MODTRAN: ITS QUIRKS AND USES.

Guest post by KEVIN KILTY

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An Atmospheric Transmission Model

Introduction

This post is not one I planned to do. It has grown from work I was doing with MODTRAN (moderate resolution transmission program) in support of my post on June 14, 2019. As I mentioned in that thread I had gotten odd results from MODTRAN, and needed to ponder them further. As in so many things on WUWT what I have learned is directly connected with a number of Willis Eschenbach’s posts (especially 08/11/2011, and 04/12/2014); but also is related to dozens of others, especially to a recent post by Nick Stokes (06/06/2019) and something Dr. Spencer said on a thread so long ago I can’t even recall the year, let alone a date. What I have to say in this post is pretty fundamental to interpreting the output of model runs from MODTRAN, as well as to understanding its relationship to models of the greenhouse effect including feedback.

1. Radiation measures

There are four quantities used to quantify radiation which are related to one another and which use standard symbols in most texts: radiant intensity (I), irradiance (G), emitted power (E or W), and radiosity (J). I used Is for solar irradiance in my previous post, but I will stick to the conventional symbols from now on. Older versions of MODTRAN use Iout as the symbol for irradiance. The definitions of these are:

  1. Intensity is the power flowing along a pencil of rays from or toward a unit area on a surface and delimited by a unit solid angle in some direction. It is what we would think of as a beam of radiation.
  2. Irradiance is intensity integrated over the entire view that a unit surface area has of incoming radiation. It is the power flux landing on a surface.
  3. Emitted power is the irradiance produced by emitting sources on a surface. The Stefan-Boltzmann law is emitted power.
  4. Radiosity is the combination of emitted power plus irradiance derived from reflection at a surface. I won’t use radiosity at all in this post, but radiation is reflected frequently on the surface or in the atmosphere. Radiosity is what one needs to handle such instances.

2. MODTRAN Input/Output

MODTRAN, like so many legacy programs written in FORTRAN, reads a formatted file, a virtual card deck if you will, that specifies the current job. It calculates radiant intensity over a specified path, and also transmission coefficients at particular wavelengths on this path. Through the input file one can define a path, what sort of surface this path reaches, and the composition, pressure and temperature structure of the atmosphere along the path. While this sounds like a great deal of flexibility, most of this isn’t available in the portal that most of us have access to.

3. The University of Chicago Wrapper

The public copy of MODTRAN that Willis uses is provided by the University of Chicago. Access to the underlying program is through a graphic interface wrapper written in some other language. This wrapper does not provide full access to MODTRAN. It allows only vertical paths in the atmosphere, it seems to have no provision for input of an arbitrary model atmosphere, but rather lets the user choose from among a number of models–U.S. standard atmosphere (1976), Tropical, Midlatitude summer, etc. and then make limited adjustments thereto, including a surface temperature offset. Its output is irradiance at a specified height, and at a view of either the upper or lower hemisphere; and makes available raw model output with a pushbutton on the wrapper.

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Figure 1. Screen output from the University of Chicago wrapper for MODTRAN. Note the range of potential input values, and the raw model output button.

How the University of Chicago wrapper turns an intensity value into an irradiance is by assuming that the intensity is the same in all directions (isotropic radiation) at which point irradiance is just π times the intensity; G = πI.

Having an unfettered input to MODTRAN would allow a person to calculate radiant intensity at a number of view angles, and integrate over a hemisphere. The restricted operation of the wrapper using a fixed value of π leads to an obvious bias.

Imagine being high in the atmosphere, 70 km above the surface, and looking down. The view is not of Earth covering the entire hemisphere, but rather includes cold, dark space at grazing angles. I figure the actual view is only 98% of an illuminated hemisphere, which produces a bias. This is a moot point, however, because we do not actually know if intensity is the same in all directions and so the factor of π is likely not correct in any case. The accuracy of irradiance values provided by the wrapper might be as much as 26W/m2 different from true values on top of uncertainty of another 10W/m2 contributed by MODTRAN itself. Despite this, difference between models can be much more accurate as long as one takes care in specifying the model.

A final point about the wrapper concerns the temperature offset it allows. Each of the model atmospheres has a default surface temperature; 299.7K for the tropical model, 294.2K for the midlatitude summer model, and so forth. By specifying an offset, though, one actually adjusts the entire atmospheric path by this offset value, and is not what one intends to do in most circumstances. It would be great if one could just adjust temperature of the boundary layer, or just the surface, but this is not possible in any easy manner.

Finally, there are two assumptions regarding water vapor that a user can choose–constant relative humidity or constant mixing ratio. The constant relative humidity choice has an interesting interaction with a negative temperature offset. This is likely to produce a relative humidity exceeding 100%, and will produce a long list of error messages in the MODTRAN raw output, which never reach the graphic output of the wrapper. Look at the raw model file. Caveat emptor!

4. The MODTRAN Oddities

In his post about the MODTRAN oddities, Willis noted that his calculation of greenhouse forcing from a doubling of CO2 using various model choices would never meet values stated by James Hansen, and he wondered why this is so.

First, Willis measured the instantaneous forcing by using the difference in upwelling irradiance of paired MODTRAN runs at 70km in the atmosphere. The only difference is CO2 concentrations–375ppm versus 750ppm. I have verified Willis’s values ranging from 3.2W/m2 in the tropics to 1.6W/m2 in the subarctic. These are much smaller than Hansen‘s stated value of 4.5W/m2. I have tried to figure out what Hansen was thinking, but I can‘t. In various places he states different values, and he applies different assumptions, and considers different end points. I don‘t think chasing Hansen around in print is very useful, but staying with the story Willis has begun is worth pursuing.

5. Calculation of the Greenhouse Forcing

CO2 Observation height View Offset T Irradiance
ppm km ˚C W/m2
280 70 km Down 0 269.35
280 Surface Down 0 382.14
280 Surface Up 0 267.18
560 70 km Down 0 266.4
560 Surface Down 0 382.14
560 Surface Up 0 270.73
560 70 km Down 0.5 268.41
560 Surface Down 0.5 384.65
560 Surface Up 0.5 272.43

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Table 1. Irradiance values calculated using the public portal for MODTRAN at the University of Chicago website.

As Willis states, the discrepancy can‘t be the result of long term feedbacks. Indeed, something a bit surprising comes from considering two necessarily nearly instantaneous feedbacks which take the disturbance of a doubling of CO2 back toward equilibrium. Table 1 shows the pertinent details of a series of experiments done with MODTRAN, which tell this interesting story.

First, let‘s use a 1976 Standard Atmosphere model with a surface temperature of 288.2K and 280ppm of CO2. MODTRAN calculates an upward irradiance at the top of atmosphere (70 km) of 269.35W/m2. Let’s call this an equilibrium baseline state. This output irradiance is exactly what is needed to balance energy considering solar irradiance of 1370W/m2 and some assumed albedo.

If we now disturb this by suddenly doubling CO2, MODTRAN keeps the surface temperature and the atmospheric temperature distribution constant, and calculates a new irradiance at the top of atmosphere of 266.4W/m2. The difference of 3W/m2 is the new forcing–or almost. MODTRAN keeps all sorts of things constant, but in a real atmosphere and surface several of these things cannot remain constant. Neglecting long term feedbacks, we still have to consider an almost instantaneous effect. Within a week or two, the atmosphere will warm slightly because of the new absorptivity the increased CO2 provides, and the surface will warm slightly because of new downwelling LW radiation.

To model this with what freedom the wrapper around MODTRAN allows, I will offset the surface temperature by 0.5C. This is not a perfect rendition of what happens, but because such a large fraction of water vapor is close to the surface, it isn’t a terrible approximation either, and it illustrates what will happen.

As Table 1 shows, this reduces the 3W/m2 difference at the top of atmosphere to 1W/m2. In the long term other feedbacks will increase the top of atmosphere value to 269.35W/m2 once again because this is what is needed to restore energy balance. At equilibrium the enhanced forcing from the doubling of CO2, in fact from any disturbance involving CO2 isn’t apparent at all at the top of atmosphere. This is an example of what Dr. Spencer meant when he stated that one can learn nothing about the feedbacks involved in a regulator through its output at equilibrium. It was also the point, partially of Nick Stokes‘s post.

Consider what Table 1 shows about goings on at the surface. The difference in downwelling between the two constant surface temperature runs is 3.5W/m2, approximately equal to what we observe at the top of atmosphere. However, the model run at slightly higher surface temperature shows the difference in downwelling radiation is enhanced by immediate feedback and now different by 5.4 W/m2. As longer term feedback kicks in the surface values will continue to adjust to eventually indicate the full enhanced greenhouse forcing. The new forcings are fully observed at the surface, not at top of atmosphere, which seems reasonable to me.

