused about the nature of radiation (light to a physicist).
You still can’t describe the GHE, can you? Adding CO2 to air does not make it hotter.
You frequently make these kinds of unsupported assertions. Unlike Kevin, you don’t bother to explain why we should believe you over him, or provide supporting citations. That is why I rarely pay much attention to what you say.
I don’t care whether you believe me or not. Do you think I should? You can believe any fairytale you like.
You can even believe that adding CO2 to air makes it hotter. Or that matter above absolute zero doesn’t emit infrared.
Why should I value the opinion of someone who refuses to accept reality? Would you?
From the above article’s eighth paragraph:
Hmmmm . . . my observation is that one always needs to exercise great care when using the word “absolutely”.
Then, following lower down, are these statements based on the apparent simplifying assumption of gas molecules existing in isolation from other gas moleccules:
There is a hint the author, Mr. Kilty, is on the right track with his statement to the effect of:
but then he appears to dismiss this important fact by later stating:
Unfortunately, that is NOT true with atmospheric nitrogen at its partial pressures and average temperature in the global lower atmosphere, where collisions with other atmospheric molecules, including other N2 molecules occurs at rates of about 1e9 times per second.
So considering nitrogen gas (like oxygen gas) as a continuum in the lower atmosphere, even “non-polar” N2 will develop momentary dipole moments during collisions with other gas molecules, including other N2 molecules. IOW, while nitrogen is indeed considered in isolation to be a non-polar molecule with no permanent dipole moment due to its symmetrical structure and equal sharing of electrons, the rapid movement of electrons within the molecule due to acceleration forces during molecule-molecule collisions will cause a temporary, uneven distribution of charge. This temporary shift in electron density creates an instantaneous dipole, which is then able to interact (e.g., absorb) EM radiation across a wide range of frequencies, including LWIR radiated from Earth’s surface.
Moreover, when two gas molecules collide, the instantaneous dipole in one molecule can induce a temporary dipole in the neighboring molecule, creating an induced dipole. This interaction, where a temporary dipole in one molecule induces a temporary dipole in another, is the basis of Van der Waals forces, the weakest type of intermolecular attraction. They occur between all atoms and molecules, regardless of whether they are polar or nonpolar. These forces arise from temporary, fluctuating dipoles that arise due to the movement of electrons in adjacent molecules.
Therefore, even though N2 molecules are nonpolar under normal circumstances, collisions and the close proximity of other molecules can lead to temporary and induced dipoles that enable absorption of IR/LWIR radiation.
Notwithstanding the above, it is true that the rapid molecule-molecule collisions that occur in the lower atmosphere mean that the re-distribution of energy from LWIR-excited molecules (predominately the “greenhouse gasses”) to the preponderance of weakly IR-active gases (mainly N2 and O2) occurs via collisional exchange of mechanical modes of energy (translational motions, and vibration and rotation of molecular bonds), with this energy transfer overwhelming the direct absorption of LWIR via temporary, collision-induced dipole moment radiation absorption.
It is also true that the photon-relaxation time for an LWIR-directly-excited molecule (one having a permanent or temporarily-induced dipole moment) is on the order of 10^-6 to 10^-9 times that of the average molecule-molecule collision time, meaning that re-emission of LWIR-originated energy in the lower atmosphere is basically an insignificant consideration for energy transport, but becomes important in the upper stratosphere where the mean free path between gas molecules is many orders of magnitude higher.
You know, before I sent this to Charles the Moderator, I took out a sentence about any argument presented having to have significance for climate before we should bother to consider it. I took it out because I thought it would just generate disputes. The Oxygen magnetic dipole complex at 60GHz is a good example of something that might be an exception to what I say here, but which has no real significance for our purposes — it occurs near 5,000 micrometers wavelength. It shows up in MODTRAN calculations, but it is a million times weaker than the sorts of IR signals one would consider weather or climate impacts, although you can measure temperature with them.
No doubt you can point to Raman scattering of Nitrogen, which is what I think you are getting at here, but once again it isn’t significant from a heat transport point of view and the Raman lines of nitrogen are well off the right side of Figure 3.
There is a difference between laboratory measurements and analytic technique on one hand and significance to climate and weather on the other.
The IPCC accepts that the atmosphere is a chaotic system.
There is no minimum disturbance to the inputs of a chaotic system which may result in completely unpredictable outputs. You may have a religious conviction that you can safely ignore things that don’t fit your view, just like Willis Eschenbach.
Good luck – as Feynman said “Nature can’t be fooled.”
Thank you for your response.
First, I admit to accidentally making a numerical sign mistake in my last paragraph of my post above; my statement “. . . photon-relaxation time for an LWIR-directly-excited molecule (one having a permanent or temporarily-induced dipole moment) is on the order of 10^-6 to 10^-9 times that of the average molecule-molecule collision time . . .” should have stated 10^6 to 10^9 times that of the average molecule-molecule collision time for CO2 molecules.
I am basing this ratio on the data presented for the photo emission “decay time” for LWIR-excited CO2 as given in the attached chart extracted from Dr. William Happer’s slide presentation “Why Has There Been No Global Warming For The Past Decade?“, Princeton University’s Physics Department Colloquium at UNC Chapel Hill, 8 September 2014.
I don’t have comparable data for the various radiation decay modes of LWIR-excited nitrogen or oxygen gases, but expect they would be faster, although perhaps still 10^3 to 10^6 times the time constant of molecular collisions at STP.
Lastly, no, I am not trying to explain or imply Raman scattering in my prior post that discussed the collision-induced momentary dipoles inevitably occurring in atmospheric gases (even those having no permanent dipole moment). Instead, I am asserting that this is the mechanism by which all gases in any continuum at any temperature above absolute zero are able to both absorb and emit “thermal” radiation across a wide range of EM frequencies.
There are other subtleties of the mechanisms by which gases, even those not in a continuum, can radiate energy (albeit a low energy level per photon). Astronomers long ago detected and have measured and studied the 21-cm spectral line (H I) emitted by “cold” neutral, interstellar hydrogen atoms. This radiation is created by a change in the energy state of solitary, electrically neutral, non-polar hydrogen atoms. It is produced by a spin-flip transition, which means the direction of the electron’s spin is reversed relative to the spin of the proton . . . that is, it really has nothing to do with the presence or absence of a dipole field.
Since cold neutral hydrogen atoms account for about 90% of interstellar gas, it may not be important to climate on Earth, but it is still scientifically significant.
As someone who has worked in the field of spectroscopy for many years I agree with the information presented here. Its accurate, however, what I have observed is ever greater levels of complexity in the discussion over AGW which serve more to obscure the issue than illuminate it. The complexity of an issue is very dependent on how one chooses to approach it and in this regard I would like to point out 2 views.
Enstein said about a century ago “no amount of experimentation can ever prove me right but one experiment can prove me wrong”. In this context AGW claims CO2 prevents surface radiation at some wavelengths (around 14.7 microns) escaping to space and thus reduces Earth’s energy loss to space both due to total CO2 and incrementally rising CO2. This reduced energy loss is why Earth will warm. NASA has been measuring Earth’s energy loss to space for around 50 years and it has been steadily rising not falling. The fundmental core prediction of the AGW theory is explicitly contradicted by experimental observation. This one experiment proves the theory wrong. Warmists try to explain this away by claiming that the initial impact is indeed reduced energy loss but the feedbacks quickly reverse that. Feedback is a RESPONSE to stimulus, it can increase the initial stimulus (positive feedback) or reduce it (negative feedback) but it cannot reverse it. If it did so the feedback would also reverse.
