Can anyone explain the REAL reason why CO$_2$ increases global temperatures (not the simplistic greenhouse analogy provided for public consumption)?

Question

The greenhouse effect analogy of global warming is that atmospheric carbon dioxide CO$_2$ absorbs some of the infrared radiation emitted by the Earth, and redirects a portion of that radiation back down to the Earth's surface, thereby heating the surface more than it would have done if that radiation had been able to escape into space.

Global warming is then simplistically explained to the general public by the idea that as atmospheric CO$_2$ concentrations rise, more infrared radiation is absorbed by CO$_2$ and re-emitted back down to Earth, causing increased heating of the Earth.

However, this explanation is not technically correct, because at present atmospheric CO$_2$ concentrations, just one kilometer of atmosphere is sufficient to fully absorb all the infrared radiation emitted by the Earth, at the wavelengths at which CO$_2$ absorbs.

Carbon dioxide absorbs infrared at the wavelengths of 2.7, 4.3 and 15 µm, and the CO$_2$ in the first kilometer of atmosphere alone is able to completely absorb all infrared at these wavelengths.

So the infrared absorption process is already fully saturated, and thus further increases in atmospheric CO$_2$ will not lead to any additional absorption. This is why the simplistic explanation provided for the general public does not seem to be technically correct, even though it roughly outlines the idea.

I found one blog article by Clive Best that tries to explain the actual process behind CO$_2$'s ability to cause global warming. Judging from that article, the actual process is more complex than the simple explanation provided for public consumption. However, I don't fully understand the explanation given in the article (and from what I did manage to understand, I am not sure if it is fully correct).

So I wonder if anyone here can provide an easy to understand explanation of the actual mechanism by which increased atmospheric CO$_2$ leads to global warming. Or perhaps if you know any good articles that explain it, please can you post the links.

I tried to find some info on the actual mechanism of global warming via Google, by using search terms such as "mechanism of greenhouse effect in global warming", but was surprised to find very little information available. Even the Wikipedia page on global warming provides only the simplistic and technically incorrect public consumption explanation.

As suggested in the comments below, I have also asked this question on earthscience.stackexchange.com — see here.

Answer 1

So the infrared absorption process is already fully saturated, and thus further increases in atmospheric CO2 will not lead to any additional absorption.

1. There is no such thing as a fully saturated absorption process of the $\mathrm{CO}_2$ line.

Let $I_{\nu}(\nu)$ be the spectral distribution of the near-infrared emission that an observer would measure after it crossed a layer of atmosphere.

If a specific intensity $I_{\nu, 0}$ is emitted by the ground at a frequency $\nu = \nu_0$, the transmission of the molecular line is described by the radiative transfer equation:

$I_{\nu}(\nu) = I_{\nu, 0}(\nu) \left( 1 - \mathrm{exp}(-\tau(\nu)) \right)$

Where $\tau(\nu)$ is the optical depth associated with the layer of gas, depending both on the quantity of absorbants along the line of sight (in our case $\mathrm{CO_2}$ molecules) and the frequency. The frequency dependance of the line opacity can be written as follows:

$\tau(\nu) = \tau_0 \mathrm{exp} (-(\nu - \nu_0)^2 / 2\sigma^2)$

Where $\tau_0$ is the central opacity (at $\nu = \nu_0$), directly related to the column density of $\mathrm{CO_2}$ ($\tau_0 \propto N_{\mathrm{CO_2}}$ [$\mathrm{cm^{-2}}$]), $\nu_0$ is the central frequency of the absorption line and $\sigma$ is the intrinsic velocity dispersion of the gas. This gaussian distribution of the line opacity is valid for a gas with in intrinsic Maxwellian velocity distribution. (For more information, see Doppler effects in emission line of molecules)

On the figure below I produced the spectral distribution of the quantity $\left( 1-I_{\nu}(\nu) \right) / I_{\nu, 0}$ to represent the amount of radiation that can be absorbed by a layer of atmosphere. The only parameter that varies between the different curves is $\tau_0$, the opacity at $\nu = \nu_0$.

We can clearly distinguish two regimes:

• Optically thin regime ($\tau_0 \ll 1$) (not saturated): these are the red curves obtained with $\tau_0 = 0.1, 0.25, 0.5$. In this regime the amount of radiation absorbed by $\mathrm{CO_2}$ grows linearly with $\tau_0$, i.e. with the column density of $\mathrm{CO_2}$.

• Optically thick regime ($\tau_0 \gg 1$) (saturated): these are the black curves obtained with $\tau_0 = 10, 30$. As you can see, the absorption is effectively saturated at the line center (at $\nu = \nu_0$) but the opacity broadening of the line wings allows the amount of absorbed radiation to grow in a non-linear way.

