Why J, J and J are Wrong

(c) 2009 by Barton Paul Levenson

How does the greenhouse effect work? If the Earth's surface had the same albedo (reflectivity) as the present Earth with an atmosphere, sunlight could only heat it to a temperature of 254 K. But with an atmosphere that contains greenhouse gases, thermal radiation from the Earth is absorbed, the warm atmosphere radiates, and some of that radiation goes back down to the ground and heats it further. The Earth's mean global annual surface temperature is 288 K because it receives both sunshine and "atmosphere shine." Here's an explanation in more detail for those who are interested:

Greenhouse 101

All wrong, say jae, Jan Pompe, and JamesG on the landshape blog. Back-radiation from the atmosphere is a mirage, they say. It can't happen, because the atmosphere is cooler than the Earth's surface, and you can't transfer heat from a cooler body to a warmer one.

I'll demonstrate why they're wrong.

Let's examine two objects in isolation, each with an internal heat source. They are granite blocks, each one meter square, with 10 meters distance between them. They radiate only to each other; their backs are insulated and their sides are too thin to matter. Each is a perfect blackbody, their absorptivity and emissivity is a = ε = 1.

[ A ] => <= [ B ]

Start off with A at a temperature of absolute zero. Each of these objects being perfect blackbodies, they radiate power (infrared light) by the Stefan-Boltzmann law:

F = ε σ T4 (1)

where F is the flux density in watts per square meter, ε the emissivity, σ the Stefan-Boltzmann constant (5.6704 x 10-8 W/m2/K4) and T temperature (K). A radiates zero watts per square meter, B radiates 1 x σ x 2504 or about 221.5 watts per square meter. Let's say B is supplied continuous power in this amount by a resistance coil inside it, which in turn is plugged into a wall outlet. Energy is conserved, so it doesn't cool down or heat up.

The inverse-square law is:

I = F / r2 (2)

where illumination I is in watts per square meter, luminosity (source flux density) F is in W/m2, and distance r is in meters. The blocks are separated by ten meters. A therefore receives 2.215 watts per square meter from B, which heats it up slightly. Its temperature rises to (2.215 / σ)0.25 or 79.1 K. A perfect blackbody, it radiates 2.215 W/m2 right back at B.

B is a perfect blackbody. The distance between A and B is ten meters. B receives 0.002215 watts per square meter from A as well as 221.5 from its internal power source. Its incoming energy is now greater than its output, so it must heat up. Its new temperature is (221.502215 / σ)0.25 or 250.000625 K. Not a great difference, but A has succeeded in raising B's temperature just a bit.

Let's give A an internal heat source, like B's, so that its initial temperature is 200 K. This requires a power source of 90.7 watts. A, being a blackbody, has to radiate this much.

New radiation budgets for each object:

Input 90.7 watts internally, 2.215 watts from B, total 92.915
Output 90.7 watts.
Temperature 200 K to start.

Input 221.5 watts internally, 0.907 watts from A, total 222.407 watts.
Output 221.5 watts.
Temperature 250 K to start.

Each is taking in more than they are radiating, so they will have to heat up and radiate more. Recalculate with temperatures based on total input, output rises due to increased thermal radiation, etc. An iterative model (code on request) stabilizes with the following budgets:

Input 90.7 internal + 2.224 B = 92.924
Output 92.924
T 201.2 K

Input 212.5 internal + 0.929 A = 222.429
Output 222.429
T 250.3 K

So A managed to raise B's temperature even though A is cooler than B. How is this possible?

It's possible because B doesn't "know" that A is cooler than it is. Radiation is radiation is radiation, as Gertrude Stein said. B absorbs the infrared photons from A, they increase the internal energy of B's molecules, those molecules increase their root-mean-square velocity through the laws of statistical mechanics, and B increases its temperature.

Now, J, J and J maintain that A can't radiate to B. Back-radiation, they say, violates the second law of thermodynamics, which states that "heat cannot flow from a cooler body to a warmer one."

Except that it doesn't state anything of the kind. What it states is that "NET heat cannot flow from a cooler body to a warmer one."

What is the net heat flow in this situation? It's 2.224 watts from B to A, minus 0.929 watts from A to B, for a net heat flow of 1.295 watts from B, the warmer body, to A, the cooler body. Nothing is being violated, but cooler A is still radiating to warmer B and raising its temperature. In the absence of the cool block A, the warm block B would be at a temperature of 250.0 K. In the presence of the cool block A, B's temperature rises to 250.3 K. Radiation from a cooler source is heating a warmer source.

And that's how the greenhouse effect works. The warmer Earth is receiving radiation from the warmer sun, and a lot more radiation from the cooler atmosphere. In the absence of an atmosphere, if the same albedo were retained, Earth would be at a temperature of 255 K. With an atmosphere providing back-radiation, its temperature is 288 K. That's the greenhouse effect, and that's how it works. And that's why J, J and J are WRONG.

Page created:01/13/2009
Last modified:  02/09/2011