Mathematics of the Environment (Part 3)

This week I’ll release these notes before my seminar, so my students (and all of you, too) can read them ahead of time. The reason is that I’m pushing into topics I don’t understand as well as I’d like. So, my notes wrestle with some ideas in too much detail to cover in class—and I’m hoping some students will look at these notes ahead of time to prepare. Also, I’d appreciate your comments!

This week I’ll borrow material shamelessly from here:

• Seymour L. Hess, Introduction to Theoretical Meteorology, Henry Holt and Company, New York, 1959.

It’s an old book: for example, it talks about working out the area under a curve using a gadget called a planimeter, which is what people did before computers.

It also talks about how people measured the solar constant (roughly, the brightness of the Sun) before we could easily put satellites up above the Earth’s atmosphere! And it doesn’t mention global warming.

But despite or perhaps even because of these quaint features, it’s simple and clear. In case it’s not obvious yet, I’m teaching this quarter’s seminar in order to learn stuff. So, I’ll sometimes talk about old work… but if you catch me saying things that are seriously wrong (as opposed to merely primitive), please let me know.

The plan

Last time we considered a simple model Earth, a blackbody at uniform constant temperature absorbing sunlight and re-emitting the same amount of power in the form of blackbody radiation. We worked out that its temperature would be 6 °C, which is not bad. But then we took into account the fact that the Earth is not black. We got a temperature of -18 °C, which is too cold. The reason is that we haven’t yet equipped our model Earth with an atmosphere! So today let’s try that.

At this point things get a lot more complicated, even if we try a 1-dimensional model where the temperature, pressure and other features of the atmosphere only depend on altitude. So, I’ll only do what I can easily do. I’ll explain some basic laws governing radiation, and then sketch how people applied them to the Earth.

It’ll be good to start with a comment about what we did last time.

Kirchoff’s law of radiation

When we admitted the Earth wasn’t black, we said that it absorbed only about 70% of the radiation hitting it… but we still modeled it as emitting radiation just like a blackbody! Isn’t there something fishy about this?

Well, no. The Earth is mainly absorbing sunlight at visible frequencies, and at these frequencies it only absorbs about 70% of the radiation that hits it. But it mainly emits infrared light, and at these frequencies it acts like it’s almost black. These frequencies are almost completely different from those where absorption occurs.

But still, this issue is worth thinking about.

After all, emission and absorption are flip sides of the same coin. There’s a deep principle in physics, called reciprocity, which says that how X affects Y is not a separate question from how Y affects X. In fact, if you know the answer to one of these questions, you can figure out the answer to the other!

The first place most people see this principle is Newton’s third law of classical mechanics, saying that if X exerts a force on Y, Y exerts an equal and opposite force on X.

For example: if I punched your nose, your nose punched my fist just as hard, so you have no right to complain.

This law is still often stated in its old-fashioned form:

For every action there is an equal and opposite reaction.

I found this confusing as a student, because ‘force’ was part of the formal terminology of classical mechanics, but not ‘action’—at least not as used in this sentence!—and certainly not ‘reaction’. But as a statement of the basic intuition behind reciprocity, it’s got a certain charm.

In engineering, the principle of reciprocity is sometimes stated like this:

Reciprocity in linear systems is the principle that the response R_{ab} measured at a location a when the system is excited at a location b is exactly equal to R_{ba}, which is the response at location b when that same excitation is applied at a. This applies for all frequencies of the excitation.

Again this is a bit confusing, at least if you’re a mathematician who would like to know exactly how a ‘response’ or an ‘excitation’ is defined. It’s also disappointing to see the principle stated in a way that limits it to linear systems. Nonetheless it’s tremendously inspiring. What’s really going on here?

I don’t claim to have gotten to the bottom of it. My hunch is that to a large extent it will come down to the fact that mixed partial derivatives commute. If we’ve got a smooth function f of a bunch of variables x_1, \dots, x_n, and we set

\displaystyle{ R_{ab} = \frac{\partial^2 f}{\partial x_a \partial x_b} }

then

R_{ab} = R_{ba}

However, I haven’t gotten around to showing that reciprocity boils down to this in all the examples yet. Yet another unification project to add to my list!

Anyway: reciprocity has lots of interesting applications to electromagnetism. And that’s what we’re really talking about now. After all, light is electromagnetic radiation!

The simplest application is one we learn as children:

If I can see you, then you can see me.

or at least:

If light can go from X to Y in a static environment, it can also go from Y to X.