6. Conclusions

Whether this fully explains the departure between Willis‘s and Hansen‘s instantaneous values I cannot say. One can argue that the quick reaction of the atmosphere to a doubling of CO2 is a feedback that should be excluded from consideration, but I would respond that it is so quick as to be completely different from something like water vapor feedback or melting glaciers. Also, people may argue about the equilibrium values of surface temperature or top of atmosphere radiation I apply. However, one has to begin with some assumptions and recognize that MODTRAN cannot fully handle conservation of energy except at the top of atmosphere because it is not equipped to account for the full range of heat transfer mechanisms at the surface. It is not a full-fledged heat transfer code.

In some future installment I plan to return to these points in connection with feedback because what one can determine about the internal workings of a system, at equilibrium or otherwise, depends to great degree on what one measures. The different response of upwelling and downwelling radiation at the surface versus at the top of atmosphere to a doubling of CO2 illustrates this.

7. Notes:

The definitions and standard symbols for radiation can be found in the text Fundamentals of Heat and Mass Transfer, by Incropera and DeWitt, John Wiley and Sons, in any of its eight editions. In his well known text Physics of Atmospheres, Houghton uses the symbol F for irradiance rather than G, and appends an up or down arrow to indicate upward or downward flux. He also uses a radiant intensity for the blackbody function, calling it B. This is therefore the Stefan-Boltzmann law divided by π.

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June 24, 2019 10:39 am

In case anyone would like to play around with the U.Chicago’s MODTRAN web interface, here are the links:

Old link was:
http://geoflop.uchicago.edu/forecast/docs/Projects/modtran.html
This link no longer works.

Newer link is:
http://forecast.uchicago.edu/Projects/modtran_form.html
This link still works, but I think it’s for an out-of-date version.

Even newer link is:
http://climatemodels.uchicago.edu/modtran

Newest link is:
http://lorelei.uchicago.edu/modtran/

The last two might be equivalent, I don’t know. The “lorelei” link might be for beta testing.

I used MODTRAN to calculate the approximate “amplification” effect of water vapor feedback, by comparing results with fixed absolute water vapor concentration to results with fixed relative humidity. I did that exercise two different times (I think it was a couple of years apart), and got very different results:

https://sealevel.info/climate/MODTRAN_etc.html

I have one hint about the user interface for the two newest versions: at first glance there doesn’t appear to be any way to select between constant water vapor concentration and constant relative humidity. I was afraid they had removed that feature! However, the option is still there, it’s just hidden. It’s the “Holding fixed” option, which only appears if you set the Temperature offset to non-zero.

BTW, in case you skimmed that article and wondered about this sentence, it’s a formatting glitch:

   “Hansen was thinking, but I can’t.”

The correct full sentence was:

   “I have tried to figure out what Hansen was thinking, but I can’t.”

Kevin kilty
Reply to  Dave Burton
June 24, 2019 12:15 pm

Funny!

Reply to  Kevin kilty
June 24, 2019 5:10 pm

Yes!

Thank you, Charles-The-Moderator, for fixing it.

kevin kilty
June 24, 2019 10:42 am

Garbled sentence, My fault:

and is available to a user as a raw model output pushbutton on the wrapper.

is more clear as

“…and makes available a raw model output with a pushbutton on the wrapper.”

June 24, 2019 10:59 am

I’ve never understood why climate scientists didn’t build instrumentation to actually measure CO2 “back” radiation (but then again, if I hit my head with a softball bat it actually does become clear).

I currently practice science in a few different fields, and often have the need to have other professionals build accurate custom equipment for measuring “things”. I’m pretty certain that they could build an instrument that would measure “back” radiation of CO2 on any level of water vapor for a million bucks. So, let’s say 10 million bucks, or a 100 million even (and save humanity a few trillion). The original Modtran paper, which I’ve lost, was actually very good, but still a model on the outer water vapor/CO2 overlap edges. I kinda lost interest in thinking about this when I realized it had nothing to do with science (not the original paper, which Ferdinand Engelbeen referenced in a WUWT post many years ago).

Good post. Maybe this issue will be addressed experimentally one day soon (Kilty, lots more authors and Happer, 2021, manuscript in preparation?)

Reply to  Joel O'Bryan
June 24, 2019 5:30 pm

Well thanks for the responses which clearly made me realize that in trying to be succinct in my question, I was a bit too succinct. I was aware of all of this and and also Feldman at al., so let me try again.

The chamber would be capable of containing CO2 at all concentrations from zero to 560 ppm and then some if required, and measurable to one decimal place. The chamber would also be capable of containing, simultaneously with the varying CO2 concentrations, from zero (as at the poles) to 30 – 40,000 ppm of water vapor (as in the tropics). The back radiation would be measured as in not blustered about. The overlap of water vapor bands and the effect of the Beer-Lambert Law would be measured for all combinations of water vapor/CO2 concentrations and temperatures, and you could probably throw in some convection too. This equipment could easily be built for less than $10 Million, probably way less. It’s not even my field, but I could get it built.

Vicus
Reply to  Joel O'Bryan
June 27, 2019 1:31 pm

Roy ignored the surface radiating heat at night, assuming it’s downwelling IR.

unka
Reply to  philincalifornia
June 24, 2019 12:04 pm

“why climate scientists didn’t build instrumentation to actually measure CO2 “back” radiation”. – Because it can’t be done.

commieBob
Reply to  unka
June 24, 2019 3:46 pm

It’s actually dead easy.

Place a transparent* container of CO2 in an anechoic chamber. Excite the CO2 with a beam of radiation. Examine the signal emitted by the CO2 from multiple angles to produce a radiation pattern.

The tricky bit is that you might have to get the temperatures within the anechoic chamber as close to absolute zero as possible.

*transparent at the wavelengths under consideration.

Kevin kilty
Reply to  philincalifornia
June 24, 2019 12:07 pm

I called van Wijngaarden more than a year ago inquiring about the promised paper, but nothing yet. The poster session summary of their paper is available online. I’d forgotten all about it until you mentioned it.

A decade ago I proposed a research project of using FTIR to monitor downwelling radiation. There are lots of challenges involved. The proposal went nowhere. I think we can do the same job from satellites. There are lots of posts at WUWT as I mentioned, but Willis’s popped up on a search first, and seemed like an interesting illustrative example.

Geoff Sherrington
Reply to  philincalifornia
June 24, 2019 12:11 pm

“I’ve never understood why climate scientists didn’t build instrumentation to actually measure CO2 “back” radiation”
.
http://asl.umbc.edu/pub/chepplew/journals/nature14240_v519_Feldman_CO2.pdf

Kevin kilty
Reply to  Geoff Sherrington
June 24, 2019 1:54 pm

Interesting reference, thanks.

Robert W. Turner
Reply to  Geoff Sherrington
June 24, 2019 5:22 pm

“Schematic. The schematic (Extended Data Fig. 1) shows how CO2 surface forcing
is derived from differencing Atmospheric Emitted Radiance Interferometer (AERI)
observations with a model calculation based on coincidental temperature and
water vapour profiles”

Yeah, that’s “science” for the climate field.

Geoff Sherrington
Reply to  Robert W. Turner
June 25, 2019 7:50 am

You have a problem with observations that show the theory sound?

Robert W Turner
Reply to  Geoff Sherrington
June 25, 2019 12:58 pm

“model calculation based on coincidental temperature and water vapour profiles”

The results are based on a model with the back radiation hypothesis already built in. They are taking observations, and then calculating a “CO2 surface forcing” based on a hypothesis. This is science in reverse. Typically measurements would be used to test a hypothesis, not to determine results based on assuming a hypothesis is already true.

Geoff Sherrington
Reply to  Geoff Sherrington
June 25, 2019 1:32 pm

Robert W Turner says: “This is science in reverse.”

Hardly

If we follow your logic, then you have to throw out Spencer’s UAH, and Mears’ RSS temperature time series, because they are based on the same principles.

Feldman et. al’s study verifies that back-radiation increases with increasing CO2 concentrations.

Geoff Sherrington
Reply to  Geoff Sherrington
June 25, 2019 1:35 pm

What is really fascinating about Feldman’s results is that you can see the seasonal variation in the back-radiation that results from the seasonal variation in CO2 concentrations. More evidence that Feldman’s work is very good.

Geoff Sherrington
Reply to  Geoff Sherrington
June 25, 2019 1:53 pm

So, tell me Mr. Robert W Turner, if you go out into a dry desert, on a cloudless night, and point an IR thermometer up at the stars, how do you explain the reading you get from it?

Robert W. Turner
Reply to  Geoff Sherrington
June 25, 2019 4:34 pm

This is not the same as AMSU based temperature measurements.
The satellite temperature profile is derived from microwave sounding units and calibrated against empirical data from radiosondes. It is an active system using microwave sounding, AERI is a passive system simply looking at the spectra of the sky from the surface and supposedly can differentiate the source of emission. AERI data is controlled by
“Quality control for the spectra was achieved by ensuring that valid sky spectra were being observed by the instrument with the hatch open, and that unphysical radiance values, anomalously low variances in brightness temperature across the infrared window band (800–1,200 cm21) and scenes with substantial variability in the view of the hot blackbody were removed”
Seems like a little bit of balderdash to me. Unphysical radiance values, scenes with substantial variability in the view of the bot blackbody? Perhaps you’d like to translate.

