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Black body problem

Someone (anonymous http://en.wikipedia.org/wiki/Special:Contributions/81.151.55.49) reverted a contribtion identifying Earth as a 'black body'. There is nothing about planet Earth that makes it behave in anyway like a thermodynamic black body.--Damorbel (talk) 15:05, 19 December 2010 (UTC)

It's true the earth doesn't behave quite like a blackbody, but it seems odd to call a guy's edit vandalism (as you did in your revert edit summary) when he is just removing something that he had just added. Only the first sentence was about right; the rest was nonsense, and he probably thought better of it. Dicklyon (talk) 16:09, 19 December 2010 (UTC)

Dicklyon, you write :"It's true the earth doesn't behave quite like a blackbody" Indeed, then should any part the albedo be included in the temperature calculation? And if so, why so? --Damorbel (talk) 17:12, 19 December 2010 (UTC)

The incorporation of albedo and differences of absorption and emission at different wavelengths is discussed in several related articles and the sources that they cite; you can incorporate some of that here if you think it's relevant. The brief discussion in this article looks like it's got albedo correctly incorporated already. Dicklyon (talk) 17:19, 19 December 2010 (UTC)

I don't remember seeing a link; care to remind me? This really is not serious, you would think all Nobel prize winners from Planck onwards just got the 1st & 2nd Laws of thermodynamics wrong!.

There is more. Tell me does the mirror finish to a Dewar flask radiate heat like the reflective part of the Earth's 'black body'? Or do some reflective bodies radiate 'like a black body' and some not. If so, how do you tell the difference? --Damorbel (talk) 18:20, 19 December 2010 (UTC)

The brief discussion in this article that I referred to is in the section you were editing: Black_body#Temperature_of_Earth; I haven't chased down the related article, but I expect you can find them. As for the rest, I'll avoid feeding the troll. Dicklyon (talk) 03:51, 20 December 2010 (UTC)

"I'll avoid feeding the troll." I suspect you are being abusive, posting remarks like this. Would you care to revise it? I have raised a technical question in connection with the article, your response does not add anything useful. --Damorbel (talk) 09:51, 20 December 2010 (UTC)

Posting a sort of technical question with the intent of stirring up conversation, even when you know the result won't be useful, is trolling, and it was my estimation, based on your history, that that was what you were doing. I didn't want to engage in that here. I subsequently read on your talk page your explanation of your model, which led me to believe that maybe there was a chance you could indeed be simply confused, so then I tried that hypothesis, taking the discussion there to keep it from bothering others who are not so confused. That doesn't seem to be working either. I'm doing my best to assume good faith. Dicklyon (talk) 11:01, 20 December 2010 (UTC)

I think you should tackle my question: "should any part the albedo be included in the temperature calculation? And if so, why so?" The reason I put the qustion is because it is where the article disagrees with the well established law of Kirchhoff on thermal radiation, a law that this article should reflect. --Damorbel (talk) 12:54, 20 December 2010 (UTC)

And I think you should stop trolling, since I already answered yes and explained on your talk page why your assertion is wrong. Kirchhoff did not make such mistakes as you do. Dicklyon (talk) 15:22, 20 December 2010 (UTC)

"And I think you should stop trolling" I am unhappy with this kind of remark, I ask you once more to consider withdrawing; it is a personal attack on me and I suggest you should consider what part of Wiki policy justifies it.

Also I am very unhappy with your explanations of the thermodynamics of Black bodies. I am familiar with Kirchhoff's text in which he describes the calculation of temperature through radiative heating as being 'independent of the nature of the body under consideration'. This is your contribution on my talk page you are making assertions about Kirchhoff's 1862 paper, he started by considering if there was wavelength sensitivity, disproved it, and expanded his conclusions to the general case. --Damorbel (talk) 22:02, 3 January 2011 (UTC)

I haven't read the Kirchhoff, but I thought we had agreed that he was speaking of systems in thermodynamic equilibrium, and I thought we already agreed that the discussion here is not about a system in equilibrium. And you have a long history of trolling the blackbody- and greenhouse-effect-related articles, don't you? Your article and talk edits based on amateur thermodynamics in contradiction to all reliable sources has been in progress since 2008. I'm sorry that my attempts to help come up short. Maybe you can find a physics tutor and learn your way around your confusion. Dicklyon (talk) 01:38, 4 January 2011 (UTC)

Per unit surface area per unit solid angle?

In the Equations governing black bodies section, it is stated that:

"I(ν,T) dν is the amount of energy per unit surface area per unit time per unit solid angle emitted in the frequency range between ν and ν + dν by a black body at temperature T;".

It is my understanding that per unit surface area and per unit solid angle mean the same thing in this case? Therefore this statement is misleading, suggesting a dependence on both, not either.
Also, in the Planck's law article:

"This function represents the emitted power per unit area of emitting surface in the normal direction, per unit solid angle, per unit frequency"

is stated so there seems to be a contradiction between articles.
I just wanted to post this here before removing "per unit surface area" from this article just in case I am missing something.
Toomuchrockcankill (talk) 13:14, 18 January 2011 (UTC)

Two different things: The surface area refers to the surface area that is emitting the radiation, the solid angle is the solid angle into which the radiation is emitted. The radiation is isotropic, so no matter what the angle it is emitted at, its the same amount being emitted into a given solid angle, so I would say that the second sentence is not right, the "in the normal direction" statement is correct, but too restrictive. PAR (talk) 16:00, 18 January 2011 (UTC)

Ok, that makes sense, but is not immediately clear in the article - I'll probably edit that at some point shortly unless anyone else gets there first. I think that "normal direction" in the other article is just put there as it is the direction the surface is cut in for summing the surface area rather than anything to do with the directivity of the radiation.Toomuchrockcankill (talk) 21:21, 18 January 2011 (UTC)

"normal direction" is typically included when it is necessary to multiply the value by the cosine of the angle for radiation that is not normal. These are known as Lambertian objects. I agree that this is very confusing, but even text books make this complicated. Q Science (talk) 22:13, 19 January 2011 (UTC)

That's right and I was wrong above. A black body is a Lambertian radiator, which means that "in a normal direction" is correct, and should be noted in the black body formula. The radiation from the radiating area element falls off as the cosine of the angle from the normal. The "projected area" also falls off as the cosine, and thats why the area element has the same brightness when viewed from any angle. PAR (talk) 05:27, 20 January 2011 (UTC)

Thanks - that's kind of what I meant, but put properly... I'll edit this now then. Toomuchrockcankill (talk) 14:49, 20 January 2011 (UTC)

Temperature relation between a planet and its star - This is wrong

Basically this derivation says that by conservation of energy, incident = reflected + emitted.

