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Discussion > Simon Abingdon/Jonathan Jones/Radiative transfer

RKS, you misuse the term 'false alternative' and fail to grasp my point. Richard Lindzen believes that within a traditional greenhouse picture there's no evidence of high climate sensitivity that is remotely likely to lead to problems. mdgnn believes that the traditional greenhouse picture is bunk and that there's no evidence of high climate sensitivity that is remotely likely to lead to problems. Spot the common element?

So if mydog's made some elementary mistakes it's not a big deal. It's strange he's not come across and accepted the challenge set by Jonathan and that doesn't increase my desire to try and decipher what he's on about. But because of what I've set out above it doesn't really matter. The only thing that would matter was if anyone thought that you had to think like mydog to be a climate sceptic. That would be very damaging. I'm sure nobody would stoop so low as to try and manipulate the sceptics at BH into such an intellectual straightjacket as a propaganda device.

Apr 14, 2012 at 6:24 PM | Unregistered CommenterRichard Drake

I think this thread has now run its useful course, at least as far as I am concerned, so I am going to stop monitoring it at this point.

Thanks to everyone who has contibuted, especially Simon Abingdon for starting the whole thing off.

Apr 14, 2012 at 8:18 PM | Registered CommenterJonathan Jones

De nada, simon

Apr 14, 2012 at 10:23 PM | Unregistered Commentersimon abingdon

Thanks J, S et al - a breakthrough thread for BH in my view.

Apr 15, 2012 at 4:34 AM | Unregistered CommenterRichard Drake

Yes, very helpful to have simple examples analysed in a way that is not open to misinterpretation and explained step by step.

One key point for me was that a black body behaves by definition as a black body - so it does not make sense to argue that it does not.

MDGNN has often stated that any chemical engineer will confirm what he says. That has puzzled me because textbooks on heat transfer by radiation seem to me to differ from his understanding of the mechanisms of generation and capture of radiation.

The website Science of Doom has a page devoted to a similar topic. Science of Doom makes a point of not accepting postings that question the correctness of established physics. I imagine that that site has had its share of postings from people who use Slaying the Sky Dragon as their physics textbook.

The popularity of the false meme that "second law of thermodynamics means photons from a hot body cannot be intercepted by a cooler body" is what, I imagine led Science of Doom to post a page Amazing Things we Find in Textbooks – The Real Second Law of Thermodynamics in which he shows the formulas for radiative transfer between two bodies, taken from around a half-dozen textbooks on heat transfer. He concludes:

There are some obvious explanations:

1. Professors in the field of heat transfer write rubbish that is easily refuted by checking the second law – heat cannot flow from a colder to a hotter body.

2 (...)

Apr 15, 2012 at 1:00 PM | Registered CommenterMartin A

so the basic pricnciple of backradiation is established but under strict conditions. How do these relate to the atmosphere of our planet? My guess is that there is no reason to suppose they would not still apply but, if so, mdgnn and others have some work to do....

Apr 16, 2012 at 12:24 AM | Unregistered Commenterdiogenes

Q. What do you call a 'private thread' where nobody else is allowed to comment?

A. Email

If you want a 'private' thread with a public audience, be prepared for heckling and talking amongst the natives, otherwise, take it to email. If people have learned anything about the 'skeptic' position, it's that the view that "we're the experts, watch us but don't talk" simply won't cut it.

Apr 16, 2012 at 8:22 AM | Unregistered CommenterTheBigYinJames

Martin A

Mdgnn is actually rather unusual among the group who complain that textbook physics is wrong about heat transfer by radiation. In the original thread he stated that disc1 would remain at the same temperature after the introduction of disc2. This is not an uncommon view amongst those who have never studied any physics but is not normally claimed by those who have. It is unfortunate that all the experiments that deal with this kind of heat transfer were performed long ago. I doubt that many fundamental investigations were undertaken much later than 1920 or thereabouts. By then it was pretty much accepted that the transfer between two radiating surfaces was a dynamic exchange of energy with zero net flow if the surfaces were at the same temperature. Prévost had worked out this idea of a dynamic exchange long before, even though he thought the surfaces were exchanging caloric.

