Thursday, January 13, 2022

91. The Schwarzschild equation and scattering

In Post 86 I showed how the Greenhouse Effect can be viewed as resulting from a combined process of absorption of infra-red radiation by molecules of carbon dioxide (CO2), followed by re-emission of the same wavelength of radiation, but in a different random direction. The law that leads to this result is the principle of detailed balance. I also showed how this process resembled a scattering process, but with a much greater strength due to the significantly greater absorption cross-section (σa) of CO2 compared to its cross-section for Rayleigh scattering, σs. Therefore it could be modelled as a scattering process, but one with an appropriately large cross-section.

Then in Post 88 I showed how this combined process of absorption and re-emission resulted in absorption that decreased linearly with distance x into a gas of uniform density, rather than exponentially as predicted by the Beer-Lambert law. The importance of this result is that it demonstrates that the transmission of radiation through an absorbing gas will be greater than that predicted by the Beer-Lambert law, and so changes to the concentration of the absorbing gas will have a larger impact on the overall transmission than the Beer-Lambert law would predict. 

Sadly this result has been questioned by some who have claimed that the scattering (or combined absorption and re-emission) approach is not valid as it does not follow from, or agree with, Schwarzschild's equation. This equation is possibly named after the German physicist Karl Schwarzschild (shown below), although it could actually be named after his son Martin Schwarzschild who worked in the field of stellar evolution. In this post I will show that this claim of non-validity is not true and that Schwarzschild's equation leads to the exact same results as I outlined in Post 88.

 Fig. 91.1: Karl Schwarzschild (1873-1916)

The Model

In order to understand the Greenhouse Effect we need to consider the radiation flows through a thin layer of gas of thickness δx at some altitude x, and then extend this to model to the gas as a whole. In particular, we need to consider radiation flows through the gas from opposing directions. This is because each thin layer of the gas only absorbs a small fraction of radiation passing through it. Most of the rest is then absorbed by subsequent layers above the initial layer. However, after absorption the radiation is re-emitted and the re-emission process occurs equally in both directions (up and down). This means that some of this emitted radiation will travel in a downwards direction and reheat the first layer from the opposite direction. This is why the Beer-Lambert law fails.

In order to model this process we need to consider radiation fluxes in both directions as shown in Fig. 91.2 below. Let's assume that the gas has a concentration or particle density n(x), and each molecule of it has an absorption cross-section σa. If radiation of intensity Iu(x) travels in an upward direction from a hot surface and enters the gas layer from below, as shown in Fig. 91.2 below, a proportion of it will be absorbed by the gas. This in turn will heat the gas and cause it to emit radiation in both an upwards and a downwards direction. The result will be that the intensity of upward radiation leaving the gas layer will change by an amount δIu(x).

Fig. 91.2: A schematic illustration of how scattering, absorption and thermal emission within each thin layer of the atmosphere alters the intensities of the upwelling (Iu) and downwelling (Id) radiation.

The upwelling radiation will then go on to interact with gas above the layer shown in Fig. 91.2 and thus heat this gas as well. This gas will then re-emit radiation, some of it in a downward direction, and this downwelling radiation Id(x) will then also pass through our original layer of gas. It in turn will be partially absorbed, heating the gas and causing it to radiate. The net result is that the intensity of the downwelling radiation leaving the gas layer will also change, in this case by an amount δId(x). So far this model is identical to the one outlined in Post 88. The next step is to consider the changes of intensity, δIu(x) and δId(x), using Schwarzschild's equation and the Planck function, B(λ,T), where λ is the wavelength of the radiation and T is the temperature of the gas in kelvins at this height x. In the following discussion I will consider the absorption and re-emission of radiation at a fixed wavelength λ.

The Proof

As the upwelling radiation Iu(x) at a fixed wavelength λ passes through the thin layer of gas of thickness δx a small amount of it will be absorbed by the gas. This amount will be proportional to the number of molecules encountered per unit area, n(x)δx, their absorption cross-section, σa, and the amount of radiation in the original flux, Iu(x). The result is that the transmitted intensity will decrease by an amount equal to Iu(x)σan(x)δx.

At the same time the thin layer of gas will be emitting radiation in both the upward and downward direction due to its temperature T. This temperature will in turn vary with height x, so both B(λ,T) and Iu(x) will vary with x. The intensity emitted in each direction is given by the Planck function, B(λ,T), multiplied by the number of molecules per unit area in the layer, n(x)δx, and the emission cross-section, σe. The result of this emission process is that the transmitted intensity above the layer will increase by an amount equal to B(λ,T)σen(x)δx.

Combining these two effects of emission and absorption gives the total change in intensity of the upwelling radiation

δIu(x) = [B(λ,T)σe - Iu(x)σa]n(x)δx


This is Schwarzschild's equation. However, it is only half the solution because there is a similar equation that can be derived for the changes to the downwelling radiation Id(x) in the thin layer at the same wavelength λ. Thus the resulting equation will be

δId(x) = [B(λ,T)σe - Id(x)σa]n(x)δx


Finally, there is one other consideration we must take into account: energy conservation at wavelength λ. The total energy flowing into the layer of gas must match the energy flowing out. This means that 

δIu(x) + δId(x) = 0


and so

[ Iu(x) + Id(x) ]σa = 2B(λ,T)σe 


We can now eliminate B(λ,T)σe from Eq. 91.1 and Eq. 91.2 and turn both equations into first order differential equtions. However, because the change in downwelling radiation Id(x) in Fig. 91.2 is defined to be positive for changes of x in a negative direction the δx term must be negative, and therefore the differential term dId/dx = - δId/δx.This means that


It then turns out that Eq. 91.5 is identical to Eq. 88.5 in Post 88. So if the equations are the same and the boundary conditions are the same, then the solutions must be the same as in Post 88. So even using Schwarzschild's equation as our starting point we end up with the same result. 

Finally, we can write Schwarzschild's equation as two differential equations, one for Iu(x) and one for Id(x). From Eq. 91.1 we get the obvious result


The equivalent equation for Id(x) is similar, but requires some explanation.


In Eq. 91.7 the order of the I(x)σa and B(λ,T)σe terms to the right of the equality sign have been reversed compared to Eq. 91.6. This is because the downwelling radiation, Id(x), is flowing in the negative direction through the thin gas layer. So for Iu(x) the radiation term B(λ,T)σe is directed in the same direction as Iu(x) so it must be positive, while the absorption in the gas layer must leads to a lower value of Iu(x) above the layer than below. In the case of Id(x) the reverse it true in both cases, hence the reversal of sign for both terms.

Finally, it should be noted that conditions of thermal equilibrium generally require the cross-sections σa and σe to be equal. This is particularly true for black body radiation where a black body at constant temperature must emit what it absorbs, otherwise it will gain or lose energy, and thus not be at constant temperature.



If we now apply the above results to a real system and consider the case of a column of gas of thickness L, and uniform density n that is independent of x, the model outlined above means that the upwelling radiation will vary with x through the gas as (see Post 88)


where Io is the intensity of radiation entering the gas at x = 0, while the downwelling radiation will vary as


The difference between Iu(x) and Id(x) will be constant (ΛIo), as required by energy conservation, and will be equal to the amount transmitted at the end of the column (i.e. at x = L), so 


In Fig. 91.3 below I have suggested what Iu(x) and Id(x) might look like through a 500 metre long horizontal column of air with 420 ppm of CO2 when it is irradiated with 15 µm infra-red radiation from one end. It should be noted, though, that this theoretical data is based on an estimated value for the scattering cross-section of the CO2 molecules that could be almost a factor of ten too small. Nevertheless, it still illustrates qualitatively how the transmission of 15µm infra-red radiation changes through the gas, and also how the Greenhouse Effect (as defined by the relative intensity of the reflected radiation at x = 0) is related to this transmission (see the red curve in Fig. 91.3).