Are you also saying that the temperature we feel at the surface is due to this IR emission from the sky?

The AERI is no doubt looking at the spectra looking skyward from the surface, mostly from air and dust within tens of meters away.

Butch123
Reply to  Geoff Sherrington
June 25, 2019 4:29 pm

“The trends in forcing are significantly (P,0.003) different from zero only in the P- and R- branches of the CO2 band.”

In These regions of the outer bands, IR travels quickly through the atmosphere and is not readily absorbed. The minor amount of IR detected by these facilities is irrelevant to the theory of AGW.
Of course when downwelling IR was detected in these far bands it was touted as proof that CO2 in the atmosphere re-radiated in every direction. However no 15 micron band radiation was ever observed.

Jean Parisot
June 24, 2019 10:59 am

Georgia Tech maintained an updated set of atmospheric models for years for the government, not sure if they still do.

RayG
Reply to  Jean Parisot
June 24, 2019 12:16 pm

Suggest that you ask Judith Curry at her Climate Etc. blog at judithcurry.com about this. If you are not familiar, she was chair of Georgia Tech’s School of Earth and Atmospheric Sciences, 2002-2018 so if anyone is familiar, she should be.

June 24, 2019 11:11 am

‘These are much smaller than Hansen‘s stated value of 4.5W/m2.”

The explanation is easy. Hansen started with the needed outcome surface temperature (~+3 C) and worked backwards to get what forcing he needed.
It’s been all downhill for climate atmospherics of GHGs from there.

michael hart
June 24, 2019 11:13 am

“This is likely to produce a relative humidity exceeding 100%, and will produce a long list of error messages in the MODTRAN raw output, …”

Yet in the real world, supersaturation does occur, frequently and persistently. Especially at radiatively important locations in the troposphere. I’ve never seen any discussion of how they treat this in climate models, or whether supersaturation is affected by changed CO2 levels which radiates in some bands that water vapor does not. My suspicions are too obvious to need stating…

J Mac
June 24, 2019 11:48 am

Those are surprising results from MODTRAN and, perhaps, reflect limitations of the software model. With 30 years of practical engineering experience however, I was surprised more than once by the outputs from PATRAN and other engineering modeling programs to not discount the possibility that a ‘surprising result’ from the model was illustrating something ‘real’ that required further dedicated mechanical property testing to verify or deny. Similarly, there were occasions where the initial PATRAN runs showed a particular part geometry/material to be acceptable but a subsequent run with a finer ‘mesh’ would highlight higher than acceptable local strain conditions. Models are useful… but require dedicated real world tests to sort out ‘surprising’ results.

wsbriggs
June 24, 2019 12:05 pm

For those wishing to attack the problem of back radiation, the files on the Gemini Observatory Website for IR Sky Background radiation might prove interesting. They are available for Mauna Kea and Cerro Pachon (the twin observatory in Chile). There is information on the assumed H2O vapor content as well.

Mark.R
June 24, 2019 12:24 pm

OT
Darwin Australia may have set an all-time min night temp last night
Copied from their page..

Darwin
Now
13.0°
http://www.bom.gov.au/

Last
Coldest Ever 13.8° 27/05/1990
Says this page.
http://www.meteorology.com.au/local-climate-history/…/darwin

Another Ian
Reply to  Mark.R
June 24, 2019 2:12 pm

Currently showing 15 degrees as minimum on Weatherzone

Mark.R
Reply to  Mark.R
June 24, 2019 11:53 pm

So looked on this site 12 hours later (other site has gone offline since) copied this

Min temp history
Darwin minimum temp history (12.4613°S, 130.8419°E, 46m AMSL)
Coldest Ever 12.1° 23/06/1963
Coldest This Year 12.7° 25/06/2019
Coldest This Month 12.7° 25/06/2019

https://weather.mla.com.au/climate-history/nt/darwin

Robert W Turner
June 24, 2019 12:57 pm

The mean free path of a 15 micrometer photon emitted at surface conditions is only 33 meters, not in kilometers. The back radiation hypothesis is pseudoscience. The temperature of a gas is entirely due to the average translational kinetic energy of the gas molecules, not from IR emissions. Energy feedback of this kinetic energy back to the surface occurs when the air is warmer than the surface and explains the so-called “greenhouse effect.”

http://www.feynmanlectures.caltech.edu/I_39.html

Reply to  Robert W Turner
June 24, 2019 3:24 pm

Rob,
SB equation between two parallel surfaces is
“Q= aConstant x (Thot^4 – Tcold^4)”
The “aConstant x Thot^4” is the “fore-radiation” and you can think of the….”aConstant x Tcold^4” part as the “back-radiation”. The sky has a “Tcold” causing the IR “back-radiation” to the ground. Hold your face close to the wall. You radiate about 450 W/sq.M to the wall. Wall radiates about 400 W/sq.M to your face. Sensitive skin might feel the 50 W/sq.M heat flow to the cool wall. Plug some numbers into SB and check it out…
Kindergarten stuff for Feynman….

Robert W. Turner
Reply to  DMacKenzie
June 24, 2019 5:12 pm

Okay, that has nothing to do with what I’m talking about, or are you arguing that the temperature of a gas is not from the average kinetic energy of the gas but rather from the IR emissions? That’s like hoping the incoming asteroid isn’t spinning too much because you think all the energy is in the angular momentum. And how does the SB equation apply to gases? A gas is a volumetric mixture of molecules, not a solid material with a radiating surface. The mythical balance layer called the radiative layer does not exist in the real world outside of climatstrologists’ heads.

Phases of matters have different optical and radiative properties. Gas molecules are not bound to other molecules and thus have degrees of freedom corresponding to the electromagnetic/quantum properties of the molecule. This freedom of movement means several things for gas – a couple important ones being that the vast majority of the energy of the molecule is in the form of kinetic energy and the molecule absorbs and emits radiation corresponding to resonant frequency corresponding to the molecular modes + doppler effect rather than a continuous spectrum based on temperature in a function described by the SB equation.

And I’m not convinced that the photons are actually ever absorbed, rather they ricochet off the molecule and either add or subtract from the kinetic energy of the gas depending on the vector directions of the two particles. This is how laser cooling works, otherwise where does the energy go within the supercooled gas? It was never absorbed.
https://www.youtube.com/watch?v=drnq_6ffTbo

So the net effect in the atmosphere, with mostly random motion, the effect should cancel out – if anything, since we know there is a net flux from the surface then the phenomenon should actually help advect IR active gases.

A 15 micron IR photon is emitted from the surface, it makes it an average of 33 meters before ricocheting off a CO2 molecule. The photon only has an extremely small chance of being sent back to the surface as the angle would need to be perpendicular to the surface. Any other angle means it is more likely to ricochet off another CO2 molecule before reaching the surface, and with the adiabatic lapse the favored direction is upward and out of the atmosphere so fast that the photon is gone in a blink of an eye, or about 4 milliseconds. The chances of you absorbing a photon from the pseudoscientific “radiative layer” is about nil.
http://www.biocab.org/Mean_Free_Path_Length_Photons.html

This is the real physics of how an atmosphere works. If these magical GHG molecules absorbed photons and “thermalized” that energy like you people claim then the Earth would be scorching hot because gases have extremely small emissivities.

You can agree with me or not, I don’t care. Just answer one question please, how does the Kinetic Theory of Gases conform to the heat budget diagrams which show a 300+ Wm^-2 absorption at the surface due to IR emissions of the atmosphere.

Reply to  Robert W. Turner
June 25, 2019 11:27 am

re: “And I’m not convinced that the photons are actually ever absorbed, rather they ricochet off the molecule ”

At this rate, no one can probably convince you that light is an EM wave ”phenomenon’.

“Science ” did us a disservice 100 yrs ago by inventing the mythical photon ‘particle’ too.

Robert W Turner
Reply to  _Jim
June 25, 2019 12:53 pm

Sure, I’m only talking about photons propagating as an EM wave but no one can convince me of what I’m arguing…

Photons are basically an observational fact with several different styles of experiments and theories showing them to be factual.

Geoff Sherrington
Reply to  _Jim
June 25, 2019 1:48 pm

“Photons are basically an observational fact ”

No, nobody has yet to observe a “photon.”

Photons are mental constructs built by men to explain what they are observing. To this day, we do not know if a photon is a wave, or if it is a particle. Don’t forget that “observing” something as tiny as a “photon” changes it’s very nature.

Reply to  _Jim
June 25, 2019 3:12 pm

Geoff Sherrington June 25, 2019 at 1:48 pm
“Photons are basically an observational fact ”

No, nobody has yet to observe a “photon.”

Photons are mental constructs built by men to explain what they are observing. To this day, we do not know if a photon is a wave, or if it is a particle. Don’t forget that “observing” something as tiny as a “photon” changes it’s very nature.