The normal claim of the dissenters is that there is no exchange, only a one way flow from hot to cold. This avoids breaking the misunderstood 2nd law about heat always flowing from hot to cold and why cold things can't heat hot things. As we have seen in this thread there is no question of heat transfer from cold to hot and all the energy comes from the heater. To avoid the exchange of energy the two surfaces have to communicate in such a way that the cooler surface does not really radiate at all and the hotter surface only really radiates the net amount. There are a number of ways this is supposed to happen involving energy states or electromagnetic waves. To my mind these are unnecessary complications as they have to agree with the experimental net results that are also predicted by the standard exchange theory,

Naturally, it would be interesting to know whether there really was a one way flow rather than an exchange but this is not quite so easy to determine as one might think. Nearly all the thermal methods for detecting the presence of radiation involve measuring a temperature increase/decrease and in reality this is caused by the net heat flow after the temperature of the measuring surface is accounted for. We are still none the wiser about whether there is an exchange of energy between the radiating surface and the measuring surface or a simple one way flow. Mdgnn argues that the presence of the measuring surface has disrupted the communication between the two original surfaces and we are not really measuring what the flow would have been without the measuring device in place. This is how he can claim that our measuring surface is not detecting back radiation from one of the original surfaces to the other.

As far as I know some original experimental work by Debye and Huckel involved placing a sphere inside an evacuated shell. In principle the shell could be held at a fixed temperature say To and the sphere could be set at some different higher initial temperature T. The rate of change of temperature T was calculated for all temperatures as T moved to equilibrium with To. Knowing the heat capacity of the sphere allows one to calculate the rate of heat loss for all values of T. This can be repeated for different values of To. It was Stefan who noticed that the calculated heat flows out of the sphere could be fitted with the equation P = k(T^4 - To^4) The constant k was a function of the geometry and emissivity of sphere and shell. It should be noticed that this experiment is simply measuring the net rate of heat loss by radiation from the sphere as a function of its own temperature and that of the surrounding shell. It sheds no light on the question of exchange or one way heat flows. It is not, however, a big step to consider that the expansion kT^4 - kTo^4 represents two separate heat flows. The kT^4 term representing heat flowing from the sphere towards the shell and the kTo^4 term representing the flow from the shell towards the sphere. To add confidence to this view is the theoretical calculation of Boltzmann that a black body should emit radiation in proportion to T^4 without any consideration of its surroundings.

To my knowledge, nobody has ever produced any experimental proof that the Stefan law P = k(T^4 - To^4) is invalid if proper account is taken of geometry and emissivities. This is all we need to solve the problem with our two discs and so we can skip any arguments about whether there are two heat flows or only one. I think that mdgnn is unique in the physics and engineering fraternities in believing that it is possible for 100W or any other number to pass, by radiation, from one surface to another when T = To.

Now that Jonathan has released this thread it would be very helpful if mdgnn would stop by to defend his assertion or even say he has had second thoughts about the matter and no longer disagrees with the calculations.

Jorge

Apr 16, 2012 at 2:58 PM | Unregistered CommenterJorge

Jonathan Jones 10.19 AM April 12th: I do not claim a black body ceases to be a black body when it is radiatively coupled, only that with this party trick, the boundary conditions are such that Kirchhoff's Law of Radiation ceases to apply.

This is because the assumption of infinitely high thermal conductivity for both discs, a perfectly insulated back to Disc 1 [the heated disc] and radiation to empty space [perfectly absorbing enclosure] from the outside face of Disc 2 means that the inside face of Disc 1 has emissivity = 1, absorptivity = 0, the reverse for the inside face of Disc 2.

