One of the goals of this blog is to dissect contrarian debate tactics, which is why I want to post this dialogue between David Appell and Bob Armstrong about Bob's Revelation that physicists have misunderstood planetary heat processes all along. It's a good example of the game of obfuscation and avoidance that under-educated contrarian types depend on to make their arguments. The original exchange can be found here.
Bob starts with a couple grandiose letters demanding rather basic information from an extremely busy professional David Crisp at NASA's JPL then Bob get's all in a huff because his puffed-up request/demand went unanswered. David Appell stepped in to try and explain, followed by a revealing exchange. I've added some supplement information.
Subject: Radiative balance of Venus
Succinctly , I assert :
Tinternal = Tgray * ( dot[ Psource ; AEobj ] % dot[ AEobj ; Planck Tobj ] ) ^ % 4where Psource is the power spectrum of the source and AEobj is the absorption=Emission spectrum of the object of See my powerpoint if this expression in http://kx.com/'s K is not immediately understood .
I welcome your "review" .
The graphs at Skeptical Science are too fuzzy for me to be sure what I'm reading . Is the paper itself available online ? The curves appear to be radiated power rather than AE . I will be very interested to learn just how such spectra are measured . Are the data tables available ? Then , I can trivially see how they match the implications of my quantitative assertion above .
Looking forward to your reply .
The world must be ruled by theoretically understood , experimentally confirmed quantitative science , not junk no matter how ideologically desired .
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Bob Armstrong Mod Bob Armstrong •
Subject: Radiative balance of Venus
David ,
What's a good time to call ?
I'm working to construct a succinct open APL model of Venus's temperature profile . I feel that quantitatively understanding Venus's extremes is a fast path to understanding planetary temperature statics .
I emailed you thru this JPL form last Thursday : cosy.com/Science/Heartl... .
I'm quite interested in your spectral data and how it is collected .
Back when I was grad student , we had form post cards requesting academic courtesy copy of journal articles . I've got file drawers full of some great and some better forgotten papers from that decade .
I would assume now days that it's just a matter of a few URLs .
What would be a good time for a few minutes chat ?
I understand you think there is some way that Venus's surface temperature can be explained based solely on the energy it absorbs from the Sun . I don't . It appeared you have outgoing power spectra over at least the IR . If I'm right , the integral of that power over the sphere should be greater than that absorbed over the solar wavelengths .
I'm really more interested in programming language structure and notation than working thru this physics, but I find no satisfactory executable explication on the web . So I'm inviting the people I've met at Heartland who've worked at measuring and understanding this domain all their lives to collaborate on this effort .
Let me know the best way to communicate .
Peace thru Freedom ,
Bob A
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DavidAppell writes:
Bob Armstrong, I'm not surprised Crisp didn't reply -- most scientists don't want to get involved with wackos who think Venus's extreme temperatures come only from internal heat. David Crisp has, in his papers, repeatedly written that they come from a strong greenhouse effect.
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Bob Armstrong writes to DavidAppell:
Well , he's provably wrong (provably wrong by someone who admittedly doesn't even understand the basics of the science he's attacking.).
And if he or you or any physicist you want to bring to the conversation can disprove the distinctly undergraduate level computations , then we will all learn a lot . (Crisp already spent the hours and sweat learning this stuff - it's up to you Bob the student to do your own homework, there are plenty of books and online resources available.)
Otherwise he just looks like another arrogant nebbish like Michael Mann .
("nebbish" = a pitifully ineffectual, timid, or submissive man.
"arrogant" = exaggerated sense of one's own importance or abilities
But Michael Mann remains a respected scientist among his peers.)
Too arrogant to even reply with a couple of URLs to answer a mathematical programmer's request for his data .
(Amazing. Armstrong makes puffed up demands of an important, very busy, full time expert in the field with stuff like: "Then, I can trivially see how they match the implications of my quantitative assertion above .
Looking forward to your reply .
The world must be ruled by theoretically understood , experimentally confirmed quantitative science , not junk no matter how ideologically desired .
I understand you think there is some way that Venus's surface temperature can be explained based solely on the energy it absorbs from the Sun . I don't." ~ "I'm really more interested in programming language structure and notation than working thru this physics, but I find no satisfactory executable explication on the web."
