Wednesday, April 17, 2013

This Is Where Eli Came In

One of the useful things the Rabett used to do was to explain what happens to the energy when a molecule, say CO2 (carbon dioxide) although you could also say H2O (water vapor) or CH4 (methane) absorbs light. For the purpose of this post, the photon would be in the infrared region of the spectrum.  This is an evergreen for two classes of bunnies

  1. Bunnies who don't realize that the molecule can also emit light.  This is a popular one amongst organikers and analytical chemists whose experience with IR spectroscopy is in an absorption spectrum for analysis of samples
  2. Bunnies who think that the only way that an excited molecule can get rid of the energy is to emit a photon.  
 For every CO2 molecule there are roughly 3000 2500 other molecules in the same volume of air.  When a CO2 molecule collides with one of the other molecules, almost certainly an oxygen or nitrogen molecule, energy transfer occurs.  Each CO2 molecule can be described as having translational, vibrational and rotational energy and the same is true of the collision partner.  Any collision can in principle change the amount of any of these forms of energy by any amount subject to conservation of energy and momentum.  The probability of this happening depends on the relative translational energy of the collision, the relative orientation of the molecules, their distance of closest approach and the distribution of energy in each of the collision partners prior to the collision.  The detailed study of such effects is called collision dynamics or molecular dynamics. 

Fortunately, we can take thermal averages over many of these variables, either theoretically or experimentally which makes life, theory and experiments much simpler and a hell of a lot less expensive and time consuming.  That sort of thing usually goes under the rubric of reaction (when there is one) kinetics or energy transfer studies when there isn't.

The fate of the energy in a vibrationally  excited CO2 molecule can be thought of as a race between emission of a photon taking essentially all of the vibrational energy away and a collision that does the same.  Here Eli is going to tell you who wins the race, something that is generally handwaved, correctly handwaved, but handwaved none the less.

For the lowest vibrational level, the 667 cm-1 doubly degenerate bend, labelled as (0,10,0), the only lower state is the unexcited ground state (0,00,0)

 CO2(0,10,0) + M --> CO2(0,00,0) + M

where M is any other molecule.  The probability of this happening depends on M, but it can be measured by exciting a small amount of CO2 mixed into a large bath of M and then watching how long it takes for the excited CO2 to decay away.   This behavior is captured by the rate equation

d[CO2*]/dt = - kR [CO2*] - kM[CO2*][M]    (1)

where kR is the radiative emission rate in the absence of collisions and kM is the collisional quenching rate, the rate at which collisions with M de-excite the CO2.  The square brackets indicate the concentration of whatever species is inside.  If [M] is very large compared to [CO2] we have what is called pseudo first order conditions, because the effect of the collision on the quantum state of M will be very small and Eq. 1 is a first order differential equation whose solution is

[CO2*] =  [CO2*]t=0 exp (-{kR  + kM[M]}t)

where t is time and we are assuming that the CO2 is being excited during a very short time so we can neglect the competition between excitation and de-excitation which simplifies the kinetic analysis.

The best study of this process for CO2 was The vibrational deactivation of the (00o1) and (0110)Modes of CO2 measured down to 140 K by Siddles, Wilson, Simpson, Chemical Physics 189 (1994) 779-91.  They measured collisional quenching cross-sections for CO2 (0,10,0) and the higher lying (0,00,1) levels against a number of gases, including carbon dioxide, the rare gases, nitrogen, oxygen and deuterated methane and deuterium.  Oxygen and nitrogen have quenching rate constants of 5.5 and 3.1 x 10-15 cm3/molecules-sec respectively at 295K, decreasing roughly linearly to about a factor of five less at 140K.  To make sense of this number, we can compare it to what is known as the gas kinetic quenching rate, e.g. what the rate would be if every collision were 100% effective, of the order of 10-10 cm3/molecules-sec.  More to the point, at atmospheric pressure where the concentration is ~2.6 x 1019 molecules/cc (Loschmidt's number)
the rate of collisional de-excitation for CO2(0,10,0) by N2 will be

kM[M] ~ 5 x 10-15 cm3/molecules-sec x 2.5 x 1019 molecules/cm3 ~ 105 s-1

Note the ~ meaning oom or order of magnitude.  It's not that Eli can't afford a calculator, it's that who needs one in this sort of calculation when toes and ears are available.  The lifetime will be the reciprocal of this, 10-5 s or 10 us.