A LOT of BS related to mythical “photons” exists in most on-line discussions. Reputable, well-performed dual-slit experiments with EM waves do not show ‘particle’ behavior with light (OR radio waves, which are JUST lower-frequency EM waves.) Photons seem to have been “invented” in relation to work by Einstein and the discovery that light could knock electrons loose in metals. I propose that his work needs revisiting in light of what we know today, in order to clarify terms, and isolate the true mechanism in-play that exhibits as “The Photo-Electric Effect”.

Now, the problem is, physicists and on-line debate ‘artisans’ have fallen into a pit they cannot extricate themselves from. Better to review “the basics” in a well-performed “dual-slit” experiment rather then jump in with both feet into abstract theory unsupportable by experiments. Here is one such series of experiments at uWave and radio waves, performed at an MIT lab. This in the vein of a dual-slit experiment, but uses a pair of radiating antennas instead:

https://youtu.be/tY3_78ONkmI?t=1073

Geoff Sherrington
Reply to  _Jim
June 25, 2019 3:34 pm

_Jim says: “Reputable, well-performed dual-slit experiments with EM waves do not show ‘particle’ behavior with light.

WRONG, no matter what experiment you configure, you cannot resolve the problem of wave-particle duality. It’s both. https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality

Robert W. Turner
Reply to  _Jim
June 25, 2019 4:47 pm

A photon is more than a mental construct. If it has physically measurable effects, like momentum, then it must be physical as far as our perceived physical reality is concerned.

I’m picturing someone being cut in half by an intense laser beam screaming “it’s not real, it’s just a mental construct!” Believe what you want, it doesn’t change the physical reality which decoheres from quantum information.

Geoff Sherrington
Reply to  _Jim
June 25, 2019 5:34 pm

_Jim, Wikipedia is perfectly suitable for presenting you with facts. When Wikipedia deals with math and science, it does not have the problem of topic that are politically charged. So, again, waves have zero mass. Photons have momentum. A wave cannot have momentum, because zero mass makes the momentum zero. Insisting that photons are only waves and not both waves and particles shows your inexperience with physics.
….
Read the Wikipedia entry on duality: https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality It will help you a lot.

Reply to  Robert W. Turner
June 25, 2019 12:25 pm

Robert W. Turner wrote, “And I’m not convinced that the photons are actually ever absorbed, rather they ricochet off the molecule and either add or subtract from the kinetic energy of the gas depending on the vector directions of the two particles. This is how laser cooling works, otherwise where does the energy go within the supercooled gas? It was never absorbed.”

Nope, it is absorbed, not reflected.

If it were reflected, then the amount of LW IR apparently-emitted would be the same as the amount apparently-absorbed. It isn’t. The amount emitted has almost nothing to do with the amount absorbed.

The only way that the amount of LW IR absorbed affects the amount emitted is by being one of the many things which affects the temperature of the gas (atmosphere). The amount emitted is determined by the temperature (and, of course, the makeup of the atmosphere).

I’m no expert on laser cooling, but my understanding is that it takes advantage of the fact that the movement of the molecules adds or subtracts from the energy of photons absorbed and emitted by those molecules. So even in the (rare!) case of a molecule absorbing a photon, and then losing that energy by emitting another photon (rather than collisionally transferring the energy to another air molecule), the wavelength of the emitted photon is likely to be slightly different from the wavelength of the absorbed photon.

E.g., suppose that a molecule absorbs a 15.08417 µm photon (which is the center of one of CO2’s absorption lines). Then suppose that same molecule happens to be moving north when it emits another photon, in a random direction. If the photon happens to be emitted in the same direction that the molecule was moving (north), the photon is blue-shifted slightly (slowing the molecule). But if the photon happens to be emitted in the opposite direction (south), then the photon is red-shifted slightly (speeding up the molecule).

Laser cooling takes advantage of this. You tune the laser(s) to just slightly below a strong absorption line of a pure substance that you’re trying to cool (i.e., tune it to a slightly longer wavelength). Put the substance in a vacuum chamber. Hit the substance with the laser(s), and wait for the “same” wavelength to be re-emitted by the target substance.

The re-emitted light will be centered on the substance’s absorption line, which is slightly above the wavelength of your laser. But the laser light will be preferentially absorbed by molecules which are moving in the direction opposite the direction of travel of the laser’s photons (which, from the perspective of the molecules, are blue-shifted). The net effect of absorbing photons which average slightly lower energy than the emitted photons is a slowing of the molecular motion, i.e., a cooling of the substance.

Here are links to some relevant resources:
https://www.sealevel.info/learnmore.html#physics

The first of those links is to a very interesting 2014 UNC Physics Colloquium by Princeton atmospheric physicist Prof. Will Happer, which I attended. Afterward I asked him a follow-up question, by email, which he very kindly answered. I think you would find that email exchange educational, as well (I certainly did!); here it is:
https://sealevel.info/Happer_UNC_2014-09-08/Another_question.html

Robert W Turner
Reply to  Dave Burton
June 25, 2019 1:55 pm

Thanks for the email exchange, that’s interesting. Though where are the papers showing that absorption of these photons actually increases the temperature of the gas in any measurable way? Doesn’t the even broadening of the absorption bands around the lines suggest that photons absorbed/emitted are both increasing and decreasing the momentum of the molecules with around zero net effect?

So assuming all of that is true, imagine you have a thermometer sitting on the surface. It is reading a temperature which you know is just a function of the heat energy being transferred to it. What % of that heat is from kinetic bombardment from gas molecules and what % is from IR photon bombardment originating from gas molecules? Conventional science says that the % of heat from the later is negligible, though looking at the heat budget diagrams and reading through the papers that they originate from suggest that it is the other way around.

Also, if the heat is being transferred to the bulk of the atmosphere from IR absorption, mostly N2, then what is the emissivity of N2. This massively important variable seems to be so hard to find that one would think that N2 has no emissivity, which we know is not true. If the GHGs are the only effective way that heat can actually emit from the atmosphere, then doesn’t this also balance any net absorption from the surface?

Reply to  Dave Burton
June 27, 2019 2:59 pm

Robert W. Turner wrote, Thanks for the email exchange, that’s interesting. Though where are the papers showing that absorption of these photons actually increases the temperature of the gas in any measurable way?”

You can’t publish a paper for something so obvious. Anything that absorbs radiation heats up. It doesn’t matter whether it’s a gas, a liquid, or a solid. That’s true for LW IR, and also for radiation with longer or shorter wavelengths than LW IR:

Longer: Do you own a microwave oven? What happens when your coffee absorbs microwave radiation?

Shorter: Have you ever stepped from a light-colored sidewalk onto a dark asphalt road in the summer sun, with bare feet?
 

Robert asked, “Doesn’t the even broadening of the absorption bands around the lines suggest that photons absorbed/emitted are both increasing and decreasing the momentum of the molecules with around zero net effect?”

No. You’re assuming that the number of approx. 15 µm photons emitted is the same as the number absorbed. As I just explained, it usually isn’t.
 

Robert asked, “…a thermometer sitting on the surface… is reading a temperature which you know is just a function of the heat energy being transferred to it….”

To it and from it. If it receives more energy that it loses, its temperature will be rising. If it loses more energy than it receives, its temperature will be falling.
 

Robert continued, “What % of that heat is from kinetic bombardment from gas molecules and what % is from IR photon bombardment originating from gas molecules?”

It depends on how much radiation it receives and absorbs, and the temperature and pressure of the air, etc. If it’s sitting in the sunlight a higher percentage of the energy it receives will be from radiation absorption. Shield it from the sunlight and a lower percentage will be from radiation absorption. Surround it with hot, dense air, and a higher percentage of the energy it receives will be from the air. Surround it with cold, thin air, and a lower percentage of the energy it receives will be from the air.
 

Robert continued, “…what is the emissivity of N2.”

At typical atmospheric temperatures, it is extremely low, compared to CO2, H2O, Methane, etc. It is considered radiatively inactive.

If you get it hot enough, it’ll radiate, but at much shorter wavelengths:
https://www.reed.edu/physics/courses/Physics332.s17/pdf/N2%20Spectrum.pdf

Robert W Turner
Reply to  Robert W. Turner
June 25, 2019 1:24 pm

Though it’s fair to say I’m misunderstanding the mechanics behind laser cooling. I don’t know if the same phenomenon works for non-ionizing energy and all gas molecules. It doesn’t really matter if the photons are absorbed or not, it’s typically referred to as scattering in this process.

https://www.sciencedirect.com/science/article/pii/0030401875901595

Reply to  Robert W. Turner
June 25, 2019 4:00 pm

Geoff Sherrington
_Jim says: “Reputable, well-performed dual-slit experiments with EM waves do not show ‘particle’ behavior with light.

WRONG, no matter what experiment you configure, you cannot resolve the problem of wave-particle duality. It’s both. https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality

Please; I give you an experiment in which you can observe the actual *wave nature* of light/EM/radio waves (with results instantly converted such that your primary senses can indicate the phenom) and you give me a wiki cite. Please, give something specific in return, something which can be challenged and possibly falsified on a technical basis, preferably a live-taped lab demo, and not an artist’s depiction or ‘cartoon drawing’.