But we want something that sounds a bit different. Namely:

The tendency of a substance to absorb light at some frequency equals its tendency to emit light at that frequency.

This is too vague. We should make it precise, and in a minute I’ll try, but first let me motivate this idea with a thought experiment. Suppose we have a black rock and a white rock in a sealed mirrored container. Suppose they’re in thermal equilibrium at a very high temperature, so they’re glowing red-hot. So, there’s red light bouncing around the container. The black rock will absorb more of this light. But since they’re in thermal equilibrium, the black rock must also be emitting more of this light, or it would gain energy and get hotter than the white one. That would violate the zeroth law of thermodynamics, which implies that in thermal equilibrium, all the parts of a system must be at the same temperature.

More precisely, we have:

Kirchoff’s Law of Thermal Radiation. For any body in thermal equilibrium, its emissivity equals its absorptivity.

Let me explain. Suppose we have a surface made of some homogeneous isotropic material in thermal equilibrium at temperature T. If it’s perfectly black, we saw last time that it emits light with a monochromatic energy flux given by the Planck distribution:

\displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }

Here \lambda is the wavelength of light and the ‘monochromatic energy flux’ has units of power per area per wavelength.

But if our surface is not perfectly black, we have to multiply this by a fudge factor between 0 and 1 to get the right answer. This factor is called the emissivity of the substance. It can depend on the wavelength of the light quite a lot, and also on the surface’s temperature (since for example ice melts at high temperatures and gets darker). So, let’s call it e_\lambda(T).

We can also talk about the absorptivity of our surface, which is the fraction of light it absorbs. Again this depends on the wavelength of the light and the temperature of our surface. So, let’s call it a_\lambda(T).

Then Kirchoff’s law of thermal radiation says

e_\lambda(T) = a_\lambda(T)

So, for each frequency the emissivity must equal the absorptivity… but it’s still possible for the Earth to have an average emissivity near 1 at the wavelengths of infrared light and near 0.7 at the wavelengths of visible light. So there’s no paradox.

Puzzle 1. Is this law named after the same guy who discovered Kirchhoff’s laws governing electrical circuits?

Schwarzschild’s equation

Now let’s talk about light shining through the Earth’s atmosphere. Or more generally, light shining through a medium. What happens? It can get absorbed. It can get scattered, bouncing off in different directions. Light can also get emitted, especially if the medium is hot. The air in our atmosphere isn’t hot enough to emit a lot of visible light, but it definitely emits infrared light and microwaves.

It sounds complicated, and it is, but there are things we can say about it. Let me tell you about Schwarzschild’s equation.

Light comes in different wavelengths. So, can ask how much power per square meter this light carries per wavelength. We call this the monochromatic energy flux I_{\lambda}, since it depends on the wavelength \lambda. As mentioned last time, this has units W/m2μm, where μm stands for micrometers, a unit of wavelength.

However, because light gets absorbed, scattered and emitted the monochromatic energy flux is really a function I_{\lambda}(s), where s is the distance through the medium. Here I’m imagining an essentially one-dimensional situation, like a beam of sunlight coming down through the air when the Sun is directly overhead. We can generalize this later.

Let’s figure out the basic equation describing how I_{\lambda}(s) changes as a function of s. This is called the equation of radiative transfer, or Schwarzschild’s equation. It won’t tell us how different gases absorb different amounts of light of different frequencies—for that we need to do hard calculations, or experiments. But we can feed the results of these calculations into Schwarzschild’s equation.

For starters, let’s assume that light only gets absorbed but not emitted or scattered. Later we’ll include emission, which is very important for what we’re doing: the Earth’s atmosphere is warm enough to emit significant amounts of infrared light (though not hot enough to emit much visible light). Scattering is also very important, but it can be treated as a combination of absorption and emission.

For absorption only, we have the Beer–Lambert law:

\displaystyle{  \frac{d I_\lambda(s)}{d s} = - a_\lambda(s) I_\lambda(s)  }

In other words, the amount of radiation that gets absorbed per distance is proportional to the amount of radiation. However, the constant of proportionality a_\lambda (s) can depend on the frequency and the details of our medium at the position s. I don’t know the standard name for this constant a_\lambda (s), so let’s call it the absorption rate.

