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.
The reason why a cold body can’t raise the temperature of a warmer is simply that the temperature frequency and radiation of the hot is relatively high frequency, whereas the cold is lower frequency, and the cold cannot raise the frequency and hence the temperature of the hot.
A very nice article. I would just like to add, that the effect of the symmetry properties of molecules have an effect on their radiation absorbing or scattering properties. For example CO2 is a centrosymmetric molecule whereas water is not. Water and CO2 can also intervene with what’s called Raman, or inelastic scattering. For CO2 (with a center of symmetry) either a vibrational mode is active in IR or in Raman. As for water or methane (CH4) (not centrosymmetric), a vibrational mode can be active in both IR and Raman. The dipolar moment of a molecule is also quite important. This is particularly important for water because this means that also rotational modes can interact to absorb or release energy. Nitrogen (N2) and oxygen (O2) molecules on the other hand have no dipole moment, and thus interact very little with IR radiation, but can collide with exited GHG molecules to take some of their energy away as kinetic (or thermal) energy. See https://www.youtube.com/watch?v=hAncKwQ-ZRg&t=61s (The Raman Effect).
I enjoyed your article. It is very understandable. However, I have a problem with this statement at the end of your essay.
I am a great proponent of studying Planck’s Theory of Heat Radiation. He shows that a body radiates at its own temperature regardless of surrounding bodies. This is a necessary condition for reaching thermodynamic equilibrium. If a hot body must raise its temperature to account for radiation from a cold body, equilibrium can never be achieved. One would have a constantly rising exchange of energy between the hot and cold bodies.
Planck explains it this way.
One can not postulate that a warm body must raise its temperature to account for “reflected” energy and at the same time assert that equilibrium results in the equal exchange of energy.
The end result is that the surface cools, albeit at a slower rate, thereby introducing time into the equation. The lack of inclusion of time and the use of calculus is a major fault line in the prediction of radiative aspects of the earth system. The sun’s insolation at any point on the surface follows a sine function and is confirmed by soil temperatures. The surface stores some of that energy for later use, and time is involved again as well as diffusion.
As I said already, you article is enlightening, just beware making a generalization about the greenhouse effects.
“The end result is that the surface cools, albeit at a slower rate, thereby introducing time into the equation. The lack of inclusion of time and the use of calculus is a major fault line in the prediction of radiative aspects of the earth system.”
flux intensity is related to temperature. heat loss is related to flux intensity. Therefore temperature is related to heat loss. The amount of heat loss that occurs is related to the integral of temperature curve. A slower temperature decay curve means more heat loss per unit time. I would think that any heat absorbed by BodyA (warmer) from BodyB (cooler) is a part of that extra heat that gets radiated away due to a slower decay. Heat loss from the earth continues even during the day. If it starts at a higher minimum temp at sunrise then its still going to radiate away that “extra” heat even during the day. What that extra heat does to the atmosphere is pretty complicated with PV = nRT and convection playing major roles. It’s a total chaotic mess that, to me at least, can’t be predicted using a simple linear projection of temperature.
You are correct. Planck calls it compensation. If a body is a source, i.e., a constant temp, it just keeps on radiating based on its temp regardless of what an additional cooler body does. The cold body’s radiation is absorbed and instantly radiated away. However, this does change the net radiation between the two bodies. In the limit, as the net approaches zero, equilibrium is reached.
Not necessarily. Ice can emit 300 W/m2, at below freezing, a polished metal kettle full of boiling water can also emit 300 W/m2. You can use your hand to detect the temperature difference by touch, if you are stupid.
That’s why temperature is not specified in W/m2. It would be completely pointless, like most “climate science”.
You missed the point entirely. Temperature *IS* related to heat loss through the associated flux intensity *at that temperature*. This has nothing whatsoever to do with the absolute value of the flux intensity. It doesn’t matter if the Earth is at 0C or 100C. Anything that slows the temperature decay curve results in *more* heat loss per unit time, i.e. compensation for the process that is slowing the temperature decay curve.
I’m not sure what you are trying to imply. Whether the Earth is cooling from 0C or 100C, faster or slower, it’s still cooling.
Some pious GHE believers think that if the rate of cooling of a glowing hot ball of rock (the Earth) is slowed, the glowing hot ball of rock will get hotter! How ridiculous is that?
Maybe you meant something else.
The issue is using temperature as a proxy for heat. Any “back radiation” affecting temperature causes more heat per unit time to be dumped. The back radiation can’t increase temp because it gets radiated away by the increased heat loss per unit time.
Tim, sorry. I misunderstood. And, as someone has already (inadvertently?) pointed out, Planck said that a body losing energy cools.
The surface, having lost energy, cools. If 100% of this energy was magically returned to the surface, all it would do is restore the lost energy. But the alleged “back radiation” (just radiation from the atmosphere) cannot even do this, as the atmosphere, warmed by surface radiation, is constantly losing energy to space.
So the surface, in the absence of sunlight more than replacing the lost energy for part of the day, cools. Slow, fast – cools.
The GHE cultists have convinced themselves that adding CO2 to air makes it hotter – in spite of experimental results from Tyndall onwards, showing the complete opposite, if surface thermometers are involved.
Religion, not science.
Again, I apologise for my misunderstanding.
True, but possibly misleading. An ice cube totally immersed in water emits “a quantity of energy”, but does not experience a “decrease of the heat in the body”. That is cooling, and ice submerged in water gets hotter. Energy emitted by the ice is not absorbed by the soup, otherwise the soup would experience “an increase of heat” – that is, heating.
shows an incorrect translation, or a misunderstand of physics. There are no “cold rays”. Bodies (including bodies of gas) cool by emitting photons. If this energy is not replenished by absorbing photons of equal or shorter wavelengths, cooling continues. In the complete absence of an external energy source, the matter concerned will cool all the way to absolute zero (in theory).
