How long ago did you decide for yourself that you have it all figured out and decided to stop trying to learn anything new?
From:
William Happer Thu, Nov 13, 2014 at 11:29 AM
To: David Burton
Dear David,
Some response are entered below in square brackets and upper case. Thanks for your interest!
Will
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:
http://scied.ucar.edu/carbon-dioxide-absorbs-and-re-emits-infrared-radiation
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. ]
Talk to Dr. Happer about that zero percent. I suppose he might explain to you how terribly wrong you are....In fact, the above email from him does just that.
