A Simple Math Model for Global Warming

The model developed here is not a rigorous one like the global circulation models used by climate scientists. Nevertheless, it contains enough elements to allow a nonexpert to gain a deeper understanding of how greenhouse gases work to warm the climate and a crude idea of how water vapor and clouds affect it. At its most basic, global warming results from an imbalance between the amount of energy the Earth receives from the sun and how much radiates away into space. Radiation increases with higher temperature, and so the equilibrium temperature of the Earth is the temperature at which incoming energy equals outgoing energy. How the Earth radiates energy depends on properties of its surface and atmosphere. Here is where the complexity comes in because many things affect the ability of the Earth to radiate energy and therefore affect global average temperature.

Elementary climate dynamics

The ultimate driver of climate is the sun or, more specifically, the energy bestowed upon the Earth as sunshine. At Earth’s orbital distance, the sun’s power output works out to about 1370 watts per square meter of surface perpendicular to the sun’s rays. This energy flux is called the solar constant, or S. The Earth presents a circular barrier to this power output of area πR², where R is the radius of the Earth. It follows that the amount of energy intercepted by the Earth is then πR²S. The Earth reflects about 30% of incident energy, the rest is absorbed. The fraction reflected is called the albedo (A). With this I can write the following expression for the rate of energy absorption by the Earth:

(1)   Energy absorption rate = πR² S (1‒A)

The Earth also radiates energy into space. This radiation is in the form of infrared (IR) radiation, not visible light as is much of the sun’s energy. All bodies radiate energy as a function of temperature. The amount of radiation that is emitted by one square meter of surface at a given temperature is given by the Stefan-Boltzmann equation:

(2)   Radiated energy flux = ε σ T⁴

Here, T is the absolute temperature in degrees Kelvin (K). The Kelvin temperature scale is simply the familiar Celsius temperature with 273.15 added to it. It measures absolute temperature; 0 K is absolute zero, the coldest temperature possible. The parameter σ is the Stefan-Boltzmann constant, and is equal to 5.67x10⁻⁸ watt m^-2·K^-4. Finally, ε is the average emissivity of the earth. Emissivity is a property of the material emitting radiation that relates a material’s actual radiative properties to that of an ideal “black body” which radiates all wavelengths equally. Real materials have values of ε between 0 and 1 as a function of wavelength. The ε in equation 2 refers to an average value over the relevant wavelengths and various surface types of the Earth.

Equation 2 refers to an energy flux, which is the energy transfer rate (watts) per unit area (square meter). To determine the total radiation emitted by the Earth, I multiply this flux by the surface area of the Earth, which can be approximated as the area of a sphere: 4 πR². Multiplying equation B.2 by this area gives the rate that energy is radiated by the Earth:

(3)   Energy radiation rate = 4 πR² ε σ T⁴ 

Over time, the Earth’s average temperature will adjust so that energy absorbed equals energy radiated. Setting equation 1 and 3 equal to each other and rearranging gives the following equation related Earth’s average temperature, ε, the solar constant, and albedo:

(4)   ε σ T⁴ = ¼ S (1-A) = I = 240 watts/m²

For a solar constant of 1370 watts/m²and albedo of 0.3, the right-hand side of equation 4 works out to 240 watts/m². This quantity is called insolation (I). It refers to the average amount of solar energy absorbed per square meter of the Earth’s surface. The actual average surface temperature of the Earth is about 15º C (288 K). If this value is inserted into equation 4 and the result solved for ε, one obtains:

(5)   ε = I / (σ T⁴) = 240 / (5.67 x 10^-8 · 288⁴) = 0.615

This value is the apparent emissivity of the Earth as viewed from space. Table 1 lists emissivity for a variety of terrain types. Emissivity for many materials does not vary much with wavelength, making the use of an average value reasonable. As Table 1 shows, the average emissivity of various types of terrain are not much different from each other either, making the use of a single average value to describe the Earth’s surface emissivity a good approximation. The emissivity of the Earth’s surface (ε_S) appears to be around 0.95. A value of 0.96 is used in some models and will be used here. With an emissivity of 0.96, equation 4 yields a temperature of 258 K or −15 C. An Earth surrounded by an IR-transparent atmosphere would be a frozen ice ball. The Earth is not frozen because the clouds and atmosphere are not transparent to IR radiation. Some of the radiation emitted from the Earth is absorbed by the clouds and atmosphere and re-radiated. This reduces the Earth’s apparent emissivity from the 0.96 value for the surface to an effective value of 0.615.

