Although the cornerstone of the "greenhouse effect" is so-called "back-radiation" from "greenhouse gases" in the atmosphere, the foundation stone is the Stefan-Boltzmann law. The reason for all the quotation marks in the sentence is because without a correct application of the law, all those principles are without basis, and so are worthless.

From: Earth’s Annual Global Mean Energy Budget, J. T. Kiehl and Kevin E. Trenberth, February 1997, Bulletin of the American Meteorological Society 78, 197-208 (http://www.geo.utexas.edu/courses/387h/PAPERS/kiehl.pdf)

"For the outgoing fluxes, the surface infrared radiation of 390 W sq.m corresponds to a blackbody emission at 15°C"

At face value, this looks fine, except it assumes an ideal radiator, and radiation into surroundings at absolute zero (-273°C). The Stefan-Boltzmann law defines the radiation of a body as proportional to the 4th. power of the absolute temperature, multiplied by the emissivity (1 for an ideal radiator) for radiation into a vacuum. (see http://hyperphysics.phy-astr.gsu.edu/hba...tefan.html) However, the earth's surface is not radiating into a vacuum, but into the atmosphere. For radiation into surroundings above absolute zero, the radiation is proportional to the absolute difference between the fourth powers of the two temperatures. This implies that if the temperature difference is zero or negative, there is no outgoing radiation. This is not surprising; a cooler body cannot heat a hotter body, neither can two bodies at the same temperature exchange heat or infrared radiation. These effects would violate the Second Law of Thermodynamics. The linked page has a calculator; emitter and surroundings temperatures and emissivity can be input to calculate radiation in Watts per sq.metre if the area of the emitter is set to 1. Given a surface temperature of 15°C with emissivity .9 and an atmosphere at 10°C, the surface would emit 23.79 W sq.m. This is far from the figures quoted on the "energy balance" diagrams of 390 or 396 W sq.m. In any case, the cooler atmosphere can't possibly heat the hotter surface.The Stefan-Boltzmann law doesn't allow radiative flux in the "negative" direction. A straight calculation using the emitter/surroundings temperatures gives a negative result!

Kiehl and Trenberth (and others) back-calculate the temperature of the top-of-atmosphere (TOA) from the measured (by satellite) outgoing infrared radiation of 235 W sq.m as -19°C, assuming an ideal emissivity of 1. They then conclude that the surface would emit the same without an atmosphere despite the surface having a smaller area, and neither surface nor atmosphere having the ideal emissivity (likely to be more like 0.9 for land/water), and even then, not identical emissivities. The TOA isn't even a surface, but a layer. I can find no quoted figure for the emissivity at TOA, so I have no idea what the correctly calculated temperature at TOA might be. The assumed surface temperature of -19°C "without greenhouse effect" has to be wrong in any case. The Ideal Gas Law can be used to calculate the air temperature at the surface, assuming a nitrogen/oxygen/water-vapour atmosphere at a pressure of 1 bar. This gives about 15°C with no requirement for any input from the sun. A calculation for the atmosphere of Venus (surface pressure 93 bar) gives a result close to the observed surface temperature. The difference can be explained by the unique atmospheric conditions on the planet; there is no need for any "greenhouse" effect calculations. Indeed, with an atmosphere similar to that on Earth, the calculated temperature is around 200°C hotter.

The "greenhouse effect" is a failed hypothesis; if its proponents employed real physics instead of misapplied laws and bad assumptions it would fall apart. The panes of the greenhouse are cracked; those that aren't cracked are missing.

From: Earth’s Annual Global Mean Energy Budget, J. T. Kiehl and Kevin E. Trenberth, February 1997, Bulletin of the American Meteorological Society 78, 197-208 (http://www.geo.utexas.edu/courses/387h/PAPERS/kiehl.pdf)

"For the outgoing fluxes, the surface infrared radiation of 390 W sq.m corresponds to a blackbody emission at 15°C"

At face value, this looks fine, except it assumes an ideal radiator, and radiation into surroundings at absolute zero (-273°C). The Stefan-Boltzmann law defines the radiation of a body as proportional to the 4th. power of the absolute temperature, multiplied by the emissivity (1 for an ideal radiator) for radiation into a vacuum. (see http://hyperphysics.phy-astr.gsu.edu/hba...tefan.html) However, the earth's surface is not radiating into a vacuum, but into the atmosphere. For radiation into surroundings above absolute zero, the radiation is proportional to the absolute difference between the fourth powers of the two temperatures. This implies that if the temperature difference is zero or negative, there is no outgoing radiation. This is not surprising; a cooler body cannot heat a hotter body, neither can two bodies at the same temperature exchange heat or infrared radiation. These effects would violate the Second Law of Thermodynamics. The linked page has a calculator; emitter and surroundings temperatures and emissivity can be input to calculate radiation in Watts per sq.metre if the area of the emitter is set to 1. Given a surface temperature of 15°C with emissivity .9 and an atmosphere at 10°C, the surface would emit 23.79 W sq.m. This is far from the figures quoted on the "energy balance" diagrams of 390 or 396 W sq.m. In any case, the cooler atmosphere can't possibly heat the hotter surface.The Stefan-Boltzmann law doesn't allow radiative flux in the "negative" direction. A straight calculation using the emitter/surroundings temperatures gives a negative result!

Kiehl and Trenberth (and others) back-calculate the temperature of the top-of-atmosphere (TOA) from the measured (by satellite) outgoing infrared radiation of 235 W sq.m as -19°C, assuming an ideal emissivity of 1. They then conclude that the surface would emit the same without an atmosphere despite the surface having a smaller area, and neither surface nor atmosphere having the ideal emissivity (likely to be more like 0.9 for land/water), and even then, not identical emissivities. The TOA isn't even a surface, but a layer. I can find no quoted figure for the emissivity at TOA, so I have no idea what the correctly calculated temperature at TOA might be. The assumed surface temperature of -19°C "without greenhouse effect" has to be wrong in any case. The Ideal Gas Law can be used to calculate the air temperature at the surface, assuming a nitrogen/oxygen/water-vapour atmosphere at a pressure of 1 bar. This gives about 15°C with no requirement for any input from the sun. A calculation for the atmosphere of Venus (surface pressure 93 bar) gives a result close to the observed surface temperature. The difference can be explained by the unique atmospheric conditions on the planet; there is no need for any "greenhouse" effect calculations. Indeed, with an atmosphere similar to that on Earth, the calculated temperature is around 200°C hotter.

The "greenhouse effect" is a failed hypothesis; if its proponents employed real physics instead of misapplied laws and bad assumptions it would fall apart. The panes of the greenhouse are cracked; those that aren't cracked are missing.

Ernest Rutherford: "If your experiment needs statistics, you ought to have done a better experiment."