(Note: this is because we are populating the high-velocity wings of the Maxwellian velocity distribution of $\mathrm{CO_2}$ molecules within the atmosphere. Indeed, the enhancement of $\mathrm{CO_2}$ column density provides a proportionally greater number of molecules that are allowed to have high velocity deviations, thus having heavy doppler-shifted emission. The photons emitted at $\nu = \nu_0 + \Delta \nu_{\mathrm{Doppler}}$ by these high-velocity $\mathrm{CO_2}$ molecules are propagating in an optically thin medium, because $\tau(\nu = \nu_0 + \Delta \nu_{\mathrm{Doppler}}) \ll \tau(\nu)$.)

If we sum the area under the different curves of the previous figure, we obtain the following relation between the amount of absorption and the line center opacity $\tau_0$:

The black markers corresponds to the area of each line profile in the first figure. As you can see, in the saturated regime an enhancement of the column density (leading to an enhancement of $\tau_0$) does increases the area, i.e. the amount of absorbed near-infrared radiation by the layer of atmosphere. The only difference between the optically thin and saturated regimes is the rate at which the enhancement takes place (linear vs logarithmic). The slope is indeed much lower in the saturated regime, which is very fortunate for us.

However, this explanation is not technically correct, because at present atmospheric CO2 concentrations, just one kilometer of atmosphere is sufficient to fully absorb all the infrared radiation emitted by the Earth, at the wavelengths at which CO2 absorbs.

2. I believe there is a misunderstanding of how the greenhouse effect works here.

We cannot reason on a single layer of atmosphere of a given width at a given altitude. The heat budget of the whole atmosphere has to be taken into account. That 1km of atmosphere fully absorbing all the infrared radiation emitted by Earth that you are talking about, is not only absorbing heart, it loses it as well. All the energy that is absorbed at some point by the medium is re-emitted via radiative losses, in both direction (toward the Earth and toward the upper layers of atmosphere where it is transferred). If the lower troposphere is saturated in $\mathrm{CO_2}$ as you claim, it is not necessarily the case of the upper troposphere, which is continuously exchanging heat with the latter. What governs the energy balance in the atmosphere is actually the radiative balance in the upper troposphere, where radiative losses are emitted toward space where it can escape. I would say that the $\mathrm{CO_2}$ concentration is much more decisive there than in the lower troposphere.

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^{For an in-depth presentation of the opacity broadening process in an astrophysical framework, see the section 2. of this paper https://arxiv.org/pdf/1603.08521.pdf. This is a paper on the opacity broadening of the CO line in molecular clouds but the physics are the same.}

^{See also http://www.pas.rochester.edu/~ebubar/CurvesOfGrowth.pdf for a presentation of the curve of growth in the framework of stellar atmospheres.}

Answer 2

The Earth's temperature is set by energy balance: the Earth must be hot enough so that the outgoing energy balances the incoming energy from the sun. If the Earth behaves as a black body with albedo $\alpha$, then the Earth's temperature is given by $$ \sigma T^4 = \frac{S_0}{4}(1 - \alpha) $$ where $S_0$ is the solar constant. Taking $\alpha = 30\%$, this gives $T = 255\mathrm{K}$. This is clearly to cold. One way around this is to introduce an emissivity caused by the presence of $\mathrm{CO_2}$. However as you correctly point out, this can't be the whole story as the atmosphere is essentially already totally opaque to infrared radiation in the relevant wavelengths, yet adding more $\mathrm{CO_2}$ clearly has a warming effect on the Earth.

This is because the greenhouse effect (in an optically thick atmosphere like ours) is caused by the vertical structure of the atmosphere.

As we rise through the atmosphere, the atmosphere becomes less and less dense. The less dense the layer of atmosphere, the more radiation emitted from that layer reaches space. From the surface, essentially no radiation reaches space, but this is less true higher up. We can consider an 'effective radiating layer' (this roughly corresponds to the stratosphere) in which all the radiation that reaches space comes from this layer. As a result, this layer has temperature $255\mathrm{K}$.

The second fact about vertical structure we need is the lapse rate. This tells you how quickly temperature drops with height. It's one of the first calculations you do in an atmospheric thermodynamics class, and it turns out that for a dry atmosphere $$ \frac{\mathrm{d}T}{\mathrm{d}z} = -\Gamma = - \frac{g}{c_p} \approx -10 \mathrm{K\,km^{-1}} $$

Now combining these facts, if it is $255\mathrm{K}$ high in the atmosphere it must be hotter at the surface - this is the greenhouse effect. If you now dump a load of $\mathrm{CO_2}$ into the atmosphere you make the atmopshere denser, and raise the effective radiating level, but the lapse rate is unchanged, so the Earth must be correspondingly hotter!

Andrew Dessler has a nice video on this.