Geoff Sherrington
Reply to  _Jim
June 25, 2019 4:28 pm

Waves do not have momentum which is demonstrated by Crookes radiometer: https://en.wikipedia.org/wiki/Crookes_radiometer

Geoff Sherrington
Reply to  _Jim
June 25, 2019 4:34 pm

How do explain Compton scattering if photons do not have momentum?

Geoff Sherrington
Reply to  _Jim
June 25, 2019 4:37 pm

_Jim, the Crookes radiometer is not the best example of radiation pressure, try the Nichols radiometer. https://en.wikipedia.org/wiki/Nichols_radiometer

Again, waves do not have momentum.

Geoff Sherrington
Reply to  _Jim
June 25, 2019 4:40 pm

_Jim, waves cannot exert pressure: https://en.wikipedia.org/wiki/Radiation_pressure

Reply to  _Jim
June 25, 2019 5:16 pm

re: Geoff Sherrington June 25, 2019 at 4:28 pm

More wiki cites? I’m beginning to think you have no valid refutation of the points raised. Let’s take another look at a proof showing light as a wave only, exhibiting no ‘particle’ nature to it.

Here is a proof, making use of polarizers, which demonstrate that light (EM wave), when propagating, propagates transversely only, and as a wave, and not longitudinally, as a would be expected of a ‘particle’.

https://youtu.be/tY3_78ONkmI?t=550

Adding a third polarizer allows “rotation” (via re-radiation of the EM energy) of the polarized light to the 90 degree angle and beyond, once again demonstrating light as a wave and not a particle:

https://youtu.be/AVn49LbYoB8?t=213

I await that definitive demonstration, Geoff, showing light as a particle. Please, no more “hand-waving” via the simple posting of links to wiki tropes.

Reply to  _Jim
June 25, 2019 5:37 pm

Geoff Sherrington June 25, 2019 at 4:34 pm
How do explain Compton scattering if photons do not have momentum?

I suggest you are mis-attributing the effect.

How about you supply some additional rationale on this aspect, too?

Reply to  _Jim
June 25, 2019 5:41 pm

Geoff Sherrington June 25, 2019 at 4:37 pm
_Jim, the Crookes radiometer is not the best example of radiation pressure, try the Nichols radiometer.

Mis-attribution of the effect to a so-called ‘particle’; you are probably not aware the radiation, EM radiation, can be shown to exert something called radiation reaction?

Radiation reaction is the backaction on an accelerated electron from the radiation it emits. Ordinarily the effect is small and can be described classically by adding a term to the equation for the Lorentz force.

Geoff Sherrington
Reply to  _Jim
June 25, 2019 6:41 pm

Light being waves cannot explain the photoelectric effect.

Butch123
Reply to  Robert W. Turner
June 25, 2019 4:35 pm

Einstein won the 1921 Nobel Prize for describing how a molecule absorbs energy. The belief that he was wrong has hoist many a dissenter on his own petard.

Kevin kilty
Reply to  Robert W Turner
June 24, 2019 3:36 pm

What modtran does is solve the equation of transfer for all radiation within the parameter limits of its run. The U of Chicago wrapper considers 4.5 micrometer and longer wavelengths. Thus it covers the spectrum over which the majority of emitted power occurs for a gray or black body at temperatures of 100C and less.

If you are insisting that no radiation ever reaches the ground from a few km above because its mean free path is only 33 meters, then you are not considering the full equation of transfer. You are considering only extinction of a beam that starts somewhere and ends at ground. There is also radiation scattered into a path that adds to the beam, at every point along the path, as well as new radiation emitted into that path. The full equation of transfer is a lot of work to calculate.

Robert W. Turner
Reply to  Kevin kilty
June 24, 2019 6:16 pm

On your last post, I did the same method for Mars and found an estimated emissivity of 0.997. Mar’s atmosphere has a partial pressure of over 500 pascals of CO2 compared to Earth’s 40. This so called back radiation apparently has zero impact on Mars.

Kevin kilty
Reply to  Robert W. Turner
June 25, 2019 7:17 am

You haven’t made clear what you did, but what you are telling me seems exactly what I would expect on Mars for two reasons. First the Martian atmosphere has nothing but CO2 and isn’t very deep, optically, at IR wavelengths. Thus, one should see that the effective emissivity and actual surface emissivity are about the same. Second, the Martian atmosphere is so tenuous that I wonder if local thermodynamic equilibrium (LTE) ever actually applies to it. In other words, the source function for the equation of transfer on Mars has to consider something other than the Stefan-Boltzmann law.

Robert W Turner
Reply to  Kevin kilty
June 25, 2019 11:04 am

I simply found the effective emissivity of the Mars surface based on the average data available, like what you did for Earth in your previous post. If the average

Your response is utter arm-waving, it has no scientific bases or meaning. Science works the opposite of this, you don’t adhere steadfast to a hypothesis and make excuses for when it doesn’t work. You need to show theoretically or empirically how IR photons being emitted from the surface of Mars are not interacting with a similar concentration of GHGs in its atmosphere like what purportedly occurs on Earth – CO2 on Earth purportedly being responsible for 20-25% of the “GHG effect” at 40 pascals whereas there is no GHG effect on Mars at all with 500 pascals partial pressure CO2.

Has anyone contributing here actually ever read
https://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v4.pdf ?

P.S. The troposphere of Mars is ~45 km in height.

Robert W Turner
Reply to  Kevin kilty
June 25, 2019 12:28 pm

Furthermore, the lack of a “GHG effect” from CO2 is exactly what we see on Earth from CERES. The Sahara Desert as a whole actually loses more heat into space than it receives from the sun, as do many of the Earth’s regionally extensive deserts. So not only does this effect not show up on Mars at over 500 pascals of CO2, but the effect isn’t seen on Earth where water vapor is lacking.

I don’t know of a physical based explanation for this based on back radiation hypothesis. You can, however, explain the phenomenon based on well established physics.

Reply to  Robert W Turner
June 24, 2019 5:06 pm

Robert W Turner wrote, “The mean free path of a 15 micrometer photon emitted at surface conditions is only 33 meters…”

Do you have a source for that figure?

I’ve seen various people say it is “about 1 meter,” “a few meters,” “25 meters,” and now “33 meters.” I assume that someone has actually measured it, but if so I can’t seem to find it.

Robert W. Turner
Reply to  Dave Burton
June 24, 2019 6:33 pm

http://www.biocab.org/Mean_Free_Path_Length_Photons.html

Not measured unfortunately. Measuring that would probably take some clever quantum entanglement experiment.

Reply to  Robert W. Turner
June 24, 2019 7:53 pm

It doesn’t sound hard, to me.

Start with a big, empty room, with walls, floor & ceiling that do not reflect 15µm IR very well. Fill the room with dry nitrogen+oxygen+argon in approximately the usual ratio, but containing no CO2 or water vapor. (It needs to be at a well-regulated temperature; and make sure the walls/ceiling/floor are at the same temperature as the air.)

(Note: the special GHG-free atmosphere is for your calibration run. If you can’t avoid walking around in the room, then you’ll need to use some sort of respirator arrangement, to avoid contaminating the CO2-free air with your breath.)

Shine a strong (though invisible), well-collimated 15µm IR beam the length of the big room. (A laser would be ideal, but not strictly necessary.)

Using a narrowband, highly directional detector, tuned to measure 15µm, and aimed at the beam source, measure the beam strength at, say, one meter intervals, the length of the room.

Since there are no GHGs in your room, you should see little diminishment of the beam strength with distance. (If you’re not using a laser, you’ll presumably see some spreading.)

Turn off the beam and repeat the experiment (to measure the “background” 15µm, from the walls/ceiling/floor).

If the measured 15 µm IR intensities are not at least two orders of magnitude lower when the beam is switched off, then your beam was not strong enough, or your sensor is insufficiently directional, or both.

With the beam switched back on, measure the µm IR intensities a few inches outside the beam. If the measured intensities outside the beam are not similar to the measurements your observed when the beam was switched off, then your walls/ceiling/floor are too reflective.

Now, replace the special atmosphere with a more realistic one, and repeat the experiment.

If the mean free path is 33 meters, then you should see a halving of the beam intensity at about 26 meters from the source.

Easy, eh?

Robert W. Turner
Reply to  Dave Burton
June 25, 2019 7:10 am

I think the scale is probably what has prevented it from being done, you’d need a very large space to conduct this experiment.

Here’s a paper that I will digest some more when I get time.
https://arxiv.org/pdf/1703.07114.pdf

Reply to  Dave Burton
June 25, 2019 12:28 pm

You could do a “scaled” version of the experiment in a smaller space, by increasing the concentrations of the radiatively active gases.

Reply to  Robert W. Turner
June 24, 2019 7:58 pm

Typo correction:

    “the µm IR intensities”

should be:

    “the 15µm IR intensities”

tty
Reply to  Dave Burton
June 25, 2019 10:04 am

As I heard it this was the original reason MODTRAN was developed. The military use IR extensively both for reconnaisance, night vision and homing missiles and needed a program that could estimate how far it is possible to “see” in various parts of the IR band.