(And because that individual doesn't have the time to jump to Bob Armstrong's demanding letter and instead ignores it, rather than recognizing how farfetched his own expectations are - then determining to pursue his quest in a more realistic fashion, Bob get's all pissy and declares the busy expert is arrogant. The lack of introspection is astounding.
Then as if to underscores his rejection of serious science, Bob goes after Michael Mann calling him "nebbish" and "arrogant" at the same time. If one were to point out that Dr. Mann's work continues and is highly respected by the greater scientific community, Bob would simply consider that more proof of some grand conspiracy. Never considering that perhaps Dr. Mann actually is a excellent scientist, who continues to do first rate science, and that the following accolades were well deserved.)
Awards https://en.wikipedia.org/wiki/Michael_E._Mann
Mann's dissertation was awarded the Phillip M. Orville Prize in 1997 as an "outstanding dissertation in the earth sciences" at Yale University. His co-authorship of a scientific paper published by Nature won him an award from the Institute for Scientific Information (ISI) in 2002, and another co-authored paper published in the same year won the NOAA's outstanding scientific publication award. He was named by Scientific American as one of fifty "leading visionaries in science and technology." The Association of American Geographers awarded him the John Russell Mather Paper of the Year award in 2005 for a co-authored paper published in the Journal of Climate. The American Geophysical Union awarded him its Editors' Citation for Excellence in Refereeing in 2006 to recognize his contributions in reviewing manuscripts for its Geophysical Research Letters journal.[63]
The IPCC presented Mann, along with all other "scientists that had contributed substantially to the preparation of IPCC reports", with a personalized certificate "for contributing to the award of the Nobel Peace Prize for 2007 to the IPCC", celebrating the joint award of the 2007 Nobel Peace Prize to the IPCC and to Al Gore.[64][65][66][67]In 2012, he was elected a Fellow of the American Geophysical Union[2] and awarded the Hans Oeschger Medal of the European Geosciences Union for "his significant contributions to understanding decadal-centennial scale climate change over the last two millennia and for pioneering techniques to synthesize patterns and northern hemispheric time series of past climate using proxy data reconstructions."[3][63]
Following election by the American Meteorological Society he became a new Fellow of the society in 2013.[68] In January 2013 he was designated with the status of distinguished professor in Penn State's College of Earth and Mineral Sciences.[69]In September 2013, Mann was named by Bloomberg Markets in its third annual list of the "50 Most Influential" people, included in a group of "thinkers" with reference to his work with other scientists on the hockey stick graph, his responses on the RealClimate blog "to climate change deniers", and his book publications.[70][71] Later that month, he received the National Wildlife Federation's National Conservation Achievement Award for Science.[72]On 28 April 2014 the National Center for Science Education announced that its first annual Friend of the Planet award had been presented to Mann and Richard Alley.[73]
... Bob have you ever considered that what you're asking for doesn't exist in the form you want it because your approach to the problem is plain wrong?)
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DavidAppell writes to Bob Armstrong •
He's not too arrogant -- he just knows it's not worth getting involved with a crackpot who thinks his lame attempt at science trumps decades of scientific knowledge by professionals the world over.
Find your spectrum some other way.
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Bob Armstrong Mod writes to DavidAppell •
If you knew any of the relevant physics , your opinions would be less a waste of time .
(As it turns out David does understand the physics - here's a look at his bio:
I'm a freelance writer living in Salem, Oregon, specializing in the physical sciences, technology, and the environment. My work has appeared in Scientific American, Physics World, Audubon, New Scientist, Wired, Salon, Popular Science, Nature, Discover, The Boston Globe, The San Francisco Chronicle, Physical Review Focus, Discovery Channel Online, Science, and many other publications, and on the syndicated radio program The Weather Notebook . I'm a Regular Contributor to Yale Climate Connections.
I have a B.S. in mathematics and physics from the University of New Mexico, and an M.A. and Ph.D. in physics from the State University of New York at Stony Brook. I've also done graduate work in the creative writing department at Arizona State University.