I've skated over a lot of the detail here, but conceptually this is correct.

Answer 3

Contrary to claims that the greenhouse effect is complicated, it's really quite simple. The IPCC's definition in the glossary of AR5 WG1 is fine, but can be made even clearer with the help of the fifth figure at http://clim8.stanford.edu/Images/ . All greenhouse gases work this way, there is no need to consider particular absorption lines unless you're using them to estimate no-feedback climate sensitivity for a particular species of a particular molecule, which is extremely difficult to confirm experimentally.

ZeroTheHero suggested I expand on this, so here goes. Though now that I look at jobal6's answer I don't see that my answer below adds much to it.

Longwave radiation in the atmosphere relevant to the greenhouse effect (GHE) divides into three kinds, downwelling, upwelling, and outgoing, respectively DLR, ULR, and OLR.

OLR is what escapes to space. It is 100% responsible for keeping Earth in thermal equilibrium with the Sun, the state in which OLR equals the absorbed solar irradiance (the fraction 1-A not reflected back to space where A is Earth's albedo, nominally 0.3).

Ignoring lensing effects of the kind that causes mirages, all longwave radiation within the atmosphere divides into DLR and ULR according to whether it points to below or above the horizon respectively. (At least that's how I would define it, your mileage may vary.)

ULR is a candidate for becoming OLR, which it does when it isn't absorbed by a greenhouse gas (GHG) molecule or an aerosol particle. Otherwise it either heats the absorbing molecule or particle, thereby returning it to Earth's heat, or is reradiated in a random direction (e.g. stimulated emission?).

DLR differs from ULR only in that none of it can become OLR directly. 100% of DLR is either absorbed by the surface or otherwise behaves in every respect just like the fraction of ULR that doesn't become OLR.

Since that may seem surprising it should be pointed out that although DLR increases with increasing GHGs, as you will read in the many accounts of back radiation, ULR also increases, which people not trained in physics tend to overlook. What matters for heating the surface is not DLR alone but DLR - ULR since the ULR removes heat at the same time DLR is contributing heat.

As a consequence of (i) the Stefan-Boltzmann relation F = σT⁴, (ii) the fact that ULR at any given point originates from a hotter place than DLR at the same point, and (iii) lapse rate maintains a constant temperature difference between any two altitudes, ULR increases faster than DLR, whence the net downward flux DLR - ULR actually decreases with increasing CO2.

So that can't be the reason the surface warms with increasing GHGs. (The IPCC knows this and you won't find anything about back radiation in AR5 WG1. It's also not in Figure 7 of Kiehl & Trenberth 1997, which shows ULR > DLR, frequently overlooked.)

The reason the surface warms is because increasing GHGs make it harder for ULR to become OLR. This traps more heat, which warms not just the surface but the atmosphere and the oceanic mixed layer, OML, enough to increase ULR until the fraction that becomes OLR is restored to the level needed to keep Earth in thermal equilibrium with the Sun.

The thermal inertia of the surface, atmosphere, and especially the OML is enough for thermal equilibrium to take centuries. This is why equilibrium climate sensitivity, ECS, is higher than transient climate response, TCR, which is only allotted 70 years to warm up.

One would also expect the deep ocean to be a further impediment to equilibrium. The tricky part here is that the deep ocean has far more thermal inertia than the OML, while also being better connected to the ice caps than to the OML, namely via the great ocean conveyor belt's thousand-year journey. This makes the ice caps themselves the real further impediment to equilibrium. Basically the deep ocean stays cold, leaving the OML to achieve the appearance of thermal equilibrium.

Eventually the ice caps melt, which I believe is what at least some people mean by Earth System Sensitivity (but I'm a bit vague on that, maybe climate modelers can kick in here). Obviously that equilibrium takes far longer than the equilibrium associated with ECS, there being a heck of a lot of ice to melt.

I think this covers the basic greenhouse effect, GHE. There's a lot more to climate than that but the GHE is the part fundamentally responsible for global warming.

"not the simplistic greenhouse analogy provided for public consumption"

Joseph Fourier is responsible for the analogy. It's a lot better analogy than it's usually given credit for these days. For starters Earth's gravity prevents Earth's atmosphere from convecting to space just like a greenhouse's glass prevents its air from convecting to the environment. And secondly the reason greenhouses need windows that can be opened is that without them the heat trapped by the glass during the day would overheat the plants, which would not be a problem in the case of no greenhouse but completely motionless (non-convecting) air.

The Louvre in Paris has the greenhouse effect problem in spades, both in the main entrance under I.M.Pei's glass pyramid (I've sweated it out there myself, it's horrible on a sunny day) and in the glass-covered Cour Marly inside.

