Reading one good text book on climate science can save 100’s of hours of reading rubbish on the internet. And there is a lot (of rubbish). Well-meaning people without the baggage of any knowledge of the subject writing rubbish, then repeated by other well-meaning people.
Text books cost money. But depending on which country you live in and whether you have an income, the “payback” means that not buying it is like working for $1/hr. That assumes reading rubbish isn’t a hobby for you..
And depending on where you live you can often join a university library as an “outsider” for anything ranging from $100/year up – and borrow as many books as you like.
Learning can be like a drug. In which case, other justifications aren’t necessary, you have to feed the habit regardless. So pawn family jewelery, sell your furniture, etc. Well, as an addict you already know the drill..
Just some ideas.
Global Physical Climatology – by Dennis Hartmann
Academic Press (1994)
Amazon for $88 (reduced from $118, the price at the normally amazing bookdepository.co.uk).
Why am I recommending such an old book? This covers the basics very thoroughly. When someone covers a lot of subjects there is inevitably a compromise. To cover each of the subjects “properly” would be 4,000 pages or 40,000 pages – not 400 pages. What I like about Hartmann:
a) very readable
b) very thorough
c) enough detail to feel like you understand the basics without drowning in maths or detail.
Maths is the language of science, and inevitably there is some maths. But without any maths you can still learn a lot.
Now, a few samples..
From Chapter 4:
From chapter 11:
As you can see, there is some maths, but if you are maths averse you can mostly “punch through” and still get 80% instead of the full 100%.
Elementary Climate Physics by Prof. F.W. Taylor
Oxford University Press (2005)
bookdepository.co.uk for $44 with FREE shipping lots of places in the world, unbelievable but true.
Amazon has it for $60 plus shipping.
This is an excellent book with more radiative physics than Hartmann, but also more maths generally. For example, in the derivation of the lapse rate there is some assumed knowledge. That’s par for the course with textbooks. They are written with an audience in mind. The audience in mind here is people who already have a decent knowledge of physics, but not of climate.
However, even with a tenuous grasp of physics you will get a lot out of this book. Here’s the downside though – quite some maths:
Well, he is teaching physics.
A First Course in Atmospheric Radiation – Grant Petty
Sundog Publishing 2006
Amazon from $48
Thanks to DeWitt Payne for recommending this book, which is excellent. This is the best place to start understanding radiation in the atmosphere. Goody & Yung 1989 is comprehensive and detailed – but not the right starting point.
Radiative physics is no walk in the park. There is no way to make it astoundingly simple. But Petty does a great job of making it five times easier than it should be:
Now onto “not climate science”:
An Introduction to Thermal Physics – Daniel Schroeder
Published by different companies in different countries.
Amazon from $45 plus shipping and Bookdepository for $56 free shipping.
A book that is nothing to do with climate science, but quite brilliant in explaining very hard stuff – heat and statistical thermodynamics – so it sounds really easy. Not many people can explain hard subjects so they sound easy. Most textbooks writers make slightly difficult stuff sound incomprehensible until after you understand it – at which point you don’t need the textbook.
It wasn’t until I read this book that I realized that Statistical Thermodynamics was actually interesting and useful.
The Inerrancy of Textbooks?
Are textbooks without error and without flaw?
No
So what’s the point then?
The people who write textbooks usually have 20+ years of study in that field behind them. And until such time as E&E start a line of textbooks, the publishers of textbooks, with their own reputation to protect, only ask people who have a solid background in that field to write a textbook.
So even if you are intent on demonstrating that climate science has no idea about basic physics – how are you going to do this?
You could follow the path of many other brave bloggers and commenters who write about the “paltry understanding” of climate science without actually knowing anything about climate science.
But if you choose to do it the old-fashioned way then you should at least find out what climate science says.
The Science of Doom 
For those who want a textbook with a formal treatment of radiative physics:
Atmospheric Radiation: Theoretical Basis, by Goody & Yung (1989 2nd ed) – this follows the 1st edition from 1964, accepted as the standard text in the field.
Radiation & Climate, by Vardavas & Taylor (2007) – equally good overall.
Here’s an extract from Goody & Yung to give you an idea what is in store:
The authors say at the end of that chapter:
as an aside, if you take a look around at the many internet “debunkings” of the “greenhouse” effect, or at the many claims such as:
– “CO2 is saturated”
– “CO2 can’t have any effect because it is 360ppm”
– etc
You will notice that none of them are apparently aware of this important theory called Theory of Radiative Transfer.
Not much of a debunking, when you don’t know what it is you are debunking.
And it’s not “all new”, it has been around for over 60 years.
Thanks. Hartman’s book is available at a good discount as a Kindle version from Amazon. I’ve downloaded a sample and will try it out.
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Petty is cheaper if you order direct from Sundog Publishing, $36 instead of $48.
http://www.sundogpublishing.com/AtmosRad/
Good idea for a post. Definitely true that sifting through a good textbook or two on climate can separate you from 95% of the blogosphere who just like to argue memorized talking points.
Petty’s book is extremely good. He doesn’t make a lot of connections to climate (his field is largely remote sensing, and UW-Madison where he is a professor is one of the best is radar and satellite meteorology) but it’s a great introduction to applied radiation physics for anyone at an upper level undergraduate or even graduate level. There’s also plenty of opportunities to interpret spectra (from space or from the surface, and learning the distinction). As DeWitt noted, sundog publishing gives you a discount. If you become an Atmospheric Science student at Madison though, you can just stop by his office and he’ll give you a copy for $10. For a bit more applications to climate, Vardavas ‘Radiation and Climate’ is worth it too. I wouldn’t touch Goody and Yung unless you want to specialize directly in atmsopheric radiation, and even so it’s not the greatest for direct application to thinking about global warming stuff. It’s the most complete and authoritative radiation book though in the field.
Hartmann is great too, and I also recommend Houghton’s ‘Physics of Atmospheres.’ Around the same difficulty level as Hartmann is Marshall and Plumb’s “Atmosphere, Ocean, and Climate dynamics”. It’s not great IMO for radiation and pure climate stuff, but you’ll learn a lot of fluid dynamics, both in the ocean and the atmosphere, understanding geostrophic & hydrostatic flows, important circulations, Ekman transport, momentum balance equations, etc.
The best book out there if you don’t want to touch anything above pre-calculus math is David Archer’s ‘Global Warming: Understanding the Forecast.” A fair amount of simple algebra, exponential-log functions, graphs, and some chemistry, but it’s very readable and is a complete guide to the global warming problem for the casual reader.
Finally, I think the best climate book out there is Ray Pierrehumberts new “Principles of Planetary Climate.” It’s the most complete in terms of covering radiation in good detail but applying it quickly to planetary atmospheres. You’ll also be well-equipped to not only handle all the blogospheric objections about radiation related topics, but is a good start even for researching in the field.
I’m an interested layman, and have both Houghton and Archer´s books. I’m happy with both. They answered a lot of my questions.
Archer’s book is indeed very readable, especially if we watch his lectures at the Chicago University website.
Houghton is not the kind of book I read for myself. I keep it as a reference source, and a very good one at that (IMHO).
Chris Colose:
I’ve been having a read of Marshall and Plumb’s “Atmosphere, Ocean, and Climate dynamics” – it’s very good.
Another thing….for good radiation books, sometimes it’s actually good to turn to astrophysics texts on the subject rather than atmospheric texts. I find that they often tend to do a better job. SOD mentioned Chandrasekhar above who came from that background.
Has anyone looked at Ray Pierrehumbert’s new book yet?
http://geosci.uchicago.edu/~rtp1/PrinciplesPlanetaryClimate/index.html
Paul Middents
My post crossed with Chris’ input. Thanks SoD both for this post and your efforts to explain the basics to the unwashed (a group in which I place myself).
Chris, your efforts along these lines are also appreciated.
Chris Colose on February 26, 2011 at 5:03 pm:
I’ve read Houghton, Physics of Atmospheres (1982).
For me a good textbook is both accurate and provides insight – a conceptual understanding of the key mechanisms and processes.
I’m sure Houghton is accurate.
From memory the first part of the book explained reasonably well, but the 2nd half of the book didn’t seem to provide any insight – until the point where the student understands the subject already..
My subjective assessment, and others are welcome to add their comment.
thanks for these reviews…
I would also reccommend a classic “Physics of Climate”, by Peixoto and Oort. Their treatment of the global energy cycle is good, in my opinion. Additionally, it covers other topics such as the existence of oceanic or cryosphere subsystems, and so on, extending coverage over pure atmosphere. The theoretical foundation of this book is impressive, in my opinion.
Just my two cents.
SoD,
Two notes.
1.
You say:
>>
… if you take a look around at the many internet “debunkings” of the “greenhouse” effect, or at the many claims such as:
– “CO2 is saturated” …
You will notice that none of them are apparently aware of this important theory called Theory of Radiative Transfer.
<<
One of the forthcoming EGU2011 presentations says:
"The line-by-line calculations for sun light from 0.1 – 8 micrometer (short wavelength radiation) as well as those for the emitted earth radiation from 3 – 60 micrometer (long wavelength radiation) show that due to the strong overlap of the CO2 and CH4 spectra with the water vapour lines the influence of these gases is significantly reducing with increasing water vapour pressure, and that with increasing CO2-concentration well noticeable saturation effects are observed limiting substantially the impact of CO2 on the warm-up of the atmosphere."
2.
I would call your readers' attention also to some items in the Bibliography at the end of Chapter 2 of Goody-Yung.
A good place to start with Physical Meteorology on line is Rodrigo Caballero’s Lecture Notes here:
Click to access PhysMetLectNotes.pdf
It’s a work in progress and somewhat heavy on math, especially Chapter 6 on the Atmospheric Boundary Layer. But, so far at least, it’s free.
Yes I’d like to endorse DeWitt Payne recommendation.
Rodrigo Caballero’s Lecture Notes have been further improved.
The notes now include the analysis around the Carnot Cycle.
Anyone hoping to understand the second law must carefully study its implications.
Clausius based his famous “heat only moves from a higher temperature surface to a lower temperature surface, never the reverse” on the insight of Carnot.
And I forgot about this excellent free resource:
Introduction to Physical Oceanography by Robert H. Stewart.
I find this very nice and also free. High level, not for readers without a solid background but nice.
http://kiwi.atmos.colostate.edu/group/dave/at605.html
Has anybody read Ray Pierrehumbert’s “Principles of Planetary Climate” and is willing to write a lenthy review?
I’ve checked at Amazon. They’ve only got one review and that one refers to the PDF Document that was available for free at some time or other.
I don’t have a lot of mathematics and bought Taylor’s “Elementary Climate Physics”. I was very disappointed. I felt it was a lazy book – splashing equations on the page without derivation or explanation.
I also have Grant Petty’s book which is excellent. It seriously aims to teach which Taylor’s I think does not.
I was an academic in another subject before I retired so I do have a feel for textbooks and teaching.
Hi, Steve.
I wonder (I know close to nothing about wordpress) whether it might be easy to convert your posts (with or without comments) to a single PDF (kind of a single click somewhere in the main page) …
Sometimes it would be nice to read blog entries in the eBook reader during travels and so on. Comments are sometimes interesting, sometimes not at all. In general I would say that it would be nice to have them too, but if only the posts could be saved, it would be an advantage.
If it were easy to do so (kind of a single command/configuration option in wordpress interface), it would be nice to get a frozen version of the blog in PDF to take with us without printing it each time we go on a travel.
Just a thought. Your blog deserves careful reading.
Miklos Zagoni on February 27, 2011 at 9:32 pm:
The criticisms of the internet “debunkings” still stands. They don’t demonstrate a flaw with the RTE. They don’t demonstrate an alternative solution to the RTE.
This paper might be interesting. Or it might be flawed. But it is yet to be revealed.
Many many papers have calculated the results of the RTE for CO2 increases using the HITRANS database with a variety of conditions – different tropopause heights, different latitudinal resolutions, different vertical resolutions.
Getting a totally different result means that everyone else made the same serious mistake. Or.. the new paper is making a serious mistake.
Which ones? There’s 3 pages in the bibliography. Don’t do a G&T:
– from p 21 of this immortalized operatic piece. (They never said which explanations were falsified).
S0D
Talking about Opera your version is of the comic type.
For many months now I have been tirelessly trying to correct your ignorance about the direction of heat flow.
G&T included reference to the error you and other believers in the “greenhouse theory” often make.
The classic error is to contradict Clausius and say;
Heat can move from a lower temperature to a higher temperature without the expenditure of work.
So here we have another cut and paste library selection.
The important thing is not to have a stack of books but to try to understand one of them.
Perhaps you noticed;
Physical Meteorology on line is Rodrigo Caballero’s Lecture Notes here:
Click to access PhysMetLectNotes.pdf
Reccomended by DeWitt Payne and myself.
Read page 36 top paragraph and see that Rodrigo agrees with Clausius;
….”heat can only flow from the warmer to the colder body”…..
It looks like a great number of your posts will have to be editted to remove this glaring error.
Bryan,
I’m a bit of a newbie to the whole topic of AGW and have been reading a lot the past few weeks, both pro and con. I’m not a scientist or expert of any type, just a BS in ChemE from Berkeley. I’ve seen a lot of your posts on this specific top of heat flow and hope you can clarify your concerns a bit.
For my part, I’m in complete agreement with your statements that heat can only flow from a warmer body to a colder. At the same time, I’m in complete agreement with SoD and others who outline how emmissive radiation (i.e. photons) are exchanged in a two-way flow between bodies of different temperatures. I don’t see any contradiction at all. All of my textbooks were consistent with both propositions.
Based on all I learned, whenever I hear or read that heat can only flow from warmer to colder (in terms of radiation), I take it for granted that the context or meaning is in regards to NET transfer of energy over some measureable timeframe. But I’ve seen quite a few posts in some places where the author interprets the 2nd Law to mean that no energy (i.e. no emmissive photons) can travel from a colder body to a warmer body, or if they do make the journey, they can’t be absorbed at a molecular level and excite a new energy state in the absorbing atom/molecule.
Can you clarify your position in this context, and explain what aspects you find in conflict, and why it’s important (what are the important implications) in terms of the AGW discussion? I’d find that very helpful. Thanks.
