# A Third Look at Radiation versus Temperature

Guest post by KEVIN KILTY

### Introduction

In a post on June 8, 2019, Willis Eschenbach showed an interesting plot of monthly average surface temperature against total irradiance absorbed at the ground surface. He has updated this original post. His original figure is at the top of this post. He made no hypothesis about the meaning of the plot. In the thread which followed no one suggested a reason for the relationship. However, it did lead a number of commenters to suggest that it demonstrated a doubling of atmospheric CO2 would produce only 0.38°C to 0.5°C of surface warming. I am going to provide a tentative explanation for the relationship. My tools will be some simple physics such as Newton‘s law of cooling or the Stefan-Boltzmann law, known to many who post here, and a block model of heat transfer at Earth’s surface.

### 1. A Word on Equilibrium

Energy balance is my primary analytical tool in this post. I will use a block model of Earth’s surface known in engineering as a control volume. In this case my control volume is a thin layer at the surface known as the skin. There is material above and below this layer, and so to do a complete analysis we will have to account for energy transfer between this skin and materials above and below. Often, to simplify such an analysis we resort to postulating equilibrium. However, at Earth’s surface we observe: evaporation, rainfall, temperatures that rise and fall, ice and snow melt and so forth. Each of these observations imply or demand non-equilibrium. Thus, the Earth is practically never in equilibrium. However, over a suitably long time period Earth keeps passing through nearly the same state, and we speak of it being in steady state. If we use average values over a suitable time period, our steady state quantities obey a sort of equilibrium even though we are out of equilibrium at any instant. Willis averaged data over each month to produce the graph.

### 2. Our Block Model

Figure 1. Control volume and heat flows for the present model.

Figure 1 shows the block model with all energy flows. Arrows indicate the typical direction of transport. The control volume is Earth’s skin characterized by a temperature T. Surface absorbed energy is a combination of solar irradiance plus longwave irradiance from the atmospheric greenhouse effect. It is total irradiance. I will use the symbol (Ia) to represent it. The other heat transfer mechanisms are:

(1) Emitted power is blackbody radiation from the surface which we can conveniently calculate because we know that materials at the skin surface are nearly black at infrared wavelengths.

(2) Heat transfer to the substrate. Over the oceans this will occur through bulk mixing or convective transfer. Over land it will occur primarily through thermal conduction. It goes back and forth.

(3) Heat transfer to the overlying air. This occurs as convective transfer through a film. Engineers would model it using Newton’s law of cooling. I think it is primarily away from the skin although there are times and places it goes the other way.

(4) Heat transfer to the overlying air through evaporation, both as sensible and latent heat, and through precipitation which lands on the skin with a different temperature. This is just about exclusively outward.

One might also consider heat stored within the skin. However, on the sea surface the skin is something like 10 micrometers thick, and in soil it is about 5 centimeters thick. This represents very little thermal capacity to consider.

### 3. Willis’s Data

The data came from the CERES compilation. A number of posts on the original thread were wondering if the temperature involved is at the Earth’s surface or 1.5m above it. The CERES surface temperatures are a skin value, derived from GEOS data. The method of producing these temperatures is complex, and involves both parameterization and maximum likelihood inverse methods. Comparison against independent temperature data, satellite and ground, suggests a precision of about 0.5°C. The amount of analysis and scientific work that goes into these efforts is well described in a series of technical documents, available online, that are well worth reading for a background in the subject.

What is pertinent about Willis’s plot are two things. First, the slope of the regression line, which is 0.38°C per 3.77W/m2 absorbed energy. Second, one will notice that the range of temperature is small, which suggests that linearizing any non-linear relationships involving temperature is appropriate.

### 4. Analysis

I collect all of the heat transfer mechanisms 2-4 in my list and call them Q. Willis would likely refer to them as parasitics. They reduce the temperature variation that would occur in their absence. I treat them as a lump because I can’t calculate them from first principles, and I don’t need to for what I plan to do. The energy balance for my control volume is:

(1) Ia = σT4 + Q

Variations in temperature are very small. As a result I will linearize Q and write it as h·(T Ts) where h is an over all transfer coefficient, and Ts is temperature of the environment surrounding the skin. The surroundings might be at different temperatures in different directions, but what is important is that we are just transferring through a small temperature gradient and transfer is proportional to rising T. The slope of Willis’s graph is:

(2)

To compare how the model explains this slope we need a partial derivative with respect to Ia from our model. Unfortunately our model has Ia as a function of T. However, the relations are monotonic in T and so we can just invert a partial derivative of Ia with respect toT to obtain what we need. From our model:

(3)

(4) = 4σT3 + h

At 288K, the term from blackbody radiation is about 5.4W/m2K, and the coefficient h is unknown, but available from a comparison between Willis’s slope and ours. This amounts to:

(5) 0.1 = 1/(5.4 + h)

This is an equality if h is about 4.6W/m2K. What this says in plain terms is that Willis’s graph implies that added irradiance gets divided between emitted power from the surface and heat transfer by other mechanisms in about equal amounts. Seems reasonable.

What about climate sensitivity? The effect of CO2 is in the down welling longwave (LW) radiation which is implicit in the total irradiance that Willis uses. There is no way, short of introducing new models, assumptions, and measurements to separate the effect and find climate sensitivity.

### 5. Conclusion

Our discussion demonstrates that the relationship between skin temperature and absorbed radiation from Willis’s post has no direct bearing on climate sensitivity. Instead it shows something very interesting about the relative magnitude of various heat transfer mechanisms which must operate at the skin of our planet. It implies that mechanisms I collectively called Q, are roughly as important to maintaining a skin temperature as is blackbody radiation. There is no way to take Willis’s diagram and determine how much warming a doubling of CO2 will produce.

Many people who post comments here seem fond of Richard Feynman. I have a quote of his that is not only appropriate to the topic at hand, but is appropriate to all analyses, scientific in particular. In his series The Feynman Lectures on Physics he devoted one chapter to the phenomena of para- and diamagnetism. Here he says

If you start a [classical] argument in a certain place and don’t go far enough, you can get any answer you want.

I have placed brackets around the word classical to make his quotation apply more broadly than the context he was using it in. People understandably thought the slope pertained to climate sensitivity but just didn’t take their argument far enough.

6. Notes:

One reference I found useful for explaining the recovery of sea surface temperature is: GOES-R Advanced Baseline Imager (ABI) Algorithm Theoretical Basis Document for Sea Surface Temperature, Alexander Ignatov, NOAA/NESDIS/STAR. ( https://www.goes-r.gov/products/ATBDs/baseline/baseline-SST-v2.0.pdf)

One can find the Feynman quotation in The Feynman Lectures on Physics, Volume II, Chapter 34, Section 6, first paragraph.

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icisil
June 15, 2019 10:52 am

“…the skin surface are nearly black at infrared wavelengths”

I would like some clarity. When people say that the earth emits as a black body are they referring to IR spectrum or IR energy? I don’t see how the latter could be true. As I understand it, a black body surface exists in contact with a vacuum and can only lose energy via radiation. The earth’s surface exists in contact with atmosphere, and can lose energy by other means, i.e., conduction. So I don’t see how it can be said that the earth emits the same IR energy as a black body would.

Stephen Wilde
June 15, 2019 11:07 am

icisil,

Good point.
Energy leaving the surface has to be apportioned between radiation upward and conduction/convection upward.
Within descending air that energy previously taken into conduction/convection during uplift returns to the surface and reduces the rate of surface cooling.
The greenhouse effect is a consequence of convective overturning and not a consequence of the presence of radiative material in the atmosphere.
A greenhouse roof prevents convective cooling just as does a column of descending air within a region of higher that average surface pressure.
Descending air dissipates clouds to let in more radiation as does the transparent roof of a greenhouse.

June 15, 2019 1:20 pm

The surface emits BB radiation. Non BB radiation entering the atmosphere via latent heat, conduction or convection is not ’emitted’ and can only be returned to the surface, as only energy transported by photons can leave the planet. The non radiant energy is transported by matter and matter can’t leave the planet. To the extent that Earth’s surface looses energy via non radiant means, it doesn’t displace the energy radiated away from the surface by photons. None the less, the final state of the surface is it’s average temperature and the NET energy leaving the surface is the BB radiation corresponding to the average temperature and whatever effect non radiant heat leaving the surface and subsequently returning can have on the surface temperature has already been manifested by the surface temperature and otherwise has a zero sum influence on the RADIANT balance and the sensitivty.

You’re not alone being confused about the radiation emitted by the surface and the planet relative to energy transported by matter. If main stream climate scientists had this right, CAGW would have gone the way of a flat Earth decades ago. Sure, the planet is not an ideal BB, but all non ideal emitters of radiation into the vacuum of space can be completely characterized by a dimensionless emissivity between 0 and 1, which for Earth is about 0.62 +/- 5%. Gray bodies are not ideal themselves, but are non ideal black bodies which can still be precisely characterized.

icisil
June 15, 2019 2:11 pm

“Non BB radiation entering the atmosphere via latent heat, conduction or convection is not ’emitted’ and can only be returned to the surface”

I disagree with this. Energy imparted to air molecules by conduction can be and is radiated. All matter with temperatures radiates energy.

June 15, 2019 3:43 pm

Air molecules, with the exception of energized GHG molecules, do not contribute to the radiant balance. O2 and N2 neither absorb or emit any significant energy at the relevant wavelengths.

icisil
June 15, 2019 5:59 pm

” O2 and N2 neither absorb or emit any significant energy at the relevant wavelengths.”

What are the IR fluxes of CO2, O2 and N2 (W/m2, pure gas), and how are they measured?

David L. Hagen(@hagendl)
June 15, 2019 7:08 pm

Oxygen and Nitrogen ARE weak Greenhouse Gases – They DO radiate and contribute to Blackbody radiation – though not as much as H2O and CO2. They ARE included in detailed Line By Line radiation models. See:
Höpfner, M., Milz, M., Buehler, S., Orphal, J. and Stiller, G., 2012. The natural greenhouse effect of atmospheric oxygen (O2) and nitrogen (N2). Geophysical Research Letters, 39(10).
https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2012GL051409

David L Hagen(@hagendl)
June 15, 2019 7:34 pm

Oxygen (O2) and Nitrogen (N2) ARE Greenhouse Gases though weak ones.
See slide 6 Absorption of the Sun’s incident electromagnetic energy in the Region from 0.1 to 30 mm by various atmospheric gases
https://topex.ucsd.edu/rs/Lec06.pdf
For details see:

The effect of collision-induced absorption by molecular oxygen (O2) and nitrogen (N2) on the outgoing longwave radiation (OLR) of the Earth’s atmosphere has been quantified. We have found that on global average under clear-sky conditions the OLR is reduced due to O2 by 0.11 Wm2 and due to N2 by 0.17 Wm2. Together this amounts to 15% of the OLR-reduction caused by CH4 at present atmospheric concentrations. Over Antarctica the combined effect of O2 and N2 increases on average to about 38% of CH4 with single values reaching up to 80%.

Höpfner, M., Milz, M., Buehler, S., Orphal, J. and Stiller, G., 2012. The natural greenhouse effect of atmospheric oxygen (O2) and nitrogen (N2). Geophysical Research Letters, 39(10). https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/2012GL051409

Max
June 15, 2019 8:34 pm

“Air molecules, with the exception of energized GHG molecules, do not contribute to the radiant balance. O2 and N2 neither absorb or emit any significant energy at the relevant wavelengths.”

The forced air heater in my house would disagree with you. The oxygen and nitrogen heat the home just fine.
When warm air goes up, it expands and cools off immediately. If you insulate the air and prevent the expansion like in a hot air balloon, there is your greenhouse. When Air comes down, it heats up due to compression. 5.4° for every thousand feet. (unless it is wet air, it begins to evaporate which always cools offsetting the heating affect) weather balloon data is consistent confirming temperatures heating on dissent, day or night.

Samuel C Cogar
June 16, 2019 8:17 am

co2isnotevil – June 15, 2019 at 3:43 pm

Air molecules, with the exception of energized GHG molecules, do not contribute to the radiant balance. O2 and N2 neither absorb or emit any significant energy at the relevant wavelengths.

O2 and N2 don’t have to “absorb n’ emit” any LWIR radiated thermal “heat” energy from the surface or from atmospheric H2O or CO2 molecules …… but they are still the 2 “thermal “heat” energy transporting elephants in the atmosphere” ….. and they do it by “conduction & convection”.

The thermal “heat” energy of the solar irradiated surface is “conducted” to the N2 and O2 when they make contact with the surface. Their newly absorbed “heat” causes them to rise in the atmosphere, (like a “hot air” balloon) and when they do, more air molecules will “flow in” to replace the “uprising” ones, (nature abhors a vacuum) which causes the per se “surface wind” ….. and the process keeps repeating until the “temperatures” normalize.

“DUH”, …… 98% of all indoor/outdoor (near-surface) thermometers are measuring the temperature (in C or F) of the O2 (21%) and N2 (78%) that surrounds them, ……… not the itty bitty punitive amounts of H2O (+-2%) and CO2 (0.041%) that is mixed in with the oxygen and nitrogen.

Ole proverb: “How can I soar like an eagle …. when I have to fly with the turkeys?

June 16, 2019 8:33 am

“O2 and N2 neither absorb or emit any SIGNIFICANT energy at the relevant wavelengths.”

I used the word significant for a reason. The combined GHG effect of O2 and N2 is such a tiny fraction of the total GHG effect, it can be safely ignored. BTW, clouds warm the surface much like GHG’s, except that the water in clouds is a broadband absorber and emitters of photons, while GHG’s are narrow band absorbers and emitters and yet pedantic climate science ignores the warming effects of clouds in favor of GHG’s! Of the 620 mw/m^2 of surface emissions per W/m^2 of forcing in excess of the 1 W/m^2 per W/m^2 characteristic of an ideal black body, more than half is replenished by cloud emissions returning to the surface.

Warm air heating a house does so by conduction, not radiation and conduction plays absolutely no role in the RADIANT balance of the planet. Conduction stops at TOA since you can’t conduct heat to the vacuum of space.

June 16, 2019 4:09 pm

Samuel,

Yes, thermometers measure the temperature of the air, which is heated by the surface, which is heated by solar radiation and radiation returned by GHG’s and clouds. The warmed O2 and N2 does not further warm the surface, nor do the photon emissions from energized GHG molecules or the water in clouds directly warm the O2 and N2 in the atmosphere. All they can warm is the surface and the water in clouds which in turn heats the N2 and O2 by conduction. The best way to think of the N2 and O2 in the atmosphere is as adding to the thermal mass of the planet being heated, representing only a small fraction of the total thermal mass. It’s basically no different than a rock, relative to the planets energy balance and the sensitivity.

Samuel C Cogar
June 17, 2019 4:19 am

@co2isnotevil – June 16, 2019 at 8:33 am

I used the word significant for a reason.

And I used the word “significant” for a different reason …… and I wasn’t critiquing your use of it.

and yet pedantic climate science ignores the warming effects of clouds in favor of GHG’s

Shur nuff they do, even though ….. clouds are GHGs.

Warm air heating a house does so by conduction, ….

The thermal “heat” source” conducts its “heat” to the air molecules that come in contact with it and then convection transports that “heat” throughout the house and then conduction heats things up. So, its conduction, …. convection, ….. conduction.

Samuel C Cogar
June 17, 2019 4:27 am

co2isnotevil – June 16, 2019 at 4:09 pm

The warmed O2 and N2 does not further warm the surface,

If the surface is cooler/colder than the O2 and N2 are, they sure as hell WILL warm it or keep it warm.

June 17, 2019 9:53 am

Sam,

Yes, but in the grand scheme of things, atmospheric conduction and convection have no impact on what the radiant balance must be, even if the Joules involved happened to be emitted into space or back to the surface. The radiant balance, the surface temperature and the corresponding sensitivity can be completely and accurately characterized based on the transport of energy by photons alone. It works within a few percent and there’s no need to account for anything else. Doing so only adds unnecessary complications providing the wiggle room to feign support for what the laws of physics can not.

BTW, if the air is warmer than the surface, then it’s just redistributing energy from one part of the surface to another and has no impact on the NET radiant emissions of the surface and its corresponding average temperature.

This illustrates the second biggest error in climate science after the feedback fubar, which is conflating the energy transported by photons involved in the radiant balance and sensitivity with the energy transported by matter that simply redistributes existing energy.

AndyHce
June 17, 2019 7:34 pm

co2isnotevil wrote:
“Air molecules, with the exception of energized GHG molecules, do not contribute to the radiant balance. O2 and N2 neither absorb or emit any significant energy at the relevant wavelengths.”

O2 emits energy at microwave frequencies, as demonstrated by satellite temperature measurements. Being lower frequency, longer wavelength photons, they are less energetic than IR radiation but there is quite a bit more O2 in the atmosphere than CO2 (or CO2 + other IR emitters). No doubt the relative quantity of energy emitted by O2 is less than that of CO2 but what about the total emitted/radiated energy? Microwaves carry energy away from the planet just as surely as does IR. Why is microwave radiation not part of the radiant balance?

Samuel C Cogar
June 18, 2019 3:52 am

co2isnotevil – June 17, 2019 at 9:53 am

Yes, but in the grand scheme of things, atmospheric conduction and convection have no impact on what the radiant balance must be,

co2isnotevil, ….. HA, …. I wasn’t aware of Mother Nature having a “grand scheme of things” pre-planned for anything. And iffen SHE doesn’t have a “grand scheme of things” pre-planned for anything then why in the natural world would SHE need to “balance” the radiant energy that’s a coming n’ a going?

co2isnotevil – June 17, 2019 at 9:53 am

BTW, if the air is warmer than the surface, then it’s just redistributing energy from one part of the surface to another and has no impact on the NET radiant emissions of the surface and its corresponding average temperature.

