Why Different Explanations of Global Warming All Mean the Same Thing

(c) 2023 by Barton Paul Levenson

I mean, of course, all competent explanations. Many commentators have noted that explanations such as "carbon dioxide traps heat in the atmosphere" or "the atmosphere acts like a glass greenhouse" are poor ones.

Dr. Peter Eirich, an engineer, maintains that there are at least five different, mutually incompatible "theories" of how the greenhouse effect works, which, of course, proves that climate scientists don't know what they're talking about, or perhaps are trying to hoodwink the public. He has advanced this idea several times on ResearchGate.net. This is an entirely respectable venue for scientists to post their papers on, so that the papers can discussed by other scientists and by students. Sadly, a number of threads have recently been invaded by global warming deniers.

Here are two explanations for the greenhouse effect commonly given by climate scientists when trying to educate the public, or students:

1. The emission height explanation. The presence of greenhouse gases in a planet's atmosphere raises the average height of emission of infrared radiation to space. Since planetary tropospheres exhibit a lapse rate--that is, they tend to get colder with altitude--the higher the mean emission altitude, the warmer the surface.
2. The back-radiation explanation. Infrared radiation from the ground warms greenhouse gases in the air. These radiate infrared radiation of their own in all directions. Half of these directions are back down toward the ground, so the ground has both sunshine and atmospheric back-radiation heating it, and consequently is warmer than it would be from sunshine alone.

I will now demonstrate that these explanations are mathematically equivalent and give the same numerical answers for the same input values.

1. Using the mean emission height.

We begin by positing a planet with no atmosphere, or no greenhouse agents (gases or clouds) in its atmosphere. The planet has the same insolation and albedo as Earth, so the ground is absorbing 240 watts per square meter of sunlight. Assume an emissivity of 1. Consider the Stefan-Boltzmann law:

F = ε σ T⁴ . . . . . . . . . . [1]

where F is the flux density of electromagnetic radiation emitted by a warm body, in watts per square meter; ε is the emissivity, or radiative efficiency (which must be between 0 and 1); σ is the Stefan-Boltzmann constant (5.670373 x 10^-8 W m^-2 K^-4), and T is the absolute temperature (K). In radiative equilibrium, the ground must radiate as much as it receives. If it radiates less, it will heat up until it vaporizes. If it radiates less, it will cool down to absolute zero. We therefore assume radiative equilibrium. Using equation [1], the temperature of the ground must be 255 K. Since all the outgoing energy is coming from the ground, the mean height of emission is 0.

We now assume the atmosphere holds greenhouse gases, or clouds, or both--the identity of the greenhouse agent is irrelevant--and that as a result, the atmosphere absorbs 77% of the radiation coming from the ground. Instead of 240 W m^-2 getting out to space from the ground, only 55.2 W m^-2 gets out. But of course the absorption of energy by the atmosphere heats the atmosphere, as well, which also radiates. It is intercepting 184.8 W m^-2 and therefore must radiate an equal amount. On average, half of this will head out to space and half back down to the ground.

So the climate system is receiving 240 W m^-2 from the sun, but is only returning 55.2 W m^-2 (ground) plus 92.4 W m^-2 (atmosphere) = 147.6 W m^-2 to space. More is coming in than going out, so the system must heat up.

Equilibrium will be restored when as much is getting out to space as is coming in--a consequence of conservation of energy. At equilibrium:

240 = 0.23 (σ Tg⁴) + (1/2) 0.77 (σ Tg⁴) . . . . . . . . . . [2]

which gives us a ground temperature of Tg = 288 K. The ground is radiating 390.2 watts per square meter, but only 240 is getting out to space, so we have a 150.2 W m^-2 greenhouse effect.

The atmosphere is absorbing 77% of the 390.2 W m^-2, or 300.5 W m^-2. It's radiating 150.2 W m^-2 in each direction. We know the atmosphere has an absorptivity of 0.77, so by Kirchhoff's Law it must have an emissivity of 0.77 as well, and using equation [1], its temperature is 242.2 K.

Earth's mean tropospheric lapse rate is 6.5 K km^-1. The atmospheric radiating height is thus (288 - 242.2) / 6.5 = 7.0 km. Space is getting 150.2 W m^-2 (62.6% of the total) from the atmosphere and 89.8 W m^-2 (37.4%) from the ground, so the mean radiating height of the climate system is 0.374 x 0 + 0.626 x 7 = 4.4 km. The presence of greenhouse agents in the atmosphere has raised the effective radiating height.

2. Using back-radiation.

We match the conditions of the first example, so nothing is different but the method of explanation. We write energy balances for space, the atmosphere, and the ground.

Space is giving the climate system F = 240 W m^-2 and getting 240 W m^-2 back out. We write

F = 0.23 Fg + Fa . . . . . . . . . . [3]

where F = 240 W m^-2, Fg = 390.2 W m^-2, and Fa = 150.2 W m^-2.

The atmosphere is getting nothing from space, but has 0.77 x Fg or 300.5 W m^-2 coming in from the ground. It is emitting Fa = 150.2 W m^-2 both up and down. We write:

0.77 Fg = 2 Fa . . . . . . . . . . [4]

where Fg = 390.2 W m^-2 and Fa = 150.2 W m^-2.

Lastly, the ground is getting 240 W m^-2 from space and 150.2 W m^-2, the back-radiation, from the atmosphere. It is emitting Fg = 390.2 W m^-2. We write:

F + Fa = Fg

In the absence of the back-radiation, the ground would be receiving 240 W m^-2 and would have a temperature of 255 K. In the presence of the back-radiation, the ground receives 240 W m^-2 plus 150.2 W m^-2 = 390.2 W m^-2, and its temperature is 288 K. The back-radiation has thus raised the temperature of the ground from 255 K to 288 K, a difference of 33 K.

All layers are in radiative equilibrium. Energy is conserved at all levels. No laws of thermodynamics are violated.

Conclusion.

The temperatures and flux densities are the same in both examples. Thus they are mathematically equivalent and give numerically identical answers. Q.E.D.