$E(T_{S})=\alpha \,E(T_{S})+B(T_{E})$

where $E(T_{S})$ is the radiation falling upon the earth from a black-body sun at temperature $T_{S}$, $\alpha$ is albedo and $B(T_{E})$ is the black body radiation of the earth at temperature $T_{E}$. This equation violates Kirchhoff's law of thermal radiation. A black body cannot be reflective. A better analysis would be to assume that the Earth is a grey body with absorptivity $A$ and emissivity $\epsilon$, which are equal by Kirchoffs Law. Albedo is simply part of the reflectivity which is $1-A$.

By conservation of energy:

$E(T_{S})=(1-A)\,E(T_{S})+\epsilon B(T_{E})$

or

$E(T_{S})=B(T_{E})$

Solving for the Earth's temperature yields

$T_{E}=T_{S}{\sqrt {\frac {R_{S}}{2D}}}$

which, using the given parameters gives an Earth temperature of 5.517 degrees C. Greenhouse effects will, of course, raise this temperature. Note that this analysis cannot be used for the moon, because of the wide swings in temperature on the light and the dark side. Since black body radiation is not linear in temperature, but rather the fourth power of temperature, this means that the moon will be radiating (losing) much more energy on its light side than its dark side, causing the effective temperature to be lower.

I don't want to replace the present referenced (but incorrect) analysis with an unreferenced one without consensus. PAR (talk) 21:15, 9 January 2010 (UTC)

The uv-vis absorptivity of the Earth is not equal to the long wave IR emissivity. In addition, ice and clouds are highly reflective. You must also consider that water (the oceans) loses energy slower than solid surfaces and that a lot of solar energy is directly absorbed by the atmosphere. Q Science (talk) 23:41, 9 January 2010 (UTC)

"The uv-vis absorptivity....", perhaps so but this is no reason for the frequently made assumption that "the Earth radiates like a black body in the infrared". A non-black body is a less efficient radiator than a black body because it has non-black properties i.e. it reflects, it is coloured, it is (partially) transparent, all of which take away from its "blackness", hence it has a lesser ability to radiate in total, in which ever part of the spectrum the radiating takes place, thus it is incapable of "radiating like a black body" i.e. with the equivalent efficiency of a black body. --Damorbel (talk) 12:51, 10 January 2010 (UTC)

"like a blackbody" means that the total energy is proportional to the fourth power of the temperature AND that the spectrum is similar to the spectrum of a perfect black body. With that as a starting point, other parameters are defined to explain exactly how the object is NOT like a blackbody. Absorptivity and emissivity are simply two of those parameters.

"... a blackbody" means that the total energy is..." This is quite insufficient. The relevant BB equ. is ^{$P=\epsilon \cdot \sigma \cdot A\cdot T^{4}$} where, for a non-black body Є = 1- reflectance (or albedo if you will). Yes the spectrum may have a similar shape to a black body (don't forget that the BB spectrum at different temperatures has not only different T⁴ energy but also a different shape). All this means that a body that reflects (or transmits) some of the incident radiation i.e. it is not black cannot match a true BB in quantity of emitted radiation for a given temperature i.e. it will always be warmer than a measurement that assumes BB properties would indicate. This is a well known effect, see page 10 of [1] (it takes a long time to load). This is the nonsense behind the Earth's 255K equilibrium temperature and its supposed 33°C warming by the greenhouse effect .

The assumption that the Earth can "emit in the infrared" with the same efficiency as a black body would mean that the material reflecting the Sun's radiation (i.e. the reason we can see the Blue Planet) is also able to radiate energy. It cannot do both, if it reflects radiation it cannot simultaneously absorb it, so where does this supposed emitted energy come from - nowhere? This is the Perpetual Motion aspect of the GHE.--Damorbel (talk) 07:28, 11 January 2010 (UTC)

Different frequencies. The Earth does not emit blue light. It's blue light emissivity emission is zero. However, the Earth is very good at absorbing blue light. It also reflects a lot of blue light. In the infrared (IR), the Earth emits about 85% of the number of photons that a blackbody would emit. (Some surfaces emit more.) The solar spectrum (at the top of the atmosphere - TOA) contains a lot of blue light but almost no IR photons. Almost all the IR photons at the TOA are emitted by the tops of the clouds. This is what weather satellites detect.

Clarification: Emissivity refers to what would be emitted is the temperature is high enough. Emission refers to what is emitted at the current temperature. Q Science (talk) 17:11, 12 January 2010 (UTC)

Your equation is not complete because it does not integrate over the complete frequency spectrum. The integration is absolutely critical. Q Science (talk) 08:55, 11 January 2010 (UTC)

There have been many measurements of the energy from the surface of the Earth. Some are made from towers, others from aircraft at various altitudes, and a bunch from orbit. Those from short towers look a lot like standard blackbody spectra. As the measurements are made from higher up, they look more like a lot of different spectra added together. In fact, that is part of the basis of one of the alternate theories on how Greenhouse gases work. Q Science (talk) 04:02, 11 January 2010 (UTC)

"water (the oceans) lose energy slower than solid surfaces" With respect, what do you mean by this? Water loses heat mainly by evaporation; a wet surface under radiation does not get hot until is dry, ja oder nein? --Damorbel (talk) 12:51, 10 January 2010 (UTC)

You are absolutely correct, evaporation is more important than blackbody radiation. I was only referring to the radiative loses. Sorry for the confusion. Q Science (talk) 04:02, 11 January 2010 (UTC)

The reference for that development is from "Planetary science: the science of planets around stars" by Cole and Woolfson. I don't have a copy of that book, do you happen to have a copy? PAR (talk) 03:40, 10 January 2010 (UTC)