Puzzle 2. Assuming the Beer–Lambert law, show that the intensity of light at two positions s_1 and s_2 is related by

\displaystyle{ I_\lambda(s_2) = e^{-\tau} \; I_\lambda(s_1) }

where the optical depth \tau of the intervening medium is defined by

\displaystyle{ \tau = \int_{s_1}^{s_2} a_\lambda(s) \, ds }

So, a layer of stuff has optical depth equal to 1 if light shining through it has its intensity reduced by a factor of 1/e.

We can go a bit further if our medium is a rather thin gas like the air in our atmosphere. Then the absorption rate is given by

a_\lambda(s) = k_\lambda(s) \, \rho(s)

where \rho(s) is the density of the air at the position s and k_\lambda(s) is its absorption coefficient.

In other words, air absorbs light at a rate proportional to its density, but also depending on what it’s made of, which may vary with position. For example, both the density and the humidity of the atmosphere can depend on its altitude.

What about emission? Air doesn’t just absorb infrared light, it also emits significant amounts of it! As mentioned, a blackbody at temperature T emits light with a monochromatic energy flux given by the Planck distribution:

\displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }

But a gas like air is far from a blackbody, so we have to multiply this by a fudge factor. Luckily, thanks to Kirchoff’s law of radiation, this factor isn’t so fudgy: it’s just the absorption rate a_\lambda(s).

Here are we generalizing Kirchoff’s law from a surface to a column of air, but that’s okay because we can treat a column as a stack of surfaces; letting these become very thin we arrive at a differential formulation of the law that applies to absorption and emission rates instead of absorptivity and emissivity. (If you’re very sharp, you’ll remember that Kirchoff’s law applies to thermal equilibrium, and wonder about that. Air in the atmosphere isn’t in perfect thermal equilibrium, but it’s close enough for what we’re doing here.)

So, when we take absorption and also emission into account, Beer’s law gets another term:

\displaystyle{  \frac{d I_\lambda(s)}{d s} = - a_\lambda(s) I_\lambda(s) + a_\lambda(s) f_\lambda(T(s)) }

where T is the temperature of our gas at the position s. In other words:

\displaystyle{  \frac{d I_\lambda(s)}{d s} =  a_\lambda(s) ( f_\lambda(T) - I_\lambda(s))}

This is Schwarzschild’s equation.

Puzzle 3. Is this equation named after the same guy who discovered the Schwarzschild metric in general relativity, describing a spherically symmetric black hole?

Application to the atmosphere

In principle, we can use Schwarzschild’s equation to help work out how much sunlight of any frequency actually makes it through the atmosphere down to the Earth, and also how much infrared radiation makes it through the atmosphere out to space. But this is not a calculation I can do here today, because it’s very complicated.

If we actually measure what fraction of radiation of different frequencies makes it through the atmosphere, you’ll see why:


Everything here is a function of the wavelength, measured in micrometers. The smooth red curve is the Planck distribution for light coming from the Sun at a temperature of 5325 K. Most of it is visible light, with a wavelength between 0.4 and 0.7 micrometers. The jagged red region shows how much of this gets through—on a clear day, I assume—and you can see that most of it gets through. The smooth bluish curves are the Planck distributions for light coming from the Earth at various temperatures between 210 K and 310 K. Most of it is infrared light, and not much of it gets through.

This, in a nutshell, is what keeps the Earth warmer than the chilly -18 °C we got last time for an Earth with no atmosphere!

This is the greenhouse effect. As you can see, the absorption of infrared light is mainly due to water vapor, and then carbon dioxide, and then other lesser greenhouse gases, mainly methane, nitrous oxide. Oxygen and ozone also play a minor role, but ozone is more important in blocking ultraviolet light. Rayleigh scattering—the scattering of light by small particles, including molecules and atoms—is also important at short wavelengths, because its strength is proportional to 1/\lambda^4. This is why the sky is blue!

Here the wavelengths are measured in nanometers; there are 1000 nanometers in a micrometer. Rayleigh scattering continues to become more important in the ultraviolet.

But right now I want to talk about the infrared. As you can see, the all-important absorption of infrared radiation by water vapor and carbon dioxide is quite complicated. You need quantum mechanics to predict how this works from first principles. Tim van Beek gave a gentle introduction to some of the key ideas here:

• Tim van Beek, A quantum of warmth, Azimuth, 2 July 2011.

Someday it would be fun to get into the details. Not today, though!

You can see what’s going on a bit more clearly here:


The key fact is that infrared is almost completely absorbed for wavelengths between 5.5 and 7 micrometers, or over 14 micrometers. (A ‘micron’ is just an old name for a micrometer.)

