I have a nasty feeling I’m going to get a science lesson. Serve me right…

It is also collisions which provide the energy for emissions, which is why the intensity is temperature and concenntration dependant. So for the majority of emissions the process is:
1. collision(s) excites molecule
2. If molecule is not again collided with, it may emit a photon. In general this is a low probability outcome – usually the molecule is stripped of its energy by another collision before it has time to radiate. However there are lots of molecules and lots of collisions, so emissions do occur.

At the top of the atmosphere, there are fewer potential absorbers between an emitter and space. So if a CO2 molecule emits a photon, it is more and more likely that it will leave the planet the higher up you are (ie the less CO2 there is to absorb it).

The question I posed was – what depth of CO2 will cause total extinction (or near enough to it) – and the answer I got was about one tenth of the atmosphere (mass) for the 15um band. So the lowest level from which 15um radiation can escape the planet is above 90% of the atmosphere, ie well into the Tropopause.

For the wings (eg Wavenumbers 600, 750) over 50% seems to be coming from above this level as well.

So virtually all the radiation from CO2 originates from the Tropopause and above…if my use of the absorption tables is correct…

One thing you need to keep in mind is that radiation from CO2 will arise from the temperature of the atmosphere itself as opposed to the surface temperature. That is, a CO2 molecule which absorbs IR or a water vapor molecule which absorbs IR will be hit by other molecules and lose the energy to the bulk of the atmosphere. Only the top layer of the atmosphere will be sufficiently thin to allow the back IR radiation to be appreciably less than the emission to space.

The standard absorption tables for CO2 imply that at most 10% of the atmosphere is required for complete absorption. So any 15um band radiation from below the 90% height of the atmosphere, will be absorbed by the overlying CO2 and not make it out into space. I make this firmly above the tropopause. Even in the wings of the band, the majority of the radiation emanates from above the tropopause.

This implies that any increase in concentration of CO2 will tend to increase the temperature of the radiation in this band – ie a NEGATIVE forcing.

[The table I consulted for absorption by CO2 is as follows:
(Figures are % transmitted)

Wave Number Concentration, atm cm
0.2 0.5 1 5 10 100 1000

600 98/ 96/ 93/ 78/ 69/ 28/ 2
650 74/ 61/ 48/ 16/ 8/ 0.1/ 0
700 86/ 76/ 64/ 33/ 22/ 2/ 0
750 99/ 98/ 97/ 88/ 82/ 45/ 10 ]

All the best,

w.

There’s this controversy where a minority of heretics have difficulty in accepting the theory that EMR from any source is always absorbed when it strikes an absorbent material, even when the target is hotter than the source. This appears to offend the second law of thermodynamics, but the standard explanation is that law 2 does not apply to EMR, because there are two different opposing flows, not 1. (in the simplest case)

1) Despite this claimed uniqueness, “opposition of flow” has equivalences in all translatable forms of energy, without offending law 2, and here follow some examples. However, before listing them, and you dismissing it with “so what”, there is this strongly related consideration:

It is not necessary to invoke in the calculation of net radiative heat transfer, that the radiation from a colder source is first absorbed by the warmer body. It is simply a matter of difference in potential, and is similar to conductive heat transfer, a matter of T1 – T2, and all of the translatable energy transfer calculations contain a similar term. (in EMR typically Q1 – Q2….or T to the fourth can be substituted for Q).

Analogy 1: Consider a simple electrical circuit with a DC power supply of 12v. Apply an opposite supply of 3v, and the net voltage will be 9v. There is no transition point between 3v and 12v; implying that the 3v passes through the 12v, and subtracts from it. The 3v is not absorbed by the higher potential source, and, the 12v source continues to exert 12v. The resultant of V1– V2 is translatable to power, W1 – W2, and is also converted to heat across a resistor. (“absorbed” as heat just like EMR and translatable as Q1 – Q2).

Analogy 2: Consider a water reservoir with a hypothetical perfect flow outlet pipe. The rate of flow will be proportional to the fluid pressure, or the difference in height between the water surface and the outlet. (potential) If we then place say a restrictive grille across the outlet, a back-pressure will be created, and we end-up with the familiar calculation P1 – P2 where P = pressure, in opposite directions, which is the equivalent of EMR. The PE of the water that is converted to KE, is the equivalent of heat.
This example can also be adapted to show equivalence of EMR between matter of equal T: If we now close the outlet and divide the reservoir into A & B, the pressures each side of the divider are equal but opposite. Remove the divider, and there is no flow of water, with equal opposing pressures. This is the equivalent of no flow of heat between matter at equal T, or equal opposing EMR

Other analogies may be derived, (assume hypothetical perfection in each case), such as:

Analogy 3: Consider a block of material sliding down an inclined plane uniformly …(F1 – F2) x d = KE

Analogy 4: Consider processes within opposing magnetic fields.

Analogy 5: Consider opposing heat flows in a conductive material at equilibrium.

As step 1 in the discussion, (more to come), do you agree that these ARE equivalences to EMR, and its translation into heat.

Gary, you are correct that you can wrap a blanket of some kind around a source of radiant heat and achieve any temperature. In addition you can do it with or without radiant heat exchange. Just cut down on the conduction, no problem.

However, in the atmosphere, there is very little heat transferred by conduction (other than the initial transfer by conduction from the surface). Almost all of the heat is transferred by convection (or radiation). This is because air is a very, very poor conductor of heat, viz:

THERMAL CONDUCTIVITIES
copper___________224
aluminium________117
steel_____________27
ice______________1.3
water___________0.32
air____________0.014

Thus, we are justified in ignoring conduction in the atmosphere.

John, I took a look at your model, and I can only shake my head in awe. That way lies madness … but a stupendous work of madness, indeed. I salute you.

For both of you, I was pushing the envelope the other direction — what is the minimum toy model sufficient to investigate the effect of a change in radiative properties? What can we do without? How simple a model can I make that is realistic enough to be useful?

w.

The effect of more radiation being emitted towards the earth then away is shown. However, although I tried to improve on Willis’s model I sill don’t think the model I arrived at is very realistic.