Sorry, no I don't have it (but it is available online). A search for the word "emissivity" indicates that it appears only once. I suggest that you find another source. (Perhaps Satellite thermal control for systems engineers, Robert D. Karam, page 158, table 6.1) In better books it explains that that both absorptivity and emissivity are functions of frequency. For many substances (like ice) the solar absorptivity is very low while the IR emissivity is fairly high. As a result, the average temperature is based on the ratio of a/e and that ratio is never equal to one (except for a black body). For instance, the absorptivity of white paint is about 0.17 and its emissivity is 0.86. Q Science (talk) 08:31, 10 January 2010 (UTC)

Thanks, I found the online source, but the page that gives the development for the planetary temperature is not included. Also, the ratio of emission to absorption is also unity for "grey" bodies, in which the absorption and emission coefficients are constant with respect to wavelength. I understand that emission and absorption are functions of wavelength. I'm trying to understand your objection to the analysis given, however. To say that the earth reflects as a non-black body (because of the albedo, or reflectance) yet emits as a black body is a strong violation of Kirchoffs law. Are you saying that there is a strong violation? I mean, forget my analysis that assumes the Earth is a grey body, what I am saying is that the present analysis is a strong violation of Kirchoff's law. PAR (talk) 14:53, 10 January 2010 (UTC)

Ok, I get it. If you do everything as a function of frequency, then you get an average absorptance, averaged over the black body spectrum of the sun, and an average emissivity, averaged over a black body emission at Earth temperature and they are not necessarily equal. If ${\overline {A}}$ is the sun-averaged absorptivity of the earth and ${\overline {\epsilon }}$ is the earth-averaged emissivity, then the Earth temperature is

$T_{E}=T_{S}{\sqrt {\frac {R_{S}{\sqrt {{\overline {A}}/{\overline {\epsilon }}}}}{2D}}}$

This is about the same as in the present development using ${\overline {A}}=1-\alpha$ where $\alpha$ is albedo, or averaged reflectance. The problem now is the average emissivity, which was just assumed to be unity in the present development. Does anyone know by what logic we assume that the average emissivity of the Earth is unity, assuming no greenhouse effect? PAR (talk) 14:37, 11 January 2010 (UTC)

Try the MODIS (Moderate Resolution Imaging Spectrometer) UCSB Emissivity Library. The area of interest is 3um to 65um. Unfortunately, the plots only go to 14 um. Q Science (talk) 18:24, 11 January 2010 (UTC)

Don't want to spoil your party but there is no way the reflecting Earth with its albedo can have an emissivity Є = 1, to do that the whole planet would have to behave like a black body, a body that absorbs all radiation that falls on it and reflects nothing. Q Science, you seem to believe it can, how?--Damorbel (talk) 22:00, 11 January 2010 (UTC)

It might help if you look at the MODIS graphs before you comment on them. These are based on actual measurements, not theory. Q Science (talk) 22:58, 11 January 2010 (UTC)

Emissivity of 1 is not an assumption, nor an approximation, but just a reference condition that's used for what's called the "effective temperature". Dicklyon (talk) 22:21, 11 January 2010 (UTC)

"just a reference condition"? The only reference condition with Є = 1 is a blackbody thus it is an assumption, an assumption that has no justification for a body that partially reflects incident radiation. Black bodies absorb all incident radiation thus they cannot reflect any, the Earth reflects radiation so any calculation that relies on Є = 1 will not give a thermodynamically consistent temperature i.e. it is off along the well trodden perpetual motion path. --Damorbel (talk) 07:59, 12 January 2010 (UTC)

Damorbel - strictly speaking, you are right and the present article is wrong to say "If we assume the following that the Sun and the Earth both radiate as spherical black bodies". I think its a matter of the original author transcribing an argument, but not understanding what they were writing, because the equation presented is roughly correct. The earth reflects about 30% of the sunlight striking it. That's because the sun is at a very high temperature and it emits mostly in the visible and ultraviolet (VUV). So the earth is definitely not behaving like a black body in the VUV. The earth absorbs 70% of that sunlight and is warmed by it. It emits this same amount of energy according to its own temperature, not that of the sun, which means it emits in the infrared. In the infrared, the earth is a very good emitter, emitting almost the same as a perfect black body. So the "albedo" or reflectance is an average for the Earth in the VUV, and in the infrared, it has been assumed that the average emissivity of the Earth is 1. I will rewrite the argument hopefully to make this clear. PAR (talk) 15:04, 12 January 2010 (UTC)

Hi Dicklyon - the present development is looking for a definition of the greenhouse effect - the temperature increase due to the earth's atmosphere, so I think what we want is the emissivity of a no-atmosphere earth. Which, to QScience, I would ask, is it fair to base the emissivity on ice, since without an atmosphere there would be no ice or water? And to Damorbel, I am starting to understand QScience's statement that it is a function of frequency. The earth can be like a black body in some wavelengths and not in others. The albedo is about how it behaves in the visible frequency range, the black body is about how it behaves in the infrared frequency range. QScience is saying that because of ice and vegetation and stuff, the earth is like a black body in the infrared. PAR (talk) 02:20, 12 January 2010 (UTC)

"The earth can be like a black body in some wavelengths and not..." No it can't, as I noted above if it reflects (or transmits) at any wavelength it is necessarilly non-black, just the same as if it reflected a little at all wavelengths (the "grey body" hypothesis). Just think about it, if it reflected (or transmitted) in the yellow and you told someone this bright yellow object was "a black body" you would get some rather strange looks. If it was a job interview maybe some"body" with different ideas gets the job. --Damorbel (talk) 07:59, 12 January 2010 (UTC)

To see the Earth without ice, ocean, or atmosphere, look at the Moon. Q Science (talk) 04:08, 12 January 2010 (UTC)

I cannot find a source that nails it down, but a number of sources suggest an average emissivity of 0.95, so saying its unity is not too far off. I will try to rewrite the present analysis a bit more clearly. PAR (talk) 05:55, 12 January 2010 (UTC)

There was some confusion as to the "effective temperature" and the estimated temperature given present conditions. I also added an estimate of the temperature without atmosphere by inserting albedo and emissivity of the moon. PAR (talk) 16:48, 12 January 2010 (UTC)

I'm afraid it only confuses it further, as it presents in the same paragraph the temperature of the moon vs the effective temperature of the earth. The estimates are already without atmosphere, since greenhouse effects are ignored (although it still refers to atmosphere twice for some reason); the only effect attributable to the atmosphere is the increased albedo. The article should say it only applies for a planet without atmospere, and refer to a greenhouse article for estimates in the presence of a atmosphere.