In other incarnations I've been a systems engineer at AT&T Bell Laboratories and MCI Communications, a business partner/software developer/whatever-it-took at a startup telecommunications software company, Gold Systems, in Boulder, Colorado, and an assistant editor of technology at Laser Focus World magazine.
http://davidappell.com)
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DavidAppell writes to Bob Armstrong •
I know much more physics than you do. If you knew any, you'd provide proof, write this meme-breaking proof in a scientific paper, and submit it to a peer-reviewed journal. If published, you'd be famous in about a week.
But you're afraid to do that.
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Bob Armstrong Mod writes to DavidAppell •
Oh ? Then what's the error my computation of radiative balance ? It produces the commonly parroted 255k value when fed the relevant hypothesis . So the error must be subtle . It's very definitely undergraduate stuff .
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DavidAppell writes to Bob Armstrong •
Your errors are many. Besides using a dumb notation that is confusing and completely unnecessary, your biggest error is assuming Venus is a "gray ball" while ignoring the composition of its atmosphere.
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Bob Armstrong Mod writes to DavidAppell •
Unless you can present an alternative equation , or point out specific errors in mine , this is the last I will respond to you , and I may delete any further non-quantitative posts .
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DavidAppell writes to Bob Armstrong •
Your specific error: Assuming that Venus is an "Irradiated Opaque IsoSpectrum Ball." It's not. If it were, it would have monochromatic outgoing radiation. It doesn't:
In fact, it clearly shows the decline in the greenhouse gases CO2 and H2O.
Since your theory predicts the wrong outgoing spectrum for Venus, it can't be correct. Because you have failed to utilize the composition of its atmosphere.
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DavidAppell writes to Bob Armstrong •
Your PPT page titled "Spectra of Sun and Gray Ball in our Orbit" shows your error explicitly. In effect, you assume Venus has no atmosphere.
You wrote pn that page, "The areas under these two curves are equal." That's false. Integrating the Planck function shows that the total area under its curve is sigma*T^4, and since the T's aren't equal, the areas aren't equal. The brightness temperature of the Sun's photosphere is 5,777 K; for Venus it is 184 K.
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DavidAppell writes to Bob Armstrong •
PS: There is not an "alternative equation" for Venus's surface temperature. The absorption spectrums of CO2, H2O, etc. are far too complicated to yield an analytic solution for the surface temperature. So scientists have done the calculations numerically.
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DavidAppell writes to Bob Armstrong •
Why did you use that weird notation for formulas, anyway? Did you really expect anyone would understand it?
Why not use the simple algebraic notation that *everyone* knows about?
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Bob Armstrong Mod writes to DavidAppell •
I just added a PayPal donate button below . I have several days of accounting that I've put off way too long . So I will try to force myself to stay away from this until Monday .
A couple of items :
o IsoSpectrum and monochromatic are not at all the same thing .
Do you have a source for the tables making that graph ? That would allow plugging
them into the computations . By IsoSpectrum , I mean uniformly colored .
o You have to understand how to calculate the temperature of an opaque ball first .
That is all that is presented but it is experimentally testable .
If you disagree that that is the correct algorithm for a uniformly colored opaque ball , what is your equation ?
o Forget about Venus for the moment . Let's get a simple ball correct first ..
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DavidAppell writes to Bob Armstrong •
1) No, I don't have a table for Venus's outgoing radiation. You could always digitize the graph.
2) Everyone understands how to calculate the temperature of a gray ball that's assumed to be a blackbody. The question is, why do you assume Venus is one? What's the justification for ignoring the fact that Venus has a think atmosphere and is ~96% CO2?
3) The title of your presentation is "How to Calculate the Temperature of a Radiantly Heated Colored Ball ( Like our Earth )." What is the justification for assuming Earth to be a heated colored ball, and ignoring its atmosphere? That assumption clearly does not get the right answer for Earth's surface temperature (288 K); instead it gives 255 K, the Earth's "brightness temperature." Where is the missing 33 K?
Sure, in the beginning everyone assumes Earth is a blackbody, and that calculation gives 255 K for the surface temperature. But since that's obviously not right, they go look for the missing 33 K, and find it in the greenhouse effect. You would say instead that shows the Earth must have enough internal heat to make up the difference. Right?