What about the radiative lifetime?  Radiative lifetimes of vibrationally excited states are long.  Insight into why this is so can be found by looking at the relationship between the Einstein A and B coefficients.  Roughly put, the A coefficient is proportional to the rate of spontaneous emission and the B coefficient to the rate at which photons can be absorbed by the molecule.  assuming that the transitions are between two isolated levels

A/B = 8 π h ν3/c3

 The ν3 is what makes emission from low lying vibrational levels so slow.  The isolated radiative lifetime of CO2(0,10,0) is ~1.1 s.  The (0,00,1) lifetime is 2.4 x 10-3, but remember it occurs at 2350 cm-1 where the ν3 factor can do its work.

Comparing the radiative rate kR (the inverse of the lifetime) to the collisional deactivation rate kM[M], provides a quick estimate that only one out of 100,000 CO2 molecules excited into the (0,10,0) by collision or absorbing a photon, will emit.


Anonymous said...

Thanks Eki, good explanation.

When this question was asked by the watties a while back I had to sit quietly for a while to allow ancient knowledge to percolate to the surface. I explained it to them but it did no good. They can't actually tell the difference between people who know what they are talking about and people who don't. The clues are just too subtle.

There is a chemist over at watt's now being simultaneously wrong and patronising on extremely basic stuff. I suspect he would not be able to correctly answer this simple CO2 question.

The climate ferret

Martin Vermeer said...

Put into astrophysics jargon, the lower Earth atmosphere is in LTE (Local Thermodynamic Equilibrium), AKA "temperature means something". Right?

toto said...

Am I naive in inferring that the strength of the "greenhouse" effect is proportional to the "radiative lifetime" (or inversely proportional to "kR"), all other things being equal?

Geoff said...

Eli, I have to admit this is beyond me, so maybe you answered this in the post and I just didn't follow it....

I'm confused about the excitation state of the electrons. As I understand it absorbing a photon pushes an electron into a higher state; when the electron decays back to the original state an equivalent photon is emitted. But when a collision imparts energy to the molecule, any transfer of energy manifests itself in a change in momentum (billiard ball analogy) and the electrons do not change state.

Maybe this is too naive an understanding, but assuming it has some validity, I don't see how energy absorbed by a change in electron state can be transferred in a collision without the electron changing state again and hence emitting a photon.

Anonymous said...

Note that in the upper stratosphere, where collisions are much less frequent, the difference between collision lifetime and radiative lifetime is not nearly so stark, so that excited CO2 molecules DO sometimes emit radiation... combine that with the fact that much of the upper stratospheric heat comes from absorption of solar radiation by ozone, and increasing CO2 levels can lead to increased cooling. E.g., Gilbert Plass, 1956.


Anonymous said...

Geoff: There is a tighter coupling between vibrational states and electron states than you might realize. For example, the reason why GHGs are GHGs is because they have asymmetric vibrational states that can interact with photons due to creation of dipole moments.

I used to work with fluorophores: when a fluorescent molecule absorbs a photon and excites an electron, for quantum reasons which I could have explained back when I was a real chemist, you can't actually excite an electron directly from the ground state into the lowest excited state, but rather into a slightly higher energy version of the lowest excited state. Then, the molecule almost instantly relaxes into that lowest excited state (by transferring some energy to translational/rotational modes, or even by reorientation of the solvent to make the excited state more comfortable). This is why fluorophores are excited with one wavelength (common ones we are used to are excited by UV light) and re-emit at a slightly longer (lower energy) wavelength (again, common ones we encounter re-emit in the visible) (this chemistry is used for whitening agents for white shirts!).