Bryan:
As part of your laudable and tireless work to rehabilitate G&T you can find out from them –
This valuable piece of information will help correct our impression of G&T as a pair of bumbling amateurs who have no idea what theory has actually been proposed by atmospheric physicists.
Or confirm it.
Sod
G&T need no help from me to be rehabilitated.
All the smart anti G&T critics now say “their Physics foundations are quite sound ” and they raise a number of important points however they did not deal with TOA “greenhouse effect” and that’s the critical point.
Which critic are you?
Do you still subscribe to the Halpern(gang of six) comment that apparently attacks G&T for things they didn’t say?
Or do you thank them for reminding you of Clausius and the restrictions of his famous Second Law.
Bryan:
As expected, no answer to my question.
I realize it is not an easy answer to give for supporters of G&T.
The atmospheric radiation that “reaches the surface” and then..
.. a mystery.
All part of their entertaining approach to physics.
Jarras
Thank you for your post.
I will reply with a more considered response later.
Your BS in ChemE should enable you to form a clear view of the debate.
I see a lot of errors creeping in to Climate Science because of a free and easy disregard to science fundamentals.
Stick to the fundamentals, demand clear definitions of words like “heat” and “work”, that’s on the Physics side.
From the Chemistry side, Gibbs Thermodynamics gives useful insights.
Bryan,
Yes, I’d agree it makes sense to stick to the basics. I just pulled out my old freshman chem text which says
“The concept of heat is clouded by … a tendency to think of heat as ‘something’ that ‘flows.’ In truth, heat is not a substance. It is, like work, a method by which systems exchange energy… work is energy transferred by virtue of a mechanical link between systems, and heat is energy transferred due to a temperature difference.”
Does that sound right?
Oops. Didn’t watch my name when posting. Jarras and Jim A are one and the same person: me
Jim A
“The concept of heat is clouded by … a tendency to think of heat as ‘something’ that ‘flows.’ In truth, heat is not a substance. It is, like work, a method by which systems exchange energy… work is energy transferred by virtue of a mechanical link between systems, and heat is energy transferred due to a temperature difference.”
Yes that sounds correct.
Nowadays it is importtant to add the transfer is from the higher temperature to the lower temperature object never the reverse.
People often omitted this last sentence as being self evident but recently some have thought otherwise.
Jim A
…….”But I’ve seen quite a few posts in some places where the author interprets the 2nd Law to mean that no energy (i.e. no emmissive photons) can travel from a colder body to a warmer body, or if they do make the journey, they can’t be absorbed at a molecular level and excite a new energy state in the absorbing atom/molecule.”…….
The idea of the photon was proposed by Planck and Einstein to avoid the implications of the “ultraviolet catastrophe” that the classically derived formula of Rayleigh – Jeans implied
Unfortunately it meant a clear departure from classical wave theory.
Claes Johnson has recently written about this and is attempting to restore the classical interpretation.
If you accept(as I do) that at the moment photon exchange is the best description then a two way interaction is involved.
So we can rule out
..”no energy (i.e. no emmissive photons) can travel from a colder body to a warmer body”…..
Two way Photon interaction
Look at blackbody spectrum of a solid at 250K (say), then look at the blackbody spectrum of the body at 350K.
Superimpose the spectra and compare.
We notice two things.
1. The higher temperature object produces short wavelengths that are absent from the lower temperature spectra.
2. Now look at any wavelength shared by both.
The higher temperature object produces much more radiation of that wavelength than the lower temperature object.
To make it even more concrete lets say colder object radiates 1000 photons of wavelength 15um.
The hotter object will radiate perhaps 1500 photons of 15um in the same time period.
All the colder object does is it insulates to some extent the warmer object and reduces its heat loss.
Net result, 500 photons move from higher temperature object to lower temperature object.
This is what we call heat transfer
I will go further.
Lets say the warmer surface was in thermodynamic equilibrium with its surroundings if then a colder object is brought near and it has a lower temperature than the surroundings then the colder object will INCREASE the heat loss from the warmer object.
Interestingly for wavelengths > 3um the Raleigh – Jeans and the Planck radiation formulas give the same result.
Work through to the Planck formula and pick the wavelength option
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
So it seems for most problems in atmospheric science either approach gives the same result.
Bryan,
Since you appear to think highly of Caballero’s Lecture Notes, I suggest you read chapter 5 carefully, particularly 5.10 Emissivity and Kirchhoff’s law starting on page 118 of the pdf or 117 actual page number. Also the definitions of radiance or intensity vs irradiance or flux in section 5.6 on page 113 of the pdf are important. Apparently G&T don’t understand that the arrows in energy balance diagrams represent fluxes through a plane with the arrows representing the normal to the plane rather than a vector (section 3.7.2 of G&T).
Bryan,
Your description is very clear and I think we’re on the same page. Energy in the form of photons are always in a two way exchange between bodies of different temperatures in such a way that the net transfer of radiant energy (heat) both at any instant and over time is from the warmer body to the colder.
So far, my understanding of SoD’s writings puts him in agreement as well. I’m not sure where the point of contention lies, and you’ve noted above to be cautious that some would say heat flows the other direction. Can you elaborate? What form does this argument take in terms of climate? Thanks.
Jarras
There is one other hurdle that the”cold” radiation suffers from.
It flows up hill.
The respective blackbody spectra of the hot and cold objects will be centred around different maxima characteristic of their temperature.
To simplify the discussion we will use a single wavelength to represent average wavelength.
I will also invent magnitudes rather than calculate accurate ones to save time.
Lets say that initially there is only the colder object present.
Lets say that some way out from the cold object placed in a vacuum 100 Joules of IR radiation of wavelength 15um pass through one metre square area in one second.
Now let a hotter object radiating at 250 Joules/s with average wavelength 5um fill the same square metre.
So what happens to the radiation from the colder object?
Three viable solutions.
1. Classical wave theory approach.
A single Pointing vector of magnitude 150W/m2 with direction hot to cold.
2. Subtraction of photon streams will result in heat flow of 150W/m2
3. Absorption of 100Joules/s of 15um radiation and emission of 100 J/s of 5um.
So the consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um radiation into 100J of 5um.
The increase in the “quality” of the radiation appears to violate the second law.
Now all three methods will give about the same result if properly applied.
I would guess SoD would pick 3.
I would pick 1 or 2 for reasons above.
The misdirection of heat flow is a feature of a number of “Greenhouse theories”
Gerhard Gerlich and Ralf D. Tscheuschner falsified a large number of them in:
Click to access 0707.1161v4.pdf
Once you have read the paper there is a comment paper criticising G&T and a further G&T reply paper available free.
Bryan,
Two questions or concerns about #3 scenario
A) You speak of the increase of “quality” of the radiation. I’m not sure what you mean by that. Can you explain that? 100J of either wavelength seems like more a conservation of energy issue (1st Law). The 5um photons carry more energy per photon than 15um photons. Is that what you mean by “quality”, or is it something else? If just energy per photon, then balancing 100J/s of 5um with 100J/s of 15um just means that the 15um stream will have more photons/sec, but still the same energy/sec in sum total. I don’t see an obvious violation of the 2nd Law.
B) Scenarios 1 & 2 fit the instantaneous change in the system fairly well. However, #3 wouldn’t be the immediate response of the hotter body. It would be similar to your earlier post. The warmer body emits 250J/sec by virtue of its temperature. Placing it in front of some other body doesn’t immediately change the temperature, so it doesn’t immediately change the rate of emission – it stays at 250J/s. However, since it’s now absorbing 100J/s more than previously (I’m assuming before jumping in front of the colder body it was radiating to empty space at 0K), the net outflow is now 150J/s (as in #1 & #2). So the effect is a slowing down of the cooling rate of the warmer body. The 100J/s absorbed doesn’t turn into an additional 100J/s emitted. And the warmer body doesn’t warm up, it just cools more slowly. Net heat still flows from warmer to cooler, and the 2nd Law is still happy.
Regards
Jarras
100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.
The first law regards them as equivalent but the second law does not.
If you had some procedure for changing without loss the wavelength of radiation, say converting radiation around 15um into radiation around 1um you could extract energy from seawater to power a boat.
Unfortunately reality and the second law forbids it.
Bryan,
Sorry for the somewhat random names I’ve used (Jim A, Jarras). I just found they belong to others, so hopefully this third one will be the charm.
I think we’re getting closer to the real crux of the issue, at least insofar as I didn’t really understand your latest post, so maybe an opportunity for me to learn a little more.
I’d certainly agree that if you could find a way to freely convert a 15um photon into a 1um photon, you’d have an infinite energy source. But I don’t see how our discussion has suggested that kind of magic, and haven’t seen where that’s a necessity or implication of any of the climate issues discussed here. Can you help me see what I’m missing? The more specific the better. Thanks.
Jarras
The point I Was Making is that the subtraction method 2 above exchanges thermodynamically equal radiation.
Method 3 implies that the energy arriving is upgraded to a higher quality.(shorter wavelength).
SoD has given a view and on the topic and I have replied with extra detail.
Bryan,
You’re almost there. If you believe that, then you should also believe that bringing an object that is warmer than the surroundings but colder than the warm surface near the warm surface, the warmer than the surroundings object will DECREASE the heat loss from the warmer object.
To be more precise, we have a warm object with temperature T1 and surroundings at T2. We define T1 > T2. If we place an object with temperature T3 where T3 < T2 then heat loss from the surface at T1 will increase. We’re all on the same page here. Now raise T3 to > T2 but still < T1. Heat loss from the surface at T1 will decrease.
DeWitt Payne
Yes perhaps I didn’t make clear my position.
A colder object if it has a higher temperature than the surroundings will reduce the heat loss from a warmer object.
The effect is a radiative insulation of the warmer object.
Long time reader, first time poster. Let’s see if I’ve learned anything. ;D
My understanding of Bryan’s critique is that heat can only flow from hot to cold without outside interference, in layman’s terms. So far so good and all are in agreement.
He seems to extrapolate from this that a cold atmosphere can never have a warming affect on a warmer planet. This is where viewpoints diverge.
All agree that the net transfer will be from hot to cold.
But energy is flowing both directions. Including from cold to hot.
Picture if you will, a dodgeball court. On one side of the court are 100 players, on the other are 20.
All players are told to pick up any balls they see lying around and throw them as fast as they can onto the other side of the court.
We can see that the ‘net rate’ of ball transfer must be from the side with more players to the side with fewer. (I know that to REALLY make this analogy work we would need a mechanism to actually change the number of players as the balls are thrown, but I think this will illustrate a point).
So the number of balls on the side with 100 players will deplete over time, and the side with 20 will increase over time. The ‘net transfer’ is from 100 to 20.
But without the 20 players throwing some of those balls BACK, the rate of loss of the balls from the 100 player side would be much FASTER. This has the effect of slowing the loss of balls from the 100 side, even though the net transfer is still from 100 to 20.
Now if we have a system wherein balls are continuously being added to the 100 players side, and then being thrown towards the 20 player side, and some of the balls from the 20 player side are rolling “out of bounds” over time, we would eventually reach a steady state, with a given number of balls on the 100 side and the 20 side.
But this steady state will have a higher number of balls on the 100 player side if the 20 players are throwing balls BACK then if those 20 players were not there at all.
That is my stupid analogy for radiative transfer. Feel free to correct any mistakes.
Purgatus,
I’ve been trying to think of a similar analogy. I think yours works well. I’d just suggest a couple of changes. Think of the number of players as the number of atoms in the material. The number of balls is the amount of energy. So start with equal number of players on both sides (similar sized objects), but vastly different number of balls (different temperatures). For instance, both sides have 50 players, but one side has 200 balls, the other side only has 20. Then the analogy should follow the physics very well. Initially, the transfer from the cold side is slower, because there’s a shortage of balls to throw, whereas the hot side is slinging them fast and furious. The rate of net transfer is fast when the imbalance in quantity is high. Eventually both sides have an excess of ammo compared to players, but the side with more finds it easier and quicker to retrieve, reload and fire, so they’re rate will continue to be higher until both sides eventually have equal number of balls, then the net rate becomes zero.
Bryan,
There’s lot’s more evidence that the concept of a photon is valid than black body radiation. There’s the photo-electric effect for one. Photomultiplier tubes, gas proportional counters and solid state detectors like CCD’s used in digital cameras literally count photons. In a gas proportional counter used in x-ray spectroscopy you get discrete pulses with a peak height proportional to the photon energy. A pure wave theory has the energy of a beam of light proportional only to the intensity, not the frequency. Yet solid state devices like CCD’s and photovoltaic cells don’t work at frequencies lower than the band gap regardless of the intensity. I don’t see Claes Johnson addressing that issue.
DeWitt Payne
Yes I think Claes still has a lot to prove.
I admire his attempt to take a fresh look at radiation.
If he had refrained from a direct criticism of the “greenhouse theory” I’m sure he would has been spared a lot of flak.
DeWitt, I have found your posts to be very informative and helpful. Could you please comment on my above understanding of the very basic levels of radiative transfer?
That’s actually not a bad analogy. I did a similar thing only with the players kicking any ball that gets close to them in a random direction and walls on the sides that reflect balls back on the court. You feed in balls at the bottom and remove balls at the top that cross the end line. More players, more balls in play at any given time.
I can make it more complicated and realistic too. Suppose you have some players that can catch and throw balls, but most of the players can’t catch and throw. However, if a non-throwing player bumps into a player with a ball before he throws it, he must pass the ball to them. Conversely, if a non-throwing ball carrying player bumps into a thrower, he can pass the ball. Non-throwers can also pass balls to other non-throwers. Any player carrying a ball runs faster. Most of the time a throwing player will be bumped and have to pass the ball before he can throw it. Add more throwing players and the average running speed goes up. Analogies are fun, but they’re never entirely accurate.
Thanks for the review, just what I have been looking for. So many books are either too simple or too complex, I hope this one finds a balance. Except that balance is different for each reader.
Even if it turns out too complex for me, avoiding the way too complex is still good. If I read it a few times maybe some understanding will seep through.
So Hartman book ordered. Don’t hold your breath for any improvement in my responses, it won’t arrive until early April.
Bryan on March 4, 2011 at 6:45 pm:
And lo, there was much rejoicing.
This is the inappropriately-named “greenhouse” effect.
The atmosphere absorbs thermal radiation emitted from the surface, re-radiates in all directions – some of it down, and therefore causes a higher surface temperature than without the radiatively-active atmosphere.