You get this by looking at how the activated density of states is filled. For Disc 1, it is filled by conduction from the interior so there are no empty sites to receive radiated photons,. For Disc 2, it is filled by received photons so there are no empty sites to receive kinetic energy.

This means zero impedance for conduction of [electrical] heat to the emitting face of Disc 1 so no internal temperature gradient, zero impedance for IR radiation from the inner face of Disc 1 to the inner face of Disc 2, no temperature difference, zero impedance for conduction from that face to the outer face.of Disc 2, no temperature difference, and finally zero impedance of IR radiation from the outer face of Disc 1 to the perfectly absorbing container walls so the isolated body S-B equation applies.in the same way as for the inner face of Disc 1 - no radiation in the reverse direction.

Therefore the answer to the problem is perfectly obvious;

1. There is no temperature difference between Disc 1 and Disc 2 so the temperature of Disc 1 must be the same as that of Disc 1; 204.9 K.

2. The thermal impedance from the outer face of Disc 2 to the electrical heater surface of Disc 1 is infinite. this is why I compare it to two diodes in series with a zero impedance gap across which EM radiation can travel only one way - Disc 1 to Disc 2.

As a rider, I am astonished that apparently highly trained individuals in academia have forgotten or were never taught that Kirchhoff's law of Radiation only applies at radiative equilibrium, and this problem is as far from such equilibrium as you can get.

And now we come to the 'coup de theatre'. Manabe and Wetherald 1967 started on the right tracks [they assumed as did Manabe and Strickland 1964 that SW IN = LW OUT at the Earth's surface] but some time later someone made a drastically wrong assumption that the radiated flux from the Earth's surface had to be at the S-B level for a black body in a vacuum and that the [presently 63 W/m^2 measured IR] is supplemented by 333 W/m^2 from the atmosphere as 'back radiation'.

This is total and complete lunacy because at best the atmosphere has an average emissivity of ~0.4 so it would have to be an average temperature of ~75 °C to generate 333 W/m*2. Furthermore, because 'back radiation' only appears because the radiometer to a cooler body is shielded from radiation from the hotter body, another damned obvious point to someone with the remotest scientific thinking capability, I would like to know when and who made this stupid assumption and whether they did so out of ignorance or whether it was scientific fraud.

DO I MAKE MY POINT....:o)

Apr 16, 2012 at 4:11 PM | Unregistered Commentermydogsgotnonose

As a point of explanation, two discs in a vacuum with insulated backs close together so view factor = 1 and the same emissivity, have on average half the activated states filled from inside by kinetic energy as are filled by photons from the outside. The Principal of Indistinguishability means those states decay randomly inside or outside with no history so no net energy transfer, emissivity = absorptivity and no temperature change.

Reduce the temperature of one disc and because Wien's Displacement Law means a lower upper cut-off energy, there are always empty states at the hotter surface which can be filled from the interior, so net energy transfer is to the cooler body AND emissivity > absorptivity, vice versa at the cooler body. In the limit with no free states, either 'conduction push' or 'conduction pull', you get the above case.

Now add a gas. Adsorbed molecules can receive the energy from activated states as well as those states decaying by emitting photons or into internal kinetic energy. Thus the gas reduces absorptivity and emissivity for photons by competing with the activated states.

This is why in air the radiation can never be at the S-B level in a vacuum. Also GHG molecules are likely to absorb more strongly so will have a greater effect. In air with natural convection a steel plate has to be ~100 °C before radiation > convection. For aluminium it's ~300 °C.

The convective boundary layer dominates heat transfer because if it's thicker you get less conduction across the laminar flow layer. The ultimate case in nature is a flat desert and the mirage. Here the boundary layer is very thick so most energy transfer is by radiation, effective emissivity is high because hotter molecules adsorb less frequently.and for a shorter time.

Back radiation does not exist in terms of being able to perform thermodynamic work because it is annulled by radiation from the hotter body.

