In Support of the A in AGW

So you do believe downward radiation is back radiation...downward radiation is just downward radiation...back radiation is radiation absorbed by an object and then emitted back to the object from which it was emitted....
That is not a belief. That is the definition of back radiation when it is applied to the the GHGs of the atmosphere. Your last sentence is a bit awkward. The earth radiates the full black body spectrum. The GHGs radiate some back to the earth only at their resonance spectra.
and the downward radiation was only measured because it was being received by an instrument cooled to a temperature lower than the atmosphere.
The downward radiation would go down whether there is a detector or not.
.radiation doesn't care about direction..only that it it is moving always from warm to cool
Yes radiation doesn't care about direction. So what principle stops back radiation from going to a cooler object if it doesn't care about direction.
...from less entropy to more entropy...
The entropy of the system of all objects will increase as it should by the second law since during radiation exchange more radiation goes from the hotter objects to the cooler objects. The net radiation imbalance insures the second law is preserved.


The atmosphere produces (imperfect) blackbody radiation. It has some elevated GHG spikes but it still resembles the Planck curve. It would still return energy even if it contained no GHGs.
and yet you can't prove that can you? I've asked and still you can't explain why that doesn't happen above a desert at night. The hottest places during sunlight and extremely cold when the sunlight is gone. hmmmmmmmm.
 
I've asked and still you can't explain why that doesn't happen above a desert at night. The hottest places during sunlight and extremely cold when the sunlight is gone. hmmmmmmmm.

That's easy enough to explain. One of the most potent GHGs is water vapor. That, however, is scarce in desert climates, which is why IR radiation from the desert surface escapes without notably warming up the troposphere, resulting in very low back radiation, which means deserts cool down quickly at night.
 
I've asked and still you can't explain why that doesn't happen above a desert at night. The hottest places during sunlight and extremely cold when the sunlight is gone. hmmmmmmmm.

That's easy enough to explain. One of the most potent GHGs is water vapor. That, however, is scarce in desert climates, which is why IR radiation from the desert surface escapes without notably warming up the troposphere, resulting in very low back radiation, which means deserts cool down quickly at night.


It isn't back radiation...it is the humidity itself heating up as a result of absorbing IR leaving the surface...
 
I've asked and still you can't explain why that doesn't happen above a desert at night. The hottest places during sunlight and extremely cold when the sunlight is gone. hmmmmmmmm.

That's easy enough to explain. One of the most potent GHGs is water vapor. That, however, is scarce in desert climates, which is why IR radiation from the desert surface escapes without notably warming up the troposphere, resulting in very low back radiation, which means deserts cool down quickly at night.
dude, I know about water vapor if you've read other threads. But it still doesn't explain the CO2 molecule not working.

Humidity can be measured, and yet you don't see a CO2 temperature measurement. hmmmmmmmm I'm still waiting.
 
It isn't back radiation...it is the humidity itself heating up as a result of absorbing IR leaving the surface...

And the energy from a warmed up "humidity" at, say, 100 to 500 meters above ground reaches the surface how?

dude, I know about water vapor if you've read other threads. But it still doesn't explain the CO2 molecule not working.

The CO2 molecule is "working", which is why the desert doesn't go into deep freeze upon sunset.

Look, back radiation is combined radiation coming back from all warmed-up GHGs throughout the troposphere. Take one important GHG (water vapor) out, and you get less back radiation, and thus a cooler surface at night.
 
It isn't back radiation...it is the humidity itself heating up as a result of absorbing IR leaving the surface...

And the energy from a warmed up "humidity" at, say, 100 to 500 meters above ground reaches the surface how?

dude, I know about water vapor if you've read other threads. But it still doesn't explain the CO2 molecule not working.

The CO2 molecule is "working", which is why the desert doesn't go into deep freeze upon sunset.

Look, back radiation is combined radiation coming back from all warmed-up GHGs throughout the troposphere. Take one important GHG (water vapor) out, and you get less back radiation, and thus a cooler surface at night.
I guess my term for warm and yours are two different things. There will still be LWIR leaving the surface, it is much less without the humidity in the air. The sand would freeze if there wasn't some LWIR coming from it.
 
It isn't back radiation...it is the humidity itself heating up as a result of absorbing IR leaving the surface...

And the energy from a warmed up "humidity" at, say, 100 to 500 meters above ground reaches the surface how?

It doesn't...energy doesn't move from cool to warm.