But, but, but, co2is, ….. iffen the surface conducts “heat” to the air, warming it up, and the air re-conducts it back to the surface, ….. cannot the surface then radiate that energy as LWIR? That energy transfer “from n’ back to” the surface happens pretty darn quick, so how do the “experts” determine that …. “Who is on 2nd”?

Samuel C Cogar
June 18, 2019 4:03 am

AndyHce – June 17, 2019 at 7:34 pm

No doubt the relative quantity of energy emitted by (molecule of) O2 is less than that (a molecule) of CO2 but what about the total emitted/radiated energy?

AndyHce, ….. great question given the fact the O2 is 210,000 ppm whereas CO2 is only 410 ppm.

June 28, 2019 8:34 am

Sam,

“iffen the surface conducts “heat” to the air, warming it up, and the air re-conducts it back to the surface, ….. cannot the surface then radiate that energy as LWIR? ”

If the surface conducts heat to the air, the surface cools. If that heat is returned back to the surface, the surface warms, but will never be warmed more than it cooled from the original conduction. Whether those specific Joules are then emitted as LWIR is irrelevant, as this becomes an independent transaction emitting photons that need to be emitted anyway. Beside, no one Joule is any different from any other relative to the energy balance, so it really wouldn’t matter anyway.

The point I was making was that atmospheric gases don’t re-radiate as photons the energy entering the atmosphere by conduction.

Stephen Wilde
June 15, 2019 10:57 pm

The energy converted to PE from KE during uplift cannot be radiated out because PE is not heat and does not radiate.
It can only be recovered as heat during the subsequent return to the surface during compression and so it reduces surface cooling to create the greenhouse effect.

June 16, 2019 8:45 pm

How about the PE of lifting water against the force of gravity. Although, it can produce hydroelectric power that can be turned into radiant heat …

Interestingly enough, the PE of water lifted against gravity is the right order of magnitude to offset the latent heat of evaporation.

tty
June 15, 2019 2:15 pm

You completely miss that much of the heat leaving the surface convectively does NOT return. It is radiated away from higher levels in the atmosphere. This is obvious when looking at the Earths IR spectrum as seen from space.

June 15, 2019 4:03 pm

The photons emitted by the planet have 2 basic sources. There are the photons emitted by the surface and the water in clouds that passes directly though the transparent windows in the atmosphere. The remaining photons are emissions from energized GHG molecules, mostly near TOA. These were energized by emissions from other GHG molecules lower in the atmosphere and so on and so forth where the ones at the bottom of the atmosphere were energized by absorbing a photon emitted by the surface. GHG absorbed energy pretty much stays with GHG molecules until it’s eventually emitted into space or back to the surface. The only exception is when an energized GHG molecule condenses upon or is absorbed by the water in clouds. Note that collisions may cause an energized GHG molecule to emit a photon and return to the ground state and that this is being conflated with the concept of ‘thermalization’.

The concept of ‘thermalization’ where energy absorbed by GHG’s is nearly instantly converted into the translational kinetic energy of molecules in motion by collisions is demonstrably wrong and the evidence is in the emitted spectrum. If this was the case, we would see no energy at all in the absorption bands of GHG’s and th2 energy in the transparent bands would be boosted. We don’t see more emissions in the transparent bands and we do see attenuated emissions in absorption bands, but the attenuation is only about 3db below what those emissions would be in the absence of GHG absorption. The remaining 3db is returned to the surface (or clouds) and comprises the GHG effect.

.

tty
June 16, 2019 3:43 am

One word: convection.

And about thermalization: compare the relaxation time for metastable states of a CO2 molecule with the average time between molecular collisions in the troposdphere.

EdB(@sailboarder)
June 16, 2019 5:31 am

Interesting.. you are schooling W Happer. Good luck.

June 16, 2019 9:28 am

tty,

What so many people fail to understand is that the ‘relaxation’ returning a GHG molecule to the ground state after a collision occurs by the emission of a photon and not by converting the state energy into the energy of translational motion. Another thing that can cause the emission of a photon is the absorption of another photon, which significantly increases the probability of spontaneous emission.

As I said before, the observed spectrum at TOA is clear on this, otherwise, there would be no possible source of absorption band photons at TOA and that while absorption band energy is certainly attenuated, it’s only attenuated by about 50% relative to what it would be without absorption. There can be only one possible source for these photons and that is the emissions of GHG’s returning to a lower state. BTW, most GHG molecules in the atmosphere are not in the ground state, thus the near immediate re-emission upon absorption of another photon is very likely.

Some may claim that little bits at a time are transferred to rotational states, but if you examine the fine structure, this works both ways in an equal and opposite manner. Certainly, energy from a vibrational state can be transferred to a rotational state in conjunction with the emission of a photon of slightly less energy. Energy is also moved from rotational states to a vibrational states by the absorption of a photon of slightly less energy and in equal proportions. The reverse also happens with higher energy photons where a slightly higher energy photon is absorbed and the energy is split between a vibrational state and a rotational state or emitted as the combination of relaxing a rotational state and a vibrational state.

TruthMatters
June 17, 2019 9:27 am

tty June 16, 2019 at 3:43 am
One word: convection.

nope
phase change
water gas is lighter than air without convection

tty
June 17, 2019 3:41 pm

“water gas is lighter than air without convection”

Not a strong enough effect. By the same reasoning CO2 should stay at ground level because it is heavier than oxygen and nitrogen. There is a slight tendency for this but not much.

Michael Hammer
June 15, 2019 7:12 pm

Kevin kilty
June 16, 2019 6:08 am

You have brought up a good point. When I was tinkering around with the ideas I present here, I used the MODTRAN code available at the University of Chicago to figure out an answer to the following question: “If surface temperature rises by One degree centigrade, how much increased down welling radiation is there from the greenhouse effect?”

In the neighborhood of 288K, 50% relative humidity and no adjustment to water vapor mixing ratio, the answer is about 3.9 W/M^2, about 75% of the black body value at unit emissivity. I find it surprisingly large, but it is what modtran provides.

Wim Röst
June 16, 2019 11:01 am

From the link below, Wenyi Zhong and Joanna D. Haigh 2013, Table 1: ” The second row in the table shows the impact of the individual gases on downward radiation incident at the surface. This component is dominated by H2O due to the very strong emission of radiation by the near-surface atmosphere through the H2O continuum described above.”

WR: The numbers on the second row tell that water vapor H2O is responsible for 91.1% of the at the surface (!) downwelling radiation and CO2 for 7.0%.

According to H. Douglas Lightfoot and Orval A. Mamer 2017 water vapor molecules are 29 times more abundant than CO2 molecules. The content of water vapor in the air is very temperature dependent: therefore we find most water vapor in the lowest part of the atmosphere: near the surface. Given the very short path length most of the surface (!) greenhouse effect must be caused by the abundant water vapor very near to the surface.

The most important question for surface warming seems to be: how sensitive is the Earth for a rise in the quantity of water vapor in the lowest layers of the atmosphere.

As we know, changes in the behavior of the oceans have a direct effect on local temperatures and so on local (and regional and latitudinal) changes in water vapor in the lowest atmosphere. Given the low content of water vapor at/near the poles, changes in surface water vapor must have the most effect near the poles: part of the polar amplification. *

Of course other [natural] factors (as changes in the wind pattern) also play a role in the quantity of water vapor near the surface.

https://wattsupwiththat.com/2018/06/15/how-the-earth-became-a-hothouse-by-h2o/

Wenyi Zhong and Joanna D. Haigh 2013: The greenhouse effect and carbon dioxide
https://rmets.onlinelibrary.wiley.com/doi/pdf/10.1002/wea.2072

Douglas Lightfoot and Orval A. Mamer 2017: Back radiation versus CO2 as the cause of climate change
https://journals.sagepub.com/doi/abs/10.1177/0958305X17722790?journalCode=eaea

June 16, 2019 9:06 pm

Kevin,

3.9 W/m^2 is the approximate incremental absorption of clear sky surface emissions by the standard atmosphere upon an increase of 1C (5.4 W/m^2 emissions increase) and only about half of this is returned to the surface to offset additional emissions while the remaining half contributes to the planets radiant balance at TOA. The cloudy sky actually absorbs significantly more. The average of cloud and clear skies is about 4.2 W/m^2. Half of this is 2.1 W/m^2 which means that 5.4 – 2.1 = 3.3 W/m^2 of solar forcing is required for a 1C increase which is a sensitivity of about 0.3C per W/m^2.

For the amount returned to the surface to be 3.9 W/m^2, the atmosphere must absorb 7.8 W/m^2 which is about 144% of the BB emissions which is obviously impossible.

Roger Clague
June 17, 2019 6:35 am

A black body is black because it absorbs all radiation and then emits all again.
The oceans are not black body radiators.

They conduct heat inwards and to the atmosphere
The reflect
The loss heat by evaporation

June 17, 2019 9:36 am

Roger,
The macroscopic behavior of the combination of the oceans, surface and atmosphere is definitely a gray body, otherwise the planet would heat or cool without bound, moreover; at TOA, only radiation enters and only radiation leaves. There’s no other possible way to characterize the macroscopic behavior of the planet. And of course, there’s no such thing as an ideal BB, but non ideal BB’s like the Earth can be completely characterized with a non unit emissivity which for Earth is about 0.62. So while in principle, ideal BB’s don’t actually exist, all matter absorbing and emitting radiation is exactly quantifiable as a gray body.

If you think otherwise, then enumerate the other laws of physics you believe can quantify how matter absorbs and emits energy,

Stephen Wilde
June 15, 2019 10:59 am

Convection is shown as a cooling effect going up.
So where is the warming effect coming down ?
Descending air warms at the dry adiabatic lapse rate does it not?
What goes up must come down.

rbabcock
June 15, 2019 11:44 am

What comes down is almost always cooler than what goes up. Stand outside on a hot day as a line of thunderstorms approach and when the first downdrafts hit, it doesn’t get warmer.

About the only thing I know when descending air heats up is due to compression heating. Air flow down a mountain will warm, but it is transient as when the air decompresses it cools back down.

Stephen Wilde
June 15, 2019 11:49 am

Of course what comes down is cooler than what went up but that is not the issue. The air that comes down is warmer than it otherwise would be in the absence of compression heating so it delivers energy to the surface which reduces surface cooling so that the surface is warmer than it otherwise would be.
That is what causes the greenhouse effect.

Alan Tomalty
June 15, 2019 12:47 pm

“Comparison against independent temperature data, satellite and ground, suggests a precision of about 0.5°C”

This means that the precision limit is larger than the trend. Therefore the whole argument is pointless. Also attempted measurements of back radiation in extremely dry places without clouds have failed to detect any CO2 back radiation.

EdB(@sailboarder)
June 15, 2019 2:14 pm

Excited CO2 molecules lose their energy by collision before they have a chance to radiate. Your comment suggests that. Upward LWIR withing the band of CO2 capture gets captured, then transformed into heat bu collision. More CO2 should lower the average capture height from say 5 meters to 3 meters. No extra heat is created. Convection happens anyway and the slight change of temperature profile does little.

We are chasing rainbows.

WXcycles
June 15, 2019 6:30 pm

old construction worker
June 16, 2019 11:59 am

‘Also attempted measurements of back radiation in extremely dry places without clouds have failed to detect any CO2 back radiation.’ I’m not saying you are right or wrong. I do know in Yuma, AZ the day time highs cab reach 3 digits but the low that night can be be quite cool. I believe that is do to the lack of water vapor.

DMacKenzie
June 15, 2019 2:02 pm

Steve, that isn’t a very technically correct description. Compression heating is NOT the cause of the radiative GHE, which is caused by IR radiative gases in the N2 and O2 atmosphere. The convective rise and fall of air In the troposphere to different atmospheric pressure (compression in your jargon) causes the “lapse rate”.

tty
June 15, 2019 2:26 pm

“The air that comes down is warmer than it otherwise would be in the absence of compression heating so it delivers energy to the surface which reduces surface cooling so that the surface is warmer than it otherwise would be.
That is what causes the greenhouse effect.”

Nonsense!

The sinking air is less energetic than what went up. That it is sometimes warmer than it was when rising is due to condensation at altitude. Dry air has much lower specific heat than wet air. Note that precipitation is NOT compressionally heated, since neither rain, nor snow is a gas.

Roger Clague
June 17, 2019 6:37 am

“compression heating”

Compression does cause heating
Heating does increase pressure ( Gay Lussac Law )

tty
June 15, 2019 2:19 pm

“What goes up must come down”

True for matter. Emphatically not true for energy. Much of the sensible heat, and even more the latent heat of rising air is radiated away into space. The total energy of the sinking air and precipitation is much less than what went up.
As a matter of fact this is the dominant mean of energy transport from the surface, larger than radiation, but this for some reason seems impossible for a lot of people to grasp.

DocSiders
June 15, 2019 5:07 pm

Similarly, we often see the assertion that the latent heat of evaporation and the subsequent heat released of during condensation cancel each other out in earth’s energy balance equations. They do not.

They do not cancel each other out because WHERE the evaporation occurs differs from WHERE the condensation occurs. Evaporation occurs primarily at the surface…producing lighter humid air that is (often enough) transported to high altitudes…where the water condenses and/or freezes resulting in cloud or precipitation formation). At altitude, the energy released during these one or two phase changes is more readily radiated into space than it would have been at the surface where the evaporation occurred. And any resultant cloud formation blocks much of the direct solar radiation that would have reached the surface.

This is one of earth’s most powerful one way energy transfer engines and this engine is switched on (emerges) by a threshold level of water heat content (temperature) at the surface REGARDLESS OF HOW THAT TEMPERATURE IS ACHIEVED. The physical properties of water act as a thermostat switch that turns on the convection engine that shunts energy into space (and forms shielding clouds) whenever things get hot enough.

Earth is a “Water World”. CO2 is a significant contributor, not the driver.

Chaswarnertoo
June 16, 2019 11:14 pm

This 👍🏼. Doc is correct.

beng135
June 16, 2019 9:51 am

tty says:
As a matter of fact this is the dominant mean of energy transport from the surface, larger than radiation, but this for some reason seems impossible for a lot of people to grasp.

Right. Restrict transport of air/heat into the atmosphere, and you get a closed car in the hot sun.

Gerald Machnee
June 15, 2019 6:38 pm

Descending air “warms” but does not gain any energy.
The parcel coming down has a certain amount of energy. The parcel gets smaller due to compression and the temperature increases, but the total energy in the parcel is the same.
Similarly ascending air cools due to expansion, but the total energy is the same in the parcel.

Heat or energy is carried up by water vapor. When the gas (water vapor) condenses to water droplets, energy is released thereby causing warming aloft.

tty
June 17, 2019 3:43 pm

Actually the air parcels do lose and gain energy of position by moving in a gravity field.

RACookPE1978(@racookpe1978)
Editor
June 15, 2019 11:29 am

Trenberth’s Diagram of his infamous flat-earth, single average flux, single average earth albedo, single average atmosphere model leaves a few things waving idlely in the mix.

For one, atmospheric absorption of inbound IR radiation, and outbound LW radiation.

One classic simplification says the atmosphere is clear in the IR spectrum of inbound solar radiation, and so “negligible” IR radiation is absorbed by the modeled atmosphere. 30% is inbound radiation is reflected from the ground, 70% is absorbed by the “average flat earth” … and then through the equations above to yield some calculated average emission-related temperature for the ground. Etc, etc.

But in the real world, the real atmosphere of nitrogen, oxygen, argon and CO2 and dust and water vapor and larger particles DO absorb significant solar energy, particularly at low solar elevation angles typical a few hours every day, and average for the arctic and antarctic all half of every year. So, what equations of theoretical gas LW radiation equilibrium between a naked earth and a greenhouse atmosphere accounts for the energy “appearing” in the atmosphere that is NOT radiated from the earth, and NOT from near-magical CO2 molecules absorbing outbound LW energy?

And “low absorption” of IR by the laboratory gasses at STP does not mean “no absorption” of IR.

June 15, 2019 12:44 pm

>>
So, what equations of theoretical gas LW radiation equilibrium between a naked earth and a greenhouse atmosphere accounts for the energy “appearing” in the atmosphere that is NOT radiated from the earth, and NOT from near-magical CO2 molecules absorbing outbound LW energy?
<<

Kiehl & Trenberth 1997 show 67 W/m^2 being absorbed by the atmosphere directly from the incoming solar flux. It’s about 20%, so it’s not nothing.

Jim

June 15, 2019 1:29 pm

Most of the solar energy absorbed by the atmosphere is by the water in clouds which is tightly and rapidly coupled to surface water by the hydro cycle. To the extent that we’re talking about averages spanning intervals of time longer than the nominal hydro cycle, the absorption and emission by the water in clouds can be considered a proxy for the absorption and emission by surface waters. BTW, most latent heat enters the atmosphere as a gas and is returned as liquid water that’s warmer than it would be otherwise. The rest goes on to drive the weather.

June 15, 2019 4:19 pm

When you say “solar energy” I’m guessing you mean the UV leftovers which didn’t get dumped into heating/replenishing stratospheric ozone and a bit of the longer IR and visible wavelengths that get down to/through the tropopause, yeah?

eyesonu
June 16, 2019 7:49 am

co2,

” …. BTW, most latent heat enters the atmosphere as a gas and is returned as liquid water that’s warmer than it would be otherwise. ”

Would explain this claim?

Stephen Richards
June 15, 2019 11:30 am

longwave irradiance from the atmospheric greenhouse effect.