Fig. 91.3: The relative intensities of transmitted (Iu) and reflected (Id) radiation along a 500 metre column of air containing 420 ppm of CO2 with an absorption cross-section of 1.6 x 10-24 m2.

The graph in Fig. 91.3 suggests that a 500 metre column of air will transmit less than 18% of the incident radiation and reflect back the rest. The question is: is this more or less than that predicted by the Beer-Lambert law? The answer to this can be seen in Fig. 91.4 below.

Fig. 91.4: The relative intensities of radiation transmitted  through a column of air of length L containing 420 ppm of CO2 with an absorption cross-section of 1.6 x 10-24 m2 based on two different models: (i) a backscattering model described by Eq. 91.10 (blue curve), and (ii) the Beer-Lambert law (red curve).

The blue curve in Fig. 91.4 demonstrates how the transmission through a column of air would decrease with the length of the column, L, if half of the absorbed radiation is backscattered or re-emitted in the reverse direction and half is re-emitted in the forward direction. The result is a curve that decays with L in accordance with Eq. 91.10. The red curve on the other hand shows how the transmission would change if it followed the (commonly accepted) Beer-Lambert law where the transmitted intensity varies with L as


What is clear is that the Beer-Lambert law underestimates the transmission for large values of L, and in effect predicts that there is no significant transmission beyond 400 metres. The blue curve shows that this is not the case with over 10% transmission for a column of length 1000 metres. In practice this means that the Beer-Lambert law will massively underestimate the impact of future increases in CO2 levels because it will assume that the backscattering is already operating at close to 100% or saturation with 420 ppm of CO2. Yet clearly it is not. In fact even at 10 km (the effective thickness of Earth's atmosphere) there is still more than 1% transmission. While this is small, it is not insignificant, particularly when considering the relatively small changes to the overall size of the Greenhouse Effect that are needed to create a warming of 1.5°C.


The only condition I have applied to the proof and discussion described here is that set out in Eq. 91.4, which basically demands that each layer of the gas radiates the same amount of energy at each wavelength as it absorbs from the two fluxes, Iu(x) and Id(x). This will be true provided all the heat of wavelength λ entering the gas originates from the Earth's surface at the bottom of Fig. 91.2, or at the very top of the atmosphere. In which case, the exact functional forms of Iu(x) and Id(x) will depend only on the functional form of n(x) and the boundary conditions. However, at the top of the atmosphere this condition will not be satisfied as some of the incoming ultraviolet radiation from the Sun is absorbed there before it reaches the ground and so gets converted to infra-red radiation within the upper atmosphere. This means that Eq. 91.4 is no longer satisfied for each and every infra-red wavelength of interest within the gas.

In the following blog posts I will consider a number of distinct and important examples to demonstrate how dependent the transmission is on the density profile of the gas, n(x), and also on the location of the heat source(s). These examples will also show that even when it sometimes looks like there is no Greenhouse Effect in operation, it is still there.

Tuesday, January 11, 2022

90: Lateral thoughts #5: Without fossil fuels there would be no Champagne


Yes, amazing as it seems, without coal we might not have Champagne. Well, possibly.

This tale arises out of the English energy crisis in the early 17th century (see here). After over one hundred years of shipbuilding for the new Royal Navy the country ran out of wood. There were of course other reasons such as population growth and increased urbanization that were equally to blame, just as they are today. But the net result was there were just not enough trees. So much for renewable energy.

In response a worried King James I banned the use of wood for non-essential purposes. And one of the areas to feel the heat was glass-making. In response they switched to using coal, which had previously been considered a dirty and ungodly fuel even though it was abundant. Yet needs must when the Devil drives. And there was an unexpected bonus, as there often is when the Devil is in play. By using coal English glass-makers were able to achieve higher kiln temperatures. This meant they could make glass bottles that were thicker and stronger - just what you need to make Champagne. Otherwise the bottles have a tendency to explode.

So, three cheers for fossil fuels. And yet another reason why the French love the English.

Of course this does raise an important philosophical question. Can you be a Champagne socialist and still fight for climate justice? I don't know. And until I do I'll probably stick to the Château Haut-Brion.

Thursday, December 30, 2021

89. The Greenhouse Effect on Mars

In my previous three posts I have explained how the Greenhouse Effect (GHE) works on Earth, and how it is affected by changes to the carbon dioxide (CO2) concentration in the atmosphere. The problem with studying the GHE on Earth, though, is that its operation is complicated by the presence of large amounts of water vapour in the atmosphere. As water vapour also has a broader absorption band and higher atmospheric concentration than CO2, this means that changes in the CO2 concentration are less important than they would be otherwise. If we want to understand and measure the GHE just due to carbon dioxide, then we need an environment with high levels of CO2 but low levels of other greenhouse gases. In this respect one of the best places to study is Mars

The Atmosphere of Mars

The Martian atmosphere has some similarities with Earth but many differences. It contains many of the same gases (nitrogen, oxygen, water vapour, argon, CO2), but the proportions are vastly different. The atmosphere of Mars is 96% CO2 with about 2% nitrogen and 2% Argon (although different pages on Wikipedia give slightly different values such as 95% CO2 and 3% nitrogen). There are also trace levels (< 0.1%) of other gases such as water vapour (210 ppm) and oxygen (0.15%). 

The other main difference in terms of the atmosphere is the pressure at the surface. At 610 Pa this is only 0.60% of the surface pressure on Earth, and as about 96% of this is CO2, this means that there is a surface density of 3610 mol/m2 of CO2 on Mars compared to only 150 mol/m2 on Earth. So any outgoing radiation from the surface of Mars has 24.06 times as much carbon dioxide gas to penetrate, in order to escape into outer space, compared to on Earth. What this means in practice is that the Greenhouse Effect in the Martian atmosphere should be easier to analyse because it can only have one source - CO2.


The Energy Balance for Mars

Mars is also approximately 52% further from the Sun than is Earth, so one might expect that to mean that its surface is colder. This is true, but not as much as it should be based on distance alone. 

The solar radiation flux entering the Martian atmosphere is only 586 W/m2 compared to the 1360 W/m2 that irradiates the Earth. As this radiation is spread over a surface area (4πr2) that is four times greater than the cross-sectional area of the planet (πr2) in each case, this means that the average radiation flux at the Martian surface is 143.5 W/m2. Yet this is only 22% less than the 184 W/m2 that reaches the Earth's surface. This is because nearly 50% of incident radiation on Earth is either reflected by clouds or is absorbed by ozone and water vapour in the upper atmosphere. But there are no clouds, ozone or significant water vapour on Mars.

Then there is the issue of surface albedo or Bond albedo. This is the proportion of incident radiation that is reflected by the planet back out into space without being absorbed, either from clouds or the surface. For Earth this is about 31%; for Mars it is only 25%. This means that the average surface absorption on Mars is 108 W/m2 compared to 161 W/m2 on Earth. So while Mars only receives 43% of the solar radiation that Earth does, after absorption and reflection the Martian surface receives 67% of the radiation that the Earth's surface does. That is a relative increase of more than 50% for Mars which partially compensates for its greater distance from the Sun.

Calculating the Surface Temperature of Mars

If we now invoke the Stefan-Boltzmann law (see Eq. 13.1 in Post 13),

I = σT4


where I is the surface radiation flux, σ is the Stefan-Boltzmann constant, and T is the surface temperature in kelvins, we see that a surface radiation flux of 108 W/m2 equates to a mean surface temperature on Mars of 209 K (or -64°C). Yet the true mean temperature is thought to be about 215 K (or -58°C). The difference is due to the Greenhouse Effect. 