By the way it is certainly possible to build a “FLIR” working in the CO2 band, but it would be like walking in a fairly dense fog.

Reply to  Robert W Turner
June 25, 2019 8:39 am

The mean free path of a 15 micrometer photon emitted at surface conditions is only 33 meters, not in kilometers.

Perhaps in the thermosphere.
At the surface the mean free path of air molecules is on the order of microns, not meters.

tty
Reply to  Johanus
June 25, 2019 9:55 am

Photon not molecule

bwegher
June 24, 2019 1:10 pm

Actual solar power reaching Earth’s sunlit side is 1360 watts per square meter.
About 400 watts per square meter is reflected (aka Albedo)
The amount absorbed by Earth is 960 watts per square meter, that is only on the sunlit side
That 960 over 24 hours for the total sphere of Earth is divided by 4. (Disk area is 1/4 of surface sphere area)
960 divided by 4 is 240 watts per square meter.
NOAA satellite data is showing 240 watts per square meter. About 238 before 1998, jumped to 242 since 1998.

This is plotted from the KNMI explorer
http://climexp.knmi.nl/selectfield_obs.cgi?someone@somewhere

Scroll down to the “OLR” line and select the 1979-now mark. Then scroll back up and hit “Select field”
The NOAA/UMD OLR page will appear, just select the “Make time series” button to see the plots.
At one time Humlum’s climate4you site kept this plot updated.

Farmer Ch E retired
June 24, 2019 1:24 pm

Note to self: don’t trust output from a model that has difficulty accounting for water vapor.

tty
June 24, 2019 1:27 pm

“there are two assumptions regarding water vapor that a user can choose–constant relative humidity or constant mixing ratio. ”

Both are unrealistic in the real world. The amount of water in the atmosphere does, on average, increase with temperature but not, on average, enough to keep relative humidity constant.

This assumption of constant relative humidity with rising temperatures sneaked into climate science back in the seventies and has somehow survived to the present, though a cursory look at real climatology immediately shows it to be incorrect.

Kevin kilty
Reply to  tty
June 24, 2019 3:58 pm

Let me state my take on this. People will tell you that they condition all their efforts on constant relative humidity because of Clausius-Clapeyron. However, Clausius-Clapeyron is a relationship that obtains at thermodynamic equilibrium.

At practically no place does thermodynamic equilibrium pertain to the real atmosphere. If it did there would be no lateral or vertical heat transport. For another thing, the atmosphere is almost never in equilibrium with open bodies of water, as the vapor pressure a short distance above the surface is below saturation. If it weren’t there would be no evaporation. The actual RH of the atmosphere, in a manner similar to temperature, is a result of transport state that varies all over the planet, and Clausius-Clapeyron is not a transport equation. There is no physical basis for the insistence on constant RH. People are insisting on applying a equilibrium thermodynamic state to a planet not in thermodynamic equilibrium.

Michael S. Kelly LS, BSA Ret.
Reply to  Kevin kilty
June 24, 2019 7:17 pm

The whole idea behind “global warming” is a buildup of energy in the earth system due to a reduction in the amount of infrared radiation that the atmosphere will allow out of the system. The data used to substantiate the theory are, ultimately, atmospheric temperature measurements. The presence of water in this system renders temperature alone as meaningless – just ask any HVAC engineer.

Say that a weather station measured 80 F on July 1st in 1919, and 90 F on July 1st 2019. One might conclude that the globe was warming at a rate of 0.1 F per century. But if the humidity in 1919 had been 50%, and in 2019 had been 20%, the energy content of the air on those two days would have been exactly the same. The fact that we have not been tabulating both wet and dry bulb temperatures at the same time every day for all of these years renders the entire “science” of global warming meaningless. And there is no way to go back and correct for it.

Reply to  Kevin kilty
June 24, 2019 9:02 pm

Kevin,

The atmosphere must be in thermodynamic equilibrium, otherwise, it would heat or cool without bound. Local regions can be warming or cooling, but averaged across the planet, these offset and the average across the planet is in thermodynamic equilibrium. Calculating a sensitivity or any other average metric when assuming any part of the system is not in thermodynamic equilibrium is meaningless.

Thermodynamic equilibrium of the atmosphere means that the input Joules are equal to output Joules. The atmosphere has 2 inputs and 2 outputs, where one input and output is to/from space and the other pair is to/from the surface. How the Joules stored within the atmosphere are dynamically rearranged doesn’t affect what the equilibrium state must be, even as the rearrangement of energy within the atmosphere may change the rate progressing towards that equilibrium state.

The only factor that matters is the ratio between the steady state net flux to/from space and the steady state net flux to/from the surface. The flux to.from space is purely radiant and about 240 W/m^2. The flux to/from the surface has both radiant and non radiant components. If all we care about is the surface temperature, then the only component of the surface flux that matters are the radiant emissions of that surface at its temperature per the SB Law.

In the steady state, we need to offset these SB emissions which means that the non radiant flux entering the atmosphere fom the surface must be returned to the surface, either non radiantly or radiantly, in order for the surface/atmosphere boundary to be in balance.

The relevant ratio is about 1.62W/m^2 of surface emissions entering the atmosphere per W/m^2 of post albedo solar input entering at TOA. This must apply to all W/m^2, including the next one, as the planet can’t tell one W/m^2 from any other, moreover; the data clearly shows that this ratio is independent of the surface temperature, the flux at TOA or the flux at the surface.

The 1.62 W/m^2 of surface emissions per W/m^2 of forcing is is the net average result after clouds, GHG’s, aerosols, feedback and whatever else we don’t know about have had their complete average influence on the result. In order to see this measured behavior with MODTRAN, you must account for clouds.

Kevin kilty
Reply to  co2isnotevil
June 25, 2019 6:47 am

You are confusing energy balance with thermodynamic equilibrium. You also admit that conditions vary from one place to another. That is not thermodynamic equilibrium.

Reply to  Kevin kilty
June 26, 2019 7:39 am

Kevin,

There’s no confusion on my part. The energy balance is only relevant when the system is in thermodynamic equilibrium. You seem to be confusing the average state of the system with the instantaneous state of a small part of it. The average atmosphere is most definitely in thermodynamic equilibrium and the average state is all that matters relative to the average energy balance and the average sensitivity. For all intents and purposes any transient imbalance in a small part of the atmosphere is irrelevant to the steady state equilibrium. Besides, the transient imbalances tend to cancel when all of the instantaneous imbalances across the planet are averaged together.

If you send a sine wave to an RC circuit, all of the nodes always changing and by your definition is not in equilibrium, but its behavior is the very definition of being in a STEADY STATE equilibrium. You’re missing the concept that a steady state thermodynamic equilibrium doesn’t mean that all parts of the system are static and unchanging. For the Earth, what it means is that when averaged over periods of the stimulus (whole years for the climate system), the average input and output energy to/from the system are the same.

tty
Reply to  co2isnotevil
June 25, 2019 10:23 am

No, the atmosphere is not in thermodynamic equilibrium, neither in any particular place, nor as a whole. If it was there would be no weather and no climate change.

And on a water planet like Earth it could only be in thermodynamic equilibrium if the ocean also is. Which it isn’t.

It would also require a planet in an absolutely circular orbit, no inclination and a bound rotation, and an absolutely stable sun.

Reply to  tty
June 26, 2019 7:50 am

tty,

You’re also confusing the static equilibrium of a system with an unchanging input with the dynamic steady state equilibrium of a system forced by a periodic stimulus.

The former has constant state, while the later has a constant average state when integrated over the period of the stimulus. BTW, the elliptical orbit of Earth has no effect on this. The energy from the Sun is still constant when averaged across multiples of its period (years).

Adapting to small changes in the Sun or small changes in the system when superimposed on the very large seasonal response per hemisphere generally occurs within a single period of the much larger seasonally variable stimulus and has little, if any effect, on the actual steady state equilibrium.

Farmer Ch E retired
Reply to  tty
July 7, 2019 8:08 am

co2isnotevil & tty – nice discussion – reminds me of micro vs. macro economics (my undergrad minor). This disagreement between two knowledgeable individuals about basic thermodynamic equilibrium does not give me a “warm and fuzzy” about the models.

RW
June 24, 2019 1:54 pm

Kevin,

There is indeed so much confusion about this. Before you double CO2, you have a fraction of the Planck Flux emitted by the surface which is absorbed by the atmosphere, which we’ll call absorption ‘A’. The remaining is transmittance ‘T’, which is the fraction of the Planck flux emitted from the surface that is transmitted directly into space (the same as if the atmosphere wasn’t even there). Instantaneously doubling CO2 results in an increase in ‘A’, which when converted into watts, amounts to about 3.7 W/m^2, or an instantaneous TOA deficit of -3.7 W/m^2. There is no change in IR intensity at the surface applied to system for an instantaneous doubling of CO2.