Factors affecting Earth’s albedo and emissivity: the effect of clouds

Table 2 shows radiative properties of clouds and the fraction of the Earth’s surface covered by cloud at various heights. Low cloud is defined as clouds with a base height below 3 km and high clouds have base heights above 6 km. Medium-high cloud is in between. Clouds cover about 63% of the Earth’s surface on average, implying 37% of the earth’s surface on average is under clear skies. The albedo of the Earth is given by the weighted sum of surface average albedo (A_SURF) and average albedos of clouds at low, medium and high altitude (A_LC, A_MC, A_HC). Overall radiation measurements provide the information needed to determine these albedos. The difference in albedo between clouds and the surface also affects climate. In the section above, I showed that an Earth with an IR-transparent atmosphere would have a temperature of −15 C. This calculation ignored the effect of clouds on albedo. An Earth with an IR-transparent atmosphere would not have any clouds; its albedo would be the surface albedo of 0.14 rather than 0.3. With A = 0.14 and ε = 0.96, equation B.4 gives T = 271 K or −2 C, not −15 C.

Those readers who have ever used a carport in cold climes know that, despite the lack of enclosure, no frost collects on the car’s windows overnight. Frost is caused by radiation from the car cooling it below the air temperature, allowing frost to form. With the carport trapping this radiation and radiating it back, the car fails to cool below air temperature and frost is avoided. Carports serve as radiation shields. Clouds serve a radiation shield function for the surface beneath them. Energy radiated from the surface below clouds is trapped by them. Some is radiated out into space while the rest is radiated back to the surface, warming it from what it would otherwise be. The energy radiated into space comes from a cloud surface which has a lower effective emissivity than the surface below it. The lower emissivity mostly reflects the lower temperature of clouds due to their elevation. Clouds act to reduce the effective emissivity of the surface below them from 0.96 to a lower value depending on their elevation.

Table 3 shows the effective emissivity of clouds as a function of their elevation. Higher clouds are more effective shields because they re-radiate at lower temperatures. Using data from Table 2, the average emissivity of Earth with clouds in an IR-transparent atmosphere would be 0.37·0.96 (cloudless surface) + 0.26·0.68 (low cloud) + 0.18·0.55 (medium-high cloud) + 0.19·0.33 (high cloud) = 0.694 (apparent emissivity of Earth into space). With a 0.694 emissivity, equation 4 yields a temperature of 279 K or +6 C. With clouds in an IR-transparent atmosphere, the Earth would be 8 degrees warmer than without clouds, but it still would be colder than its actual temperature of +15 C. This extra warming comes from the fact that the atmosphere is not completely IR-transparent, the apparent emissivity of the surface unobscured by clouds in not 0.96, but some lower value because of “screening” provided by the atmosphere itself.

Note that cloud cover is treated as an opaque screen in this calculation, meaning no radiation from the surface is assumed to penetrate through the cloud layer; it is all absorbed and radiated into space or back to the surface. The clouds radiate into space with a lower effective emissivity than the ground they cover. Thus, the 0.96 emissivity of the cloud-covered surface is reduced to the weighted sum of the cloud emissivities as shown in Table 3. In actuality, clouds are not strictly opaque, and the interaction between clouds and radiation is more complex than this simple shield concept suggests. Nevertheless, this approach gives a reasonably good semi-quantitative picture of the effect of clouds. For example, an Earth completely covered by high cloud would be warmer: with A = 0.49 and ε = 0.33, equation 4 gives a temperature of 311 K (100 F). In contrast, an Earth covered by low cloud would be cooler: with A = 0.35 and ε = 0.68, equation 4 gives a temperature of 276 K (36 F). These results are consistent with more sophisticated models that also show warming from high cloud and cooling from low cloud. The exact amounts predicted are different, of course―a high degree of precision would not be expected from such a simple model.