For comparison, on Earth a solar flux at the surface of 161 W/m2 would equate to a mean surface temperature of 231 K (or -42°C), yet the true mean temperature is about 289 K (or +16°C). So the GHE on Earth adds 58°C of warming while on Mars it only adds 6°C. Yet there is almost twenty-five times more carbon dioxide on Mars (3690 mol/m2) than on Earth (150 mol/m2), so you would expect the greenhouse effect due to CO2 to be stronger. But how much stronger? The answer is: not very. In fact it is significantly less than the actual measured value on Earth.

Calculating the Strength of the GHE on Mars

In Post 87 I showed how the width of the CO2 absorption band at 15 µm can be determined using the known concentration of CO2 (No), its scattering or absorption cross-section (σs), the quantized frequency of rotation of the CO2 molecules (B), and the temperature (T). From this it is possible to estimate the critical CO2 concentration needed to absorb most of the infra-red radiation (Nth). The term Nth can be estimated to be 0.5 mol/m2 based on the value of σs, while No will be about half the total CO2 concentration, so No = 1845 mol/m2. From this we can estimate the maximum number of excited rotational states in the absorption band, Jth, using (see Post 87)


where k is Boltzmann's constant, h is Planck's constant and Z is a normalization constant (see Eq. 87.2 in Post 87) equal to 185.5 in this case. The ratio term kT/hB has the value 185.3 (as hB = 0.1 meV and T = 215 K) and is always approximately equal to the Z value. 

The result we obtain is that Jth = 36.3, which, given that the spacing of the bands is 0.2 meV, means that the absorption band has a width of 14.52 meV and extends from a wavelength of 13.78 µm to 16.44 µm. This compares to a calculated range of 14.00 µm to 16.14 µm for the same band on Earth, although the measured width on Earth is actually found to be from about 13.35 µm to 17.35 µm. So even though the calculated width of the 15 µm absorption band for Mars is slightly larger than the equivalent for Earth, it is not significantly greater. But it is significantly less than the measured band width on Earth.

Fig. 89.1: The electromagnetic emission spectrum for the surface of Mars at a mean temperature of 215K (blue curve) together with the absorption profile due to CO2 between 13.78 µm and 16.44 µm (red curve).

The impact of the high Martian atmospheric CO2 concentration on the radiation feedback is demonstrated in Fig. 89.1 above. The red curve indicates the proportion of the outgoing infra-red radiation (blue curve) that is reflected by the CO2 molecules and it amounts to f = 12.5%. This is slightly more than the 10% seen for the GHE due to backscattering from CO2 on Earth (see Post 87) despite the absorption band being further from the peak in the emission spectrum due to the lower surface temperature on Mars.

Calculating the Temperature Rise

The radiation feedback of  f = 12.5% shown in Fig. 89.1 equates to a 14.3% increase in the surface radiation and also of T4 (because of Eq. 89.1), which then equates to a 3.4% increase in T. So if the initial surface temperature on Mars without the GHE was 209 K, the temperature with the GHE included will be 3.4% greater, or 216.1 K. This means that the temperature rise at the surface due to the Greenhouse Effect is expected to be 7.1 K, which is pretty close to the observed value of around 6 K (or 6°C). The reason for the small difference could be the limited availability of accurate Mars temperature data, or the uncertainty in the value of the CO2 absorption cross-section, σs.

We know how much radiation from the Sun is arriving at Mars to high accuracy, but knowing how much is being absorbed by the planet surface is more difficult as this depends on an accurate measurement of the Bond albedo. However, conventional astronomical telescopes should be able to measure the reflected radiation to pretty good accuracy as well. That allows us to estimate the expected mean surface temperature without the GHE to fairly high precision. The problem is knowing what the actual surface temperature is with the GHE in operation. This is a difficult enough calculation to do on Earth where we have over 16,000 weather stations measuring the surface temperature on a daily basis, and numerous satellites in orbit. Sadly, none, or very little, of this exists for Mars.

So far in this blog post I have assumed a value of 215 K for the mean surface temperature of Mars, but some reports have put it as high as 225 K (or as low as 210 K). In which case Z = 194.15 and Jth = 37.0. This leads to a 15 µm band stretching from 13.76 µm to 16.47 µm, and a feedback factor of f = 13.0%. Under these circumstances the warming from the Greenhouse Effect increases slightly, but only to 7.7°C. 


A Comparison with Earth

For comparison, it is instructive to hypothecate the extent of warming on Earth if its atmosphere also contained 3690 mol/m2 of CO2. In that case Z = 249.3 and Jth = 41.6, which leads to a 15 µm band stretching from 13.62 µm to 16.67 µm, and a feedback factor of f = 14.1%. The resulting predicted temperature rise due to CO2 would be 10.74 K, which is 3.26°C less than the 7.48°C rise currently predicted for Earth as was shown in Post 87). So a twenty-five fold increase in the CO2 concentration would only result in a 3.26°C temperature increase, although as I showed in Post 87, masking by water vapour would probably reduce this by 75% to only 0.8°C. 

It is a point of note that the density of CO2 molecules on Mars (3690 mol/m2) is more than double the combined density of all the greenhouse gases on Earth (1560 mol/m2), yet it results in a temperature rise of 5°C - 7°C that is almost ten times less than the 58°C observed for Earth. This is mainly because most of the GHE on Earth is due to water vapour as the width of the CO2 absorption band is so much less than that for water vapour. Even increasing the concentration of CO2 on Mars by a factor of twenty-five cannot appreciably change this.

Summary and Conclusions

What I hope I have shown in this post is that Mars is a good test bed for studying the Greenhouse Effect (GHE). Knowing only its albedo, the atmospheric concentration of CO2, and the intensity of radiation arriving from the Sun, it is possible to accurately predict the temperature rise due to the Greenhouse Effect. This I have predicted to be about 7°C, in close agreement with the current estimate based on observational data (6°C). And this is despite the significant uncertainty over the true measured value of the mean surface temperature on Mars.

The reduced GHE on Mars relative to the Earth occurs despite its much higher (i.e. 24 times greater) atmospheric CO2 concentration. This in turn suggests that the increasing levels of atmospheric CO2 we are currently seeing on Earth will produce only slight temperature increases in the future. 

Yes, Mars has its own complicating factors. Heat retention on Mars is limited by the thin atmospheric blanket compared to Earth. This means that the planet does not retain heat very well, but conversely it means that the atmosphere will warm quickly when heated by the Sun. For this reason it may be better to consider Mars under direct solar illumination in daytime. Under these conditions the peak solar flux at the surface of Mars will be four times greater than stated above, or 432 W/m2. This will equate to a peak surface temperature of 295 K (or +22°C). Yet the actual maximum temperature is reported to be around 303 K to 308 K (or +30°C to +35°C). So on this measure the warming from the Greenhouse Effect in daytime near the equator appears to be in the range 8°C to 13°C. Yet the predicted value based on a calculation of the width of the 15 µm absorption band is found to be 10.9°C, in other words in the mid-range of the observed values. Once again this is still much less than the total warming seen on Earth and comparable to the contribution to Earth's GHE seen just from CO2.

Tuesday, December 28, 2021

88. A scattering model of The Greenhouse Effect

In my two previous posts I analysed the structure of the 15 µm absorption band for carbon dioxide (CO2) in order to demonstrate how increasing the concentration of CO2 in the atmosphere increases the the band's width. This widening of the absorption band then slightly increases the proportion of infra-red radiation that is reflected back to the Earth's surface and thus increases the strength of the Greenhouse Effect, thereby increasing the surface temperature. As a result I showed that this enhancement is unlikely to contribute more than about 0.5°C to the strength of global warming, and probably contributes less than 0.2°C; i.e. much less than the 1.2°C that is claimed.