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Bob Armstrong Mod writes to DavidAppell •
1) No, I don't have a table for Venus's outgoing radiation. You could always digitize the graph.
- It would be much more precise to get the actual data . That is why I have tried to contact Crisp .
2) Everyone understands how to how to calculate the temperature of a gray
ball that's assumed to be a blackbody.
- This makes me wonder if you appreciate the fact that it doesn't matter how light or dark the body is -- one of the most first most basic poorly appreciated points I make .
The question is, why do you assume Venus is one?
- I DON'T . The computation is specifically for non-gray , ie , colored balls .
Once again it makes me question your critical reading ability .
What's the justification for ignoring the fact that Venus has a think atmosphere and is ~96% CO2?
- I DON"T . We need to understand the computation of a ball of any spectrum before we get to specific cases . Once again you make me wonder if you have actually taken any quantitative hard science courses .
(What do mean you don't? When your next sentence dismisses and side steps that point!)
--
At this point your failure to understand what I have done my damnedest to make clear , and for that mater to directly ignore points I spend a lot of time to explain -- like the fact that the 255k number is NOT the temperature of a black ( or gray ) ball in our orbit ; it is created by as step function hypothetical spectrum .
(Here it starts getting interesting. Bob is loosing his cool and proceeds to attack and denigrate the education of his "opponent" rather than focusing on better explaining his point. Perhaps because he doesn't have a case.)
It may be harsh , but I think maybe you just do not have the abilities and discipline to understand actual quantitative physics . Perhaps you had some "eco" class where you never actually had to compute heat transfers or anything else .
In any case , I have no more time for you unless you figure out how analytical quantitative physics is actually done and you read what I say more accurately .
(Then, like Henry the 8th waving off a knave with his pinky, the feigned sense of superiority is a joke from the outside looking in, but it's all they have.)
--
3) The title of your presentation is "How to Calculate the Temperature of a Radiantly
Heated Colored Ball ( Like our Earth )." What is the justification for
assuming Earth to be a heated colored ball, and ignoring its
atmosphere.?That assumption clearly does not get the right answer for
Earth's surface temperature (288 K); instead it gives 255 K, the Earth's
"brightness temperature." Where is the missing 33 K?
Sure, in the beginning everyone assumes Earth is a blackbody, and that calculation
gives 255 K for the surface temperature. But since that's obviously not
right, they go look for the missing 33 K, and find it in the greenhouse
effect. You would say instead say that shows the Earth must have has
enough internal heat to make up the difference. Right?
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DavidAppell writes to Bob Armstrong •
"I DON"T . We need to understand the computation of a ball of any spectrum before we get to specific cases."
This has been understood FOR DECADES, and is in the 1st or 2nd chapter of every planetary science textbook. You haven't done anything new, except for interpreting the results as a realistic description of Venus (and Earth).
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Bob Armstrong Mod writes to DavidAppell •
All of this was understood by some a century ago . YOU don't understand any of this in a manner which allows you to compute anything .
(Again, using insults rather than focusing on explaining his side of the issue.)
I and others with independent minds (Here again, that self-certain superiority complex) demand to understand it ourselves (Then put in the time to learn it - however your phrasing and notions expose that lie.). And in any quantitative class , the students are required to provably, quantitatively understand the computations themselves .
You reveal yourself again and again to be a dupe , incapable or unmotivated to understand physical reality for yourself . (Where's that come from? If your interested in understanding computations why not focus on that rather than hide behind malicious insults and such juvenile crap?)
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DavidAppell writes to Bob Armstrong •
Look, what you're doing is trivial stuff, done long ago. See Pierrehumbert's textbook, section 3.3, "Radiation Balance of Planets."
If you want to get into realistic models of planetary temperatures, read his chapter 4.
4.1 Overview
Our objective in this chapter is to treat the computation of a planet’s energy loss by infrared emission in sufficient detail that the energy loss can be quantitatively linked to the actual concentration of specific greenhouse gases in the atmosphere. Unlike the simple model of the greenhouse effect described in the preceding chapter, the infrared radiation in a real atmosphere does not all come from a single level; rather, a bit of emission is contributed from each level (each having its own temperature), and a bit of this is absorbed at each intervening level of the atmosphere. The radiation comes out in all directions, and the rate of emission and absorption is strongly dependent on frequency. Dealing with all these complexities may seem daunting, but in fact it can all be boiled down to a conceptually simple set of equations which suffice for a vast range of problems in planetary climate.