But in general, energy is fungible, and can be transferred from vibrations to rotations to translation to electron excitation under the right circumstances: so excited CO2 molecules with photons can lead to an increase in translational energy of the gas mix the CO2 is in, and heating up the gas mix can sometimes lead to a jump in excitation state of a CO2 molecule... which could lead to emission of a photon if the CO2 can't get rid of the energy some other way...

I don't know if this helps, I mostly wrote this as an exercise for myself to see how much of this stuff I could remember...


Gator said...

Geoff, CO2 is a molecule, not an atom. So there are additional ways for the C02 molecule to store energy than simply bumping an electron into a higher energy level. In this case Eli is describing "a vibrationally excited CO2 molecule..." The energy absorbed by the molecule goes into making the individual atoms vibrate relative to each other, like masses on either end of a spring.

It's not just that the molecule bounces away like a billiard ball, it's that the molecule has internal states that can take and give energy too.

Geoff said...

MMM, Gator, thanks for the replies.

So my original description was wrong: collisions or absorption of photons can cause a change in electron energy level or a change in momentum or vibration (or presumably a mixture of these). And a decaying electron does not necessarily emit a photon. It might increase the level of vibration (but presumably not momentum) instead. Did I understand correctly?

Is it different for an isolated atom? Can such an atom only do the photon/energy-level thing I first proposed? Or can collisions also bump an electron's energy level up? And can the energy from a decaying electron in an isolated atom be converted into momentum (conservation of momentum implies not, to my naive mind)?

chris said...

Absorption of a photon doesn’t necessarily induce electronic transitions. It depends on the energy. Photons with energies in the UV region of the EM spectrum induce electronic transitions. Photons with IR energies induce changes in vibrational states.

So IR absorption by CO2 doesn’t involve electronic transitions – it induces excited vibrational states which can relex by emission or by non-radiative collisional transfer. Eli has told us that the non-radiative transfer (to N2 and O2) greatly dominates relaxation of excited vibrational states in the atmosphere (until one gets to v. high altitudes).

How about electronic transitions? Absorption of UV light by chromophores results in promotion of electrons to excited electronic states. Anonymouse was thinking of the Franck-Condon (no tittering at the back) principle which states (paraphrasing) that since bond vibration is slow (v heavy nuclei) relative to the virtually instantaneous elevation of electron to excited state, the electron will most likely find itself in a higher energy vibrational substate of the excited electronic state. If the excited state is sufficiently stable then it will relax to a lower vibrational substate (losing some energy to its surrounds presumably) before the electron drops back to the ground state by emitting a photon (this is fluorescence as Anonymouse described).

However the excited electronic state may be quenched in various ways resulting in non-radiative (no photon emitted) relaxation to the ground state. Quenching by molecular collisions often with electron rich quenchers, can depopulate the excited electronic state without photon emission. Likewise fluorescence energy transfer can transfer excited electronic state energy to acceptor chromophores by dipole interactions so long as the latter is sufficiently close in space and the emission (donor) and excitation (acceptor) transition energies overlap. This is also radiationless so far as the donor is concerned, ‘though the acceptor might itself emit a photon.

In a nutshell:
(i) electronic transitions not relevant to IR absorption by CO2 (ii) where one does have electronic transitions (UV absorption) the electron must always return to the ground state, but this may occur by emission of a photon or non-radiative transfer of excited state energy to surroundings (or mixture of both).

KAP said...

Clarification: that means 1 out of 10,000 CO2 molecules will emit, at surface pressure.

Way up in the stratosphere, the ratio is a lot greater, because the time between collisions is greater.

Gator said...

Yes, a sole atom is different from a molecule. The sole atom has to conserve momentum and energy, and it has no internal vibrational or rotational states to allow for lower energy photon interactions.

Cyrus said...

