Monday, April 22, 2013

Eli Grabs Another Envelope

Bunnies are not really into complex calculations where the meaning is hidden in the math.  For one thing there is great opportunity to fool yourself, for another, it is pretty simple to mislead others.  To Eli math is something you grab after you have figured out what is happening, and then you grab it in shells, with the easy parts first.  That way if you mess up, you always can look at the inner shell to figure out where you went wrong.  EO Wilson had some recent thoughts about the role of mathematics

Fortunately, exceptional mathematical fluency is required in only a few disciplines, such as particle physics, astrophysics and information theory. Far more important throughout the rest of science is the ability to form concepts, during which the researcher conjures images and processes by intuition.
and simple models help with forming those concepts.  In a recent previous life the Bunny Collective discussed the relative probabilities for emission by a single vibrationally excited  CO2 molecule that has been vibrationally excited by absorbing a photon.  There are two fates for the little guy, emission of an IR photon.

[1]  CO2* -->  CO2 + hν          Rate constant = kR

and collisional de-excitation by another, say nitrogren or oxygen molecule

[2]  CO2* + M -->  CO2 + M   Rate constant = kM

Plugging in the numbers, at atmospheric pressure the Rabett showed that only one in a hundred thousand excited molecules would emit an IR photon.  As several pointed out, high up in the stratosphere this decreases in proportion to the pressure, but, for example, at ~17 km it is one in ten thousand, and at ~ 25 km one in a thousand.

On the other hand, thermal excitation of CO2, the reverse of reaction [2] is just

[3]  CO2 + M -->  CO2* + M   Rate constant = k-M

so now, in addition to absorbing an IR photon CO2* can be excited by collisions, converting thermal translational energy into vibrational energy. 

How fast a reaction happens is called the reaction rate, and is proportional to the concentration of each reactant and a number called the rate constant which hides all the dynamics, e.g. the quantum chemistry, the probability of a collision having enough energy to actually cause molecular changes, etc.  If the rate of the forward reaction, equals the rate of the reverse reaction,  an equilibrium requires that as many molecules react in the forward direction as in the reverse

[4]  (kM [M] + kR )  = k-M  [CO2*][M]

If  kR < < kM[M] it can be neglected and then [2] and [3] form a simple equilibrium

[5]  CO2* + M = CO2 + M

which can be rearranged to yield

[6]    [CO2*]/[CO2] = k-M /kM = Keq

The next step is to show how thermodynamics allows a simple estimation of Keq and [CO2*].  The emission rate in photons per second per unit volume will then simply be

[7]   kR [CO2*]


tonylearns said...

Once again you post juggling. being a professional juggler I must comment.
I also retaught myself calculus 5 years ago after a 30 year hiatus. it was confusing at first, but then i got the hang of it.
I have since completely forgotten everything again

Anonymous said...

Bunnies are not really into complex calculations where the meaning is hidden in the math...

...he said before the blizzard of algebra.


Anonymous said...


In his Angus Miller lecture, Matt Ridley quotes a IPCC report:

"In the idealised situation that the climate response to a doubling of atmospheric CO2 consisted of a uniform temperature change only, with no feedbacks operating…the global warming from GCMs would be around 1.2°C.” Paragraph"

I've read a lot of rebuttals to various points in Ridley's speech, but I've never seen anyone dispute this. Is this a robust finding or are there other subtle points relating to this?

EliRabett said...

No, it's a perfectly good statement, except there really are feedbacks. The strongest one is the water vapor feedback, e.g. that the vapor pressure of water in the atmosphere increases if the temperature goes up from some positive forcing. So on the one hand it is a perfectly good statement, on the other hand it is perfectly irrelevant.

Anonymous said...

I'm having a very hard time understanding what #4 above is supposed to represent (and can't see how in the world you get from #4 to #6 (other than by magic)

Note that in my comment, “ksubM “ is just “kM” and I have represented the "-" subscript with an underbar "_")

[4] (kM [M] + kR ) = k_M [CO2*][M]

In #3 you specified "k_M" as the rate constant for a 'thermal excitation of CO2" , so why wouldn't the quantity on the right be k_M [CO2][M]?

Why would you multiply by [CO2*] as you have actually done if you are specifying the reaction rate for excitation of CO2?

It seems that you would want to multiply by [CO2*] only for the reaction rate of the de-excitation of CO2*.

In #1 you gave "kR" as the rate constant for the de-excitation of CO2* by photon emission, so why wouldn't the corresponding quantity on the left hand side in #4 be "kR[CO2*]" instead of just "kR"?

Surely, if the CO2* concentration ([CO2*]) were zero, the reaction rate for the deexcitation of CO2* by photon emission would also be zero, so it seems there needs to be a [CO2*] factor multiplying kR on the LHS of #4

It also seems that there should be a [CO2*] factor multiplying the first quantity on the LHS of #4 (kM[M]) because presumably, the first quantity on the LHS of #4 is supposed to represent the reaction rate for "thermal deexcitation of CO2*" which also depends on [CO2*]. Again, if [CO2*] were zero, the reaction rate would also be zero, would it not?

Finally, if you are writing an equation for the equilibrium excitation/deexcitation of CO2, why wouldn't you also include a term for the excitation of CO2 due to photon absorption? Call that rate k_R and you would add the term k_R[CO2] on the RHS, would you not?

Putting all of this together, I would rewrite #4 as the following

kM[M] [CO2*] + kR[CO2*] = k_M[M][CO2] + k_R[CO2]

or, one can factor out [CO2*] on the LHS and [CO2] on the RHS

and get

(kM[M] + kR)[CO2*] = (k_M[M] + k_R)[CO2]

NOW, if kR << kM[M] and k_R << k_M[M]

we would end up with
kM[M][CO2*] = k_M[M][CO2]

which means

[CO2*]/[CO2] = k_M/kM

which is equal to what you called Keq

That's the equation #6 that you ended up with above, but it is not at all clear how you got to that from your #5 above,
[5] CO2* + M = CO2 + M

(an equation which contains neither rate constants nor concentrations) to your equation #6 which contains BOTH rate constants and concentrations.

Clearly, something (in fact, more than 1 something) is amiss in # 5, which actually does not follow from your number 4 (an equation which contains rate constants multiplying concentrations, but no [CO2] factor, which seems to magically appear in step 6)

Or perhaps I am just completely mistaken? (and/or lost my mind?)

n-g said...

What Anonymous said.

Also, how does [3] deal with the fact that not all M's have enough energy to kick CO2 up a notch? Is that subsumed in k-M or is the M in [3] different from the M in [2]?

Anonymous said...


My guess is the rate constant k_M accounts for the fact that only some of the "M" molecules have enough energy to "kick CO2 up", or more generally, that only a certain fraction of collisions between CO2 and M will result in excitation of CO2 (having enough energy is a necessary but not sufficient condition.)

Eli: only crickets on the algebra mistakes?