The CO2 molecule is "working", which is why the desert doesn't go into deep freeze upon sunset.

No...the atmospheric thermal effect is working...climate sensitivity to CO2 is zero or less.

Look, back radiation is combined radiation coming back from all warmed-up GHGs throughout the troposphere. Take one important GHG (water vapor) out, and you get less back radiation, and thus a cooler surface at night.

There is no back radiation.....energy detected coming from the atmosphere to a cooled instrument isn't back anything...it is energy moving from warm to cool.
 
Mass markets consumer electronics only measure the amount of, and rate of temperature change in an internal thermopile...

And since the temperature changes only because radiation from a colder object is hitting it, your attempted evasion there makes no sense, and you're still left with your crap theory getting debunked.

Idiot....the temperature change resulting from measuring a cooler object is due to the thermopile cooling off at a faster rate...not because energy from the cool object is radiating to the thermopile...
 
I'm not sure what you mean by "very few interactions". The earth radiates in the far IR. N2 and O2 are transparent to that IR. It seems that if it were black body radiation in the IR it could only be through the GHGs. Another way of looking at it is that the emissivity of N2 and O2 is near zero in the IR, whereas the emissivity of the GHGs are high in parts of that region. That's what makes them GHGs. Also I don't think that black body radiation is a term that can be used casually in simple gasses. There are very few vibrational states. There are almost a continuum of vibrational states in a solid because the neighboring atoms are intimately coupled. That coupling promotes a flatter emisivity curve as a function of wavelength which in turn promotes truer BB radiation only for solids or liquids.

What he means is explained in this conversation between Dr Happer, a physicist who has specialised in the study of atomic physics, optics and spectroscopy. He is the Cyrus Fogg Brackett Professor of Physics at Princeton University, and a long-term member of the JASON advisory group, where he pioneered the development of adaptive optics.... Dr Happer's responses are in ALLCAPS

From:David Burton
Sent: Wednesday, November 12, 2014 10:49 PM
To: William Happer
Subject: Another dumb question from Dave

Dear Prof. Happer,

At your UNC lecture you told us many things which I had not known, but two of them were these:

1. At low altitudes, the mean time between molecular collisions, through which an excited CO2 molecule can transfer its energy to another gas molecule (usually N2) is on the order of 1 nanosecond.

2. The mean decay time for an excited CO2 molecule to emit an IR photon is on the order of 1 second (a billion times as long).

Did I understand that correctly? [YES, PRECISELY. I ATTACH A PAPER ON RADIATIVE LIFETIMES OF CO2 FROM THE CO2 LASER COMMUNITY. YOU SHOULD LOOK AT THE BENDING-MODE TRANSITIONS, FOR EXAMPLE, 010 – 000. AS I THINK I MAY HAVE INDICATED ON SLIDE 24, THE RADIATIVE DECAY RATES FOR THE BENDING MODE ALSO DEPEND ON VIBRATION AND ROTATIONAL QUANTUM NUMBERS, AND THEY CAN BE A FEW ORDERS OF MAGNITUDE SLOWER THAN 1 S^{-1} FOR HIGHER EXCITED STATES. THIS IS BECAUSE OF SMALL MATRIX ELEMENTS FOR THE TRANSITION MOMENTS.]


You didn't mention it, but I assume H2O molecules have a similar decay time to emit an IR photon. Is that right, too? [YES. I CAN'T IMMEDIATELY FIND A SIMILAR PAPER TO THE ONE I ATTACHED ABOUT CO2, BUT THESE TRANSITIONS HAVE BEEN CAREFULLY STUDIED IN CONNECTION WITH INTERSTELLAR MASERS. I ATTACH SOME NICE VIEWGRAPHS THAT SUMMARIZE THE ISSUES, A FEW OF WHICH TOUCH ON H2O, ONE OF THE IMPORTANT INTERSTELLAR MOLECULES. ALAS, THE SLIDES DO NOT INCLUDE A TABLE OF LIFETIMES. BUT YOU SHOULD BE ABLE TO TRACK THEM DOWN FROM REFERENCES ON THE VIEWGRAPHS IF YOU LIKE. ROUGHLY SPEAKING, THE RADIATIVE LIFETIMES OF ELECTRIC DIPOLE MOMENTS SCALE AS THE CUBE OF THE WAVELENTH AND INVERSELY AS THE SQUARE OF THE ELECTRIC DIPOLE MATRIX ELEMENT (FROM BASIC QUANTUM MECHANICS) SO IF AN ATOM HAS A RADIATIVE LIFETIME OF 16 NSEC AT A WAVELENGTH OF 0.6 MIRONS (SODIUM), A CO2 BENDING MODE TRANSITION, WITH A WAVELENGTH OF 15 MICRONS AND ABOUT 1/30 THE MATRIX ELEMENT SHOULD HAVE A LIFETIME OF ORDER 16 (30)^2 (15/.6)^3 NS = 0.2 S.