Apr 16, 2012 at 6:17 PM | Unregistered Commentermydogsgotnonose

OK, for those not familiar, Kirchoff's law of radiation says that if a body is at the same temperature as its surroundings, it emits radiation at each wavelength equal to the coefficient of absorption as a function of wavelength times the Planck black body spectrum. The coefficient of absorption is the proportion of incoming radiation absorbed (as opposed to being reflected, scattered, or transmitted).

The reason for specifying that the body be at the same temperature as the surroundings is that emmisivity is temperature dependent. At a given temperature, emission and absorption coefficients are equal, but the coefficient at one temperature is not necessarily the same as the coeficient at another. This is in effect saying that the colour of an object may change with temperature.

As a rule, coloured objects are far more complicated than black bodies and grey bodies, which is why they are usually avoided in textbooks trying to teach the basics. The ability to absorb light depends on whether there are pairs of energy levels at the right separation, with the lower one occupied. The electronic band structure of solids can have strong effects here. As temperature rises, higher energy bands are occupied and the availability of allowed transitions changes. Things like thermal expansion changing the inter-atomic separations can change the band structure as well.

However, these effects are very material-specific, and you're not going to get any general effects applying in all cases. And it still doesn't affect the point that you do get back-radiation, it does get absorbed, and there are plenty of materials that are sufficiently black across the thermal waveband that such considerations are negligible.

Jonathan Jones's discussion is specifically for black bodies, and a black body is by definition 'black' at all wavelengths and all temperatures, its absorptivity and emmisivity both equal to 1. So Mdgnn's claim that the disk surfaces can have one-way absorptivity/emmisivity equal to zero is a startling claim.

Kirchoff's law does apply to black bodies (and grey bodies if you are careful about how you define them) even if temperatures are not equal. It's only coloured bodies where you have to be careful.

How does Mdgnn do this? He speaks of the "activated density of states" which I don't understand, but I'm guessing is something to do with the Fermi level; he says this is "filled by conduction from the interior" which again I don't understand, there is always a fuzzy boundary of partially-filled levels around the Fermi level, and it makes no difference how they're filled; and for the second disc he says it is "filled by received photons so there are no empty sites to receive kinetic energy". This seems to be an attempt to say that the photons have saturated the upper level of every pair of levels, so heat from disc 2 cannot do so and lead to emission. I can make no sense of this. You won't be able to saturate all the energy levels (think about reducing the intensity of the incoming radiation). You can only fill the upper level of every pair if you have a 100% population inversion up against the bottom of a bang gap. And even if the energy levels are supplied by photons, they can still drop down and emit.

The next step leaps from the lack of internal thermal impedence within each disc to saying there's no thermal impedance across the vacuum between the plates. This would appear to imply that vacuum flasks don't insulate very well. And even given the above one-way emission, I've no idea how this is implied.

I'm afraid I gave up at that point. I assume that not understanding the start of the argument, that's going to cascade through the rest of it.

Probably the best place to start would be to clarify what is meant by "the activated density of states". That's where I lost it.

Apr 16, 2012 at 9:11 PM | Unregistered CommenterNullius in Verba

Hi NiV: what you must appreciate is that with my materials' scientist hat on and defining the activated state intermediate between kinetic energy and emitted or absorbed photon, I am trying to develop a simple idea explaining why emissivity and absorptivity are not constant, but variable depending on the divergence from thermal equilibrium. Also to explain why Prevost Exchange Energy or what physicists call the photon gas communicates between the two bodies in radiative thermal exchange to control heat transfer. In the end it's simple statistical thermodynamics.

It is of course nothing like the Fermi level. In a gas it is the number of occupied GHG vibrational levels as a proportion of the total concentration. Nahle has described this in more detail with the energies: http://jennifermarohasy.com/2011/03/recycling-of-heat-in-the-atmosphere-is-impossible/

For a solid it is the surface states with broken bonds thus less tightly bound, giving the range of vibrational states that can couple with internal or external energy. Unlike any normal heat transfer problem, the disc-disc problem can never reach thermal equilibrium [there is no increase of thermal impedance] therefore according to this idea, emissivity and absorption diverge to their limits for this particular problem.