S o D
I’m glad when I am the source of rejoicing in another.
However in this instance I am a bit puzzled by they immediate cause.
Ive always said that the atmosphere insulates the Earth Surface.
All four method of heat exchange are involved.
Conduction, convection, radiation and phase change.
Bryan on March 4, 2011 at 7:05 pm:
All you need to do now is prove that.
Let’s review the equation for change in temperature of a body:
ΔT = ΔQ / mc
ΔT = change in temperature
ΔQ = change in energy
c = specific heat capacity
m = mass
So let’s consider a 1kg body with heat capacity, c=1000 J/K.kg
The 1kg body is losing heat by various means and also receiving heat so that it is in equilibrium at a temperature = 200 K.
Now an additional source of energy is introduced.
Scenario A: The body absorbs 1000 J of incident radiation of 5 μm.
ΔT = 1000 / (1 x 1000) = 1K
New temperature = 201K
Scenario A: The body absorbs 1000 J of incident radiation of 15 μm.
ΔT = 1000 / (1 x 1000) = 1K
New temperature = 201K
What’s the difference?
Nothing.
What you are getting confused with is the fact that generally a higher temperature source emits thermal radiation of shorter wavelengths (higher frequencies).
So, for example, a 5800K body emits radiation centered around 0.5μm, whereas a 290K body emits radiation centered around 10μm. The real difference between these two bodies is that the 1st emits 6.4×107 W/m2 (reduced by the emissivity at this temperature) and the 2nd emits 401 W/m2 (reduced by the emissivity at this temperature).
Now if we calculate the entropy change for radiation emitted by a 5800K body which is absorbed by a 290K body, then compare that with the emission by a 500K body absorbed by a 290K body we have a different change in entropy.
It is the temperature and energy emitted which determines the entropy change.
But, assuming equal absorptivity by the receiving body, there is no “thermodynamic” difference between 10μm radiation and 0.5μm radiation.
And your comment of March 4, 2011 at 4:51 pm makes no sense.
I recommend that you write a new example with some real calculations of two bodies at different temperatures.
This will give the opportunity to identify what you are trying to explain.
Or review an article such as The Three Body Problem.
It really depends on what you’re trying to do with the radiation. If it’s just being absorbed and thermalized then there is no difference. If you are trying to do something with a system that is transparent at 15 μm but absorbs at 5μm or the converse, then there is a difference. It also depends on whether the source is thermal or not. Non-thermal sources like 10 μm CO2 lasers or microwave klystrons can do things like cut steel or fry bacon that would not be possible with a thermal source centered at those wavelengths.
DeWitt Payne
I think in a roundabout way you are agreeing with me.
…..”100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.”…….
According to the second law they are not equivalent.
The shorter wavelength radiation has a higher “quality” than the longer wavelength radiation.
You can convert a bigger fraction of short wavelength radiation into non thermal work than you can with an equivalent magnitude of longer wavelength radiation.
It’s not agreeing so much as saying you’re not completely wrong. On balance SoD is more correct than you are. You focus on one aspect of EM radiation, the brightness temperature and the ability to do work. Sunlight has a brightness temperature of ~5800 K but an effective temperature at the TOA of 394K while emission from the atmosphere to the surface has an effective temperature of 277 K and deep space is 2.7 K. But heating something does no work. Only a tiny fraction of absorbed sunlight is converted to work as air or water circulation or is used to reduce carbon dioxide by photosynthesis. And most of that is rapidly dissipated and ends up as heat too. What’s important to the heat balance of the planet are the effective temperatures, not the brightness temperature. OTOH, life as we know it wouldn’t exist if the brightness temperature of the sun were lower or higher.
DeWitt Payne
I’m a bit disappointed by your equivocation.
The question is simple;
…..”100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.”…….
According to the second law they are not equivalent.
S o D thinks they are!
The answer should be a simple yes or no!
There are whole papers (sometimes written by supporters of the IPCC position) devoted to the “quality” of radiation.
S o D
I said
“100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.
The first law regards them as equivalent but the second law does not.”
You disagree and use an equation from the first law to make your point which I have already agreed and stated above.
This seems totally pointless.
The law it that is violated is the second law.
Now I realise that you have stated explicitly in an earlier post that this is your position.
I think you will think this one through and reconsider.
The Solar radiation(short wavelength) energy arriving on Earth is roughly equal to the long wavelength energy leaving the Earth.
So according to you it is possible to reverse the sequence with long wavelength radiation arriving and solar radiation leaving.
That there is nothing in principle against taking long wavelength energy from seawater up converting it to a suitable short wavelength to power a device to propel a ship.
You say in example above
……”Scenario A: The body absorbs 1000 J of incident radiation of 5 μm.
ΔT = 1000 / (1 x 1000) = 1K
New temperature = 201K”…….
But the reverse of thermalisation namely the 201K body radiating 1000J of 5um energy although allowed by first law contradicts the second law.
All three of the above examples contradict the second law.
Bryan,
You continue to repeat that X joules of wavelength A is different than X joules of wavelength B, and say it’s in regard to the “quality” (your quotes) of the radiation. I’ve searched and can find no definitions of what this quality might be. I’ve asked at least a couple of times for an explanation, or to be pointed to an explanation. That would be really helpful. I know that photons have wavelength, energy, speed. What is “quality”? Really, until you can answer this, I’m afraid I’m not going to have any chance of understanding your problems with SoD’s posts.
Bryan on March 5, 2011 at 2:46 pm:
You are claiming that the example given complies with the first law and violates the 2nd law?
Please show how it violates the 2nd law. Don’t claim. Prove.
The equation of entropy change is given by δS = δQ / T.
As I already explained following my example (March 5, 2011 at 7:02 am) you need to know the temperature of the source. And once you start writing out the movement of thermal radiation between bodies at different temperatures you will realize that what causes a reduction in entropy is net heat moving from cold to hot.
See The Real Second Law of Thermodynamics.
You have made lots of claims over many months that my various examples violate the 2nd law of thermodynamics and yet you have never once provided an entropy calculation to prove it.
So go ahead and demonstrate that my example of a 200K body absorbing 15 μm radiation and 5 μm radiation to give the same temperature rise is a violation of the 2nd law.
SoD,
Above you state “… what causes a reduction in entropy is net heat moving from cold to hot.” This seems incorrect. What am I missing?
I think I understand now. But it’s almost like a double negative with reference to a prior example. I would expect something more like “an increase in entropy due to net heat going from hot to cold”. I missed “reduction” when saying entropy change is result of net heat going from cold to hot.
Jimwit:
Have a read of The Real Second Law of Thermodynamics.
If you still have that question – or others – feel free to ask it here.
Our comments crossed. You already have your answer.
Thermal energy which is sometimes called heat energy differs from other forms of energy.
Electrical energy, Kinetic Energy and Gravitational Potential Energy can be readily interchanged with almost 100 % efficiency.
All these energies can be converted to work like lifting a load or spinning a wheel with near 100% efficiency.
Thermal energy cannot be transferred into other forms very easily.
A very efficient thermal power station will perhaps achieve 50% at best conversion.
The second law explains why this is so.
Thermal energy likewise has different grades of “quality”.
1000 Joules of 1um radiation will be transformed into work with a much higher efficiency than 1000 Joules Of 15um radiation.
According to SoDs analysis above 100 Joules Of Electrical Energy will produce a temperature rise as calculated.
So according to SoD 100J of Electrical Energy is thermodynamically equivalent to 100J of thermal energy of any wavelength.
It appears that SoD has no idea of what the second law is all about.
My impression is that Bryan forgets that there is not the same amount of 5 μm and 15 μm photons in two streams of 100 W m^-2. Otherwise, I don’t understand what he tries to explain and I would appreciate a clear explanation or a cite to a real paper explaining it. G&T is not a real paper, in my opinion.
Bryan continues to make claims without proof or references. I earlier (March 5, 2011 at 7:47 pm) asked him to provide a calculation of entropy – to prove his claim. Instead, another claim.
The clue is often in the units.
The units of entropy are J/K.
The units of entropy are not J/K.μm or J/K.Hz
Let’s hear from textbook writers.
An Introduction to Thermal Physics, Schroeder (2000):
Notice the calculation of entropy. As already noted in my comment of March 5, 2011 at 7:47 pm.
An online resource recommended by Bryan as a means to brush up my parlous knowledge of the second law of thermodynamics:
Lecture notes in Physical Meteorology by Rodrigo Caballero:
Note that J is earlier defined as heating rate per unit mass.
We will wait in vain for an entropy calculation by Bryan for my two scenarios of March 5, 2011 at 7:02 am.
The reason is simple.
It is impossible to calculate an entropy change without knowing the temperature of the source body, the temperature of the receiving body and the energy transferred between them.
And the wavelength of the radiation is irrelevant. It is not used in the calculation of entropy change – as is clearly identified by the equations cited.
In The Real Second Law of Thermodynamics Bryan claimed similar ideas to his arguments above – again without any proof.
Do a search within your browser for “Bryan” on that post.
In The Three Body Problem I provided a comparison of two scenarios:
– a sun heating an earth with a very cold background radiation
– a sun heating an earth with a much warmer background radiation
I demonstrated that this earth would be much warmer in the second scenario, and also did an entropy calculation to demonstrate that the second law of thermodynamics was not violated.
Bryan did not demonstrate a flaw with any of the calculations but instead exhorted me to review the Carnot cycle, his favorite red herring.
We all agree with the Carnot cycle. The energy powering the terrestrial “heat engine” is the sun. Yet the atmosphere affects the temperature of the earth. Hence the title of this last article – The Three Body Problem.
All around the internet on climate blogs I find comments from Bryan telling everyone I don’t understand heat transfer.
Bryan, lift your game.
Do an entropy calculation.
Provide a textbook reference of the actual problem at hand.
If you are such an expert and I have it so wrong it should be easy.
Where is the flaw in the entropy calculation in the two articles I wrote?
Do you even know what entropy is?
S o D
You could ” lift your game” by admitting you are wrong about the second law.
Rodrigo Caballero’s Lecture Notes
Click to access PhysMetLectNotes.pdf
Rodrigo uses an entropy calculation to prove that;
….”heat can only flow from the warmer to the colder body”…..
Read page 36 top paragraph and see that Rodrigo agrees with Clausius;
A great number of your posts tried to prove otherwise.
However we dont need advanced entropy calculations to prove that the wavelength determines the quality of the radiation, simple thermodynamics is sufficient.
So lets expand somewhat on an earlier example I gave.
According to S o D 1000J of 15um radiation is thermodynamically equivalent to 1000J of 2um radiation
According to Weins Law;
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
This link can be used to check my figures
Radiation centred around 15um corresponds to a temperature of 15C (about the temperature of sea water)
Radiation centred around 2um corresponds to a temperature of 1000C.
So according to S o D there is no thermodynamic obstacle to transforming sea water radiation into radiation of 2um which will produce steam if directed on a cylinder containing water.
Hence boats could simply extract the plentiful thermal energy contained in sea water to power the boat day and night.
Why didn’t some smart guy take out a patent on S o D technology.
Because the second law makes it impossible!
Bryan on March 6, 2011 at 10:55 am:
No, Bryan, you said 1000 Joules of energy at 5 μm was not equivalent to 1000 Joules of energy at 15 μm.
I said they are equivalent thermodynamically. Converting one into the other will actually produce exactly the same temperature rise, because 1000J = 1000J and the equation for temperature change is ONLY dependent on the amount of energy and not on its wavelength.
You have yet to demonstrate why my example of these two scenarios violates the 2nd law.
And as I already explained in my earlier comment from March 5, 2011 at 7:02 am:
Now you need to produce your entropy calculation to demonstrate that scenario A is different from scenario B, and that one or the other is prohibited by the second law.
Or, as I already requested:
Bryan also claims:
And once again fails to actually demonstrate any violation of the 2nd law in any of my posts.
Claiming is easy.
I have provided plenty of ammunition for people like Bryan.
Just explain in detail which one of my examples is flawed, with reference to an equation of entropy calculation.
There is a reason why Bryan has never yet done this. And it seems that most readers, even newcomers, understand why.
SoD says of me
…..”And once again fails to actually demonstrate any violation of the 2nd law in any of my posts”……
Well here is a recent post
…..”.Does Back Radiation “Heat” the Ocean? – Part Four”….
The answer S o D gives is yes.
Notice that SoD now puts inverted commas round Heat in deference to my criticism.
He considers I am making a “dull point”.
The reason SoD thinks it dull is he has no idea of heat flow and the second law.
He must pause to think why ships do not extract energy from the sea for propulsion.
Why long wavelength radiation leaves the Earth and solar(short wavelength) enters.
According to SoDs interpretation it could work perfectly well the other way round.
Bryan,
It’s not equivocation and the Second Law says no such thing. It really does depend on what you’re doing. If it’s heat flow then they are equivalent because the greater energy per photon of shorter wavelength radiation is irrelevant. 100 Joules is 100 Joules as required by the First Law. If it isn’t a thermal process then they aren’t equivalent. Your insistence that they are always and ever not equivalent is incorrect. A statement that they are always equivalent would be equally incorrect. But SoD doesn’t say that.
You appear to be assuming that there’s a temperature associated with EM radiation that is determined solely by the wavelength. That’s wrong. The brightness temperature for EM radiation at a given wavelength is determined by the radiance as well as the wavelength. You can calculate the brightness temperature by inverting the Planck function. The effective temperature is determined by the flux and only the flux. That’s why a CO2 laser at 10 μm can cut steel and you can fry bacon in a microwave oven.
Replace ‘appear to be’ with ‘are’ in the first sentence of the second paragraph.
That would be blackbody radiation. We’re not talking about blackbody radiation in this specific example. It’s narrow band radiation and the source isn’t specified. Narrow band radiation can have any brightness temperature depending on the radiance.
Strawman argument. He says no such thing.
DeWitt,
Thanks for the hint and link on inserting greek symbols. How about a hint on how to embed block quotes (with or without the big quote icon)? I’ve come up blank in my own search. Thanks!
DeWitt Payne
Its difficult to communicate if the most obscure interpretations are made of straightforward statements.
Originally I said
“100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.
The first law regards them as equivalent but the second law does not.”
I did not expect having said that that I would be given only examples where the first law finds them equivalent.
