3
$\begingroup$

As a newbie on climate science and meteorology I'm trying to understand the almost logarithmic behavior of CO2 concentration on radiation forcing

$$F(c) = A \cdot \ln \frac{c}{c_0}$$

and why there is no "saturation" due to further increasing of CO2 even when the atmosphere is already totally absorbing as a whole.

So far I learned that the air layers which contribute most to outgoing radiation are placed where optical thickness, as measured from TOA, is about 1. Let's say in an undisturbed atmosphere this location is about a height of z=2 (arbitrary units), then a CO2 doubling leads to a shift of this layer to about z=3. enter image description here

Since this higher layer has usually lower temperature, less energy is irradiated an, therefore, the planet will warm up.

This is described also in some detail in https://escholarship.mcgill.ca/downloads/db78th05j.

On the other hand, other explanations, even mentioned in the cited publication, are based on spectral effects where the wings of the absorption band become more absorbing at higher CO2 concentrations: enter image description here

Q1: Firstly, I don't understand why this last effect is, because CO2 is quite dilute in atmosphere and this effect cannot be due to pressure broadening. What is the reason for absorbing more CO2 at the wings when concentration rises up?

Q2: secondly, which of the two effects is more relevant to explain the observed log behavior? Are they related in any way or do they just happen to give the same overall result?

$\endgroup$
7
  • 1
    $\begingroup$ Have a look at the 'curve of growth' of spectral line equivalent width: spiff.rit.edu/classes/phys440/lectures/curve/curve.html Even at high optical depths, 'saturation' doesn't mean that there is no increase in absorption anymore. Rather, absorption starts depending logarithmically, rather than linearly on the absorber density, as the line core is saturated, but the line doppler wings continue to contribute. The arguments for a single line then directly translate to the line forests of molecules, which is why the overall CO2 opacity also follows the log-behaviour. $\endgroup$ Nov 10, 2022 at 15:05
  • $\begingroup$ Does it really mean that the absorption characteristics of a particular molecule, e.g. CO2 depend on the density of those molecules? Cannot imagine, because wit 300ppm CO2 molecules are very rare and they do not interact in any way. Most of the other molecules for a single CO2 are not CO2, but N2, O2, ... so if there is some modification because of collisions, most of the collisions come from other molecule types - not CO2. So I would assume some dependence on the total pressure, but not on partial pressure of CO2. Do we talk about the same thing? $\endgroup$
    – MichaelW
    Nov 10, 2022 at 15:31
  • 1
    $\begingroup$ The pressure broadening/collisional broadening does not matter on the collision partner when you are close to thermal equilibrium (we're far from the ionosphere here). Study the document I've linked carefully, it doesn't take a long time. The log(c) dependency of the total absorption depends on the number of excitations any single line gets hit with times the number of absorbers, which is just another way of encoding the optical depth and hence the partial pressure. $\endgroup$ Nov 10, 2022 at 17:45
  • 1
    $\begingroup$ I would say that is correct. $\endgroup$ Nov 15, 2022 at 20:44
  • 2
    $\begingroup$ @AtmosphericPrisonEscape It seems your comments helped to answer the question; mind writing it as an answer? $\endgroup$
    – hichris123
    Nov 21, 2022 at 5:36

2 Answers 2

3
$\begingroup$

Your confusion, I think, comes from a subtlety, and I think I can only answer this from the perspective of an astrophysicist. We don't think in terms of "forcings", the vocabulary is still alien to me (I have yet to see a sensible, formal defition of it). But I will answer in terms of a certainly related concept.

In the optically thick part of an atmosphere (i.e. the one that potentially features a greenhouse effect), we know from analytic solutions that the Temperature $T$ as function of optical depth $\tau$ is approximately $T^4=T^4_0(\frac{2}{3} + \tau)$, where $T_0$ is a measure of the energy content of the optically thin radiation field.

The total optical depth $\tau$, itself is simply computed as a sum of all opacity carrier contributing species $s$, with number densities $n_s$, opacity function $\kappa_s$, and optical ray path $ds$ as $$\tau = \sum_s \int ds \, n_s \, \kappa_s. $$ This already answers the part of your question about the partial pressure dependency: $\tau\propto n_s$, no matter how the opacity is generated (I am using pressure and number density at constant temperature interchangeably, but to first order it is $n_s$ and not $p_s$ determining $\tau$).
Now collision-induced absorption, which gives rise to the broad Lorentzian line wings, which grow logarithmically at large optical depth, is dependent on all available collision partners, hence this is where the total pressure comes into play, via $\kappa_{CO2} = \kappa_{CO2}(\sum_s n_s)$. In the last sum, the $CO_2$ contribution would be negligible, but $\kappa_{CO2}$ still grows logarithmically.

So in my understanding we'd rather have something along the lines of $\tau \propto n_s \times log(f(\sum_s n_s))$

I would still like to better understand how the 'forcing' comes about from those quantities, but it does give you an answer as to why the temperature to which the system wants to get to (the above $T^4$) has the dependencies that you are confused about.

$\endgroup$
1
$\begingroup$

Just wondering about your statement above: "Since this higher layer has usually lower temperature, less energy is irradiated an, therefore, the planet will warm up". This doesn't sound right. For one thing, temperature at any layer is not a passive variable: if radiation (absorption and emission) is happening at a higher level, then the temperature at that level should rise too.

As I understand it, the "greenhouse effect" is not that the planet warms because less energy is irradiated. There is still approximately overall radiative balance, i.e., energy radiated out balances energy absorbed coming in. The blackbody temperature of the earth as a whole remains about the same - lets say, for the sake of argument, that it's 255K. With more CO2, as you illustrate above, outgoing radiation leaves from higher altitudes, and so the 255K surface rises to a higher altitude as well to maintain radiative balance. However, the temperature below that 255K level more or less continues to follow the moist adiabatic lapse rate (in the tropics), or even just the same lapse rate that it had before, but now the temperature increases along that lapse rate for a greater vertical distance, and so the surface temperature is raised. (The atmosphere is not just in radiative balance, but in radiative-convective balance). The very simplest & crudest picture of this is to take a typical temperature profile from surface to stratosphere (i.e., with a minimum at the tropopause), and simply lift it up a bit. You'll have to extend the bottom of the curve a bit now to a higher surface temperature, while the stratosphere will become cooler.

Since clouds are good blackbodies (i.e., IR absorbers and emitters), this also explains why more high clouds (i.e., above the ~255k level) should raise surface temperatures, while more low clouds should lower them.

This warmer surface temperature and cooler (lower) stratosphere are characteristic of the greenhouse effect as explained in the classic Manabe & Wetherald JAS paper from 1967 ("Thermal Equilibrium of the Atmosphere with a Given Distribution of Relative Humidity", https://doi.org/10.1175/1520-0469(1967)024%3C0241:TEOTAW%3E2.0.CO;2) - a paper which probably helped Manabe win his Nobel prize a couple of years ago.

$\endgroup$
1
  • $\begingroup$ This doesn't answer the question, it is more of a rant. Otoh the offending sentence in OP's writing should be "re-radiation" instead of irradiation, which would make it clear that OP does understand the physics as outlined in this here text, and makes it unnecessary. $\endgroup$ Mar 23, 2023 at 11:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.