Some people are confused further because the same symbol is used for planet's temperature in the general case and effective temperature estimated for Earth. Mihaiam (talk) 05:52, 2 August 2011 (UTC)

I don't see that. T_E is used for the effective temperature (the temperature the planet would be if it were a black-body) only. Where exactly is the confusion that you see? What it says about atmosphere appears sensible to me; do you object to mentioning atmospheric effects? How would that make the example calculation better? Dicklyon (talk) 06:18, 2 August 2011 (UTC)

Oh, I see, I got confused because you changed it some time during when I was trying to look at it. The T_E did change from the "effective" to the one that includes emissivity, probably due to other recent changes. Let's see if we can patch it up; taking out what it sensibly says about atmosphere is probably not the way, nor is using T_P for both places. Dicklyon (talk) 06:31, 2 August 2011 (UTC)

As long as we ignore greenhouse effects we effectively set longwave IR atmosphere emissivity to 0, so it no longer may influence radiative transfer from the planet to space. The ramaining effect of the atmosphere, the increased albedo due mainly to the cloud cover, can be simply modeled as a whitening of the surface. At least we should explicitly mention that we only consider a transparent atmosphere in the relevant band. Mihaiam (talk) 09:06, 2 August 2011 (UTC)

Which statement about the atmosphere are you finding to be wrong, misleading, or unhelpful? It looks to me like they're OK, saying what the influence is, and that we have to ignore it go get to a simple calculation. Dicklyon (talk) 15:26, 2 August 2011 (UTC)

"The power emitted by the planet and its atmosphere is then:" may sugest to some that the atmosphere radiate energy in this model. If so, we cannot get rid of the greenhouse effect, since the atmosphere will radiate toward the planet, too. The easy way out is to consider a transparent (in IR) atmosphere. BTW, the whole example is moot for gaseous planets, so we may as well specify that we consider a terrestrial planet.Mihaiam (talk) 16:06, 2 August 2011 (UTC)

Yes, I agree that "and its atmosphere" is just a distraction at that point; take it out. But as for gas planets, I'd say the model still applies; it's just that it's not as good a model. Where it says "most important impact for the inner planets," I think it's referring more to the other sources that affect energy balance in the gas giants, including nuclear sources and maybe gravitational. Dicklyon (talk) 18:06, 2 August 2011 (UTC)

For a gaseous planet it's not clear what "planet temperature" we obtain, since it lacks a surface to refer to. I guess we should limit ourselves to computing the effective temperatures for them in this article.Mihaiam (talk) 21:36, 2 August 2011 (UTC)

PAR is correct . The discussion of spectral effects has no place in an article on black bodies . Rather , it should be pointed out that the calculated temperature applies to any flat spectrum ( gray ) body . Bob Armstrong (talk) 20:22, 11 August 2011 (UTC)

Article changes

Please put back the section on computing the blackbody temperature of a planet. If you want to produce a major rewrite, you need to create a work page and develop it there. Making major changes in the article space is not going to work.

In addition, your statement

Because of its high temperature, the sun emits almost entirely in the visible and ultraviolet (VUV) frequency range.

is not correct. The Sun emits much more IR radiation than the Earth. However, because the Sun is so far away, the amount of IR radiation at the top of the atmosphere is insignificant. This is because of "one over R squared" dimming and not what is emitted. Q Science (talk) 17:29, 12 January 2010 (UTC)

Hi - sorry about the unintended deletion. Yes, "entirely" is not a good word to use. 90 percent of the sun's radiation is to the blue side of 1 micron. so, a "large majority" might be a better term. I meant it in relative terms, not absolute terms, certainly the sun emits more IR than the earth, but as a percentage of its total, its small. Also, all the changes that I made are things that I believe we agree on. PAR (talk) 18:07, 12 January 2010 (UTC)

I still think that it is better to develop the new text in a work area. These are a few of the problems that need discussing, fixing, or clarification.

• The instruments I use refer to the solar spectrum as UV-Vis, not VUV which typically means vacuum ultraviolet.
• The solar spectrum measured in space is referred to as TOA (Top of Atmosphere). The effective temperature of the Sun is computed from that value, not the other way around.
• You dropped the abs and emt qualifiers (absorbed and emitted) from a number of variables. I think that this makes the formulas less clear.
• The Earth does not reflect "a fraction 1 − α of this energy"

There are other issues, but these are enough for now. Q Science (talk) 21:18, 12 January 2010 (UTC)

I fixed all of the above problems, plus a few typo's. (I did not use "TOA", but stressed that the solar radiation was that striking the top of the atmosphere rather than the Earth's surface). I think that it will not take a large number of iterations to get this right. PAR (talk) 21:58, 12 January 2010 (UTC)

Emissivity varies with wavelength. In the longwave (LW) the Earth (surface) is essentially "black" - ie, emissivity ~1 (actually ~0.997 if I recall correctly. Was that an average or minimum? Can't remember). In the shortwave (SW, i.e. visible) the emissivity, i.e. the (1-albedo), varies, as we can see William M. Connolley (talk) 22:11, 12 January 2010 (UTC)

"In the longwave (LW) the Earth (surface) is essentially "black"...". As I noted above, this is illogical. Look at it this way, is a blue object "black" in the red part of the spectrum? Can you justify using the Є = 1 of a 100% black body in this case? --Damorbel (talk) 07:19, 13 January 2010 (UTC)

It depends on what the definition of "black" is. With the red and blue, "black" is the absense of photons. You should interpret William's comment as saying "In the longwave (LW) the Earth (surface) emits photons with the same intensity and spectrum as a theoretical blackbody". In this context, "black" means radiating energy, not absorbing it. Think of the filament of an incandescent light bulb, it is a blackbody that gives off light. The surface of the Earth does the same thing, but, since the temperature is lower, your eyes can not see the photons. Q Science (talk) 08:11, 13 January 2010 (UTC)

