Subject: Re: AGWers , Show me the Physics !
Date: Thu, 06 Aug 2009 11:25:24 -0600
From: Bob Armstrong <bob@cosy.com>

Last night I finally had time to reply to the first 19 or so replies to my request for comments on my implementation of the basic Stefan-Boltzmann/Kirchhoff equation for radiantly heated spheres . Rather than burden everybody with all the posts and my comments , I have once again posted them to the topic on my forum . I thank everyone for the unexpected amount of feedback . I feel overall it illustrates the pathetic state of the understanding of the basic physics involved , even within the realist camp . This ignorance cries out for the simple experiments which would put much of these arguments to rest .

Perhaps the most substantive comments came from Howard Hayden . I think we are much closer together in our understanding than he thinks . He seems to think I'm saying more than I am . I'll repeat that I have only implemented the scalar , gray body absorptivity/emissivity parameter model . I am working on the couple of lines required to implement the full spectrum version . But I already spend far more time on this than is conducive to my economic well being .

Apparently a lot of people have made lots of incorrect implementations of Stefan-Boltzmann which produce ridiculous predictions . My correct implementation gives a temperature of about 278k for a uniform gray ball at this distance from a 5772k sun . I believe that's within a few percent of commonly observed mean values .

Essentially the only question I am addressing with this cut is which of the following formulas for a gray ball is correct :

The ubiquitous , eg , on the Wikipedia black body and Stefan-Boltzmann pages , and apparently a lot of texts :

        (  EarthAbsorptivity  *  SunSolidAngle * TempSun ^ 4   )  =  (  TempEarth ^ 4 )  
or
        (  EarthAbsorptivity  *  SunSolidAngle * TempSun ^ 4   )  =  (  EarthEmissivity  * TempEarth ^ 4 )     

where for a gray ball
EarthAbsorptivity and EarthEmissivity are equal . SunSolidAngle is expressed as a fraction of the total sky .

The first equation produces a result
EarthAbsorptivity^0.25
lower than the second , about 253k for the earth . From there on , all I see is word waving to fill in that gap and claim there is no upside limit to the planet's temperature .

Can we eliminate one of these two equations ?  Can there be any claim to this being a scientific pursuit if we can't ?

--
-- Bob Armstrong -- CoSy.com -- 719-337-2733 --
I reserve the right to post all communications I receive or generate to CoSy website for further reflection .




[ BA comments ]

Well , I'm glad to be getting light shed on this timely bit of physics . Here are my comments on the stream of emails I've received .


        Alan Siddons 8/3/2009 14:19 :

By the way: "a null hypothesis that gray bodies absorb as gray bodies but emit as black bodies" says it very succinctly. Wish I'd thought of those words myself!  

        Howard Hayden 8/3/2009 15:51 :
Time to be careful !!
 
The absorptivity and emission coefficients are identical at each wavelength; however this does not mean that the coefficients at (say) 550 nm and at 15,000 nm are the same.  It is perfectly possible for a (gray) body to absorb sunlight poorly (centered at around 550 nm) but to emit IR well (centered at 10 nm, corresponding to 288 K).  Indeed, it's the case that plain old white snow has an emission coefficient very near 1.0.  Funny that snow is a good "blackbody", but that's the way it is.  Almost everything except shiny metals emits well in the IR region.
 
The only definition of "gray" I can make sense of is a totally flat spectrum for all frequencies . That is certainly what is implied by using a scalar value for AE . As I have said , I'm working on implementing the Planck function , BB[ f ; T ] ,  so that each item in AE will be a spectrum to be multiplied by BB[ ; T ] to calculate the effect of color . ( NB : the K language , and perhaps J , has the useful feature that eliding a parameter essentially "projects" the function onto that parameter . )

        Alan Siddons 8/3/2009 17:33 :
As Brian has lectured me, though, emissivity for real objects also has a directional component. Change your angle of view and you’re likely to get a different emissivity reading. Another complication is albedo (reflectivity), which varies by direction as well. I suppose that’s one reason why "bond albedo" was devised, given that Saturn’s frosty moon Enceladus has a visual albedo above unity, which would imply that it can’t be radiatively heated at ALL! In some sense it seems to me that if you want to determine a real body’s temperature response to sunlight, there’s nothing like a simple thermometer...

Again , I'd refer to the actual formula for what is actually implied .  The function is 

 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

where SAsun is the solid angle subtended by the sun .

AE then assigns a gray value to each point on that integrating sphere . An integral over the hemisphere normal to each direction would be more precise , but I would bet that it would be hard to construct an angular distribution of albedo which would significantly the results .

All this aside, however, the point that needs stressing is that there’s a necessary dislocation between irradiance and temperature on a spherical surface.

