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Blog - good vibrations

Or: how big is the "greenhouse effect" really?

This page is a blog article in progress, written by Tim van Beek.

When we talked about putting the Earth in a box, we saw that there is a gap of about 33 kelvin between the temperature of a black body in Earth’s orbit with an albedo of 0.3, and the estimated average surface temperature on Earth. An effect that explains this gap would need to

1) have a steady and continuous influence over thousands of years,

2) have a global impact,

3) be rather strong, because heating the planet Earth by 33 kelvin on the average needs a lot of energy.

Last time, in a quantum of warmth, we refined our zero dimensional energy balance model that treats the Earth as an ideal black body, and separated the system into a black body surface and a box containing the atmosphere.

With the help of quantum mechanics we saw that:

This is an effect that certainly matches the points 1 and 2 above: It is both continuous and global. But how strong is it? What do we need to know in order to calculate it? And is it measurable?

Survival in a combat zone

There has been a lively - sometimes hostile - debate about the “greenhouse effect” which is the popular name for the increase of incoming energy flux caused by infrared active atmospheric components, so maybe you think that the heading above refers to that.

But I have a different point in mind: Maybe you heard about guiding systems for missiles that chase “heat”? Do not worry if you have not. Knowlegeable people working for the armed forces of the USA know about this, and know that an important aspect of the design of aircrafts is to reduce infrared emission. Let’s see what they wrote about this back in 1982:

The engine hot metal and airframe surface emissions exhibit spectral IR continuum characteristics which are dependent on the temperature and emissivity-area of the radiating surface. These IR sources radiate in a relatively broad wavelength interval with a spectral shape in accordance with Planck’s Law (i.e., with a blackbody spectral shape). The surface- reflected IR radiation will also appear as a continuum based on the equivalent blackbody temperature of the incident radiation (e.g., the sun has a spectral shape characteristic of a 5527°C blackbody). Both the direct (specular) as well as the diffuse (Lambertian) reflected IR radiation components, which are a function of the surface texture and the relative orientation of the surface to the source, must be included. The remaining IR source, engine plume emission, is a composite primarily of $C0_2$ and $H_20$ molecular emission spectra. The spectral strength and linewidth of these emissions are dependent on the temperature and concentration of the hot gaseous species in the plume which are a function of the aircraft altitude, flight speed, and power setting.

This is an excerpt from page 15 of

You may notice that the authors point out the difference of a continuous black body radiation and the molecular emission spectra of CO 2CO_2 and H 2OH_2 O. The reason for this, as mentioned last time in a quantum of warmth, is that according to quantum mechanics molecules can emit and absorb radiation at specific energies, i.e. wavelengths, only. For this reason it is possible to distinguish far infrared radiation that is emitted by the surface of the Earth (more or less continuous spectrum) from the radiation that is emitted by the atmosphere (more or less discrete spectrum).

Our Battle Plan

Last time I told you that only certain molecules like CO 2CO_2 and H 2OH_2O are infrared active. The authors seem to agree with me. In order to understand the effect that this has on the atmosphere of the earth, there are two parts that we will have to understand:

1) Why do certain molecules react to to infrared light and others not? And how do atmospheric conditions like pressure influence this reaction?

2) Can we calculate the radiation generated by the atmosphere from this information, how does it look like and how will it change when the composition of the atmosphere is changed (more “greenhouse gases”)?

Since there is a little bit more I would like to tell than fits in one blog post, I will talk about 1) in this blog post only, and defer number 2) to a follow up. So, the rest of this post will be a short tour through molecular quantum mechanics. If you either know all of this stuff already, or if it is too high brow or otherwise boring for you, you can safely skip it. In the next post I will try to summarize what you need to know in a way that will enable you to follow the discussion, even if you don’t understand all the nitty gritty details explained here :-)

When we try to understand the interaction of atoms and molecules with light, the most important concept that we need to understand it that of an electric dipole moment.

Why is the dipole moment important?

One fascinating aspect of quantum mechanics is that classical concepts are both necessary and very useful to understand it. So, let us switch for a moment to classical electrostatic theory. If you place a negative electric point charge at the origin of our coordinate system and a positive point charge at the point x\vec{x}, I can tell you the electric dipole moment is a vector p\vec{p} and that:

p=x \vec{p} = \vec{x}

For a more general situation, let us assume that there is a charge density ρ\rho contained in some sphere SS around the origin. Then I can tell you that the electric dipole moment p\vec{p} is again a vector that can be calculated via

p=xρ(x)dx \vec{p} = \int \vec{x} \rho(\vec{x}) d \vec{x}

So that is the definition of how to calculate it, but what is its significance? Imagine that we would like to know how a test charge flying by the sphere SS is influenced by the charge density ρ\rho in SS. If we assume that ρ\rho is constant in time, then all we need to calculate is the electric potential Φ\Phi. In spherical coordinates and far from the sphere SS, this potential will fall of like 1/r1/r or faster, so we may assume that there is a series expansion of the form

Φ(r,ϕ,θ)= n=1 f(ϕ,θ)1r n \Phi(r, \phi, \theta) = \sum_{n = 1}^{\infty} f(\phi, \theta) \frac{1}{r^n}

When our test charge is far away from the sphere SS, only the first few terms in this expansion will be important to it.