It was shown in Chapter 3 that there is almost invariably an order of magnitude separation in wavelengths between the shortwave spectrum at which a planet receives stellar radiation and the longwave (generally infrared) spectrum at which energy is radiated to space. This is true throughout the Solar system, for cold bodies like Titan and hot bodies like Venus, as well as for bodies like Earth that are habitable for creatures like ourselves. The separation calls for distinct sets of approximations in dealing with the two kinds of radiation. Infrared is both absorbed and emitted by an atmosphere, at typical planetary temperatures. However, the long infrared wavelengths are not appreciably scattered by molecules or water clouds, so scattering can be neglected in many circumstances. One of the particular challenges of infrared radiative transfer is the intricate dependence of absorption and emission on wavelength. The character of this dependence is linked to the quantum transitions in molecules whose energy corresponds to infrared photons; it requires an infrared-specific description.
...
We’ll begin with a general formulation of the equations of plane-parallel radiative transfer without scattering, in Section 4.2. Though we will be able to derive certain general properties of the solutions of these equations, the equations are not very useful in themselves because of the problem of wavelength dependence.
To gain further insight, a detailed examination of an idealized model with wavelength-independent infrared emissivity will be presented in Section 4.3. A characterization of the wavelength dependence of the absorption of real gases, and methods for dealing with that dependence, will be given in Sections 4.4 and 4.5.
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Bob Armstrong Mod writes to DavidAppell •
It appears to be over your head .
I looked at Pierrehumbernt's book when it was online as a manuscript . I saw a lot of word waving but no equations or calculations backing them up .
(Hmmm, looked at it, paged it, perhaps?
What about spending a semester learning it?)
2e σT4 = e OLR. ir skin ir
(3.23)
= 1(OLR)1 = 1T skin1 41rad
(3.24)
________________
Figure 3.14: The unstable pure radiative equilibrium for an optically thin atmosphere (solid line) and the result of adjustment to the adiabat by convection (dashed line). The adjustment of the temperature profile leaves the surface temperature unchanged in this case, because the atmosphere is optically thin and has essentially no effect on the OLR.
pressure and ptrop be the tropopause pressure. For the dry adiabat, the requirement is then Ts(ptrop/ps)R/cp = Tskin. Since Ts = 21/4Tskin, the result is
4R
ptrop =2−cp (3.25)
ps
Note that the tropopause pressure is affected by R/cp, but is independent of the insolation S.
____________________
The energy balance for the skin layer now reads eirσT4 = eirOLR + aswS
(3.26)
(3.27)
T = T (1 + asw S ) 1 skin 4
where Tskin is the skin temperature in the absence of Solar absorption. The formula shows that Solar absorption always increases the temperature of the skin layer. The temperature increases as the ratio of shortwave absorption to infrared emissivity is made larger. So long as the temperature remains less than the Solar blackbody temperature, the system does not violate the Second Law of Thermodynamics, since the radiative transfer is still acting to close the gap between the cold atmo- spheric temperature and the hot Solar temperature. As the atmospheric temperature approaches that of the Sun, however, it would no longer be appropriate to use the infrared emissivity, since the atmosphere would then be radiating in the shortwave range. Kirchoff’s Law would come into play, requiring a/e = 1. This would prevent the atmospheric temperature from approaching the photospheric temperature.
If the shortwave absorptivity is small, the skin layer can be divided into any number of sublayers, and the argument applies to determine the temperature of each one individually. This is so because the small absorptivity of the upper layers do not take much away from the Solar beam feeding absorption in the lower layers. We can then infer that the temperature of an absorbing stratosphere will increase with height if the absorption increases with height, making asw/eir increase with height.
