You said “No, you don’t.

DSL has already provided a number of links including one to how the pyrgeometer works.

Have you read it? Do you understand it?”

Well…my understanding, according to the primers DSL provided, the receiver temperature is reduced/subsidized. Specifically, from DSl’s link:

“To derive the absolute downward long wave flux, the temperature of the pyrgeometer has to be taken into account. It is measured using a temperature sensor inside the instrument, near the COLD junctions of the thermopile”

Please explain.

As to the radiative transfer equations and text books, in what instances do the equations not hold?

..At surface temperature equilibrium, E_in = E_out

That is, alpha*(e*S-B)a – (e*S-B)s = 0″

Now, the term (e*S-B)a can be expressed relative to the term (e*S-B)s thus:

alpha*z*(e*S-B)s = (e*S-B)s (z less than 1)

That is, alpha*z = 1

This implies alpha greater than 1, an unphysical condition.

How did you prove surface temperature equilibrium? You just assumed it.

The 1st law of thermodynamics tells you that E_in – E_out = C x ΔT, where C = heat capacity in J/K.
[Assuming no phase change like evaporation]

When the surface absorbs less energy than it loses it cools down.

If you have a condition of constant temperature then E_in = E_out.

And of course you need to consider all of the sources of energy in – so you must include solar radiation, and all the mechanisms of energy out – radiation, convection and conduction.

“The relationship between surface temperature, convection, lapse rate and emission of radiation to space from the troposphere is of course the most important determinant of the surface temperature”

Isn’t it necessary to distinguish between temperature level and variation?Climate system processes determine the variations (diurnal, seasonal and inter-glacial) in surface temperature, not its level (Ts) which reflects the internal energy of the whole planet. The internal energy of the climate system, a fraction of that of the whole, is reflected in its effective radiating temperature (Ta) which, therefore, must be less than Ts. Kramm and Dlugi’s endorsement of G&T’s finding, that the arithmetical excess of Ts over Ta is a “meaningless number” (and says nothing about the greenhouse effect), is warranted.

Here is another attempt:

From Newton’s thrid law:

“To every action there is always opposed an equal reaction:” (Google Translation)

The conservation of both linear and angular momentum can be seen as consequences of the term “opposed”.

Opposed can be seen as having two aspects which I will combine in the phrase “collinearly opposed”.

Without the (“co[l]“) the remaining “linearly opposed” is sufficient to conserrve linear momentum. The “co-” part is necessary to conserve angular momentum.

Simply put the “co-” part states that the opposed reaction is in alignment (along the line) of the action. Otherwise the forces could only be in opposed directions.

The “co-” aspect was not left to be deduced from the non “co-” aspect. Newton’s usage of opposed is read to imply both.

Here is a modern rendering courtesy of Wikipedia:

“The mutual forces of action and reaction between two bodies are equal, opposite and collinear.”

The “co-” part establishes that a closed system cannot subject subject itself to a spontaneous net torque. This prevents the need to speculate on whether such a system could convert some other internal energy into a spontaneous rotation or vice versa. Such a rotation would not of itself change the linear momentum thus statisfying its conservation.

Mentions above of Noether are correct but are not additional to Newton’s Laws. The existence of the stated symmetries are inherent to those laws.

Alex

I’m very pleased to read (both) your comments – I find myself in a very similar position. I have a fair amount of conceptual understanding but I would struggle to use equations to communicate that understanding to others. I worry too, that maybe I’m just appealing to an authority that suits my biases?

However, your second comment was an expression of something similar I’ve thought myself – a wandering body in deep space. Internal heat source, external temp a constant 223K. Giving off heat [in my layman's language] everywhere it goes. Does not reflect other radiation, merely stays at 223K.

An astronaut, cruelly abandoned in deep space has a choice. Rather rapidly cooling to death he can hail the passing body and ask it to park up alongside him, or can hope that his spacesuit will slow his cooling long enough for him to be alive when his ‘friends’ return at some point in the future.

Is he better off with the presence of the body, will it make no difference or will it make things worse? The body will remain at 223K giving out heat in all directions. He, however has a surface temperature of 273K, so the body is a depressing 50K colder than him.

In this situation I’m saying “Oi, you, over here, quick!”. And assume that my life might be saved by the heat emanating from the cooler body. It may indeed only slow my cooling but that also means keeping me warm[er than I would otherwise have been]

Energy really will move from colder to warmer.