Factors affecting the earth’s emissivity: the greenhouse effect

Certain gases in the atmosphere called greenhouse gases absorb and re-radiate IR radiation emitted from the surface, just like clouds. Unlike clouds, the greenhouse-gas containing atmosphere is not modeled as an opaque screen, but rather as a translucent screen that partially shields the ground emissions. This effect serves to reduce the effective emissivity of the cloudless sky from the 0.96 of the surface to a new value (ε_T) that takes into account this partial shielding from greenhouse gases.

Table 3 shows a summary of four kinds of IR shields lying over the Earth’s surface: the three types of cloud and the translucent atmosphere. The simple model I am developing here holds that the total emissivity of the Earth (which we know to be 0.615) is simply the weighted sum of the effects of all these shields. The weights are the percent coverage of each type of screen. The effective emissivity for three of the four screens are known (Table 2) and are reproduced in Table 3. The contributions from these three sum to 0.3385. The rest of the emissivity comes from surface radiation through the cloudless, translucent atmosphere, which has an effective emissivity of ε_T. A value for ε_T can be obtained by subtracting the cloud-provided shielding from the overall emissivity of 0.615 and dividing the difference by the cloudless fraction of the Earth’s surface: (0.615−0.3385)/0.37 = 0.747 for ε_T (see Table 3).

Once again, this translucent screen analogy to the IR-absorbing atmosphere is a simplification of a more complex situation. Absorption and re-radiation of IR radiation occurs throughout the atmosphere and interacts with clouds. A completely accurate assessment of this phenomenon requires complex radiation balance models. Nevertheless, the results obtained with this simple model can provide a reasonably good understanding of how CO₂ affects the climate.

The value of ε_T depends on the concentration of greenhouse gases in the atmosphere such as CO₂. Thus, adding CO₂ to the atmosphere affects ε_T, which affects the earth’s overall emissivity as shown by equation.6:

(6)   ε = 0.3385 + 0.37 ε_T

Changes in overall ε produced by rising CO₂ affect temperature according to eq. 4.

The analysis of the translucent atmosphere is more complex than the effect of clouds, which were treated as opaque barriers with a fixed emissivity. The translucent atmosphere both absorbs and transmits (allows to pass through) radiation from the surface. This situation can be described by a radiation balance on the translucent atmosphere:

RADIATION INTO ATMOSPHERE = RADIATION OUT OF ATMOSPHERE

(7)   σ ε_S T_S⁴            =   2 σ ε_A T_A⁴  +         (1− ε_A) σ ε_S T_S⁴

surface into atm = atm. into space & back to surface + transmission

Here, ε_S is the emissivity of the surface (0.96) and ε_A is the intrinsic emissivity of the atmosphere (a function of greenhouse gas levels). T_S and T_A are the temperature (in Kelvin) of the surface (288 K) and the atmosphere, respectively. The factor (1− ε_A) is called the transmission of radiation through an absorbing medium. The factor of 2 occurs because the atmosphere radiates both outward into space and inward towards the surface. Equation 7 can be solved for T_A as follows:

(8)   T_A⁴ = ½ ε_S T_S⁴ → T_A = (½ ε_S)^¼ ∙ T_S    = (½*0.96)^¼ ∙ 288 = 240 K

The total emissions through and from the translucent atmosphere into space are given by:

(9)   total emissions = σ ε_T T_S⁴ = σ ε_A T_A⁴ + (1− ε_A) σ ε_S T_S⁴

The left-hand side of equation 9 relates ε_T (the effective emissivity of the cloudless surface, as modified by the translucent atmosphere above it) to the actual surface emissivity (0.96) and that of the cloudless atmosphere (ε_A). Note that the factor of 2 from equation 7 is absent because only the radiation going into space from the atmosphere is considered here. Equation 8 can be substituted into equation 9 to obtain the result:

(10)  σ ε_T T_S⁴ = ½ σ ε_A ε_S T_S⁴ + (1− ε_A) σ ε_S T_S⁴  =  (½ ε_A + 1− ε_A) σ ε_S T_S⁴

Simplifying,

(11)  ε_T = ε_S (1 − ½ ε_A)

For the actual value of ε_T of 0.747 from Table 3 and ε_S = 0.96, equation 11 gives a value of 0.444 for ε_A.