The work of Simpson

The first person to give a reasonably successful explanation of how the power of radiation emitted by the Earth balances the power of the sunlight it absorbs was George Simpson. He did it in 1928:

• George C. Simpson, Further studies in terrestrial radiation, Mem. Roy. Meteor. Soc. 3 (1928), 1–26.

One year earlier, he had tried and failed to understand this problem using a ‘gray atmosphere’ model where the fraction of light that gets through was independent of its wavelength. If you’ve been paying attention, I think you can see why that didn’t work.

In 1928, since he didn’t have a computer, he made a simple model that treated emission of infrared radiation as follows.

He treated the atmosphere as made of layers of varying thickness, each layer containing 0.03 grams per centimeter2 of water vapor. The Earth’s surface radiates infrared almost as a black body. Part of the power is absorbed by the first layer above the surface, while some makes it through. The first layer then re-radiates at the same wavelengths at a rate determined by its temperature. Half this goes downward, while half goes up. Of the part going upward, some is absorbed by the next layer… and so on, up to the top layer. He took this top layer to end at the stratosphere, since the atmosphere is much drier in the stratosphere.

He did this all in a way that depends on the wavelength, but using a simplified model of how each of these layers absorbs infrared light. He assumed it was:

• completely opaque from 5.5 to 7 micrometers (due to water vapor),

• partly transparent from 7 to 8.5 micrometers (interpolating between opaque and transparent),

• completely transparent from 8.5 to 11 micrometers,

• partly transparent from 11 to 14 micrometers (interpolating between transparent and opaque),

• completely opaque above 14 micrometers (due to carbon dioxide and water vapor).

He got this result, at the latitude 50° on a clear day:


The upper smooth curve is the Planck distribution for a temperature of 280 K, corresponding to the ground. The lower smooth curve is the Planck distribution at 218 K, corresponding to the stratosphere. The shaded region is his calculation of the monochromatic flux emitted into space by the Earth. As you can see, it matches the Planck distribution for the stratosphere where the lower atmosphere is completely opaque in his model—between 5.5 and 7 micrometers, and over 14 micrometers. It matches the Planck distribution for the stratosphere where the lower atmosphere is completely transparent. Elsewhere, it interpolates between the two.

The area of this shaded region—calculated with a planimeter, perhaps?—is the total flux emitted into space.

This is just part of the story: he also took clouds into account, and he did different calculations at different latitudes. He got a reasonably good balance between the incoming and outgoing power. In short, he showed that an Earth with its observed temperatures is roughly compatible with his model of how the Earth absorbs and emits radiation. Note that this is just another way to tackle the problem of predicting the temperature given a model.

Also note that Simpson didn’t quite use the Schwarzschild equation. But I guess that in some sense he discretized it—right?

And so on

This was just the beginning of a series of more and more sophisticated models. I’m too tired to go on right now.

You’ll note one big thing we’ve omitted: any sort of calculation of how the pressure, temperature and humidity of the air varies with altitude! To the extent we talked about those at all, we treated them as inputs. But for a full-fledged one-dimensional model of the Earth’s atmosphere, we’d want to derive them from some principles. There are, after all, some important puzzles:

Puzzle 4. If hot air rises, why does the atmosphere generally get colder as you go upward, at least until you reach the stratosphere?

Puzzle 5. Why is there a tropopause? In other words, why is there a fairly sudden transition 10 kilometers up between the troposphere, where the air is moist, cooler the higher you go, and turbulent, and the stratosphere, where the air is drier, warmer the higher you go, and not turbulent?

There’s a limit to how much we can understand these puzzles using a 1-dimensional model, but we should at least try to make a model of a thin column of air with pressure, temperature and humidity varying as a function of altitude, with sunlight streaming downward and infrared radiation generally going up. If we can’t do that, we’ll never understand more complicated things, like the actual atmosphere.

12 Responses to Mathematics of the Environment (Part 3)

  1. Again, a very timely and relevant post, thanks.

    If we understand gray-body and black-body radiation principles, we should be able to come up with the equivalent radiation spectrum of a hypothetical earth covered completely by a still surface of water (at a great enough depth to not reflect anything from the bottom).

    The question has come up whether this surface of water will exhibit any kind of greenhouse effect by itself, ignoring any water vapor that it releases into the atmosphere.