I thought that gases absorp at a quatum of energy and re emit that same energy. Gases absorp at a given energy and re emit that same energy don’t they ?
Don’t the longwaves come from reflected energy where some is converted to heat thereby reducing the amount of energy for emission

June 15, 2019 2:40 pm

“I thought that gases absorp at a quatum of energy and re emit that same energy.”
Not generally the same quantum. Gas molecules are rapidly exchanging energy by collision. Sometimes a GHG molecule will have excess energy that it can emit. That is just a function of the temperature of the gas and the chance distribution of the energy exchanged. GHG molecules absorb IR energy, but that is mostly exchanged before it can be re-emitted. The probability of a GHG molecule emitting is mostly related to the gas temperature, not to its IR absorption history. IR absorption maintains the temperature of the air overall in the face of GHG emission losses.

LdB
June 15, 2019 9:27 pm

Ignore Nicks mumbo jumbo answer which is dead wrong, in his head I am sure he has the IR being absorbed and the molecule speeding up and running into other molecules. He really is silly enough to believe that.

The reality the IR absorption of CO2 is almost all V2 & V2 vibrational. You are correct you the exchange can only be made at that specific frequencies and you need a third species to get it to classical heat.

CO2 is an almost linear diatomic molecule unlike H20 … it really does look like O—-C—-O

It has a v1 vibration linear vibration the oxygen vibrate in and out together (7.5um wavelength)
It has a v2 vibration linear vibration the oxygen alternate in and out (4.3um wavelength)
It has a v3 vibration where the oxygen molecules bends or flex (15um wavelength)

In none of those does the molecule move any faster it just vibrates faster and in classical physics you would need a special setup to detect the vibration as heat because likely your device is looking for molecule actual motion speed (sort of what Nick was intimating).

It is a fact Molecular vibrations are the only ones that have IR absorption bands because they can change the dipole moment of the molecule. O2 and N2 for example have no IR absorption bands as they are fully symmetrical molecules.

So your answer is correct and rather than give you the answer I will encourage you to work on it. So you are looking for other molecules that overlap those wavelengths.

LdB
June 15, 2019 10:45 pm

So you got it right now you just need to work out what it can “bump into” to transfer that energy to molecule speed. I really hate the term “bump into” the molecules are held apart by charge the concept of bump implies a physical nature of interaction.

LOL@Klimate Katastrophe Kooks
June 16, 2019 6:29 am

Exactly so. Nick Stokes’ claim that it’s “Not generally the same quantum” highlights the fact that he really doesn’t understand what “quantum” even means in a particle physics sense.

A molecule with a non-zero net magnetic dipole will *always* absorb and emit exactly the same quanta of energy (barring the minute Doppler-shift effects)… that’s sort of the underlying premise of quantization and hence quantum mechanics itself… it’s what gives us the absorption and emission spectra of molecules… but it will not absorb and emit the same *photon* containing that quanta of energy.

Here’s my take on it:
A Comprehensive Debunking Of The Catastrophic Anthropogenic Global Warming (CAGW) Hypothesis Utilizing Particle Physics First Principles

Introduction:
——————–
First, you’ll notice the title states “the CAGW hypothesis”, rather than “the CAGW theory”… that is deliberate.

The words “fact”, “theory”, “hypothesis” and “law” have very specific definitions in science:
———-
Hypothesis: A tentative explanation of an empirical observation that can be tested, but which has little or no validation. It is merely an educated guess.

Fact: An empirical observation that has been confirmed so many times that scientists can accept it as true without having to retest its validity each time they experience whatever phenomenon they’ve empirically observed.

Law: A mathematically rigorous description of how some aspect of the natural world behaves.

Theory: An explanation of an empirical observation which is rigorously substantiated by tested hypotheses, facts and laws.

Laws explain how things behave in the natural world, whereas theories explain why they behave the way they do.

For instance, we have the law of gravity which explains how an object will behave in a gravitational field, but we’re still looking for a gravitational theory which fits into quantum mechanics and the Standard Model and explains why objects behave the way they do in a gravitational field.
———-

Climate science as it is currently studied has barely reached the level of hypothesis, but rather than testing their hypothesis via (non-cherry-picked) data and empirical observation, the climate alarmist “scientists” have defenestrated the Scientific Method, choosing instead to cherry-pick data, outright manufacture data, skew existing data to fit their hypothesis, and validate their hypothesis with computer models, which are nothing more than other hypotheses!

This is not scientific, which is exemplified by the fact that they then crow about some percentage “consensus” as an appeal to authority, while attempting to set themselves up as the very authority they seek to appeal to!

And they do this while dismissing any possibility that their “science” has become corrupted by group-think, political ideology and grant-troughing, as we saw in the ClimateGate 1.0, ClimateGate 2.0, ClimateGate 3.0 emails, and the Mann v. Steyn lawsuit discovery.

There has been an awful lot of scientific fraud perpetrated, and those responsible for that fraud need to be called to account to save the credibility of all of science.
——————–

Why Is Will Happer Leading The Climate Review Team?
——————–
The atmosphere consists of atoms and molecules. Those atoms and molecules obey the laws which particle physics has discovered. There is no magic occurring in the atmosphere and there are no exceptions to those laws, no matter what any climate “scientist” claims.

This is why Will Happer is now leading the team to review climate science. Happer is a retired particle physicist, specialized in optics and spectroscopy, and had a long research career studying physics closely related to the ‘greenhouse effect’, for example, absorption and emission of visible and infrared radiation, and fluid flow dynamics.

Happer, while at Princeton, extended Princeton’s amazing history of scientific discovery, and has written dozens of peer-reviewed papers on the topic of particle physics. He is eminently qualified to lead the team tasked with reviewing climate science.
——————–

Backgrounder:
——————–
The basic premise behind CAGW (Catastrophic Anthropogenic Global Warming) is that the CO2 molecule absorbs ~15 micron radiation, becomes vibrationally excited in one of its bending modes, collides with a nitrogen or oxygen molecule, and imparts that vibrational energy to the translational energy of the other molecule via a process known as collisional de-excitation, thereby increasing the temperature of the atmosphere.

Quantum mechanics dictates that all atoms and molecules are harmonic oscillators (with anharmonic force constants complicating the wave equation in the case of molecules). Atoms have only electronic mode quantum states, whereas molecules have vibrational, rotational and electronic mode quantum states. These modes are activated via absorption of electromagnetic radiation at specific frequencies. For vibrational and/or rotational modes, there must exist a magnetic dipole for photon absorption to occur. The molecule is unaffected by nonresonant frequencies, and if the molecule is already excited, even radiation at the correct frequency will be rejected (scattered) by the molecule.

The time-independent Schrodinger Equation treats translational molecular movement (which we sense as temperature) differently than it treats vibrational, rotational and electronic mode quantum states because the quantum states are quantized (have discrete energy levels) whereas translational movement is not quantized.

The 2nd Law of Thermodynamics states that energy cannot flow from a lower-energy region to a higher-energy region… not directly, not via excitation of a molecule’s vibrational, rotational and/or electronic mode quantum states then transferred to translational motion of other molecules via collisional de-excitation, not ever.

Kirchhoff’s Law states that the frequency at which an atom or molecule absorbs radiation is also the frequency at which it will emit radiation. That is, however, affected by fine structure rotational mode quantum states and by Doppler broadening.

Doppler broadening of the absorption / emission spectrum is caused by two factors… translational motion and collisional frequency (which is determined by density (and thus partial pressure) and temperature).

If a molecule is moving in the same direction as the incoming radiation, that radiation is red-shifted in the reference frame of the molecule, thus a higher-frequency photon could be absorbed, broadening the absorption spectrum.

If a molecule is moving in the opposite direction to the incoming radiation, that radiation is blue-shifted in the reference frame of the molecule, thus a lower-frequency photon could be absorbed, broadening the absorption spectrum.

The Equipartition Theorem states that energy is shared equally amongst all energetically accessible degrees of freedom of a system. “Energetically accessible” being a proviso that the energy must be sufficient to occupy the quantized energy states of rotational or vibrational modes for it to be shared in those modes. At prevalent atmospheric temperatures, collisional energy imparted to the CO2 molecule is sufficient to occupy rotational mode quantum states, whereas it is generally insufficient to occupy vibrational mode quantum states (typically ~3.6% of CO2 molecules are vibrationally excited to the lowest (v1) vibrational mode quantum state, and 1.024% of CO2 molecules are vibrationally excited to the highest (v3) vibrational mode quantum state via molecular collision at prevalent tropospheric temperature).

According to the Equipartition Theorem, molecular collisions maintain an equilibrium distribution of molecules in higher rotational mode quantum states and vibrational mode quantum states. As stated above, at prevalent atmospheric temperature, ~3.6% of CO2 molecules are vibrationally excited to the lowest (v1) vibrational mode quantum state, and 1.024% of CO2 molecules are vibrationally excited to the highest (v3) vibrational mode quantum state via molecular collision, whereas nearly all CO2 molecules are in excited rotational mode quantum states. The radiative lifetime of the vibrational mode quantum states is on the order of 200 milliseconds due to the small matrix elements for transition moments (which scales as the cube of the wavelength, or inversely as the square of the electric dipole matrix element), whereas the mean time between collisions is on the order of only 3.45 ns.

The momentum of photons in the atmosphere is insufficient to directly affect translational kinetic energy of the atmospheric molecules absorbing them, and thus cannot directly raise atmospheric gas temperature. That energy goes into changing the vibrational mode and rotational mode quantum states of the molecules.

The amount of IR radiation the planet’s surface receives from the atmosphere and the amount of IR radiation the planet’s surface emits are not closely correlated. The amount of IR the planet’s surface receives from the atmosphere is almost entirely dependent upon atmospheric molecular composition and temperature. The two are loosely coupled in that the IR emission from the planet’s surface affects air temperature, which in turn affects IR emission from the atmosphere, but the linkage is weak, other factors such as convective heat transfer affect air temperature much more.

Only a molecule’s vibrational and rotational modes can interact with infrared radiation, and this interaction can take place only if the molecule exhibits a non-zero magnetic dipole, which implies asymmetry in the distribution of electronic charge. For CO2, this non-zero magnetic dipole occurs when the alignment of the atoms comprising the CO2 molecule is distorted in the asymmetric stretch (v3) mode and bending (v2) modes.

All singular photons are circularly polarized either parallel or antiparallel to their direction of motion. This is a feature of their being massless, which precludes their exhibiting the third state expected of a spin-1 particle. A macroscopic electromagnetic wave is the tensor product of many singular photons, and thus may be linearly polarized if all singular photons comprising the macroscopic electromagnetic wave are not circularly polarized in the same direction.

When an atom or molecule emits a photon, there is a probability distribution of emission direction, with maximum emission probability perpendicular to the dipole moment and zero emission probability parallel to the dipole moment.

Similarly, when an atom or molecule absorbs a photon, there is a probability distribution of absorption direction, with maximum absorption probability perpendicular to the dipole moment and zero absorption probability parallel to the dipole moment.

While the angular momentum selection rules in the immediately-above two paragraphs don’t imply any overall direction of emission of photons from a molecule, they do imply that the emitted photon will be either roughly parallel or antiparallel to the vector of the absorbed photon, disregarding molecule rotation and dependent upon photon polarization direction.

Blackbodies absorb and emit at a wide range of wavelengths exemplified by the Planck curve, whereas a gas absorbs and emits at specific spectral lines. As a first approximation, atmospheric gases cannot and do not emit as a blackbody. The dominant source of blackbody radiation is transient oscillating dipoles induced by inter-molecular thermal vibrations within a material. Solids, liquids, plasma of sufficient density and gases of sufficient density can emit blackbody radiation, but our atmosphere cannot simply because the gas molecules spend the majority of their time relatively distant from each other, and thus they cannot sustain the inter-molecular oscillations necessary for blackbody radiation. As gas density increases, blackbody radiation production increases and eventually dominates the discrete emission spectra.

A body can absorb radiation from another body which is at a higher temperature but it can only emit radiation at or below its own temperature. Thus the temperature of outgoing radiation can only be less than or equal to the temperature of incoming radiation. This means that energy cannot flow from a cooler to a warmer body simply because any radiation from the cooler body incident upon the warmer body will experience a temperature already higher than that radiation’s temperature, thus that radiation will be reflected, not absorbed. Since gasses cannot really reflect radiation, for a gas the radiation would be scattered.

Blackbody radiative flux scales as the fourth-power with temperature (ie: for a doubling of temperature, radiative flux increases 16 times). So any heating at the surface of our planet is quickly damped by increased radiative flux, the main reason our planet’s temperature has been relatively stable for billions of years.

An atmospheric greenhouse gas enables a planet’s atmosphere to radiate at a temperature lower than the surface temperature if there is cold air aloft. All other factors held constant, this would cause the surface temperature in balance with a given amount of absorbed solar radiation to be higher than would be the case if the atmosphere were transparent to IR. Of course, all other factors are not held constant… radiative flux and convection in particular increase. This is why Clough and Iacono’s study (linked below) showed CO2 to exhibit a spectral cooling rate at all altitudes from surface to thermosphere, except for a slight warming at the tropopause.

The only way our planet can shed heat is via radiative transmission of energy to space. N2 and O2 (the two largest constituents of the atmosphere) are homonuclear diatomic molecules, so they have no net magnetic dipole and thus cannot effectively emit (nor absorb) infrared, and thus cannot shed energy to space. CO2 has been shown to cool the troposphere, the stratosphere, the mesosphere and the thermosphere (studies below). Indeed, CO2 is the largest contributor to upper atmosphere cooling. Without CO2, the planet would heat up due to a reduced ability to radiate energy to space. If the upper atmosphere cannot cool via CO2 radiative emission to space, the air below would not be able to convect upward, thus quickly warming the entirety of the atmosphere. Thus CO2 acts as a cooling gas, not a ‘heat-trapping’ gas. More CO2 will cause more radiative emission to space, thus the upper atmosphere will cool more, thus tropospheric air can convect more easily, thus cooling the troposphere more effectively. Indeed, this is exactly what is observed (studies below).

The Rule of Mutual Exclusion states that for molecules like carbon dioxide, which are centrosymmetric, quantum state vibrational modes that are IR-active are Raman-inactive, and vice versa. Thus carbon dioxide has 1 Raman band and two IR bands.

The total energy of a molecule equates to the sum of the translational kinetic energy, electronic mode quantum state energy, vibrational mode quantum state energy and rotational mode quantum state energy of the molecule.

The molecule’s translational kinetic energy is non-quantized. It is represented by what we classically know as temperature.

The molecule’s electronic mode quantum state energy is quantized. It is represented by the equation E = h^2 / m * a^2. Typically, visible range and ultraviolet photons are necessary to excite this quantum state.

The molecule’s vibrational mode quantum state energy is quantized. It is represented by the equation E = h^2 / a^2 * SqRt(mM). Typically, thermal and near infrared photons are necessary to excite this quantum state.

The molecule’s rotational mode quantum state energy is quantized. It is represented by the equation E = h^2 / M * a^2. Typically, microwave and far infrared photons are necessary to excite this quantum state.
——————–

Particle Physics Of CO2
——————–
The CO2 molecule (a triatomic linearly symmetric molecule with an axis of symmetry along the nuclei and a plane of symmetry perpendicular to this axis) has two rotational mode quantum states, and four fundamental vibrational mode quantum states at 3 radiation wavelength bands centered on:
———-
– 4.25695 micron (v3; 2349.1 cm-1 wavenumber) { v20(0) -> v3(1) }
Asymmetric stretch mode, this mode is very IR-active, but the dipole moment oscillates parallel to the molecule’s symmetric axis, and therefore Delta_J = 0 Q-branch transition is forbidden, making this very narrow-band. The radiance at this narrow frequency band is also minimal, falling at the minima between the Planck curves of solar (incoming) and terrestrial (outgoing) radiation. As discussed below, however, this vibrational mode quantum state is the main route for transfer of energy from vibrationally-excited N2 to CO2.

– 7.20357 micron (v1; 1388.2 cm–1 wavenumber) { v20(0) -> v1(1) }
Symmetric stretch mode; this mode is IR-inactive, it cannot absorb radiation since the molecule has no change in net magnetic dipole moment unless the molecule is perturbed via collision at the same time that it absorbs a photon.

– 14.98352 micron (v2; 667.4 cm-1 wavenumber) { v20(0) -> v21(1) }
2 degenerate bending modes.

There is also a narrow absorption band centered on 2.7 micron, but it is swamped by the v3 (asymmetric stretch) fundamental of H2O centered at 3490 cm-1, the v1 (symmetric stretch) fundamental of H2O centered at 3280 cm-1 and the {v2 + v3} band of H2O centered at ~5000 cm-1, and thus has little radiance available to it except in extremely low humidity locales:
– 2.76785 micron (v20(0) -> v22(2) + v3(1); 3612.91 cm-1 wavenumber)

– 2.69209 micron (v20(0) -> v1(1) + v3(1); 3714.59 cm-1 wavenumber)

…the only one of those vibrational modes which has any appreciable radiance available to it, and which is IR-active is 14.98352 micron, and hence this wavelength band is the largest contributor to CO2 quantum energy states from IR absorption.

Note that the lowest excited vibrational mode quantum state of N2 has a (very) slightly lower energy level than the highest vibrational mode quantum state of CO2. This energy is more than made up by translational kinetic energy during collision, as discussed below. Were this not so, CO2 lasers could not work.
Reference: Journal of Geophysical Research: Atmospheres; Vibrational-vibrational and vibrational-thermal energy transfers of CO2 with N2 from MIPAS high-resolution limb spectra, Jurado-Navarro et al.; p. 8008

The Interaction of O3, N2 and CO2:

Satellites see CO2 and (a bit of) water vapor radiating at the temperature of the lower stratosphere all over the planet. This is because ozone (O3, excited by incoming solar radiation) and collisional processes excite nitrogen (N2) to its v1 (symmetric stretch) vibrational mode, and N2 then transfers energy to the v3 (asymmetric stretch) mode of CO2 via collision as shown in the image, whereupon the vibrationally excited CO2 partially de-excites by dropping from the v3 (asymmetric stretch) mode to either the v1 (symmetric stretch) mode by emitting a 10.6 micron photon, or to the v2 (bending) mode by emitting a 9.6 micron photon.