Armed with our new understanding of the optically thin outer portions of planetary atmospheres, let’s take another look at a few soundings. The skin temperature, defined in Eq 3.24, provides a point of reference. It is shown for selected planets in Table 3.3. Except for the Martian case, these values were computed from the global mean OLR, either observed directly (for Jupiter) or inferred from the absorbed Solar radiation. In the case of present Mars, the fast thermal response of the atmosphere and surface makes the global mean irrelevant. Hence, assuming the atmosphere to be optically thin, we compute the skin temperature based on the upwelling infrared from a typical daytime summer surface temperature corresponding to the Martian soundings of Figure 2.2. The tropical Earth atmosphere sounding shown in Fig. 2.1 shows that the temperature increases sharply with height above the tropopause. This suggests that solar absorption is important in the Earth’s stratosphere. For Earth, the requisite solar absorption is provided by ozone, which strongly absorbs Solar ultraviolet. This is the famous ”ozone layer,” which shields life on the surface from the sterilizing effects of deadly Solar ultraviolet rays.
However, it is striking and puzzling that virtually the entire stratosphere is substantially colder than the skin temperature based on the global mean radiation budget. The minimum temperature in the sounding is 188K, which is fully 26K below the skin temperature. If anything, one might have expected the tropical temperatures to exceed the global mean skin temperatures, because the local tropospheric temperatures are warmer than the global mean. A reasonable conjecture about what is going on is that high, thick tropical clouds reduce the local OLR, thus reducing the skin temperature. However, the measured tropical OLR In Fig. 3.7 shows that at best clouds reduce the tropical OLR to 240W/m2, which yields the same 214K skin temperature computed from the global mean budget. Apart from possible effects of dynamical heat transports, the only way the temperature can fall below the skin temperature is
eir OLR
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DavidAppell writes to Bob Armstrong •
No calculations? Wow, you are deluded. His equation 3.7 is all you're doing.
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Bob Armstrong Mod writes to DavidAppell •
I don't have time even for this .
Could you state his equation here for all of us . Since my equation boils down to
dot[ solar ; objSpectrum ]
= dot[ Planck[ T ] ; objSpectrum ]
his should not be too complicated .
If he agrees with that , then how does he overcome the divergence theorem to claim the interior of a sphere can be hotter than the value calculated by that equation .
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DavidAppell writes to Bob Armstrong •
No, it's too complicated to "state." You'll have to take 15 seconds and look at it:
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Bob Armstrong Mod writes to DavidAppell •
Thanks for the link to Pierrehumbernt's book . So he's out at Berkley now ?
The massive difference between my algorithms and his right off the bat is that mine assume a sphere rather than reducing it to a disk . That is irrelevant here but becomes crucial when painting the sphere with actual colors , white at the poles , blue oceans , clouds , etc . See http://cosy.com/Science/Temper... for the essential partition functions , which I happened to do first . But you'd know that if you read my newsletter . (More of Bob's sense of superiority and entitlement. Bob would know a lot more if he read Pierrehumbernt's book with a eye towards learning it, rather than a conviction that they are fools and you see truth, even though you admittedly don't fully understand what they are doing.)
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DavidAppell writes to Bob Armstrong •
No.
Your "massive differnce" is false. Notice this near the top of page 115:
The planet loses energy by radiating from its entire surface, which has area 4*pi*^a2.
3.3. RADIATION BALANCE OF PLANETS 115
We are now equipped to compute the energy balance of the planet, subject to the preceding simplifying assumptions. Let a be the planet’s radius. Since the cross-section area of the planet is πa2 and the solar radiation arrives in the form of a nearly parallel beam with flux L, the energy per unit time impinging on the planet’s surface is πa2L; the rate of energy absorption is (1 − α)πa2L, where α is the albedo. The planet loses energy by radiating from its entire surface, which has area 4πa2. Hence the rate of energy loss is 4πa2σT4, where T is the temperature of the planet’s surface. In equilibrium the rate of energy loss and gain must be equal. After cancelling a few terms, this yields
σT4 = 1(1−α)L (3.6) 4
Note that this is independent of the radius of the planet. The factor 1 comes from the ratio of 4
the planet’s cross-sectional area to its surface area, and reflects the fact that the planet intercepts only a disk of the incident solar beam, but radiates over its entire spherical surface. This equation can be readily solved for T. If we substitute for L in terms of the photospheric temperature, the result is
1 1/4r
T = √2(1−α) r T (3.7)
Formula 3.7 shows that the blackbody temperature of a planet is much less than that of the photosphere, so long as the orbital distance is large compared to the stellar radius. From the displacement law, it follows that the planet loses energy through emission at a distinctly lower wavenumber than that at which it receives energy from its star. This situation is illustrated in Figure 3.3. For example, the energy received from our Sun has a median wavenumber of about 15000 cm−1, equivalent to a wavelength of about .7 μm. An isothermal planet at Mercury’s orbit would radiate to space with a median emission wavenumber of 1100 cm−1, corresponding to a wavelength of 9 μm. An isothermal planet at the orbit of Mars would radiate with a median wavenumber of 550 cm−1, corresponding to a wavelength of 18 μm.