So, after a CO2 (or H2O) molecule absorbs a 15 micron IR photon, about 99.9999999% of the time it will give up its energy by collision with another gas molecule, not by re-emission of another photon. Is that true (assuming that I counted the right number of nines)? [YES, ABSOLUTELY.]


In other words, the very widely repeated description of GHG molecules absorbing infrared photons and then re-emitting them in random directions is only correct for about one absorbed photon in a billion. True? [YES, IT IS THIS EXTREME SLOWNESS OF RADIATIVE DECAY RATES THAT ALLOWS THE CO2 MOLECULES IN THE ATMOSPHERE TO HAVE VERY NEARLY THE SAME VIBRATION-ROTATION TEMPERATURE OF THE LOCAL AIR MOLECULES.]


Here's an example from the NSF, with a lovely animated picture, which even illustrates the correct vibrational mode:

Carbon Dioxide Absorbs and Re-emits Infrared Radiation | UCAR Center for Science Education

co2_absorb_emit_infrared_anim_320x240.gif


Am I correct in thinking that illustration is wrong for about 99.9999999% of the photons which CO2 absorbs in the lower troposphere? [YES, THE PICTURE IS A BIT MISLEADING. IF THE CO2 MOLECULE IN AIR ABSORBS A RESONANT PHOTON, IT IS MUCH MORE LIKELY ( ON THE ORDER OF A BILLION TIMES MORE LIKELY) TO HEAT THE SURROUNDING AIR MOLECULES WITH THE ENERGY IT ACQUIRED FROM THE ABSORBED PHOTON, THAN TO RERADIATE A PHOTON AT THE SAME OR SOME DIFFERENT FREQUENCY. IF THE CO2 MOLECULE COULD RADIATE COMPLETELY WITH NO COLLISIONAL INTERRUPTIONS, THE LENGTH OF THE RADIATIVE PULSE WOULD BE THE DISTANCE LIGHT CAN TRAVEL IN THE RADIATIVE LIFETIME. SO THE PULSE IN THE NSF FIGURE SHOULD BE 300,000 KM LONG, FROM THE EARTH'S SURFACE TO WELL BEYOND A SATELLITE IN GEOSYNCHRONOUS ORBIT. THE RADIATED PULSE SHOULD CONTAIN 667 CM^{-1} *3 X 10^{10} CM S^{-1}*1 S WAVES OR ABOUT 2 TRILLION WAVES, NOT JUST A FEW AS IN THE FIGURE. A BIT OF POETIC LICENSE IS OK. I CERTAINLY PLEAD GUILTY TO USING SOME ON MY VIEWGRAPHS. BUT WE SHOULD NOT MAKE TRILLION-DOLLAR ECONOMIC DECISIONS WITHOUT MORE QUANTITATIVE CONSIDERATION OF THE PHYSICS.]


(Aside: it doesn't really shock me that the NSF is wrong -- I previously caught them contradicting Archimedes: before & after.)

If that NSF web page & illustration were right, then the amount of IR emitted by CO2 or H2O vapor in the atmosphere would depend heavily on how much IR it received and absorbed. If more IR was emitted from the ground, then more IR would be re-emitted by the CO2 and H2O molecules, back toward the ground. But I think that must be wrong.[YES, THE AMOUNT OF RADIATION EMITTED BY GREENHOUSE MOLECULES DEPENDS ALMOST ENTIRELY ON THEIR TEMPERATURE. THE PERTRUBATION BY RADIATION COMING FROM THE GROUND OR OUTER SPACE IS NEGLIGIBLE. CO2 LASER BUILDERS GO OUT OF THEIR WAY WITH CUNNING DISCHARE PHYSICS TO GET THE CO2 MOLECULES OUT OF THERMAL EQUILIBRIUM SO THEY CAN AMPLIFY RADIATION.]


