Tag: radiative balance

Lesson 5b: More on atmospheric absorption

800px-Atmospheric_Transmission

This image comes from the Wikipedia article on the greenhouse effect.

The red bit is the sunlight coming down. The drawn line is roughly what’s at the top of the atmosphere (there is some loss because of Fraunhofer lines, but basically it’s a perfect blackbody) and the coloured in red bit is what reaches the Earth’s surface. The missing wavelengths are absorbed – and you can see below why:

  • UV and blue are absorbed by ozone in the upper atmosphere and “Rayleigh Scattering” (the thing that makes the sky look blue in the daytime and red at sunset).
  • The near infrared (sunlight with wavelengths too long for us to see) is absorbed mostly by water vapour (and a few wavelengths by carbon dioxide) (more to come)

The blue and purple lines are what the Earth would emit at different Earth temperatures (the one most to the left is for a blackbody at 310 K – or about 35 ºC – the one furthest right for temperatures around 210 K – or about -63 ºC). The solid blue bit is the only bit that gets through the atmosphere. All other wavelengths are absorbed by water molecules (“water vapor” plot) or by carbon dioxide or other greenhouse gases.

The diagram is drawn for a normalised blackbody curve – so you can see the thermal infrared one and the solar one on the same picture. In reality the solar one would be much “higher” as well – and the blue ones would not only shift left with higher temperatures, but also get taller.

Because so much of the output spectrum is absorbed, the Earth will heat up until it’s output is equal to its input: it needs to be at a hotter temperature for the energy in the coloured in blue bit to be equal to the area under the whole curve at a lower temperature.

This is known as the ‘greenhouse effect’ – but that’s actually a poor name. Yes, there is some real “greenhouse effect” in a greenhouse: the sunlight gets through the glass, but the thermal radiative energy of the surfaces in the greenhouse, emitting thermal infrared radiation, can’t get back out again … but actually the main reason real greenhouses warm up is that hot air can’t escape… ah well!

Lesson 5: Atmospheric absorption

So in Lesson 4, we learnt that if the Earth had no atmosphere, but still reflected about the same amount of sunlight as it does now, it would be at about -15 °C to -20 °C on average to be in “thermal equilibrium” where the energy coming in from the sun matched the energy coming out through the Earth’s own, thermal infrared, blackbody radiation.

Of course, we all know from our personal experience that the average temperature of the Earth (averaged over the whole Earth, whole day, whole year) is a lot hotter than that. So what is it that the atmosphere does?

To think about that, let’s start with a revision of Lesson 3 about light being absorbed and emitted by atoms. First, the “electromagnetic spectrum” is what I drew in lesson 1: it is the “rainbow” in the visible, and extends that to other wavelengths of electromagnetic radiation. If you look at the visible spectrum (the rainbow) of the sun, you see black lines in the spectrum. These are known as Fraunhofer Lines after the scientist who first described them (see lesson 3b).

Light coming from inside the sun “excites” an atom in the outer parts of the sun, which means that an electron goes to a higher orbital. Then, when the atom returns to its lower state, it releases light with the same wavelength: but it does so in a random direction. So the amount of light heading towards us decreases at that wavelength and we see a black line in the solar spectrum.


In the Earth’s atmosphere the same thing happens – both on the way down and on the way up. Every atom has its own set of lines where it absorbs. But additionally, molecules can absorb lines too. In the atom case, the absorbed energy from the light is used to move a very light-weight electron up to another orbital inside the atom. With molecules, the absorbed energy from the light makes the molecules vibrate in new ways. Since in molecules the things moving are much heavier atoms (rather than very light electrons), all this happens with a lower frequency – and molecular absorptions are in the thermal infrared.

Incoming light from the Sun reaching the top of the atmosphere is in the UV, visible and near infrared spectral region. The UV is absorbed by atoms (and some molecules like ozone) this light gets re-emitted but in all directions, including out of the atmosphere, and is lost. That’s how our ozone layer protects us from harmful UV. Other visible wavelengths are absorbed by the atmosphere too – some Fraunhofer lines are due to atoms in the Sun, others are due to atoms in our atmosphere. This means that some wavelengths do not make it down to Earth.  But this absorption is only a few lines, and it doesn’t affect the overall amount of energy reaching the surface very much.