In this post I will consider a different impact of an increase to the atmospheric concentration of CO2 by calculating the strength of the backscattering at the centre of the 15 µm absorption band. This will involve describing the mathematical and statistical basis for the Greenhouse Effect, and how it arises from the backscattering of photons by greenhouse gas molecules. As a result I will show just how little infra-red radiation within the 15 µm band can escape through the atmosphere, and so also show just how insensitive the backscattering and transmission of infra-red radiation are to further increases in the CO2 concentration. Finally, I will calculate the expected temperature change for an increase in atmospheric CO2 concentrations from 280 ppm to 420 ppm, and show that it is significantly less than 0.1°C.

How the scattering process works

The origin of the Greenhouse Effect is the interaction of infra-red photons with air molecules, specifically those of carbon dioxide (CO2) and water vapour (H2O). While all air molecules, including those of both oxygen and nitrogen, are capable of scattering infra-red photons via the process of Rayleigh scattering, this scattering is so weak at infra-red wavelengths that it is unimportant. In fact in Post 86 I showed that Rayleigh scattering at wavelengths of 15 µm is a million times weaker than that seen for visible light and which gives the daytime sky its blue colour.

Instead, the dominant process is one of absorption and re-emission of photons where air molecules absorb radiation, and then re-emit it in a random direction. This is only important at wavelengths that match both the peak in the thermal emission spectrum of the Earth's surface (from about 5 µm to 100 µm) and the excitation energies of the molecules. As only CO2 and H2O of the main components of the atmosphere have excitation energies that match the peak region of the thermal emission spectrum of the Earth's surface, they are the most important greenhouse gases.

As I showed in Post 86, the combined absorption and re-emission process results in radiation received from one direction (i.e. the Earth's surface) being redistributed in all directions with equal probability. This is because the time delay between absorption and re-emission means that the orientation of the emitting molecules could be in any random direction. The result is a process that looks just like random scattering, and because the direction of scattering is random it is determined by statistical probability.

One way to visualize the backscattering process is by considering the photons of outgoing infra-red radiation as particles that, every so often, collide with greenhouse gas molecules, and bounce off them at random angles. We can define the average distance photons travel between these collisions as the mean free path. At each collision half the phonons will be reflected back in the opposite direction from which they came, while the other half will be scattered forward. But eventually those that are scattered forward will collide with another molecule and half of those will also be backscattered. This process continues as the remaining forward scattered photons travel further into the medium or gas with more and more being reflected. The result is that, eventually almost all the photons have been reflected back, and virtually none are travelling in the original forward direction. This is how the backscattering process works.

It is important to acknowledge, however, that the real process that is occurring is not a backscattering one. The infra-red photons are actually being absorbed by CO2 molecules which then become excited. They may then transfer this energy to other air molecules via collisions, but eventually those other air molecules will transfer the energy back to the CO2 and the CO2 molecules will re-emit it in some random direction. The time this process takes could be quite long, or very short, but it is happening so often that it will appear almost instantaneous. And therefore it will look just like scattering. 

The mathematics of scattering

In order to gauge the importance of this process we need to quantify it. That means defining the process mathematically. By doing so we will be able to see which factors affect the amount of backscattering, and in particular, how it depends on the thickness of the atmosphere and its CO2 concentration.

In order to do this we must first construct a model of the atmosphere and consider a thin horizontal layer of the atmosphere of thickness δx at some arbitrary altitude x as shown in Fig. 88.1 below. In this model the density of carbon dioxide molecules is n(x), the intensity of the upwelling (or outgoing) radiation at a height x is Iu(x), and the intensity of the downwelling (or backscattered) radiation at the same height is Id(x). 

Fig. 88.1: A schematic illustration of how scattering within each thin layer of the atmosphere alters the upwelling (Iu) and downwelling (Id) intensities.


In thermal equilibrium the atmosphere will emit the same number of photons each second that it absorbs. This is the origin of the scattering process. Conservation of energy then dictates that the difference between Iu(x) and Id(x) must be the same at all heights, x, and equal to the total intensity transmitted at the top of the atmosphere. It therefore follows that 

Iu(x) - Id(x) = IoΛ


where Λ is the effective total transmission coefficient of the atmosphere and Io = Iu(0) is the total intensity of radiation emitted at the surface.

Now we consider the change in intensities of Iu(x) and Id(x) as the radiation passes through the thin layer of thickness δx at height x (see Fig. 88.1 above). In each case there is a probability δp that each photon will be scattered where



and σs is the scattering (or absorption) cross-section of the CO2 molecule at the particular wavelength of the radiation being considered. For each photon that is scattered, half will be scattered forwards and half will be scattered backwards. This means that the value of Iu(x) will decrease by ½Iu(x)δp but increase by ½Id(x)δp due to the backscattering of the reflected photons in Id(x). For Id(x) the opposite will be true. It therefore follows that the change in Iu(x) for a positive increase in x of δx will be

δIu(x) = ½[Id(x) - Iu(x)]n(x)σsδx


while the change in Id(x) for a positive increase in x of δx will be (note the extra negative sign)

δId(x) = -½[Iu(x) - Id(x)]n(x)σsδx


These results can be written in differential form as



The solution to Eq. 88.5 will depend on the distribution of the CO2 molecules within the atmosphere, n(x), and the boundary conditions. The first boundary condition is Iu(0) = Io and sets the amount of radiation that initially enters the atmosphere from the surface. The second is Id(h) = 0 where h is the height of the atmosphere, and this states that once the radiation leaves the top of the atmosphere there can be no more backscattering. The final piece of information we need is the dependence of the CO2 concentration on altitude, n(x).

The simplest model is to assume n(x) is constant (and equal to no) and thus independent of x. This is a fairly good approximation for the lower part of the troposphere and it is one I shall use as a starting point for the analysis. 


Transmission coefficient for a gas of uniform density

From Eq. 88.1 we can see that Iu(x) - Id(x) must be a constant. As n(x) is a constant, Eq. 88.5 dictates that the first differentials of both Iu(x) and Id(x) must be constant as well. It therefore follows that


The solution for Id(x) must then be of the form Id(x) = a(h - x) to satisfy both Eq. 88.6 and the boundary condition Id(h) = 0, while Eq. 88.5 demands that a = ½ΛIonoσs. So

Id(x) = ½ΛIonoσs(h - x)


The only unknown term is Λ, but we can determine it from Eq. 88.1 and the boundary condition for Iu(0) = Io

From Eq. 88.1 we see that Iu(x) = a(h - x) + IoΛ, which means that at x = 0 the condition Io = ah + IoΛ  must hold. This means that the fraction of radiation that is transmitted at the top of the atmosphere, Λ, is given by


If noσsh >> 2, then Λ ≈ 2/noσsh.

It also means that Iu(x) will have a negative linear dependence on x when x < h as follows


while for x > h we find that the transmitted intensity is Iu(x) = IoΛ. This negative linear dependence is very different from the exponential decay that is usually associated with pure absorption or attenuation of radiation and described by the Beer-Lambert law.

Knowing Λ allows us to estimate the percentage of infra-red radiation within the 15 µm absorption band that escapes through the atmosphere for different CO2 concentrations, no. A CO2 concentration of 280 ppm (as existed before 1750) equates to a CO2 molecular concentration of no = 0.0118 mol/m3. The effective thickness of the atmosphere (h) at uniform density is approximately 10 km, while the scattering cross-section (σs) was estimated at around 1.6 x 10-24 m2 in Post 87. This gives a value for Λ of 1.67%. But if we instead consider a CO2 concentration of 420 ppm (as exists today), no = 0.0177 mol/m3 and Λ = 1.12%. So an increase in the CO2 concentration from 280 ppm to 420 ppm leads to the proportion of radiation being reflected by the 15µm CO2 absorption band increasing from 98.33% to 98.88%. This will lead to a (small) increase in the surface temperature which we can now estimate.

