PgrayBall : { +/ AE * SfeerPart * Psb x }  / total radiated power from shaded gray ball

where my perhaps unfortunately named SfeerPart is the only parameter which induces any specific geometry . And it induces spherical geometry only in that it is composed of a partition of the celestial sphere . 

  : SfeerPart : ( ( ( :: ; pi2 - ) .\: SAsun ) , pi2 ) % pi4    />/  5.4113742e-006 0.49999459 0.5

Absolutely. But look at it this way while playing with simplistic climatology assumptions. Say a body is 0.7 absorptive, i.e., it reflects 30% of the incident light. Whereas 1000 W/m² would raise a blackbody to 364 K, then, this body will absorb 700 and reach 333. Fine so far. Maybe. Ah, but absorptivity equals emissivity, remember, and climate models make no wavelength distinctions. It's watts per square meter from beginning to end, which is the ONLY factor you have to worry about when a blackbody is involved. So yes, a blackbody at 333 K will emit 700 W/m². But a 0.7 emissive body AT THE SAME TEMPERATURE will radiate 490!

What I’m getting at is that the famous "255 K" earth estimate simply regards the target as a uniform, semi-reflective plate and divides net radiation by four to compensate for a sphere’s greater surface area. Yet many different substances with different emissivities are creating that 30% reflection. Look at water alone.

http://www.monarchserver.com/TableofEmissivity.pdf

http://www.ib.cnea.gov.ar/~experim2/Cosas/omega/emisivity.htm

Keeping in mind that the earth is 70% covered with water and that laboratory-derived figures tend to be higher than in the real world, can you say with fair assurance that a 333 K graybody will radiate at the same intensity as a blackbody? Will a 288 K "average earth" actually radiate 390 W/m²?

Again , I'd like to stress that all I am demonstrating with the monocromatic implementation is that albedo , uncorrelated with the sources or sinks of heat  doesn't matter . And it is a total howler to use equations which simply ignore Kirchhoff's insight . 

Now for the more difficult stuff.

G&T are perfectly incorrect in saying that a body cannot radiate IR to a hotter body.

Aye, there’s the rub. It is more precise to say that a cooler body radiates IR toward a hotter body. But does this radiation raise the hotter body’s temperature? No, because it can’t. I realize that such an assertion is controversial, but the second law of thermodynamics wouldn’t make sense otherwise. Here's how professor M. Quinn Brewster, for instance, expresses it in his book Thermal Radiative Transfer and Properties:

Like conduction, thermal energy is in harmony with the second law of thermodynamics such that, in the absence of work, thermal energy is radiated spontaneously from higher temperature to lower temperature matter.

FROM and TO.

To frame that a bit more tightly, if radiation TO a body is less than the radiation FROM that body, this energy does not raise that body’s temperature. If radiation to a body is greater than radiation from that body, this energy does raise that body’s temperature. Conductive and radiative heat transfer obey the same rule, that of thermal energy flowing to a zone of lesser energy. That is my own formulation of the second law as it applies to radiative heat transfer.

Were a cool body to lend heat to a warm body, as a warm body lends heat to a cool body, then logically both would be heating each other. So when would this reciprocal heating process STOP? Fact is, it wouldn’t. A hot radiating ball suspended inside a cooler (and well-insulated) shell, for example, would exchange heat between them infinitely, both of their temperatures mounting minute by minute.

In real-life, however, the hot ball will transfer heat to the cool shell until both of their temperatures are the same. Just as in a conductive transfer context, the cooler shell contributes nothing of its own; it is merely a receiver. Because transfer literally means transfer: the heat flows only one way, from greater to lesser. So the shell gains at the ball’s expense. A transfer is not an exchange. It is from and to.

But here’s the upshot: If I am wrong, simply prove it! I’ve read about "reciprocal heat transfer via radiation" till my eyes bug out of their sockets. Yet no formula in any physics textbook describes this alleged phenomenon, as Gerlich and Tscheuschner have also mentioned.

Question: With a radiative input of 1000 watts per square meter and an output of 50% (meaning 50% back-radiation from a re-emitter), what temperature will a blackbody reach?

There SHOULD be an answer to such an elementary question. But there isn’t. The formula is missing from textbooks because the phenomenon itself does not exist. Neither conductively nor radiatively can a cool body transfer heat to a warmer body. So says me. But if I’m wrong, show me.

 Looking at your statement here Cork 

“Do you contend that your heated object will remain at the same temperature as it was?”

Cork, you’re not suggesting that there are situations for which heat is transmitted from cold -> hot, are you? 

        Alan Siddons 8/4/2009 11:06 : 
Howard,

I haven’t changed the subject anywhere; I’ve merely addressed the points you raised.

1. The emissivity of an entire planet cannot be assumed to be 1 after factoring in its overall albedo. The method of estimating the earth’s temperature and radiation budget is therefore faulty.