Exercise 3.3.1 A planet with zero albedo is in orbit around an exotic hot star having a photo- spheric temperature of 100,000K. The ratio of the planet’s orbit to the radius of the star is the same as for Earth (about 215). What is the median emission wavenumber of the star? In what part of the electromagnetic spectrum does this lie? What is the temperature of the planet? In what part of the electromagnetic spectrum does the planet radiate? Do the same if the planet is instead in orbit around a brown dwarf star with a photospheric temperature of 600K.
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Bob Armstrong Mod writes to DavidAppell •
But YOU cannot map a spectrum to each direction over the sphere .
You have yet to SHOW US YOUR EQUATION ! (Is that better?)
You posted some loop equivalent to +/ . That's a start .
But I've had it with your determined troll stupidity .
I'm about to start doing a lot of deleting of your redundant crap .
(What redundant nonsense, David is doing his best to explain. Science is about a mutual desire to learn, you are very one-sided, thinking everyone is supposed to "get" what you are thinking. But in the real world these things have already been figured out - many modern marvels including communication wouldn't be possible without that thorough understanding. But you are screaming it's bs in favor of your TRUTH.)
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DavidAppell writes to Bob Armstrong •
I read one of your newspapers. You insist on obfuscating the science with obscure notation, when all the calculations can be done analytically.
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Bob Armstrong Mod writes to DavidAppell •
How the hell do you take an observed spectrum , say 1000 data points , or a million and do anything "analytically" with it .
But you have not demonstrate that you even know how to calculate the temperature of a uniformly colored ball analytically . If you know how
SHOW US YOUR EQUATION !
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DavidAppell writes to Bob Armstrong •
"YOU don't understand any of this in a manner which allows you to compute anything ."
Even a freshman physics student can calculate a planet's brightness temperature. That's literally as far as your presentation went.
PS: Sentences shouldn't get a space after the last word and before the period.
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DavidAppell writes to Bob Armstrong •
Are you going to address your claim that "the area under the two curves are equal?" Because they're not. Obviously.
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Bob Armstrong Mod writes to DavidAppell •
Look , get an APL and implement the functions yourself . THEN maybe you'll understand what I've expressed in multiple ways in multiple places .
Alternatively , give us YOUR computation for the equilibrium temperature for a uniformly colored ball .
(Notice that Bob avoids engaging any of the substantive points made, such as the request to explain his claims: "the area under the two curves are equal?")
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DavidAppell writes to Bob Armstrong •
I don't need APL. I can do algebra and calculas.
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Bob Armstrong Mod writes to DavidAppell •
Let me see you do a dot product between two 10,000 value data sets then .
I am looking to crunch real data , not just word wave without reality checks .
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DavidAppell writes to Bob Armstrong •
That would be trivial to code. But I don't need to do a dot product numerically, because the integrals are easy.
What are your units for "A" and "E"?
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Bob Armstrong Mod writes to DavidAppell •
Bull Shit !
(juvenile deflection)
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DavidAppell writes to Bob Armstrong •
[Read in vectors x and y]
sum = 0
for i in range (1,10000):
sum += x(i)*y(i)
print(sum)
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Bob Armstrong Mod writes to DavidAppell •
Ok , you've got a dot product ( for exactly 10000 pairs ) .
Now SHOW US YOUR EQUATION !
(Specifically which equation are you asking for?)