In Post 86 I showed that the width of the 15 µm band typically extends from about 13.35 µm to 17.35 µm and thus reflects about 18.1% of all the infra-red radiation emitted by the Earth's surface. But that assumes that the reflectivity of the band is 100%. In fact it is only 98.88%, having risen from 98.33% due to the increase in atmospheric CO2. This means that the increase in CO2 will, in effect, have caused the feedback factor, f, to rise from 18.0% to its current value of 18.1%. We can then use this information, together with the Stefan-Boltzmann law, to determine the change in the surface temperature as we know that 1-f is inversely proportional to T4. What we find is that the temperature rise associated with this increase in reflectivity (and hence increase in f) is less than 0.09°C. In fact the actual rise will be probably be less than half this value due to the masking effect of the water vapour.

Result: Increasing the CO2 concentration from 280 ppm to 420 ppm results in a global temperature increase of less than 0.09°C.


Transmission coefficient for a real atmosphere

The analysis so far has assumed that the atmosphere has a finite thickness of constant density that is independent of altitude. This, though, is not true for Earth. The increasing gravitational potential energy with increasing altitude means that fewer molecules have sufficient energy to reach high altitudes. The result is that the density of air decreases with height, and that dependence on height is exponential.

As carbon dioxide molecules are about 50% heavier than those of oxygen and nitrogen one might expect even fewer CO2 molecules to be present by percentage at high altitudes, but this is not the case. The mixing ratio does not appear to change significantly with altitude. So if there are 420 molecules of CO2 in every one million molecules of air at sea level, the same is true at heights of 10 km, 25 km, and 50 km. It is just that there are fewer numbers of all molecules at those altitudes.

Over the bottom 80 km of the atmosphere the molecular density, n(x), varies with altitude (x) as

n(x) = no e-kx


where k = 1.4 x 10-4 m-1. The same dependence is seen for CO2 with no = 0.0177 mol/m3 for an atmosphere containing 420 ppm of CO2, and no = 0.0118 mol/m3 for an atmosphere containing 280 ppm of CO2. The dependence of molecular density on altitude changes some of the mathematics outlined in the previous section, but both Eq. 88.1 and Eq. 88.5 will remain unaffected. The first change is to substitute Eq. 88.10 into Eq. 88.5 to give


It therefore follows that Iu(x) is given by


where c is a constant of integration. In which case Id(x) will be


The values of c and Λ will be dictated by the boundary conditions, Iu(0) = Io and Id(h) = 0, which are the same as before. Consequently, the results for c and Λ are




It can be seen that if kh << 1, then Eq. 88.15 reduces to the same approximate form as Eq. 88.8, and so both the reflectivity and the temperature rise will be the same as before (see previous section). The more realistic scenario, though, is for kh >> 1, particularly if h > 80 km (in which case kh = 11.2). Then Eq. 88.15 reduces to


which if noσs >> 2k, then becomes Λ ≈ 2k/noσs. We can now calculate the expected temperature rise by following a similar procedure to that outlined at the end of the previous section. 

If no = 0.0118 mol/m3 (equivalent to a CO2 concentration of 280 ppm), then Λ = 0.0232 and so 97.68% of the radiation at 15 µm is being absorbed and then reflected. However, if no = 0.0177 mol/m3 (equivalent to a CO2 concentration of 420 ppm), then Λ = 0.0156. This means that an increase in the CO2 concentration from 280 ppm to 420 ppm leads to the proportion of radiation being reflected by the 15µm CO2 absorption band increasing from 97.68% to 98.44%. Given that the feedback factor for the 15 µm band is currently f = 0.181, this suggests that the previous value was 0.1796.

As the current mean temperature is 289 K, and the temperature must have increased by a factor of [ (1 - 0.1796)/(1 - 0.181) ]1/4 due to the change in feedback factor, then the corresponding temperature rise must be only 0.12°C. Once again the masking effect of absorption by water vapour in the same frequency range as CO2 will probably halve the real value of f from CO2 . So a large increase in the CO2 concentration will probably have less than half the effect on the mean global temperature than is stated here.

Result: Increasing the CO2 concentration from 280 ppm to 420 ppm results in a global temperature increase of less than 0.12°C.


1) Increasing the atmospheric CO2 concentration from 280 ppm to 420 ppm results in the infra-red transmission through the 15 µm absorption band reducing from 2.32% to 1.56%.

2) This leads to the 15 µm band reflecting back an extra 0.76% of the radiation within the borders of the band. But as the band only interacts with 10% of the outgoing infra-red radiation, this amounts to a total net increase of only 0.076% of all radiation being reflected.

3) This increase in reflection is only sufficient to raise global temperatures by 0.06°C.

NOTE: The above analysis assumes a scattering cross-section of σs = 1.6 x 10-24 m2. This unfortunately is just an estimate, and the true value could be up to an order of magnitude greater. Given the importance of this value to the whole theory of climate change one might think its value would be easily found and widely known, but unfortunately it is not. If σs is indeed greater by a factor of ten, then both Λ and the temperature increase due to the CO2 concentration rising from 280 ppm to 420 ppm will be a factor of ten smaller.

Wednesday, December 22, 2021

87. How the Greenhouse Effect on Earth changes with increasing carbon dioxide concentration

In my previous post (Post 86) I explained how infra-red photons emitted by the Earth's surface interact with carbon dioxide (CO2) in the atmosphere to create the Greenhouse Effect. I also showed that increasing the temperature of the planet and increasing the concentration of carbon dioxide in the atmosphere will both lead to an increase in the width of the 15 µm absorption band of CO2. This in turn will increase the amount of radiation that is backscattered by the CO2, and therefore increase the amount of radiation heating the surface of the planet. 

In this post I will attempt to quantify the temperature increase for different increases in the CO2 content of the atmosphere using the results presented in Post 86 and Post 85. What I will show is that the increase in atmospheric levels of CO2 from 280 ppm in 1750 to almost 420 ppm today can only be responsible for at most a 0.5°C increase in average temperatures. This is only about 40% of the 1.2°C claimed by the IPCC and climate scientists. In fact the actual temperature rise due to CO2 is likely to be less than half the calculated value of 0.5°C due to the masking effects of water vapour, and could be as little as 0.1°C. To put this into context, this is less than the values I have calculated for urban heating effects from waste heat (see Post 14 and Post 29) which would persist even without the use of fossil fuels.

The maths and physics

The starting point for this analysis is the quantum structure of the absorption band. This is shown in Fig. 87.1 below and was discussed in detail in Post 86. The key issue is the height of the various absorption lines in the P and R branches. These are identified by their angular momentum quantum number, J, which is numbered for each branch from the centre of the band, Q. 

Fig. 87.1: The detailed structure of the 15 µm absorption band for CO2 showing the absorption peaks associated with rotational transitions.

In Post 86 I also showed that the width of the 15 µm band is determined by the value of J that satisfies the following equation (see also Eq.86.9), this value being denoted as Jth.


In this equation T is the thermodynamic temperature in kelvins, k is the Boltzmann constant, h is Planck's constant and B is the frequency of the rotational angular momentum states. For the rotational transitions shown for CO2 in Fig. 87.1, hB = 0.1 meV and is equal to half the energy separation of the lines in the spectrum in Fig. 87.1. The other terms will be explained below.


The Z term

The term Z is a normalization term equal to the total number of possible rotational states per molecule in the R (or P) branch as follows. 