Lastly it should be pointed out that although Earth's atmosphere is thousands of times thicker than a sheet of glass, CO2 at 1000 ppm if chilled to dry ice would cover the Earth with a layer one centimeter thick. So 400 ppm is an excellent approximation to a sphere of glass 4 mm thick suspended somehow over Earth.

Answer 4

FWIW, here is my answer from the Earth Sciences SE, the key is to focus on how energy leaves the planet, rather than the fate of IR photons emitted from the surface:

Borrowing an explanation from one of my other answers, the basic mechanism of the greenhouse effect is roughly as follows (note this is also a simplified model)

The Earth is in (to all intents and purposes) a vacuum, so it can only gain or lose heat via radiation. The sun emits most of its radiation at visible and UV wavelengths. The Earth's atmosphere is fairly transparent at these wavelengths and so the Sun's radiation mostly passes through it and hits the surface. Some of this radiation (determined by the Earth's albedo) is reflected from the surface back out into space, but the rest is absorbed by the surface, which causes the surface to be warm. The surface loses heat by radiating in infra-red wavelengths. Greenhouse gases absorb some of the IR radiation, which causes the atmosphere to warm up (the GHG molecules transfer some of this heat to non-greenhouse gasses by collisions, but heat is also transferred upwards by convection). The warm atmosphere re-radiates some of this energy both upwards into space and downwards back to the surface. The part that is radiated downwards is also known as "back-radiation" (and is directly observable). Now the important factor is not the amount of outbound IR radiated from the surface that is absorbed, but the altitude at which there are not enough greenhouse gases above to absorb the IR radiated upwards from that layer, so that it can escape out into space. The lapse rate means that the temperature of the atmosphere decreases with increasing height. This means that the more CO2 we put into the atmosphere, the higher this emitting layer becomes, and the colder it is. As the amount of IR radiated depends on the temperature of this layer, if this height increases then the amount of IR radiated from the planet falls, leading to an energy imbalance, with the planet absorbing more of the sun's radiation than it emits as IR, and so the planet warms up. This continues until the radiating layer warms up enough for the outbound IR to be in balance with the incoming radiation from the sun. So the more CO2, the warmer the mean surface temperature, all things being otherwise equal.

Thus it is irrelevant that most IR emitted by the surface is absorbed by the atmosphere, it is the height from which IR is not absorbed that matters. If it is colder than the surface, it will emit less IR than the surface is emitting and hence less energy is radiated into space.

For a more detailed explanation of why absorption at the surface is of little relevance, see this RealClimate article by Spencer Weart.

Just to add a historical note, this explanation goes back at least as far as Ekholm's paper of 1901 "On The Variations Of The Climate Of The Geological And Historical Past And Their Causes":

The atmosphere plays a very important part of a double character as to the temperature at the earth’s surface, of which the one was first pointed out by Fourier, the other by Tyndall. Firstly, the atmosphere may act like the glass of a green-house, letting through the light rays of the sun relatively easily, and absorbing a great part of the dark rays emitted from the ground, and it thereby may raise the mean temperature of the earth’s surface. Secondly, the atmosphere acts as a heat store placed between the relatively warm ground and the cold space, and thereby lessens in a high degree the annual, diurnal, and local variations of the temperature.

There are two qualities of the atmosphere that produce these effects. The one is that the temperature of the atmosphere generally decreases with the height above the ground or the sea-level, owing partly to the dynamical heating of descending air currents and the dynamical cooling of ascending ones, as is explained in the mechanical theory of heat. The other is that the atmosphere, absorbing but little of the insolation and the most of the radiation from the ground, receives a considerable part of its heat store from the ground by means of radiation, contact, convection, and conduction, whereas the earth’s surface is heated principally by direct radiation from the sun through the transparent air.

It follows from this that the radiation from the earth into space does not go on directly from the ground, but on the average from a layer of the atmosphere having a considerable height above sea-level. The height of that layer depends on the thermal quality of the atmosphere, and will vary with that quality. The greater is the absorbing power of the air for heat rays emitted from the ground, the higher will that layer be, But the higher the layer, the lower is its temperature relatively to that of the ground ; and as the radiation from the layer into space is the less the lower its temperature is, it follows that the ground will be hotter the higher the radiating layer is.”

[Ekholm, 1901, p19-20]

(h/t Steve Easterbrook's blog article)

Answer 5

The average time for an excited CO2 molecule to emit an LR infrared wave is about 1 second or greater. And excited CO2 molecule will collide with another oxygen or nitrogen molecule in a couple of microsecond and lose it s energy in the collision. This will occur about 99.8% of the time versus releasing a LR photon. With only one seal two molecule per 2500 other molecules in the atmosphere, it's seems very unlikely that any significant back radiation to the Earth will occur. CO2 in the atmosphere cannot be the cause of climate change.