My point was to examine cases where the second law finds them not equivalent.
I could easily have said
“100Joules of electrical energy is not thermodynamically equivalent to 100Joules of 5um radiation.
The first law regards them as equivalent but the second law does not.”
And all your and SoDs replies would be equally applicable.
However such a statement would lead a reader to doubt the understanding of anyone claiming they are equivalent in terms of the second law.
Bryan,
Your blanket statement: “100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation.” isn’t specific enough so it isn’t strictly true. If you said that 100 Joules of blackbody radiation centered at 15 μm is not thermodynamically equivalent to 100 Joules of blackbody radiation centered at 5 μm you would be correct. It’s not. The brightness temperature is different as are the total flux and radiance. But an individual photon at a given wavelength doesn’t have a defined brightness temperature.
But you generalize from what might be a correct statement if phrased properly to draw conclusions about the greenhouse effect and heat transfer that are incorrect.
Why not use the greek letter mu (μ or µ) instead of u? It’s only a few more keystrokes.
http://htmlhelp.com/reference/html40/entities/symbols.html
Make that ‘with peak intensity’ rather than ‘centered’ in the second sentence.
Bryan,
If your earlier statements carry the meaning proposed by DeWitt “100 Joules of blackbody radiation with peak intensity at 15 μm is not thermodynamically equivalent to 100 Joules of blackbody radiation centered at 5 μm” then I finally understand and would also agree. Is that what you mean?
Jimwit
As far as Im concerned DeWitt Payne’s formulation and mine are both correct.
Equal magnitudes of radiative energy at different wavelengths are equivalent for the purposes of the first law but not of the second.
Bryan,
The potential for confusion is that when you speak of X Joules of 15μm radiation, it’s not clear by your formulations whether you mean X Joules of monochromatic radiation at 15μm and only 15μm, or X Joules distributed across a range of wavelengths consistent with a blackbody radiating with a peak intensity at 15μm. Your answer to the previous question about this isn’t very clear.
DeWitt Payne
……”That’s why a CO2 laser at 10 μm can cut steel and you can fry bacon in a microwave oven.”………
The electrical energy in both the laser and microwave case ends up being thermalised.
The second law is not involved.
It is easy to turn all energy forms, electrical, kinetic, gravitational potential and chemical into thermal energy with near 100% efficiency.
However turning thermal energy into one of the other forms is very problematic.
The second law is involved here and understanding the Carnot Cycle is helpful as Rodrigo explains in his notes.
Thanks for the link about obtaining μm
Bryan.
SoD has always talked about thermal energy in his posts. Therefore, your remark above:
“However turning thermal energy into one of the other forms is very problematic.”
does not apply since he is not talking about work or any other kind of chemical reaction or photoelectric effect.
You can study Lorenz’s energy cycle and so on if you are interested in the way the atmosphere converts thermal energy into work (kinetic energy). That is outside the scope of this thread. You can try with Wiin-Nielsen and Chen’s Fundamentals of Atmospheric Energetics to learn about the details.
Bryan
I have read a bunch of atmospheric physics books in my life and I find yor statement absolutely empty. I can think that you refer to a difference in terms of the ability of the photons to produce some effects related to quantum processes but then your allusion to thermodynamics distracts our attention from your point.
Given what I have seen so far, it is my impression that SoD is right and you are wrong. I can accept I am wrong but I would appreciate if you could give us a thought experiment showing that:
* the streams of monochromatic radiation you refer to violate the second principle
OR
* you make a correction to the energy fluxes so that they are “thermodinamically equivalent” in terms of the first and second principles
OR
* you explain us in detail or give us a reference explaining exactly what you mean
I would appreciate a detailed explanation, if you can provide one
As I see it, Bryan’s argument isn’t so much “wrong” as it is incoherent. As Pauli said, it is “not even wrong.” Bryan is trying to apply thermodynamic reasoning to thought experiments that cannot be analyzed by thermodynamics.
Bryan writes:
” The question is simple;…..”100Joules of 15um radiation is not thermodynamically equivalent to 100Joules of 5um radiation. According to the second law they are not equivalent. S o D thinks they are! The answer should be a simple yes or no!
OK Bryan, I will give you a simple answer: NO But I expect that you will not like my explication. They are not “thermodynamically equivalent” because there is nothing to compare: Neither the 15 um radiation nor the 5 um radiation has any “thermodynamic properties”, so there is no way to compare them. They are not “thermodynamically equivalent” because they are not describable by thermodynamics. Your entire exercise is thermodynamically nonsensical.
Classical thermodynamics – the science created by Clausius, Carnot, and Kelvin, and applied to electromagnetic radiation by Kirchhoff, Boltzmann, Wien, and Planck – deals with systems that are in thermal equilibrium. Every question that can by solved by a classical thermodynamic argument involves comparing some final equilibrium state to its initial equilibrium state. For a system to be in thermal equilibrium, its quantum states must be populated according to the appropriate distribution function: Boltzmann for most material objects, Fermi-Dirac or Bose-Einstein for material objects at low temperatures, Planck (a special case of Bose-Einstein) for electromagnetic radiation. We can apply classical thermodynamics to the exchange of energy between any of these sorts of objects in a straightforward way. We cannot apply classical thermodynamics to monochromatic 15 um , or 5 um, radiation, because such systems correspond to ensembles of photons that are very far from equilibrium – they simply does not fall within the domain of systems that classical thermodynamics describes. We can apply classical thermodynamics to systems that have been forced to absorb such radiation, once we assume that they have attained a new equilibrium system, but that is all.
As for which wavelengths are more effective for converting the electromagnetic energy into “work”, rather than heat, that depends entirely upon what absorbs the radiation. Suppose that the absorber is grass. Chlorophyll-a, the primary pigment in green plants, absorbs strongly between 625 and 675 nm (orange and red) but almost not at all between 600 and 400 nm. Absorption of radiation by chlorophyll-a is the primary step in photosynthesis. So, 1 Joule of 650 nm radiation produces much more “work”, when it hits my front lawn, than 1 Joule of 500 nm radiation. The 650 nm radiation is absorbed by chlorophyll and, eventually, used to synthesize molecules like ATP that can supply energy to power chemical reactions. The 500 nm radiation is either dissipated as heat or reflected. So, contra Bryan, the supposedly “higher quality”, 500 nm radiation yields much less ‘chemical energy’ than the longer wavelength 650 nm radiation.
Robert P
You certainly get the prize for packing the maximum number of big names into a piece of text.
The block of text could be used in a multi purpose way to stop any discussion on thermodynamics anywhere.
Because we have strayed from thermal equilibrium indeed.
Well, here’s some news for you, we never have complete thermal equilibrium, full stop.
This like blackbody radiation is an ideal abstraction.
Try to read the comment build up from my reply to
Jarras March 3 8.49
“There is one other hurdle that the”cold” radiation suffers from.
It flows up hill.
The respective blackbody spectra of the hot and cold objects will be centred around different maxima characteristic of their temperature.
To simplify the discussion we will use a single wavelength to represent average wavelength.
I will also invent magnitudes rather than calculate accurate ones to save time.”
So when I say in this instance 15um it stands in for blackbody type spectrum centred around 15um.
That often trouble with gatecrashers wanting to “butt in” they dont take the time to read before mouthing off.
Bryan’s confusion is that for all his talk of the second law of thermodynamics he doesn’t know what it is really about.
It is about entropy. (Take a look at Bryan’s explanations of entropy..)
Bryan said 1000 Joules of energy at 5 μm was not thermodynamically equivalent to 1000 Joules of energy at 15 μm.
Apparently if one is converted to the other you can power a boat from sea water. I can’t see how this would work because if you convert 1000 J to 1000 J you have created zero energy.
I said that you cannot calculate the entropy of these two scenarios because we do not have the source temperature.
And I provided textbook examples.
As is usually his way, Bryan does not confirm or deny – for obvious reasons.
Bryan still refuses to provide any entropy calculations – for obvious reasons.
Or to demonstrate a flaw with any of my earlier cited entropy calculations – for obvious reasons.
Bryan’s confusion is that he has related some magic properties of wavelength to the second law instead of the important property:
Change in entropy, δS = δQ / T
Because of his confusion he believes that somehow 1000 J of radiation has a different effect on the temperature of a body that absorbs it depending on its wavelength. (See his comments in The Real Second Law of Thermodynamics).
You don’t find his ideas in a textbook.
Instead, Bryan will invent ideas that he thinks I have.
Bryan is still trying to invent a method by which the atmosphere can have less effect (or no effect) on the surface temperature of the earth than is claimed by atmospheric physicists.
You can find very clear explanations of what I think in:
The Real Second Law of Thermodynamics
The Three Body Problem
In radiation, the reason that more energy is transferred from a hotter to a colder body is quite simple. The hotter body radiates more energy. The colder body radiates less energy.
The result is conservation of energy and an increase in entropy.
There’s no other mysterious mechanism by which entropy is increased.
The hotter body still absorbs energy from the colder body and this energy is applied in the equation of temperature change in exactly the same way as energy from the hotter body received by the colder body.
The second law does not prohibit certain wavelengths of radiation from affecting temperature of a body.
Hopefully, no one else is confused about the science – if you are, please ask.
No doubt everyone is confused about Bryan’s magical claims.
SoD, thanks for the reply. I simply hope in good faith that Bryan makes a complete explanation of his comments. Once we have a complete explanation, it can (or can not) be refuted. Without an explanation, there is no way to try a refutation. That’s the reason I am insisting that I want a clear explanation. I have no doubt that 100 J are 100 J. I want to know what exactly does he imply when stating that 100 J are not the same as 100J.
I think that what Bryan is getting confused with is that moving 1000 Joules of energy from a body at 280K (for example) to a body at 1000K (for example) will allow work to be done.
If someone can do this – without an equivalent “cost” – they have reduced entropy.
This is a violation of the second law.
Once again we see that temperature of the source and receiving bodies are the important parameters.
SoD
You said
…..”moving 1000 Joules of energy from a body at 280K (for example) to a body at 1000K (for example) will allow work to be done.
If someone can do this – without an equivalent “cost” – they have reduced entropy.
This is a violation of the second law.”……
Is halfway to agreeing with me.
Lets try the full version,
If we had a “black box” into which 1000J of of blackbody spectra type radiation centred around 15um was the input and the output was 1000J of radiation centred around 2um.
The transformation was accomplished without the expenditure of work.
Would this violate any thermodynamic laws?
I say that although the first law is not violated the second law is.
Bryan,
I won’t presume to speak for SoD, and my reply is unlikely to be as rigorous or correct as his, but in answer to your question, I can’t think of a way the black box scenario you’ve proposed wouldn’t be a violation of the 2nd Law.
Having said that, I still look forward to your pinpointing precisely where and how you think something like this is happening in relation to the “greenhouse” effect (I think it would be better described as a “carport” effect, but that’s a different discussion.)
Bryan
SoD has already explained that temperature matters and he has even written the definition of entropy above. He has already presented you a simple but powerful argument (units of entropy do not include wavelength) and you have answered with this paragraph which simply does not contradict SoD’s previous posts about the topic. Your “centered around 2 μm” is just a way of talking about Wien’s displacement law without mentioning it.
I agree with Jimwit that your position has no relation to the atmospheric greenhouse effect. Surface and atmosphere are not the only sources of heat fluxes in the system. The secret (you have already been told) is that you must consider the NET heat fluxes, not every single heat flux forgetting the rest of the fluxes. It is not a matter of a cold atmosphere warming the planet, as you pretend to simplify the problem. Long-wave radiation from the atmosphere to the surface exists but there are other energy fluxes, some of them non-radiative ones (such as the latent heat flux). No climatologist on Earth says the atmosphere warms the surface in isolation from the other bodies of the solar system and without additional energy fluxes. Your example can not be applied to the problem being studied as one of the hypothesis can not be applied. You say “The transformation was accomplished without the expenditure of work.”. This is not what happens in the real system, since there ARE other energy fluxes acting. Your idealized experiment fails to illustrate the violation of the second principle because you fail to recognize this.
Your position implies that a blackbody at 300 K does not radiate energy towards a blackbody at 2000 K, because it “somehow knows” that the second one is warmer, and this is clearly in contradiction with Planck’s law. Blackbodies radiate according to their temperatures and not depending on the temperatures of blackbodies placed next. I find surprising that you still insist in this idea.
The atmosphere radiates energy towards the surface and the space. The atmosphere receives energy from the sun, the surface and the sensible and latent heat fluxes. The NET balance of the surface when all the energy fluxes are considered implies that the surface is warmer than it would be without an atmosphere, and this does NOT violate the second principle. Do people living in New Jersey use thermal isolators in water pipes? Do these isolators violate the second principle? Are they in the outside colder than the water inside? So … how can the “cool isolators” warm the warm water inside? The answer is … they don’t. They just change the net energy fluxes in the system.
I hope this helps
Jon
You say
“Your position implies that a blackbody at 300 K does not radiate energy towards a blackbody at 2000 K, because it “somehow knows” that the second one is warmer,”
Could you indicate anything Ive written that gives you this impression?
Bryan
Please, see the next three examples in this thread.
Bryan February 28, 2011 at 10:06 pm
[…]
The notes now include the analysis around the Carnot Cycle.
Anyone hoping to understand the second law must carefully study its implications.
Clausius based his famous “heat only moves from a higher temperature surface to a lower temperature surface, never the reverse” on the insight of Carnot.
Bryan March 3, 2011 at 1:48 pm | Reply
[…]
The classic error is to contradict Clausius and say;
Heat can move from a lower temperature to a higher temperature without the expenditure of work.
Bryan March 4, 2011 at 9:24 am
[…]
Nowadays it is importtant to add the transfer is from the higher temperature to the lower temperature object never the reverse.
I get the impression that you say exactly that several times (without being exhaustive in the search). Pay attention to your “never” word. Never menas … never.
Regards
jon
Jon
I think I know what is confusing you.
You are assuming that electromagnetic radiation is the same as heat.
It is not.
If you pick up the dialog at Jarras first question it should help explain the difference.
Sorry, Bryan
It’s you who have been talking about radiation at 5 or 15 micrometers during the last one and a half day
If you talk about wavelength, then, you are talking about energy fluxes by means of radiation
I am not confounding it, you mentioned micrometers first
Regards
jon,
What would be wonderful would be to get Bryan’s worked example of the three body problem – the sun, the earth and the atmosphere – with a calculation of the alleged increase in entropy that he believes is implied by the claims of atmospheric physics.