In the case of Earth where the mean surface temperature T = 289 K, the term Z = 249.3. The energy term EJ = J(J+1)hB. As the degeneracy term (2J + 1) is the differential of the J component of the energy term J(J+1), it follows that for large T the summation in Eq. 87.2 reduces to an integral over all J states, in which case ZkT/hB.
The No term
The term No in Eq. 87.1 is equal to the total number of CO2 molecules per unit surface area found in the R branch. This can be estimated as being equal to approximately half the molecules, with the other half being in the P branch which is assumed to be the mirror image of the R branch (but is not really as was explained in Post 86). This also neglects the significant number of CO2 molecules (particularly at low temperatures) found in the Q peak. Nevertheless, this approach does at least set an upper limit to the width of the R branch, and thus the width of the 15 µm band as a whole. And as will be shown below, it does give results that are remarkably accurate. As the number of CO2 molecules per unit surface area found on Earth is 150 moles per square metre, it therefore follows that No is equal to 75 mol/m2.

Calculating Nth and Jth.

The final remaining parameter to calculate is Nth. Ideally, if the absorption band edge had vertical edges, it would be the threshold number of CO2 molecules per unit area that are just sufficient to completely block the radiation and would be equal to the reciprocal of the scattering cross-section, σs. As σs for CO2 molecules is estimated to be between 10-24 m2 and 10-23 m2 in the 15 µm band, that would imply a value for Nth of about 1 mol/m2. In practice, however, the band edge is curved so the usual definition of the edge is to take the position of the half maximum. This means using a value of Nth = 0.5 mol/m2 is more appropriate.

With all the parameters now set we can calculate Jth using Eq. 87.1 above. The result we get is 29.4, which when multiplied by the line spacing, 2hB, gives the width of the R branch as 47.4 cm-1 in wavenumbers. Assuming the P branch is identical means that the 15 µm band will extend from 619.6 cm-1 to 714.4 cm-1, or from 14.00 µm to 16.14 µm. This is remarkably close to the 14.2 µm to 16.2 µm that is generally observed for the peak in the absorption.

Having calculated the width of the 15 µm band with an atmospheric CO2 concentration of 420 ppm, we can also repeat the procedure for any other CO2 concentration of our choosing. For example, an atmospheric CO2 concentration of 280 ppm that is characteristic of global conditions in 1750 leads to a value for Jth of 27.3, which means that the width of the R branch would be 44.0 cm-1.

The temperature rise

In Post 85 I showed how the reflection of a fraction f of outgoing infra-red radiation would reheat the Earth's surface and cause the radiation it absorbed to increase from Io to a higher value IT as follows


Then in Post 86 I showed how the width of the 15 µm absorption band could be used to determine the value of f by calculating the relative area of this band under the absorption spectrum (see Fig. 86.1). This can then be used to infer a temperature rise due to the absorption by utilizing the Stefan-Boltzmann law,

 I = σT4


If IT is the intensity of radiation emitted by the Earth's surface normally (i.e. 396 W/m2), and f is the fraction of radiation reflected back by the CO2, then the intensity of radiation emitted by the Earth's surface without the CO2 greenhouse effect will, according to Eq. 87.3, be Io = (1-f)IT. We can then use Eq. 87.4 to calculate the respective surface temperatures Tf and To for each radiation emission intensity, IT and Io. The difference in the two temperatures will be the warming due to the CO2.

I showed above that an atmospheric CO2 concentration of 420 ppm leads to an absorption band that extends from 14.00 µm to 16.14 µm. Combining this with the black body spectrum of the Earth at Tf = 289 K (see Fig. 87.2 below) allows us to determine f to be f = 10.0%. This in turn implies that Io = 356.1 W/m2 (where Io is the radiation intensity without CO2 feedback), and thus To = 281.52 K. So the temperature rise due to CO2 is 7.48 K.

Fig. 87.2: The electromagnetic emission spectrum for the Earth's surface at a mean temperature of 289K (blue curve) together with the absorption profile due to CO2 between 14.00 µm and 16.14 µm (red curve).

Now if we reverse this calculation but use an atmospheric CO2 concentration of 280 ppm, we find that the absorption band now extends from 14.06 µm to 16.05 µm, so f = 9.28%. The value of Io = 356.1 W/m2 will be the same as before but the addition of a different amount of CO2 will change IT and Tf because f is different. The new values will be IT = 392.6 W/m2 and Tf = 288.46 K. So the temperature rise from CO2 is now only 6.94 K. This implies that the temperature rise since 1750 due to the atmospheric CO2 concentration increasing from 280 ppm to 420 ppm is only 0.54°C (i.e. 7.48°C - 6.94°C). 

If we repeat this process for other past or future (potentially) CO2 concentrations we can calculate a theoretical temperature rise for each. This is shown in the graph in Fig. 87.3 below with the temperature changes all measured relative to the 1750 value when the atmospheric CO2 concentration was 280 ppm.

Fig. 87.3: The theoretical effect of an increasing atmospheric CO2 concentration on the contribution of CO2 to global warming.

What Fig. 87.3 demonstrates is that the expected temperature increase from an increase in atmospheric CO2 has a logarithmic dependence on the CO2 concentration. However, the trend for concentrations between 280 ppm and 420 ppm is fairly linear and leads to a 0.5°C increase. Overall it appears that doubling the CO2 concentration leads to about 1°C (or 1 K) of warming.

The interpretation of the data

While the logarithmic trend shown in Fig. 87.3 is in general agreement with climate models, the magnitude of the temperature changes are not. Whereas Fig. 87.3 suggests that 420 ppm of CO2 leads to about 0.52°C of warming, the IPCC and climate science are claiming the rise is much greater at about 1.2°C. The difference, I suspect, is probably down to the impact of water vapour. Unfortunately, there are many ways that water vapour can impact the temperature trend.

The conventional view from climate scientists is that water vapour is a positive amplifier; the theory being that a warmer climate causes the amount of water vapour in the troposphere to increase, thus creating even more warming. As I pointed out in Post 86, the total feedback factor for infra-red radiation, f, is about 59%, while CO2 alone can only account for about 18%, assuming that the 15 µm CO2 absorption band width is measured at its half-maximum points from 13.35 µm to 17.35 µm. So the assumption is that water vapour is responsible for the rest, and that as its atmospheric concentration is dependent on the temperature, its concentration will increase as the level of CO2 increases. So this will make the feedback, f, increase about three times faster than from CO2 alone and so massively increase the temperature change to 1.2°C. The problem with this theory is that it ignores two major snags. 

First, the 15 µm CO2 absorption band overlaps with the H2O absorption band, as shown in Fig. 87.4 below. At the high wavelength edge of the 15 µm CO2 band (17 µm) the water vapour will absorb almost 100% of the outgoing radiation while at the low wavelength edge (13 µm) it will absorb about 50%. This means that about 75% of any increase in the width of the 15 µm CO2 absorption band will be masked from the outgoing radiation by the water vapour. In which case the temperature rise will only be 25% of the predicted value, or about 0.13°C.

Fig. 87.4: The absorption bands of carbon dioxide and water vapour at sea level.

Secondly, the window in the H2O absorption band extends from 8 µm to about 15 µm (using the half maxima points). This window allows only 41% of the Earth's outgoing infra-red radiation to escape. As we know that the feedback factor f = 59%, this means that water vapour could be responsible for absorbing and reflecting almost all the outgoing infra-red radiation that is absorbed and reflected. In other words, the 15 µm CO2 band is not needed, and in fact is probably, largely redundant because it is hiding behind the water vapour. So again, a small change to the width of the CO2 band is unlikely to cause any major temperature changes. This is why so many eminent physicists have so many serious reservations regarding the global warming predictions coming out of climate science.


The conclusions

1) Increasing the atmospheric CO2 concentration will increase the width of its 15 µm absorption band.