It depends on what the definition of "black...". This may be the difficulty. A black body in the thermal sense neither reflects nor transmits any incident radiation - at all. --Damorbel (talk) 09:22, 13 January 2010 (UTC)

Right, but it does emit radiation. It emits its own, internally generated thermal radiation. For a body at room temperature, all that emission is in the infrared, so you cannot see it, so it looks black. But when it gets heated up, that thermal radiation moves into the visible. A hot black body glows red, but physicists still call it by the name "black body" because it doesn't reflect or transmit any radiation. Also, for non-black bodies, the emissivity can be different for different wavelengths. For any wavelength where the emissivity is one, physicists say it is "black" at that wavelength or you could say "it acts like a black body at that wavelength". By Kirchoff's law, it won't reflect or transmit any radiation at that wavelength. But any body that did not have its emissivity equal to one at every wavelength could not be called a black body. PAR (talk) 17:14, 13 January 2010 (UTC)

"or you could say "it acts like a black body at that wavelength" I am afraid that just doesn't add up. In this case the formula for black body radiation just doesn't describe the radiated output. Look at it this way, suppose your non-black body emitted at a green band, in some circumstances it could be that the output had the intensity of a black body at that wavelength but it would not have the energy of a black body. Think further, green radiation doesn't have a temperature because it doesn't have a blackbody spectrum. Your green radiation could have the same energy as a black body but this would be extremely difficult to do, for starters you would have to produce narrow band radiation from heat with the same efficiency as a black body i.e. the maximum possible. --Damorbel (talk) 21:44, 13 January 2010 (UTC)

Maybe a better way to say it is this - suppose you had a filter that only passed red light, like 600 nanometers. If you looked through that filter at a hot black body and some other body with the same shape and size and distance, with no outside light falling on them, and both at the same temperature, and they looked the same, same brightness, then that other body is black at that red wavelength. If you looked at them both through a green filter, and they looked different, then that other body is not black at that green wavelength. And that could happen. PAR (talk) 22:44, 13 January 2010 (UTC)

"a hot black body and some other body..."; are they both at the same temperature? "And they looked the same", but there is "no outside light falling on them" so there is nothing for the non-black to reflect (or transmit). "And they looked different". Assuming they are at the same temp. etc., then the one that was non-black would seem cooler using an infrared thermometer (green filter) because its non-black property would make it a less efficient emitter, even though the two bodies were at the same (contact) thermometric temperature. (Have I understood your question?) --Damorbel (talk) 22:21, 14 January 2010 (UTC)

Right - both at the same temperature, neither one having any light to reflect. But an infrared thermometer measures in some infrared wavelength band, and we are only looking at them thru a green and a red filter. An IR thermometer will of course give a correct reading for the black body, but who knows what the other one will read? We have not specified whether the other body is black or not in the infrared range of the IR thermometer, only that it is black in the wavelength range of the red filter, and yes, it will look dimmer thru the green filter and therefore is not black in the wavelength range of the green filter. PAR (talk) 23:03, 14 January 2010 (UTC)

"only that it is black in the wavelength range of the red filter". I think the problem lies here. Do you mean that the emission strength of a particular waveband for a non-black body is the same as that of a black body? You are making very particular cases with filters etc. and "how things look", we are far from the general case of black and non-black, for example a non-black body can be (partially)transparent in the infrared i.e. only absorb in part of the spectrum, it might be completely transparent in which case it would look "black" in this waveband at any temperature. I'm sorry about my need for detail but if you are not happy about the general case and need to resort to filters and "how things look" I suggest you should look carefully at you qualifications for rewriting the article. --Damorbel (talk) 07:31, 15 January 2010 (UTC)

PAR is correct . The discussion of spectral effects has no place in an article on black bodies . Rather , it should be pointed out that the calculated temperature applies to any flat spectrum ( gray ) body . Bob Armstrong (talk) 20:19, 11 August 2011 (UTC)

I don't think you understood (but I think I agree that PAR is correct). One could indeed compute a temperature for a "gray" (spectrally flat) planet. But what was done here was to include the albedo (reflectance) at short wavelengths where the Sun's energy is, but to treat it as "black" at the long wavelengths where it emits. This is not a bad approximation, and the calculated temperature is known as the "effective temperature", the temperature of a blackbody that would emit the same power. It depends on albedo; if you let albedo be zero, it will the give the same result as the flat gray body you're talking about. Dicklyon (talk) 07:07, 12 August 2011 (UTC)

Reasons for edits that were reverted for no properly stated reason

Dear Dicklyon, your reversion is based on a question that you ask, not on your properly stating a reason of physics. I will explain my edits here, and this will answer your question.

The rigidity of the walls of the cavity is irrelevant. What matters is the opacity of the walls, and that they should not be perfectly reflecting. Because they are not perfectly reflecting, some of the incident radiation that hits them is not reflected by the interface between the body and the contiguous medium in the cavity, but penetrates into their skin, where it is at least partly absorbed, and is absorbed completely if they are opaque. This complete absorption is the factor that correlates with their black body emission, which hits the interface and, because of the Stokes-Helmholtz reciprocity principle, is partly reflected back into the interior and partly penetrates and gets out into the contiguous medium where it is detectable as emitted radiation. The person who wrote about rigidity was making it up off the top of his head. Planck's 1914 book explains the situation as I have just written it. The reference to the article by Robitaille is unsuitable as a Wikipedia reference, because though Robitaille has done a nice bit of historical research, and he writes persuasively, ultimately his physical understanding is wrong and his article is seriously misleading. That the walls of a cavity should be made of graphite is a well established tradition that precedes Kirchhoff's 1860 paper and it would need some more historical research to give an original reference for it.

Kirchhoff's law is generally assumed to refer only to the universal thermal radiative spectrum recognized by Kirchhoff in his 1860 paper, and eventually properly characterized by Planck. Therefore radiative equilibrium is not really the right kind of equilibrium for a general application of Kirchhoff's law as usually understood: the right kind of equilibrium is local thermodynamic equilibrium. This is a matter of basic physics.