If 99.9999999% of the IR absorbed by atmospheric CO2 is converted by molecular collisions into heat, that seems to imply that the amount of ~15 micron IR emitted by atmospheric CO2 depends only on the atmosphere's temperature (and CO2 partial pressure), not on how the air got to that temperature. [YES, I COULD HAVE SAVED A COMMENT BY READING FURTHER.] Whether the ground is very cold and emits little IR, or very warm and emits lots of IR, will not affect the amount of IR emitted by the CO2 in the adjacent atmosphere (except by affecting the temperature of that air). Is that correct? [YES, PRECISELY. WE HAVE BEEN TALKING ABOUT WHAT CHANDRASEKHAR CALLS AN “ABSORBING ATMOSPHERE” AS OPPOSED TO A “SCATTERING ATMOSPHERE.” ASTROPHYSICISTS ARE OFTEN MORE INTERESTED IN SCATTERING ATMOSPHERES, LIKE THE INTERIOR OF THE SUN. THE BLUE SKY DURING A CLEAR DAY IS AN EXAMPLE OF SCATTERING ATMOSPHERE. VERY LITTLE HEATING OR COOLING OF THE AIR OCCURS WITH THIS “RAYLEIGH SCATTERING.”]


Thank you for educating a dumb old computer scientist like me! [YOU ARE HARDLY DUMB. YOU GET AN A+ FOR THIS RECITATION SESSION ON RADIATIVE TRANSFER. ]

What the professor is saying is that in other words; insofar as moving energy out of the atmosphere, convection rules....radiation is a bit player of such minute proportions that it hardly rates mention.
 
Idiot....the temperature change resulting from measuring a cooler object is due to the thermopile cooling off at a faster rate...not because energy from the cool object is radiating to the thermopile...

According to your stupid theory, everything colder than the detector should display the same "shade", since the thermopiles "reject" all radiation from any colder object. Whether it's a 1 degree or 100 degree difference, your theory doesn't differentiate. It's colder, so the photons don't get absorbed.

Yet that doesn't happen. We see different "shades" of cold on the output of the thermal scanner. The thermopiles are very clearly absorbing different amounts of energy from cold objects that are different temperatures.

Hence, your theory fails.

Oh, that wasn't a "conversation" with Dr. Happer. That was some loon writing a crazy letter. Conversations require that both people talk.

And if you want to know the flaw in your brilliant theory there ... "vibrational modes". That's how GHG molecules can store energy. Collisions between molecules rarely result in molecule losing the energy of that vibration.
 
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What say you wuwei-? Although Happer's numbers are orders of magnitude different than what I have seen elsewhere, the basic principle agrees. Surface IR warms the atmosphere, the atmosphere sends some energy back by blackbody radiation (which was not discussed)
 
Idiot....the temperature change resulting from measuring a cooler object is due to the thermopile cooling off at a faster rate...not because energy from the cool object is radiating to the thermopile...

According to your stupid theory, everything colder than the detector should display the same "shade", since the thermopiles "reject" all radiation from any colder object. Whether it's a 1 degree or 100 degree difference, your theory doesn't differentiate. It's colder, so the photons don't get absorbed.

Yet that doesn't happen. We see different "shades" of cold on the output of the thermal scanner. The thermopiles are very clearly absorbing different amounts of energy from cold objects that are different temperatures.

Hence, your theory fails.

Oh, that wasn't a "conversation" will Dr. Happer. That was some loon writing a crazy letter. Conversations require that both people talk.

And if you want to know the flaw in your brilliant theory there ... "vibrational modes". That's how GHG molecules can store energy. Collisions between molecules rarely result in molecule losing the energy of that vibration.


So what's your answer to that SSDD? If you fall back to only the net energy gets transferred from the SB equation, then what happens to the momentum and entropy that doesn't get created by the gross radiation 'not counted'?
 
What the professor is saying is that in other words; insofar as moving energy out of the atmosphere, convection rules....radiation is a bit player of such minute proportions that it hardly rates mention.

The professor is right for the one narrow aspect of it he covered. Yes, a CO2 molecule will probably loose its excited vibration state through a collision rather than an emission.

What the professor didn't cover is the equipartition theorem as it involves the energy states of CO2 and H2O and other GHGs. For a simple explanation see
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html

There will be equal amounts of energy divided among all the GHG vibration states, rotation states and kinetic energy. Most of the vibration states as shown in the animation will arise from the churning energy of the atmospheric molecules hitting the GHGs and not from the earth's upward LWIR. The upward IR will increase the population of the GHG vibration part of the total energy, and that energy will dissipate through collisions and some LWIR in arbitrary directions.