The effect of IR absorbers on emissivity is described by Beer’s Law, which states that for a given wavelength, the logarithm of the transmission through a mixture is equal to the negative of the sum of the absorbance (A_i) of each IR-absorbing species present:

(12)  ln(1─ ε_A) = − ∑ A_i for each wavelength

The symbol ∑ means “the sum of, for all i”. The absorbance of species i is given by A_i = a_i∙b∙C_i, where C_i is the concentration of species i, b is the length of light path through the i-containing medium and a_i is the molar absorptivity, which is a property of that species. The molar absorptivity is a function of η (wave number per cm, or 10000 divided by wavelength in microns); that is, a_i should be written as a function, a_i(η). For the application here, b is dependent on the thickness of the atmosphere, which is a constant independent of C_i. Hence a_i(η) and b can be combined into a single function k_i(η) and equation 12 becomes:

(13)  ln(1─ ε_A) = − ∑ k_i(η)∙C_i  →  ε_A = 1 − exp[−∑ k_i(η)∙C_i] for each η

We have been using average emissivities throughout this simple analysis, so I replace the function k_i(η) with a constant k equal to its average value over all relevant η. I will consider only two greenhouse gases, water vapor (H₂O) and CO₂. For these assumptions, equations 12 and 13 can be written for the entire spectrum of emitted radiation as follows:

(14)   ε_A = 1 − exp[−A_H2O − A_CO2] = 1 − exp(−k_H2O∙C_H2O − k_CO2∙C_CO2)

where C_H2O and C_CO2 are the concentrations of water vapor and CO₂ in the air. For ε_A = 0.444, the sum of A_H2O and A_CO2 is equal to 0.587. Assume w is the fraction of total absorbance of 0.587 that is due to water vapor at the base condition. Equation 14 can then be written for the base condition as:

(15)  ε_A  = 0.444 = 1 − exp[ −0.587w − 0.587(1−w) ]  = 1 − exp(−k_H2O ∙H2O_B − k_CO2 ∙CO2_B)

Here, H2O_B and CO2_B are the average water vapor and CO₂ concentrations in the atmosphere for the base case used for all previous calculations. Comparing terms shows that k_H2O = 0.587w/ H2O_B and k_CO2 = 0.587(1−w)/ CO2_B. Substituting these into equation 14 gives:

(16)   ε_A = 1 – exp[ −0.587w ∙ C_H2O/H2O_B − 0.587(1-w) ∙ C_CO2/CO2_B]

The CO₂ level associated with these calculations was 350 ppm. To obtain a value for C_H2O/H2O_B I assume that changing global temperature does not affect average relative humidity, only absolute humidity. With this assumption, the effect of temperature on C_H2O/H2O_B is governed by the Clausius-Clapeyron equation:

(17)   C_H2O/ H2O_B = exp[ −ΔH^VAP/R∙(1/T − 1/288)]

Here, ΔH^VAP is the heat of vaporization of water and R is the gas constant. ΔH^VAP/R has a value of 5294 K. Global temperature for the base case is 288 K, as previously noted. With these values equation 16  becomes a function of T:

(18)   ε_A = 1− exp[−0.587w ∙ exp(18.382 − 5294/T) – 0.587 (1-w) C_CO2/350]

The greenhouse model is now complete. For a given value of C_CO2, ε_A as a function of T is obtained from equation 18. This value of ε_A is used in equation 11 to obtain ε_T. This value is used in equation 6 to obtain ε for the Earth as a function of T. This value of ε is then used to calculate temperature using equation 4. The value of temperature obtained is a function of T and must equal T. This will be true for a particular value of T which can be obtained using trial and error (the goal seek function in Microsoft Excel™ makes this easy to do).

Before this can be done, a value is needed for w. As noted earlier, this parameter reflects the amount of greenhouse warming due to water vapor relative to CO₂. Its value serves to calibrate the sensitivity of temperature to CO₂ levels. A w value of 0.79 provides results that agree with more sophisticated models. This procedure was used to calculate the change in temperature from the base case (ΔT), produced by changes in CO₂ level (Δ%CO2). The results are plotted in Figure 1.