    Since the time when Fourier formulated the heat conduction equation, a non-linear boundary condition describing radiative transfer of a globe with a Sun-side and a dark side has never belonged to the family of solvable heat conduction problems, even in the case of a non-rotating globe. — Gerlich and Tscheuschner

Dr Hayden mentioned this to me when we met at the 1st Heartland conference . While I appreciate the elegance of an analytical answer , for good or ill from a modern programmer's perspective it's a big "so what" . What matters is if the error function converges uniquely . And integrals over T^4 very definitely do .

That is, our spherical earth can NOT be regarded as a flat plate receiving an "average irradiance" in the first place. A flat plate model gives you a 255 K estimate. But a half-and-half model (one round hemisphere sunlit, the other one dark) gives you 153 K at best! Incorporating a meridional solar zenith irradiance will yield yet another figure. In other words, the very assumption that you can derive a temperature from the average irradiance on a sphere is wrong. Among prominent AGW skeptics, only Vincent Gray and (to a lesser extent) Richard Courtney understand that the Kiehl-Trenberth radiation model is inherently invalid, therefore. On a mathematical basis alone, average temperature does not conform to average irradiance due to the 4th power relationship. Still "more" impossible is trying to make those two parameters get along on a non-blackbody sphere. Bottom line: the widely-cited average earth temperature of 255 Kelvin is bogus.

I'm not clear on just how you are doing your calculations , but my PgrayBall  matches the value calculated by the  equation on the Wikipedia Black Body page before it was screwed up . And both give temperature ratios of  about 0.0478 and 0.0486 for an object at aphelion and perihelion in our orbit relative to that of the sun . These give mean temperatures of about 276.4 and 281.0 for a temperature for the sun of 5778 . That leaves at most a few percent error to be explained . The perihelion - aphelion variation is an obvious effect to be looked for in the historical data to test the computation .

       
Howard Hayden 8/3/2009 18:31 :
Alan:
 
Right.  The moon, for example, has highly directional reflectivity; the light from the moon is down by nearly 50% one day after or one day before full moon.
 
G & T are perfectly correct that you should not treat a sphere as a flat plate.  They also comment correctly that the average temperature is a meaningless quantity.

So , is it meaningless to assign a mean temperature to every astronomical body ? I think this is a total cop out .
 
G & T are perfectly incorrect in saying that a body cannot radiate IR to a hotter body.  (At least, that's the way I read what they said.  Correct me if I'm wrong.)  We receive lots of microwave radiation from outer space that is at 2.8 degrees above absolute zero.  It has a minuscule, but non-zero, warming effect on the earth.

Correct . I have the cosmic background radiation at 3k in my calculations , but with that  T^4 relationship , the 0.9999946 of the sky at that temperature makes little difference .
 
But all of that is beside the point that I was making.   I was commenting on this sentence:  "I'm left astounded that such a howler as having as a null hypothesis that gray bodies absorb as gray bodies but emit as black bodies could get past any competent peer review ."  It is incorrect to equate the absorptivity at one wavelength to the emissivity at another wavelength.
 
We must be very careful, because the warmers are wont to jump onto every misplaced comma to demonstrate that "deniers" are all a bunch of incompetents.

Again , that's why I am working on "unfolding"  AE across spectra . But I won't be able to get to it til late this month . And then , I will be looking for a convenient source of all relevant spectra .

        Alan Siddons 8/3/2009 21:08 :

It is incorrect to equate the absorptivity at one wavelength to the emissivity at another wavelength.

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/m 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/m. 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/m?

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.

Actually , what you can't do is add the temperature of layers of an object  heated by a source to subseqent layers because the energy in those layers has come from that source . All it can do is add back the energy it has absorbed . However , this is another example of the common failure I see to consider that we are a sphere heated by only 5 millions of the sky . In all other directions , we are the hotter body . Here's what the equation gives for a uniform blackbody with background temperatures of  3k , 0k , and 100k
 : SfeerPart : ( ( ( :: ; pi2 - ) .\: SAsun ) , pi2 ) % pi4 />/ 5.411374e-006 0.4999946 0.5 
 AE : 1 1 1 ; Tcs : 5778   3    3.0   ; ?[ Tdif ; 0.0 ; 300. ] />/ 278.6791
 AE : 1 1 1 ; Tcs : 5778   0    0.0   ; ?[ Tdif ; 0.0 ; 300. ] />/ 278.6791
 AE : 1 1 1 ; Tcs : 5778 100  100.0   ; ?[ Tdif ; 0.0 ; 300. ] />/ 279.8271


As I said above , the 3k background becomes totally insignificant when raised to the 4th . However , if the background temperature were 100k in all directions , it would raise the temperature of the earth about 1.15 degrees .

Energy is always exchanged in both directions but the net is always from hotter to colder . That the cooler radiates toward the warmer is reflected in the fact that the rate of energy flow is
( Th - Tc ) ^4 . The closer in temperature the source and sink , the slower the rate of convergence of temperature .


        Howard Hayden 8/3/2009 22:12 :
Alan:
 
"Absolutely. But look at it this way while playing ..."

    That's changing the subject.  I'm simply saying DO NOT support strong cases with incorrect arguments.  Remove the incorrect "howler" sentence and get on with the correct argument.