This is the same method by which a CO2 laser works… the laser filling gas within the discharge tube consists of around 10–20% carbon dioxide (CO2), around 10–20% nitrogen (N2), and a few percent hydrogen (H2) and/or xenon (Xe), and the remainder helium (He). Electron impact vibrationally excites the N2 to its first vibrational mode quantum state {v1(1)}, the N2 collides with CO2, the CO2 becomes excited in the asymmetric stretch vibrational mode quantum state {v3(1)}, and de-excites to its v1 or v2 vibrational modes by emission of 9.6 micron or 10.6 micron radiation as described above. The helium is used to fully de-excite the N2 to the ground state after it’s collided with CO2 to maintain population inversion, but this is unimportant to the process of energy transfer from vibrationally excited N2 to CO2 in the atmosphere. The process by which the N2 becomes vibrationally excited (in the case of a CO2 laser via electron impact; in the atmosphere via translational-to-vibrational collisional processes and via vibrational-to-vibrational collisional processes with solar-excited O3) is similarly unimportant… the concept of energy flowing from N2 to CO2 is the same. Laser wavelength can be tuned by altering the carbon and oxygen isotopes comprising the CO2 molecules in the discharge tube. Similarly, in the atmosphere, we can determine the ratio of anthropogenic CO2 (which consists primarily of 13C) and natural CO2 (which consists primarily of 12C) via spectral analysis.

You’ll note the immediately-above two paragraphs describe the energy flow from vibrationally-excited N2 to CO2, which then emits at either 9.6 microns or 10.6 microns, both of which are in the Atmospheric Infrared Window. Thus this radiation has a nearly unfettered path out to space.
———-

Remember, N2 is a homonuclear diatomic, thus it has no net magnetic dipole, and thus it cannot emit (nor absorb) infrared radiation. So once it is vibrationally excited, its vibrational states are metastable and relatively long-lived… much longer-lived than the average emission time of CO2. Thus CO2 will emit radiation and break Local Thermodynamic Equilibrium, whereupon there are more N2 molecules waiting to excite it again. Thus energy must flow from N2 to CO2, just as it does in a CO2 laser.

The Equipartition Theorem states that energy is shared equally amongst all energetically accessible degrees of freedom of a system… in this case it means that just as much energy flows from CO2 vibrational mode quantum states to O2 or N2 molecule translational energy as flows from O2 or N2 molecule translational energy to CO2 vibrational mode quantum states (barring violation of Local Thermodynamic Equilibrium conditions via radiative emission… and in this instance, CO2 is the only molecule which can radiatively emit, so net energy must flow from N2 to CO2)… in other words, at best it’s a wash and therefore CO2 cannot cause atmospheric warming… which is why, despite years of looking, scientists haven’t been able to find the mid-tropospheric ‘hotspot’ their models predicted.

They thought they had found it, but it was satellite drift error:

Santer and Sherwood completely ignored the millions of weather balloons with direct-recording thermometers, and attempted to use wind speed as a proxy for temperature, then reported that they’d ‘found’ the missing hotspot hidden in the temperature noise… they were soundly rebutted and laughed at for their desperate and transparent ploy.
http://joannenova.com.au/2008/10/the-missing-hotspot/

On the left is the data collected by millions of weather balloons. On the right is what the climate models say was happening. The theory (as per the climate models) is incompatible with the observations. In both diagrams the horizontal axis shows latitude, and the right vertical axis shows height in kilometers. Image from Dr. David Evans

Indeed, as CO2 concentration increased, OLR (Outgoing Longwave Radiation) increased. The computer climate models predicted it would decrease, thereby heating the atmosphere (their claim that CO2 ‘traps’ heat in the atmosphere).

Further, the very graphic the climate alarmists use to claim that CO2 causes warming proves the exact opposite:
———-
Brightness Temperature vs. Wavenumber graphic

Brightness temperature is the temperature a black body in thermal equilibrium with its surroundings would have to be to duplicate the observed intensity of a grey body object at a frequency v (nu) .

A brightness temperature lower than the equivalent blackbody temperature implies that energy is flowing from that equivalent blackbody temperature to the matter with that brightness temperature, which is shedding energy via radiative emission.

Matter emits according to the law: B=Sigma*T^4
Sigma is the Stephan-Boltzmann constant, B is the brightness and T is the absolute temperature

Therefore a lower Brightness Temperature implies an absolute temperature below the temperature of the environment from which CO2 is attaining its energy. In this case, it means CO2 is radiatively cooling the atmosphere, as explicated in the studies below.
———-

The average kinetic energy of CO2 molecules at prevalent atmospheric temperature (288 K) is given by:
KE_avg = [1/2 mv^2] = 3/2 kT
… which gives an average kinetic energy of 0.03722663 eV and a mean CO2 molecular translational speed of 372.227941 m/s. This thermal energy is equivalent to the energy of a 33.3283159 micron photon. You’ll note the thermal energy is LESS THAN the energy necessary to excite a CO2 molecule’s vibrational mode quantum states. So one would simplistically assume that the opposite applies, that the vibrational mode quantum state energy of CO2 is greater than the translational energy of N2 or O2 molecules (which would be approximately the same as calculated above, due to the Equipartition Theorem) and therefore a photon-excited CO2 molecule will de-excite via a thermalizing collision with N2 or O2, thereby raising atmospheric temperature… except that assumes N2 and O2 are in their ground vibrational mode quantum states; it neglects the energy in vibrational mode quantum states of N2 and O2.

The wavenumber of any transition is related to its corresponding energy by the equation:
1 cm-1 = 11.9624 J mol-1
667.4 cm-1 = 667.4 * 11.9624 / 1000 = 7.9837 kJ mol-1
The Boltzmann Factor at 288 K has the value exp(-7983.7 / 288R) = 0.03607 which means that only 3.607% of the CO2 molecules are in the lowest state of the lowest vibrationally excited mode {ie: v21(1)}. These are the molecules that form the lower energy state for the next higher transitions which have an even lower population.

The v2 vibrational (bending) mode of CO2 in its lowest vibrational state {v21(1)} requires ~0.08279 eV, equivalent to an ~14.98576 micron photon (per VR Molecules Pro molecular modeler). The first vibrational mode quantum state of N2 has quantum energy of ~0.291929 eV, more than enough to activate CO2’s v2 (bending) vibrational mode quantum state. Thus, given that the Equipartition Theorem indicates that the thermal (kinetic, translational) energy of both molecules is similar, during a collision the vibrational mode quantum state energy of a vibrationally-excited N2 molecule will flow to the non-excited CO2 molecule, not the other way around.

1 cm-1 = 11.9624 J mol-1
2349.3 cm-1 = 2349.3 * 11.9624 / 1000 = 28.1032 kJ mol-1
The Boltzmann Factor at 288 K has the value exp(-28100.8 / 288R) = 0.010247 which means that only 1.024% of the CO2 molecules are in the lowest state of the highest vibrationally excited mode {ie: v3(1)}.

The v3 (asymmetric stretch) mode quantum state of CO2 requires ~0.291477 eV, equivalent to an ~4.256 micron photon. The lowest vibrational mode {v1} of N2 in its lowest energy state {v1(1)} has, surprisingly enough, nearly exactly the same energy as the highest vibrational mode {v3) lowest quantum state {v3(1)} of CO2, differing by only 4.3 cm-1 (equivalent to a very low-energy 2325.58 micron photon, or 0.0005335 eV; equivalent to a blackbody temperature of -271.9038 C or 1.2463 K, which is colder than the temperature of outer space at ~3.74 K) when accounting for N2 anharmonicity, centrifugal distortion and vibro-rotational interaction, which is more than covered by the distribution of translational kinetic energy as calculated above. So N2 exhibits strong collisional coupling to this vibrational mode quantum state of CO2, and during a collision the vibrational mode quantum state energy of a vibrationally-excited N2 molecule will flow to the non-excited CO2 molecule, not the other way around (simply for the fact that CO2 can radiatively emit, whereas N2 cannot, so the net energy flow must be from N2 to CO2). If this were not so, CO2 lasers could not work.

This, of course, assumes that N2 in the atmosphere is vibrationally excited to at least its first vibrational mode quantum state. And a good percentage of it is…
———-
https://www.osapublishing.org/DirectPDFAccess/5CCF1401-BEE0-71DE-4128232482B99888_303623/oe-22-23-27833.pdf?da=1&id=303623&seq=0&mobile=no

Vibrationally Excited Molecules In Atmospheric Reactions
“It follows from the solar ultraviolet intensities quoted by Watanabe and Hinteregger that the production of N2* through Eq. 21 will be of the order of 10^10 cm-2 sec-1. Most of the N2* will be in low vibrational levels.”

{Comment: That’s 10,000,000,000 per square centimeter per second)
———-

We can again use the Boltzmann Factor to determine the vibrationally excited population of N2 due to collisional processes. While the N2 molecule is IR-inactive due to no change in magnetic dipole, it is Raman-active. Further, it can be vibrationally excited by a UV-excited O3 (ozone) molecule in the stratosphere, and via collision with another molecule with sufficient translational energy.
N2 v1(1) (stretch) mode at 2345 cm-1 (4.26439 micron), correcting for anharmonicity, centrifugal distortion and vibro-rotational interaction
1 cm-1 = 11.9624 J mol-1
2345 cm-1 = 2345 * 11.9624 / 1000 = 28.051828 kJ mol-1
The Boltzmann factor at 288 K has the value exp(-2805.1828 / 288R) = 0.102667 which means that 10.2667% of N2 molecules are in the N2 v1(1) excited state.

—–
“But wait!”, you may exclaim, “The energy differential in your calculations between N2 and CO2 isn’t 18 cm-1 as most graphics show!”

{Sigh} I just covered that above. Because I’ve accounted for N2 anharmonicity, centrifugal distortion and vibro-rotational interaction, this lowers the quantum state energy differential between N2 and CO2 to a mere 4.3 cm-1.

And you should be glad I’ve accounted for anharmonicity, centrifugal distortion and vibro-rotational interaction… without doing so, the N2 molecule total energy is 18 cm-1 higher energy than CO2. And that would destroy your CAGW argument. I’m giving your whacky hypothesis every chance I can, but it still fails.

Considering the total molecular energy E_tot:
CO2 {v3(1)}: E_tot = 1174.7 + 2349.3 = 3524 cm-1
N2 {v1(1)}: E_tot = 1176 + 2345 = 3521 cm-1
… gives a differential of only 3 cm-1. This is equivalent to a 3333.333 micron photon, or 0.0003722 eV with an equivalent Wien’s Displacement Law temperature of only 0.8694000008694001 K.

That’s why N2 {v1(1)} and CO2 {v3(1)} are so closely collisionally coupled, they are nearly perfectly resonant. Random thermal energy variations (ie: the speed distribution of molecules) completely overwhelms the energy differential. A mere 20.453 m/s speed differential (for CO2, moving at an average of 372.227941 m/s at 288 K) or 25.194 m/s (for N2, moving at an average of 458.5489 m/s at 288 K) is enough to overwhelm the energy differential. The Maxwell Speed Distribution Function gives a far wider speed distribution than that. Thus, since CO2 is the only molecule of the two which can emit radiation (and thus break Local Thermodynamic Equilibrium conditions), the net energy flow is FROM N2 TO CO2.
—–

When the molar mass of any gas is divided by the density of that gas at a pressure of 1 atmosphere and a temperature of 288 K, the value 23.633 L/mol is obtained. So when looking at any 23.633 liter volume of the atmosphere, there will be one mol of N2 and one mol of CO2, when assuming that CO2 is a well-mixed gas.

The mol of N2 in that 23.633 liter volume will contain 28.051 kJ of energy, whereas the mol of CO2 in its v2 mode quantum state will contain 7.98 kJ of energy.
Similarly, the mol of N2 in that 23.633 liter volume will contain 28.051 kJ of energy, whereas the mol of CO2 in its v3 mode quantum state will contain 28.1 kJ of energy.

Given that CO2 constitutes 0.041% of the atmosphere (410 ppm), and N2 constitutes 78.08% of the atmosphere (780800 ppm), this means that 14.7969 ppm of CO2 is excited to its v1 mode quantum state via collisional translational-to-vibrational (t-v) processes, whereas 80162.3936 ppm of N2 is excited via the same t-v processes. This is a ratio of 1 vibrationally excited CO2 to 5417 vibrationally excited N2. You’ll note this is 2.845 times higher than the total CO2:N2 ratio of 1:1904, and 195 times more excited N2 molecules than ALL CO2 molecules (vibrationally excited or not).

Similarly, 4.1984 ppm of CO2 is excited to its v3 mode quantum state via collisional translational-to-vibrational (t-v) processes, whereas 80162.3936 ppm of N2 is excited via the same (t-v) processes. This is a ratio of 1 vibrationally excited CO2 to 19093 vibrationally excited N2. You’ll note this is 10.028 times higher than the total CO2:N2 ratio of 1:1904, and 195 times more excited N2 molecules than ALL CO2 molecules (vibrationally excited or not).

N2 has a molar mass of 28.014 g / mol. Thus 4.518 grams of N2 in that 23.633 liter volume will be vibrationally excited in its v1 mode quantum state.

CO2 has a molar mass of 44.009 g / mol. Thus 1.587 grams of CO2 in that 23.633 liter volume will be vibrationally excited in its v2 mode quantum state, and 0.451 grams will be vibrationally excited in its v3 mode quantum state.

As you can see, the quantity and mass of excited N2 molecules in any given 23.633 liter volume of the atmosphere swamps the quantity and mass of excited CO2 molecules (indeed, it swamps the total number of CO2 molecules in that volume, excited or not). Hence, energy flows FROM N2 TO CO2.

Now granted, the ratio of excited CO2 molecules which will radiatively de-excite rather than collisionally de-excite is on the order of 1 in 100,000 at low altitudes… but consider how many CO2 molecules there are, and you’ll quickly conclude that this still gives a lot of radiative de-excitation, which is why we see CO2’s radiation signature in the atmosphere (at 9.6 micron, 10.6 micron and 15 micron). And as altitude increases, that probability increases until nearly all CO2 emits rather than undergoing collisional de-excitation. At the same time, the mean free path length of the emitted radiation (at the altitude at which CO2 radiative emission becomes prevalent over collisional de-excitation) becomes longer as altitude increases. This is why CO2 molecules collisionally-deexciting with ground-state vibrational mode N2 molecules cause the air at low altitudes to convect upward… they (barely, considering the Equipartition Theorem) heat the low-altitude air and (significantly) cool the high-altitude air. Indeed, without “greenhouse” gases, there would be very little convection! Without “greenhouse” gases, the only way IR-inactive gases (O2 and N2, for instance) can convect upward is if they gain heat via conduction by contacting the planet’s surface. So in fact, CO2 helps to *cool* the atmosphere via convection at low altitudes and via radiative emission at high altitudes… as we see in many studies below. Without CO2, our planet would heat up. The upper atmosphere could not effectively radiatively shed heat to space and would thus heat up, making the air below it unable to convect upward, which very little of it would want to do because it would only be able to pick up energy by directly contacting the surface of the planet.

We should see an increase in convective airflow as a signature of increasing CO2…
{{{ STILL UNDER RESEARCH }}}}
——————–

CO2 Cools The Troposphere, The Stratosphere, The Mesosphere And the Thermosphere
——————–
“Based on all the foregoing discussion, of the log-dependence of CO2 forcing (Myhre et al., GRL, 1998, vol. 25, doi: org/10.1029/98GLO1908) and its possible climate-cooling effect, I have a simpler hypothesis on the ineffectiveness of CO2 in warming the climate. I realize that this explanation is unacceptable to the IPCC and to many climate-warming advocates. I believe that the ‘gap’, now 40 years long, according to Christy, has existed throughout the Industrial Revolution — and probably during the whole of the Holocene. In other words, I consider that the ‘pause’ may be permanent.”

The Thermosphere Has Cooled:

The Stratosphere Has Cooled:

The graph shows multiple analyses of data from radiosondes that have measured stratospheric temperature for several decades. Graph adapted from Figure 2.7 in Bulletin of the American Meteorological Society, State of the Climate, 2011.

Cooling of Atmosphere Due to CO2 Emission
“Abstract: The writers investigated the effect of CO2 emission on the temperature of atmosphere. Computations based on the adiabatic theory of greenhouse effect show that increasing CO2 concentration in the atmosphere results in cooling rather than warming of the Earth’s atmosphere.”

How increasing CO2 leads to an increased negative greenhouse effect in Antarctica
https://agupubs.onlinelibrary.wiley.com/doi/full/10.1002/2015GL066749

Why CO2 cools the middle atmosphere – a consolidating model perspective
https://www.earth-syst-dynam.net/7/697/2016/esd-7-697-2016.pdf

Observations of infrared radiative cooling in the thermosphere on 2 daily to multiyear timescales from the TIMED/SABER instrument
https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20100011897.pdf
“Abstract:. We present observations of the infrared radiative cooling by carbon dioxide (CO2) and nitric oxide (NO) in Earth’s thermosphere.”

A Guide to CO2 and Stratospheric Cooling
https://climatephys.wordpress.com/2015/05/22/a-guide-to-co2-and-stratospheric-cooling/

Cooling of the mesosphere and lower thermosphere due to doubling of CO2
The sensitivity of the mesosphere and lower thermosphere (MLT) to doubling of CO2 has been studied. The thermal response in the MLT is mostly negative (cooling) and much stronger than in the lower atmosphere. An average cooling at the stratopause is about 14 K. It gradually decreases to approximately 8 K in the upper mesosphere and again increases to about 40–50 K in the thermosphere.

https://phys.org/news/2012-11-atmospheric-co2-space-junk.html
The enhanced cooling produced by the increasing CO2 should result in a more contracted thermosphere, where many satellites, including the International Space Station, operate. The contraction of the thermosphere will reduce atmospheric drag on satellites and may have adverse consequences for the already unstable orbital debris environment, because it will slow the rate at which debris burn up in the atmosphere.