My expectations are sadly quite low in this regard.
After all, I’ve already done the calculations and he has yet to find a flaw.
Bryan,
I’m really trying to understand your objections. I don’t know that I’ve got it, but it seems that you think SoD and others imply that net heat spontaneously flows downward from the colder atmosphere and warms up the even warmer ocean. That would be a clear violation of the 2nd law. Heat can be transferred from a colder body to a warmer body (refrigeration and air conditioning come to mind), but that takes the additional input of work. And if you include the generation of that work as part of the system, then overall the entire system experiences a net increase in entropy, and the 2nd Law is satisfied. That’s why air-conditioners can actually function and remove heat from the inside of our relatively cool houses and exhaust it outside into the much warmer atmosphere.
If you really think greenhouse theory implies or demands spontaneous net transfer of heat from cool atmosphere to warm surface, I honestly can’t see where that comes from. Greenhouse theory states that upward radiation from the surface is GREATER in magnitude than downward radiation from the atmosphere. So the NET energy transfer is upward from the warmer surface into the colder atmosphere, and eventually from colder atmosphere into even colder space. The net rate of radiative heat transfer is from warmer to colder in all the treatments here on SoD.
If you believe SoD or others have claimed more energy coming down from the atmosphere than going up from the surface, please point that out with specifics. I don’t think that’s there, but if you can find it, I’d really like to see it. And it won’t be much use to say it’s everywhere or it’s obvious. I’ve already looked pretty much everywhere and it hasn’t been obvious yet. Thanks.
Jimwit and DeWitt Payne
When I said to Jimwit
“The respective blackbody spectra of the hot and cold objects will be centred around different maxima characteristic of their temperature.
To simplify the discussion we will use a single wavelength to represent average wavelength.”
It turns out that this hastie abbreviation has led to needless confusion.
I just did it to save typing the full version each time.
I will reply to your other points when I get some time.
Bryan,
Thanks for the clarification. Reading back and finding the original posting quoted, I realize I missed the statement about using a single wavelength to represent a BB distribution. That did cause a bit of confusion on my end.
Jimwit
You asked originally about differences between myself and SoD regarding heat transfer.
I’m waiting for an answer from SoD to confirm his attitude to the second law.
I have my ideas but it would be incorrect of me not to let him set the record straight.
Once we have answer all will be revealed.
Bryan,
A restatement of SoD’s view of the 2nd Law shouldn’t really be necessary to answer my original request. I simply asked “If you believe SoD or others have claimed more energy coming down from the atmosphere than going up from the surface, please point that out with specifics.” Hasn’t SoD written enough already to pin down a specific example where he indicates this type of impossible heat transfer? Or, for instance, take the IPCC cartoon on heat balance, and tell which heat flow is contrary to the 2nd law. It should be as simple as that. Thanks
Jimwit
You previously asked
…..”So far, my understanding of SoD’s writings puts him in agreement as well. I’m not sure where the point of contention lies,”……
SoD previously asked on multiple occasions
…….”I realize it is not an easy answer to give for supporters of G&T.
The atmospheric radiation that “reaches the surface” and then..
.. a mystery.”…….
So I’m trying hard to answer both at once if SoD replies yes or no to a question asked twice
Bryan,
Your statement about electrical energy being thermalized in a laser or microwave doesn’t make sense, but I’ll ignore that for the moment. Let’s not use a laser or microwave then. How about a source at a temperature of 1000 K with a filter that only transmits radiation at a particular wavelength. Set the filter to 5 μm with a bandwidth of 0.1 μm and let the beam impinge on an object at a temperature of 300 K. Then set the filter to 15 μm with the same bandwidth. Will the temperature of the object increase in both cases? For the same temperature increase, is the net energy transfer the same? My answer is yes in both cases. 5μm and 15μm at at brightness temperature of 1000K will cause a net transfer of energy to an object at 300K. They are for this experiment thermodynamically equivalent.
Let’s calculate the effective temperature of a CO2 laser at 10.6μm. To do that we need to know the power level and the spot size. For a power level of 50W and a spot diameter of 300μm the radiant flux is 50/(π*0.00015^2) = 7.074E08 W/m2 for an effective temperature of 10,569 K. And that’s a very low power laser. Steel cutting lasers are in the kilowatt range.
[…] Comments « Find Stuff Out and Book Reviews […]
I explained ( March 6, 2011 at 8:50 pm) that moving 1000 J from a 280K body to a 1000K body (as an example) without any “cost” would violate the 2nd law of thermodynamics.
This is because entropy is reduced.
Bryan says (March 7, 2011 at 2:34 pm):
Even though it probably isn’t what Bryan means, the “halfway” is a nice summary of the problem. Like half agreeing about the scores in a football game.
Claim away that Man Utd won. It won’t change the result.
Bryan is still confused about the 2nd law and obviously not interested in resolving his problem because he still refuses to do an entropy calculation of a real problem or to find fault in my entropy calculations (previously cited).
Once he does either of these it will be clear exactly why he is considering only half of the 2nd law of thermodynamics. And therefore, not considering the real 2nd law of thermodynamics at all.
There is little point in explaining it again.
SoD.
I like the football analogy, well done!
Here’s another.
What a neat body swerve!
You deftly managed to avoid answering my question, so here it is again.
If we had a “black box” into which 1000J of of blackbody spectra type radiation centred around 15um was the input and the output was 1000J of radiation centred around 2um.
The transformation was accomplished without the expenditure of work.
Would this violate any thermodynamic laws?
I say that although the first law is not violated the second law is.
Do you agree, a simple yes or no will suffice.
Now here’s me trying without stint to answer a question that has troubled you for a number of years as to why G&T dont give explicit advice as to what happens to the “cold” radiation when it reaches the warmer body.
The question I ask here is a component of the answer.
Yes. That’s a no-brainer.
But Bryan, you already admitted that the rate of heat transfer changes with the temperature of the colder body. For parallel planes the rate of heat transfer E = εσ(T1^4-T2^4). So heat transfer goes to zero as T1 and T2 approach each other. It is obvious that that equation is the difference of the Stefan-Boltzmann equations for each plane. Therefore the flux from the colder plane must be absorbed by the warmer plane. Photons don’t have id tags identifying the temperature of the surface that emitted them. For infinite parallel planes, ε is irrelevant as long as it’s not identically zero as emitted photons will eventually all be absorbed.
Strictly speaking it should be E = σ(ε1T1^4-ε2T2^4)
DeWitt Payne
…..”Yes. That’s a no-brainer.”……
I’m not so sure SoD will say yes.
Suppose T1 = T2, e1=1 and e2=0.5. Would there not then be a flux from T1 towards T2 even though the bodies were at the same temperature? And presumably even if T1 were a wee bit less than T2?
Actually I am following this thread for comments on various texts rather than disagreements about the 2nd Law. It seems to me that physical laws are equations and that verbal formulations serve only a heuristic purpose in facilitating understanding. Disagreements need to be expressed in different equations. Otherwise the discussion is non-terminating. Which this one threatens to be 😦
Let’s examine Bryan’s problem to see what has been created.
We will call the first radiator (15μm), body A, and the black box, body B.
Radiation centered around 15μm means (I assume) that body A is at a temperature of 193K.
Radiation centered around 2μm means (I assume) that body B is at a temperature of 1449K.
For the purposes of this discussion let’s assume the emissivity of both Body A and Body B = 1.
Every second, A is radiating 79 W/m2.
Every second, B is radiating 250,000 W/m2.
Clearly there is a view factor to consider – i.e. not all of the radiation from A will reach B and vice-versa.
However, it is defined that 1000 J from body A (at 193K) reaches B and only that body B (at 1449K) emits 1000 J in total.
You haven’t defined it, but I assume that body B has no other source of energy.
Well, if my more complete statement of your unphysical problem is correct, then this is an impossibility.
The actual example is an impossibility because body A, the colder body, is radiating much more energy than body B.
This is why the second law of thermodynamics is true. If hotter bodies radiated less than colder bodies then the world would be a different place and there would be different laws of thermodynamics.
But hot bodies radiate more than cold bodies.
This is why entropy of a system never decreases.
Unfortunately, this simple fact is something Bryan will never understand and instead will keep making bizarre claims about energy from colder bodies being prevented from fulfilling the first law of thermodynamics by an imaginary second law.
I still await Bryan’s explanation of what is wrong with my entropy calculations in The Three Body Problem.
I will wait for a very long time for obvious reasons.
Correction, in the statement ‘The actual example is an impossibility because body A, the colder body, is radiating much more energy than body B.‘ instead of “much more”, it should read “more”.
I can’t say whether it is “much more” without knowing extra details.
And even if we said “the same energy as” the statement would still be valid.
S o D
I take you rather lengthy response is to agree with me that such a clever black box is an impossibility.
…”If we had a “black box” into which 1000J of of blackbody spectra type radiation centred around 15um was the input and the output was 1000J of radiation centred around 2um.
The transformation was accomplished without the expenditure of work.”….
Its a rational question.
I make no assumptions about how the radiation arrives or the workings of the black box.
You can have a continuing 1000J/s if that helps
Your question to G&T is also a rational question.
Jarras question is also straightforward.
If you do not reply I will assume you agree with me about the black box and will answer Jarras on that basis.
Let’s put the two objects at the focal points of a large evacuated ellipsoidal cavity with high eccentricity and with perfectly reflecting walls. That way we can ignore the view factor, I think. We’ll assume that each body has the same mass, surface area, emissivity/absorptivity and heat capacity. We’ll start by putting the 1449 K body in first and watch it for a while. Nothing happens, the temperature remains constant because there’s no place for the heat to go. Now remove the 1449 K body and replace it with the 193 K body. Again nothing happens. Now put both bodies in. The 1449 K body starts to cool and the 193 K body warms. The radiation from the 1449 K body is being absorbed by the 193 K body rather than passing through the focal point and the 193 K body is absorbing far more energy than it emits while the 1449 K body is emitting far more than it absorbs.
Eventually both bodies will be at 821 K. At that point, both bodies will again be emitting and absorbing the same amount of radiation, 25,761 W/m2. The rate of cooling and the final temperature depend on the initial temperatures. That can only happen if the radiation emitted by the cooler body is absorbed by the warmer body at all times so the net heat transfer is the difference between the amount radiated and the amount absorbed. The initial rate of heat transfer will be 249,000 W/m2 = 250,000-1,000. The final rate will be zero = 25,761-25,761.
Bryan,
Have your read Willis Eschenbach’s Steel Greenhouse post at WUWT?
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
Before you make an objection, be sure to check the comments as your objection has probably been made and answered there. In the end, we did manage to convince one commenter that his initial opinion was wrong and Willis was indeed correct.
Bryan:
I wrote this in plain English and bolded the text, so well done for finding it.
So clearly you don’t understand my answer at all.
Your example has a very cold body radiating the same – or more – energy than a much hotter body.
That’s why it is flawed. It doesn’t happen.
Once you have grasped this essential point then you will understand the second law of thermodynamics.
SoD
I have the difficulty of suggesting a test for a box which I think is impossible by the second law in order to elicit a straightforward answer from you.
Remember we are dealing with electromagnetic radiation.
Lenses and mirrors can be used to shape the radiation.
There is no problem arranging for the input; its the output of equal energy at a shorter wavelength that’s the impossibility.
Poor Jarras will be pulling his hair out.
I had already prepared an answer for him assuming we were in agreement.
I am fresh from Judith Currys site where consensus is the dominant trend.
Jarras thinks we are saying the same thing – are we?
Bryan,
If I’m pulling my hair out, I don’t think it’s why you think. Here’s my earlier post under the name “Jarras”
“Bryan,
Your description is very clear and I think we’re on the same page. Energy in the form of photons are always in a two way exchange between bodies of different temperatures in such a way that the net transfer of radiant energy (heat) both at any instant and over time is from the warmer body to the colder.
So far, my understanding of SoD’s writings puts him in agreement as well. I’m not sure where the point of contention lies, and you’ve noted above to be cautious that some would say heat flows the other direction. Can you elaborate? What form does this argument take in terms of climate? Thanks.”
At the time, I thought we had gotten to the same page. But as of yet, I’m still waiting for you to specify where greenhouse theory contradicts this understanding. I haven’t been able to find it myself. You’ve been very vocal about saying the contradiction exists so I’d think you could point me to it.
Jimwit
….”I’m still waiting for you to specify where greenhouse theory contradicts this understanding. I haven’t been able to find it myself. You’ve been very vocal about saying the contradiction exists so I’d think you could point me to it.”……
Which greenhouse theory are you referring to?
There are several or them.
Just like your user name!
Bryan,
As for which greenhouse theory, the SoD topical posts would work well.
As for the multiple names I’ve used, yes sorry about that and any confusion caused. Those were my first posts on this or any other WordPress site. When I later tried to register the name, it was already taken, so I came up with the new one and have been using it consistently since.
Jimwit
Thanks for the reply.
Ive never bothered registering on any of climate sites I visit.
I just put in my real first name and e-mail address and it works every time.
Peter2108,
Nope. That could only happen if the emissivity was not equal to the absorptivity. That probably means that the equation I used was incorrect. If the emissivity/absorptivity is less than 1, then the transmissivity and reflectivity need to be specified as well to account for all the radiation. Absorptivity + transmissivity + reflectivity must sum to exactly one. Suppose we assume that the emissivity of plane 2 is 0.5 and the reflectivity is 0.5 and T1 = T2 = 300 K.
Plane 1 emits 459.27 W/m2 and plane 2 emits half of that 229.635 W/m2 and absorbs half the energy emitted by plane 1 229.635 W/m2 so it’s in balance. Plane 2 sees the 229.635 W/m2 emitted by plane 2 and also 229.635 W/m2 reflected by plane 2 so it’s in balance too. No net heat transfer. If instead plane 2 had transmissivity of 0.5, then plane 1 is indeed losing heat, but not to plane 2. For convenience in these sorts of things, the planes are normally considered to be opaque and what isn’t absorbed is reflected.