The Earth’s emitted radiation is in the thermal infrared. This longer wavelength (lower energy) light gets absorbed by molecules to make them vibrate in lots of ways.

Wikipedia has some great images of water molecules vibrating:

The yellow ball is the oxygen. The blue balls (which should really be much smaller than the yellow ball) are the hydrogen atoms (H2O!). Imagine you were holding a model of this with springs for the bonds and balls for the atoms. You can imagine that there are lots of ways for the molecule to vibrate and rotate. Each transition from one way of vibrating to any other way of vibrating requires just the right amount of energy supplied through light at “just the right wavelength”. So you can imagine there are lots and lots of thermal infrared wavelengths that get absorbed by all the water molecules in the atmosphere. And, while that light can also be re-emitted, that will be in any direction – including straight back down to Earth and into the path of another molecule.

[Actually, because the water molecules aren’t cold themselves, they are already doing some vibrations of their own – this actually leads to even more wavelengths being “just right” to create transitions between different vibrational modes.]


There are some difficult concepts in here, so I’ll stop and add give space for questions. 

Lesson 4b: Further thoughts on the temperature of the Earth

The ideas developed in Lesson 4 are the basic “radiative balance” that make up the Earth’s “energy budget” (look up radiative balance or Earth’s energy balance on Wikipedia as a starting point). This is all based on a very simple concept – if an object has more energy coming in than going out, it will heat up, until the energy in balances the energy out. Similarly, if an object has more energy going out than coming in, it will cool down until the energy out balances the energy in. All physical systems try to maintain an equilibrium. In the Earth system, the energy “in” is coming from the Sun, the energy “out” is coming from the Earth’s own blackbody radiation – and the hotter it is, the more it is radiating out.

I’ve considerably over-simplified the problem in the example in lesson 4 (and not just by ignoring the greenhouse gas effect – which we’ll come onto next).

First, I’ve ignored any heat generated by the Earth’s core, which does work its way up to the surface (hot springs and geysers). This is ok, though, the heat that comes up from the core is about 0.03 % of the energy that comes from the sun. That is much smaller than the size of the approximations I’m giving above.

Second, to improve this calculation you’d have to properly know the average reflectance in the solar reflective spectral region (from UV to short wave infrared). That’s called the “Earth’s albedo” and will vary from something very high (clouds, snow) to something very low (deep ocean, dark forests). It’s generally assumed that the average albedo is somewhere between 0.2 and 0.4 (search it yourself if you want).

You also need to know the Earth’s emissivity in the thermal infrared. Generally natural surfaces have a high emissivity in the thermal infrared – around 0.8 (low end of shiny snow ice) – 0.96 (deep water), with stone and mud around 0.9. (see: https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA18833). So my calculation should be more like 340 × (1 – 0.3) = 0.9 × σT4. That gives a temperature of 261 K, or -12 ºC.

Third, I’ve ignored the effect of atmospheric and ocean circulations that move the energy around the Earth (though that’s ok with my “no atmosphere” approximation).

But my basic premise holds: without greenhouse gases (next lesson we’ll talk about what they do), the Earth would be really, really cold – with average temperatures in negative teens.  

Before we leave this simplification, it’s worth thinking about what you’d do to do this simple calculation better. You’d probably split the Earth up into little boxes. In each box you’d work out what the average energy (over a day, over a year) coming in from the Sun would be (higher at the equator than at the poles). And you’d work out how much was reflected (more over ice and sand than over ocean or forest) and what the thermal infrared “emissivity” is (i.e. how well that type of surface emits thermal infrared radiation). Then you’d do the energy balance equation in each box and then add that all up for the whole Earth. That would be the beginning of a climate model (more later!)

 

Lesson 4: the temperature of the Earth

temperature of the earth

In this lesson, we’re going to do what physicists like to do – we’re going to over-simplify the Earth and do a “thought experiment”.