While the cavity is large compared with the hole, that is already implicit I think in the use of the term small hole. If not, then I agree it needs to be added. But what really matters about the cavity is that it have opaque and not perfectly reflective walls, as explained by Planck and as noted above.

What does not matter is the particular opaque non-perfectly-reflective materials of which the walls are made. What does matter is their opacity and non-perfect reflectivity.

The wording geometrical structure is not quite physically helpful. What matters is the fine structure of the interface, admittedly with geometrical features, but I think that the word geometrical is not quite as physically explicit and illuminating as words that refer to the surface properties. Incomplete absorption means 'at least one of partial reflection or partial transmission', and the latter is contrary to the requirement for opacity. In effect, opacity makes a body's skin and interior effectively black, and the interface with the contiguous medium must be imperfectly reflective, so that some radiation can actually penetrate into the skin and interior where it can be absorbed. The interface itself is a mathematical fiction, a two-dimensional surface, and cannot emit or absorb. All this is carefully explained by Planck 1914.

I trust that this explanation will answer your questions and will allow you to accept my edits as valid and beneficial.Chjoaygame (talk) 09:35, 20 February 2011 (UTC)

I wasn't questioning the physics; but you took a statement that was supported by three sources, and left it supported by none. If you have sourced material to add, or you question what's there, this should not a side effect of trying to fix it. And if you're questioning the reliability of a source, say so here, instead of just removing it. Also, be aware that when you make a practical source of blackbody radiation, the material does matter. If you use something black like graphite, then you can support a much bigger hole size to relative to cavity size for a given spectral accuracy; if you start with something with a colored reflectance spectrumm, it take a bigger cavity and smaller hole to approach the appropriate spectrum, right? I think that's what the sourced statement was about. Dicklyon (talk) 21:02, 20 February 2011 (UTC)

Dear Dicklyon, Thank you for this reply. Others have also complained about Robitaille as a Wikipedia source. As I was the one who put in those three references, and I agree that I was mistaken to include Robitaille, my starting point was to remove the reference to him. The problem with him is that he is direct and very emphatic in saying that without a carbon-like body, which is very nearly black in itself, the cavity will never get very near radiative thermodynamic equilibrium. He does not provide direct evidence for this claim, but seems to be making inferences from remarks by Kirchhoff and Planck that do not really support those inferences. He puts the view that the radiation gets to be of the Planck kind only because the detector has a near black surface and this supplies the black radiation by way of the observing hole. He does not understand that the opacity of the wall material combined with imperfect reflectivity is enough to make the cavity radiation of the Planck kind. He exaggerates the importance of the theoretical case, discussed by Planck, of perfectly reflective walls. Supposing walls of given non-black emissivity, for a spherical cavity one can calculate the relative size of the hole required to get a specified degree of blackness; the result is equation (3-22) of Hottel and Sarofim 1967, pages 80-81. It is surprisingly small. Robitaille does not actually consider this question of relative size because he does not understand that one key factor is the closeness to local thermodynamic equilibrium; this point is also missing from the article; the combination means that Robitaille is a misleading reference for a Wikipedia article. This absence from the article of reference to the need for local thermodynamic equilibrium brought itself to my attention as I made the removal of the Robitaille citation. I take your point that it is usually found helpful to use graphite walls, but Robitaille is wrong to claim that this is the main physics. If you can cite a sound defence of Robitaille's claims, I would be interested to read it. As I remarked above, I take your point that the Wikipedia article would do well to make it explicit that the cavity is better large, and I am happy to put that in. The key physics discovered by Kirchhoff was not really made quite clear in his 1859 and 1860 articles, that it is the opacity combined with non-perfect reflectivity that really matters; and Planck only makes it fully explicit in a footnote: one has to follow Planck's logic carefully to appreciate its importance. The need for opacity and non-perfect reflectivty is not mentioned in the present version of the Wikipedia article. To read the present article, one might conclude that the walls of the cavity could be of glass provided it is a rigid kind of glass. It is not the colour of the spectrum that matters; it is the opacity. The opacity is practically obtainable without too much trouble, but by itself it it not enough; in addition, the skin and interior-body of the wall material must be accessed by the cavity radiation, and that would be prevented by a perfectly reflective interface, a point thoroughly made by Planck. I agree that it is customary to repeat Kirchhoff's original claim that the chemical nature of the wall material does not matter; and that this is misleading unless it is pointed out, as does Planck, that the walls must be opaque and not too reflective, and with those features provided, then the chemical nature does not matter.Chjoaygame (talk) 23:01, 20 February 2011 (UTC)

In summary I think you are saying that I ought not remove the Robitaille citation without some adequate replacement; or that I need your permission to remove it. I reply that the citation is so bad that it should be removed even with no replacement. The article went for long without it before I put it there. Perhaps the calculation by Hottel and Sarofim would be suitable, but I don't have the book right here to fully check the context, and would take some time to get it from the library, because it is in the stack store. If you have a good replacement citation, it would be good to know of it.Chjoaygame (talk) 00:08, 21 February 2011 (UTC)

I think you need to write less, and read more. Don't attribute to me what I did not say. Dicklyon (talk) 00:19, 21 February 2011 (UTC)

Dear Dicklyon, Thank you for this reply. Thank you for your helpful personal advice. I am sorry that it seems I have misinterpreted what you wrote as your reason for reverting my edit: "huh? ... why?". Do you have further thoughts about my Wikipedia edit which you reverted?Chjoaygame (talk) 03:38, 21 February 2011 (UTC)

Nope, I pretty well expended all my thoughts in my comments above. But I can repeat it for you: "you took a statement that was supported by three sources, and left it supported by none." Dicklyon (talk) 04:06, 21 February 2011 (UTC)

Dicklyon's edit

Dicklyon has made an information-deleting edit (00:23, 28 March 2011) on page references. Is it Wikipedia policy to make cuts like this? Is there a Wikipedia policy against page numbers in references? What if more page references are needed in future to Planck? How does one decide how many can fit into one reference? Will the editor who puts them in have to go back and recover the deleted information?Chjoaygame (talk) 02:53, 28 March 2011 (UTC)

No reply?Chjoaygame (talk) 02:55, 29 March 2011 (UTC)

You could have reverted the edit if you felt that Dicklyon's edit was unfair. 204.134.45.36 (talk) 22:04, 2 August 2011 (UTC)

