From jbh@IDT.NET Thu Nov 20 08:43:55 EST 1997
Article: 159639 of sci.environment
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From: Joshua Halpern <jbh@IDT.NET>
Newsgroups: sci.environment,sci.physics
Subject: Re: radiative forcing of CO2
Date: 20 Nov 1997 05:49:26 GMT
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In sci.environment Onar Aam <onar@hsr.no> wrote:
: >While the most detailed discussions of collisional energy transfer 
: >requires specifying the molecular system, effectively
: >thermalization of IR excitations requires 10-100 collisions.
: >at most.  At room temperature the average collision time 
: >per Torr is about 1E-7 second.  This is much faster than 
: >the time characteristic of IR emissions

: But the transition to equilibrium is not complete when the CO2 has 
: lost all its surplus energy. The heat doesn't just disappear. 
: The cooling of CO2 naturally causes nearby molecules to warm up. 

Well... First of all what is the equilibrium concentration of
vibrationally excited CO2?  Small, eh.  How fast is this dumped
into the bath.  Fast, eh.  What is the ratio of emission to
quenching.  About 1/1000, small, eh.  This means that although
you could treat the problem as a single problem, you can also
accurately treat the radiative part of the problem in the 
following manner:

1.  Measure the light intensity as a function of wavelength into 
    a unit volume element.  
2.  Calculate how much of that light is absorbed by H2O, CO2, etc.
3.  Assume that the energy is thermalized in that element
    (since everything nearby is undergoing the same processes,
    we will neglect energy flow between neighboring elements
    for this problem.  Eddy diffusion will be considered separately)
4.  Calculate the equilibrium concentration of molecules in
    vibrationally excited states as a function of T
5.  Based on measured radiative lifetimes, determine the emission
    per unit time per unit wavelength from the unit volume.
6.  Determine temperature by finding T at which energy in 
    equals energy out.
7.  Now add heat flow from the bottom and the top by standard
    methods and do same.

: How long does it take before ALL neighboring molecules have reached 
: equilibrium after an excitation? This is the relevant figure here, not
: how long it takes for the single CO2 molecule to cool. 

No, because the heating of the system due to excitation of one CO2
molecule is small compared to the heat content of the system.  For
example at the total heat content of the system is 7/2kT 
per molecule for diatomics (N2 and O2, which is essentially
everything).  For each CO2 molecule there are about 3000 N2 and
O2 molecules.  About 4% of the CO2 molecules will be thermally
excited at 300 K.  Less at lower temperatures. (The vibrational
frequency of the bend is 667 cm-1, kT at 300 K is about 200 cm-1,
The relative population will thus be exp(-667/200)).  Thus, for
each excited CO2 molecule there will be about 75000 O2 and N2 
molecules.  The total heat energy content of those molecules will
be 7/2 kT x 75000 = 5.25 e6 cm-1.  The relative amount of deposited
energy per excited CO2 is 1.27 e-5 of the total energy.  The
temperature rise per CO2 excited is negligable, which is 
another reason why the various processes can be decoupled.
(I use cm-1 as an energy unit which is standard for spectroscopy
and kinetics.  To obtain the energy in J multiply by hc, and
find out why I use cm-1).

: Furthermore, is
: this total transition time short enough to prevent IR emissions outside
: the CO2 band from heated nearby water molecules? 

Confusing statement.  Thermal degradation is fast enough, and
vibrationally excited molecules are rare enough that one
can neglect vibrational collisional energy transfer between
molecules with allowed vibrational emission. Excited CO2
for all practical purposes cannot vibrationally excite H2O 
by radiation or collisions.  See below for some details.  

Excitation and heating by other greenhouse gases 
is also accounted for in the models, O3, CH4, CFCs, etc.  
Stop trying to teach atmospheric chemists how to suck eggs. 

: Another possible factor
: is the rate of CO2 excitation, or more relevant: the probability as a 
: function of time that a water molecule collides with an excited CO2
: molecule, or something like that. 