The gap is not a thermos flask. It is emissivity = 1 opposite absorptivity = 1, zero impedance to radiative heat transfer one way, infinite the other. A vacuum flask is metallised to give low starting emissivity and absorptivity, also the glass is not a particularly good conductor and heat transfer the other side is poor.

This explains what happens in a vacuum. Now, if you go to any Heat Transfer handbook you will see tables of empirical heat transfer coefficients for coupled convection and radiation, usually as a function of dimensionless numbers. I have measured convection and radiation many times in designing industrial processes. Radiative flux does not exceed convective flux at ambient temperatures even for a good black body.

So, the Trenberth cartoon's claim is totally unrealistic. You can't get S-B B-B radiation in the atmosphere for a solid until it's >~600°C. As for the 333 W/m^2 'back radiation', that is also ludicrous because you'd need air at >~75°C for emissivity = 0.4. I think it was a desperate guess by someone who had never actually measured convection in association with radiation and like most people apparently, failed to understand Kirchhoff's Law of Radiation for which Prevost Exchange Equilibrium is the corollary.

If I have not explained things well enough, please ask again. Remember, all process engineers are taught what I was taught. We have all looked with amazement at what climate science claims. Had a very interesting chat with a metallurgist from Wollongong steel plant who had analysed things exactly as had I, even to pointing out that the pyrgeometers are glorified thermometers. If you have ever worked in a steel plant, you have used many different pyrometers operating on different principles.

To summarise; back radiation is bunkum, a perpetual motion machine of the 2nd kind. Radiation is much less than S-B unless you go to very high temperatures. And here is an article by someone else on this same subject, except he's got some things wrong: http://wattsupwiththat.com/2011/10/26/does-the-trenberth-et-al-%E2%80%9Cearth%E2%80%99s-energy-budget-diagram%E2%80%9D-contain-a-paradox/

Apr 16, 2012 at 10:16 PM | Unregistered Commentermydogsgotnonose

Hi MyDog,

"what you must appreciate is that with my materials' scientist hat on and defining the activated state intermediate between kinetic energy and emitted or absorbed photon"?

I asked what is meant by "the activated density of states". Are you saying you meant the density of activated states? And what is an activated state "intermediate between kinetic energy and emitted or absorbed photon"?

Apr 17, 2012 at 7:03 AM | Unregistered CommenterNullius in Verba

Oh dear, mydog. I fear we'll have to agree to disagree.

As NiV says, black bodies have - by definition - absorptivities and emissivities of 1. Experimentally, many objects approach this behaviour. I can't see any precedent for the behaviour you describe. To me it looks as though you proceed backwards from the supposed non-existence of back-radiation and then redefine terms at will to maintain this.

Too bad!

Apr 17, 2012 at 7:52 AM | Unregistered CommenterJeremy Harvey

From what I can glean from dog's posts...

You get this by looking at how the activated density of states is filled. For Disc 1, it is filled by conduction from the interior so there are no empty sites to receive radiated photons,. For Disc 2, it is filled by received photons so there are no empty sites to receive kinetic energy.

Is he actually saying there's no more 'room' (i.e. unfilled 'sites' to receive kinetic energy) which just sounds... unlikely.. to me, sorry. The idea that a body already at maximum emission has no more capacity to heat up because all available 'spaces' are being used in the creation of emitted photons? Have I read that right?

What is supposed to happen to the photons which fall on this unaccepting surface? Reflected?

Apr 17, 2012 at 8:16 AM | Unregistered CommenterTheBigYinJames

NiV: my nomenclature may be out of order [literally] but I hope the meaning is clear.