Climate “Science” on Trial; Evidence Shows CO2 COOLS the Atmosphere
https://co2islife.wordpress.com/2017/01/29/climate-science-on-trial-evidence-shows-co2-cools-the-atmosphere/

Spectral Cooling Rates For the Mid-Latitude Summer Atmosphere Including Water Vapor, Carbon Dioxide and Ozone

Note the CO2-induced spectral cooling rate (positive numbers in the scale at right) extend right down to the surface of the planet, whereas CO2 shows just a slight bit of warming (negative numbers in the scale at right) only at the tropopause (ie: just above the clouds, where it absorbs a greater percentage of reflected solar insolation).
——————–

——————–
For an earlier take on this topic, refer to:
https://wattsupwiththat.com/2010/08/05/co2-heats-the-atmosphere-a-counter-view/

For a comprehensive treatise on heat transfer via infrared radiation in the atmosphere, refer to the standard reference on the subject, the 1942 Harvard Meteorological Studies No. 6 tome, entitled “Heat Transfer By Infrared Radiation In The Atmosphere” by Walter Maurice Elsasser.

Feynmen Lectures:
http://feynmanlectures.caltech.edu/III_18.html

A discussion of angular momentum selection rules:

Vibrational-vibrational and vibrational-thermal energy transfers of CO2 with N2 from MIPAS high-resolution limb spectra
https://agupubs.onlinelibrary.wiley.com/doi/epdf/10.1002/2015JD023429
Because of the strong collisional coupling of CO2 with N2(1) via V-V energy transfer, vibrational populations of N2(1) are also included in the non-LTE model. The SEE (Statistical Equilibrium Equation) for N2(1) is coupled into the SEE for the main CO2 isotope. The CO2 levels are connected by 695 radiative transitions, 39 of them considering full radiative transfer in the atmosphere. These 39 transitions involve vibrational states up to (4,1) for the main isotope; (3,0) and (2,1) for the 636 isotope; and (1,1) for the 628 and 627 isotopes.

Georgia State University – Department of Physics and Astronomy
HyperPhysics thermal energy, molecular speed and photon-equivalent energy calculations:
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/kintem.html#c4
http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/eqpar.html
http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html#c3

VR Molecules Pro molecular modeler:
https://profs-perso.teluq.ca/mcouture/www/Fodar99/Hzdemo_EN.htm
In the simulation, all vibration frequencies and atom equilibrium positions are “exact” (v.g. not calculated by a molecular modeling software but obtained from handbooks or public databases), while displacements of atoms in the various vibration modes are calculated using the simple (classical) models used in most of the literature.

Absorption Spectra of Water Vapor and Carbon Dioxide in the Region of 2 .7 Microns
https://nvlpubs.nist.gov/nistpubs/jres/46/jresv46n3p246_A1b.pdf

Maayke, Huisman, Stal, Matthijs – Colorful niches of phototrophic microorganisms shaped by vibrations of the water molecule.

Lidar-measured atmospheric N2 vibrational-rotational Raman spectra and consequent temperature retrieval
https://www.osapublishing.org/DirectPDFAccess/5CCF1401-BEE0-71DE-4128232482B99888_303623/oe-22-23-27833.pdf?da=1&id=303623&seq=0&mobile=no

A. Dalgarno, Department of Applied Mathematics, The Queen’s University of Belfast
Planetary Space Science, 1963, pp. 19 – 28
Vibrationally Excited Molecules In Atmospheric Reactions

Steven DeWitte, N. Clerbaus – Decadal Changes of Earth’s Outgoing Longwave Radiation
In this paper, decadal changes of the Outgoing Longwave Radiation (OLR) as measured by the Clouds and Earth’s Radiant Energy System from 2000 to 2018, the Earth Radiation Budget Experiment from 1985 to 1998, and the High-resolution Infrared Radiation Sounder from 1985 to 2018 are analysed. The OLR has been rising since 1985, and correlates well with the rising global temperature. An observational estimate of the derivative of the OLR with respect to temperature of 2.93 +/- 0.3 W/m^2 K is obtained.

https://www.sciencedirect.com/science/article/pii/B9780124095489103963
https://www.sciencedirect.com/referencework/9780128032213/comprehensive-remote-sensing
C. Crevoisier – Use of Hyperspectral Infrared Radiances to Infer Atmospheric Trace Gases
“Infrared channels are intrinsically sensitive to temperature; greenhouse gases, such as CO2, CH4, or N2O, have a significant but minor impact, difficult to separate from this dominant signal. For instance, a variation of 1% (~4 ppm) of CO2 only induces a variation of 0.1 K at 670 cm- 1. Moreover, the typical radiometric noise of the IASI channels is 0.25 K at this wavelength (more than twice the CO2 signal), and the channels are mostly sensitive to atmospheric temperature variations, with a moderate variation of 1 K of the atmospheric temperature profile inducing an 0.8 K variation of brightness temperature!”
——————–

Quotes:
——————–
“For present-day Earth, the only important continuum is the water vapor continuum in the window around 1000 cm-1. Carbon dioxide continua are unimportant for conditions which have prevailed on Earth during the past several billion years, but they are important for plugging the gaps in the line spectra for the dense CO2 atmospheres of Venus and early Mars. Diatomic homoatomic molecules like N2, which are transparent to IR in Earthlike conditions, have collisional continua that become important in cold, dense atmospheres. For example, the continuum makes N2 one of the most important greenhouse gases on Saturn’s largest moon, Titan.
– Infrared Radiation And Planetary Temperature, Raymond T. Pierrehumbert (Louis Block Professor in Geophysical Sciences at University of Chicago), Physics Today, January 2011, p. 35 (https://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf)
——————–

Joe Campbell
June 17, 2019 11:48 am

To “LOL@Klimate Katastrophe Kooks”: Whew! What a great bunch of info to digest. I’ll need to read it over, again, tonight so I can ask a question or two tomorrow. Thanks…

LOL@Klimate Katastrophe Kooks
June 17, 2019 9:28 pm

That’s not even taking into account the direct conversion of translational movement energy into radiation via collisional processes.

https://apps.dtic.mil/dtic/tr/fulltext/u2/725111.pdf
CO2 (000) + CO2 (000) CO2 (000) + C02 (010) – 667 cm-1 (33)
The absorbed energy in Reaction (33) once again comes from translation. Two reactions of type (33) must occur for every one of the type indicated by Reaction (32) to maintain the CO2 in thermal equilibrium. The removal of energy from the translational modes by Reactions (32) and (33) cools the CO2 molecular system, and, concomitantly, the air.

Or, to put it into modern scientific vernacular:
CO2 {v20(0)} + CO2 {v20(0)} CO2 {v20(0)} + CO2 {v21(1)} + 667 cm-1 photon

Or, in plain english:
Two CO2 molecules in their ground vibrational state collide. The translational energy of the collision excites one of the molecules to the v21(1) vibrational state, whereupon it emits a 14.98352 micron photon. This directly cools the atmosphere by removing translational mode energy.

Note that this takes nearly all the translational energy of both molecules, meaning those molecules are then very close to absolute zero, whereupon they absorb energy by colliding with other atmospheric constituents… it’s a significant effect.

LOL@Klimate Katastrophe Kooks
June 17, 2019 9:41 pm

Further:

The radiative cooling of air via solely translational mode energy converting to radiation:
CO2 {v20(0)} + CO2 {v20(0)} CO2 {v20(0)} + C02 {v21(1)} -> CO2 {v20(0)} + CO2 {v20(0)} + 667 cm-1

You’ll note this is a direct conversion of translational motion (which we perceive as temperature) to 14.98352 micron radiation. This directly cools the air, and the effect is significant, since nearly all the translational mode energy is converted to radiation, leaving the CO2 molecules at a very low temperature, whereupon they absorb energy
by colliding with other atmospheric constituents. The effect begins taking place significantly at 288 K, the temperature at which the majority of the CO2 molecules will have sufficient translational mode energy to convert to vibrational mode energy. 288 K also happens to be the stated average global temperature… that is not a coincidence, it is a mechanism long known.

As CO2 concentration increases, this effect will become more pronounced.

Joe Campbell
June 20, 2019 7:39 am

As noted in my response above (to your original post), I did reread your post – several times, not just overnight. I do have some history in this area having worked on CO2 gas dynamic lasers (GDLs) during the late 1970s and early 1980s while at UTRC (United Technologies Research Center). So, I can read and follow your discussion at some level (you lose me from time-to-time in the detailed molecular physics). But, CO2 GDLs really did work: We put out a “bunch” of photons from GDL devices operating both as amplifiers and unstable resonators, culminating, as far as my participation, in the Airborne Laser Laboratory (ALL) demonstrations. So, thanks. I remain baffled how the “CO2 causes global warming” meme continues to hold sway.

Joe Campbell
June 20, 2019 8:26 am

Has anyone done an analysis on the H2O molecule and its energy interactions with its fellow atmospheric partners?

June 20, 2019 10:35 am

>>
This means that energy cannot flow from a cooler to a warmer body simply because any radiation from the cooler body incident upon the warmer body will experience a temperature already higher than that radiation’s temperature, thus that radiation will be reflected . . . .
<<

The energy from the CMB is at 2.7K. That means (according to you) that any Earth-based antennas cannot receive those photons because of the temperature difference. However, Penzias and Wilson did manage to discover the CMB. Earth-based antennas easily detect the CMB. There’s no temperature assigned to photons.

Your black body and second law statements are not correct.

Jim

LOL@Klimate Katastrophe Kooks
June 20, 2019 5:28 pm

Further, while Penzias and Wilson did cool their receiver front-end (LNA, antenna probe) with liquid helium to ~4K as means of reducing surface plasmon interference and thereby increasing sensitivity, that ~4K is still warmer than the effective temperature of the CMB.

So how does an antenna pick up radiation with an effective temperature below the temperature of the material comprising the antenna? The antenna is resonant with that radiation.

Finally, as regards your “There’s no temperature assigned to photons” statement… then how, exactly, did we arrive at a CMB temperature of 2.72548 K, with nothing received except the *photons* emitted by the CMB?

Oh, that’d be via the Planck Law Of Blackbody Radiation, and Wien’s Displacement Law… as I’d stated in my original text.

At thermal equilibrium temperature T is defined by:
T=∂U/∂S
…where:
U = total energy
S = entropy

In physics, a photon gas is a gas-like collection of photons which has many of the same properties of a conventional gas… including pressure, *temperature*, and entropy.

Joe Campbell
June 21, 2019 8:17 am

I don’t know, Jim; not really my area of specialty. But, using Wien’s displacement law, you get the value of the maximum wavelength at 3.5K (my value of the background radiation temperature) of 828 micrometers. Using that value in the SB relationship (and if anything is a blackbody, space surely is) you can find that emissive power is 8.5X10-6 W/meter sq, over all directions and all wavelengths. If you use something like the GB radiotelescope, with its 2+ acres of surface are, you’ll find 70 millivolts at the detector, which should be readable…

Wrt the 2nd Law, it is what it is: Energy cannot be moved from a lower energy level to a higher one without work being done on the system…

June 22, 2019 5:28 am

Well, let’s go back to basics. The energy of a photon is described by the Planck-Einstein relation:

$\displaystyle E=h\cdot \nu$

where $\displaystyle h$ is Planck’s constant and $\displaystyle \nu$ (nu) is the frequency. We can also use angular frequency which is

$\displaystyle E=\hbar \cdot \omega$

where $\displaystyle \omega =2\cdot \pi \cdot \nu$ and $\displaystyle \hbar$ (h-bar) is Planck’s constant divided by $\displaystyle 2\cdot \pi$.

We also have the velocity of a wave which is equal to the product of the wavelength and the frequency or

$\displaystyle v=\lambda \cdot \nu$

In a vacuum, photons travel at the speed-of-light or $\displaystyle c$, and we have

$\displaystyle c=\lambda \cdot \nu$.

This gives us another equation for photon energy

$\displaystyle E=\frac{h\cdot c}{\lambda }$.

If we know a photon’s energy then we know its frequency-wavelength, and if we know a photon’s frequency-wavelength then we know its energy. But that’s all we know about a photon.

Just for grins, let’s compute the energy of a 10 micron photon. I get 1.9864e-20 joules. That is an incredibly small amount of energy.

I don’t know what Feynman said about black bodies, but Kirchhoff is the one who invented them. A black body absorbs all radiation that strikes it. At no time does a black body reflect some radiation–that would violate its definition. A black body absorbs at all frequencies-wavelengths, and a black body radiates at all frequencies-wavelengths.

The Sun can emit a 10 micron photon, the Earth can emit a 10 micron photon, you can emit a 10 micron photon, the red spot on Jupiter can emit a 10 micron photon, M-31 can emit a 10 micron photon, and even the CMB can emit a 10 micron photon. It’s impossible to tell the difference among those photons, where they came from, or determine what temperature produced them.

Black bodies radiate according to Planck’s law. There are two forms (among others) of that law:

$\displaystyle {{B}_{\nu }}(\nu ,T)=\frac{2\cdot h\cdot {{\nu }^{3}}}{{{c}^{2}}}\cdot \frac{1}{{{e}^{\frac{h\cdot \nu }{k\cdot T}}}-1}$

$\displaystyle {{B}_{\lambda }}(\lambda ,T)=\frac{2\cdot h\cdot {{c}^{2}}}{{{\lambda }^{5}}}\cdot \frac{1}{{{e}^{\frac{h\cdot c}{\lambda \cdot k\cdot T}}}-1}$

Notice we can factor out $\displaystyle h\cdot \nu$ from the first version and $\displaystyle {}^{h\cdot c}\!\!\diagup\!\!{}_{\lambda }\;$ from the second version. That’s just the energy of a single photon. The rest gives us the number of photons.

A 10 micron photon has a frequency of about 29.979 Terahertz. For a 288 K body radiating at that frequency, we get roughly 4.08e21 photons per second per square meter per steradians. There are $\displaystyle 4\cdot \pi$ steradians in a sphere, so that increases the number to about 5.13e22. It’s a lot of photons.

Now to determine the temperature, we need to find multiple points on one of those Planck curves. Trying to curve-fit an exponential function requires at least 3 or more points, which means we need photons at 3 or more frequencies-wavelengths, and billions upon billions of photons at each frequency-wavelength.

One photon is not going to tell you anything about temperature–PERIOD!!!!

Therefore, there’s no law about black bodies reflecting photons at a higher temperature–that’s just nonsense.

As for the second law–you aren’t even close. The law only applies to isolated systems, or systems and their surroundings (which is essentially an isolated system). In such a case, the entropy must always increase or equal zero if at equilibrium. Open and closed systems need not obey the second law. If a closed system obeyed the second law, then it could never cool down.

Jim

LOL@Klimate Katastrophe Kooks
June 22, 2019 6:56 am

Your disingenuous attempt at conflating a ‘blackbody’ with a ‘body’ (you’ll note I specifically stated that radiation from a cooler *body* incident upon a warmer *body* will be reflected, not absorbed) fails from the outset.

Of course a hypothetical perfect blackbody will absorb all radiation incident upon it… your task is to provide at least one example anywhere in the universe of a perfect blackbody which actually DOES absorb all radiation incident upon it. Not even black holes meet this requirement… they are considered the nearest to being a perfect blackbody, but not a perfect blackbody.

Likewise, your attempt at conflating a single photon’s energy with temperature fails. The Second Law is very clear on this, especially in the Clausius Statement sense… a photon with a lower energy than the surface upon which it impinges will not be absorbed, it will be reflected or scattered. Hence photons from a cooler body will not be absorbed by a warmer body. Your claim that photons of lower energy incident upon a body of higher energy will be absorbed violates conservation of momentum, for the simple fact that photons interact electromagnetically and hence experience inelastic scattering. A higher electromagnetic energy than the photon will inelastically scatter (reflect) that photon.

If the photons from a cooler body incident upon a warmer body were not inelastically scattered (reflected), they would experience a red-shift of photon energy of the amount E/mc^2. We don’t empirically observe this because you’re wrong.

LOL@Klimate Katastrophe Kooks
June 22, 2019 8:24 am

Now Jim will likely counter with something akin to:
“But photons are massless! There CAN’T be any “E/mc^2″ loss!”

The mass-energy equivalency principle:
E^2 = p^2 c^2 + m^2 c^4, Jim.
E=mc2
E=hf
E=pc

Thus, f = (mc2)/h

h is Planck’s constant, 6.626^-34 joule-second.

Einstein’s equation proves that invariant-mass matter and energy are fungible.

LOL@Klimate Katastrophe Kooks
June 22, 2019 8:54 am

Jim Masterson wrote:
“As for the second law–you aren’t even close. The law only applies to isolated systems, or systems and their surroundings (which is essentially an isolated system). In such a case, the entropy must always increase or equal zero if at equilibrium. Open and closed systems need not obey the second law. If a closed system obeyed the second law, then it could never cool down.”

Open and closed systems need not obey the second law only for the fact that *external* *energy* can do work upon the system… where’s the external energy coming from to do work such that (according to you) radiation of a lower energy is being absorbed by a body with a higher energy, Jim?