Geoff, atomic helium doesn't adsorb much (any) in the IR so I think you are on the right track with your thought experiment. Of course there is an adsorption election for He with peaks in the UV for electronic transitions.

I like the chemistry, but do We have deniers who deny spectroscopy? Yikes that is sad. Nice post Eli.

Phillip said...

I have a question (somewhat related to the topic) - how do you calculate the mean free path (MFP) for an IR photon? I understand that the photon MFP varies with GHG concentration and pressure but my physics knowledge is pretty limited. My attempts at calculating the MFP gave me values of less than a meter assuming sea level pressure, 400 ppm CO2, and saturated water vapor. The MFP increases with altitude, particularly in the upper stratosphere where pressure is low and CO2 is the principal GHG. An IR photon there can travel much farther, on average, before encountering a GHG molecule and being absorbed. Or the photon can escape to space as outgoing radiation.

My reason for asking is to improve my understanding, and to help me better explain the GHE to friends and family. Why nights with clear skies and low humidity get colder than cloudy and humid nights, for example.

Any help you can give me will be greatly appreciated.


Martin Vermeer said...


you need quantum theory to do that (no, I wouldn't manage either). You compute the absorption cross section of the molecule, and after that it's classical billiard ball physics.

For the lazy, there is David Archer's applet:

You can roughly deduce the level where the radiation escapes to space; there the mean free path of the photons equals the atmosphere's scale height, about 8 km. In the centre of the CO2 660 cm^-1 band there is an "anti-spike" that may originate as high as 30 km above the ground.

Downward translating this to the ground gives MFP = 8 km * e^{-30/8} = under 200 m, ignoring pressure broadening. So, only two orders of magnitude off, which in astrophysics would be considered promising :-)

For other wavelengths it will be longer (yes, it varies with wavelength too!). This is for all GHGs together.

Phillip said...


Thank you for your response - it helps. My naive approach to the calculation was clearly wrong.

Oh well - my humility needs a booster shot every now and then.

EliRabett said...

Rabett Run did the antispike a couple of years ago:)

Martin Vermeer said...

Very good, Eli.

Those who think this is only about the Earth, no, Solar astronomers have been doing this too for a while: looking at the chromosphere in the hydrogen alpha spectral line. So, you may write 'AGW theory' as long as you pronounce it 'astrophysics'.

The long version:

hydrogen has a number of spectral lines in the visible range, of which the alpha line of the Balmer series at 656 nm in the red part of the spectrum is most useful.

Now if you take a spectrum of the Solar disc, this is an absorption line. According to Kirchhoff-Bunsen, H is good at emitting at this wavelength, but equally good at absorbing. And as within the Solar atmosphere there is a negative vertical temperature gradient (sound familiar?), the higher layers are cooler, and the hydrogen there absorbs light from the continuum radiation below, creating an absorption line in the spectrum. (But then, at some level, temperatures start going up again, transitioning to the corona with its millions of degrees.)

This is where the H-alpha Solar filter comes in: through it, you can look, not at the Solar surface or photosphere, but at the higher, thinner, cooler layers called the chromosphere, where much of the interesting stuff happens. Just like a satellite looking down at Earth at 667 cm^-1, and seeing the stratosphere.

Heck, you can even build a tunable spectral filter (called a Fabry-Perot interferometer) and move up and down the chromosphere, studying the local conditions...

BTW where the name "chromosphere" come from will not be a secret to anyone who has witnessed a total Solar eclipse :)

Martin Vermeer said...

> a couple of years ago

2013-2009 = ?

This ageing bunny understands Eli so well... those Olympics do seem to happen every year nowadays

Anonymous said...

The #2 misapprehension above ( Bunnies who think that the only way that an excited molecule can get rid of the energy is to emit a photon) is actually understandable, given that nearly everything one reads about the greenhouse effect (including here at RR) helps to perpetuate that idea.

"Below that level [10km], energy emitted by a greenhouse gas molecule is soon absorbed by another relatively nearby one. Thus the energy simply cannot be radiated to space to balance the incoming solar energy."