    The greenhouse effect in this case is that visible light will preferentially propagate through the water layers, while any incoming infrared and fraction of near-infrared will get absorbed only near the skin layer.

    So, just like the atmosphere, there is a stratification of wavelength absorptivity and emissivity with depth. Should this lead to a change in the surface temperature, with a deviation from the Stephan-Boltzmann law similar to that which occurs for the atmosphere?

    And if this temperature increase does occur, does the effect simply get factored into the effective albedo of the earth’s surface?

    I am curious because I have not been able to find the effective black-body or gray-body radiation characteristics of a body comprised completely of water. Since the earth is 70% covered by water, I think this is important and shouldn’t be swept under the rug for any first-principles physics calculation.

    I can entertain the possibility that the reason the calculation is not done is because it is impractical to hypothetically to sweep away the effects of latent heat of evaporation at the water surface, which would tend to cool the surface. On a hot bright summer day, I would much rather touch the surface of water than walk barefooted on a black lava field, that’s for certain.

    Ask ourselves this question. What happens if there was a dense layer of water suspended in the atmosphere at some elevation? This would generate a greenhouse effect just like a real greenhouse, would it not? So why is having this layer at the surface of the earth any different? The effect is there, I just haven’t seen it quantified, which is puzzling to me.

    • Tim van Beek says:

      This is an interesting question which has been puzzling me for quite a while now. The problem is that books about physical oceanography that I know don’t say anything about radiative transfer, but concentrate on fluid dynamics. And the books about radiative transfer for climate science concentrate on the atmosphere. So the question about radiative energy transfer in the oceans seems to fall in one of those dreary nasty black spots between established academic disciplines :-)

      There is one book that could help us our here that I know about (unless there is an expert who would like to help us out, which would be most welcome), which is:

      • Gary E. Thomas and Knut Stamnes, Radiative Transfer in the Atmosphere and Ocean, Cambridge Atmospheric and Space Science Series, Cambridge University Press, Cambridge, 1999.

      It has a lot of interesting fromulas, but browsing through it, I did not find a simple rule-of-thumb explanation explaining the most important effects one has to understand.

      • Thanks Tim for your consideration.

        The stumbling block to me is the evaporative cooling. We might be able to get the emission spectra correct but then can’t balance the latent heats between evaporation and condensation properly.

        Maybe a thought experiment could apply. Find a visibly transparent glass material with the same radiative properties as water. Ask how much does this heat up if placed on the surface of a pool of water on a bright summer day.

  2. Matt West says:

    You are quite correct that reciprocity in engineering is essentially just equality of mixed partial derivatives. An example is Betti reciprocity [1] for elastic structures. The Wikipedia page only states this for linear systems, but it holds generally for infinitesimal responses of nonlinear systems, where the system state is a stationary point of an action functional and the responses are thus related to the second derivatives of the action. Symmetry of these derivatives is then reciprocity. This is an old result, but is restated in a relatively clean form in Section 6.3 of an old paper of mine [2], if you’ll excuse the self-citation.

    An interesting extension is to look at the statement of reciprocity for time evolution. Here symmetry of second derivatives of an action functional recovers the fact that the evolution of Lagrangian (or Hamiltonian) dynamics is symplectic. This is discussed in Section 6.4 of the previous paper [2], as well as in the context of Hamiltonian mechanics in another paper [3] of mine, which connects it to the statement of sensitivity in the optimization literature.

    All of this can be understood in the framework of Lagrangian submanifolds, which provides a nice unification [4].

    Thanks for the engaging blog posts,
    Matt.

    [1] http://en.wikipedia.org/wiki/Betti's_theorem

    [2] http://lagrange.mechse.illinois.edu/mwest/pubs/LeMaOrWe2003/ or http://dx.doi.org/10.1007/s00205-002-0212-y

    [3] http://lagrange.mechse.illinois.edu/mwest/pubs/LaWe2006/ or http://dx.doi.org/10.1088/0305-4470/39/19/S11

    [4] Marsden and Hughes, “Mathematical foundations of elasticity”, Dover, 1994.

    • John Baez says:

      You’re welcome, and thanks for these references! I was hoping that nonlinear systems would still display some form of reciprocity, since mixed partial derivatives are equal even for functions that aren’t quadratic. (In other words, all functions are quadratic to second order at a local minimum.) I’ll have to read your work—the great thing about self-citations is that you can be sure the author actually read those references.