2) In the absence of water vapour this could raise global temperatures, with an increase in CO2 concentration from 280 ppm to 420 ppm resulting in a 0.5°C increase in global temperatures. 

3) The projected temperature increase has a logarithmic dependence on CO2 concentration (see Fig. 87.3).

4) Water vapour masks most of the CO2 15 µm absorption band and so dominates the infra-red absorption. It can also account for almost all of the radiation feedback on its own.

5) The impact of water vapour means that an atmospheric CO2 concentration rising from 280 ppm to 420 ppm could result in as little as a 0.13°C increase in global temperatures. This is ten times less than is currently claimed by climate science.

The caveats and discrepancies

In this analysis there are major uncertainties over the value of the CO2 scattering cross-section, σs, and the widths of the CO2 and H2O absorption bands. This is also due to the difficulty in estimating the parameters No, Nth and Jth. However, the overall level of agreement between this analysis and real data and existing theory is encouraging.

The major discrepancy is between the measured width of the CO2 15 µm absorption band at its half maximum, where it extends from 13.35 µm to 17.35 µm, with the predicted width based on Jth where it extends from 14.00 µm to 16.14 µm. This difference could be due to line broadening from pressure broadening and temperature.

Sunday, December 19, 2021

86. How photons interact with carbon dioxide molecules

In my previous post (Post 85) I countered a number of myths surrounding the greenhouse effect, and outlined how it really works. Many people falsely believe that it is caused by a heating of the atmosphere by outgoing infra-red radiation, but this is not the case. The key concept at the heart of the greenhouse effect is photon scattering, or more accurately, the absorption and re-emission of infra-red photons by molecules of the greenhouse gases. There are three main gases that perform this role: carbon dioxide (CO2), water vapour (H2O), and methane (CH4). Each of these compounds absorbs electromagnetic radiation at its own set of unique wavelengths, but it is the absorption in the wavelength range between 6 µm and 90 µm that is key as this is where 96% of the thermal emission from the Earth's surface takes place. As CO2 is seen as the most important player in terms of anthropogenic greenhouse gas emissions I will consider its role in greatest detail.

The emission and absorption spectra

In Fig. 85.1 of Post 85 I showed the main absorption bands of carbon dioxide in the infra-red part of the electromagnetic spectrum. In total there are four main bands at 2 µm, 2.7 µm, 4 µm and 15 µm, but only the 15 µm band is of any importance as the other three have energies that are way beyond the peak of the infra-red emission spectrum for Earth's outgoing radiation. This is illustrated in Fig. 86.1 below, where the blue curve represents the Earth's emission spectrum at 289 K and the red curve shows which frequencies are absorbed by CO2. The area under the blue curve thus represents the total power of the radiation emitted at the Earth's surface, while the area under the red curve is the amount of radiated heat that can be absorbed and then reflected by CO2.


Fig. 86.1: The electromagnetic emission spectrum for the Earth's surface at a mean temperature of 289K (blue curve) together with the absorption profile due to CO2 between 14.2 µm and 16.2 µm (red curve).

The graph in Fig. 86.1 shows that only the 15 µm band is important as it is the only one close to the radiation peak. It is centred at 667 wavenumbers. Note: wavenumbers are the number of wavelengths of a radiation type that can fit in a distance of one centimetre. They are used because they are a measure of frequency, they result in much smaller numbers compared to other units of frequency like hertz, and are proportional to the photon energy of the radiation. The absorption bands at 2 µm, 2.7 µm and 4 µm correspond to wavenumbers of 5000, 3704 and 2500 cm-1 respectively. Consequently, they can only interact with the tail of the frequency spectrum where there is very little energy being emitted (i.e. less than 1% of the total).

What the curves in Fig. 86.1 show is that only about 9% of the Earth's outgoing radiation can be absorbed and then reflected back by the carbon dioxide molecules in the atmosphere. In other words, the area under the red curve is only 9% of the area under the blue curve. Yet in Post 85 I showed that in reality 59% of the radiation is reflected back. The difference in these two numbers is largely due to water vapour which not only accounts for most of that difference (i.e. 50%), but also overlaps the 15 µm absorption peak of CO2 (see Fig. 85.1 in Post 85), and so may render much of its function redundant. This is one reason why the claim that increasing the amount of CO2 in the atmosphere will necessarily increase the strength of the greenhouse effect is disputed by many.


Fig. 86.2: The absorption bands of carbon dioxide and water vapour at sea level.


The other complicating factor is the actual width of the 15 µm band. The peak of maximum absorption generally stretches from about 14.2 µm to 16.2 µm, or from about 700 to 620 wavenumbers (or a width of 10.8 meV), and results in a feedback factor f = 9.2% and a temperature rise of 6.88°C. 

The band width, though, is dependent on both temperature and pressure. The full band at its tails can stretch from 12.5 µm to 18.5 µm (800 to 540 wavenumbers or a width of 32.2 meV), but absorption in the outer parts of the band decreases dramatically (see Fig. 86.2 above). If we were to use the full width of the band to determine the backscatter it would result in 26.8% of the outgoing radiation being reflected (see Fig. 86.3 below) and a temperature rise of 21.7°C; still much less than the 59% backscatter and the 58°C temperature rise that is actually seen.

Fig. 86.3: The electromagnetic emission spectrum for the Earth's surface at a mean temperature of 289K (blue curve) together with the potential absorption profile due to CO2 between 12.5 µm and 18.5 µm (red curve).

A more realist estimate would be achieved by considering the width at the half maximum points of the band. This will be from about 13.35 µm to 17.35 µm (749 to 576 wavenumbers or a width of 21.4 meV) and would result in an absorption and reflection of 18.1%. This equates to a temperature rise at the surface of the Earth of 14.1°C.

Backscattering by Rayleigh scattering

The photons of infra-red radiation emitted by the Earth's surface interact with the carbon dioxide in two ways. One is by elastic scattering and the other via absorption and re-emission.

In elastic scattering the photons basically bounce off the CO2 molecules like the collision of two billiard balls. This is the basis of what is known as Rayleigh scattering, named after the nineteenth-century British physicist Lord Rayleigh (John William Strutt), and it is a result of the electromagnetic interaction of the photon with the charged particles (electrons and protons) in the molecule. After collision the photons will end up travelling in a different direction. Some will be reflected back while others will carry on in a forward direction. It is the ones that are reflected that contribute to the greenhouse effect. The intensity of photons at a distance R from the molecule after being scattered through an angle θ is given by the equation


It can be seen that the overall scattering probability depends on the wavelength of the radiation (λ) and the strength of the polarizability of the molecule doing the scattering (α). The strong inverse dependence on wavelength explains why the sky is blue in daylight. This happens because blue light photons with wavelengths of about 400 nm scatter sixteen times more often as they travel through the Earth's atmosphere than red light photons of wavelength 800 nm do. It also means that Rayleigh scattering from carbon dioxide at wavelengths of 15 µm is about a billion times less likely to occur than the scattering of visible light. This is because the wavelengths of the photons being scattered are at least twenty times greater, so the scattering rate of 15 µm photons by CO2 molecules is at least 160,000 times less than for visible light. But in addition, the proportion of CO2 in the atmosphere is only 0.042% of all the molecules. This equates to about 150 moles of CO2 per square metre of the Earth's surface.

Another way of describing the strength of the scattering is via a scattering cross-section, σs. This is a measure of the effective cross-sectional area of each molecule that a photon will see, or alternatively how effective that molecule is at blocking the radiation. The scattering cross-section is usually (but not always) much less than the physical dimension of the molecule. 