Figure 1. Temp. change and forcing as function of CO₂ level relative to 350 ppm

Example of use of the simple model: converting Earth into Venus

As an exercise, let’s see what would happen if we moved Earth into Venus’s orbit. The solar constant for Venus is 2600 watts m^-2, much higher than Earth’s 1370 watts m⁻², reflecting its closer proximity to the sun. Let’s cover the Earth with maximum-cooling low cloud. What would be the temperature be according to the simple model? With A = 0.35, S = 2600, and ε = 0.68, equation 4 gives T = 324 K, or 123 F. This calculation suggests that a Venus with liquid water and life could exist, although would be a pretty hot place by Earth standards. This was the rationale for pre-1950’s science fiction stories that depicted a steamy-hot, swamp-like Venus.

But how realistic is it for all the cloud to be low-lying cloud? If, here on Earth, there is high cloud cover equal to 75% of low cloud cover, shouldn’t this be the case for a warmer Earth? A 100% low-cloud covered Earth would most likely have at least 75% high cloud coverage as well, which would lower ε from 0.68 to around 0.4. This would raise the temperature to 350 K (171º F), at which point water vapor would make up a substantial portion of the atmosphere. With such a wet atmosphere, cloud cover would likely be complete at low, medium and high elevations. With 100% high cloud cover, ε would be 0.33, A = 0.49 and equation 4 gives the temperature as 365K (198 F). At this temperature evaporation would greatly increase the water content of the atmosphere increasing surface atmospheric pressure by 75%.

Increasing the amount of atmosphere (surface pressure) has a powerful effect on climate. A thicker atmosphere would expand the elevation at which clouds can form. On Earth, the 0.33 emissivity for high cloud provides a measure of the temperature of the atmosphere at which high cloud forms. This temperature is 218 K because (218/288)⁴ = 0.33. On Earth, atmospheric pressure in the troposphere (the thick portion of the atmosphere under 11 km elevation) roughly follows this relation:

(19)                        P = P_S(T/T_S)^5.26

Here the subscript s refers to surface. I note that at the level of high cloud T/T_S = ε^¼. With this I can write:

(20)    P_HC = P_S ∙ ε_HC^1.32 = 1.0 ∙ 0.33^1.32 = 0.23 atmospheres

Thus, high cloud is associated with a pressure (P_HC) of about 0.23 atmospheres (in this model). At pressures below this, clouds apparently do not form, perhaps the air is too thin to keep ice crystals in suspension. If I assume that high cloud occurs where P_HC = 0.23, then I can use equation B.18 to predict ε for high cloud in atmospheres thicker than the Earth’s. The relation is:

(21)    ε_HC = (0.23/ P_S)^0.76 = 0.33 P_S^−0.76

For P_S = 1.75 atmospheres, ε becomes 0.22, for which equation 4 gives a temperature of 268 F. This value is above the 241ºF boiling point of water at 1.75 times atmospheric pressure. The oceans would boil away, vastly increasing the surface pressure P_S and increasing the elevation of the uppermost cloud layers, resulting in a reduction in the value of ε_HC according to equation 21. The mass of the oceans is more than 100 times the mass of the atmosphere, enough to form a much, much denser atmosphere than Earth has now. It is probably more appropriate to use the atmospheric pressure and albedo for Venus itself, which is covered by high clouds, and has a surface pressure P_S of 93 atmospheres and an albedo of 0.77. With P_S of 93, equation 21 gives a value for ε of 0.0105. Using A = 0.77, ε = 0.0105 and S = 2600 in equation 4 gives T = 708K (814 F). The hot Earth would gradually lose its water to photolysis by solar ultraviolet rays, producing hydrogen and oxygen. The former would escape and the latter would oxidize surface rocks and sulfur compounds released by volcanism (producing sulfuric acid). The heat would bake out CO₂ from surface carbonates, maintaining a high atmospheric pressure despite loss of water by photolysis. The end result would be a bone dry, extremely thick atmosphere composed mostly of CO₂ with very high temperatures and covered with thick high clouds composed of sulfuric acid, which is exactly what Venus is today. I note the average surface temperature of Venus is 847° F, close to the value estimated by the simple model. Note that CO₂ was not even considered in this calculation. According to the simple model it is the sheer thickness of the Venusian atmosphere and the complete cloud coverage that makes the planet so hot. Given this, one would expect a planet with very little atmosphere, like Mars, to have little greenhouse warming even if what little atmosphere it had were entirely composed of a greenhouse gas like CO₂, which is indeed the case for Mars.