Now, if I happened to have some super-combusto-carb food that was just enough to keep me at about 273K, the arrival of the good-natured, colder body might warm me enough for me to reach a survivable temperature allowing me to plan my revenge on my fellow astronauts should they return.

That’s the clearest way I can conceptualise energy moving from colder to hotter. I’d be glad to know if that is an explanation that is meaningful for anybody else..

P.S. Having said that, I’ve now forgotten why the ‘lagging on a hot water pipe’ isn’t an even simpler idea. The lagging doesn’t need to be a heat source, but ‘heats’ the surface of the pipe nonetheless. And is always colder than the pipe.

I do have only a vague idea of where most outgoing OLR originates (except for the 40 W/m2 that goes through the window). The wonderful calculations you have shown before show how the outward flux originating at the surface is changed by absorption and emissions as it moves up through the atmosphere. By the time we get to the top of the atmosphere, few of the original photons emitted from the surface are left and I don’t have a good feeling for what has been happening to the photons emitted by the atmosphere. Unlike the surface, which emits at both strongly- and weakly-absorbed wavelengths; the atmosphere emits most strongly at the wavelengths that it absorbs best, and never emits wavelengths it can’t absorb. On the K&T diagram, 40 W/m2 escape directly to space, 26 W/m2 eventually make it to space and the rest returns to the ground to try again. How many cycles of emission and absorption are need to get that measly 26 W/m2 through the atmosphere?

Could you calculate what happens to a photon emitted upward or downward from a CO2 molecule from given altitudes, for example 0.2, 0.5, 1, 3, 6, and 10 km? One way to set up the problem is to imagine a box of air at each of these altitudes that is 1 m2 wide and tall enough to weigh 100 kg, or 1% of the atmosphere above every m2 of surface.

For the box at each altitude, we might calculate:
1) % of photons reaching the surface and W/m2 carried by them.
2) % of photons emitted downward absorbed before reaching the earth, average vertical distance traveled (preferably weighted by energy carried), and W/m2 carried this average distance.
3) % of photons emitted upward but absorbed before reaching space, average vertical distance traveled (preferably weighted by energy carried), and W/m2 transported this average distance.
4) % of photons reaching space and W/m2 carried by them.
5) Height of the box that contains 100 kg.

..In order to “measure” DLR the receiver temperature must be cooled below ambient…must be colder then radiation source, in this case the atmosphere. Do I have this right?

No, you don’t.

DSL has already provided a number of links including one to how the pyrgeometer works.

Have you read it? Do you understand it?

Internally the pyrgeometer measures a temperature difference. It can measure a positive difference or a negative difference.

..However, I question the conclusions drawn from the calculated answers…the conclusion that radiation from the cooler atmosphere will increase the temperature of a warmer surface or object..

If you have the same equation as me, then what happens when the atmosphere radiates 1 W/m² more than currently?

Let’s say the surface absorptivity, α = 0.9. This means that the surface now absorbs 0.9 W/m² more than before.

Where does this energy go?

All thermodynamics textbooks that I have read are quite clear. The absorbed radiation increases the internal energy. This is an inevitable consequence of the first law of thermodynamics. And internal energy is related to temperature via the parameter called heat capacity (as shown in an earlier comment).

So temperature goes up and as a result all kinds of consequences follow. The surface emits more radiation, the convection increases.. at this stage we are not interested in the feedbacks or the ultimate new equilibrium temperature.

Only in what happens to the surface when the incident radiation (irradiation) increases by 1 W/m² and absorbs 0.9 W/m².

The result is exactly the same as if some change had resulted in the surface absorbing 0.9 W/m² more solar radiation.

Exactly the same, even though the first case the absorbed energy is from a colder source and in the second case the absorbed energy is from a hotter source.

If you think differently provide your equations of absorption of energy by the surface, or your equations of temperature change due to changes in energy. With these equations we can do some calculations.

Your equations are no different then mine…I think. However, I question the conclusions drawn from the calculated answers…the conclusion that radiation from the cooler atmosphere will increase the temperature of a warmer surface or object. Case in point, a pyrgeometer. In order to “measure” DLR the receiver temperature must be cooled below ambient…must be colder then radiation source, in this case the atmosphere. Do I have this right?