Not to mention the fact (pointed out in the above post) that the vast majority of the energy "simply cannot be radiated" because, at low altitude, the CO2 molecules give up that energy to surrounding molecules (mainly N2 and O2) through collisions before they ever even have a chance to radiate.

Einstein once said "Everything should be made as simple as possible, but not simpler."


EliRabett said...

Stay tuned for the next chapter (btw, Eli had some simple explanations of what you are talking about and a rather excellent analogy

Susan Anderson said...

Thanks for the simplicity, particularly the links. Still a mite beyond me but I think I have a better idea now. Knowing what I don't know goes a long way, but not far enough. Who'd be an aging amateur with a sense of urgency?!

Anonymous said...

That quote above was yours Eli, from your "Simplest Explanation" post (which I linked to just above the quote).

"Below that level [10km], energy emitted by a greenhouse gas molecule is soon absorbed by another relatively nearby one. Thus the energy simply cannot be radiated to space to balance the incoming solar energy."

But there was no mention of the fact that emission and re-absorption (by GHG's) in the lower atmosphere is actually not the dominant process involved.

Hence, perpetuation of wrong idea #2 (which is what I was pointing out)

The second part of Einstein's saying "...but not simpler" is also important (and all too often overlooked)


Unknown said...

There's only one problem with this analysis and its implications for climate alchemy. This is the assumption that there is experimental proof of direct thermalisation. It goes right back to John Tyndall.

Every experiment that uses external energy to create activated GHG molecules in an otherwise LTE gas mixture also incorporates the interaction of the gas with the walls of the containing vessel. In the case of Tyndall it was a brass tube.

What really happens is that the excited molecules do decay by collision but to maintain LTE, the first thing that happens is that GHG molecules emit photons outside the LTE volume to eject the energy >LTE. These thermalise at cooler heterogeneities, in the case of the atmosphere, clouds, bare aerosols and space.

This is a form of pseudo-diffusion driven by the excess energy above LTE, so will increase as the atmosphere cools! No normal lab experiment can measure it.

The bottom line is that the 23 W/m^2 average surface IR absorbed by the atmosphere goes to space either directly after a random walk or via second phases which frequency shift it to a grey body spectrum. The claim that 157.5 W/m^2 is absorbed is bunkum from incorrect boundary conditions. What's more, CO2-AGW is kept strictly zero by the control system in the atmosphere which allowed ice ages when we had 12 times as much CO2 as now.

a_ray_in_dilbert_space said...

Bullshit. Your analysis ignores the lapse rate and the fact that the ghg must be in equilibrium with the radiation field--which cannot happen when there are photons in its absorption band.

Anonymous said...

I don't see that unknown has even done any "analysis".

All I see are bare claims, and totally nonsensical ones at that.

What does it even mean to claim that "GHG molecules emit photons outside the LTE volume to eject the energy >LTE."?

is that supposed to mean that all GHG molecules that have been excited by absorption of IR photons are somehow motivated (magically?) move outside some "volume" (unspecified) to dump the energy they just absorbed?

How far do they move to do that?

More importantly, how do they avoid collisions with other molecules (N2 and O2) on their way to the dump?

What the hell is "energy > LTE" supposed to mean anyway?

LTE is not an "energy."

TE is a state of a system in which "there are no net flows of matter or of energy, no phase changes, and no unbalanced potentials (or driving forces), within the system."

And the statement "These thermalise at cooler heterogeneities, in the case of the atmosphere, clouds, bare aerosols and space" is just gibberish.

As is

"This is a form of pseudo-diffusion driven by the excess energy above LTE, so will increase as the atmosphere cools! "

That whole comment is a form of pseudo-scientific nonsense.

It's simply not possible to have a scientific "discussion" about such stuff.

It's utterly meaningless.


EliRabett said...