Now, when you add GHGs to the atmosphere, like instantaneously doubling CO2, each atmospheric layer absorbs a little more from above and below and subsequently emits a little more upwards and downwards, which then changes the intensities at the surface and the TOA. However, this effect can’t be simulated (in the instantaneous steady-state) without creating new energy from nothing, so the result of the simulation actually increases the DLR at the surface and decreases the OLR at the TOA (which violates COE). The net difference is then applied to the system to satisfy COE. If for example, it’s -6 W/m^2 at the TOA (i.e. an increase in ‘A’ of 6 W/m^2) and +2.3 W/m^2 at the surface, the net difference is an upward absorption increase of 3.7 W/m^2, or an increase in ‘A’ of 3.7 W/m^2. This is often referred to as the ‘net absorption increase’ or the ‘net increase in IR optical thickness from the surface looking up to the TOA’.

As Dick Lindzen once said regarding the 3.7 W/m^2 from 2xCO2, ‘it’s simply a change in the up of IR’ (or something like this).

Kevin kilty
Reply to  RW
June 24, 2019 3:26 pm

If 269W/m^2 what is required to achieve conservation of energy at the top of the atmosphere, then no matter what you do to the system internally (more GHGs or whatever) when equilibrium is achieved again the upwelling LW is going to be 269W/m^2. MODTRAN cannot calculate conservation of energy within the atmosphere or on the ground because it is not a heat transfer code. However, use a standard u.s. atmosphere 1976 version, use a ground surface temperature of 288.2 and a CO_2 proportion of 280, and you get 269 W/m^2. That’s it. You can postulate all you want about this being too much irradiance, or that the surface is too warm or whatever. But you are now dealing with different problems than I considered.

RW
Reply to  RW
June 24, 2019 5:33 pm

Kevin,

The instantaneous case for 2xCO2 is done in the steady-state, i.e. no heating or cooling taking place. This sets a COE limit that the energy going into the atmosphere cannot exceed the amount of energy exiting the atmosphere, whether radiant or non-radiant (at the surface).

There are no significant sources of energy to the system aside from that which is entering the system radiantly from Sun, and the surface radiates back up into the atmosphere the same amount of (net) flux entering the surface from the bottom of the atmosphere (assuming an emissivity of 1 or near 1); and EM radiation is all that can pass across the system’s boundary between the atmosphere and space. This means the entire energy budget of the Earth-atmosphere system is all EM radiation.

This is why the Planck flux emitted directly from the surface is the measure of upward IR opacity through the whole of the atmosphere, including the emission that originates from the atmosphere. It’s established by COE. In the steady-state, balance at the surface already exists (however it’s being manifested); therefore you can’t have additional upwardly absorbed IR from the surface and additional DLR at the surface, for the instantaneous 2xCO2 case, as this violates COE. That is, it creates energy from nothing, and therefore has to be subtracted to get the net change in absorption.

RW
Reply to  RW
June 24, 2019 6:26 pm

Kevin,

This would be kind of analogous to light passing through a semi-opaque medium in the steady-state in that simultaneously more light can’t be absorbed by the medium and transmitted through the medium than is supplied into the medium in the first place, as that would violate COE (assuming the medium has no alternate energy source). The difference is the atmosphere is effectively both supplied light externally and internally re-emits absorbed light at the same time, but it none the less can’t have more light, i.e. energy, absorbed and transmitted out its opposite side (the TOA) than is being supplied in from the one side (the surface) at the same time, i.e. ‘instantaneously’ or in any one instant.

Therefore, you can’t have additional upward IR absorption in the atmosphere along with increased DLR at the surface, for the instantaneous case, because more energy is now coming out of the atmosphere and going into the surface than was prior to the change, which directly violates COE. This is why difference between the two, or the ‘net absorption increase’ is taken to arrive at the final calculation that is applied to the system for the 2xCO2 case.

The critical point is, in the steady-state (for 287K), all power in excess of 385 W/m^2 incident on the surface has to be exactly offset by power in excess of 385 W/m^2 leaving the surface, and that the surface specifically emits 385 W/m^2 of radiant power solely due to its temperature (and emissivity, which is really close to 1). Moreover, all non-radiant power leaving the surface has to be in excess of the 385 W/m^2 directly radiated from the surface, otherwise the surface temperature would be higher, where as there is no such requirement for the proportions of radiant and non-radiant power incident on the surface from the atmosphere.

June 24, 2019 2:15 pm

“The new forcings are fully observed at the surface, not at top of atmosphere, which seems reasonable to me.”

That might make sense if you were discussing a planet without an atmosphere. But asking for sense – is in my experience a complete waste of time as almost everything in this “subject” is just what someone fancied doing rather than anything thought out.

June 25, 2019 1:16 am

The vast majority of energy into Earth’s climate system is via tropical oceans.

The average OLR inferred by top of atmosphere measurements is 240W/sq.m. in significant portion of the tropical western Pacific. The SST is around 300K.

Have a go at getting MODTRAN to produce 240W/sq.m above a 300K surface. The folly is relating any OLR calculation to a US Standard Atmosphere. Such conditions occur over land and land has negligible contribution to the energy in the global climate system.

The global surface temperature is primarily a function of the extent and distribution of water over the globe; its connectedness and circulation providing heat distribution. There are very powerful negative feedbacks to control surface temperature that make any influence from CO2 negligible.

kribaez
June 25, 2019 2:46 am

Kevin,
A thoughtful and well-written article.
I will prefix my comments by saying that I do not know for certain why this version of MODTRAN yields such a low TOA value, but I think I can explain a large part of your discrepancy with Hansen’s estimate.

There are three IPCC definitions of forcing type:-

Instantaneous Radiative Forcing (IRF or Fi) which evaluates the instantaneous net flux change AT THE CLIMATOLOGICAL TROPOPAUSE while keeping the temperature fixed everywhere.
Adjusted Radiative Forcing (RF or Fa) which evaluates the net flux change at the climatological tropopause while keeping the tropospheric temperatures fixed but allowing stratospheric temperatures to stabilise.
Effective Radiative Forcing (ERF) which evaluates the net flux change at TOA while allowing temperatures and saturations to change throughout the atmosphere, but keeping sea-surface temperatures and sea-ice conditions held fixed and allowing the run to re-establish a steady-state.

Generally Fi is greater than Fa is greater than ERF.

So the first thing I would note is that definitionally the reference elevation for Fi is at the tropopause and not the top of the atmosphere. The IR from surface is inhibited by the additional CO2 in the RTE calcs upto the top of the troposphere, but the additional CO2 in the stratosphere in your MODTRAN runs has the effect of increasing radiative flux out to space in a relative sense from the tropopause upwards. Hence, if you rerun your calcs but with a lookdown value at the top of troposphere, you will find the net forcing value should be greater than the net TOA value obtained from your 70km lookdown. A quick test on MODTRAN gives me a forcing of 3.5 for the Standard US Atmosphere, and 4.6 for “Tropical” (using 11kms and 18kms respectively as estimates of tropopause elevation).

Secondly, Hansen’s estimate of 4.52 W/m2 was unambiguously an IRF or Fi value, and you therefore cannot play around with rapid temperature adjustments in an attempt to explain the apparent discrepancy.

The third thing I would note is that there are various papers out there dealing with estimation of ERF values for a doubling of CO2 which make direct use of a “Gregory plot”. One such paper is the 2013 paper by Forster et al, called “Evaluating adjusted forcing and model spread for historical and
future scenarios in the CMIP5 generation of climate models”. [To add some confusion, Forster uses the term “adjusted forcing” or “AF” values, whereas his estimates, since they incorporate fast feedbacks, are probably much closer to IPCC’s definition of the ERF forcing species.] The interesting thing is the spread of values obtained from the MODTRAN calculations run by each of the GCMs. These range from 2.66 to 4.31 W/m2 for a doubling of CO2, a massive range given that they should all be sourced from basically the same calculation procedure benchmarked to LBL code!!

Fourthly, I would just note that the GISS-E series of models (Hansen’s basis for calculation) come out towards the top end of the range for estimates of Fa values and Fi values. I doubt if a value of 4.5 W/m2 is a median estimate of Fi.

I wish you well in your further researches.

You wrote:-
“If we now disturb this by suddenly doubling CO2, MODTRAN keeps the surface temperature and the atmospheric temperature distribution constant, and calculates a new irradiance at the top of atmosphere of 266.4W/m2. The difference of 3W/m2 is the new forcing–or almost. MODTRAN keeps all sorts of things constant, but in a real atmosphere and surface several of these things cannot remain constant. “

Kevin kilty
Reply to  kribaez
June 25, 2019 7:36 am

I suppose there are even more versions of forcing one can define. I swear that climate scientists are their own worst enemies. They proliferate definitions and complicate the message. Some are also obnoxious.

I don’t think the output from my models is low. Another person posting indicated they were too high. They are what they are considering how I conditioned the problem.