So, we’re going to imagine that the Earth doesn’t have an atmosphere and we’ll work out what temperature it “should be”. This builds on the lesson on blackbodies.

First, the Sun is sending light towards the Earth. The Sun is very hot and emitting light in all directions, but the amount of energy coming directly towards the Earth (the solar irradiance) is 1360 W / m2 (ish – we’ll come back to how we measure this later). But that’s the light coming towards the Earth and, of course, half the Earth doesn’t get hit at any one moment (it’s the night time) and towards the poles, that 1360 W gets spread over a much bigger area of Earth.

To understand that consider 1 square metre rings in a row in front of the Earth (top picture): over the equator the light going through those rings forms a circle on the Earth; but over the poles, it would be spread over a much bigger ellipse. So – the average power falling on a square metre of Earth’s surface at any one time (averaged over the whole Earth) is about 1360 / 4 = 340 W / m2 (watts per square metre). That’s like having 4-old fashioned lightbulbs on every square metre of the Earth.

Now, the Earth can’t get hotter and hotter and hotter! It will reach an “equilibrium” where the heat in equals the heat out. (Equilibria are very common in physics). The way it releases energy into space is via its own blackbody radiation. You may remember from our lesson on blackbodies that everything that is hotter than absolute zero radiates energy with a blackbody curve. And that is true of the Earth too. As the temperature of the Earth is quite low (compared to the Sun!), it will radiate most of its blackbody radiation in what we call the “thermal infrared” (long wavelengths).

We can work out the total power of the blackbody by working out the area underneath that curve. There’s a simple calculation there. The total power of a blackbody in a square metre of its surface emitted into space is sigma times Temperature to the power 4. (σT4. Sigma (σ) is the Stefan-Boltzmann constant and is 5.670 367 × 10-8 W m-2 K-4 .

If the Earth were perfectly black at both short wavelengths (visible, near infrared – the wavelengths the Sun emits) and at long wavelengths (thermal infrared – wavelengths the Earth emits), then we could write:

Incoming power in a square metre = outgoing power in a square metre

340 = σ × T4

So the temperature = 278 K = 5 ºC.

(To do this calculation yourself, remember that the Stefan-Boltzmann constant can be written with the decimal place moved 8 places, so 0.000 000 056 704 and to get from Temperature to the power 4 is something to Temperature is something you can press the square root button twice)

If, as is more realistic, the Earth has an average reflectance in the visible of 30% (so it reflects about 30% of the light from the sun straight back to space and absorbs 70%) but it is still perfectly black in the thermal infrared (not unreasonable), then

Incoming power in a square metre = outgoing power in a square metre

70% × 340 = σ × T4

So Temperature = 254 K = -18 ºC

Now, we have made A LOT of approximations here. The actual average reflectance of the Earth might not be quite 30%, and it’s not quite perfectly black in the thermal infrared – but the basic picture holds. If the Earth had no atmosphere at all, the average temperature across the whole world would be something close to -15 ºC to -18 ºC.

And just in case you think I’ve pulled the wool over your eyes, I thought I’d find out the average temperature of the moon – after all, it’s about the same distance from the sun as us and it’s about the same sort of reflectance. So I searched the internet for “average temperature of the moon” and found an answer here: https://socratic.org/questions/what-is-the-average-surface-temperature-of-the-earth-s-moon

I was most amused that it said:

“You could take an average of the mean maxima and minima to get a mean surface temperature of -23 °C, but it wouldn’t be very meaningful.”

I beg to differ – that is very meaningful – as it’s very close to what I calculated with my overly simplistic thought experiment. (The moon reflectance is a bit different and it’s fraction pointing towards the sun is perhaps a bit different and the average temperature is not necessarily the average of the minimum and maximum temperatures)

Of course, the reason that the average temperature of the Earth is not that cold – and is nearly 35 ºC hotter – is that we do have an atmosphere. But you might be surprised to know that if our atmosphere was 100% oxygen, nitrogen and argon (and not 99.9 % oxygen, nitrogen and argon) it would still be -18 ºC. I’ll explain why in the next lesson.