2nd and far more important view. Over 200 years ago Carnot described the way heat engines work – systems that convert thermal energy to mechanical energy. There is no doubt that wind and rain falling from high altitude are forms of mechanical energy and also no doubt the original source of that energy is thermal energy from the sun. This means Earths atmosphere is the working fluid of a giant heat engine the operation of which we call weather. Carnot pointed out that 100% conversion efficiency is impossible. The working fuid must gain energy at the high pressure/high temperature point in the cycle and lose energy at the low pressure/low temperature point of the cycle. For Earth’s atmosphere the high temperature point is quite obviously the surface mainly in the tropics. The low pressure temperature point is not the poles as many assume but the tropopause. The poles are not a low pressure point and anyway the atmosphere does not cycle from equator to poles only to about +-30 degrees latitude – Hadley cell. The atmosphere rises up from the surface to the tropopause where it cools and then descends again completing the convective loop. So the atmosphere must lose energy at the tropopause (to space the only colder place) but at the tropopause there is only gas, so atmospheric gases MUST be radiating energy to space at this altitude. A gas that can radiate energy in the thermal IR is by definition a green house gas. If there were no green house gases the atmosphere could not lose energy at the tropopause so it could not cool and thus could not descend again. The heat engine could not function and there would be no weather. No convection, no wind, no rain, no clouds a completely vertically isothermal atmosphere. The surface would gain and lose energy as a more or less black body. Every day over much of the planet the surface temperature would rise to around 100C (with the low albedo caused by a transparent static atmosphere incoming solar radiation at noon is around 1340 watts/sqM which by SB equation yeilds an equilibium temperature of over 100C) and every night it would fall way below freezing – similar to the situation on the moon. We sort of see that already to a very limited extent in desert areas where the main GHG – water vapour – is low. So GHG’s do not increase temperature extremes as claimed, rather the opposite they reduce them dramatically and thereby make Earth livable. An incremental increase in CO2 would incrementally reduce the extremes further making the climate even more equitable.
It is interesting to look at the further implications (all largely positive) but this post is already too long.
Your first point I disagree with as you cite a measurement, not an experiment. Even if I accept your statement that NASA has been measuring an increasing amount of energy loss to space, that’s not an experiment. An experiment would require a second earth with no increase in CO2 as a comparison. Sadly, we don’t have one available. But there’s nothing to say that the increase isn’t natural, and would have been even higher without increased CO2.
Your second point however, deserves an article and thread of its own. Even a limited understanding of Stefan-Boltzmann yields no other conclusion than that the bulk of warming will be observed in night time lows, in winter, at high elevation and at high latitudes. Once you start including all the processes which, in all their complexity, drive weather, I for one keep coming to the same conclusion. The differential from day time highs and night time lows being decreased, and the same for temperature differential from winter to summer, high elevations to low, high latitudes to low, the only possible outcome is an incrementally calmer (for lack of a better word) world. Less fierce storms, less droughts and less floods. The less temperature differential there is, the less vigorous the processes are to move energy from high to low. The processes strive for uniformity, the day/night cycle and seasons ensure that will never be achieved. But the processes are driven by differentials, which SB Law insists must be reduced as the earth warms.
The alarmists cling to their claims that a warmer world will feature more frequent and more intense severe weather, droughts and floods. We’re not seeing it in the data, and for hurricanes in particular, we’re seeing a trend in the opposite direction. In my mind that’s exactly what we should be seeing in a warming world. It would be nice to have the reasons why documented in as much depth and detail as has gone into the article above. Thank you for your comment, and I hope you submit your thoughts in this as an article for WUWT..
Story Tip
Not so much a tip, but worth drawing attention to mh’s comment. We see intense debates on WUWT about what the GHE can and cannot do, we see many articles on severe weather, flood, and drought data refusing to conform to the alarmist narrative. But I don’t recall (not that I’ve read every article) much discussion of what a warming world implies for weather. Many comments that allude to that, or state it, but a thorough analysis of what the physics suggests should happen I do not recall seeing. I think we’ve we’ve beaten the GHE discussion to death. I’d like to see as much focus on what happens to the strength of weather processes when the temperature differentials that drive them are reduced.
What GHE is this? Obviously a figment of your imagination, because there is no consistent and unambiguous description of this “GHE”.
Maybe you are simply confusing religion with science?
I think it hilarious that down thread you argued that jungles don’t get as hot during the day as deserts because the deserts have no GHG’s while also arguing that the the GHE doesn’t exist.
Maybe you could explain what you mean by GHGs. I believe that GHE cultists think that CO2 and H2O are “GHGs”.
Correct me if I’m wrong. The GHE cult “definitions” tend to change from moment to moment.
Michael, as far as I am concerned a GHG is any gas that can absorb and radiate energy at some wavelengths in the thermal IR range (4-50 microns). Both CO2 and H2O do that so by definition they are GHG’s. I claim the primary action of GHG’s in the atmosphere is to absorb surface emission at these absorption wavelengths and replace it with emissions at those wavelengths from the top of the GHG gas column (which is primarily the tropopause). Since the tropopause is colder than the surface, overall energy loss to space is reduced. So GHG’s in totality do reduce energy loss to space (although the magnitude of the incremental impact is very questionable) but by far the greater impact is to firstly to couple energy into the working fluid of the atmospheric heat engine (only energy coupled into the working fluid can participate in the heat engine operation) and even more importantly to allow energy loss from the cold junction of this heat engine which enables the heat engine (and thus weather) in the first place. This causes many other effects such as clouds which have far more impact than the reduction in energy loss to space.
All matter above absolute zero radiates infrared. All matter absorbs infrared.
A mixture of gases does not change its temperature if one gas is removed. For example, removing CO2 and H2O from air does not lower its temperature.
Conversely, adding CO2 or H2O to air does not make it hotter.
The Earth loses energy at about 44 TW.
Four and a half billion years of continuous sunlight has not prevented the Earth cooling.
It is not possible to predict future states of the atmosphere any better than a 12 year old.
No GHE. No “greenhouse gases”.
Like the Antarctic plateau? Sorry, not much warming to be observed.GHE believers even created a “negative GHE” to explain the lack of warming that you describe.
Just as various religious sects have various views on aspects of their religion.
I’m not querying your religious beliefs.
No experimental support – speculation remains speculation. Not even a disprovable hypothesis is possible.
David I absolutely don’t agree. The AGW theory claims rising co2 warms the Earth by reducing energy loss to space. The experiment is to test that theory by looking energy loss to space over time. This is no different to say testing Einsteins time dilation prediction except in Einsteins case the experimental finding agreed with his predictions. Your comments could just as well be applied to that situation or for that matter any test of a prediction made by a proposed theory. In essence it means you are defining AGW predictions as unstable / unverifiable which takes them out of the realm of science. Anyone can float a theory and use it to make a prediction. Then when the prediction is contradicted by experimental observation they still claim they are right, the impact is simply being swamped by something else, something undefined. That way lies anarchy and conspiracy theories.
If there is another effect which raises energy loss to space more than the predicted drop due to rising co2 then Earth should be cooling but the further claim is its warming. So what is this other effect and since its impact is clearly stronger than that claimed for co2 maybe we should be focusing on that instead.
I could go further and point out that if we apply the claimed thermal sensitivity of Earth and multiply by the claimed temperature rise, the rise in energy loss to space is pretty well exactly what NASA have measured. Suggests that the rise is simply due to the observed warming (SB law) leaving no impact for rising co2. But fundamentally since around 1970 observation shows Earth has warmed (satellite measurements) and as it did so energy loss to space rose so the warming cannot be driven by suppression of energy loss to space which is the basis of the AGW theory. The theory is falsified!
On my second issue, I have written up a much more detailed (and longer) article but have not found anyone interested in it.
“For example, Figure 6 shows a spectrographic measurement of Earth’s surface measured from a satellite.”
It doesn’t particularly affect the discussion, but it isn’t from satellite, and isn’t from Harde (2017) – in fact, it doesn’t appear there at all. It is actually taken from 20 km over an icefield at Barrow. That is important, because of the uniformity of the surface emissivity. It appears in Grant Petty’s textbook and is frequently shown. Here is a 2010 example:
One thing that does matter a little is that the top profile, which you marked as 290K is actually 280 K. The spectrum without absorption is following ice temperature, not Earth’s average.
Well, FWIW, attached are spectra from the Nimbus-4 Earth-orbiting satellite that was looking down through the atmosphere over several different areas of Earth.
Notice that Figure (b) in the attachment, the spectrum looking down over “Antarctic ice sheet” falsifies your assertion above referencing your presented Figure 8.2 that
Ooops . . . here is the attachment that I just referenced.
No, all it shows is that the Antarctic ice sheet is much colder than the Barrow ice sheet at thaw time (273K).
But it is interesting in that it shows the Antarctic inversion, where the surface is colder trhan the air above. So the dip in the other plots becomes a rise.
