PeerReview_AlanSiddons.html   [ BA comments ]

See Peer Review : AGWers , Show me the Physics !  for the initial note to which this is a reply .

Subject: Re: AGWers , Show me the Physics !
Date: Fri, 17 Jul 2009 13:27:59 -0400
From: alan
To: Marc <>, 'Bob Armstrong' <>

My immediate post @ 13:37 :
Thanks for your quick response .
I haven't had time to read your whole post and need to run out and get some horse and duck and dog food , but you seem to be saying the same think that I am . For a uniform ball Kirchhoff balances the 0.7 absorptivity with an equal reduction in emissivity leaving the equilibrium temperature of any gray ball the same as the black body temperature . It's not the "divide by 4" that's a problem , altho my implementation is far more flexible and will handle any shading of a ball , It's the ignoring of Kirchhoff .
Anybody with a vacuum chamber , a light bulb and a thermometer should be able to experimentally prove which notion is correct .

Even on the realist side, the discussion always sounds like the Sun is just another "forcing" whose effect is still open to question.

Well, because the actual physics involved opens a can of worms that neither side of the debate is willing to deal with. This is why both sides ignore Gerlich and Tscheuschner, for instance.

arguments as to why the [Stefan-Boltzmann] equation doesn't apply to earth (despite the fact that it clearly does).

Okay, here are a few. Assume that the method of dividing irradiance by four to obtain the temperature of a spherical, reflective Earth is valid (although it isn’t). Thus, with 1366 watts per square meter available but with 0.7 absorption, you divide by 4 and get 239 W/mē, which, via Stefan-Boltzmann, corresponds to about 255 Kelvin on a blackbody. The accepted method also assumes that this 255 K body will then emit 239 W/mē. But Kirchhoff says it won’t, for emissivity is equal to absorptivity. Given an absorptivity of 0.7, then, this semi-smooth body at 255 K will emit 167 W/mē. Since it can’t absorb as well as a blackbody, it can’t emit as well either. In short, the accepted method of obtaining the Earth’s base temperature incorporates absorptive but not emissive reduction. No body radiates as efficiently as a blackbody. This means that a graybody necessarily retains its heat longer than a blackbody, which thereby invalidates the initial 255 K assumption, that of dividing irradiance by four.

The main thing I took away from a first browsing of Gerlich and Tscheuschner was Wood's 1909 experiment . They get too detailed too quickly to answer my questions . Let me understand the lumped system , looking at our 6370 + ~ 100 atmosphere km sphere as a whole , before getting into details . Thus by Stefan-Boltzmann/Kirchhoff only it's spectrum as seen from the outside that determines its temperature as a whole . Let's get that algorithm correct first . And you confirm what I found on Wikipedia that it is ubiquitous to ignore Kirchhoff's fantastic insight 150 years ago this year that absorptivity and emissivity are in fact a single parameter , AE , and treat planets as absorbing as a gray body but radiating as a black body . This creates an error to the down side of AE ^ %  4  ( fourth root of AE  ) . You confirm below that  the temperature of every planet is above this erroneous value . Are they in fact much closer to the black body calculation ? It appears the earth is quite close thus not immune to the correct classical physics which holds for every other object in the universe .  It seems "climate science" is working with an incorrect null hypothesis . I don't understand what the nature of the theory can be after this other than epicycles attempting to get back to that  correct classical hypothesis . 

An interesting attribute of array programming languages is that geometry is not not specified beyond the shape of the parameters . For instance , While an integral expresses a summation over every minute piece of a domain which thereby thoroughly defines the domain , with finite arrays , one is more inclined to think in terms of large scale partitions of the domain . Thus the variable SfeerPart
in my implementation just must satisfy  1 = +/ SfeerPart . The most complicated  partition I've used divides SfeerPart into the solid angles of the nightside , the sun disk , and the dayside minus the sun disk . 

That partition of the celestial sphere is matched with temperatures , Tcs , and Kirchhoff parameters , AE , for each element Given the Stefan-Boltzmann law :

    sb : 5.6704e-008          / Stefan-Boltzmann constant in  ( W % m ^ 2 ) % K ^ 4
    Psb : {[ T ] sb * T ^ 4 }     / Stefan-Boltzmann Law . returns W % m ^ 2  

the mean temperature of a gray ball can be calculated by setting the sum of its radiance in each direction

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

with the temperatures in each matching direction :

 Tdif : { -/ PgrayBall @/: ( Tcs ; x ) }    / Difference between celestial sphere & earth

That is , find  scalar temperature ,T , for the ball , such that  0.0 = Tdif T . With respect to temperature , the sphere is a point  or heat-superconductor  instantaneously averaging its temperature in all directions . It displays no storage . I thank Mike Day for translating these expressions into Iverson and Hui's freely available J and Dyalog's APL so anyone can elaborate the partition as they wish , for instance , specifying grayness as a function of  latitude . Even at this level one can think of a lot of predictions testable either thru experiment or data analysis . I've seen the statement that the sun's temperature varies by only a ten of a percent from max to min of its sun spot cycle . But we we're only dealing with variations of a few tenths of a percent at most  in mean temperature .

