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No longer naked: IR selection rules and greenhouse gases

Posted on September 23, 2021 by Anshul Kogar | Leave a comment

In the previous post, I walked through the simple “naked planet” model to calculate the average temperature of planets. This resulted in an excellent approximation to the average temperature of Mercury and was a little off when it came to the earth. What I didn’t say was how horrendously this model performs for Venus. In that case, our model predicts a temperature of 232 K while the observed temperature is 735 K. Put more terrestrially, we predicted that the temperature would be like a Siberian winter (-40 degrees C (or F)), whereas the real temperature is hot enough to melt lead. Why is the model so far off in this case?

Simply stated, we failed to take into consideration the atmospheric greenhouse effect. Mercury happens to lack a substantial atmosphere, so the greenhouse effect doesn’t play a role. It turns out that Venus’ atmosphere is a hot, dense gaseous stew of carbon dioxide (~96% $CO_2$ ) and other gases (including sulfuric acid ( $H_2SO_4$ )!) — the greenhouse effect is extremely dramatic there. However, on Earth, the greenhouse effect leads to a comparatively small warming of the surface, but this small warming effect is decisive for the biology of the planet.

So, what is the greenhouse effect, anyway? Essentially, it is a one-way light (and thus heat) absorber. Visible light from the sun passes through the atmosphere and heats up the surface of the earth. Because Earth is much cooler than the sun, the blackbody radiation emitted from the earth is in the infrared part of the spectrum. So while the atmosphere is almost transparent to visible light from the sun, it absorbs the infrared light from the earth. Energy thus gets in, but has trouble leaving, resulting in a heating effect.

There are a lot of questions that one can ask about the greenhouse effect, and in this post I’ll be addressing a couple of the them: (1) What makes a gas a greenhouse gas? and (2) Can we model the greenhouse effect to get an idea about how it warms the earth?

In our atmosphere, there are many gases — nitrogen (~78% $N_2$ ), oxygen (~21% $O_2$ ), argon (~1% $Ar$ ), and trace amounts of other gases (~0.04% $CO_2$ and <1% water vapor). So why is it that we are so concerned with trace amounts of $CO_2$ (and water vapor)? Why are these greenhouse gases? Is there a way to predict whether a gas will be a greenhouse gas?

Greenhouse gases consist of molecules that possess “infrared active” vibration modes that absorb light at wavelengths that the earth emits. Infrared active modes vibrate asymmetrically, and these asymmetric vibrations are the only ones capable of absorbing a significant amount of light. (A couple posts ago, I described the symmetry principles describing how infrared active vibrations interact with light’s electric field.) Gases like $N_2$ and $O_2$ only possess symmetric vibration modes and are thus not capable of absorbing light (at least to leading order, i.e. in the electric dipole approximation). Now let’s take a look at the vibration modes of the $CO_2$ molecule to see why it is critical to the climate.

The four modes of vibration of the $CO_2$ molecule. The asymmetric elongation and the bending vibrations are infrared active. The bending vibrations are primarily responsible for the greenhouse effect because the frequency falls within the happens to coincide with the peak of Earth’s infrared emission.

The four modes of vibration of the $CO_2$ molecule. The asymmetric elongation and the bending vibrations are infrared active. The bending vibrations are primarily responsible for the greenhouse effect because the frequency falls within the happens to coincide with the peak of Earth’s infrared emission.

$CO_2$ possesses four modes of vibration, as shown in the image above. Of these four, three of them are infrared active, because they are inversion asymmetric vibrations (again, see this post if that doesn’t make sense.) Most important are the bending vibrations, because their frequencies are close to where most of the light is emitted from earth. Below is a simulation (using this applet) showing the emission spectrum from Earth as viewed 70 km from the earth’s surface. On the left hand side, a few greenhouse gases are included in the atmosphere, but no $CO_2$ . On the right, 0.042% (or 420 ppm) of $CO_2$ is added to the atmosphere. (Also included in the plot are the blackbody spectra at various temperatures.) As you can see, the effect is very dramatic — a lot of light is now absorbed by the bending vibration of the $CO_2$ molecule at 666 $cm^{-1}$ . About half of this absorbed light will be re-radiated back down towards the earth. Water vapor and methane also have dramatic effects on the spectrum, which is why much of the radiation is absorbed above 1400 $cm^{-1}$ . (Below about 500 $cm^{-1}$ , it is mostly the water molecule’s rotational degrees of freedom that are responsible for absorption.)

*Earth’s simulated emission spectrum as viewed from 70 km above the surface with (left) no $CO_2$ in the atmosphere and (right) 0.042% of $CO_2$ in the atmosphere.*

*Earth’s simulated emission spectrum as viewed from 70 km above the surface with (left) no $CO_2$ in the atmosphere and (right) 0.042% of $CO_2$ in the atmosphere.*

These plots suggest that the presence or absence of trace amounts of $CO_2$ (and water vapor) can potentially have quite a dramatic effect on the earth’s climate. To see whether this is the case, it helps to see this in a simple model; we thus need to update the naked planet model from the previous post and dress it up. One way to do this is to consider a medium surrounding the earth that transmits visible light and absorbs infrared light — something like in the image below (adapted from here). In this crude model, the atmosphere absorbs all the infrared light being emitted from the earth and then re-emits it isotropically.

*A schematic of the one-layer model for the atmosphere*.