But all reflected photons are eventually absorbed and the photon gas between the planes when the temperatures are equal will have a pure blackbody spectrum regardless of the emissivities of the planes as long as one of the planes has a non-zero emissivity. That’s why a box with a hole in it is a good approximation of a blackbody. The only reason it isn’t perfect is the hole. Radiation will leak in or out of the hole.
“that probably means that the equation I used was incorrect.” I was only querying the equation.
I was right the first time. For opaque parallel planes the rate of heat transfer is εσ(T1^4-T2^4) where ε = ε1*ε2. Or at least I think that’s now correct.
Steve
Please, keep up posting, you are making a very valuable contribution to the blogosphere and I personally know other people (besides me) who strongly appreciate this blog.
I understand it may be tiring sometimes (last two days are probably a good example) but … other people read your posts to get information
Time to move to next thread 😉
Jon
Who is Steve?
Is he perhaps Captain SoD ?
Bryan
Yes Steve is SoD. SoD has never tried to remain anonymous, you can find his name in several places in this blog and several other blogs, from WUWT to RealClimate.
I do not know him personally and my reasoning is not affected due personal reasons, if that is what you suspect. 😉
Best regards.
jon
Thanks Jon.
No I had no idea he was Steve Sod.
As I said to Jimwit above I use my name also in the past I said I have a Physics Degree.
I dont know what SoDs background is.
He publishes so much stuff that I was beginning to think he was a committee.
Jarras your original question.
….”For my part, I’m in complete agreement with your statements that heat can only flow from a warmer body to a colder. At the same time, I’m in complete agreement with SoD and others who outline how emmissive radiation (i.e. photons) are exchanged in a two-way flow between bodies of different temperatures. I don’t see any contradiction at all. All of my textbooks were consistent with both propositions.”…
…………………………………………………………………….
SoD previously asked on multiple occasions
…….”I realize it is not an easy answer to give for supporters of G&T.
The atmospheric radiation that “reaches the surface” and then..
.. a mystery.”…….
Here sustitute atmosphere = lower temperature surface and Earth surface = higher temperature surface.
……………………………………………………………………
Editted part of my original answer to Jarras
So what happens to the radiation from the colder object?
Three viable solutions.
1. Classical wave theory approach.
A single Pointing vector of magnitude with direction hot to cold.
2. Subtraction of photon streams will result in heat flow of the resultant
3. Absorption of XJoules/s of 15um centred radiation and emission of X J/s of 5um centred emission.
So the consequence for that square metre of having the hotter object there is the up-conversion of 100J of 15um centred radiation into 100J of 5um cenrted radiation.
This also raises the possibility of increasing the temperature of the hotter object
The increase in the “quality” of the radiation appears to violate the second law.
Now all three methods will give about the same result if properly applied.
I would guess SoD would pick 3 however this is now in some doubt.
I , and I would guess G&T would pick 1 or 2 because you get the correct answer without speculating about quantum mechanical effects of photon absorption.
What do the textbooks say.
(This partly mirrors approach of DeWitt Payne post on March 7, 2011 at 10:40 pm )
University Physics Young and Freedman Pg 484.
“If a body of absolute temperature T is radiating and its surroundings (at temperature Ts )is also radiating and the body absorbs some of this radiation.
If it is in thermal equilibrium with its surroundings then
T = Ts .
For this to be true the rate of absorption must be
= AεσTs^4 ”
If however T > Ts
Flow of heat from body = AεσT ^4 – AεσTs^4
H = Aεσ(T ^4 – Ts^4 )
“In this equation a positive value for H means heat flow out of the body.
Equation 15-26 shows that for radiation as for conduction and convection the Heat current depends on the temperature difference between the bodies.”
So it looks like option 2, mirrors the equation above.
.
Bryan,
Yes, to my unexpert eye, the equation you end with is right, and does correlate to option 2.
I’ve been looking around more on SoD site (I keep finding more and more – it’s quite extensive), and I still find it to be consistent with this same understanding.
So I’m still wondering – since we seem to be in agreement about this two way interchange of energy and net heat flow from hot to cold, do you still see SoD (or greenhouse theorists in general) advocating a mechanism that violates this flow (and thereby the 2nd Law)?
For instance, in the simple diagram which I’ve seen a number of places, the surface (hot) radiates 350 W/m^2 to the atmosphere (cold). The atmosphere radiates back 324 W/m^2. The net flow is 350-324 = 26 W/m^2 upward from the warmer surface to the colder atmosphere.
Granted, it’s a simple representation of a complex phenomenon, but it seems as consistent with the 2nd Law as the other example discussed above. This is the essence of what I understand to be the “greenhouse” effect, and I can’t see how it’s thought to be a violation of the 2nd Law.
There are ways to find good prices with meta-search engines, like gettextbooks.com. For the Hartmann book:
http://www.gettextbooks.com/search/?isbn=9780123285300
Beforehand, one might also wish to look into Worldcat. For the same example:
http://www.worldcat.org/title/global-physical-climatology/oclc/29356562&referer=brief_results
Jon,
Thanks for the encouragement.
Quality of Radiation
There has been recent comment on this thread about the quality of Radiation and more generally the quality of energy.
Some have doubted that such an entity exists.
It exists all right, several papers have been written about it.
The QUANTITY of energy can be expressed by the number of Joules and is generally covered by the first law of thermodynamics.
The QUALITY of radiation or energy is about how much work can be extracted from it in the given situation.
This is definitely the preserve of the second law.
If you google some key words you will find a number of hits.
Most are behind pay walls unfortunately.
However here are two sources for those interested;
Click to access ECOS2008_Pons_Part1_Web.pdf
http://www.scribd.com/doc/31017446/The-Second-Law-of-Thermodynamics
Bryan,
Thanks for the links. I took a quick look at the first, got interrupted, then tried the second. I now understand what you mean by “quality” of energy. I don’t remember any discussion of that term back in the stone ages when I went to school.
I may also have a better appreciation of your struggle with the “greenhouse” effect and the 2nd Law. If I do understand the problem, I’m afraid I don’t yet have an answer, but I’m fairly certain one exists. I don’t know if this is the same question you have. If so, great. If not, it’s now rattling around in my brain. Maybe SoD or someone can clear it up. I’ll try to make it an extreme example for clarity.
Question: Start with two bodies, one really hot (the sun) and one really cold (Pluto). Pluto is radiating very low energy photons. Some of those make it all the way to the sun. Assuming those low energy photons get absorbed at the sun, what kind of atoms absorb them, and what kind of energy state changes do they undergo? For instance, the photon hits a Helium atom at 6000K. What can this pitiful little low energy photon do to amp the Helium atom up even more? I know this is extreme, and maybe the answer is different than what’s going on with back radiation in the greenhouse effect. But it illustrates the question. What’s happening on the molecular level that can allow “hot” atoms/molecules to absorb photons from a colder body, and then typically radiate higher energy photons associated with their black body temperature? I’m convinced that must be happening, I just don’t know the specifics of the mechanisms.
Hi, Jimwit
I am not an expert on quantum mechanics or statistical physics, but I would say that:
* Not all the molecules are at the same energy levels, despite the average (most likely) temperature is, let’s say 6000 K at the Sun. There will be a distribution function properly describing the energy distribution at the Sun. Therefore, your low-energy photon might be absorbed by a molecule at a lower energy.
Additionally:
* Not all the transitions a molecule suffers must involve the highest energy levels, there will also exist vibrational and rotational transitions which need lower energy levels.
You can check sections 2.7 and 2.8 in Bohren-Clothiaux (Fundamentals of Atmospheric Radiation) for the details. The details are not easy, but I don’t find a problem really exist in the example you propose. Perhaps a real specialist in spectroscopy might be more useful to answer your question. However, I would say that specialists in atmospheric radiation are using the absorption coefficients used and measured by spectroscopists, so, I am quite sure that there is not a big error there.
Regards
Hi, Jimwit.
I am far from a specialist in spectroscopy but I think there are two questions that can be raised to explain that your “problem” is a no-problem.
The first one is that not all the molecules in the Sun are at the same temperature
The second one is that the fact that a molecule is excited at the electronic level does not mean that it can not suffer rotational or vibrational transitions (longer wavelengths…)
You can read sections 2.8 and 2.9 in Bohren – Clothiaux (Fundamentals of atmospheric radiation)
Therefore, I don’t think that what you present is actually a real problem. Additionally, my common sense tells me that specialists in atmospheric radiation take the absorption coefficients from spectroscopists’ data bases. Therefore, the values are probably OK and the theory is the same in both cases.
Regards
Jimwit
I pointed you to a paper written by G&T which analysed many of the problems with “Greenhouse Effect”.(GE)
I think that paper summarise my own objections so to list them again would be repetitious.
Some highlights from the paper.
1.The experiment by R W Wood
2. The figure of 33C calculated as due to GE is “a meaningless number wrongly calculated”
3. Average temperature of Earth Surface of 15C has no statistical rigour.
Two comment papers and one reply to comment exist in the public domain.
Read all of them plus threads on this site and form your own opinion.
However to make the matter even more complicated there is a further view.
A consistent argument could be made for a G E along these lines.
Perhaps everything G&T said was correct however since they wrote almost entirely about the troposphere it is limited to that extent.
Some hold that the real GE is above the troposphere(TOA) and G&T hardly mention it.
This is called the TOA effect.
Your other question
…”For instance, in the simple diagram which I’ve seen a number of places, the surface (hot) radiates 350 W/m^2 to the atmosphere (cold). The atmosphere radiates back 324 W/m^2. The net flow is 350-324 = 26 W/m^2 upward from the warmer surface to the colder atmosphere.”….
G&T wondered in their paper what such diagrams were meant to represent.
However in light of my answer above perhaps it doesn’t matter who is right or wrong on this issue.
I will point out my own opinion for what its worth.
The numbers seem to be calculated rather than experimentally derived.
The numbers are based on the Stephan Boltzmann Equation(SBE) and idealised black body emitters.
Instruments are used to measure radiation having been calibrated using SBE so the problem of circular reasoning might apply.
My opinion is then that the numbers are on the large side but again my answer above might mean that it makes little difference to GHE who is right or wrong on this issue
They represent total fluxes, something which G&T claim can’t be calculated either, if I remember correctly. G&T is so painful to read that I’m not going to bother trying to confirm my memory.
Flux definition for radiation:
R. Caballero, Lecture Notes in Physical Meteorology, section 5.6, p112
Click to access PhysMetLectNotes.pdf
The total flux is then the integral of the flux over all wavelengths.
Energy balance diagrams also include the total flux from convective heat transfer usually separated into latent and sensible. Latent heat is evaporation and condensation of water and sensible is movement of air with different temperature.
That G&T do not comprehend this simple fact is sufficient reason to disregard the whole paper. A substantial part of the paper is devoted to explaining why ‘greenhouse’ is a misnomer. Another part is semantic analysis of public statements attempting simple explanations to the general public. So what. If you know anything at all about how the atmospheric ‘greenhouse’ effect actually works, you’ll be tearing your hair and screaming after reading a few pages of G&T. It’s nearly as bad as some of those books described in fantasy fiction, P.C.Hodgell’s Book Bound In Pale Leather e.g., that will drive you mad or worse if you read them.
The arrows in the diagrams represent the normal the to plane of the hemisphere.
DeWitt Payne
Ive not had enough time to read carefully Rodrigo’s notes on this topic.
I will get round to it.
The diagram in G&T is not exactly the same as the K&T diagram.
Perhaps its a German version of a similar type.
Even supporters of the IPCC position distance themselves from such diagrams calling them cartoons and so on.
G&T likened them to graphs found in the financial press which seem to show something or other dramatic.
They could not find any similar diagrams in the scientific literature.
I dont think their criticism of the possible ambiguity of such diagrams is a fundamental part of their paper.
I’m still a supporter of the G&T viewpoint.
I find that this is true for people coming from a traditional physics background.
I often challenge opponents to be specific about their criticism of the paper.
So far as I’m aware there have been no fundamental errors found in the paper.
Jimwit declares himself new to the debate but asks some significant questions.
You are very well read perhaps you could help him by suggesting a reasonably compact case for the IPCC position.
Similarly a well grounded robust critical assessment of the IPCC position
Bryan,
As a newbie to climate science, I’m trying to focus on the basics as a starting point. The additional items you raise may be very worthwhile for discussion, but are irrelevant if the greenhouse effect violates the 2nd Law. That’s why I’ve been a little bull-headed about trying to settle that issue in my mind. I still feel there’s no contradiction, but your questions about quality of energy and “upconverting” have led me to conclude my basic understandings of absorption and emission needs some filling in of detail. Gotta run
Jimwit
Your strategy of focusing on thermodynamics is a good one.
Revise over your old degree level books.
Particularly the second law.
Then read G&T.
G&T say that only certain of the greenhouse theories fail the second law test.
Others may be quite sound on thermodynamics but place to much emphasis on radiative transfer.
There are a whole number of reasons why the theory might be wrong apart from thermodynamics.
Then again the greenhouse theory might be correct.
Look for experimental evidence.
Jimwit,
You could try starting with this article here:
https://scienceofdoom.com/2009/11/28/co2-an-insignificant-trace-gas-part-one/
Then read the rest of the articles in the series. It may not be all that compact, but it’s a good introduction. Skip the comments for now.
Bryan,
Referring to the energy balance diagram as a cartoon is not derogatory or distancing in this case. It’s not a graph or a table, it is a drawing or cartoon. It’s no more derogatory than calling a simple model of the atmosphere as a single slab, a toy model. It illustrates the point.
Jimwit,
The Kiehl et al energy balance papers are available online. Google Kiehl Trenberth 1997. The first and third links when I did it were for the pdf’s of the 1997 and 2009 papers. The diagrams or cartoons are just a quick summary of the data in the tables in the papers.
For a review of Gerlich and Tscheuschner, see
On the Miseducation of the Uninformed by Gerlich and Tscheuschner (2009).
As the only thing that Gerlich and Tscheuschner prove is that they haven’t actually read any papers – or even textbooks – on the last 50 years of climate science.
And so, they don’t address any real theories.
But they are handy for a refresher course in how to do double integrals and for a good demonstration of missing the point.
Jimwit:
Bryan has invented something called “quality of energy”.
The 2nd law is about something called entropy, a quite different topic.
The temperature of the source is what determines the “usefulness” of energy.