Unfortunately for you, the 2010 reference that you hyperlinked (https://moyhu.blogspot.com/2010/05/greenhouse-effect-and-adiabatic-lapse.html ) specifically states for the same spectrum that you referenced as Figure 8.2(a): “Note that in the dip it tracks the BB curve for about 225K, and in the window, about 268K – the spectrum is taken over a polar ice sheet near Barrow AK.”
Nowhere in that reference URL is there mention of the polar ice sheet at Barrow, AK, being at the “thaw time” or being at a melting point of fresh water, 273 K.
Nice try, but FAIL!
The 273 K is evident from the spectrum. That is the profile followed when there is no obstruction. That is the temperature of thawing ice.
As to whether it is Barrow, he says it is a polar ice sheet. The 2010 reference is mine; I knew at the time it was Barrow, but the online source has vanished. Nothing hangs on whether it was Barrow or not, but it was. Here is another Petty plot taken at Barrow a couple of months earlier:
Uhhh . . . the title of the plot you posted immediately above states “. . . .as seen from ground level looking up.”
The previous graphs that we discussed were specific to looking downward on the ice sheet near Barrow, Alaska.
Big difference.
As to your statement:
That is just laughable . . . one simply cannot read the presented graphs to an accuracy of 1 K. Besides, I cited the stated claim from the originator(s) of the published graph that it was for a temperature being “about 268 K” when viewing the ice sheet looking downward in the “atmospheric widow”.
Finally, are you really going so far as to claim that the entire ice sheet near Barrow, Alaska was at a uniform melting temperature (i.e., at 273 K) at the time of obtaining the cited “looking down” temperature data?
Knowledgeable scientists will state that melting of ice sheets occurs mostly at the ice-water peripheral edges and, for floating sea ice, at the underneath ice-water surface since the heat capacity of liquid water is so much higher than the heat capacity of air. Significant temperature gradients are present across the horizontal and vertical dimensions of ice sheets.
A couple of things..
Not quite! Water has a hemispheric, spectral emissivity of 0.91, meaning a substantial deviation from 1. Next to my own work you can get this figure in the good literature – Baehr, Stefan “heat and mass transfer”, or Huang et al 2016.
https://greenhousedefect.com/what-is-the-surface-emissivity-of-earth
Also modtran has a surface emissivity of 0.975. One can see that by lowering the altitude to 0km. It is just that it also considers some amount of N2O, which can not be altered.
As before, the surface is NOT a blackbody. If satellites looked through the atmosphere, projecting the specific emissivity of water over different wavelengths, that green line is what they would see..
No! The cloud contribution to the GHE far exceeds their albedo effect. One could make a simple calculation: average cloud top altitude 8km, average cloud emission temp. thus 288 – 8 * 6.5 = 236K, and (288^4-236^4)*5.67e-8 = 214W/m2. So per effective cloud cover, they provide a 214W/m2 of GHE. Assume 1/3 effective global cloud cover, it would be 71W/m2 in total. Clouds are actually warming!
No! These are seperate things and must not be mixed up. Surface is surface, atmosphere is atmosphere. Surface emissivity is NOT determined by atmospheric properties.
Really? Stand in the direct Sun in the tropics. Cloud comes between you and the Sun, temperature drops. No cloud? Use a parasol, hat, or Stevenson screen. Same physical laws apply.
The hottest places on Earth have the least clouds and the least GHGs (water vapour).
Deny reality all you like – it won’t help.
Stand on the Canadian prairie in January on a clear windless night so you know what 40 below feels like. If you’re the one (and I was) that had to get up at 6:00 AM to feed the cattle, you prayed for clouds. On a cloudy night, its more like 5 to 10 deg C and the water pumps aren’t so frozen solid that you have to thaw them out with a tiger torch.
Same physical laws apply sir.
Clouds during the day at low latitude do not have the same effect as clouds at night at high latitude.
What particular effect are you referring to? I think you may be trying to be too clever for your own good. You do know the difference between day and night, clouds, latitude, temperature and so on, do you? Or apples and oranges? Chalk and cheese, perhaps.
Maybe you are just trying to disguise your ignorance and gullibility. There is no GHE, and the Earth’s surface has cooled in spite of four and a half billion years of continuous sunlight.
Where you get “far exceeds”?….A Cloud top can reflect 70% of incoming solar SW back to outer space. That’s a couple of hundred watts/sqM using average incoming….yes, clouds block the view of -273 K outer space from the 288 K surface….so they are “warming” by that criterion…..but one can’t forget the “cooling” by reflecting solar back into space.
-273K do not exist
70% of 342W/m2 would be 240W/m2. But on average clouds do not reflect 70%, but rather like 50%.
Hmmm…ok -273.15 does not exist cuz of background microwave of about 3 C…but we are talking approximations here….
No, we are talking about 0K being the absolute minimum temperature. THERE IS NO -273K, GOD GRACIOUS!!! Are you senile?
Awesome article and a great thread.
Very interesting essay, With regard to this point:
“It is absolutely correct that nitrogen does not emit IR radiation”
The first part of the essay is vague. I assume you are not equating IR with thermal radiation, they are different things and not all thermal radiation is in the IR spectrum. I just wanted to clarify this. Under the right circumstances, as I understand it, nitrogen, Argon and oxygen will emit, just not in the IR spectrum.
A second point, I realize that the essay is titled “Radiation and Matter,” but all energy stored and circulated on Earth’s surface (atmosphere + land + oceans) is not in the form of radiation. Much of it is sensible and latent heat, and much of the rest is put to work as kinetic or potential energy eg convection. In other words, radiation arriving at one location may not be emitted to space at the same location. As a result, the residence time of absorbed energy can, and does, vary. Residence time matters.
My point is that looking at Earth’s surface and atmosphere only from a radiative perspective is misleading. More here:
https://andymaypetrophysicist.com/2025/02/01/energy-and-matter/
Convection varies over time and makes a difference in the energy budget. One should not equate radiation with energy. Other than that, excellent essay.
Radiation is radiation. Photons, energy in transit. Light, to a physicist.
From Wikipedia –
The “thermal” is superfluous, but you don’t believe Wikipedia anyway do you? Otherwise you would have to agree with me.
For those who believe oxygen and nitrogen do not emit infrared, the supposed “back radiation” comes from the atmosphere – even at night, and over the deserts. Is it the same regardless of H2O content? The 340 W/m2 of “thermal” “back radiation” emitted by the atmosphere must come exclusively from CO2 above the desert, as according to nutty GHE believers, no other gases in the atmosphere are capable of emitting infrared!
About 4 molecules in every 10000 in the atmosphere are CO2. Obviously emitting vast amounts of radiation for their size – would have to be emitting well into the ultraviolet to achieve that energy density! Apparently, H2O in the atmosphere doesn’t emit IR either, or the “thermal back radiation” would be much less in desert regions!
Belief in the GHE is religious, not scientific. Even its followers are confused by the unclear language employed by some of its proselytisers. Not surprising, really.
MF believes N2 and O2 emit IR but can’t produce Hitran runs showing it, but can produce a Wiki page that talks about thermal radiation from “matter” that might not be talking about “gases”….some Hitran runs would clarify that for us MF…
HiTrans? Would these “runs” be computer model output? You don’t have to believe anything that contradicts your religion, and I don’t expect you to.
If you can show reproducible experimental output to show that adding CO2 to air makes it hotter, I will certainly look at it.
The Earth cooled despite four and a half billion years of continuous sunlight. I hope your HiTrans “runs” agree. If they don’t, HiTrans is wrong.
Michael thermal radiation can be at any frequency, IR is a specific frequency band: from Wikipedia: 780 nm (380 THz) to 1 mm (300 GHz). These are not frequencies that O2 and N2 or Argon can emit. That was my only point. When these molecules are at the right temperature and conditions, they emit their characteristic radiation, but that radiation is not in the IR.
Andy, no, infrared is all radiation with longer wavelengths than visible red. All matter above absolute zero emits infrared radiation. The peak wavelength is determined by temperature – nothing else, at atmospheric temperatures.
You might be thinking about the specific wavelengths emitted by excited gases – the characteristic colour of neon tubes, sodium vapour lamps.
Or possibly absorption spectra, where light is shone through the gas, and specific absorption lines are evident.
All fairly irrelevant, though, if you are trying to imply that adding CO2 to air makes it hotter. It doesn’t. Adding CO2 to a mixture of N2 and O2 at the same temperature just results in a mixture of all three, with no change in temperature.
