A Toy Model of Climate Change

Introduction

A screenshot of the online book Math for the People.

In Introductory Physics for Medicine and Biology, Russ Hobbie and I make use of toy models. Such mathematical descriptions are not intended to be accurate or realistic. Rather, they‘re simple models that capture the main idea without getting bogged down in the details. Today, I present an example of a toy model. It’s not related to medicine or biology, but instead describes climate change. I didn’t originally derive this model. Much of the analysis below comes from other sources, such as the online book Math for the People published by Mark Branson and Whitney George.
Earth Without an Atmosphere

First, consider the earth with no atmosphere. We will balance the energy coming into the earth from the sun with the energy from the earth that is radiated out into space. Our goal will be to calculate the earth’s temperature, T.

The power density (energy per unit time per unit area, in watts per square meter) emitted by the sun is called the solar constant, S. It depends on how far you are from the sun, but at the earth’s orbit S = 1360 W/m². To get the total power impinging on our planet, we must multiply S by the area subtended by the earth, which is πR², where R is the earth’s radius (R = 6.4 × 10⁶ meters). This gives SπR² = 1.8 × 10¹⁷ W, or nearly 200,000 TW (T, or tera-, means one trillion). That’s a lot of power. The total average power consumption by humanity is only about 20 TW, so there’s plenty of energy from the sun.

We often prefer to talk about the energy loss or gain per unit area of the earth’s surface. The surface area of the earth is 4πR² (the factor of four comes from the total surface area of the spherical earth, in contrast to the area subtended by the earth when viewed from the sun). The power per unit area of the earth’s surface is therefore SπR²/4πR², or S/4.

Not all of this energy is absorbed by the earth; some is reflected back into space. The albedo, a, is a dimensionless number that indicates the fraction of the sun’s energy that is reflected. The power absorbed per unit area is then (1 – a)S/4. About 30% of the sun’s energy is reflected (a = 0.3), so the power of sunlight absorbed by the earth per unit of surface area is 238 W/m².

What happens to that energy? The sun heats the earth to a temperature T. Any hot object radiates energy. Such thermal radiation is analyzed in Section 14.8 of Intermediate Physics for Medicine and Biology. The radiated power per unit area is equal to eσT⁴. The symbol σ is the Stefan-Boltzmann constant, σ = 5.7 × 10^–8 W/(m² K⁴). As stated earlier, T is the earth’s temperature. When raising the temperature to the fourth power, T must be expressed as the absolute temperature measured in kelvin (K). Sometimes it’s convenient at the end of a calculation to convert kelvin to the more familiar degrees Celsius (°C), where 0°C = 273 K. But remember, all calculations of T⁴ must use kelvin. Finally, e is the emissivity of the earth, which is a measure of how well the earth absorbs and emits radiation. The emissivity is another dimensionless number ranging between zero and one. The earth is an excellent emitter and absorber, so e = 1. From now on, I’ll not even bother including e in our equations, in which case the power density emitted is just σT⁴.

Let’s assume the earth is in steady state, meaning the temperature is not increasing or decreasing. Then the power in must equal the power out, so 

(1 – a)S/4 = σT⁴. 

Solving for the temperature gives

T = ∜[(1 – a)S/4σ] .

Because we know a, S, and σ, we can calculate the temperature. It is T = 254 K = –19°C. That’s really cold (remember, in the Celsius scale water freezes at 0°C). Without an atmosphere, the earth would be a frozen wasteland.

Earth With an Atmosphere

Often we can learn much from a toy model by adding in complications, one by one. Now, we’ll include an atmosphere around earth. We must keep track of the power into and out of both the earth and the atmosphere. The earth has temperature T_E and the atmosphere has temperature T_A.

First, let’s analyze the atmosphere. Sunlight passes right through the air without being absorbed because it’s mainly visible light and our atmosphere is transparent in the visible part of the spectrum. The main source of thermal (or infrared) radiation (for which the atmosphere is NOT transparent) is from the earth. We already know how much that is, σT_E⁴. The atmosphere only absorbs a fraction of the earth’s radiation, e_A, so the power per unit area absorbed by the atmosphere is e_AσT_E⁴.

Just like the earth, the atmosphere will heat up to a temperature T_A and emit its own thermal radiation. The emitted power per unit area is e_AσT_A⁴. However, the atmosphere has upper and lower surfaces, and we’ll assume they both emit equally well. So the total power emitted by the atmosphere per unit area is 2e_AσT_A⁴.

If we balance the power in and out of the atmosphere, we get 

e_AσT_E⁴ = 2e_AσT_A⁴. 

Interestingly, the fraction of radiation absorbed by the atmosphere, e_A, cancels out of our equation (a good emitter is also a good absorber). The Stefan-Boltzmann constant σ also cancels, and we just get T_E⁴ = 2T_A⁴. If we take the forth root of each side of the equation, we find that T_A = 0.84 T_E. The atmosphere is somewhat cooler than the earth.