In order to completely fix the series expansion, we need to choose an orthonormal basis for the coordinates ϕ\phi and θ\theta, that is an orthonormal basis of functions on the sphere. If we choose the spherical harmonics Y lm(ϕ,θ)Y_{l m}(\phi, \theta), with proper normalization we get what is called the multipole expansion of the electric potential:

Φ(r,ϕ,θ)=14πϵ 0 l=0 m=l l4π2l+1q lmY lm(ϕ,θ)r l+1 \Phi(r, \phi, \theta) = \frac{1}{4 \pi \epsilon_0} \sum_{l = 0}^{\infty} \sum_{m = -l}^{l} \frac{4 \pi}{2 l +1} q_{l m} \frac{Y_{l m}(\phi, \theta)}{r^{l+1}}

ϵ 0\epsilon_0 is the electric constant.

The l=0l = 0 term is called the monopole term. It is proportional to the electric charge qq contained in the sphere SS. So the first term in the expansion tells us if a charge flying by SS will feel an overall net attractive or repulsive force, due to the presence of a net electric charge inside SS.

The terms for l=1l = 1 form the vector p\vec{p}, the dipole moment. The next terms in the series Q ijQ_ij form the quadrupol tensor. So, for the expansion of the potential we get

Φ(r,ϕ,θ)=14πϵ 0(qr+pxr 3+12Q ijx ix jr 5+) \Phi(r, \phi, \theta) = \frac{1}{4 \pi \epsilon_0} (\frac{q}{r} + \frac{\vec{p} \cdot \vec{x}}{r^3} + \frac{1}{2} \sum Q_{ij} \frac{x_i x_j}{r^5} + \cdot \cdot \cdot)

For atoms and molecules the net charge qq is zero, so the next relevant term in the series expansion of their electric potential is the dipole moment. This is the reason why it is important to know if an atom or molecule has states with a nonzero dipole moment: Because this fact will in a certain sense dominate the interactions with other electromagnetic phenomena like light.

If you are interested in more information about multipole expansions in classical electrodynamics, you can find all sort of information in this classical textbook:

In quantum mechanics the position coordinate x\vec{x} is promoted to the position operator; as a consequence the dipole moment is promoted to an operator, too.

Molecular Emission Spectra or: Only Greenhouse Gases are infrared active? Really?

A rough estimate of energy levels for molecules shows that

Tim van Beek: Heuristic explanation for the different energy levels?

For atoms and molecules interacting with light, there are certain selection rules. A strict selection rule in quantum mechanics rules out certain state transitions that would violate a conservation law. But for atoms and molecules there are also heuristic selection rules that rule out state transitions that are far less likely than others. For state transitions induced by the interaction with light, a heuristic transition rule is

Transitions need to change the dipole moment by one.

This selection rule is heuristic: For one, it is valid when the radiation wavelength is bigger than the molecule, which is already true for visible light.

Secondary, transitions that change the dipole moment are far more likely than transitions that change the electric quadrupole moment only, for example. But: If an atom or molecule does not have any dipole transitions, then you will maybe still see spectral lines corresponding to quadrupol transitions. But they will be very weak.

Molecules that are infrared active need to have vibrational modes that have a nonzero dipole moment. But if you look close enough you will find that molecules that are not “greenhouse gases” can indeed emit infrared radiation, but the amount of radiation is insignificant compared to that of the greenhouse gases.

If you take a look at molecules consisting of two atoms of the same species like O 2O_2 and N 2N_2, you will find that such molecules can never have any vibrational states with a dipole moment at all, which means that more than 99% of all molecules in the atmosphere have an insignificant contribution to infrared radiation.

Tim van Beek: Compare black body radiation to the emission spectrum of CO2 and H2O.

If you would like to learn more about molecules, have a look at:

If you speak German and are interested in a very thorough and up to date treatment you could try:


So, if we try to calculate the DLR effect for the atmosphere of the Earth, it is sufficient to focus on molecules with vibrational modes with a nonzero dipole moment.

Pressure, Temperature and all that

The next important point is mentioned by the air force authors in this statement:

The spectral strength and linewidth of these emissions are dependent on the temperature and concentration of the hot gaseous species...

Of course the temperature, pressure and concentration of atmospheric components are not constant throughout the whole atmosphere, so this is a point that is also important for us when we investigate the radiation properties of the atmosphere.

But why is this important? Temperature and pressure change molecular emission spectra by line broadening mechanisms. The concentration of gases is important because saturation effects lead to a non linear dependence of absorption and emission on the concentration.

The most important mechanisms of line broadening are:

Tim van Beek: List of line broadening mechanisms.

We will need to calculate thermodynamic properties of the atmosphere, at least approximately, to determine the molecular emission spectra.

category: blog, climate