By the way, “thermal radiation” is just “radiation”. Light. Any frequency, from infinite to approaching zero. Photons.
Sorry to sound picky.
You start out quite well but then fall apart conceptually. This phrase shows a fundamental lack of understanding:
There is a no “passing” electric field. There is a passing disturbance (wave train). There is but one EM field in this knowable universe. The energy in that field varies depending on the local strength of the EM field.
Cold stuff does not emit radiation to warm stuff. The EM field is affected by the presence of the two objects and the energy transfer is always unidirectional in the field from warm to cold. Warm is high potential and cold is lower potential.
Like the gravity field, the EM field equilibrates at the speed of light. The interesting aspect is why objects in the gravity field communicate at the same speed as light when that speed is solely a function of the electric permittivity and the magnetic permeability of the space. Answer that and you are on the way to the unified theory.
Read the linked paper by Michael Mischenko and you will begin to understand a bit more about fields:
Open the PDF for in another window
Part of the summary is worth stating here:
Cold stuff does not emit radiation to warm stuff.
A body at a given temperature emits radiation, period. The radiation does not know or care what temperature anything in its path is. It could be outer space at near absolute zero or it could be the sun at millions of degrees, the radiation emitted does not change It does not cease to exist because something hotter is in its path.
I guess you are trying to say something relevant, but don’t know how to say it. You are possibly ignorant enough to believe that water can absorb energy from ice, and get warmer!
Or that the photons of visible light are absorbed by solid matter that they encounter. Not if the solid is transparent, like glass! You can’t explain what happens to the IR photons emitted by ice totally submerged in water, can you?
The ice gets warmer, and melts, The water doesn’t freeze. Where did the energy radiated by the ice go?
Adding CO2 to air does not make it hotter. There is no GHE. Religious belief in the heating properties of CO2 is not science.
The ice gets warmer, and melts, The water doesn’t freeze
If you put a large cold chunk of ice into a container with a small amount of water, the water absolutely freezes and then the ice melts faster than if there is no water. You can do this experiment in your kitchen.
You can’t explain what happens to the IR photons emitted by ice totally submerged in water, can you?
Well I can but the topic is atmospheric physics so its not relevant.
Or that the photons of visible light are absorbed by solid matter that they encounter.
So you admit that photons travel toward matter! What happens to them when they reach the matter is another discussion. They might be reflected, pass through, or be absorbed. If they are not reflected and do not pass through (because not everything is a mirror or glass and not all photons emitted are in the visible spectrum) then the only possibility left is that they are absorbed. The energy they carry cannot simply disappear.
Well you can’t, and the religious belief that ice can warm water (or a colder atmosphere can make a warmer surface even hotter) is central to the GHE religion.
Well yes it can, Energy that leaves the Earth (currently 44 TW net) disappears. Gone. Kaput. Never to be seen again.
We were not talking about a photon headed for outer space, we were talking about a photon that encounters matter.
You really are hilarious. You totally lose the argument so you simply come up with a completely irrelevant example in response.
OK, a visible light photon that encounters transparent matter. Or an infrared photon from ice which encounters liquid water. Or a photon from a cellphone tower which encounters wood or brick.
I’m not arguing – that’s your fantasy. I’m not even questioning your GHE religious beliefs. I leave the stupid gotchas to you.
OK, a visible light photon that encounters transparent matter. Or an infrared photon from ice which encounters liquid water. Or a photon from a cellphone tower which encounters wood or brick.
What happens to the photon and the energy it carries when it encounters matter and is neither reflected nor passes through the matter.?
I ask you specific questions, you change the subject then, accuse me of religious dogma. You aren’t answering because it is in fact a gotcha question. You’ve argued yourself into a corner from which you cannot escape except by changing the subject and hurling insults.
You do nothing to help the skeptic side. You only discredit us.
Don’t you know? Just how ignorant are you? You say
You may boast about your ignorance and gullibility all you like. I won’t interfere, or try to stop you.
Adding CO2 to air doesn’t make it hotter. The surface has cooled, in spite of being subjected to four and a half billion years of continuous sunlight. There is no GHE.
As a kid in Maryland, I used to capture fireflies in early summer. The temperature of the light sensitive cells in my retina are at 98.6 F. The light from the fireflies are much colder–definitely not 98.6 F. How did I see them?
With your eyes, surely. Look up “bioluminescence”, if you are truly as ignorant and gullible as your “question” implies.
You are probably dim enough to believe that adding CO2 to air makes it hotter! Feel free to tell me you are not silly enough to believe something as ridiculous as that.
You said that colder radiation can’t be absorbed by warmer radiation. Your ignorance is obvious. My eyes saw colder radiation. Please explain, instead of calling me ignorant.
If you don’t know the answer, you are ignorant, aren’t you? Lacking knowledge. That’s why you’re asking me to improve your knowledge, aren’t you? To become less ignorant?
Radiation is not “hot” or “cold”. Visible light may result from radiation as a result of temperature – say white hot steel, or from atomic excitation – “cold light”, neon tubes, bioluminescence etc.
If you can’t be bothered finding out for yourself, remain ignorant. See if I care.
Adding CO2 to air doesn’t make it hotter, and the radiation from ice won’t warm water.
“Radiation is not “hot” or “cold”. ”
Alas you contradicted your position. So radiation from a colder object can be absorbed by a warmer object. Your position is nonsense.
No it can’t. The contents of your tortured imagination are not reality. Next thing, you’ll be saying that adding CO2 to air makes it hotter!
That would just demonstrate that you are ignorant and gullible – or totally delusional.
“No it can’t. ”
So by your argument, I can’t see fireflies. Yet, I can.
So what’s the problem? Don’t you believe your own eyes?
Radiation from a colder object can be absorbed by a warmer object. The light from the firefly is from an object colder than my eyes. In order to see that light, my retinas must absorb it. You have to explain why a warmer object can’t absorb radiation from a colder object.
That’s a tough question. However, I would point out that you are no longer dealing with IR (and heat radiation) when discussing light.
Okay I’ll play this silly game.
The Earth’s IR window ranges from 8 microns to 13 microns. I like round numbers, so let’s pick 10 microns which is clearly inside the window. For reasons I’ll explain, I want to know the frequency of this wave length. To find frequency, you divide the wave velocity by wave length. The wave velocity is the speed-of-light, so we get 30 terahertz. The IR band is from 0.3 terahertz to 400 terahertz, and we do have an IR frequency.
For grins, I’d like to calculate the energy of a 30 terahertz photon. For that we need the Einstein-Planck formula. That formula in its simplest form is E = h*nu; where E is energy, h is Planck’s constant, and nu is the Greek letter (lower case) for frequency in physics. Plugging 30 terahertz into the formula gives 1.98e-20 joules which isn’t an impressive amount.
Main sequence stars are reasonably good black bodies. The Sun is a main sequence G2 spectral class star with a surface temperature of 5,800 K. The Sun acts like a reasonably good black body with a temperature of 5,800 K. The brightest star in the night sky is Sirius (Sirius A actually). Sirius is a main sequence A0 spectral class star with a surface temperature of 9,800 K. Sirius acts like a reasonably good black body with a temperature of 9,800 K.
Black bodies have two basic properties (among others): they absorb all frequencies, and they radiate at all frequencies. Clearly, Sirius could emit a 30 terahertz photon, and that photon could travel to the Sun. Now we have a quandary. The Sun sees two identical photons (energy-frequency wise). How does the Sun know that one photon comes from Sirius which it CAN absorb, and the other comes from the Earth (at roughly 300 K, more or less) which it MUST refuse?
In the same way that photons know if they have been counted going through a particular slit, in the double slit experiment. If they have, no interference pattern emerges. If not, interference is seen.
Or the same way that photons know if a substance is transparent to that particular wavelength, and how to travel! For example a glass sheet is transparent to visible light. Visible light photons impinging at an angle, are very smart, deviating at an angle dependent on refractive index, and the deviating at an exactly equal angle, parallel to their original path.
Yes, Jim, photons are clever indeed. The photons emitted from ice emitting 300 W/m2 know better than to try and make a kettle full of near boiling water hotter!
The photons from the Sun at the same intensity, have no trouble boiling water, or even making the collector of a solar plant like Ivanpah blinding white hot!