One way of looking at it is that CO2 can absorb IR, but CO2 excited by collisions will emit most of the IR.

In short, the animation shows a single CO2 molecule absorbing IR and emitting the same energy. In reality an large statistical ensemble CO2 molecules will absorb IR and the statistical ensemble will emit IR, although an individual molecule most likely won't do both in a small time window. Remember through the equipartition theorm there is a tremendous amount of energy in the vibrational states that scatter the LWIR.
 
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What say you wuwei-? Although Happer's numbers are orders of magnitude different than what I have seen elsewhere, the basic principle agrees. Surface IR warms the atmosphere, the atmosphere sends some energy back by blackbody radiation (which was not discussed)
I thought I answered the black body aspect in post 379. Happer's comments were answered in the previous post to SSDD.
 
What the professor is saying is that in other words; insofar as moving energy out of the atmosphere, convection rules....radiation is a bit player of such minute proportions that it hardly rates mention.

The professor is right for the one narrow aspect of it he covered. Yes, a CO2 molecule will probably loose its excited vibration state through a collision rather than an emission.

What the professor didn't cover is the equipartition theorem as it involves the energy states of CO2 and H2O and other GHGs. For a simple explanation see
http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/eqpar.html

There will be equal amounts of energy divided among all the GHG vibration states, rotation states and kinetic energy. Most of the vibration states as shown in the animation will arise from the churning energy of the atmospheric molecules hitting the GHGs and not from the earth's upward LWIR. The upward IR will increase the population of the GHG vibration part of the total energy, and that energy will dissipate through collisions and some LWIR in arbitrary directions.

One way of looking at it is that CO2 can absorb IR, but CO2 excited by collisions will emit most of the IR.

In short, the animation shows a single CO2 molecule absorbing IR and emitting the same energy. In reality an large statistical ensemble CO2 molecules will absorb IR and the statistical ensemble will emit IR, although an individual molecule most likely won't do both in a small time window. Remember through the equipartition theorm there is a tremendous amount of energy in the vibrational states that scatter the LWIR.


I don't have time to look at your link right now but the problem with the equipartition theory is that there is a temperature gradient to the atmosphere. The type and the quantities coming out of the top are different than what is going in at the surface.
 
What say you wuwei-? Although Happer's numbers are orders of magnitude different than what I have seen elsewhere, the basic principle agrees. Surface IR warms the atmosphere, the atmosphere sends some energy back by blackbody radiation (which was not discussed)
I thought I answered the black body aspect in post 379. Happer's comments were answered in the previous post to SSDD.


I re-read 379. I did not realize that you don't believe that objects made of gas, or objects made up of a single element or molecule were exempt from the law that every object emits radiation according to its temperature. Interesting.
 
I don't have time to look at your link right now but the problem with the equipartition theory is that there is a temperature gradient to the atmosphere. The type and the quantities coming out of the top are different than what is going in at the surface.
I don't see a problem. The energies of the various gases will still be partitioned equally. At lower temperatures there will still be equal amounts of energies in each degree of freedom, but of course the energies will be lower.
 
I re-read 379. I did not realize that you don't believe that objects made of gas, or objects made up of a single element or molecule were exempt from the law that every object emits radiation according to its temperature. Interesting.
I believe that a gas will radiate according to it's temperature. A higher rate of collisions will keep the vibration states more strongly populated.

The temperature of a single element or molecule doesn't make sense. You can describe it to have kinetic energy and internal energies, but you need a bunch of them to define temperature. A gas at a particular temperature has a statistical distribution of velocities given by the Maxwell-Boltzmann distribution. You can't define a statistical distribution of one molecule.
 
What say you wuwei-? Although Happer's numbers are orders of magnitude different than what I have seen elsewhere, the basic principle agrees. Surface IR warms the atmosphere, the atmosphere sends some energy back by blackbody radiation (which was not discussed)

Wasn't discussed because it doesn't happen....if it did it would certainly be a central issue...
 
What the professor didn't cover is the equipartition theorem as it involves the energy states of CO2 and H2O and other GHGs. For a simple explanation see
Equipartition of Energy

He clearly stated that while he could't lay his hands on H2O studies at that moment, he was quite certain that the numbers weren't that different for H20...and there is no reason to believe that the rate of decay for the other so called greenhouse gasses is significantly different.

Back radiation does not exist and even if it did, it would be so small as to be unmeasurable...
 

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