JH: back radiation defined as the transmission by radiation of heat from a cooler to a hotter body is a 'perpetual motion machine of the second order', forbidden by the 2nd Law of Thermodynamics.

BY: The physicists to whom I have talked imagine a photon nearing a filled site is backscattered. They also talk of Prevost Exchange as the 'photon gas', bouncing between the bodies but not being converted into heat. The rest is simple statistics: when there is no temperature change of a body in radiative equilibrium, by definition half the 'free sites' are filled from kinetic energy [heat], half from external energy therefore when you impose extreme non-equilibrium, you bias that statistic up or down.

To all; I am trying to do is to go beyond where Planck left off and mechanistically explain why Kirchhoff's Law of Radiation only applies at thermal equilibrium. It is up to the proponents of the original 'mathematical trick' to explain why they are absolutely certain that a black body with two surfaces connected to a perfectly thermally conductive sink/source meaning no resistance to transfer of absorbed incident radiative energy by internal kinetic energy to radiate from the other side, could as was claimed output twice as much energy as was being received!

I corrected the area error but rejected 'back radiation' because of it breaching the 2nd Law. Ultimately, as I argued with BY, if you accept 'back radiation' heading back to the emitter then adding to the energy it emits, you must also explain why the Stefan-Boltzmann constant is not a constant. rather an operational parameter which increases when the radiating body is surrounded by gas molecules.

This constant is related to Boltzmann and Planck's constants and the speed of light so seems immutable. Therefore back radiation cannot do thermodynamic work.

Apr 17, 2012 at 9:39 AM | Unregistered Commentermydogsgotnonose

Apr 17, 2012 at 7:52 AM | Jeremy Harvey

You say "To me it looks as though you proceed backwards from the supposed non-existence of back-radiation and then redefine terms at will to maintain this."

I say "You might very well think that; I couldn't possibly comment"

Jorge

Apr 17, 2012 at 10:05 AM | Unregistered CommenterJorge

My Dear Jorge: I did not the non-existence of back radiation, it is expressly forbidden by the 2nd law of Thermodynamics and is classified as a 'Perpetual Motion Machine of the 2nd Kind'.

I then went on perhaps foolishly to pander to the mob by arguing why the mathematical trick you postulated [in error] wasn't valid if you consider, as is correct, that Kirchhoff's Law only applies at thermal equilibrium. But then, it has been argued that I do have an over-generous personality!

It is up to you to prove a mechanism by which back radiation is thermodynamically possible. No offence meant but I have detected a possible debating trick to divert attention from the flaw in your argument.

Do a search for Zlop who believes he/she or the computer automaton it could be, can beat the 2nd Law!

And as a rider, no process engineer or past physicist has ever measured 'back radiation' defined as a multiplier of thermodynamic work above the input level and this goes back to the physics' greats!

Apr 17, 2012 at 10:18 AM | Unregistered Commentermydogsgotnonose

I did not assert the non-existence of back radiation.......

Apr 17, 2012 at 10:29 AM | Unregistered Commentermydogsgotnonose

mydogsgotnonose

Do a search for Zlop who believes he/she or the computer automaton it could be, can beat the 2nd Law!

`Twas brillig, and the slithy toves
Did gyre and gimble in the wabe;
All mimsy were the borogoves,
And the mome raths outgrabe.

Apr 17, 2012 at 12:20 PM | Unregistered CommenterRichard Drake

Apr 16, 2012 at 4:11 PM | mydogsgotnonose

Hi again,

Of all those who have responded I am probably the one who is least committed to the "real" existence of back radiation. However albeit theoretical, it does seem to lead to calculations that are in accord with observations. My evidence for making this assertion is that I have never heard of any experimenter claiming that these kinds of Stefan/Boltzmann/Planck calculations have been falsifed by real measurements. Perhaps the two disc experiment, if performed, would be the first time!

As you say, it was never my intention to propose some kind of trick question so I am interested in which of the many assumptions involved turned it into one. Perhaps we could take them one at a time.