If you have 1000J in a source which has a temperature of 500’C, it is a lot more useful than the 1000J in a source with a temperature of 0’C.
This is because heat flowing from a hotter source to a colder sink can do useful work.
Gerlich and Tscheuschner have left the uninformed with the idea that energy from the atmosphere to the surface is in violation of the 2nd law of thermodynamics.
Either they don’t understand the second law of thermodynamics or they don’t understand what climate science says.
The choice is incompetence or laziness. I like to think they did it for a laugh (3rd choice), thus getting them off the hook.
In their follow up they demonstrate their operatic credentials by agreeing that the energy from the atmosphere “reaches the surface” but not explaining what happens to it after that. Not their problem to answer this little question apparently.
A joke this time only for more discerning readers..
SoD
…”Bryan has invented something called “quality of energy”. .
That’s very clever of me.
To have preplanted all those references in books and papers so as to mislead the truebelievers.
Well here’s another book I invented and preplanted.
The Physical Universe
An Introduction to Astronomy
Frank H. Shu
University of California, Berkeley
Page 80 contains a discussion on“quality of energy”. .
…..”The choice is incompetence or laziness.”….
Were you talking about yourself here?
…..” laziness.”…..
All you had to do was google the phrase and you would find lots of references particularly around solar energy capture systems.
……” incompetence.”…..
Well almost all your comments about the second law.
So in your case its incompetence and laziness.
Jimwit:
The answer is very simple.
Solids and liquids have continuous but varying absorptivity.
Gases have absorption lines across a very wide range of bandwidths.
It’s possible to come up with a scenario where no energy from a colder source can be absorbed by a hotter source but this would be very unusual (where the wavelengths of the colder source were “greeted” by a zero absorptivity at the hotter source.
And in the case of the earth’s surface and atmosphere this is impossible, as you can see in Absorption of Radiation from Different Temperature Sources.
In general, the paltry amount of energy from a very cold source (compared with the hot target) has a paltry effect on the temperature of the hot target. But it is still real.
First, a certain proportion is absorbed – defined by the absorptivity for those wavelengths.
Second, the energy absorbed has an effect on temperature – the first law of thermodynamics is still true – energy cannot be lost or destroyed.
I would add that, speaking in terms of mechanism, I see no problem in Jimwit’s example. A molecule in a highly excited electronic state can also suffer a vibrational or rotational transition of a lower energy level.
Jimwit, you can check sections 2.8 and 2.9 from Bohren-Clothiaux Fundamentals of atmospheric radiation for a longer and detailed explanation.
Re: Bryan comment 10297
They didn’t look very hard. The existence of the neutrino was postulated to explain the missing energy, momentum and angular momentum in beta decay. One can construct a balance diagram for any conserved quantity. Material balances are done routinely for chemical processes. Energy balance is fundamental to dieting, calories in and out must balance.
And that’s the problem. If you accept that it’s possible to construct an energy balance diagram for the surface of the planet and that K&T97 or KT&F2009 are reasonable approximations of such balances, then G&T’s conclusion must be wrong.
There are only multiple greenhouse theories in the minds of G&T. They spend a lot of time deconstructing explanations meant for a lay audience as if they were the actual theory and mechanism of the greenhouse effect. They aren’t. They do lay out some of the fundamental equations used in calculating the greenhouse effect but then claim they can’t be solved so, apparently, the effect itself can’t happen. But obtaining an analytical solution to a complex problem is actually quite uncommon in physics. Any gravitationally bound system containing more than 2 bodies has no general analytical solution. Does that mean we can’t calculate orbits. No. We use numerical analysis. The same goes for atmospheric radiative and convective energy transfer. The quantities like DLR are real and can be measured. The measured quantities are in good agreement with calculated quantities in both magnitude and structure (the spectrum of atmospheric IR radiation, e.g.).
G&T is fit only for lining bird cages or wrapping fish.
DeWitt Payne
….”And that’s the problem. If you accept that it’s possible to construct an energy balance diagram for the surface of the planet and that K&T97 or KT&F2009 are reasonable approximations of such balances, then G&T’s conclusion must be wrong.”……
The K&T numbers are highly suspect.
The diagram is not one I can find anywhere in a physics textbook.
It was not a big issue as I remember for G&T more a throw away comment.
However I was going to ask you to clarify a point you made in an earlier post.
Someone asked how big a greenhouse would have to be to have a significant radiative component.
You replied (if my memory is correct) that it would have to be the size of the troposphere.
This could be interpreted as meaning it has very little effect in the troposphere.
Is this your position?
Leonard Weinstein was of the opinion that the radiative effect (greenhouse) was essentially above the troposphere.
Do you agree?
Which numbers and suspect how? Have you read the papers or just seen the graphic? The papers, especially K&T97, list the sources and uncertainties in the numbers. Most of them are measured values or averages of measured values.
K&T97:
Click to access KiehlTrenbBAMS97.pdf
KT&F09:
Click to access TFK_bams09.pdf
A Feynman diagram is a form of conservation diagram that should be in most modern physics textbooks. You have particles in and out and you have vectors representing the momentum of the particles.
Energy and material balance is done more in engineering than physics and chemistry. In essence, though, it’s just balancing equations but using a diagram to illustrate rather than just equations.
I didn’t agree with Leonard Weinstein and don’t now. A radiative flux in the troposphere is necessary to maintain the lapse rate. The height of the tropopause, which affects radiative forcing, is determined by the surface temperature and relative radiative cooling in the troposphere, which are affected by greenhouse gas concentrations, compared to the warming in the stratosphere from UV absorption by oxygen and ozone created by UV absorption by oxygen. Convection does stop at the tropopause because there’s a temperature inversion there.
DeWitt Payne
…….” A radiative flux in the troposphere is necessary to maintain the lapse rate.”…….
What exactly does this mean?
The adiabatic lapse rate has nothing to do with radiation.
Bryan doesn’t believe the K&T numbers because he doesn’t believe it is possible for gases with such a low atmospheric concentration to emit so highly.
And so he has been trying to find a defect in fundamental physics to support this belief.
But let’s be clear – as DeWitt Payne has already pointed out – the numbers come from measurements.
Bryan doesn’t believe the measurements.
Bryan doesn’t have his own alternative measurements. Or any scientific source for alternative measurements. Or any scientific source for not believing the measurements.
I hope Bryan can accept that this is a science blog not a blog for opinions of people who believe their conceptual confusion over the science that has been carried out over the last 100+ years.
And I hope Bryan will take this to heart and return with evidence. Not claims with zero support.
And for anyone who wants to claim:
– they are welcome to do so but this claim requires evidence.
Kiehl and Trenberth provide plenty of evidence in their paper.
SoD says
…”Bryan doesn’t believe the measurements.”….
Which SoD knows to be untrue.
However now that your paying attention;
I believe mostly what I see.
Over at Claes site you informed him that no serious climate scientist thinks that the lapse rate is affected by radiation.
Now, your going to have to explain yourself to DeWitt Payne above who evidently does!
Bryan:
Time for you to clear up the confusion.
Please state which numbers of K&T are suspect, and your evidence for this claim.
Bryan said:
Over the last year I hoped you might learn something so that you are able to understand what people are writing.
Your understanding level is well below that physics degree you claim.
I’d like to assume good motives, which means that you are genuinely unable to understand the subject.
The alternative is that you are not interested in making the effort.
Either way, not much point explaining.
scienceofdoom
Instead of explaining to DeWitt Payne your belief that radiation does not influence the lapse rate; you choose to mount a personal attack on me for politely pointing out your recent post there.
Do I have to go to Claes site and dig out your quote?
Nothing much changes here at climate pseudo science.
I think you are an ideological hack and the only reason I am back it to reply to a post from DeWitt Payne.
Bryan:
The past year has demonstrated your lack of interest in substantiated your claims, or in trying to understand atmospheric physics.
1. As proof of this – we will never see any substantiation of your claim that the K&T numbers are suspect. See my earlier request.
I would love to be proved wrong.
2. Why do I need you to “dig out my quote from Claes Johnson’s website”?
I have already stated the same in Things Climate Science has Totally Missed? – Convection:
You don’t understand that the lapse rate being independent of radiation is not the same thing as DeWitt Payne is stating.
It is a different proposition. To understand the difference requires spending 5 minutes reading what he has written, rather than “word matching” to try and find imaginary flaws.
Bryan,
Answer my question and I’ll answer yours. Your use of misdirection to avoid answering inconvenient questions is becoming very annoying.
I say again: What numbers in K&T97 or KT&F09 are suspect and why?
DeWitt Payne
….”Your use of misdirection to avoid answering inconvenient questions is becoming very annoying.”…….
Well I think I asked you about why you appeared to think that you would need a greenhouse the size of the troposphere to show significant radiative results several months ago without reply.
Not that I disagree with your conclusion as it fits in very well with R W Wood and G&T.
However my next project it to look at the backradiation figure of 324W/m2 and compare it to the 390W/m2 surface radiation figure.
I will integrate the Plank function for the filtered line radiation and compare it to that of a continuous T^4 curve to find any discrepancies.
However I have no results as yet.
I have an enviable reputation for accuracy which I intend to keep.
Not for me the ‘shoot from the hip’ careless approach of SoD and others.
[…] the always excellent Grant Petty, A First Course in Atmospheric Radiation, the solar heating of the atmosphere, for a standardized tropical atmosphere: From Petty […]
The defense for Kiehl & Trenberth 1997 rests.
The prosecution, wounded by the unjust claim that it would present no evidence, retires without presenting evidence.
Oh, cruel barbs, oh, ad hominem defense.
Bryan,
I’ll answer the greenhouse question anyway as someone else may be interested.
Take a greenhouse where the vertical dimension is much larger than the horizontal dimensions so we can ignore horizontal variation in temperature at the surface. I’ll also assume constant illumination at the annual average Earth surface level and a perfectly transparent top. If we take the conditions of the US 1976 standard atmosphere with a surface temperature of 288.2 K, a lapse rate of 6.5 K/km and CO2 at 375 ppmv (MODTRAN default), then a greenhouse 500 m high would have a surface temperature ~0.4K higher than if the greenhouse were filled with a perfectly transparent gas. At 1 km , the difference would be ~2K. At 20 km the difference would be ~30K. The temperature difference would be larger if we looked at higher illumination, like the sun directly overhead ~1000 W/m2, but you would still need a very tall greenhouse to reliably see the difference.
[…] Global Physical Climatology, Hartmann, Academic Press (1994) – reviewed here […]
As a late addition to the list may I add ;
Heat and Thermodynamics by Zemansky.
This was the book we used at university and has the honour of being recommended by Feynman.
DeWitt Payne
As I indicated I was examining the K&T paper as against G&Ts criticism.
Firstly I wanted to check if G&T were correct in page 21 of their paper regarding a filtered spectrum.
They claimed that a filtered spectrum does not obey the T^4 law.
My method of checking was to integrate the Planck radiation function for a series of temperatures from 200K to 400K but to subtract from each one the radiation from 10um to 15um.
Then to plot the reduced remaining total intensity against a T^4 in a graph.
The result confirmed the G&T observation.
This means that any claims made regarding the atmosphere behaving as a black body or even a grey body must be false.
Straw man. No one who actually knows anything about the science claims that the atmosphere radiates as a black or gray body. Nor does the calculation of radiative forcing depend in any way on that assumption.
DeWitt Payne
My second examination of the K&T 97 paper was to see if the backradiation figure was physically possible.
To do this I compared the atmosphere spectrum profile from Petty(2004) as given in page 127
Physical Meteorology on line is Rodrigo Caballero’s Lecture Notes
Click to access PhysMetLectNotes.pdf
With the surface continuous spectrum.
My calculation is that all the radiation shown in this chart is reduced by 24.4% compared to the surface up radiation.
So the maximum possible back radiation would be 75.6% of the surface radiation!
K&T on the other hand have the reduction of only 8%.
Thus they claim a figure of 92% of radiation is surface bound.
Um, page 127 in my copy of Caballero is concerned with OLR, not DLR. You’re going to have to be more specific in exactly how you did your calculation. I know I wouldn’t be able to take Figure 5.14, if that’s the one you’re referencing, and convert it to actual radiance.
75.6% is actually about right for clear sky. But you’re ignoring that at any given time 60% of the Earth’s surface is covered by clouds. The lower surface of a cloud is a black body for all practical purposes. For low clouds Teff ≅ Tsurf and the ratio of back radiation to surface radiation is very close to 1.
I don’t see how you get 8% from 396 W/m2 up and 333 W/m2 down in TFK09. That looks like 16% to me.
Numbers (MODTRAN 100-1500 cm-1)
US76 atmosphere default conditions
clear sky; 0 km altitude
looking up: 258.673 W/m2
looking down: 360.472 W/m2
Cumulus Cloud: base 0.66 km, top 2.7 km; 0 km altitude
looking up: 359.216 W/m2
looking down: 360.472 W/m2
For 288.2 K, the surface temperature used as default for US76 and emissivity 0.98 for the surface, total emission = 391.16 W/m2. Leaving off the tails of the spectrum less than 100 cm-1 and greater than 1500 cm-1 loses 30.69 W/m2. If we add that to the cloud bottom emission we get 389.904 W/m2. Teff for the cloud bottom assuming an emissivity of 1 is then 288.0 K.
Since clear sky emissivity for the tails of the spectrum is effectively 1, we can add the same 30.69 W/m2 to the calculated clear sky emission and get 289.36 W/m2. And a quick and dirty average of clear plus cloudy is then:
0.4*289.36 + 0.6* 389.90 = 349.69 W/m2
For a very crude calculation ignoring that all clouds aren’t low clouds and that the cloud layers may overlap, that’s pretty close.
Oh, and 289.36/391.16 = 0.74.
Bryan,
With your “enviable reputation for accuracy” (self-declaration), why not identify where this claim:
..is made by atmospheric physicists. Not statements made by others about atmospheric physicists. If you get the important distinction..
And not simple models in introductory textbooks leading up to the Schwarzschild equation.
G&T are having a field day with you. You spend a long time carefully defending the claims that.. everyone agrees with.
Most of their statements are actually true.
They state, with huge gravitas – and sometimes double integrals which are their favorite party piece:
Then they claim or imply that no one in climate science knows that 1+1=2.
Then you rush around saying, I’ve checked, G&T are correct:
1 plus 1 does equal 2. G&T are right.