LOL@Klimate Katastrophe Kooks
June 22, 2019 9:29 am

Submitted edit for Jim’s approval:

The v3 (asymmetric stretch) mode quantum state of CO2 requires ~0.291477 eV to populate that state, equivalent to an ~4.256 micron photon. The lowest vibrational mode {v1} of N2 in its lowest energy state {v1(1)} has, surprisingly enough, nearly exactly the same energy as the highest vibrational mode {v3) lowest quantum state {v3(1)} of CO2, differing by only 4.3 cm-1 (equivalent to a very low-energy 2325.58 micron photon, or 0.0005335 eV; equivalent to the Planck curve peak of a blackbody temperature of -271.9038 C or 1.2463 K, which is colder than the temperature of outer space (the Cosmic Microwave Background, CMB) at ~2.725 K… so even the frigid temperature of outer space would be enough to overwhelm the CO2 / N2 energy differential) when accounting for N2 anharmonicity, centrifugal distortion and vibro-rotational interaction, which is more than covered by the distribution of translational kinetic energy as calculated above. So N2 exhibits strong collisional coupling to this vibrational mode quantum state of CO2, and during a collision the vibrational mode quantum state energy of a vibrationally-excited N2 molecule will flow to the non-excited CO2 molecule, not the other way around (simply for the fact that CO2 can radiatively emit, whereas N2 cannot, so the net energy flow must be from N2 to CO2). If this were not so, CO2 lasers could not work.

Happy now, or is there another nit you’d like to pick? :p

June 22, 2019 12:08 pm

So magical temperature properties of photons that aren’t even available to perfect absorbers/emitters like black bodies somehow appear because real bodies aren’t perfect? That’s nonsense of course. You still haven’t explained why a 10 micron photon (which is identical to every other 10 micron photon) knows what the temperature is of the body it’s leaving and knows what the temperature is of a body that might absorb it and then get reflected.

Also, if you place one or more black bodies in the confines of an isolated system (the only place where the second law applies) they will obey the second law too–no magical reflection required.

I repeat, your black body and second law statements are not correct.

Jim

LOL@Klimate Katastrophe Kooks
June 22, 2019 2:01 pm

I can’t figure out if you’re being purposely obtuse because you just want to argue, if you truly cannot grasp the concept, or if you’re trying to discount the 2nd law in the Clausius Statement sense because that destroys the whole ‘backradiation’ argument.

You know about the Planck curve and Wien Displacement Law, you know that blackbody photon emission is related to temperature, you know we can calculate the photon wavelength at the peak of the Planck curve from the temperature… yet you persist in trying to claim that I’m using “magical temperature properties of photons “, as you conflate concepts and misrepresent what I’ve written, in some sort of seemingly desperate attempt at disproving a long-known and well-proven physical law.

So, Jim… what exactly is your fundamental misunderstanding or major mental malfunction as regards this simple concept? ;p

LOL@Klimate Katastrophe Kooks
June 22, 2019 3:28 pm

Edit submitted for Jim’s approval:

A body can absorb radiation from another body which is at a higher temperature but it can only emit radiation at or below its own temperature. Thus the energy of outgoing radiation can only be less than or equal to the energy of incoming radiation. This means that energy cannot flow from a cooler to a warmer body simply because any radiation from the cooler body incident upon the warmer body will experience an energy already higher than that radiation’s energy, thus that radiation will be reflected, not absorbed. Since gasses cannot really reflect radiation, for a gas the radiation would be scattered.

Happy now, Jim?

Now, Jim… your task is to provide proof of your claim that a warmer body can absorb radiation from a cooler body. One real-world example will be sufficient.

June 23, 2019 1:02 am

I have two cases for you. In the first one, we place a black body inside a larger sphere that has a totally reflective inner surface. All the radiant heat from the black body is reflected back to the same black body. The temperature of the black body is $\displaystyle T$. The energy leaving the black body is $\displaystyle -Q$ and the energy entering the black body is $\displaystyle Q$. The change in entropy is:

$\displaystyle \Delta S=\frac{-Q}{T}+\frac{Q}{T}=0$

This isolated system satisfies the second law in two ways. First, the change in entropy is less than or equal to zero, and because the system is in equilibrium, the change in entropy is zero.

Now let’s do the second case. We place two black bodies inside our spherical reflector. To make it simple, both black bodies are of the same size. One black body I’ll call $\displaystyle A$, and the other I’ll call $\displaystyle B$. The temperature of black body $\displaystyle A$ is $\displaystyle {{T}_{A}}$, and the temperature of black body $\displaystyle B$ is $\displaystyle {{T}_{B}}$.

There are four energy flows to consider. All the energy that leaves black body $\displaystyle A$ and returns to $\displaystyle A$ I’ll call $\displaystyle {{Q}_{1}}$. Like in case one, the change in entropy of $\displaystyle {{Q}_{1}}$ is zero.

All the energy that leaves black body $\displaystyle B$ and returns to $\displaystyle B$ I’ll call $\displaystyle {{Q}_{2}}$. Again, the change in entropy of this flow is zero.

The energy flow from $\displaystyle A$ to $\displaystyle B$ I’ll call $\displaystyle {{Q}_{A}}$, and the energy flow from $\displaystyle B$ to $\displaystyle A$ I’ll call $\displaystyle {{Q}_{B}}$.

Notice if $\displaystyle {{T}_{A}}>{{T}_{B}}$ then $\displaystyle {{Q}_{A}}>{{Q}_{B}}$. Now let’s compute the various entropies. I’m still following the convention that heat leaving a system is negative and heat entering a system is positive.

$\displaystyle \Delta {{S}_{A}}=\frac{{{Q}_{B}}-{{Q}_{A}}}{{{T}_{A}}}$

Because $\displaystyle {{Q}_{A}}>{{Q}_{B}}$, $\displaystyle \Delta {{S}_{A}}$ is negative. However, black body $\displaystyle A$ is a closed system and it need not obey the second law. And now for $\displaystyle \Delta {{S}_{B}}$

$\displaystyle \Delta {{S}_{B}}=\frac{{{Q}_{A}}-{{Q}_{B}}}{{{T}_{B}}}$

The entropy of black body $\displaystyle B$ is positive.

Now we compute the total entropy of the isolated system.

$\displaystyle \Delta S=\Delta {{S}_{A}}+\Delta {{S}_{B}}=\frac{{{Q}_{B}}-{{Q}_{A}}}{{{T}_{A}}}+\frac{{{Q}_{A}}-{{Q}_{B}}}{{{T}_{B}}}$

$\displaystyle =\frac{{{T}_{B}}\cdot {{Q}_{B}}-{{T}_{B}}\cdot {{Q}_{A}}+{{T}_{A}}\cdot {{Q}_{A}}-{{T}_{A}}\cdot {{Q}_{B}}}{{{T}_{A}}\cdot {{T}_{B}}}$

$\displaystyle =\frac{{{Q}_{B}}\cdot ({{T}_{B}}-{{T}_{A}})+{{Q}_{A}}\cdot ({{T}_{A}}-{{T}_{B}})}{{{T}_{A}}\cdot {{T}_{B}}}$

$\displaystyle =\frac{-{{Q}_{B}}\cdot ({{T}_{A}}-{{T}_{B}})+{{Q}_{A}}\cdot ({{T}_{A}}-{{T}_{B}})}{{{T}_{A}}\cdot {{T}_{B}}}$

$\displaystyle =\frac{({{T}_{A}}-{{T}_{B}})\cdot ({{Q}_{A}}-{{Q}_{B}})}{{{T}_{A}}\cdot {{T}_{B}}}$

Because $\displaystyle {{T}_{A}}>{{T}_{B}}$, the first quantity is positive, and because $\displaystyle {{Q}_{A}}>{{Q}_{B}}$, the second quantity is positive. That makes the total change in entropy $\displaystyle \Delta S>0$.

Therefore, this isolated system obeys the second law. For some reason, you think that $\displaystyle {{Q}_{B}}$ violates the second law, but it doesn’t. Energy flow $\displaystyle {{Q}_{B}}$ must exist, because both $\displaystyle A$ and $\displaystyle B$ are black bodies. Because $\displaystyle A$ is hotter than $\displaystyle B$, more energy flows from $\displaystyle A$ to $\displaystyle B$ than flows from $\displaystyle B$ to $\displaystyle A$. This will cause $\displaystyle A$ to cool and $\displaystyle B$ to warm. Eventually, $\displaystyle {{T}_{A}}={{T}_{B}}$ and $\displaystyle {{Q}_{A}}={{Q}_{B}}$ and at that time this system will reach equilibrium. Then $\displaystyle \Delta S=0$, which is what the second law requires at equilibrium.

>>
Now, Jim… your task is to provide proof of your claim that a warmer body can absorb radiation from a cooler body. One real-world example will be sufficient.
<<

Easy, the Earth radiates in all directions. Some of those photons must travel to the Sun. There is nothing preventing the Sun from absorbing those photons–it doesn’t violate the second law.

Jim

June 23, 2019 1:07 am

>>
First, the change in entropy is less than or equal to zero,
<<

That statement should read: the change in entropy is greater than or equal to zero.

Jim

LOL@Klimate Katastrophe Kooks
June 23, 2019 7:44 pm

Further, the sun is composed of plasma. No radiation emitted by Earth will be absorbed by the sun, as the sun’s plasma has far too high an energetic barrier for the photon to be absorbed. This is the same reason low-frequency radio signals are *reflected* off the plasma in the ionosphere, Jim.

For a photon frequency below the plasma frequency, we get reflection. Hence the incident photon energy would have to be *higher* than the energy at the optical surface of the sun in order for it to be absorbed. And there are no photons emitted from the planet with that high an energy.

So your example of a cooler body’s photons being absorbed by a warmer body fails.

{dances victory dance, twirls, bows}

LOL@Klimate Katastrophe Kooks
June 23, 2019 9:41 pm

Right about now, Jim’s likely furiously Googling “plasma frequency”, and learning that for a photon frequency below the plasma frequency, we don’t get “just” reflection, we get *total* *reflection*. He was likely hoping that at least some of the photons were absorbed by the plasma, but alas, physics isn’t Jim’s friend.

Thanks for providing the clearest example of what I’m talking about, Jim. LOL

Clyde Spencer
June 24, 2019 11:24 am

Jim Masterson
You said, “Therefore, there’s no law about black bodies reflecting photons at a higher temperature–that’s just nonsense.” A black body is a theoretical construct that is convenient to explain many things. However, real objects are going to be grey bodies. Real objects will exhibit a dispersion with wavelength of the complex refractive index. That is, some photons will be absorbed strongly, others will be reflected. That explains how photons impinging on a real object will behave differently depending on their energy.

LOL@Klimate Katastrophe Kooks
June 23, 2019 6:24 pm

You’re going on about hypothetical blackbodies and perfect reflectors, Jim, while I’m operating in the real world of quantum physics.

The net energy flow is always going to be from warmer to cooler because photons interact electromagnetically, a higher-energy body has higher electromagnetic potential, and thus will reflect photons from a lower-energy body incident upon that higher-energy body.

Looking at it from another angle, ‘backradiation’ cannot warm its own source, for the same reason. I’m sure even you would agree with that statement.

Looking at it in yet another way, heat doesn’t flow from your cold refrigerator to the surrounding (warmer) room.

It requires *WORK* to move energy to a higher potential, Jim. There is an energetic barrier which must be overcome. Just because open and closed systems can violate 2LoT (because *external energy* can do *work* upon the system, Jim) doesn’t free them to magically move energy to any potential without regard to that energetic barrier. Where is the *WORK* coming from for your lower-energy photons to be absorbed by a higher-energy surface, Jim?

If you consider light as a wave phenomenon, a lower frequency wave (from a colder body) produces destructive interference with a higher frequency / energy / temperature wave (of a warmer body), thus the lower frequency wave cannot increase the frequency / amplitude / energy / temperature of the warmer body.

If you consider light as a quantum phenomenon, a lower energy photon from a colder body cannot increase the energy of a higher energy / temperature body because all of the lower energy quantum states in the higher energy / warmer body are already filled. A lower-energy photon cannot raise a quantum state above its own energy level, thus that lower-energy photon has no effect on the quantum state / energy / temperature of the higher energy / temperature body.

In other words, there is an *energy* *barrier* which the lower-energy photon must overcome.

This ties into the Equipartition Theorem, which states that energy is shared equally amongst all energetically accessible degrees of freedom of a system. “Energetically accessible” being a proviso that the energy must be sufficient to occupy the quantized energy states of electronic, rotational or vibrational modes for it to be shared in those modes (you’ll note that molecules and atoms are open systems and thus can violate 2LoT… but they still must obey the laws of physics as regards energetic barriers). This implies that the energy must be sufficient to overcome any energetic barriers to populate those quantum states… and that correlates to photons of a lower energy incident upon a surface of higher energy being *not* “equally shared amongst all energetically accessible degrees of freedom of a system” due to that *energetic* *barrier*.

And therein lies your fundamental misunderstanding, Jim. You apparently believe that since open and closed systems don’t have to obey 2LoT (because *external* *energy* can do work upon the system, Jim), they also don’t have to obey energy barriers and thus any energy level can flow to any other energy level without any work being done on the system… your premise is wrong, therefore your conclusions are wrong, therefore *you* *are* *wrong*.

June 24, 2019 10:41 pm

Well LOL,

The problem stems back to your first, incredibly long comment that no one in their right mind would read in its entirety. I only scanned a few paragraphs. I repeat (for the third time), your statements about black bodies and the second law are not correct. I tried to enlighten you on why they are incorrect. I find it strange that you think I should drop discussing black bodies when it was you who first brought the subject up.

You also have a non-standard view of reflection physics. There are several reasons why photons reflect, almost none care about temperature.

There’s Snell’s law. If you look at a rectangular aquarium filled with water, you’ll notice that light travels from front to back easily, but the sides and the underneath surface of the water look like mirrors. Look in at a side, and now the front and back look like mirrors. It’s the same principle used by optical fibers. You have total internal reflection and temperature plays no role.

Look at a calm lake with a low Sun angle, and you’ll see the Sun’s reflection in the water. According to your idea, the water–being cooler than the Sun–should absorb those rays. Reflection and refraction are explained by Fresnel’s equations. Again temperature doesn’t appear in any of those equations.

Metals have a reflective property due to free electrons. Those free electrons are what give metals their luster. The act of reflection is more complex than just photons bouncing off. They are actually absorbed by the free electrons which then re-emit similar photons. That’s essentially what the ionosphere does to radio waves. That alone destroys your idea that reflection doesn’t involve absorption.

Reflection also occurs in systems where there is an impedance mismatch, (It’s similar to what happens when light travels from one index of refraction to a different one.) Signals traveling down a transmission line will get partially reflected at points of impedance mismatch. Sometimes impedance depend on the temperature, but it’s not the temperature that’s causing the reflection.

And then there’s the Sun. Radio astronomers have bounced microwaves (radar) off of Venus, Mercury, the Moon, Mars, asteroids, the moons of Jupiter, the rings of Saturn, and Saturn’s moon Titan. But when they tried to bounce them off the Sun, the signals came back weaker than expected. The strongest returns were from CMEs (coronal mass ejections) and almost none from coronal holes. If the Sun was as reflective as you say, then it should have easily reflected the radar back. The Earth radiates mostly in the infrared. If the Sun can absorb lower energy microwaves, then it should easily absorb higher energy infrared photons.

Jim

LOL@Klimate Katastrophe Kooks
June 24, 2019 11:51 pm

Sorry to ruin your day (and now your night), Jim, but you’re wrong yet again. Perhaps, if you’d actually read the entire original text, you’d understand… alas, you went off half-cocked and now you’re embarrassing yourself by promulgating a non-standard view of reflection physics due to your misinterpretation of the underlying actual physics.

The band gap of a solid and the wavelength of light that is reflected by it are related to one another via the Planck-Einstein Relation.

Energy bandgap in joules = h (Planck’s constant) * nu (frequency in Hz).

The energy bandgap of a solid tends to decrease as the temperature is increased. This behaviour can be better understood if one considers that the interatomic spacing increases when the amplitude of the atomic vibrations increases due to the increased thermal energy. An increased interatomic spacing decreases the potential seen by the electrons in the material, which in turn reduces the size of the energy bandgap

And as the bandgap decreases, so too must the frequency (per the Planck-Einstein Relation) and hence the energy of photons which that material will absorb.

And if that photon can’t be absorbed by the material, it must be reflected.

Reflection does not entail absorption of the photon, it is reflected. It is strictly an electromagnetic phenomenon.

For instance, let’s go back to your example… a plasma. When a photon below the plasma frequency encounters a plasma, the electrons in the plasma can move fast enough to cancel out any electric field caused by the photon. The light is perfectly reflected (which is not the same thing as being absorbed).

When the photon frequency is high enough that the electrons can not respond to the changes in the electric field then the plasma become transparent to that frequency.

In most materials, absorption occurs when an electron is excited to a higher energy state, or when a molecule/atom absorbs a photon and vibrates (photon / phonon interaction). In order for this to occur, there has to be an energy state available for the electron or phonon to reach after it absorbs the photon. The warmer body has no available energy states to accommodate that lower-energy photon, Jim.

{dances victory dance yet again, twirls, bows, catches roses}

Ain’t physics fun, Jim? LOL

LOL@Klimate Katastrophe Kooks
June 25, 2019 12:29 am

Jim Masterson wrote:
“And then there’s the Sun. Radio astronomers have bounced microwaves (radar) off of Venus, Mercury, the Moon, Mars, asteroids, the moons of Jupiter, the rings of Saturn, and Saturn’s moon Titan. But when they tried to bounce them off the Sun, the signals came back weaker than expected. The strongest returns were from CMEs (coronal mass ejections) and almost none from coronal holes. If the Sun was as reflective as you say, then it should have easily reflected the radar back. The Earth radiates mostly in the infrared. If the Sun can absorb lower energy microwaves, then it should easily absorb higher energy infrared photons.”