Next, let’s reanalyze the power into and out of the earth when surrounded by an atmosphere. The sunlight power per unit area impinging on earth is still (1 – a)S/4. The radiation emitted by the earth is still σT_E⁴. However, the thermal radiation produced by the atmosphere that is aimed inward toward the earth is all absorbed by the earth (since the emissivity of the earth is one, e_E = 1), so this provides another factor of e_AσT_A⁴. Balancing power in and out gives

(1 – a)S/4 + e_AσT_A⁴ = σT_E⁴ .

Notice that if e_A were zero, this would be the same relationship as we found when there was no atmosphere: (1 – a)S/4 = σT_E⁴. The atmosphere provides additional heating, warming the earth.

We found earlier that T_E⁴ = 2T_A⁴. If we rewrite this as T_A⁴ = T_E⁴/2 and plug that into our energy balance equation, we get

(1 – a)S/4 + e_AσT_E⁴/2 = σT_E⁴ .

With a bit of algebra, we find

(1 – a)S/4 = σT_E⁴ (1 – e_A/2) .

Solving for the earth’s temperature gives

T_E = ∜[(1 – a)S/4σ] ∜[1/(1 – e_A/2) ] .

If e_A were zero, this would be exactly the relationship we had for no atmosphere. The fraction of energy absorbed by the atmosphere is not zero, however, but is approximately e_A = 0.8. The atmosphere provides a dimensionless correction factor of ∜[1/(1 – e_A/2)]. The temperature we found previously, 254 K, is corrected by this factor, 1.136. We get T_E = 288.5 K = 15.5 °C. This is approximately the average temperature of the earth. Our atmosphere raised the earth’s temperature from –19°C to +15.5°C, a change of 34.5°C.

Climate Change

To understand climate change, we need to look more deeply into the meaning of the factor e_A, the fraction of energy absorbed by the atmosphere. The main constituents of the atmosphere—oxygen and nitrogen—are transparent to both visible and thermal radiation, so they don’t contribute to e_A. Thermal energy is primarily absorbed by greenhouse gases. Examples of such gases are water vapor, carbon dioxide, and methane. Methane is an excellent absorber of thermal radiation, but its concentration in the atmosphere is low. Water vapor is a good absorber, but water vapor is in equilibrium with liquid water, so it isn’t changing much. Carbon dioxide is a good absorber, has a relatively high concentration, and is being produced by burning fossil fuels, so a lot of our discussion about climate change focuses on carbon dioxide.

The key to understanding climate change is that greenhouse gasses like carbon dioxide affect the fraction of energy absorbed, e_A. Suppose an increase in the carbon dioxide concentration in the atmosphere increased e_A slightly, from 0.80 to 0.81. The correction factor  ∜(1/(1 – e_A/2) ) would increase from 1.136 to 1.139, changing the temperature from 288.5 K to 289.3 K, implying an increase in temperature of 0.8 K. Because changes in temperature are the same if expressed in kelvin or Celsius, this is a 0.8°C rise. A small change in e_A causes a significant change in the earth’s temperature. The more carbon dioxide in the atmosphere, the greater the temperature rise: Global warming.

Feedback

We have assumed the earth’s albedo, a, is a constant, but that is not strictly true. The albedo depends on how much snow and ice cover the earth. More snow and ice means more reflection, a larger albedo, a smaller amount of sunlight absorbed by the earth, and a lower temperature. But a lower temperature means more snow and ice. We have a viscous cycle: more snow and ice leads to a lower temperature which leads to more snow and ice, which leads to an even lower temperature, and so on. Intermediate Physics for Medicine and Biology dedicates an entire chapter to feedback, but it focuses mainly on negative feedback that tends to maintain a system in equilibrium. A viscous cycle is an example of positive feedback, which can lead to explosive change. An example from biology is the upstroke of a nerve action potential: an increase in the electrical voltage inside a nerve cell leads to an opening of sodium channels in the cell membrane, which lets positively charged sodium ions enter the cell, which causes the voltage inside the cell to increase even more. The earth’s climate has many such feedback loops. They are one of the reasons why climate modeling is so complicated.

Conclusion

Today I presented a simple description of the earth’s temperature and the impact of climate change. Many things were left out of this toy model. I ignored differences in temperature over the earth’s surface and within the atmosphere. I neglected ocean currents and the jet stream that move heat around the globe. I did not account for seasonal variations, or for other greenhouse gasses such as methane and water vapor, or how the amount of water vapor changes with temperature, or how clouds affect the albedo, and a myriad of other factors. Climate modeling is a complex subject. But toy models like I presented today provide insight into the underlying physical mechanisms. For that reason, they are crucial for understanding complex phenomena such as climate change.

Source: http://hobbieroth.blogspot.com/2024/10/a-toy-model-of-climate-change.html