For example I am quite happy to allow disc 2 to have some thermal resistance so that, for example, it needs 1º between faces to transmit the 100W. In my mind this just means the inner face will have to be at 205.9K instead of 204.9K. According to my sums this will cause the inner face of disc 1 to be at 244.3K instead of 243.7K. This is less than 1º which is not surprising in view of the T^4 relationship. Is it your view that the two inner faces will both be at 205.9K? More generally, do you think that both inner surfaces will be at identical temperatures regardless of how much we increase the inner face temperature of disc 2 by arbitary changes to the thermal resistance of disc 2?

Jorge

Apr 17, 2012 at 1:57 PM | Unregistered CommenterJorge

Apr 17, 2012 at 7:52 AM | Jeremy Harvey:

As NiV says, black bodies have - by definition - absorptivities and emissivities of 1. Experimentally, many objects approach this behaviour. I can't see any precedent for the behaviour you describe.

I shall be grateful if you would tell me, what is the nature of these "objects"?

I mean, are they made of exotic, fabulously expensive rare materials or are they commonplace materials, cheap and readily available?

If I may ask a second question, on Apr 17, 2012 at 1:57 PM | Jorge said:

Perhaps the two disc experiment, if performed, would be the first time!

I assume that Jorge means the set up described in the first comment in this thread:

Let's try running through this very slowly from first principles. Start with disc 1. This is assumed to have one side perfectly insulated and non radiating, and the other side a perfect black body (an ideal radiator) which is exposed to space (which for convenience we will take as having a temperature of 0K, corresponding to truly empty space, rather than the 2.7K of local space, which is filled with the cosmic microwave background; the difference is negligible so let's stick to the simple case). Disc 1 also has a heater, but we'll start with that turned off.

To your knowledge, has such an experiment (at least with conditions as close as practical to those described) been performed in the laboratory with a temperature sensing device attached to disc 1?

I am asking this because, on Apr 11, 2012 at 10:53 PM | simon abingdon said:

An experiment will confirm the predicted result beyond dispute.

To me this implies that this particular experiment has not yet been performed.

Thank you.

Apr 23, 2012 at 8:17 AM | Unregistered CommenterBrownedoff

I think I showed in my discussion post 'Where has all my Back Radiation gone?', without a lot of navel gazing, how the effect on the atmosphere of back radiation from CO2 is virtually zero - or in fact zero if mydog's assertion it performs no work is correct.

Apr 23, 2012 at 10:08 AM | Unregistered CommenterRKS

I shall be grateful if you would tell me, what is the nature of these "objects"?

Black silicone paint tested by NASA had reflectivity less than 5% over a wide range of wavelengths.


I remember being told that a stack of razor blades bolted together is about the blackest thing you can imagine. I found just this: BLACK REFLECTIONS.

Laser beam dumps have relectivity <10^-4 therefore, presumably, emissivity > 0.9999

My son, then a student of molecular physics, remarked that very few photons ever escaped the fur of out black Rex rabbit.

Apr 23, 2012 at 2:32 PM | Registered CommenterMartin A

Apr 23, 2012 at 2:32 PM | Martin A

Thank you very much.

So, to do the 2 disc experiment, we need:

2 sets of razor stacks, call them RS1 and RS2,

3 small pieces of fur from a Rex rabbit to affix to three "sides" of RS1,

1 vacuum tank (an old propane gas tank?) either coated internally with NASA paint or lined with Rex rabbit fur, with access hatches and instrumentation ports,

1 small heater to attach to RS1,

2 thermocouples for RS1 and RS2.

Anybody reading this with access to a university engineering department willing to give it a go?

What could possibly go wrong?

Or, has anybody reading this witnessed the execution of this exact disc1 - disc2 proposal?

Thank you.

Apr 23, 2012 at 3:23 PM | Unregistered CommenterBrownedoff