You likely got that from here:
https://forum.cosmoquest.org/archive/index.php/t-68862.html
A Proposal to Establish a Solar Radar at Arecibo – July 16, 2001 (http://www.lund.irf.se/helioshome/combined.pdf) (PDF)
———-
It is possible to probe the solar corona with radar at frequencies between 18 MHz and 50
MHz. The transmitted wave will be coherently reflected between 1.1 and 2.5 RS. In addition to providing the ability to probe a very important region, radar provides one of the few methods of observing coronal mass ejection’s (CME’s) directed Earthward.
Solar echoes were first detected at 26 MHz at Stanford in 1959, and an extensive series of
38 MHz measurements were made at El Campo in 1961-1969 [James, 1970]. These have
never been adequately explained: the echoes were much weaker than expected and highly variable; the doppler broadening was much larger than expected and also quite variable; anomalous ranges (≈ 0.4 RS); and anomalous doppler shifts (≈ 200 km/s) were observed. We have since learned a great deal about the corona and the solar wind, and the radar observations could undoubtedly be explained if we had complementary observations from Yohkoh, Soho, and Trace. It is likely that the echoes were weak because coronal holes reflect weakly, and the anomalous observations were probably due to CME’s. It would be very valuable to repeat the experiments with modern radar technology, and complementary optical, UV, and X-ray observations of the corona. A dual-polarization, tunable radar would provide estimates of the density and magnetic field, and their radial gradients in this very interesting region. Unfortunately no existing radar can do this work.
———-

Might that be where you got the idea that the microwave reflections were weaker than expected? Perhaps you just didn’t read that entire paragraph, especially the part that reads:

“The transmitted wave will be *coherently* *reflected* between 1.1 and 2.5 RS.” (emphasis mine)

…and the part that reads “It is likely that the echoes were weak because *coronal* *holes* *reflect* *weakly*, and the anomalous observations were probably due to CME’s.”. (emphasis mine)

Now, I’m not going to accuse you of being disingenuous in your attempt at conflating coronal holes with the optical surface of the sun, Jim, but it’s not looking good for you thus far. Perhaps you just misread (or, given your admission that you didn’t even read the entirety of my original post, skipped the majority of) that text, eh?

Perhaps you should spend more time reading, and less time pontificating and attempting to prove yourself ‘not wrong’, Jim. LOL

LOL@Klimate Katastrophe Kooks
June 25, 2019 12:44 am

This:
And as the bandgap decreases, so too must the frequency (per the Planck-Einstein Relation) and hence the energy of photons which that material will absorb.

And as the bandgap decreases, so too must the frequency (per the Planck-Einstein Relation) and hence the energy of photons which that material will reflect.

June 25, 2019 10:05 am

From “QED: The Strange Theory of Light and Matter” by Richard Feynman, page 76:

“I’ll show you how photons do nothing but go from one electron to another, and how reflection and transmission are really the result of an electron picking up a photon, “scratching its head,” so to speak, and emitting a new photon.”

I guess you need to do some more reading.

So I repeat (for the fourth time now), your statements about black bodies and the second law are not correct.

Jim

LOL@Klimate Katastrophe Kooks
June 25, 2019 11:19 pm

If you insist on being wrong, Jim, there’s not much I can teach you.

Photon interaction with electrons does not imply ‘absorption’, Jim. Absorption implies a change of energy state of the absorbing atom or molecule, whereas an electromagnetic interaction can be (relatively) long-range and doesn’t affect energy state. You’re conflating again.

Feynman was overly-simplifying the concept (he also says that light “goes in straight lines”, which is incorrect… it follows the path of least time, which in an n-dimensional Riemannian manifold is not necessarily a straight line; and he says that if light is ‘squeezed’ too much, it doesn’t obey the rules anymore (it always ‘follows the rules’, only the conditions have changed); and he says that light doesn’t always travel at c… it does in the medium known as ‘vacuum’, and in all other transparent mediums, except those other mediums increase the path length of the photon, reducing the apparent photon speed, which explains why electron frequency doesn’t change between mediums, but apparent wavelength does, without a change in photon energy (remember, E ~= 1.2398/wavelength… if the wavelength actually did change, so would the photon energy)), and you’re apparently talking about specular reflection of visible range photons.

We’re talking about reflection from a step in potential which the photon energy cannot overcome. This is the same reason *why* ‘backradiation’ cannot warm its own source, Jim. The photons are not absorbed, they are reflected.

Radiation is an oscillation with the equation Phi = A.exp (kx-wt) which, when incident upon an equipotential body or a body of greater potential, is reflected.

https://en.wikipedia.org/wiki/Rectangular_potential_barrier

Energy cannot magically flow from a lower potential to a higher potential, Jim. Again, where is the *WORK* coming from to force your lower-energy photons to be absorbed by a higher energy body?

https://quantummechanics.ucsd.edu/ph130a/130_notes/node149.html

There is a (very minute, exponentially decaying with increasing energy barrier height and physical barrier width) probability of quantum tunneling, but that’s not absorption, because the wave packet ‘borrows’ energy from its surroundings to tunnel (then pays that energy back afterward in accordance with the Heisenberg Uncertainty Principle), therefore it cannot increase the energy of the material through which it is tunneling.

LOL@Klimate Katastrophe Kooks
June 26, 2019 5:02 pm

An addition / edit to the above text. The first paragraph remains unchanged and is included only for positioning purposes of the added second paragraph.
———-
Indeed, as CO2 concentration increased, OLR (Outgoing Longwave Radiation) increased. The computer climate models predicted it would decrease, thereby heating the atmosphere (their claim that CO2 ‘traps’ heat in the atmosphere).

The two decade (and counting) “global warming pause” gave us the perfect opportunity to test the CAGW hypothesis as pertains to CO2 ‘trapping’ heat energy in the atmosphere… in the range of OLR measured by satellite, if CO2 was warming the planet there would be a reduction in OLR due to the climate alarmists’ purported ‘trapping’ of energy by CO2, but it would be more than offset by a generally increased brightness temperature at all wavelengths due to the increased global temperature. That OLR has increased as CO2 increased, while at the same time the global temperature has not risen, empirically nullifies the CAGW hypothesis… the result is exactly opposite to what the climate alarmists claim. Hence the entire CAGW ‘industry’ is build upon a nullus resultarum… making the entirety of the CAGW ‘industry’ an unscientific sham.
———-

LOL@Klimate Katastrophe Kooks
June 26, 2019 7:53 pm

An addition / edit to reflect further research into CAPE (Convective Available Potential Energy):
———-
We should see an increase in convective airflow (Convective Available Potential Energy, CAPE) as a signature of increasing CO2… and indeed, that is exactly what we have empirically observed the world over (studies below).
———-

Studies:
Trends of Convective Available Potential Energy over the Indian region and its effect on rainfall
https://rmets.onlinelibrary.wiley.com/doi/pdf/10.1002/joc.2359
“Here, we analyse daily radiosonde data of 32 stations over the Indian region from 1984 to 2008 (25 years), to study the climatology and long-term changes in CAPE, lower troposphere moisture content and upper tropospheric temperature. The increase in CAPE is found to be due to increase in low-level moisture content and decrease in upper level temperature over the last 25 years.”

Multidecadal trends in tropical convective available potential energy
“Time series of convective available potential energy (CAPE) calculated from 15 tropical radiosonde stations indicate mostly positive trends in CAPE during 1958-1997.”

Global climatology of Convective Available Potential Energy (CAPE) and Convective Inhibition (CIN) in ERA-40 reanalysis
https://www.sciencedirect.com/science/article/pii/S0169809508002706
“Significant trends in CAPE and CIN give the following results: (i) In general, a CAPE increase is noted during all seasons while, in particular, in autumn CIN shows a decrease over continents. (ii) Splitting of the time series reveals a sign change in trend commencing at the end of the 70s; this is observed in parts of the tropical continents and North America.”

LOL@Klimate Katastrophe Kooks
June 27, 2019 9:29 pm

An addition / edit. The last paragraph is added, the rest remains unchanged.

Further, the very graphic the climate alarmists use to claim that CO2 causes warming proves the exact opposite:
———-
Brightness Temperature vs. Wavenumber graphic

Brightness temperature is the temperature a black body in thermal equilibrium with its surroundings would have to be to duplicate the observed intensity of a grey body object at a frequency v (nu).

A brightness temperature lower than the equivalent blackbody temperature implies that energy is flowing from that equivalent blackbody temperature to the matter with that brightness temperature, which is shedding energy via radiative emission.

Matter emits according to the law: B=Sigma*T^4
Sigma is the Stephan-Boltzmann constant, B is the brightness and T is the absolute temperature

Therefore a lower Brightness Temperature implies an absolute temperature below the temperature of the environment from which CO2 is attaining its energy. In this case, it means CO2 is radiatively cooling the atmosphere, as explicated in the studies below.

https://imgur.com/fPkPALO
The image above shows the CO2 brightness temperature versus altitude. Note that as atmospheric temperature decreases and altitude increases, brightness temperature decreases. Thus, the decreased brightness temperature is due to the lapse rate of our atmosphere, or conversely the lapse rate of our atmosphere is due to the progressive radiative emission of energy to space with increasing altitude. Chicken and egg.
———-

LOL@Klimate Katastrophe Kooks
June 27, 2019 9:39 pm

And a further addition to the above:
———-
So while the alarmists claim that increasing CO2 concentration in our atmosphere has caused that ‘dip’ in brightness temperature to deepen because CO2 is ‘trapping’ more energy in the atmosphere, in reality the ‘dip’ has deepened because the upper atmosphere has experienced a long-term cooling trend (and ironically enough, that long-term cooling trend is because of the increased CO2 concentration emitting more radiation to space).
———-

June 28, 2019 9:37 pm

Now that LOL has stopped flooding the internet with more words than anyone can possibly read, let’s get back to the original argument. The following is pure LOL:

>>
LOL@Klimate Katastrophe Kooks
June 16, 2019 at 6:29 am

A body . . . can only emit radiation at or below its own temperature. Thus the temperature of outgoing radiation can only be less than or equal to the temperature of incoming radiation. This means that energy cannot flow from a cooler to a warmer body simply because any radiation from the cooler body incident upon the warmer body will experience a temperature already higher than that radiation’s temperature, thus that radiation will be reflected, not absorbed. . . .
<<

Notice the theme repeated in the paragraph is that radiation has a temperature. Now if LOL could stop dancing and patting himself (or herself) on the back, maybe LOL could answer one question for me. What’s the temperature of a 10 micron photon? There’s no need to cause another broadcast storm of words–just tell me the temperature (preferably in Kelvin), and how you determined it.

Thanks,
Jim

Stephen Wilde
June 15, 2019 11:43 am

“What this says in plain terms is that Willis’s graph implies that added irradiance gets divided between emitted power from the surface and heat transfer by other mechanisms in about equal amounts. Seems reasonable.”

If one varies that apportionment one way in sunlit regions and the equal and opposite way in unlit regions then you can get a greenhouse effect from conduction and convection alone as shown in the recent posts here by Philip Mulholland:

https://wattsupwiththat.com/2019/06/02/modelling-the-climate-of-noonworld-a-new-look-at-venus/

The logic is that conduction and convection delay radiative emission to space and thereby lead to a raised surface temperature.

LdB
June 15, 2019 9:34 pm

You can do the same trick in space, simply spin the object. For example done near the ISS the sun facing side wants to go to 250deg C the dark side wants to go to -250 degC. So place a ball there and spin it and inside the ball you get a temperature somewhere between the two. The exact temperature depends on the balance between how fast it absorbs to how fast it radiates.

June 18, 2019 12:56 am

Stephen in 2015 you said: “26C is the point at which the warmth of the ocean surface plus the greater buoyancy of humid air reliably defeats the weight of the mass of the atmosphere above so as to allow extensive convection” can you tell me how to find out more about this?
(You should be able to contact me via Scottish Sceptic website)

Joe Born(@jhborn)
June 15, 2019 11:55 am

Even without the parasitics, moreover, the response could be a very small fraction of the equilibrium response if relatively high-frequency components predominate in the input, deviation-from-seasonal-average signal.

Sweet Old Bob
June 15, 2019 11:57 am

Condensation ? Is that a possible reason for increased winter temperatures in the Arctic ?

Gary Pearse
June 15, 2019 4:58 pm

Sweet Old: increased temperatures in winter in the Arctic are from freezing the water thereby releasing latent heat, at least until the ocean is covered. In Winnipeg, Siberia, my father used to put a zinc tub of water in the cellar to keep the potatoes we harvested from freezing in the winter.

J Mac
June 15, 2019 12:40 pm

Kevin Kilty,
Very interesting! I understand your model and followed the assumptions, algebra and derivation. I had noted the 0.1 slope of Willis’s plot previously but hadn’t the wit to untie that knot into meaningful understanding. You have teased the first part of the knot apart and quantified cumulative heat transfer by various mechanisms (2)-(4) across the skin [h(2-4) is about 4.6W/m2K at 288K] but those contributor mechanisms remain confounded.

RichDo
June 15, 2019 1:40 pm

Thanks Kevin. In case anyone is interested here is a link to the Feynman lecture that you reference.
http://www.feynmanlectures.caltech.edu/II_34.html
That’s a great quote with meaning well beyond the realm of physics.

Alasdair
June 15, 2019 1:41 pm

Am I crazy?
Figure 1 shows a trend of 0.38 C for a 3.7 W/sq.m increase., giving s slope of 0.103 C/(W/sq.m). which ,to me, maybe described as a sensitivity factor.

The IPCC. uses the Planck energy equation dF = K*dT to convert the GHE of 1.6 W/sq.m to an equivalent temperature increase. The value of “K” being the sensitivity coefficient, of much debate.
If I divide the “dF” by the the “dT”. as per figure 1, I get a figure of 9.74 (W/sq.m)/C for “K” as a sensitivity factor. OOPS ! Where have I gone wrong? Or have I?.
Can anyone help?

ferd berple(@ferdberple)
June 15, 2019 3:19 pm

It is total irradiance. I will use the symbol (Ia) to represent it.
========
Kevin you might consider solar irradiance a constant and see if you can isolate back radiation on the left side of the equation.

Kevin kilty
June 16, 2019 6:16 am

The observed relationship of Willis’s data implies that for each net watt of new irradiance at the surface of the Earth, the surface temperature rises by about 0.1 degrees centigrade. Your calculation is correct. One interesting consequence of this is, the observed temperature rise is not even large enough to fully convey way the new energy–ergo there must other mechanisms at play.

Kevin kilty
June 16, 2019 6:43 am

I meant to say “convey away the energy with black body radiation alone.”

Javert Chip
June 15, 2019 2:14 pm

Observation:

We get more real science in a single WUWT blog post than from Mann and all his UN/Nobel buddies.

Hmmm…

Toto
June 15, 2019 2:14 pm

“One might also consider heat stored within the skin. However, on the sea surface the skin is something like 10 micrometers thick, and in soil it is about 5 centimeters thick. This represents very little thermal capacity to consider.”

I could mention boreholes but I’m not sure I want to go down that rabbit hole. Some gardener can reply.

June 15, 2019 2:19 pm

I think the characteristics of the “skin” needs a lot of qualification.
“Convective transfer through a film” – mostly.
“I think it is primarily away from the skin although there are times and places it goes the other way.”
Yes.
“in soil it is about 5 centimeters thick. This represents very little thermal capacity to consider.”
There are huge thermal capacities to consider. Try rock, or synthetic rock, eg concrete. Not a large proportion of the skin, but significant in the context of where measurements are made.
In the tropics rock, natural or synthetic, can have so much heat absorbed that it does not disappear overnight, and so it remains warm for up to five months.
I have been involved in assessments of the amount of turbulence that occurs at 0m to 3m above the skin (land) on hot days in the tropics, using smoke generators. At midday the turbulence at ground level is substantial, and only begins to stratify at 2-3m. One reason why it is more comfortable to be on an elevated floor rather than one on the ground.
In spite of this turbulence, there are also “flat” spots usually close to “bluff bodies” such as walls where there is very little turbulence. That’s where the mosquitoes tend to gather.

DMacKenzie
June 15, 2019 2:20 pm

An excellent analysis by Kevin Kilty. There was something about the 0.38 C per 3.7W/ sq.M on Willis’ graph that people interpreted as a climate sensitivity that didn’t quite feel right, and you clarified it.

June 15, 2019 2:44 pm

Well said.

ferd berple(@ferdberple)
June 15, 2019 6:34 pm

and you clarified it.
=====
Disagree. It seems to me more like the work is incomplete. That there may still be a solution if one splits the incoming radiation into solar and back radiation.

And the beauty of using a constant for solar radiation is that the slope of a constant is zero, so when you differentiate the solar portion should disappear and all you are left with is the back radiation. Which would appear to me to indicate that Kevin is only a couple of steps away from solving climate sensitivity from Willis’ diagram.

Which would be a pretty big deal after \$100 billion dollars has been spend over 35 years with no progress to show for it. I leave it to Kevin and Willis to (hopefully) provide the final steps. For my part I’m off to Tuktoyaktuk for the summer and will be mostly sans internet.

tty
June 15, 2019 2:49 pm

“One might also consider heat stored within the skin. However, on the sea surface the skin is something like 10 micrometers thick, and in soil it is about 5 centimeters thick. This represents very little thermal capacity to consider.”

This is about as wrong as it is possible to be. The solar irradiation is absorbed by the entire photic zone of the ocean which varies by the transparency of the water, but is up to 200 meters thick, though most of the energy is absorbed by the first several meters. This has a HUGE heat capacity that dominates the climate system. You are probably confused by the fact that while water is quite opaque to IR radiation it is fairly transparent to visible light (particularily blue light/near UV):

http://www1.lsbu.ac.uk/water/images/water_spectrum_2.gif

June 15, 2019 4:31 pm

Yes! One assessment of the effective thermal capacitance of the oceans is equivalent to 110 m of depth ttps://agupubs.onlinelibrary.wiley.com/doi/10.1029/2007JD008746

June 15, 2019 4:37 pm

“This is about as wrong as it is possible to be.”
It isn’t wrong, although 5cm is probably inconsistently large. He is specifying a control volume, and it is one that is responsive to the data. There is a provision (2) for heat transfer to the substrate, which includes that photic layer.