Another Example: Forcings

More sophisticated models can give more detailed predictions. These results are expressed in terms of forcings, expressed as change in the effective solar insolation in watts per square meter. Temperature change due to changes in climate-affecting factors (ΔT) are directly related to forcing (F):

(22)   ΔT = λ F

Greenhouse gas forcings calculated from complex models are approximated by equations having a logarithmic form, as is implied in Figure 1:

(23)   F = 5.35∙ln[C_CO2/^BC_CO2]

Here, ^BC_CO2 is the base case for C_CO2. The λ parameter in equation 22 is called the climate sensitivity. Calculations for just the effect of CO₂ provide a value of 0.3 for λ. With λ = 0.3, the CO₂ level of 419 ppm in 2022 and the value of 316 ppm in 1959, equations 23 and 22 give a forcing change of 1.52 watts per sq. meter and temperature change of 0.46º C. The actual temperature increase over 1959 to 2021 has been 0.9º C, about twice the value predicted with λ = 0.3. This value for λ does not consider the effect of the higher levels of water vapor which would accompany higher global temperatures. Since water is a greenhouse gas, this factor would serve to increase temperature beyond the effect of CO₂ alone. Climate researchers call this a feedback effect. The simple model presented above includes a way to account for this feedback effect of higher water levels with higher temperatures.

Forcings as well and temperature changes can be obtained from the simple model, which was used to calculate both in Figure 1. The value of λ obtained from the simple model is 0.45. This value roughly corresponds to the ratio of the two scales in Figure 1. Using λ = 0.45 predicts that the temperature change from 1959 should be 0.68ºC.

But that is not all. Since higher temperature means more water in the air, it seems reasonable that more water in the air would mean more clouds. Where would these clouds form―in the wet lower atmosphere, or in the dry upper atmosphere? The lower atmosphere, being closer to surface water, has a higher water content, yet low cloud makes up less than half of total cloud cover (see Table 2). This suggests some factor other than water availability is limiting low cloud formation, so more water probably won’t mean more low cloud. On the other hand, more water in the atmosphere should translate into more water in the dry upper atmosphere, where it might be limiting. If so, more atmospheric water due to higher temperatures would translate into more high cloud.

To explore this idea, I employed the simple model to calculate the effects of a direct proportional relation between increased water vapor and increased high cloud. That is, a 1% increase in water vapor was assumed to produce a 1% increase in high cloud. With this assumption, the simple model predicts a climate sensitivity of 0.59, which yields a temperature increase of 0.90, same as the actual value of 0.9 C. The actual situation is far more complex, of course. Scientists employ far more complex models that deal with the physics directly, rather than using the simplistic concept of different kinds of “screens” that act like “atmospheric carports” to affect the temperature of the surface beneath them. Nevertheless, by employing very simple concepts (such as warmer air having more water in it because of higher vapor pressures with temperature, and proportionally more cloud cover forms with more atmospheric water) allows us to predict that water vapor and cloud should roughly double climate sensitivity (λ), which is what is observed.

Conclusions

No scientifically literate climate skeptic denies the direct greenhouse effect due to CO₂ (i.e., equation 23 with λ = 0.3, which predicts a temperature increase of about +1.1º K for a doubling of CO₂). Rather, their disagreement is about the value of λ (see discussion by physics professor and climate skeptic Nur Shaviv). They acknowledge that warmer temperatures produce more water vapor, which as a greenhouse gas will produce more warming, that is, a higher λ (as shown by the simple model). They argue, correctly, that we do not yet understand the role of clouds in detail. Most climate researchers have results implying higher cloud cover coming from greenhouse warming would increase λ, but, as discussed earlier, more low cloud can be cooling. However, as we saw in the Venus example, the idea that only maximally-cooling low cloud would form as atmospheric water content increases with temperature doesn’t really make sense.