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If the climate modeler says: "more carbon dioxide in the atmosphere results in less out-going longwave radiation energy emitted to space", then this too is a clear, simple, assumption which must be verified and quantified by basic experiment.
nyway, an acid test is intentionally potentially destructive: if a substance is pure gold, or an article is covered with contiguous gold plating, it will not dissolve in hydrochloric acid, but pretty well anything else will be destroyed.
Hold on! An acid test can't be a basic test!
I presume that you know that it has been verified.
Carbon dioxide absorbs IR which "traps" heat.
My studies indicate:CO2 absorbs IR, and emits IR; more or less at the same frequency and intensity.Absorption and emission vary depending on temperature. Using Planck's Law, one can calculate the equilibrium temperature for CO2 main band (15µm). It is -77.8C. Which is very cold. At, or below, -77.8C CO2 absorbs IR. Above that temperature it emits IR. There are very few places on Earth or in the atmosphere which get that cold. So it looks to me that the equilibrium position (w.r.t. temperature) for this CO2 'absorption band' is to emit. It, no doubt depends on the temperature of the molecule. An energised molecule should emit a photon, taking the activated electron back to a minimal energy orbital. (is that a 'ground state'). No IR photons will be trapped. No trapped heat.Other people, say that CO2 makes the atmosphere more opaque to IR. Can anyone show me a citation from the people who measured that?
Then look for the science behind those basic assumptions.
you think you can sell me some gold?
You want me to take that on trust ?
At, or below, -77.8C CO2 absorbs IR. Above that temperature it emits IR.
So it looks to me that the equilibrium position (w.r.t. temperature) for this CO2 'absorption band' is to emit.
Other people, say that CO2 makes the atmosphere more opaque to IR. Can anyone show me a citation from the people who measured that?
So the question is, what is the mean free path of a 15 µm photon in an atmosphere with 400 ppm of CO2 vs that with only 350 ppm? And how does that translate into time required to exit the atmosphere? (according to the paper linked below, the mean free path is on the order of a meter at 300 ppm, which means that the photon would have to be absorbed an re-emitted at least 50,000 times before getting from the ground to an altitude of 50 km--and this small number would only occur if every re-emission were up (highly unlikely)
He doesn't understand that "The acid test" was used to distinguish real gold from pyrites.
None of the components I tested was likely to be plated with FeS2, but porous gold plating showed up pretty quickly as the underlying copper or steel bubbled away. Whilst pure gold won't dissolve in HNO3, several perfectly acceptable alloys will, so we didn't use it.
In waters case it would rather liberate itsel from its earthly bonds and become gaseous that emit radiation,
the lower atmosphere can be explained by up to 3 equations of state
Immediately after 9/11 there was an absence of vapor trails over the USA
None of the components I tested was likely to be plated with FeS2