How clever is that! Cleverer than you, by the look of it.
Have you figured out why adding more heat energy in the form of ice cubes radiating 300 W/m2, doesn’t raise the temperature of your soup? No? How hard can it be?
The two slit experiment applies to most subatomic particles, some atoms and even some molecules. There’s nothing smart about the effect. When you treat a subatomic particle like a particle–it acts like a particle. When you don’t–it can act like a wave. So there’s nothing about temperature in that experiment.
I guess you’re talking about Snell’s law. The reason why there is an index of refraction is that light travels slower in air, water, and glass than it does in a vacuum.. The index of refraction is just the ratio of the two velocities. The wave front shifts as it enters the slower medium and shifts back when it exits that medium. Cherenkov radiation is where electrons (charged particles) travel faster than the speed of light in water–much like military planes traveling faster than the speed of sound in air.
You obviously have no idea about identical photons from different temperature sources. I can say one thing about photons, they are smarter than most of your comments.
That’s just silly. Are you trying to say that ice submerged in water emits identical photons to those emitted by the water? Or are you just creating meaningless word salad?
You would obviously be ignorant and gullible if you believed that adding CO2 to air made it hotter, but you wouldn’t believe anything as silly as that, would you?
I don’t blame you for not admitting to anything at all.
For example, your comment takes care not to disagree with a single thing I said. I don’t wonder why.
“For example, your comment takes care not to disagree with a single thing I said. I don’t wonder why.”
I disagree with everything you claim–obviously. And you assume some stupid nonsense about CO2 which I never discussed. You should add gin to your ice cubes and enjoy life for a change.
Unfortunately, you refuse to clearly state what it is you believe I “claim”.
Adding CO2 to air doesn’t make it hotter, It doesn’t matter whether you agree with me, or not. It’s fact – unless you can provide experimental proof to the contrary.
If you can’t, you are expressing a belief based on fantasy – that would make you ignorant and gullible.
If you could demonstrate experimentally that I am wrong, that would make me ignorant and gullible. Go for it.
That is an ideal or theoretical Black Body. Real world solids have a complex refractive index that varies with wavelength and with temperature. The imaginary component of the complex refractive index (k), commonly called the extinction coefficient, determines the absorptivity which Kevin claims is equivalent to emissivity. The normal (perpendicular) reflectivity is: R↑ = ((n – 1)2 + k2) / ((n + 1)2 + k2) although it also varies with the angle of incidence. It also varies with temperature.
I don’t know your point. Huh?
I took issue with the claim that Black Bodies absorb and radiate at all frequencies. Real world materials behave differently in different portions of the electromagnetic spectrum. Notably, in some regions a material will be highly reflective (low emissivity) and in other regions it will be highly absorptive (high emissivity). Real materials don’t have the properties of an ideal Black Body.
The S-B curve defines the shape of radiation from any solid body, especially incandescent solids. The light from the sun approximates the appearance of a firefly because the peak of the sun’s emission is near green light; Rayleigh scattering eliminates most of the blue light, shifting its appearance to more of a yellow at the bottom of the atmosphere. However, the light from a firefly is bioluminescence and is unrelated to the temperature of the luciferase in the beetle. They are different processes.
“However, the light from a firefly is bioluminescence and is unrelated to the temperature of the luciferase in the beetle.”
So your argument is that the beetle photon knows the temperature of the beetle–or doesn’t know? Photons do not carry thermometers. Photons do not know the temperature of the source that created them. If the process isn’t temperature–so what? A beetle at -50 degrees Fahrenheit is not emitting any light photons.
No argument with such a stupid statement. What the heck is a “beetle photon”? You are obviously ignorant. Are you gullible as well, or did you make up “beetle photons” all by yourself?
Adding CO2 to air doesn’t make it any hotter – any amount of “beetle photons” notwithstanding.
I guess you didn’t take my advice and mix gin with your ice cubes. I have another problem for you. Back when you could tune a TV set to an unused channel, the static on the screen was mostly from the CMB. The CMB has been measured to be slightly less than 3 K. So how do Earth based TV systems which are nowhere near 3 K receive such signals?
Unlikely. CMB is extremely weak, and around 160 GHz – 1.8mm. You may have equipped your TV with a traveling-wave maser, a low-loss (0.027-db) comparison switch, and a liquid helium-cooled reference termination, but I doubt it.
You get just as much static with the antenna disconnected. You must enjoy being ignorant and gullible.
Why would I take unsolicited advice from someone ignorant and gullible enough to believe that adding CO2 to air makes it hotter?
The firefly photons don’t know or care about the temperature of the firefly because the photon is generated by a chemical reaction in which heat plays little part. The firefly is dead at -50 deg F and can’t release the enzyme that triggers the emission of photons.
No, I don’t. The light you are seeing is being emitted as a result of bioluminescence, not temperature. You don’t believe it do you? You obviously claim you can’t see neon lights, the Aurora Borealis, or LEDS. The fact that TVs, radios, and cellphones can absorb the same radiation that penetrates walls of various materials is no doubt unbelievable to you.
Try making water warmer by completely immersing an ice block in it. You can’t even figure what happens to the photons emitted by the ice, can you? If you can’t, that makes you ignorant.
Michael you come across to me as unreasonably arrogant bordering on abusive and that’s neither necessary nor courteous. Please see my earlier comment in reply to Jim. To re-iterate, you say above that radiation from ice wont warm water. In fact it will but the water will radiate even more thermal energy to the ice which will cool it more so NET it would cool. There is real energy flowing both ways its just that the flow from warmer to colder is always greater than the flow from colder to warmer.
I have absolutely no idea why you would talk about adding co2 to air making it hotter. Seem utterly irrelevant to me.
Thank you for your unsolicited and completely pointless opinion. It’s value is precisely zero – in my worthless opinion, of course.
No it won’t.
Boasting about having no idea about something is often done by ignorant and gullible people.
Adding CO2 to air does not make it hotter. If you believe otherwise, you are obviously ignorant and gullible, because it doesn’t.
Jim; I am getting tired of people (not you) misquoting the second law claiming that cold objects cannot transmit heat to warmer objects. This is NOT what the second law states, they leave out one very important word. It states the NET heat energy flow is always from warmer to colder.
Cold objects can and do transmit heat energy to warmer objects all the time. Its just that the warmer object also transmits heat energy to the colder and since the sender is warmer it transmits more energy so the NET heat flow is from warmer to colder. Consider, you are outside in the cold and your body radiates away thermal energy to the colder environment. Now you go into a warm room, how would your body know to stop radiating energy? It doesn’t, it keeps radiating at exactly the same level but outside you got little energy back from the cold environment (so you net lost energy- got cold) whereas inside the warm room you receive more energy from the room than you radiate so you net gain energy ie” get warmer.
Thank-you mh, for your kind words.
No, matter above absolute zero emits infrared.
Infrared from colder matter does not interact with hotter matter. You can’t explain what happens to the energy emitted by ice totally immersed in water, can you? Is it destroyed? Does it vanish? The water doesn’t get hotter, does it?
If you can’t answer what happens to the photons emitted by submerged ice (even to yourself), this would make you ignorant, and gullible enough to try to overcome fact by playing semantic games such as the sentence you wrote – which is meaningless word salad.
Adding CO2 to air does not make it hotter. There is no GHE.
Your unsupported assertions are bad enough without resorting to ad hominem attacks. You are becoming tiresome.
Are you disagreeing with something I presented as fact? Or just complaining because I’m right?
If you believe my facts are wrong, just provide some experimental data to support your disagreement. You can’t can you?
Adding CO2 to air does not make it hotter. There is no GHE.
I’m disagreeing with something that you presented as fact without support.
If you believe my facts are wrong, just provide some experimental data to support your disagreement. You can’t can you?
But the net heat transferred depends on the temperature of both the hot and cold objects…you groking yet that there must be “back radiation” from the cold object.
But in the q=k(Thot^4 – Tcold ^4), the back radiation is just the -Tcold^4 part….always less than the “fore-radiation” so never breaks the 2nd law….
And yet you can’t use ice to make water hotter, so the amount of mythical “back radiation” doesn’t matter anyway, does it?
I think you are trying to say that all matter above absolute zero emits radiation, and radiation from colder bodies cannot be absorbed by a hotter one, raising its temperature.
