<![CDATA[Mathematics of the Environment (Part 5)]]>

<![CDATA[Mathematics of the Environment (Part 5)]]>We saw last time that the Earth’s temperature seems to have been getting colder but also more erratic for the last 30 million years or so. Here’s the last 5 million again:

People think these glacial cycles are due to variations in the Earth’s orbit, but as we’ll see later, those cause quite small changes in ‘insolation’—roughly, the amount of sunshine hitting the Earth (as a function of time and location). So, M. I. Budyko, an expert on the glacial cycles, wrote:

Thus, the present thermal regime and glaciations of the Earth prove to be characterized by high instability. Comparatively small changes of radiation—only by 1.0-1.5%—are sufficient for the development of ice cover on the land and oceans that reaches temperate latitudes.

How can small changes in the amount of sunlight hitting the Earth, or other parameters, create big changes in the Earth’s temperature? The obvious answer is positive feedback: some sort of amplifying effect.

But what could it be? Do we know feedback mechanisms that can amplify small changes in temperature? Yes. Here are a few obvious ones:

• Water vapor feedback. When it gets warmer, more water evaporates, and the air becomes more humid. But water vapor is a greenhouse gas, which causes additional warming. Conversely, when the Earth cools down, the air becomes drier, so the greenhouse effect becomes weaker, which tends to cool things down.

• Ice albedo feedback. Snow and ice reflect more light than liquid oceans or soil. When the Earth warms up, snow and ice melt, so the Earth becomes darker, absorbs more light, and tends to get get even warmer. Conversely, when the Earth cools down, more snow and ice form, so the Earth becomes lighter, absorbs less light, and tends to get even cooler.

• Carbon dioxide solubility feedback. Cold water can hold more carbon dioxide than warm water: that’s why opening a warm can of soda can be so explosive. So, when the Earth’s oceans warm up, they release carbon dioxide. But carbon dioxide is a greenhouse gas, which causes additional warming. Conversely, when the oceaans cool down, they absorb more carbon dioxide, so the greenhouse effect becomes weaker, which tends to cool things down.

Of course, there are also negative feedbacks: otherwise the climate would be utterly unstable! There are also complicated feedbacks whose overall effect is harder to evaluate:

• Planck feedback. A hotter world radiates more heat, which cools it down. This is the big negative feedback that keeps all the positive feedbacks from making the Earth insanely hot or insanely cold.

• Cloud feedback. A warmer Earth has more clouds, which reflect more light but also increase the greenhouse effect.

• Lapse rate feedback. An increased greenhouse effect changes the vertical temperature profile of the atmosphere, which has effects of its own—but this works differently near the poles and near the equator.

See also “week302” of This Week’s Finds, where Nathan Urban tells us more about feedbacks and how big they’re likely to be.

Understanding all these feedbacks, and which ones are important for the glacial cycles we see, is a complicated business. Instead of diving straight into this, let’s try something much simpler. Let’s just think about how the ice albedo effect could, in theory, make the Earth bistable.

To do this, let’s look at the very simplest model in this great not-yet-published book:

• Gerald R. North, Simple Models of Global Climate.

This is a zero-dimensional energy balance model, meaning that it only involves the average temperature of the earth, the average solar radiation coming in, and the average infrared radiation going out.

The average temperature will be $T,$ measured in Celsius. We’ll assume the Earth radiates power square meter equal to

$\displaystyle{ A + B T }$

where $A = 218$ watts/meter² and $B = 1.90$ watts/meter² per degree Celsius. This is a linear approximation taken from satellite data on our Earth. In reality, the power emitted grows faster than linearly with temperature.

We’ll assume the Earth absorbs solar energy power per square meter equal to

$Q c(T)$

Here:

• $Q$ is the average insolation: that is, the amount of solar power per square meter hitting the top of the Earth’s atmosphere, averaged over location and time of year. In reality $Q$ is about 341.5 watts/meter². This is one quarter of the solar constant, meaning the solar power per square meter that would hit a panel hovering in space above the Earth’s atmosphere and facing directly at the Sun. (Why a quarter? We’ve seen why: it’s because the area of a sphere is $4 \pi r^2$ while the area of a circle is just $\pi r^2$ .)

• $c(T)$ is the coalbedo: the fraction of solar power that gets absorbed. The coalbedo depends on the temperature; we’ll have to say how.

Given all this, we get

$\displaystyle{ C \frac{d T}{d t} = - A - B T + Q c(T(t)) }$

where $C$ is Earth’s heat capacity in joules per degree per square meter. Of course this is a funny thing, because heat energy is stored not only at the surface but also in the air and/or water, and the details vary a lot depending on where we are. But if we consider a uniform planet with dry air and no ocean, North says we may roughly take $C$ equal to about half the heat capacity at constant pressure of the column of dry air over a square meter, namely 5 million joules per degree Celsius.