It’s true that a large flux of sunlight passes through the ocean skin layer without interaction, so the flux to substrate is not responsive to skin temperature. That is a gap in the analysis. But the point of the analysis is not to achieve closure, but to demonstrate that the flux Willis speaks of is not to be identified with the forcing from CO2.

June 16, 2019 2:37 pm

Of course. I was misled by the mention of thicknesses. This is a static analysis so effective thermal capacitance is not relevant.

ferd berple(@ferdberple)
June 15, 2019 3:23 pm

It is total irradiance. I will use the symbol (Ia) to represent it.
========
Kevin you might consider solar irradiance a constant and see if you can isolate back radiation on the left side of the equation.

ferd berple(@ferdberple)
June 15, 2019 3:30 pm

(1) Ia = σT4 + Q
======
Ia = Isol + Iback

Iback = σT4 + Q – Isol

Solve dT/dIback (climate sensitivity)

EdB(@sailboarder)
June 16, 2019 5:15 am

Such a solution is to ask that turbulence be solved. Actual measurements over time will work, but only if natural variations are known. We are a long way from an answer.

Kevin kilty
June 16, 2019 6:23 am

EdB is right here. The model is approximate, the tools inadequate and the complications are daunting. Heat transport is very complex, even in some simple situations.

ferd berple(@ferdberple)
June 15, 2019 5:00 pm

There is no way to take Willis’s diagram and determine how much warming a doubling of CO2 will produce.
=======
The x and y axis contain the values of interest, so a single failure doesn’t imply proof that all future attempts will fail.

Offhand, I think you are on track to a solution.

LdB
June 15, 2019 9:37 pm

Correct … more technically the Co2 absorption itself it subject to subtle QM effects and it may not even be valid to project it you really need to get there first.

Kurt
June 15, 2019 5:11 pm

“As a result I will linearize Q and write it as h·(T −Ts) where h is an over all transfer coefficient, and Ts is temperature of the environment surrounding the skin.”

How can you do this when the temperature of the substrate below the skin and the temperature of the air above the skin (what you are lumping together as Ts) not only have different temperatures relative to each other, but change at different rates in response to heat transfer from the skin? The ground is essentially a heat sink that is going to change extremely slowly, if at all, as heat flows downward from the surface. The air above the surface varies a lot based on evaporation, convection, conduction etc.

Kevin kilty
June 16, 2019 6:41 am

All I postulated, Kurt, is that there are other heat transfer mechanisms which over a small temperature range can be linearized. I was just try to explain the observations which contain an interesting paradox–note I said my model is tentative.

Willis Eschenbach(@weschenbach)
Editor
June 15, 2019 9:35 pm

Kevin, most fascinating, thanks. I’ll need a bit of time to go through and consider the various thoughts and equations, and I’m working in Alaska. However, tomorrow is Fathers Day and as one of the aforementioned, I’m gonna take it easy and I might be able to find the time.

My best to you, and thanks to all the interesting commenters.

w.

Bernard Lodge
June 15, 2019 10:28 pm

You have to be careful in linking temperatures to energy. Sometimes new energy raises the temperature and sometimes it does not. For instance, energy from a colder source will not raise the temperature of a warmer source, no matter how much energy comes from that colder source.

This fact seems to be fudged in the above analyses that seem to be based on ‘averages’.
As the poster says, the earth is in fact never in equilibrium, rather it is in some form of ‘steady state’ … his excuse for then using averages.

Some energy (from a warmer source) will raise temperatures while other energy (from a colder source) will not raise temperatures. It is binary … either yes or no. There is no such thing as averages with regard to the impact of energy on temperature. You cannot average yes and no. One needs to know the temperature of both the emitting body and the absorbing body to know if the temperature of the absorbing body will increase.

EdB(@sailboarder)
June 17, 2019 10:01 am

Consider the higher temperature object(earth) to be receiving input energy(sun), and the cooling process to be partially slowed via an insulator(air with water vapor).

The surface temperature would rise, it has to. The new equilibrium would be dependent on many factors, such as air density and H2O concentration.

Otherwise, why insulate your house, or use triple pane windows with argon gas?

Kevin kilty
June 16, 2019 6:52 am

Willis, enjoy your Father’s day. I admit to being a fan of your iris model, and I think about it often. In addition to closing an iris in the visible spectrum by presenting a shadow on the surface, it opens an iris in the infrared by transporting latent heat up to a height where there isn’t much “back radiation” Then in the surrounding air I’ve noticed thunderstorms around here clear the surrounding air and dry it out–thus opening simultaneously irises in both the visible and IR. Interesting to contemplate.

Geoff Sherrington(@sherro1)
June 16, 2019 5:27 am

If there was no gas other than O2 and N2 in the atmosphere of a hypothetical planet, would that planet be rather hot because it lacked radiative gases to convey the sun’s energy back to space?
Or, do O2 and N2 have adequate cooling ability to maintain some temperature at equilibrium? Geoff

Anthony Banton
June 16, 2019 9:57 am

In order for radiative gases “to convey the Sun’s energy back to space”
They must first absorb them.
That is the GHE.
Absorption then emittance higher in the atmosphere and therefore at a colder and less efficient temp.
Therefore an atmosphere without GHGs (just O2 and N2)would let LWIR emitted from the surface through unimpeded to space and so making such a planet colder.

EdB(@sailboarder)
June 17, 2019 9:53 am

How much change in GAT if CO2 is not present at all and only H2O and methane etc.? Plse show your work.

June 16, 2019 10:39 am

All matter above Absolute Zero radiates all the time in all directions. N2, O2, and Argon radiate to space. Increasing CO2 increases the altitude at which CO2 is no longer opaque to 15-micron radiation, thus lowering the temperature at which the Atmosphere radiates freely to space, thus lowering the amount of heat transferred to Space.

This is not simple at all. My Transport of Heat and Mass class at U of M was incredibly difficult. Final exam, One problem, two hour test, 180 kids including many National Merit Scholars such as myself, NO ONE finished it.

Steve Richards
June 16, 2019 6:21 am

It is interesting that everyone still uses the SB formula for a perfect blackbody.

Looking at a BB curve, and the radiation emitted from the earths surface, with windows due to CO2 and H2O, it appears that the alleged relationship between this radiation (upwelling) and a black body curve is completely broken.

Are we sure that the T^4 formula is robust enough to be used to calculate surface temperatures when it is plain to see that the earth does not emit like a blackbody.

The surface is warm, it does emit, it does not look like a blackbody curve. Lots of emissions missing for the formula to be applied here.

Yes, you can mention emissivity and 0.3 but is that truly valid for a non blackbody like earth?

tty
June 16, 2019 10:45 am

The earth is a pretty good approximation to a black body. The fact that the atmosphere is opaque to parts of this black-body radiation doesn’t change that.

Samuel C Cogar
June 16, 2019 7:11 am

Excerpted from Kevin Kilty’s above published article:

In a post on June 8, 2019, Willis Eschenbach showed an interesting plot of monthly average surface temperature against total irradiance absorbed at the ground surface ……………….. However, it did lead a number of commenters to suggest that it demonstrated a doubling of atmospheric CO2 would produce only 0.38°C to 0.5°C of surface warming.

I am going to provide a tentative explanation for the relationship. My tools will be some simple physics such as Newton‘s law of cooling or the Stefan-Boltzmann law, known to many who post here, and a block model of heat transfer at Earth’s surface.

Kevin Kilty, …. what the ell do you hope to prove via “a tentative explanation” ….. other than the fact you don’t have a clue.

Iffen you really wanted to prove or disprove the claim of ….. “a doubling of atmospheric CO2 will/won’t produce 0.38°C to 0.5°C of surface warming” ……. then do it via a physical experiment ….. instead of all the “fuzzy” math calculations using fictious number amounts that were generated via previously calculated “fuzzy” math results, hypotheticals, imaginary numbers and/or laboratory “closed system” environment testing.

Kevin K, why not use some simple clear plastic sheeting, plastic pipe and heat sensitive thermocouples with their conjoined “time n’ temperature” recorders ….. to determine the actual effect in degrees F or C caused by atmospheric CO2’s absorption/emission of LWIR?

Give up the ….. associations, correlations, insinuations, percentagations & imaginations ….. and do it scientifically correct.

Kevin kilty
June 16, 2019 8:05 am

With due regard to the beauty of good experiment, setting up a experiment of this sort would be about as difficult an undertaking as I could imagine. The radiative properties of gases depend upon path length, partial pressure of gasses, absolute pressure of all gasses in the mixture, temperature, and interference among the gasses. How in the world could I construct an analog of the T vs. P relationship of the real atmosphere in a lab scale model?

Samuel C Cogar
June 17, 2019 3:46 am

Kevin kilty – June 16, 2019 at 8:05 am

With due regard to the beauty of good experiment, setting up a experiment of this sort would be about as difficult an undertaking as I could imagine.

Kevin, I wasa waiting for you to say something like that, because, …. here is my written commentary from several years ago with some modifications, to wit:
==============

Why is it that everyone persists in hashing, re-hashing and re-re-re-RE-HASHING the same ole, same ole “Climate Sensitivity to atmospheric CO2” question?

Why is it that no one wants to perform a physical experiment to prove or disprove said “sensitivity” claim?

It wouldn’t take much money, maybe a couple thousand dollars, max.

Just build two (2) identical size frameworks, ……. out of 1/2″ white PVC plastic pipe, ……. with the dimensions of 20 x 10 x 8 feet square, ……. outside in an area where each will be subjected to the same environmental conditions (sunshine, darkness, rain, wind), ……. place temperature sensing thermocouples inside of them which are connected to an external located recording device, ……… then cover them “air tight” (top, bottom and sides) with 4 mill clear plastic sheeting, …. and then inject enough CO2 in one of the structures to increase its 400+- ppm of CO2 to say 800 ppm.

Then, when the night time temperatures in both stabilizes and reads the same, …….. say at 3 AM, start recording the temperatures in each structure …… and again record said temperatures every hour on the hour (or every half hour, or ten minutes) ……. for the next 24, 48 or whatever hours.

And if CO2 is the “global warming” gas that all the proponents of AGW claims it is, then when the Sun rises in the morning and starts shining on the structures, the temperature in the structure containing 800 ppm CO2 ……. should start increasing sooner and faster and reach a greater temperature than in the other structure ….. and when the Sun starts setting in the afternoon, the temperature inside the structure with 800 ppm CO2 should remain higher than it is in the other structure until later in the night.

And if it doesn’t, …… then the CO2 causing AGW claims are totally FUBAR … and the re-hashing of the “sensitivity” thingy should cease among learned individuals.
———————-

Kevin, even 2,000 ppm of CO2 in the one structure shouldn’t affect the inside temperature.

But no one will know unless you conduct said experiment.

Kevin kilty
June 16, 2019 7:46 am

A couple of things I forgot to mention in this post. First, Willis’s relationship pertains strictly to clear sky conditions, I think. My reason for thinking this is it depends on a skin temperature and skin emission which one can only derive when a satellite can see the surface. A different relationship must obtain in shaded conditions under clouds. NOAA uses a cloud mask to avoid mixing cloudy conditions into the data.

Second, I spent some time with the portal to a version of MODTRAN at the University of Chicago in order to look at how down welling IR changes with surface temperature. The results I got seem not fully realistic. I know that Dr. Happer has criticized current approaches to modeling of atmospheric transmission. Perhaps HITRAN would do a better job, but I know of no convenient portal to working with this data.

Willis Eschenbach(@weschenbach)
Editor
June 16, 2019 12:45 pm

OK, Kevin, it took me a bit of time, but I see a difficulty with your model.

The problem is that you assume that all of the energy leaving the surface is turned into radiation heading back to the surface. In your terms,

Ia = σT^4 + Q

Simple, clear, and differentiable. However, in the real world, a significant percentage of the energy leaving the surface ends up going out to space. Let’s call that “L”, because it is lost to the system. This means that your initial starting point, equation 1, should be

Ia = σT4 + Q – L

It gets worse. L is a function inter alia of the temperature and the weather. Clear skies, for example, allow more radiation to space. So we have.

L = f( T, W, a, b, …)

So now we have

Ia = σT^4 + Q – f( T, W, a, b, …)

But wait, as they say on TV, there’s more. Ia itself is temperature and weather dependent because when tropical temperatures increase, clouds cut down the incoming solar energy. Nor is this simple and linear, because in snow and ice country, warming has the opposite effect. As ice and snow melt, you get more solar absorbed. This is the opposite of the tropics.

Let’s call the solar component of Ia “Is” and we’ll retain “Ia” for the longwave component shown in your diagram. This gives us

Is = f( T, W, d, e, …)

So our starting point, our first equation, is not the simple and readily differentiable equation you give:

Ia = σT^4 + Q

Instead, to have even a passing similarity to the real world, it would have to be

Ia + f( T, W, d, e, …) = σT^4 + Q – f(T, W, a, b, …)

where both temperature-dependent functions are quite complex, have differing signs at different temperatures, and involve other variables.

And that function, I would submit, is not readily differentiable …

Here’s an example. Albedo is the total reflection of sunlight by the earth. The graphic below shows the dependence of albedo on temperature.

Here’s the oddity of the earth. Where it is cold, as temperature goes up albedo goes down. This tends to warm up the cold areas.

But in the tropics, as temperature goes up, albedo goes up. In other words, the effect of natural albedo variations is to warm us where we are cold, and to cool us where we are warm …

Your model entirely ignores that hugely important thermostabilizing mechanism, with unknown effects on the trend line calculated from your model … which is OK as far as it goes. However, as someone once told us, if you don’t go far enough, you can get any answer you want …

Thanks again for a most interesting post, Kevin. As my work was the subject of your head post, with your permission, I’d like to append this comment to the head post. Let me know.

w.

June 17, 2019 3:07 pm

For years I’ve been saying: “there is evidence of positive feedbacks when temperatures lower and negative feedbacks when temperatures rise” …. I’ve not been able to say what causes them, but that clearly is what happens.

I knew there was something, somewhere, and I’ve been trying to find it for well over two years.

Now you have found that:”Where it is cold, albedo goes down [with increasing temp]- this tends to warm up the cold areas (positive feedback) But in the tropics, as temperature goes up, albedo goes up. … to cool us where we are warm (negative feedback).

I now know what stabilises the climate, I know it is already affecting the climate and that in turn means I know the climate sensitivity for warming is less than 1C.

Gordon Dressler
June 16, 2019 1:05 pm

Ooops . . . missing from the Block Model as presented in Figure 1, obviously or perhaps not-so-obviously: non-radiative, non-convective energy transport to/from the control volume surface due to precipitation (snow, sleet, rain, fog) and water condensation (dew, frost) . . . these processes add or remove substantial energy at the control volume surface depending on process temperature differentials from “T”.

And, yes, snow (at, say, -5 C) can indeed transfer heat energy to soil or ice that is at, say, -10 C.

In other words, the one-directional “Evaporation” heat flow arrow ignores the complete hydrological cycle.
Don’t glaciers and global ice extent and depth increase during global cooling periods, even over “equilibrium” periods as short as 20 years?

Put another way, any model having an averaged temperature as the primary variable for a model of Earth is not credible unless it specifically accounts for energy exchange (“heat flows”) associated with water phases changes of BOTH liquid to vapor (and vice versa) and liquid to ice (and vice versa). The presented model appears to only consider one of these four processes.

Bindidon
June 17, 2019 8:37 am

Michael Moon

“All matter above Absolute Zero radiates all the time in all directions. N2, O2, and Argon radiate to space. Increasing CO2 increases the altitude at which CO2 is no longer opaque to 15-micron radiation, thus lowering the temperature at which the Atmosphere radiates freely to space, thus lowering the amount of heat transferred to Space. ”

Yes.

This is in perfect agreement with what was written in 2011 by J.-L. Dufresne and Jacques Treiner:

http://documents.irevues.inist.fr/bitstream/handle/2042/39839/meteo_2011_72_31.pdf?sequence=1

Unfortunately it was written in French… but maybe I have overseen an English translation of it somewhere in the list of Dufresne’s papers:

http://www.lmd.jussieu.fr/~jldufres/publi/liste_publi.html

June 17, 2019 2:40 pm

The article says: “Our discussion demonstrates that the relationship between skin temperature and absorbed radiation from Willis’s post has no direct bearing on climate sensitivity”

That’s Rubbish!

What it shows is that the climate sensitivity is 0.38C / 3.7W/m2 isolation (roughly doubling of CO2) AS WILLIS SHOWED AND then you show what is causing that sensitivity is that roughly half goes into IR emitted and half goes into heat sinks like the ocean.

Also, Willis’ graph he claimed did not show a “lag” actually shows 0.25year lag between insolation and temperature which indicates that most of the change is occurring well within one year.

RW
June 17, 2019 5:42 pm

The 3.7 W/m^2 from 2xCO2 has nothing to do with and/or doesn’t say anything about surface DLR (there is no change in surface DLR from 2xCO2). It’s simply an instantaneous decrease in upward emitted IR transmittance from the surface and an increase in upwardly absorbed IR from the surface. The result is a -3.7 W/m^2 TOA deficit and no change at the surface.

This was pointed out to Willis and he just ignored or never addressed it. As far as I’m concerned, it invalidates his entire thing.

tom0mason(@tom0mason)
June 18, 2019 3:02 am

Always keep in mind the 1st and 2nd Law