If radiation from the colder ice was accepted by the hotter water, this would result in a situation where totally immersing ice in water, would result in the ice losing energy to the water, and becoming colder, if that energy was used in making the water hotter.
No free lunch, you see.
The water would rapidly start boiling, as the totally immersed ice would head towards absolute zero. Are you a fanatical GHE cultist, or just naturally ignorant and gullible?
Adding CO2 to air does not make it hotter.
“If radiation from the colder ice was accepted by the hotter water, this would result in a situation where totally immersing ice in water, would result in the ice losing energy to the water, and becoming colder, if that energy was used in making the water hotter.”
No, because you don’t understand how radiation heat transfer works. In the situation you describe radiation would also be occurring from the water to the ice!
And I should value your opinion because . . . ?
Well for a start my PhD is in Physical Chemistry and for ~20 years I ran a lab which developed measurement techniques involving lasers. It’s not an opinion anyway, it’s a statement of fact!
Phil, excellent. But you still can’t say where the photons from ice totally immersed in water go, can you?
Exactly so. So where do the photons emitted by the ice go?
Don’t you understand what the word “also” means?
Photons travel from ice to water and from water to ice.
Exactly so. So where do the photons emitted by the ice go?
You would also say that the photons emitted by the rose outside your window travelled to your window, wouldn’t you? True, but essentially meaningless.
i know you are ignorant and gullible, but I’ll expand by asking what happens to the photons from the ice when (and after) they reach the water. Before you spout irrelevant nonsense, consider the photons from the rose travelling to your window, and what happens to them
Off you go, now.
Any photons emitted by the ice in this scenario would be absorbed by the liquid water.
Just like any photons from the rose outside your window are absorbed by the glass?
What then?
You seem to be changing the definition of “back radiation”. “Back radiation” is radiation that started as from the hot object and is absorbed by cold and emitted back to hot.
In that case, it achieves nothing at all. A hot object cannot even raise the temperature of a non-contiguous object to the same temperature as itself!
And f course, if the hot object loses energy in the process, it cools.
Just like the Earth – in spite of absorbing four and a half billion years of continuous sunlight.
You might as well forget the pseudo-scientific “back” part, and call it “radiation”. Otherwise, you’ll probably start using silly jargon like “downwelling”, “upwelling”, “front”, or even “sideways twirly-whirly” radiation!
Just radiation will do fine. You can refine your meaning by saying that infrared radiation is photon frequencies lower than visible red. Then you would understand the meaning of “all matter above absolute zero emits infrared”.
It might also show you why the mythical GHE has no consistent and unambiguous description.
David, It is a little more complicated with gases as explained in the post. A solid body, with all the molecules packed tightly together does pretty much emit all the time, but it is according to its temperature as the SB equation says. Gases only emit when stimulated properly, they can go a long time without emitting anything, but when they do it is when the conditions (temperature, pressure, etc) are correct and it is at a characteristic frequency.
You are probably thinking of the phenomenon of excitation – like light emitted by a “cold” neon tube.
All matter (including gas) above absolute zero emits infrared.
You might notice that even the slightly dim Trenberth claims the atmosphere at night is emitting over 300 W/m2 – of infrared. Even he wouldn’t be silly enough to claim that this radiation is coming from “properly stimulated” CO2 molecules alone, would he?
Mind you, he might believe it. You never know.
True, however the amount of radiation emitted at a given wavelength depends on the emissivity at that wavelength. For Argon, Nitrogen and Oxygen the emissivity is close enough to zero that it can be ignored in many cases.
Is this actually true? An emitted photon from its perspective is absorbed with no time passing, so from it does in fact “care” about where it’s headed.
Another part says
“Photon gas”, eh. News to me.
Does anyone own the referenced book?
It was presented to me that way many years ago in Holman’s “Heat Transfer” text. There is a newer edition on-line here:
https://udghoshna.wordpress.com/wp-content/uploads/2013/06/143898071-holman-heat-transfer-10th-txtbk-pdf.pdf
File page 405, book page 380.
Thanks for the reference
So its a way of establishing an energy density as per
and then
But is a non-real, non-physical representation.
In quantum mechanics, what is “real” or “physical”?
For many purposes, I find using photons in the analysis the most useful way of looking at things. Explanations of how IR photons are emitted and absorbed by dipoles in molecular bonds, as this post covered, are the best I have ever seen.
And if we use photons at the beginning and end of a radiation transfer, does it really make sense to say that considering what happens in between to be a stream of photons is “non-real” or “non-physical”?
In a different vein, I have designed optoelectronic sensors. Nothing made sense on either the emitting or receiving end without quantization into photons.
Yes, there are wave properties in the transmission as well. But quantizing into photons has allowed many real-world phenomena to be explained that could not be explained by wave theory alone. So I do not have a problem with the “photon gas model”.
The problem I have with considering them to be a stream of photons is that they then lose their wave like properties in one’s thinking. I prefer to think of them as waves until such a time as they interact and then they “show” their energy. I expect that’s the moment your photosensor “sees” them.
Photon gas model sounds string theory-ish to me but I’m no expert on that.
It’s an interesting puzzle. I will point out that electrons also have wave properties – you can get an interference pattern using a double slit with electrons as well as photons. But people don’t seem to have a problem considering electrons (at least primarily) as particles.
I’m not a chemist, but I’ve studied QM some. Electrons do not orbit the nucleus in an atom. They actually exist as a wave function as described by Schrödinger’s equation.
It is refreshing to read a comment here that demonstrates an understanding the nature of radiation. The photon only exists as a mathematical construct. The electromagnetic field is not quantized. Planck’s perspective was that the energy states of matter were quantized, snd these energy states determined the “quantum” of energy absorbed from or returned to the field.
A proper perspective is provided by Loudon in his book “The Quantum Theory of Light”. He addresses the issue head on in his introduction and on page 10 explains that a photon is created when energy is absorbed from the field by matter, raising its energy state. Conversely, the photon is destroyed when the matter relaxes to a lower state and returns energy to the field.
You can read Loudon’s book by creating a free account at the link below:
https://reader.z-library.sk/read/32d023ae707b8236925ef732fe4c1ff936f204828ad83e84c5db8bcdd4ee95fc/991036/8af58f/the-quantum-theory-of-light.html?client_key=1fFLi67gBrNRP1j1iPy1&extension=pdf&signature=9a96213956a9cfc918969d96b93738d81e4f123ff176829c5f2c040b6d145861&download_location=https%3A%2F%2Fz-library.sk%2Fdl%2F991036%2F635e76
The image of the photon as a ballistic packet of electromagnetic energy evolved as a convenience and creates much confusion. The dynamics of the atmosphere can only be understood from the perspective of the interaction of IR active gases with the field. I have discussed this in other threads, and in this one most recently here:
https://wattsupwiththat.com/2025/06/07/radiation-and-matter/#comment-4081596
One of the best examples of this confusion is the never ending discussion of “upwelling” and “downwelling” radiation. The scalar radiative transfer equation, upon which Schwarzschild’s equation of radiative equilibrium is based, is not an equation of radiation, but of radiance, aka specific intensity. It is not a measure of energy, but of “brightness”, for lack of a better term. It is the measure of one scalar component in a single direction of a quantity (energy flow) that can only be determined by measuring in all possible directions and computing the vector sum.
As Planck explains in his “Theory of Heat Radiation”(1908), “Generally speaking, radiation is a far more complicated phenomenon than conduction of heat. The reason for this is that the state of the radiation at a given instant and at a given point of the medium cannot be represented, as can the flow of heat by conduction, by a single vector (that is, a single directed quantity). All heat rays which at a given instant pass through the same point of the medium are perfectly independent of one another, and in order to specify completely the state of the radiation the intensity of radiation must be known in all the directions, infinite in number, which pass through the point in question; for this purpose two opposite directions must be considered as distinct, because the radiation in one of them is quite independent of the radiation in the other.”
The FTIR spectrometer does not measure energy, but radiance. Mishchenko shows, for example, That in an omnidirectional field in which there is no net energy flow, a detector immersed in that field can detect a component in any direction. The troposphere is a good example of this. Likewise, he shows that there can be situations where there is a net energy flow that is not seen by a detector. This is because all detectors, whether an FTIR spectrometer or our eyes have a limited field of view.
Great post.