Adding CO2 to air, even at the same temperature, does not make the air hotter! That’s just silly!
This is a direct quote from Max Planck’s treatise, The Theory of Heat Radiation, as I stated in the post.
Perhaps you should reevaluate your understanding of physics before you begin to criticize an outstanding physicist like Planck, who as far as I know, has never been shown to be incorrect.
You can obtain a Kindle copy of his book for about $3.00 from Amazon. I suggest you get it and study it.
Or a translation of “Vorlesungen uber die theorie der warmestrahlung”?
In any case, there are no “cold rays”, so Planck was either mis-translated or confused. Totally immersing an ice cube in hot soup does not result in the soup heating, and the ice cube getting colder.
Which is why adding CO2 to air, even at the same temperature, does not make the air hotter! That’s just silly!
You are ignorant and gullible. Maybe you could consider accepting reality, if it conflicts with what someone wrote. Try appealing to more recent authority.
The original document is available so why don’t you translate it yourself and show us the error that was made.
As far as Planck being confused, yah, right! Keep up the good work!
Why don’t you?
Because I am not the criticizing Planck’s statement, nor the translation.
If you wish to criticize, then YOU need to back it up with evidence.
I’m not criticising anything. I’m just pointing out that putting ice-cubes in soup doesn’t make the soup hotter or the ice colder, as the comment you posted suggests.
If you choose to ignore reality, that’s religion, not science.
Calling Planck confused is a criticism!
Calling out the translator for not knowing what they are doing is a criticism.
Since you apparently have no idea of how to obtain a translation, I’ll do it for you.
From the original.
From Copilot
From Grok 3
What exactly is functionally different between any of the translations?
None of them mentioned the part you left out.
Dropping ice cubes into soup does not result in the ice gettin* colder, and the soup getting hotter.
Adding CO2 to air does not make it hotter. Cold bodies do not make hot bodies hotter.
No GHE,
You are ignorant and gullible, no matter what Max Planck or anybody else says.
I find this statement by Planck to be appropo.
“For example, if we let the rays emitted by the body fall
back on it, say by suitable reflection, the body, while again absorbing
these rays, will necessarily be at the same time emitting new rays, and
this is the compensation required by the second principle.
Generally we may say: Emission without simultaneous absorption
is irreversible, while the opposite process, absorption without emission,
is impossible in nature.”
CO2 acts as a partial reflector. Reflected rays get absorbed by the Earth and are immediately re-emitted as new rays if the Earth is considered to be a black body. Since CO2 is only a partial reflector the Earth still cools since only a part of the originally emitted rays are reflected. As the reflected rays are absorbed it keeps the Earth from cooling as fast so the new rays emitted due to the absorption are at a higher intensity than if the absorption didn’t occur. it is that higher intensity that results in the immediate re-emission of the reflected rays.
This is for black bodies only. Since Earth is not a black body time enters into the picture. The re-emission of the heat absorbed gets offset in time. Earth acts as a heat sink which doesn’t immediately re-emit heat – but it eventually does emit it, just at a later time.
To me this once again reinforces the view that the only real way to treat the biosphere is through the use of enthalpy, i.e. heat. Temperature is not a direct measure of heat. Using the average of the temperature curve as a measure of heat transfer is misleading. The heat gain/loss is actually the area under the temperature curve. And it is important to analyze the temperature curve as a function of time.
Seems fair to me. Left alone, matter cools. No storing or accumulating heat.
Agrees with observation.
Nailed it: 165 Comments
Great article – you are gifted at explaining very technical concepts in terms non-technical people can understand. Which means that you understand them very well, indeed.
It was especially interesting to me that you taught from the text by Incropera and Dewitt. Both were teaching at Purdue when I was a student in ME (1975-1980), and though I had interactions with both, I never had them as teachers. The text Purdue used back then was “Transport Theory” by Bird, Stewart and Lightfoot. I always liked that book because of the gag they put in at the end.
This has troubled me for some time. Despite what would appear to be comparable educations of what are collectively bright people, there is often disagreement over claims apparently resulting from different interpretations of basic principles. It is difficult to make progress when the ‘team members’ can’t agree on simple definitions and processes.
Your excellent description of key points is a step in the right direction. However, people are also “bickering” with you.
Bloody impossible, I’d say.
Very hard indeed to falsify something that is so vague ‘the team’ can’t even agree on a description. Falsification by experiment or observation is the foundation of the scientific method.
What amazes me is that the article has 252 (now 253) comments, but only12 people rated it. WUWT?
Thanks, Kevin, a most informative and interesting article. One small objection. You say:
You are right that there is a divergence of definitions. However, what is called “cloud feedback” by orthodox climate science is the change in the effect of clouds on surface temperature due to a small increase in surface temperature. The orthodox point of view is that if the surface warms, the cloud feedback will be positive, meaning the change in clouds will warm the surface further.
However, as I’ve shown in a number of posts here on WUWT, the actual response of the clouds to increasing surface warmth is negative rather than positive, and is much more complex and temperature-dependent. See my post “Rainergy” for a discussion of these issues.
My best to you and yours,
w.
Willis, you write the darndest things. From your “Rainergy” nonsense/fairytale –
No, Willis, temperature is measured in “degrees of hotness” – for example, Celsius or Fahrenheit.
After four and a half billion years, the Earth has cooled – to its present temperature!
Your GHE religious beliefs may cause you to disagree, but religion is not science.