1. It's not the GHG molecules that absorb the photons that emit the photons. See part II

2. At pressures of a few torr, there is not enough time for molecules in a reasonably sized cell to hit the walls before they are collisionally quenched. Using thermal velocity @ 300K, and not even considering diffusion, the velocity would be ~500m/sec, so from the middle of a 1 m chamber it would take 1 ms to hit the wall. The quenching time is more like 10 us. Of course, at higher pressures, the time it would take for molecules to hit the wall would be much longer.

3. Unknown waves hands.

Anonymous said...

Good explanation. It means this pretty NSF illustration, which shows IR photons being absorbed and re-emitted by atmospheric CO2, is wrong (though they got the vibrational mode correct):

You estimated the mean time for an atmospheric CO2 molecule, which has been excited by absorption of an IR photon, to lose that energy by collision with another air molecule, as approximately 1 µS[1], or 10 µS[2], or a few tens of µS[3]. That compares with a mean time to lose it by emission of another IR photon of approximately 1 second.[4] That gives a ratio between the two processes in the neighborhood of 10^5 to 10^6.

However, physicist Will Happer did similar calculations, and calculated the ratio as being nearly 10^9.

Y'all roughly agree about the mean time for photon emission (close to one second), but he calculated a mean time for collisional deexcitation which is measured in nanoseconds, rather than microseconds. You can read what he wrote to me about it here:

Obviously, the practical conclusion is the same. You and Prof. Happer agree that, as you put it, "Essentially NONE of the absorbed energy is re-radiated from the excited molecule before it is quenched."[5] The amount of IR emitted by the air is determined almost entirely by its temperature (and GHG partial pressures), not by the amount of IR which it absorbs (except to the extent that absorbed IR raises the air temperature).

But a difference of three orders of magnitude seems worth examination.

(BTW, I'm confused: is "Fred" the name of a mouse?)

TheChemistryOfBeer said...

Dave Burton, you seem to be a trifle confused. The IR and CO2 molecules are in a local thermodynamic equilibrium, which means both absorption and emission rates are equal. If as you say (and is true) a particular CO2 molecule does not emit a photon to lose energy, there must be emission from other molecules which did not absorb an IR photon, but obtained energy through collisions with other molecules (many of which did get that energy by absorption of IR).

Rib Smokin' Bunny

Anonymous said...

Sorry, I didn't realize that the two URLs wouldn't automatically turn into hyperlinks.

The NSF illustration is:

The link to Prof. Happer's discussion is:

Aside: I had some trouble getting this comment to post. What does this mean?
"Input error: Memcache value is null for FormRestoration"

Anonymous said...

t_p_hamilton wrote, "The IR and CO2 molecules are in a local thermodynamic equilibrium..."

I assume you mean that the CO2 molecules are in thermal equalibrium with other air molecules, because what you actually wrote makes no sense.

t_p_hamilton continued, "...which means both absorption and emission rates are equal."

No, the IR absorption and emission rates from atmospheric GHGs are not necessarily equal. The absorption rate is determined by how much IR there is to be absorbed, at wavelengths to which the GHGs are not transparent. The emission rate is determined (almost entirely) by the temperature of the atmosphere (and the partial pressure of the GHGs in it). If the two rates ever happen to be the same, that's just a coincidence.

t_p_hamilton continued, "If as you say (and is true) a particular CO2 molecule does not emit a photon to lose energy, there must be emission from other molecules which did not absorb an IR photon, but obtained energy through collisions with other molecules (many of which did get that energy by absorption of IR)."

No, there are many other ways in which air can gain, lose and transport heat, other than by the absorption and emission of IR, so there's no requirement that the rates of IR absorption and emission by atmospheric GHGs be equal.

Alastair said...


You wrote "Bunnies who don't realize that the molecule can also emit light. This is a popular one amongst organikers and analytical chemists whose experience with IR spectroscopy is in an absorption spectrum for analysis of samples."

Do the analytical chemists, who do not realize that molecules emit light, include those who drew up the HITRAN data base? In other words, are they reporting the measured net absorption as the Einstein B coefficient when they should be recording the total absorption?