This type of process has been studied to death for all sorts of 
pairs.

: If excitation is a rare event then
: we expect this to have no impact on the transitional process. However,
: if the excitation rate is high enough then a water molecule may be
: heated again before it has had the time to cool off from its previous
: high temp collision. 

Although the water concentration is much higher than the CO2
concentration, the lowest vibrational mode of H2O is at 1600 
cm-1, so 

1.  The ratio of vibrationally excited to unexcited H2O is 
3e-4 at 300 K and less at lower temperature.  In addition,
the saturated vapor pressure decreases as a function of 
temperature, so there is not much water vapor in the 
upper troposphere.  

2.  CO2 excited with one quanta in the bending mode or the
asymmetric stretch does not have enough energy to 
vibrationally excite an H2O molecule at 300 K. CO2 excited 
with one quanta in the symmetric stretching mode does,
but this is at such a high energy, that for all practical
purposes this state would not be found excited at 300 K

: In short: this is not a trivial problem. 

It is if you know a little chemical kinetics and spectroscopy.

: It's by 
: no means obvious that the transitional process is 100% efficient, i.e. 
: that no heat will be re-radiated in the window regions right outside 
: the CO2 band. 

CO2 certainly only radiates in CO2 bands (lines).  H2O certainly
radiates only in H2O bands (lines).  The "trick" here is that
you also have to understand the effects of pressure and Doppler
broadening, which lead to molecules at higher altitudes/lower 
temperature/lower pressure being transparent to some light in 
the wings of lines emitted by molecules at lower altitudes/higher
temperature/higher pressures.  One also has to account for
overlaps of absorption lines of different molecules.  You really 
don't believe in quantum mechanics, do you? 

: In fact, from the top of my head it seems pretty unlikely.

Where do you have your head?

: Unfortunately there is no simple way to quickly calculate this process.

Well, Arhennius got pretty close on the back of an envelope.

: It must probably be modeled on computers to be determined. 

Only because of the large number of lines.  However there is a
database of lines and line shapes that make the calculation 
relatively simple (just the radiative part).  

: But we can try to make some crude estimates here. We may e.g. 
: assume that on average the CO2 molecules are distributed a 
: cubic lattice.


: 			CO2----------CO2
: 			 |-            -
: 			 |  -            -
: 			 |    -            -
: 			 |      -CO2--------CO2
: 			 |        |          |
: 			CO2       |          |
: 			  -       |          |
: 			    -     |          |
: 			      -   |          |
: 			        -CO2--------CO2


: If so then every CO2 molecule is separated by 13 non-CO2 molecules along
: three three axes on average. 

For every CO2 molecule there are roughly 3000 other molecules
in the same volume of air.  The probability of any collision of
a CO2 molecule being with another CO2 molecule is simply one in
3000.  Thermalization requires many fewer collisions. (This is
a bit of a cheat. Energy transfer between like molecules tends
to be more efficient).  In general the probability of transfering
and amount of energy dE in a collision follows what is called
an exponential gap law, P(dE) = exp(-dE/kT) where T is the bath
temperature.  There are other functional forms used, but the 
variation is in detail, not in principal.

: Thus, there's definitely a chance of two excitation avalanches 
: to interfere with each other. 

No.  Because the molecules move in three dimensions not one.  
Therefore the correct comparison is not the relative number
of CO2 and other molecules along a straight line, but the
relative number of each in a volume.  

By the way, the term most people use is cascade
not avalanche.  A constant amount of energy cascades down
through the system.  It does not grow larger as would be
implied by the word avalanche. 

: The average emission rate at 15 microns is about 1E-3 seconds. 
: This should indicate that a single avalanche is not sufficient 
: to cause significant emissions from water. The crucial factor 
: then is the rate of excitation of CO2 molecules.
: If we know this factor then we can roughly calculate whether 
: infrared radiation in the +-15 micron range from water molecules will 
: increase significantly.

Forget it.  See above.  

josh halpern