The calculation proceeds in a similar way to the one from the previous post. Each tier is in a steady state — the power coming in must equal the power going out. Above the atmosphere, this means that the incoming power from the sun must equal the power leaving the atmosphere:

$P_{sun} = L(1-A)\pi R_E^2 = P_{a} = \sigma T_a^4*(4\pi R_E^2)$

Because this equation is identical to the one solved in the previous post, this gives $T_a$ = 255 K. Now, we can solve for the earth’s surface temperature by considering the second tier. Here, the incoming power from the sun and the atmosphere must be balanced by the outgoing power from the earth’s surface:

$P_{sun} + P_{a} = L(1-A)\pi R_E^2 + \sigma T_a^4*(4\pi R_E^2) = P_{s} = \sigma T_s^4*(4\pi R_E^2)$

Having already solved for $T_a$ from the previous equation, $T_s$ can be obtained. $T_s = 2^{1/4} T_a$ = 303 K. Remember that the naked Earth model yielded a temperature of 255 K. So this atmosphere, which in our model absorbs all outgoing light, increases the temperature of the earth by roughly 48 K. Now, the actual temperature of the earth’s surface is 288 K, so this model overestimates the greenhouse effect quite a bit. But considering how crude this model was, in that it absorbs all outgoing radiation, it’s not surprising that this model inflates the actual temperature.

Nonetheless, I find that this model gives a good intuition about how the greenhouse effect works. It captures the essential physics of what is going on and makes us realize that these changes we are observing in our climate actually arise from some very simple and basic fundamental principles.

Note: This model does not work very well for Venus because its atmosphere is not a single thin layer. It is a dense and thick medium. It would be more appropriate to have many layers tiled atop one another.

Naked planets

Posted on September 21, 2021 by Anshul Kogar | 1 comment

Today, there is no greater global challenge than climate change — an existential threat to most lifeforms on the planet, including to us humans. Thus, while our day-to-day work is important, it probably helps to understand and think about the basics behind the warming of the earth.

In this post, I want to give a general overview of the main scientific idea behind the temperature of most planets. Rather than focusing on data, which has been presented in various places, in this post, I’ll be focusing on the simple concept that the sun heats the planets up. We can actually get a fairly good estimate of the average temperature of some planets by considering only this effect. I’ll do this for Mercury first and then the Earth to illustrate.

To start, we treat the sun as an idealized blackbody with a surface temperature of roughly 5800 K. Using the Stefan-Boltzmann law, we can figure out the total power emitted by the sun:

$P = \sigma T^4 S$ ,

where $\sigma = 5.67\times 10^{-8} W/K^4m^2$ and $S$ is the surface area over which radiation is emitted (i.e. $S = 4 \pi R^2$ , where $R$ is the radius of the sun). Since blackbodies emit in a totally isotropic fashion, we can calculate the power per unit area, $L$ , at Mercury’s distance from the sun by dividing the power by $4\pi R_{s-to-M}^2$ , where $R_{s-to-M}$ is the distance from the sun to Mercury. It helps to appeal to the image below (adapted from here) to see where this expression comes from.

*The sun radiates throughout the full $4\pi$ solid angle. The power per unit area at the distance of Mercury is thus $P/4 \pi R_{s-to-M}^2 W/m^2$ .*

*The sun radiates throughout the full $4\pi$ solid angle. The power per unit area at the distance of Mercury is thus $P/4 \pi R_{s-to-M}^2 W/m^2$ .*

To obtain an expression for the total power received by Mercury, we need to multiply the power per unit area times the area of Mercury that receives the light. Because only the part of Mercury that faces the sun receives the light, we would need to integrate the power over the exposed part of Mercury’s surface. However, there is a simpler way to do this calculation and that is to consider the shadow behind Mercury (see the image below, which is taken from here.)

*The total power incident on Mercury can be calculated by noting that its shadow is circular.*

By considering this image, we can see that the area intercepting the light is the area of the circle that casts the shadow. The total power absorbed by Mercury will then be:

$P_{M-abs} = L (1-A) \pi R_{Mer}^2$ ,

where $R_{Mer}$ is the radius of Mercury and $A$ is the “albedo” which quantifies how much light is reflected from the surface of the planet. For instance, some of the sun’s rays incident on Earth will be reflected by clouds, ice covering land or ocean, aerosols and other reflective objects. For Mercury, the albedo is roughly 0.1, meaning that 10% of the light is reflected back into space, and is not absorbed, and thus has no heating effect on the planet.

The last idea we have to use is that of the steady state. In the steady state, the power absorbed by Mercury is the same as that emitted, so that the planet stabilizes at some temperature. (Away from the steady-state, the planet would heat up or cool with time). To figure out how much power is emitted, we treat Mercury as a blackbody, so that the power emitted is:

$P_{M-emit} = \sigma T^4 S_{Mer}$ ,

where $S_{Mer} = 4 \pi R_{Mer}^2$ . Note here that even though only the part of Mercury that faces the sun absorbs light, the whole surface area of Mercury emits radiation (in line with the definition of a blackbody). Now equating the power absorbed to the power emitted, we can solve for the temperature of Mercury. By plugging in the numbers, I get that Mercury’s temperature is 437.7 K (~165 C). The observed temperature of Mercury, by the way, is roughly 440 K — only a couple of degrees off!

Were we to perform the same calculation for Earth, we’d get a frigid ~256 K for the Earth’s temperature. However, in this case, the observed average temperature is ~288 K.

Considering the crudeness of this naked (meaning no atmosphere) planet model, it performs quite well for certain planets. However, the obvious question that arises is: Why is the discrepancy for the Earth greater than that for Mercury? This is where the atmosphere comes in (and the greenhouse effect…). I want to say that I’ll address greenhouse gases in a subsequent post, but my track record for these kinds of promises is not something I’m particularly proud of (see here and here for instance.)

For those of you who would like to carry out the calculation yourselves, I used the following numbers:

Radius of the sun, $R = 696.3$ million meters