Yes indeed I could have reverted the edit but I thought it better to discuss it rather than just chop it. I expected that Dicklyon would be willing to try to justify his edit. Without discussion, if he had felt like it, Dicklyon could have reverted my reversion and so on. Not an efficient way. Sometime, when I have the leisure, I will, I suppose, go back and put it right. In the meantime it stands as a reminder that Dicklyon did not bother to try to justify his destructive and unhelpful edit.Chjoaygame (talk) 06:00, 3 August 2011 (UTC)

I merged a bunch of refs to the same book; the justification in the edit summary needed no amplification. If you think it's important that each footnote include the specific page numbers, rather than list 4 pages together as I did, a better fix would be to use Template:rp. A revert would be a bad alternative. Dicklyon (talk) 06:24, 3 August 2011 (UTC)

I have time to put in constructive edits but not time to run around repairing Dicklyon's destructive ones, which he feels need no "amplified" "justification".Chjoaygame (talk) 20:54, 3 August 2011 (UTC)

thermal equilibrium

Thermal equilibrium usually means just that all the temperatures are time-invariant. You are interpreting it to mean that there is no net transfer heat by conduction into or out of the system but that perhaps there might be net radiative transfer of heat into or out of the system. 'Thermal equilibrium' sounds much more recondite, abstruse, and clever than just saying that all temperatures are time-invariant, and has indeed succeeded in confusing you, as it succeeds in confusing many. Thermodynamics textbooks often don't talk about radiative transfer until they come to want to talk about the Stefan-Boltzmann law. But that doesn't mean that radiative transfer of heat doesn't count as an ordinary form of heat transfer. Perhaps the term should be changed in this place in this article for this reason. For the moment I have just put in a link. What do you think?Chjoaygame (talk) 00:24, 21 May 2011 (UTC)

"Thermal equilibrium usually means just that all the temperatures are time-invariant." Not really; 'invariant temperatures' is best described as 'steady state'. 'Thermal equilibrium' means 'the temperature is uniform'; which amounts to 'no energy flow' (2nd Law of thermodynamics)

The situation with the Sun and the Earth is 'steady state' the temperatures are constant (i.e. invariant) - Sun = 5780K & Earth = 279K; this means there is a large energy flow from the Sun outwards. If there was to be thermodynamic equilibrium, Earth & Sun would have to have the same 5780K temp., obviously not the case. --Damorbel (talk) 20:58, 23 May 2011 (UTC)

Perfect black bodies do not exist in nature, nor indeed can they be manufactured artificially

Editor 99.0.104.35 changed "do not exist" to "are not known to exist". The statement was supported by a reference to very good authority, Planck. It seems most likely that editor 99.0.104.35 had not read what Planck had to say, and did not himself have the physical understanding that would have been needed to justify his edit. A perfect black body must be indefinitely large in order to be able support indefinitely long wave length radiative exchange. Also it must have a perfectly black interface with every medium including a vacuum, which is impossible because the medium interfaces with the black body and this establishes a discontinuity of density across which there must be a non-zero reflectivity, contrary to the requirement that a perfectly black interface has zero reflectivity. Editor 99.0.104.35 seems to have a good motive, that he doesn't want the Wikipedia to make unjustified extravagant statements, and so he wants to say 'not known' instead of 'not'. But the motive does not justify his mistaken edit.Chjoaygame (talk) 22:43, 2 June 2011 (UTC)

reversion of edit by editor 129.11.77.198

I have reverted the edit 455056750 of 20:56, 11 October 2011 by editor 129.11.77.198. That edit was mistaken, as may be checked by looking at Planck 1914 page 168 equation (274) for a plane polarized ray, and at page 22 of Rybicki and Lightman 2004.Chjoaygame (talk) 21:02, 11 October 2011 (UTC)

Question re: "Explanation"

"All matter emits electromagnetic radiation when it has a temperature above absolute zero. The radiation represents a conversion of a body's thermal energy into electromagnetic energy, and is therefore called thermal radiation. It is a spontaneous process of radiative distribution of entropy." This isn't really an explanation, it's really more of an exposition; an explanation would answer _why_ matter does this. Methinks what is required is simply more detail, please: what is going on at the molecular/atomic/sub-atomic level? Is the radiation due to the (directional) "acceleration" in the EM field of the charged constituents of matter, (which is why at absolute zero the radiation would stop, i.e., because absolute zero = no motion)? If this _isn't_ the explanation in terms of the atomic model of matter, then what is?

OlyDLG (talk) 08:50, 19 December 2011 (UTC)OlyDLG

The electromagnetic radiation comes about from the mechanical vibration (due to the thermal energy) of an electrical dipole. --Damorbel (talk) 21:55, 19 December 2011 (UTC)

edits by Waleswatcher

Dear Waleswatcher, you are reverting my edits faster than I can keep up with. To see the grey appearance, one must view the body in the dark. The lowest intensity is small, too small to activate colour vision; that's why it looks grey. Laws don't cause things to happen; it's perfectly good English to say that nature obeys her laws.Chjoaygame (talk) 16:15, 19 January 2012 (UTC)

I don't have a problem with "obeying", the wording before was awkward but this is fine. However regarding "viewed in the dark" the reference does not support that, nor is it true. Perhaps to see the grey color you need a dark background for a non-ideal blackbody, but for an ideal one it's irrelevant how much ambient light there is, because the body absorbs all of it and so the only light that reaches your eye from it is what it emits. Moreover, everyone that's ever seen a glowing hot piece of metal knows that you can see the other colors even in daylight. Finally, you're not allowed to add uncited content like that. I'm removing that phrase. Waleswatcher (talk) 16:30, 19 January 2012 (UTC)

Welcome to the talk page.

True, Partington doesn't say "viewed in the dark", but I would say it is implicit in his statements. Draper, cited by Partington, reported seeing grey first appearing in the dark, as I have now sourced.

As I noted above, to see the faintest light emitted from a body, one must view it in the dark, when one's eyes are to some degree dark-adapted. Then from the weakest visible source there is only grey to be seen. Brighter emission stimulates colour receptors and one sees colour, in this case red. It is ordinary experience that when it is not dark, the grey of a poorly visible object cannot be seen because one's eyes are not dark-adapted. Contrary to your above reasoning, the ambient light is very relevant, because it determines one's degree of dark adaptation.