This is what I learned. A “photon” or quanta as Planck called it, is nothing more that an absorbed/emitted amount of energy. It does not exist as a unique “bullet” of energy. It is part and parcel of a continuous function called an EM wave. Only when the EM wave is interacted with in some fashion does a “photon” come into existence. This is the duality feature of an EM wave.
Planck is one of my candidates for a generational figure in physics. He was obviously a very intelligent person capable of creating insights to heat and radiation supported by mathematical work at the highest level. His work with entropy is as insightful as his work with radiation.
I have read through his Theory of Heat Radiation numerous times and spent hours visualizing and recreating in my mind his mathematics. His depth of knowledge embraced the work of Stefan, Boltzmann, Maxwell, Kirchoff, and others. Truly a great physicist.
Kevin’s diagram utilizing springs in molecules probably arises from Planck’s work on quanta.
NASA states –
You state
With respect, Kevin, in this case NASA is right, and you are completely and utterly wrong. Your post is worthless, except as an instance of the dangers of ignorance and gullibility.
Nitrogen both absorbs and emits infrared, Solid nitrogen, below 48 K or so, does not magically become liquid, at a higher temperature, due to magic. It absorbs infrared if the environment is hotter, and heats as a result, becoming liquid. If the liquid nitrogen absorbs infrared emitted by something hotter, it becomes gaseous. If allowed to radiate energy without replacement, the gas liquifies, and eventually solidify. Basic physics.
All matter above absolute zero emits infrared.
Air has a temperature – it is radiating energy – infrared. Remove CO2 and H2O, and it remains at the same temperature, whether you believe it or not.
Yes, different types of matter have different specific heats. Nitrogen has a specific heat, and this varies slightly with temperature. And yes, its wavelength of emitted radiation varies with temperature, as does all matter.
Adding CO2 to air does not make it hotter, if that’s what your misinformational post was leading up to.
Sorry, Kevin, you’re dreaming.
If you can find an experiment to show that nitrogen neither absorbs nor radiates infrared, and remains at absolute zero (not radiating any photons at all), I’ll change my view.
Remove CO2 and H2O, and it remains at the same temperature, whether you believe it or not.
So deserts, with by definition very little water vapour, have higher highs and lower lows than jungles at the same latitude because…
They have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface. I didn’t realise you were so ignorant of physics But what has that to do with the comment you were responding to? As I said, remove CO2 and H2O, and it remains at the same temperature, whether you believe it or not
Remove CO2 and H2O from a sample of air at 25 C, its temperature changes not at all. You don’t believe it, but your belief is based in religion, not science. You can’t describe the GHE in any consistent and unambiguous way, but you still believe in it!
Religion, not science.
Got any more stupid gotchas to demonstrate your ignorance and gullibility?
They have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.
Which explains how they get colder than jungles at night how?
Talk about missing the point!
BTW, calling people ignorant adds nothing to your argument.
Basic physics. Are you really as ignorant and gullible as you appear?
Go on, tell me that you don’t understand why temperatures drop so low on the airless moon.
Basic physics.
By all means, explain these basic physics to me. Just stating “basic physics” is meaningless.
So explain to me why deserts get colder at night than jungles at the same latitude even though they got hotter during the day? If its basic physics you should be able to explain that easily.
Why should I? If you intentionally refuse to understand basic physics, I will only be wasting my time on a fool who boasts about his ignorance.
You believe that adding CO2 to air makes it hotter. It doesn’t, but you refuse to believe it.
They have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.
So you wish to argue that GHG’s are highly effective at reducing radiated energy going down, but that they do not going up.
Keep talking, you amuse me.
Do I really? I don’t believe so. Maybe you could quote me saying that. I don’t argue about anything. I leave that to the various religious GHE sects.
You absolutely did. You said, when talking about deserts and the higher daytime temps vs jungles at the same latitude:
They have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.
So you are arguing that GHG’s in fact block radiated energy from the Sun, but that they do nothing with energy radiated by the earth. Its right there in black and white for everyone to see.
No, that’s what you are saying. I said [deserts] have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.
You don’t have to accept facts. You can argue they don’t exist until you drop from exhaustion – but they still won’t go away.
No, that’s what you are saying. I said [deserts] have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.
So you’re saying, once again, that if there were more GHG’s in the desert, the desert would get less of the Sun’s radiation. But they have less GHG’s so more of the Sun’s radiation gets to the surface.
I’m curious as to how the GHG’s are active for downwelling radiation from the Sun, but inactive for upwelling radiation from the Earth? They only work one way? How does a molecule floating in the atmosphere even know which way is up and which down let alone which direction a photon came from?
I don’t know what condition you have but its not Dunning-Kruger. You seem unable to understand your own contradictions when they are repeated back to you verbatim.
The article above is predicated on experiments that have been conducted all over the world, repeatedly, for decades. The exact same principles are used in diagnostic imaging, DNA analysis, blood gas analyzers, cops carry a breathalyzer in their car, detectors for drugs, explosives and poison. Pollution monitoring. Process control in chemical plants, purity testing. Every Caterpillar lab in the world has one for analyzing oil samples to detect unusual wear particles.
If you’re right and I’m wrong, then none of those things actually work.
Except they do.
Scream your rage and insults into the void, I’m done with this thread so I will not read them. That must piss you off. But thanks for the entertainment.
It’s obvious to see why you refuse to quote me. I said what I said, and said again. If you think I meant something else, then you are delusional.
Your multiple gotchas are pointless. If you are boasting about your ignorance and gullibility, I won’t stop you.
What rage and insults do you perceive? What mental defect leads you to think that I will be affected in some way if you do not read something that doesn’t exist?
Adding CO2 to air does not make it hotter. Sorry if that insults you. Don’t blame me for your ignorance and gullibility.
Here it is summarised for you
David asked…
You responded with
David responded with
And your explanation is
And some insults. So when pressed you said
And you dont even attempt to address the obvious point that its the GHGs that keep jungles warmer at night with the equally obvious conclusion that GHGs “warmed the surface” overnight.
So in summary according to your very own argument
Adding CO2 to air does make it hotter.
No. Deserts get hotter because they lack the cooling effect of WV.
No, they get hotter because they receive more radiation from the Sun. More H2O in the air, they are not deserts any more, and don’t get as hot.
John Tyndall explains in some detail. You don’t have to believe him, either.
Here’s what I said “ [deserts] have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.”
Only a completely deranged lunatic would logically infer from this, that –
Or possibly someone exceptionally ignorant and gullible.Which one most applies to you?
No, adding CO2 to air does not make it hotter, Tim. Get someone to explain why, if you have difficulty comprehending written English.
Well you said it. You said “Basic Physics” and that is completely correct. The basic physics is that GHGs slow the rate of cooling leaving the surface warmer than it would have been.
So you effectively said “Adding CO2 to air does make it hotter”
I’m just commenting on your very own response, nothing else.
No I didn’t. Adding CO2 to air does not make it hotter. You are effectively (very effectively) boasting that you are ignorant and gullible.
Playing semantic games shows that you are stupid as well. The scientific method does not depend on consensus or argument. As Feynman said –
You don’t have an experiment, do you?
Yes you did. That’s what the basic physics is.
Unless of course you have a different idea of what you meant by “basic physics”? How about you tell us exactly what you meant by “basic physics” to clear it up.
I’m going to predict you wont do that. And you’ll just add some more insults. And we’ll be left with “basic physics” and that means you really do mean “Adding CO2 to air does make it hotter” because you wont offer anything else.
Don’t be a complete idiot. I mean what I said – adding CO2 to air does not make it hotter. If you don’t want to believe that’s what I said, that’s your choice.
Do you suffer from a mental defect? Do you lack the ability to understand English?
Adding CO2 to air does not make it hotter.
“They have less GHGs in the atmosphere. And therefore more of the Sun’s radiation reaches the surface.”
You’ve omitted part of it: more of the sun’s radiation reaches the surface and more of the surface IR emission leaves the atmosphere, any change in temperature will depend on the relative magnitudes.
You’ve noticed that there is a difference between day and night, have you?
Yes, but we’re discussing the effect of GHGs. The incoming solar radiation ranges between 200nm and 2500nm in which range CO2 has a very trivial absorption at ~2000nm so the effect on incoming solar is small.