The easiest thing to do is find equilibrium solutions, meaning solutions where $\frac{d T}{d t} = 0,$ so that

$A + B T = Q c(T)$

Now $C$ doesn’t matter anymore! We’d like to solve for $T$ as a function of the insolation $Q,$ but it’s easier to solve for $Q$ as a function of $T$ :

$\displaystyle{ Q = \frac{ A + B T } {c(T)} }$

To go further, we need to guess some formula for the coalbedo $c(T).$ The coalbedo, remember, is the fraction of sunlight that gets absorbed when it hits the Earth. It’s 1 minus the albedo, which is the fraction that gets reflected. Here’s a little chart of albedos:

If you get mixed up between albedo and coalbedo, just remember: coal has a high coalbedo.

Since we’re trying to keep things very simple right not, not model nature in all its glorious complexity, let’s just say the average albedo of the Earth is 0.65 when it’s very cold and there’s lots of snow. So, let

$c_i = 1 - 0.65 = 0.35$

be the ‘icy’ coalbedo, good for very low temperatures. Similarly, let’s say the average albedo drops to 0.3 when its very hot and the Earth is darker. So, let

$c_f = 1 - 0.3 = 0.7$

be the ‘ice-free’ coalbedo, good for high temperatures when the Earth is darker.

Then, we need a function of temperature that interpolates between $c_i$ and $c_f.$ Let’s try this:

$c(T) = c_i + \frac{1}{2} (c_f-c_i) (1 + \tanh(\gamma T))$

If you’re not a fan of the hyperbolic tangent function $\tanh,$ this may seem scary. But don’t be intimidated!

The function $\frac{1}{2}(1 + \tanh(\gamma T))$ is just a function that goes smoothly from 0 at low temperatures to 1 at high temperatures. This ensures that the coalbedo is near its icy value $c_i$ at low temperatures, and near its ice-free value $c_f$ at high temperatures. But the fun part here is $\gamma,$ a parameter that says how rapidly the coalbedo rises as the Earth gets warmer. Depending on this, we’ll get different effects!

The function $c(T)$ rises fastest at $T = 0,$ since that’s where $\tanh (\gamma T)$ has the biggest slope. We’re just lucky that in Celsius $T = 0$ is the melting point of ice, so this makes a bit of sense.

Now Allan Erskine‘s programming magic comes into play! I’m very fortunate that the Azimuth Project has attracted some programmers who can make nice software for me to show you. Unfortunately his software doesn’t work on this blog—yet!—so please hop over here to see it in action:

• Temperature versus insolation.

You can slide a slider to adjust the parameter $\gamma$ to various values between 0 and 1.

In the little graph at right, you can see how the coalbedo $c(T)$ changes as a function of the temperature $T.$ In this graph the temperature ranges from -50 °C and 50 °C; the graph depends on what value of $\gamma$ you choose with slider.

In the big graph at left, you can see how the insolation $Q$ required to yield a given temperature $T$ between -50 °C and 50 °C. As we’ve seen, it’s easiest to graph $Q$ as a function of $T:$

$\displaystyle{ Q = \frac{ A + B T } {c_i + \frac{1}{2} (c_f-c_i) (1 + \tanh(\gamma T))} }$

Solving for $T$ here is hard, but we can just flip the graph over to see what equilibrium temperatures $T$ are allowed for a given insolation $Q$ between 200 and 500 watts per square mater.

The exciting thing is that when $\gamma$ gets big enough, three different temperatures are compatible with the same amount of insolation! This means the Earth can be hot, cold or something intermediate even when the amount of sunlight hitting it is fixed. The intermediate state is unstable, it turns out—we’ll see why later. Only the hot and cold states are stable. So, we say the Earth is bistable in this simplified model.

Can you see how big $\gamma$ needs to be for this bistability to kick in? It’s certainly there when $\gamma = 0.05,$ since then we get a graph like this:

When the insolation is less than about 385 W/m² there’s only a cold state. When it hits 385 W/m², as shown by the green line, suddenly there are two possible temperatures: a cold one and a much hotter one. When the insolation is higher, as shown by the black line, there are three possible temperatures: a cold one, and unstable intermediate one, and a hot one. And when the insolation gets above 465 W/m², as shown by the red line, there’s only a hot state!

Mathematically, this model illustrates catastrophe theory. As we slowly turn up $\gamma,$ we get different curves showing how temperature is a function of insolation… until suddenly the curve isn’t the graph of a function anymore: it becomes infinitely steep at one point! After that, we get bistability:

$\gamma = 0.00$

$\gamma = 0.01$

$\gamma = 0.02$

$\gamma = 0.03$

$\gamma = 0.04$

$\gamma = 0.05$

This is called a cusp catastrophe, and you can visualize these curves as slices of a surface in 3d, which looks roughly like this picture:

from here:

• Wolfram Mathworld, Cusp catastrophe. (Includes Mathematica package.)

The cusp catastrophe is ‘structurally stable’, meaning that small perturbations don’t change its qualitative behavior. In other words, whenever you have a smooth graph of a function that gets steeper and steeper until it ‘overhangs’ and ceases to be the graph of a function, it looks like this cusp catastrophe. This statement is quite vague as I’ve just said it— but it’s made 100% precise in catastrophe theory.

Structural stability is a useful concept, because it focuses our attention on robust features of models: features that don’t go away if the model is slightly wrong, as it always is.

There are lots more things to say, but the most urgent question to answer is this: why is the intermediate state unstable when it exists? Why are the other two equilibria stable? We’ll talk about that next time!

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