It is truly the gift that keeps on giving.
Like here, where you made my day and then continued to do so, especially here!. And there was more.
Joy.
Bryan,
To paraphrase W.F.Buckley on Ayn Rand:
Where G&T are correct, they aren’t original. Where they are original, they’re wrong.
DeWitt Payne says;
…”I don’t see how you get 8% from 396 W/m2 up and 333 W/m2 down in TFK09. That looks like 16% to me.”….
I used the older K&T diagram but the numbers are approximately the same.
390 up (subtract 40 from ‘window’ as everyone agrees this radiation doesn’t come back).
This leaves 350W/m2 up and 324 W/m2 “backradiation”.
(350-324)/350 = 7.4%
Not physically possible on the evidence produced.
DeWitt Payne says;
….”Um, page 127 in my copy of Caballero is concerned with OLR, not DLR. “…..
Yes as the title says ” absorptivity compared with Planck function at temperature indicated”
The portion not absorbed cannot be backradiated.
What would be even better for a backradiation calculation would be a mid nighttime surface up compared to surface down.
The K&T diagram is presumably a derived from year long total Earth surface study.
If so then under the title ‘backradiation’ must be included Solar energy absorbed by atmosphere i.e. has not an origin from the Earth surface.
Those data are available from SURFRAD. Go here and plot to your heart’s content:
http://www.srrb.noaa.gov/surfrad/pick.html
Bryan:
Correct.
Back-radiation is radiation emitted from the atmosphere incident on the surface.
Bryan on May 1, 2011 at 2:24 pm:
“Not physically possible” – on what evidence?
Measurements by pyrgeometers confirm.
Measurements by FT-IR confirm.
The radiative transfer equations predict this result. They successfully predict the spectral results and of course the integrated flux.
Confirmed by theory, confirmed by measurement.
Bryan on May 1, 2011 at 2:24 pm:
“Not physically possible” – on what evidence?
SoD says
Measurements by pyrgeometers confirm.
Measurements by FT-IR confirm.
I said above
This means that any claims made regarding the atmosphere behaving as a black body or even a grey body must be false.
I now conclude
So any instruments calibrated to expect black body radiation will be in error.
Bryan:
As and when you have a specific claim with some evidence – feel free to make it. Your random imaginations are not worth responding to.
Measurements by pyrgeometers
Pyrgeometers are calibrated to expect blackbody radiation!
When I pointed out that ….
.. regarding the atmosphere behaving as a black body or even a grey body..
T^4 radiation does not come from the atmosphere and you agreed that no serious person thinks otherwise!
“..is made by atmospheric physicists. Not statements made by others about atmospheric physicists. ”
Then what value can be placed on pyrgeometers reading atmospheric radiation?
If you are unable to answer, I will quite understand.
Bryan,
You clearly don’t understand how a pyrgeometer works. It’s a form of bolometer. It measures the total incoming EM radiation by integration over all wavelengths that the cover dome transmits. It doesn’t care about the structure of the radiation. If it saw 300 W/m2 from a laser at 10 μm it would read the same as if it saw 300 W/m2 from a black body. It’s just convenient to calibrate it with black body sources as the power levels are know with high precision and accuracy. The AERI FT-IR spectrophotometer is also calibrated against black body sources.
DeWitt Payne
Most people do not understand how a pyrgeometer works.
Perhaps even yourself.
For instance most people do not realise there is more radiation leaving the device than entering it.
Most people do not realise that it is only ‘reliable’ when measuring clear sky conditions.
I say ‘reliable’ meaning reproducible but not necessarily accurately.
It demands almost a standard atmosphere.
It works by relying on calibration to be accurate as it subtracts from a continuous Planck curve at a known temperature an assumed downward flux.
This assumed downward flux ‘the missing bit’ is what the device is supposed to record.
The blackbody calibration is correct for outgoing flux but incorrect for incoming flux as the T^4 law is not followed.
A bolometer requires an onboard heat sink.
That may be true when the sky temperature is lower than temperature of the device, but so what? Also, if you’re just talking about the disk itself, the difference will be small. An IR thermometer, which works much like a pyrgeometer only with less precision, and a restricted field of view, measuring the temperature of a mixture of ice and water is warmer than what it’s measuring. It will still read 0 C for the ice. It will also read the effective temperature of the sky. See for example Roy Spencer’s post here:
http://www.drroyspencer.com/2010/08/help-back-radiation-has-invaded-my-backyard/
A pyrgeometer does the same thing, but more accurately and precisely.
Cite please. Obviously there can’t be tall trees, buildings or mountains blocking a significant fraction of the sky, if that’s what you mean by clear sky. If you mean that it can’t measure a cloudy sky, you’re simply wrong.
Cite please.
All temperature measuring devices must be calibrated. What’s measured is the temperature of the disk. The disk will be in approximate radiative equilibrium with the sky (or whatever it’s pointed at) if it’s well insulated. There is no ‘missing bit’. The spectrum of the radiant flux seen by the device is irrelevant. There’s only the total radiant flux integrated over the wavelength range of the device window. In the case of a pyrgeometer, that’s ~4-50 μm.
Do you actually mean chilled? If so, only if it’s measuring very low fluxes like in a telescope.
We know that an insulated object can achieve close to radiative equilibrium with the sky. That’s how the Romans made ice in Palestine. That’s why frost or dew forms on grass.
Then there’s the fact that atmospheric IR emission can be calculated by RT programs and the integrated flux from the calculation is in good agreement with the pyrgeometer measured flux. The calculated spectrum is also in good agreement with high resolution FT-IR measurements. This sort of agreement is usually taken to mean that the science is well understood.
The sensors in bolometers are quite different to the sensors in pyrgeometer.
In a bolometer the resistance of the sensor changes.
This throws a Wheatstone bridge arrangement out of balance causing a current to flow, hopefully proportional to the physical reason for the resistance change.
Traditionally a blackened strip of aluminium was used.
For sensitive work a semiconductor material is used chosen to have a large temperature coefficient of resistance (dR/dT).
Pyrgeometers on the other hand use thermocouples as sensors.
Usually several of these are connected in series forming a thermopile.
Two dissimilar metals say A and B are joined, one junction held at a constant reference temperature and the other junction at the temperature to be measured.
If a gap is place between (say A) an EMF develops.
This is called the Seebeck EMF.
This depends on the temperature being measured but in a non linear way
Thermocouples use a power series relationship
E = a + a1T + a2 T^2 + a3 T^3………
The problem of the radiation flux causing implicit temperature change has to be allowed for.
In addition the temperature of the instrument body has to be factored in
Some think that since the heat loss is small Newtons Law of cooling might give more accurate results than Stephan Boltzmann.
In general science there have been several tests where this view has been vindicated.
However as far as I can make out all the Climate Science fraternity use Stephan Boltzmann.
Which brings us up to date with the much troubled instrument, the pyrgeometer.
There have been several false dawns where it was claimed that its associated errors had been eliminated.
When 1997 data using EPLAB blackbody calibrations is compared with similar temperature and conditions but using NREL blackbody 2004 we have a difference of 12W/m2 .
The current flavor of the month is the Philipona formula with four constants as against the traditional manufacturer EPLEY with two constants.
This (Philipona) has been used by a users group to try to get different pyrgeometers at the same location to give similar readings.
The groups using the Philipona formula claim to have achieved this.
Now if several working groups from around the world had assembled at one spot and announce that their mercury thermometers were in close agreement we would not be impressed.
It should be taken as a given that any measuring instrument should give the same reading for the same conditions.
The other advantage claimed for this method is that if in the future the readings are found not to correspond with reality they can be back corrected as they all have the same consistent error.
The absolute sky scanning radiometer(ASR) is used as a reference.
However the working groups want a second ASR with a different physical design to confirm the first as circular reasoning is still thought to be a problem.
Now any device that works back from an EMF to determine a power value and hence arrive at a temperature of the radiating source must use a formula to calculate the temperature.
If aimed at clouds, then T^4 will be correct.
However if pointed to the clear sky with its filtered spectrum, then T^4 will be in error.
So it’s a pyrometer, not a bolometer. My bad. It still measures temperature by measuring irradiance.
I also think you don’t understand the concept of effective temperature.
Thought problem:
Suppose we have a black body inside a totally reflecting cavity. We cut a small hole in the cavity and shine EM radiation that has the properties of power spectral density = 1.5 W/(m&178; cm-1) from 400 to 600 cm-1 and zero everywhere else. The total irradiance is then 300 W/m&178;. What is the temperature of the black body at equilibrium?
Answer: 269.7K
That’s the Teff of the radiation by definition. But we don’t care about the value of Teff, we just care about the irradiance. So we don’t care if the atmosphere obeys
Stefan-Boltzmann. The black body in the instrument does, also by definition. So the temperature of the black body, not the effective temperature of the sky, is used to calculate the irradiance.
No measuring instrument is perfect. Your hand waving is about errors of a few percent, well within the error bounds of K&T97 and TF&K09.
Backwards. A pyrgeometer measures irradiance by measuring the temperature of its internal black body. For a black body source, the measured temperature is the same as the actual temperature. For a gray body, the emissivity correction is simple. For a non-gray body, it’s not simple. But we don’t actually care about the temperature of the atmosphere for the purpose of measuring its irradiance.
DeWitt Payne
Have a look at this rather frank analysis of the problems with readings from pyrgeometers.
It should answer a number of the points you raise.
During the calibration process the Win is supplied by a blackbody.
Then constants are set.
Some of these constants are factory presets but others K(zero) is a variable offset.
The numbers of constants vary from method to method as you can see.
On page 2 you can clearly see that the expectation of the sky is that it depends on T^4 relationship which if pointed to clouds might match the blackbody assumption with grey body scaling.
However in the clear sky filtered spectrum it does not.
On page 1 the extent of the possible errors are given.
Error bias = 12w/m2 + or – a further 5W/m2.
This indicates that some readings may have errors of as much as 17W/m2.
You say that these errors are quite acceptable in the KT diagrams and I’m sure you are correct on that point.
Which is why I will not accept the KT diagram as representing anything like reality.
I think these instruments and the measurements they produce are an unacceptable basis for the IPCC proposals to dislocate the economies of the world.
Click to access stoffel_t.pdf
You’re moving the goalposts again. We’re not talking policy here. The question is about the current energy balance not some hypothetical future balance. The point being that if TF&K09 and K&T97 are even in the same ballpark, then G&T are simply wrong. There is downwelling IR radiation and it does cause the temperature of the surface to be higher than it would be in the absence of ghg’s in the atmosphere. If we know the local temperature, humidity and other ghg concentration profiles with altitude, we can calculate with pretty good accuracy the radiative balance. Closing the energy balance is tricky because it’s a lot harder to measure horizontal and vertical convective energy transfer.
The equation on page 2 for the sky irradiance is irrelevant to the actual irradiance from the sky. Tsky in that case is not the actual temperature of the sky, it’s the effective temperature and the emissivity is the effective emissivity. Those are calculated from the irradiance, not the other way around. It says nothing about whether the instrument is more or less accurate for measuring irradiance from clear or cloudy skies.
left out the # sign from the HTML code. I’ll figure this out eventually:
1.5 W/(m² cm-1)
DeWitt Payne
…” G&T are simply wrong. There is downwelling IR radiation “…..
G&T would agree with you!
Their paper has several diagrams showing two way radiative interaction between hot and cold surfaces.
This radiation from cold to hot surfaces is best thought of as the radiative component of insulation.
Just like the convective , conductive and phase change components they slow down the heat loss from the warmer surface.
G&T take issue with the magnitude of the DIR citing work by Schact? and others.
The KT diagrams seem to be a collection of calculation and experimentally derived sources.
The unfortunate pyrgeometer has no doubt helped them create a ‘problem’ that is pure fiction.
No doubt there will be further attempts to make the greenhouse theory have a bit more credibility.
But error ranges stretching to 17W/m2 and a CO2 ‘problem’ of 1W/m2 seems to be stretching credibility a bit far.
Bryan,
G&T state specifically that energy balance diagrams are nonsense, not that they are inaccurate.
Your saying that K&T97 or TF&K09 are inaccurate is a long way from saying that they’re nonsense. G&T are wrong. The fact that they are wrong demonstrates beyond a shadow of a doubt that they do not understand radiative greenhouse theory. Their lack of understanding is clearly willful arrogance. They didn’t do their homework, as the correct information is out there. It’s also obvious to anyone with half a brain what the arrows in the energy balance diagrams mean.
Even Miskolczi makes more sense than G&T.
DeWitt Payne
As with anything else the Climatological radiation balance diagrams will be somewhere in a spectrum stretching from Useful to Useless.
Although I must say if they are inaccurate it pushes them to the Useless end.
The specific one that G&T cite has no units but in other respects it resembles the K&T diagrams.
I can find no other examples from physics textbooks that resemble these diagrams.
The diagrams take great care to balance out radiation intensities but I don’t think there is a physical law that requires this.
Radiation can be absorbed and turned into chemical energy for instance.
The incoming figure of 342W/m2 is obviously the total Radiation from the Sun divided by 4 to account for the Earth radiating as a sphere.
But is a reduced power wholly illuminated sphere the same as an illuminated plate?
The plate approach gives a Earth system radiative temperature of -18C
However the K&T diagram approach would make this -59C.
This brings to mind the G&T comment about the 33K greenhouse effect as “a meaningless number wrongly calculated”
The average surface temperature of 15C gives rise to the 390W/m2 surface up figure.
Several papers have commented lack of statistical rigour in such an approach.
Sourceless fluxes.
The lack of any horizontal components of heat transfer such as winds and ocean currents.
The list goes on on and on.
I certainly would find no use for such a diagram.
Is this not the definition of Useless?
And a hundred years of measurements of radiation went up in smoke.
“Oh goodness“, they said, “we should have realized that this product only measures blackbody radiation. What idiots we were, now we have to redo absolutely everything. Pull the textbooks from the shelfs, raise Stefan and Planck from the dead, and let’s get to work.
And let’s get Bryan on the team, he really understands this stuff!.”
You should tell a lot of people about this discovery. You will be famous. People will write songs about you.
Rodrigo Caballero has moved to Stockholm University. His Lecture Notes on Physical Meteorology can now be found here:
Click to access PhysMetLectNotes.pdf
The old link no longer works.