On the other hand the Earth emits in the range 6μm- over 30μm and CO2 absorbs near the maximum, so the effect of CO2 is mainly the absorption of Earth’s emissions not the solar irradiation.
https://seos-project.eu/earthspectra/earthspectra-c03-p06.html
As Fourier said, the Earth loses all the radiation it receives from Sun, to outer space.
Phil:
You do realise you are *talking* to the most notorious and intransigent respondent in the denial-o-sphere?
Been around for decades. And not long ago finally got banned from Roy Spencer’s blog.
It is a total and hair-pulling waste of time trying.
I explained the reason that deserts are hot at least 10 years ago to him and he is still ignorant. It is like trying to get a flat-earther to see sense.
Deserts are hot because they lie under permanent subsidence zones beneath the sub-tropical jets. Ever noticed that it gets hot when an anticyclone lies over *you* for a length of time?
Subsiding air from aloft warms in descent and creates a low level inversion which inhibits surface convection until it becomes hot enough to pass as dry convection at a much higher temp.
Additionally the dryness of the atmosphere above allows increased LWIR emission to space overnight and a surface inversion forms over the highly insulating (from ground heat flux) sand surface. That cold surface air rapidly warms out as the sun heats the surface.
The GHE has a greater affect there because of the lesser WV content and the reason why Saharan intrusions into Europe in summer have become hotter.
I understand that Antony, however I wouldn’t want someone to think that the rubbish he posts might be true if no-one contradicted it!
Understood. It is a point of view I sympathise with.
That’s what ignorant and gullible GHE believers want.
No thinking. Acceptance. Religion, not science.
No that’s what you do with your religious falsehoods such as the Earth losing 44TW to space causing cooling for 4.5 billion years! Also your religious belief that all gases absorb and emit as grey bodies!
You can make up whatever nonsense you like.
Adding CO2 to air does not make it hotter.
“You can make up whatever nonsense you like.”
Which you do!
Deserts are hot because they lack H2O in the atmosphere, allowing greater amounts of insolation to reach the surface.
The Moon gets even hotter after the same exposure time, No GHGs at all! No atmosphere, in fact.
Adding CO2 to air does not make it hotter.
Deserts are hot because they lack the cooling effect of evaporation and lack latent heat transport of energy to altitude through convection. They typically lack clouds and the cooling effect of precipitation too.
LW energy from the sun is not a significant factor because the energy is absorbed by the GHGs high up in the atmosphere (ie CO2) well before it gets to the ground.
AKA Basic Physics.
Read John Tyndall’s publications.
Adding CO2 to air does not make it hotter.
Quote John Tyndall’s publications to support your argument if you think you can.
Here’s my prediction. You wont. You may simply add insults.
No argument. Fact. If you can’t be bothered supporting your disagreement, that’s fine.
Sorry michael but you are wrong. Black bodies emit energy dependent on their temperature as defined by Planks law. However objects that are not “black bodies” emit energy at a given wavelength as defined by Planks law TIMES the emissivity at that wavelength. Solids and liquids because of the close proximity of the molecules have substantial emissivity at essentially all wavelengths. However for gases that is not the case. The emissivity of nitrogen and oxygen in the thermal IR is so close to zero any emission is insignificant which means they neither emit nor absorb at those wavelengths. They do however have significant emissivity at longer wavelengths and do radiate and absorb at those wavelengths.
Adding CO2 to air does not make it hotter, why would you think so? What is does is enable the air with added co2 to interact more with the environment so that in a warmer environment (say near earth’s surface in the tropics during the day) it can heat up and in a cooler environment (say with exposure to outer space) it can cool down.
Did I misspell a word somewhere?
Well, according to GHE cultists, the atmosphere emits over 300 W/m2 at night, and NASA agrees with John Tyndall that the atmosphere absorbs (blocks) around 30% of insolation (most of which is IR). You claim that this is “insignificant”, but I disagree.
Without this “insignificant” interaction, surface temperatures would reach 125 C or so, like the airless Moon.
Adding CO2 to air does not make it hotter, in any case. Four and a half billion years of continuous sunlight has resulted in the Earth becoming cooler, not hotter. So GHGs are just pseudoscientific jargon, and there is thus no GHE.
Sad but true.
And -171C at night averaging about -23C so in fact the atmosphere and its GHGs does make it hotter on earth on average.
Ah, the wondrous “average”. The refuge of the ignorant and gullible.
Adding CO2 to air does not make it hotter.
“Without this “insignificant” interaction, surface temperatures would reach 125 C or so, like the airless Moon.”
No it wouldn’t because the length of day on the Moon is ~28 times longer than the Earth.
Phil, after the same exposure time. Look at the Diviner data if you don’t believe me. You might believe that six months of continuous sunlight in the polar regions might make them hotter than four hours of sunlight in Death Valley, but it doesn’t seem to be so, does it?
As usual when proven wrong you change the subject! The 125ºC is the temperature at the moon’s equator at noon over rocks, after 7 days of exposure, not exposure at a pole at low angle over reflective ice.
For those interested in actual data about lunar temperatures, see my post “Fun With Lunar Temperatures“. And FWIW, discussing climate with Michael Flynn is a colossal waste of time.
w.
Yes that was a very good post Willis.
Go on, Willis, have fun with figuring out how hot the Moon’s surface gets, exposed to 1366 W/m2 for 6 hours (Earth time).
Check with Diviner’s data. Call yourself ignorant, gullible, and a fool if you get less than 110 C.
Your fantasies are not fact. Just look at your nonsensical “Steel Greenhouse”. No thermal gradient between one object which heats another at a distance? [SNIPPED—Stick to the science, leave out the childish personal attacks. w. ]
Sorry Willis, adding CO2 to air does not make it hotter, and after four and a half billion years of continuous sunlight, the Earth has cooled!
Throw a few facts my way, Willis. Or just pretend that you know basic physics. Up to you.
So the surface in the tropics heats during the day, and cools at night. Quelle surprise! In your opinion CO2 is responsible?
Sorry, it’s the sun – and lack thereof, at night.
Kevin is right for N2 at atmospheric temperatures….NASA’s webpage authors are incorrect on some of their pages…N2 needs to be a lot hotter or stimulated by much higher frequency than IR before you see any emission lines….see attached…nothing past 700 nM (red). Of course hot Nitrogen can still burn you by conduction….
No, you’re confused. The graphs refer to excited emission of radiation, and absorption of radiation from external light sources. All matter above absolute zero emits infrared radiation due to, and frequencies dictated by, its degree of hotness (temperature).
The difference between a piece of iron heated until it appears the same colour as a neon tube, is an example. One colour is due to temperature, the other is due to electrons being excited to a new orbital level, and emitting photons of a specific wavelength as the electron drops back to its previous level.
You mention hot nitrogen. Yes, nitrogen, like any other gas can be heated with IR. And, as you point out, emit IR.
John Tyndall measured the IR emitted by hot dry air, which had been purged of CO2.
The IR from hot air emitted by a heat gun or hairdryer can be perceived from a distance, off to the side of the hot air stream.
“No, you’re confused. The graphs refer to excited emission of radiation, and absorption of radiation from external light sources. All matter above absolute zero emits infrared radiation due to, and frequencies dictated by, its degree of hotness (temperature).”
You are the one who is confused, that is not true for gases! The temperature can only excite gases to certain energy levels which can then emit at those select frequencies, so the energy levels would have to exist in N2 and the temperature would have to be hot enough to excite the molecule to that level, doesn’t happen around room temperature!
[SNIPPED—Leave out the ad hominems and personal insults. w. ]
All matter above absolute zero emits IR, the frequencies of which are determined by temperature. You are talking about excitation (such as a neon sign, or the Aurora Borealis). Nothing to do with “temperature”.
Adding CO2 to air does not make it hotter. [SNIPPED—What I said above. w. ]
“Don’t be more stupid than you have to, Phil.
All matter above absolute zero emits IR, the frequencies of which are determined by temperature.”
Not true, unfortunately you do not understand elementary physics of gases!
Others will no doubt decide for themselves. Luckily, they don’t have to pay you for advice. No monetary loss at all.
Do you have a link? That could be spectra for plasma.
All this is way above my head but a simple question..
Has the ”increased warmth” in the atmosphere ”caused by” added CO2 been separated from natural variation or not?
A one-word answer is enough.
No.
It would appear that some people have seen physics books but have never bothered to read them