 
"But does this radiation raise the hotter body’s temperature? No, because it can’t...."

    Wrong.  Do you seriously believe that the background microwave radiation (as trivial as it is) has precisely the same effect on the earth's temperature as if the background were at absolute zero?  Do you believe that if the temperature of outer space were suddenly warmed to 200 K that it would not warm the earth?


Here's the calculation :
  AE : 1 1 1 ; Tcs : 5778 200  200.0   ; ?[ Tdif ; 0.0 ; 300. ] />/ 295.5635

The point, from a 2nd-law point of view, has to do with net radiation flow.  Remember, the thing (aside from emissivity considerations) that determines the amount of radiation is the temperature.  A cold body does not fail to radiate simply because a warm body is near, and the warm body does not fail to absorb the radiation that impinges upon it.  There is no net heat flow from colder to hotter.  This is not a violation of the 2nd law.

 
        Alan Siddons 8/3/2009 23:58 :
Well, I’m sorry that you disagree, sir. As I said, it’s a simple matter: prove me wrong. Take the previous back-radiation question. A blackbody absorbing 1000 W/m will, at 364 K, radiate 1000 in turn. So if 50% is being re-radiated or reflected back to it, does one calculate its "net outgoing emission" as 500 W/m and does its temperature also rise to 403 K? If, as you insist, 500 W/m can indeed heat a 1000 W/m object, just tell me by how much.

All I’m asking for is a number, you see. Also please point me to an empirical study which demonstrates that less intense radiation transfers heat to a more energetic body. Otherwise I am forced to conclude that the second law prevails and only a greater amount of energy can heat it.

Or answer your own question. Say a 288 K body IS surrounded by a 200 K universe. How much warmer will that body get?

I did the computation you want above . If it's totally surrounded by 200K , it better come to 200k also , no matter how shade or colored .


        Brian Valentine 8/4/2009 10:38 :
Ooops, sorry for butting in, I had, to,
 

 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 , 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

Now this I'd like to see the physics of . That does violate basic law .

 
        Alan Siddons 8/4/2009 12:08 :

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,

Bob


        Peden HQ 8/4/2009 12:18 :

 
        Bob Ashworth 8/4/2009 12:30 :

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, 

Bob
 

        Alan Siddons 8/4/2009 12:32 :
Agreed, Bob. However, if the emitter has a constant energy supply (as does the Kiehl-Trenberth earth model with 235 watts per square meter constantly radiating on it), it will not cool off at all. Neither will a lesser magnitude of energy via reflection or back-radiation RAISE its temperature because less cannot transfer heat to more. Thermal energy keeps moving outward from that body, never backward.  I think you agree.

 
        Alan Siddons 8/4/2009 12:32 :
This is little more than a second version of  Jim's oven (always steal from the best), but mine has a few touches that make it a better buy.

This one’s insulated and firmly sealed like a thermos, for instance, and has a quick-start radiation source, a battery-operated flashlight bulb that makes it convenient for family camping. A double-A Duracell will do. Principle of operation: Radiant energy goes in but cannot get out. Greenhouse physics dictates that without an avenue of exit, incoming photons accumulate and generate more and more heat. Ergo, cut off exiting radiation entirely and you've got the cheapest oven imaginable. Never overcook, though. The thing is liable to melt.
I think all the stuff you are talking about is the epicycles the AGWers use to compensate for their AE^%4 deficit . I've pointed out that by their incorrect equation , they could make a cyrogenic cooler just by coating the inner chamber of a vacuum bottle with MgO .


        Howard Hayden 8/4/2009 17:44 :
All:
 
Attached is a drawing for you.  It's a test.
 
G & T make a good point that the earth's surface in full sunlight cannot be understood by moronic application of the Stefan-Boltzmann law.  Such an application ignores thermal conductivity, heat capacity, evaporation, convection, directionality, and all sorts of stuff that wind up with the surface much cooler than the moronic application would suggest.

It is obvious that Stefan-Boltzmann combined with Kirchhoff explains all but a few percent of the temperature of the planet . How can that possibly be denied ? How can it possibly be denied that ignoring Kirchhoff and using equations which assign an absorptivity of .7 to the earth and an emissivity of 1.0 is simply wrong . How far off is 280k ?
All of the atmospheric processes you mention work furiously to make the lumped system stay in balance with its radiant bath . I've made absolutely no denial of various forms of storage in the system . But , damn it , you have to start with the correct equation and when you leave Kirchhoff out , it is simply wrong .
My quarrel was and is that the missive that came my way had two errors of physics.  It said that IR cannot be radiated from a colder body to a hotter body, and that a grey body (for sunlight) must be also a gray body (not a blackbody) in the IR.

Again I will repeat that I am working on the full spectrum equation . I am not denying the effect of spectrum but I won't be happy until I'm calculating it , not just word waving .

Cheers,
Cork
  

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I'd like to see the explanation . Is the claim that you can make a perpetual heat engine ?