Geometry Deniers

(c) 2023 by Barton Paul Levenson



Why we need the greenhouse effect

Earth's energy balance at the top of the atmosphere is

[1]     (S / 4) (1 - A) = σ Te4

where S is the Solar constant (1,361.5 W m-2 pace Kopp and Lean 2011), A the Earth's bolometric Russell-Bond spherical albedo (0.294 from averaging recent satellite measurements), σ the Stefan-Boltzmann constant (NIST CODATA value 5.670373 x 10-8 W m-2 K-4) and Te the Earth's radiative equilibrium temperature.

Plug in the numbers and you get Te = 255 K. This is below the freezing point of water, 273 K. Thus we need some process to raise Earth's surface temperature to its observed value of 288 K. That process is the greenhouse effect. For a description of how it works, try here.



Geometry deniers

It is no secret that many global warming deniers deny the greenhouse effect even exists (see, for instance, here and here. But a new variation on this has been introduced by Joseph E. Postma here. Postma describes himself as an astrophysicist "working under contract to the Canadian Space Agency." In fact he is an instrument calibrator for a contractor that does some work for the CSA. The man actually does have an M.S. in astrophysics, though how he managed to get it and still make elementary mistakes in thermodynamics boggles the mind. But on to his argument.

Postma maintains that to divide by 4 as in equation 1 is illegitimate. All that matters, he says, is the illumination on the day side.

If you change the "4" in equation 1 to a "2," you get a temperature of 303 K. Postma manages to reconcile this with the actual 288 K by saying some of the absorbed energy must go into "non-thermal" processes like latent heat, evaporation, and convection, apparently missing the point that the Earth would still have to radiate this energy outward eventually or you'd have more energy coming into the system than leaving it and Earth would be out of energy balance. You can't transfer energy off a planet by latent heat, evaporation, or convection. Only radiation works in a vacuum. But at any rate, in Postma's world, the problem is that temperature from sunlight alone is too hot, not too cold. No need for that pesky greenhouse effect, which Postma says violates the laws of thermodynamics, though he doesn't exactly demonstrate how.

But wait! It gets worse!

ResearchGate is a legitimate venue for page-sharing and discussion among professional scientists and students. But it lacks any serious moderation of the discussion threads. One contributor, Brendan Godwin, has posted an "energy balance" chart for the Earth's climate system that superficially resembles similar charts by Kiehl and Trenberth (2009), Stephens et al. (2012), etc. But he has either photoshopped and altered one of those, or just used similar drawing software, to put in his own numbers. He has the top of atmosphere illuminated as 1,360 W m-2. The full Solar constant. S, not S/4!

With our figures so far, that would give us T = 394 K, equal to the temperature of an airless, nonrotating Earth at the subsolar point, assuming Earth somehow kept the same albedo.



Why the mean Solar flux density at the top of the atmosphere is S/4 and not S. Part 1. The hard way.

Using the SORCE TIM satellite instrument, Kopp and Lean (2011) measured the Solar constant as 1,360.8 W m-2 at Solar minimum. They found an amplitude of 1.5 W m-2, implying a mean Solar constant of 1,361.5-1,361.6 W m-2. To simplify the math, we will assume S = 1,360 W m-2.

Now, the Earth being a sphere, half of it is sunlit at any one time ("day") and the other half is dark ("night"). The most flux density the day side could possibly be receiving--which we will see later is an overestimate--would be the Solar constant, 1360 W m-2. Meanwhile, the night side is receiving 0 W m-2. The highest the average over the whole surface would be is therefore (1,360 + 0) / 2 = 680 W m-2.

But the day side is a hemisphere. Most of it slants away from the sun, and a surface slanted away from a light source receives less light per unit area. This is easily demonstrated with a flashlight and a tile floor. Hold the flashlight so it shines straight down and each tile is well lit. Hold it at an angle and each tile is not as well lit, because the same amount of light has to spread out over more tiles.

The amount by which illumination changes with angle is described by Lambert's cosine law:

[2]     I = I0 cos(θ)

where I0 is the original or perpendicular illumination, and θ is the angle of tilt away from the light source--90° is flat on to the light source and 0° is parallel to the light beams. We see at once that the top of Earth's atmosphere only receives the full Solar constant at the subsolar point (I = S), while at the terminator, I = 0.

But we can't just integrate the cosine of the Solar altitude, because different strips or rings of Earth's surface centered on the subsolar point have different areas. To be exact, the fraction of area on a hemisphere equatorward of a given latitude (or in our daylight illumination example, terminator-ward of a given ring angle) is equal to the sine of that angle. For example, half of Earth's surface is between the 30° lines of latitude, since sin(30°) = 0.5. (In reality it's slightly more since Earth is an oblate spheroid rather than a perfect sphere.)

So we are integrating the expression

     cos(θ) sin(θ)

from θ = 0 to 90 degrees. The antiderivative is:

     sin2(θ) / 2

and applied to the limits of 0 to 90°, this works out to

     1/2

So that the day side only receives, on average, half the Solar constant, or 1,360 / 2 = 680 W m-2. The average between 680 W m-2 on the day side and 0 W m-2 on the night side is (680 + 0) / 2 = 340 W m-2, which is S/4. Q.E.D.



Why the mean Solar flux density at the top of the atmosphere is S/4 and not S. Part 2. The easy way.

The Earth collects the Solar constant, which we are taking as 1,360 W m-2, on its cross-sectional area, which is that of a circle:

[3]     Across = π R2

The total power coming in is then

[4]     Pin = S π R2

But Earth's total surface area is that of a sphere:

[5]     Atotal = 4 π R2

The average Solar illumination per unit surface area--let's call it "Q"--is therefore

[6]     S = π R2 / (4 π R2)

Quantities in both the numerator and denominator of a fraction--the top and the bottom--cancel out. We are therefore left with

[7]     Q = S / 4

If S = 1,360 W m-2, then Q = 1,360 / 4 = 340 W m2. Q.E.D.



A warning

I do NOT recommend going to Postma's blog, "Climate of Sophistry," and attempting to argue with him. A number of people have gone there and, very politely and respectfully, pointed out his errors. His response was to 1) insult them unmercifully, and 2) ban them from his blog. In one case (described here) he made up a fake post under the user's screen name claiming to have been convinced that Postma was right. This is a man whose reaction to any suggestion that he might be wrong is pathological. I strongly advise staying away.



References

Kopp, G., Lean, J.L. 2011. A new, lower value of total solar irradiance: Evidence and climate significance. Geophys. Res. Lett. 38, L01706.

Trenberth, K.E., Fasullo, J.T., Kiehl, J. 2009. Earth's global energy budget. Bull. Amer. Meteorol. Soc. 90, 311-324.

Stephens, G.L., Li, J., Wild, M., Clayson, C.A., Loeb, N., Kato, S., L'Ecuyer, T., Stackhouse, Jr., P.W., Lebsock, M., Andrews, T. 2012. An update on Earth's energy balance in light of the latest global observations. Nature Geosci. 5, 691-696.



Page created:03/23/2023
Last modified:  03/23/2023
Author:BPL