Thanks for your well-wishes. My current research has nothing to do with climate. It is now completed work from my perspective, but is apparently unpublishable. I have been rejected by four journals. It is a subject area more contentious, and even more filled with intrigue and pettiness than climate change–if such is possible. Somehow the paper keeps escaping the confidentiality of “peer-review” and getting into the hands of third parties who then begin agitating that the journal should reject the work. The academic world is about as nasty as one can be.

kribaez
Reply to  Kevin kilty
June 25, 2019 10:09 am

Kevin,
I can’t help thinking that you missed the gravamen of my previous comment. It was not to demonstrate my knowledge of definitions of forcings. It was to explain why you (and Willis previously) are comparing bananas and apples, specifically, your TOA estimate of instantaneous forcing cannot be directly compared with Hansen’s instantaneous forcing. If you didn’t get this, please read it again. My apologies if you did already understand this.

I am sorry to hear about your frustration in trying to get your work published. You must persevere if you believe you are right, and stop when you know you are wrong. Some paradigms only change when the old guard are carried out.

Kevin kilty
Reply to  kribaez
June 25, 2019 10:56 am

I understand the apples vs. oranges. It was some of my point as well. Your post just surprised me at the variety of fruit involved. Sometimes I do not acknowledge some things too explicitly. Thank you, however for the new information, the details of which I was not aware.

Re: the old guard…Planck once said, “Physics advances one funeral at a time.”

tty
Reply to  Kevin kilty
June 25, 2019 10:26 am

Arxiv?

tty
Reply to  Kevin kilty
June 25, 2019 10:41 am

“The academic world is about as nasty as one can be.”

Unfortunately too often true. That is one of the reasons I left it and got into aerospace instead. Something I have never regretted by the way. Though it has always struck me as a bit odd that people that develop, build, procure and use deadly weapons are on the whole a lot more pleasant and trustworthy than people whose job is supposedly to spread knowledge and enlightenment.

Steven Mosher
June 25, 2019 3:41 am

Thanks Kevin

Long ago MODTRAN ( the engineering version from SSI) used to be classified

here
http://modtran.spectral.com/modtran_about

See that F22 flying?

designed with Lowtran and modtran

http://modtran.spectral.com/modtran_faq

Kevin kilty
Reply to  Steven Mosher
June 25, 2019 6:58 am

Yes. I have read that MODTRAN has a technical credibility of 9, meaning it is well tested and highly reliable. Some people on this thread seemed to think I was pointing out problems with MODTRAN, which I was not. I do wish there was a more capable public portal into MODTRAN than the hobbled version at U.of Chicago, but I suppose SSI would not want to take the risk of dropping its bread butter side down.

Having said that a person can do a lot with the hobbled version to explore interesting questions.

Steven Mosher
Reply to  Kevin kilty
June 25, 2019 9:43 pm

a lot of people misuse it as a method of calculating the effect of doubling C02.

and they think they discovered something.

Reply to  Steven Mosher
June 26, 2019 8:24 am

MODTRAN is good for what it does.

Unlike GCMs, MODTRAN doesn’t try to do things which can’t be done well. It does not attempt to model the whole climate system. It does not attempt to model wind and water circulation, or biological feedbacks.

But if you want to compare the warming effect in the tropics with specified cloud conditions of an increase in CO2 to an increase in CH4, or to the same increase in CO2 or CH4 plus the amplifying effect of additional water vapor (approximated by specifying constant relative humidity), MODTRAN can do that.

Some people pooh-pooh MODTRAN because it’s a “model,” and climate models are so notoriously bad. But computer models can be either good or bad.

A “computer model” (or just “model”) is a computer program which simulates (“models”) real processes for the purpose of predicting their progression. The utility and skillfulness of models is dependent on how well the processes which they model are understood, how faithfully those processes are simulated in the computer code, and whether the results can be repeatedly tested so that the models can be refined.

Specialized models, which try to model reasonably well-understood processes, like PGR and radiation transport (like MODTRAN), are useful, because the processes they model are manageably simple and well-understood. Weather forecasting models are also useful, even though the processes they model are very complex, and not well-understood, because the models’ short-term predictions can be repeatedly tested, allowing the models to be validated and refined.

But more ambitious models, like GCMs, which attempt to simulate the combined effects of many poorly-understood processes, over time periods too long to allow repeated testing and refinement, are of dubious utility. (Worst of all are so-called “semi-empirical models,” which aren’t actually models at all.)

Max Stavros
June 25, 2019 7:35 am

Figure 1 displays planck functions and calculated OLR by wavenumber, which is known distort the longer wavelengths. Thus CO2 absorption at 667 cm-1 appears to block a significant proportion of OLR These functions should be plotted against waavelength, where maximum OLR occurs near 9 microns, close the Ozone absorption band.

Mike Crow
Reply to  Max Stavros
June 25, 2019 1:25 pm

Right in the middle of the optical window.

Kevin kilty
June 25, 2019 7:45 am

If anyone is still looking at this thread, PhilinCalifornia, reminded me of the Wijngaarden and Happer work on radiation calculations. I called Wijngaarden about a year ago inquiring about what I could read, and he sent a pdf of a poster they had made for some presentation. That poster is online here. and is well worth a look.

In effect they claim the voigt model of the wings of spectral lines is a poor approximation and leads to error of tens of percent in the calculation of radiant intensity. Considering that forcing from doubling CO2 is in the neighborhood of 5 W/m^2 these are significant issues.

bwegher
Reply to  Kevin kilty
June 25, 2019 10:14 am

Direct link to pdf
https://ams.confex.com/ams/15CLOUD15ATRAD/webprogram/Handout/Paper342816/AMSVancouver2018poster.pdf

Poster conclusion
Voigt lineshape overestimates radiative forcing by tens of percent. Earth’s surface temp. increases by ∼1.4 C due to doubling CO2, CH4, N2O & assuming constant relative humidity. Future work should include clouds which can warm or cool. Warmer surface may increase convection enhancing heat transport from Earth’s surface & is a negative forcing

tty
June 25, 2019 10:12 am

Actually that MODTRAN interface has its uses for demonstating basic radiative physics.

Run the standard version with 400 ppm CO2 as background, then double the CO2 and re-run the model. The difference is barely visible

Do the same and double the amount of H2O instead.

Quite a difference eh?

Brett Keane
June 27, 2019 12:28 am

Robert W Turner: Thanks for your well-grounded scientific expositions, not needing specious models or equatons. Pretty much what I learnt from the works of Maxwell etcwho did the experiments too.
I have regarded the simple fact of EMF being a Vector Force to see of any claims of “back radiation” and wonder if you could comment on this. They claim it ‘does not know (or care?) from whence radiation comes, as if that matters, when it is the force balance that decides what quantum oscillators accept. Same as in Mechanics, as far as I can see? Brett Keane.

Brett Keane
Reply to  Brett Keane
June 27, 2019 12:55 am

RWT oops – see off not of. Brett

Frank
July 3, 2019 2:24 am

Kevin: Modtran requires users to input the exact temperature, density, and composition of the atmosphere at all altitudes, but the UofC wrapper is does most of this work for users. Normally one inputs conditions appropriate for today’s climate which varies with latitude and season. There are several definitions of forcing: instantaneous, change at the tropopause, and at the TOA after stratosphere has adjusted to the change in CO2 (which would require about two months in the real world). The stratosphere is in radiative equilibrium with the radiation passing through it, so at equilibrium the upward flux leaving the troposphere and leaving TOA should be equal. I think this version of Modtran probably reports the instantaneous change at the TOA – which is different from the definition of forcing used by the IPCC.

Your post doesn’t mention clouds, but they are extremely important in the real world. Using a clear US Standard Atmosphere, OLR is about 270 W/m2, not the global average of 240 W/m2, partially because there are no clouds. When clouds are included, OLR presumably originates at the top of the clouds you choose and DLR originates at the bottom of the clouds you choose.

3.7 W/m2 is what the best study obtained using line-by-line methods for the change at the tropopause when you combine the output for a an appropriate variety of latitudes, seasons and cloud covers typical of the Earth.

Hansen’s value of 4.5 W/m2 for the forcing from doubled CO2 comes from an AOGCM, which calculates its own atmospheric temperature profile, composition, density and clouds at all altitudes after CO2 has doubled. This is called the effective radiative forcing (ERF) after the troposphere (and stratosphere) have adjusted to more CO2 but the surface and SSTs have not changed. ERF is a different quantity from simple calculations that don’t allow any change in the troposphere in response to more CO2. ERF for 2XCO2 for CMIP5 models ranges from 2.3 to 4.4 W/m2 and averages 3.45 W/m2. The current ERF for the GISS model is 3.8 W/m2. EBMs rely of ERF’s.

https://www.ipcc.ch/site/assets/uploads/2018/02/WG1AR5_Chapter09_FINAL.pdf Table 9.5

When amateurs like me mess around with Modtran, I have significant doubt about the meaning of the quantities I obtain. However, the trends and direction of change seems pretty reliable. How does clouds effect forcing? What happens if I raise the temperature 1 degC and keep relative humidity constant in the tropics vs temperate zones? How does a 7% increase in water vapor change OLR?