I find it interesting that since power is simply buried in the summation , I've never thought in terms of W%M^2 , just temperature . I think it useful to be able to state a fraction , determined completely by geometry , for the mean temperature of a planet heated just by the sun .

Moreover, whereas a blackbody radiates 100% of the thermal energy impinging on it, the maximum rate of heat loss, a real body has internal conductivity, allowing it to store heat. This too skews temperature estimates. The moon, for example, is considered an 89% blackbody. 1366 times 0.89 thus yields 1216 W/mē, which corresponds to about 383 Kelvin on a blackbody, which should be close to its temperature at solar noon. Does the moon’s surface actually reach that temperature, though? No. Because, as Apollo-era measurements indicated, the moon’s regolith stores some of that heat and releases it later.

This chart shows the deviation between predicted and actual lunar surface temperatures throughout the moon’s one-month "day".

The blue zone depicts the moon's thermal handicap, the orange its advantage.

A real body exposed to the sun doesn’t heat up as fast as a blackbody because it’s busy storing heat, conducting it internally into itself rather than fully radiating it. So it never gets as hot. But then it never gets as cold. Reaching its highest temperature in the solar afternoon, it begins to cool thereafter. And as it does so, the stored heat below now creeps toward the surface. In effect, a real body is a thermal battery. A blackbody has no such attributes. And this gives the moon a higher than predicted average temperature.

Your charts are great !

I absolutely agree that non-transparent mass acts like a battery or capacitor . In fact , CO2 definitely a significant role transferring heat back and forth to the atmosphere's more transparent molecules .  Storage changes the rate of heating an cooling , but under what conditions can it change the mean over a cycle ? Brian Valentine raises this point . I've not looked at conductive or convective heat dispersion so closely , but certainly Fourier's law is symmetric . My intuition is that the sign of the derivative is always in the direction of the equilibrium , it can't move that equilibrium .

Yet the theory of the greenhouse effect was concocted for the very purpose of explaining why the earth in particular is warmer than predicted.

As a final point, let me add that EVERY planet is warmer than predicted by a divide-by-four blackbody formula. 

As I have pointed out , It's not the divide by 4 that causes any error . It's the ignoring of Kirchhoff .

1. As one can see by the yellow band on this chart, something happens to a planet's gases at pressures above a tenth of a bar. In every case, air that had been getting cooler as it approached the planet now becomes progressively warmer, irrespective of what it’s made of -- hydrogen, helium, nitrogen, carbon dioxide... whatever.
2. Moreover, in every case it's apparent that air temperature would only keep rising if the planet itself (rake symbol) didn’t get in the way. As its atmospheric pressure mounts, for instance, Jupiter grows far hotter than Venus.
3. Finally, see how the heat lines extend beyond the red circles? Each circle’s position refers to the temperature assigned to that planet by the blackbody equation (see note). In every single case, then, even for Mars, the actual temperature exceeds the estimate, i.e., the scientifically predicted temperature for this planet.
I find this a most interesting piece of data and two steps beyond where I currently am in my understanding . I would first check that  all forms of energy including radiant and macro kinetic are being accounted for . I assume charts like the above refer to dayside profiles , not nightside .  I want to repeat again that at the application of the algorithm above is at the level of the lumped planet and its atmosphere . It's only equating radiant spectra between objects in space , not within the atmosphere . In a sense , I am looking at the radially lumped spectrum , but accounting for the the fact that the ~ 6000k sun occupies less than a millionth of the total angular area of the celestial sphere. It is clear it accounts for the fact that we are ~ 300k . One gedanken test for any theory is what does it predict if , instead of a single hot spot like our sun , the temperature is constant over the entire celestial sphere - like if the sun were gone and the radiant temperature were 3k in all directions . Does the theory predict a temperature for any object  to be greater or less than 3k ? If so , it is clearly broken . 

Does the Stefan-Boltzmann equation apply to the Earth, then? No. There are too many other parameters (some perhaps unknown as yet) that compromise its applicability. But does this significant discrepancy bother so-called climate realists, let alone alarmists? No. In my view, both sides of the radiative forcing debate are chasing their tails, having never verified the initial assumptions of a theory they both endorse. As you say, Bob, show me the physics.

Alan Siddons

Alan , Thank you enormously for such an informative response . 

As I've said above tho , nothing in the universe is immune to Stefan-Boltzmann/Kirchhoff . And that missing 33k is created by using a wrong equation , one which fails the gedanken experiment above .

The next step in my understanding is implementing the Planck function

 Planck : {[ f ; T ] ( 2 * h * f ^ 3 ) % c * ( _exp ( h * f ) % Boltz * T ) - 1 }

to generate black body spectra so that  AE can be expanded across spectra . While simply varying a spatially uniform flat spectrum gray parameter does not change the equilibrium temperature , a difference in the correlation of spectra between an object and its radiant heat sources and sinks will . Then the actual effect of various gas spectra can be computed .

At this point I would modify my challenge to Show me the experimental physics .

Source: NASA’s Planetary Fact Sheets

Earth 254.3 Kelvin  
Mars 210.1
Jupiter 110.0
Saturn 81.1
Titan 84.6 (my estimate based on its 0.22 albedo)
Uranus 58.2
Neptune 46.6