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DavidAppell writes to Bob Armstrong •
Re: your PPT page "Can be Simplified to A%E Ratio," you calculate
dot[ 1 ; sourceSpectrum ]
which is just the area under the Planck spectrum. That area is easily calculated analytically. It equals sigma*T^4.
Then you write
Absorptivity wrt source / A : dot[ aeSpectrum ; sourceSpectrum ]
Emissivity wrt Planck / E : dot[ aeSpectrum ; Planck objectTemp ]
which are wrong. It has no basis in physics. It's obviously wrong on dimensional grounds, because A does not have the same units of E -- A has units of (W/m2)^2, and E has units of KW/m2.
So how can you set these two dot products equal to one another?
Worse, neither is how one's calculates absorptivity or emmisivity. In fact, you CAN'T calculate them -- they come by measuring the properties of the gases in the atmosphere and the composition of the surface. They are observational values.
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Bob Armstrong Mod writes to DavidAppell •
Actually , this looks like you are trying .
Yes , I wrote that sum ( integral ) over the sourceSpectrum in the form of a dot just to make the forms the same . And that's not over a Planck spectrum , that's over the observed source spectrum .
What are you talking about with KW ( kilowatts ) being a different dimension than W ( watts ) ? THE fundamental result of Ritchie's experiment as formalized by Kirchhoff and Stewart is that A == E at any particular wavelength .
Yes , of course the spectra are observational data . And non trivial to collect .
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DavidAppell writes to Bob Armstrong •
Still, the units of A are obviously not the same as the units of E, so the big equation below them obviously cannot be right:
ObjTemp = TgrayBody * ( A % E ) ^ % 4
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Bob Armstrong Mod writes to DavidAppell •
What in the hell can Kirchhoff of Stewart mean by saying A == E at each wavelength if they are not "the same units" ?
Now in that equation , A is the absorptivity wrt the solar spectrum , ie , the dot product of the planet's color ( aespectrum ) and the solar spectrum , and E is the dot product of the planet's color and a Planck spectrum of the temperature which equates the two . This is a use of having an executable language in which you can just play with the relationships .
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DavidAppell writes to Bob Armstrong •
Since A has no units of temperature, and E does, A and E cannot possible be equal.
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Bob Armstrong Mod writes to DavidAppell •
I don't know what the hell you are talking about .
(perhaps because you are a dabbler, rather than a serious student?)
the aespectrum is , as you say , a spectrum of proportions of absorptivity ( or emissivity ) between 0 and 1 for each wavelength . It is the color of the object . A is the dot product of that spectrum with the source power spectrum , ie , the sun , and E is the dot product of that spectrum with the Planck power spectrum for the temperature of the object . Thus scalar values of power , just as , as you recognized dot[ 1 ; Planck T ] gives the Stefan-Boltzmann total power for a particular T . Note that Planck T is the spectrum for T .
I would point out again that the expressions produce the 255k result for an aespectrum which is 0.7 over the solar spectrum and 1.0 over the IR .
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DavidAppell writes to Bob Armstrong •
What are the units of your "A"?
What are the units of your "E"?
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Bob Armstrong Mod writes to DavidAppell •
I just went back and looked at the slide :
Absorptivity wrt source / A : dot[ aeSpectrum ; sourceSpectrum ]
Emissivity wrt Planck / E : dot[ aeSpectrum ; Planck objectTemp ]
Obviously both are power , W%M^2 because they are absorption=emission spectra ( dimensionless values between 0 and 1 ) times power spectra .
What is your hangup ?
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DavidAppell writes to Bob Armstrong •
You are misapplying Kirchoff's Law.
It says, "The emissivity of a substance at any given frequency equals the absorptivity measured at the same frequency." Note: OF A SUBSTANCE. One substance. But you have two substances -- the Sun, and Venus. Kirchoff's Law is only true for each individually, not for the combined system of Sun+Venus.
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Bob Armstrong Mod writes to DavidAppell •
Your thought processes are so byzantine I don't know how you ever got thru a quantitative course . You recombine shit is such bazaar ways .
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David Appell's comment was deleted.
And with that, in typical climate science contrarian fashion Bob Armstrong bans David Appell from the conversation. Seems Bob found explaining his position in a civil constructive manner an impossible assignment.