      If many forms of reciprocity boil down to equality of mixed partial derivatives, it would be nice to teach this to Wikipedia, and thus the world. Right now the articles there don’t emphasize a single unifying principle. The article on reciprocity in electromagnetism is fascinating, but it says all forms of reciprocity there boil down to the self-adjointness of a certain operator. That’s probably correct as far as it goes, but I’d like to show the self-adjointness of that operator follows from equality of mixed partial derivatives of the action, or something like that.

  3. spironis says:

    working out the area under a curve” 1) measure the length and width of a rectangle of plotting paper, 2) crumple, 3) weigh on an analtyical balance, 4) You know the area/weight. 5) Cut out the curve, 6) weigh on the balance, 7) you know the area of the curve to three signiicant figures or more.

    If engineering fails, call in a chemist. When something is dropped, he uses his foot not his hands. Never play catch-up.

  4. Arrow says:

    Can you elaborate why we can treat air as if it were in thermal equilibrium? There are huge variations in surface air temperature around the globe which can exceed 100 K.

    • John Baez says:

      I said that when I was computing how much infrared radiation was emitted by a tiny parcel of air. The fact that some air somewhere else is much hotter or colder does not affect the amount of radiation emitted absorbed by this little parcel of air:

      What about emission? Air doesn’t just absorb infrared light, it also emits significant amounts of it! As mentioned, a blackbody at temperature T emits light with a monochromatic energy flux given by the Planck distribution:

      \displaystyle{ f_{\lambda}(T) = \frac{2 hc^2}{\lambda^5} \frac{1}{ e^{\frac{hc}{\lambda k T}} - 1 } }

      But a gas like air is far from a blackbody, so we have to multiply this by a fudge factor. Luckily, thanks to Kirchoff’s law of radiation, this factor isn’t so fudgy: it’s just the absorption rate a_\lambda(s).

      Here are we generalizing Kirchoff’s law from a surface to a column of air, but that’s okay because we can treat a column as a stack of surfaces; letting these become very thin we arrive at a differential formulation of the law that applies to absorption and emission rates instead of absorptivity and emissivity. (If you’re very sharp, you’ll remember that Kirchoff’s law applies to thermal equilibrium, and wonder about that. Air in the atmosphere isn’t in perfect thermal equilibrium, but it’s close enough for what we’re doing here.)

      So, when we take absorption and also emission into account, Beer’s law gets another term:

      \displaystyle{  \frac{d I_\lambda(s)}{d s} = - a_\lambda(s) I_\lambda(s) + a_\lambda(s) f_\lambda(T(s)) }

      where T is the temperature of our gas at the position s.

      In short, the amount of radiation emitted at wavelengh \lambda and position s is very close to

      a_\lambda(s) f_\lambda(T(s))

      because this small parcel of air is, taken by itself, close to being in thermal equilibrium. It’s not exactly in thermal equilibrium, so this is just an approximation, but it’s close.

      None of this is saying that the climate system as a whole is close to thermal equilibrium.

  5. Michael Knap says:

    New Puzzle :
    Is the Beer-Lambert law partially named after the same guy for which we owe the Lambert-W function, the multi-valued function W(z) for which z = W(z)e^{W(z)} is satisifed ?

    Solution to Puzzle 2:
    The Beer-Lambert law is separable, isn’t it ? If so, it isn’t to hard to show the relationship between the intensity of light at the two positions s_2 and s_1. I can do this, but it almost seemed to easy !

    Guess to Puzzle 4:
    I have very little training in physics, so this is more of a guess than a solution. Please correct me if I am wrong! My guess is that it has to do with the air density as altitude increases. Using the ideal gas law as a model, PV=nRT, one can see that the temperature is directly proportional to pressure and volume. Then as altitude increases, pressure and volume decrease, meaning temperature decreases. This wouldn’t explain, however, the increase in temperature as altitude increases in the stratosphere, though, would it ?

    Guess to Puzzle 5:
    My guess is that it has something to do with water vapor… the effect of gravity on water vapor. The idea is that the air above the tropopause has a different chemical makeup then the air below the tropopause, perhaps less water vapor ? One reason why I am thinking in this direction is that it seems clouds do not form above the tropopause.

  6. tqcd says:

    Solution to Puzzle 1: Yes he is the same guy, they also named an Institute in Heidelberg after him.

  7. We’ve been looking at some very simple models of the Earth’s climate. Pretty soon I want to show you one that [...]

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