Using the definition of scattering cross-section Eq. 86.1 becomes

where the scattering cross-section, σs(λ,α) is a function of both the wavelength of the radiation (λ) and the polarizabilty (α) as follows

All molecules have their own unique polarizabilities; a list of some of the more common ones can be found here. Most of the common atmospheric gases have similar values for α, although the value for CO2 is about 50% higher than most. So scattering from CO2 is about twice that from oxygen and nitrogen. Nevertheless, this still means that with only about one in a billion 15µm photons being scattered by the 150 moles per square metre of CO2, the Rayleigh scattering cross-section of CO2 at this wavelength equates to only about 10-35 m2. This is a million billion times smaller than the actual size of the molecule.

The key point, therefore, is that Rayleigh scattering is of negligible importance when discussing the greenhouse effect because it is a billion times weaker than the Rayleigh scattering we see of visible light. Absorption and re-emission of photons by the 15 µm band are the more important processes, but as I will explain next, their combined behaviour in terms of scattering is very similar to Rayleigh scattering, but much stronger.

Backscattering by absorption and re-emission

Almost all backscattering by carbon dioxide occurs via a process of absorption and re-emission of the infra-red photons with wavelengths in the region of the 15 µm band (i.e. 12.5-18.5 µm). The closer these photons are in wavelength to the centre of the band the more strongly they are absorbed.

Once excited, the molecule may then collide with other molecules and exchange its extra energy with them, thereby heating the gas as a whole. But once all the gas is heated equally, thermal equilibrium will be established where some of the excited CO2 molecules will re-emit photons at the same rate that other CO2 molecules are absorbing them. This situation physicists refer to as the steady state. At this point, for every photon being absorbed, another is being re-emitted by another molecule in a random direction.

Ultimately, every molecule wants to get to its lowest energy state, so while bombarding it with radiation will force its energy state to increase, it also increases the rate at which it tries to lose energy. So eventually an equilibrium is achieved where the amount of radiation being absorbed by the gas balances the amount that is lost through re-emission. The key point to note, though, is that the re-emission is a random process that can result in the emitted photons being ejected in any direction. In contrast, most of the absorbed photons come from a single direction: the Earth's surface. This means that fewer of the re-emitted photons end up travelling away from the surface as some are reflected back. Those photons that are reflected back then provide additional heat to the surface, thereby raising its temperature even further. This is the origin of the Greenhouse Effect. 


Structure of the absorption band

The infra-red photons emitted from the Earth's surface are absorbed by carbon dioxide via a process of molecular excitation where the molecule is promoted to a higher energy state. Usually this process is in the form of electrons within the molecule moving to higher energy levels, but those transitions are usually of very high energy (several electron volts or eV) and so they only result from the absorption of visible or ultraviolet light. 

The 15 µm transition in carbon dioxide is in the infra-red part of the spectrum and so it is of a much lower energy (about 83 meV). Instead it results from a bond stretching excitation where the molecule bends, as shown in mode v2 in Fig. 86.4 below, under the action of the incoming electromagnetic wave of the photon. This bond stretching occurs because of an uneven distribution of electric charge within the molecule which then interacts with the electric field of the incoming electromagnetic wave or photon.

Fig. 86.4: The three vibrational modes of carbon dioxide. The symmetric stretch mode (v1) occurs at 7.5 µm but does not absorb because of symmetry violation. The bend mode (v2) corresponds to absorption at 15 µm, while the asymmetric stretch mode (v3) corresponds to absorption at 4.3 µm.

In addition to the bond stretching, a much smaller amount of energy from the incoming photon may go towards increasing the angular momentum or rotational spin of the molecule. This leads to "wings" or branches to the central band as shown in Fig. 86.5 below, with the lower energy branch being denoted as the P branch, and the higher energy one the R branch. It is these branches that cause the band to broaden from a single sharp peak Q at 15 µm to a broader band stretching from 12.5 µm to 18.5 µm.


Fig. 86.5: The detailed structure of the 15 µm absorption band for CO2 showing the absorption peaks associated with rotational transitions.


The branches P and R are each comprised of a set of evenly spaced discrete transitions, each one matched to a different change in rotational energy of the molecule due to a change in its angular momentum state J by an amount ∆J = ±1. This is because the incoming photon has an angular momentum quantum of J = +1 and angular momentum must be conserved in the absorption process. The R branch arises from absorption processes where the photon causes the molecule to spin faster, while the P branch arises from absorption processes where the photon causes the molecule to spin more slowly (i.e. the angular momentum of the photon is in the opposite direction to that of the molecule). The two branches are, in theory, mirror images of each other, but in practice centrifugal distortion and rotational-vibrational coupling lead to the line spacing increasing at lower energies.

As the energy of each rotational state J of the molecule is (ideally) equal to J(J+1)hB where h is Planck's constant and B is given by


with I being the moment of inertia of the molecule about its spin axis, the energy change associated with a transition from state J to J+1 when a photon is absorbed will be ±2hB(J+1) (see here). So the absorption peaks due to changes in rotation have an ideal energy separation of 2hB, or about 1.54 wavenumbers (cm-1) or 0.2 meV for lines in the 15 µm band of carbon dioxide. 

These individual line transitions are also broadened via doppler broadening due to the different speeds and directions of motion of the various molecules, and pressure broadening due to collisions with other molecules in the atmosphere. Both of these broadening mechanisms are temperature dependent and increase as the temperature increases. The result is that the width and overall scattering cross-section of each line will increase with temperature thereby narrowing or closing completely the gaps between the individual rotational lines in the spectrum. For lines in the 15 µm absorption band of CO2 the typical scattering cross-section is about 10-23 m2, which is about one trillion (i.e. 1012) times greater than the scattering cross-section of each molecule due to Rayleigh scattering at the same wavelength.

In addition, an increase in temperature will shift the position of the peak or maximum in both the P and R branches, Jmax (again see here). This is because the height of each line in the 15 µm band is dependent on both its J number and the temperature. Each line has a degeneracy of 2J+1, while the probability of exciting that rotational J mode depends exponentially on both energy and temperature. This means that the relative occupancy of each mode, NJ, is given by 


where the energy EJ of the mode J is given by J(J+1)hB, No is the total number of excited modes, and Z is a normalization term that is equal to the sum of the probabilities.


This also explains why the P and R branches in Fig. 86.5 have peaks in their distributions. The occupancy of each rotational state NJ increases linearly with J via the degeneracy term 2J+1, but for large J decreases more rapidly due to the energy term EJ in the exponential. This leads to a maximum in the distribution, the position of which relative to the band centre at Q can be determined by differentiation with respect to J. The result is given by 


What this shows is that as the temperature of the gas increases, the position of the maxima will move further from the band centre at Q, so the overall width of the band will increase with temperature as well. This is because higher temperatures will lead to a higher proportion of CO2 molecules in higher energy rotational states with larger J values. Generally, though, it takes a large temperature change to significantly widen the band. As the position of the R branch maximum relative to Q increases with temperature T as √T, it is reasonable to expect the width of the band to do the same.

Finally, if the width of the absorption band can change with temperature, then it can also change with CO2 concentration. In this case the width will be set by the highest value of J for which NJ exceeds some threshold value Nth, this being the number of modes required to almost completely absorb the outgoing radiation at that wavelength. According to Eq. 86.5 this will be the value of J such that


Rearranging Eq. 86.8 gives the following result for the threshold J value, Jth, that defines the edge of the band


So Eq. 86.9 not only confirms that the band width varies approximately as √T, but also that it varies approximately as √ln(No). 


The dominant mechanism by which CO2 molecules scatter infra-red radiation in the 15 µm band is by absorption and re-emission.

The absorption and re-emission process has a similar effect to that of Rayleigh scattering, but it is almost a trillion (i.e. 1012) times stronger. 

The width of the 15 µm band increases with temperature as √T, but increases much more slowly with CO2 concentration (see Eq. 86.9).