2. Back-radiation or lesser radiation as a mechanism for raising an emitter’s temperature is part of greenhouse-theory physics but not of real physics. Or, if it is an instance of real physics, I ask to be shown.

So please don’t accuse me of changing the subject, which implies that I’m being evasive. To the contrary, I am charging headlong into it!

Your lab scenario is a good one, a variation on my hot ball suspended inside an insulated shell. And yes, I do contend that the hotter body (activated by electrical resistance in this case) will transfer heat to its surrounding shell while the shell won’t raise the hotter body’s temperature. If radiation from the hot ball to the shell is greater than what the shell is emitting, the shell will get warmer. If radiation from each to the other is equal, neither body’s temperature will rise. If radiation from the shell to the hot ball is greater than what the ball is emitting, the ball will get warmer.

You see, I believe that the concept of "radiative cooling" is a tricky one. Yes, a heated body radiates. But by no means does this radiation compensate for the energy input. Rather, the amount of radiation is simply a result of the temperature the input has induced. That is to say, an input of energy vibrates that body’s molecules, and this vibration produces an electromagnetic disturbance which propagates through space. Confine that radiation like a thermos does, though, and the heated body simply sustains the temperature it has reached.

By direct contrast, greenhouse physics contends that with a constant input of energy (like sunlight), the suppression of outgoing radiation must raise the body’s temperature because that body can only "cool off" by radiating. Ergo, if outgoing energy is suppressed entirely, the input keeps pumping energy in and the temperature rises inexorably. In this way a 1 watt input is capable of producing the heat of a billion watts. THIS is the principle of "radiative forcing" in a nutshell. And it’s nutty.

This false notion came about because the founders of greenhouse theory misunderstood the physics involved in their experiments. As you and I know, a high interior temperature inside a glass box exposed to sunlight only occurs because the enclosure suppresses convective cooling. But the founders believed that glass was suppressing outgoing radiation (or back-radiating it) and making incoming energy accumulate. Do you get the idea? The founders thought that they were hindering radiative cooling, which resulted in heating. The founders thought that they had discovered a law of physics written in stone: Radiative disequilibrium causes a temperature increase.

With a hop, skip, and a jump, then, scientists started talking about the physics of suppressed radiative cooling. If outgoing is 1 but back-radiation is ½, well gosh, the temperature must rise, they insisted. Okay fine, I say. Just show me the equation and tell me how MUCH the temperature rises. Show me the physics.

Actually I just did above . Note Corky described how clouds keep the ground warmer at nigh . By the same token , they keep it cooler in the day time . Perhaps it's because my background is much stronger in statistics , but I am amazed that I have never seen the notion of variance mentioned in any climate discussion . While I won't know how much the spectrum of CO2 effects the mean temperature of the planet until I flesh out the spectum computation , it most certainly effects the rate of both heating anc cooling by transferring energy back and forth with the rest of the atmosphere . Over a cycle , its not unreasonable to call this a reduction in variance .


        Howard Hayden 8/4/2009 11:50 :

Egad.
 
I am NOT -- repeat NOT -- suggesting any violation of the 2nd law of thermodynamics.
 
I am saying with great certainty that IR WILL travel from the colder body to the warmer, and will be absorbed by the warmer body.  That body will be warmer than if the cold body were replaced by empty space that did not radiate heat back to the warmer body.
 
There are some interesting "paradoxes" in the non-equilibrium thermodynamics about this very topic.  One of them has two bodies at equal temperature, with Radiant heat going back and forth.  One body becomes hotter than the other.  (It's all done with mirrors, but the physics is unassailable.)
 
Cheers,
Cork

Whether radiatively or conductively, a less energetic body (cooler) amplifying a more energetic body (hotter) would violate the second law, though.

For my part, I do think it's possible that a zone can be made hotter with a mirror.

The temperature of a self-luminous body exposed to radiation equal to what it is emitting will NOT increase. But focusing a reflected beam onto this self-luminous body will raise the temperature of that zone. The atmosphere cannot be regarded as a parabolic mirror, however.

 
        Bob Ashworth 8/4/2009 12:09 :
Gentlemen: 

The heated object will not remain at the same temperature.  It will cool slower with insulation than it would if none were there.  Every body in the universe radiates and absorbs energy constantly, but the overall net heat transfer effect is always from the hotter to colder body, never vice versa.  That is it in a nutshell.  Pretty simple stuff.

Kindest Regards,

That sums it up - great visual!  Use of a chicken is very appropriate for the AGWers are like Chicken Little (same size brains I would expect) There cannot be any real scientists on the AGWers side, by definition of what they say, charlatans maybe, scientists - no way!   It is a shame we even have to spend a second of our time fighting this play station crap! 

Kindest Regards,