It is very good to see you zealously pursuing the rule against uncited content. There should be more of it. I would say, however, that I was making a good editorial explication for readers who may not have noticed for themselves what one ordinarily sees for oneself in the dark. I may say that I am the only editor to put any source into that lead, the others, including you, not bothering, and not having attracted your newly evident most admirable zeal.Chjoaygame (talk) 17:56, 19 January 2012 (UTC)

OK - then you need to make clear that while the grey color can only be seen in the dark, the other colors can be seen in normal lighting conditions (which as I said is obvious to anyone that's ever looked at a glowing piece of hot metal). That wasn't how the statement was written before.

Also, since there's considerable skepticism over "grey" (from me initially too), perhaps that part should be more explicit. The source said a "pearlescent grey", or something along those lines. You could add "viewed in the dark" to that part. Waleswatcher (talk) 17:59, 19 January 2012 (UTC)

What nonsense is this? Grey is not the color in the dark. A grey body is gray because if you illluminate with a white light source, the light it reflects will be that same neutral color (works for other colors, too, but that's less relevant). Dicklyon (talk) 18:43, 19 January 2012 (UTC)

What seems to be true is the article has undergone too much radical and nonconsensus change in the last day or so. Waleswatcher's idea of "perfect emitter" seems like a very odd one to start with; is there a source for that concept? What is it supposed to mean? Does anyone understand or support what he's trying to do? Dicklyon (talk) 18:50, 19 January 2012 (UTC)

• Dear Dicklyon, it is not nonsense. It is what is stated in reliable sources, if you like to check them. Moreover, it is what you yourself will observe if you go for a walk on a dark night in the bush where there are no street lights, get dark adapted, and await the slow coming of the dawn. First, in the feeblest light, you too will see coloured objects as grey, before it gets light enough for you to see their colours. The reason is to be found in textbooks of visual physiology, and even at Adaptation (eye).Chjoaygame (talk) 19:15, 19 January 2012 (UTC)

• Dicklyon, first, I didn't add "grey", that was there before I touched the article. Second, grey is what the cited source says, so your comment on your edit is not comprehensible. If you think it's incorrect, you must find a source that agrees with you. As it stands, your edit makes the article disagree with its own reference. Third, I don't see any reason to doubt that a hot object could glow with a perceived greyish color when the emitted light is very faint, for the reasons Chjoaygame points out. Finally, as for "perfect emitter", no such phrase appears in the article. Ideal black bodies indeed emit perfectly thermal radiation; that's a basic fact that can be sourced with hundreds of references if you insist. Since I've edited the article several times today I'll leave it alone for now, but clearly your edit needs to be reverted. Waleswatcher (talk) 19:54, 19 January 2012 (UTC)

• Partington 1949 is indisputably a reliable source.

The sources quoted by Partington are also reliable sources, at least John William Draper, Heinrich Friedrich Weber, R. Emden, and Otto Lummer are well known reliable reporters of observations.

Draper 1847/1878 looked down a rifle barrel into the dark chamber that contained his glowing specimens.

Weber 1887 wrote "führte ich die Beobachten in nahezu absoluter Dunkelheit, nämlich im Dunkelzimmer bei Nacht aus." He introduced the description "„gespenstergraues Licht“ oder „düsternebelgraues Licht“".

Emden 1889 wrote "im Dunkelzimmer und bei Nacht ein absolut dunkles Gesichtsfeld vor sich." and referred also to Weber's description.

Lummer 1897 wrote of "die Beobachtungen im Dunkelzimmer bei Nacht ausfürthe" and of "Graugluth".

It stands to reason that these reports make sense, with the obvious need to look in the dark. Perhaps you will recall that many of Newton's observations were made in a darkened room and it might be perhaps almost the default way of looking at faint lights, and so Partington could perhaps be forgiven for not explicitly mentioning it. But the average Wikipedia reader would likely gain from having this stated explicitly, or he might think it odd that the first-seen faint light should be experienced as grey.

The references are

• Draper, J.W. (1878). Examination of the radiations of red-hot bodies. The production of light by heat, pages 23–51 in Scientific Memoirs, being experimental contributions to a knowledge of radiant energy, Harper & Brothers, New York, pages 29, 34, 46, 48.
• Weber, H.F. (1887). Die Entwicklung der Lichtemission glühender fester Körper, Annalen der Physik und Chemie (Leipzig) 32: 256-270.
• Emden, R.(1889). Ueber den Beginn der Lichtemission glühender Metalle, Annalen der Physik und Chemie (Leipzig) 36: 214-235.
• Lummer, O. (1897). Ueber Graugluth und Rothgluth, Annalen der Physik und Chemie (Leipzig) 62: 14-29.

I wonder if we might exercise a little editorial discretion and fill in what I think was implicit in Partington's account, that one looks for the faint lights in the dark, without quoting explicitly all his four concordant sources that say that's how they made their observations? To say that one has made one's observations of faint lights in the dark does not imply that as the lights get brighter one will still need darkness to see them.

I prefer the vague "several hundred degrees" to the explicit number 500 degrees, because some materials glow even at 210 °C. I have not checked the claim that the Draper point is at 798 K; it doesn't seem to fit the data attributed to Pouillet by Partington, but I don't claim to have knowledge about that.Chjoaygame (talk) 20:27, 20 January 2012 (UTC)

Partington says 525 degrees Celsius. If some particular material glows at a much lower temperature, it's probably not because of black body radiation - don't you agree? We could check that by looking up the perceptual thresholds for human vision and comparing, but I suppose that would constitute "original research"... Anyway if you're convinced the number should be lower, go ahead and change it back.

Regarding the in the dark stuff, I don't have a problem with that so long as it clearly applies to the grey glow only. I know for a fact (from direct personal experience) that the red and "hotter" colors are visible in daylight, at least indoors. I've seen it on a cast-iron coal burning stove many times, as well as on metals heated in a forge. The way the sentence was written before, it sounded like the in the dark requirement applies to all those colors, which is false and could be very confusing to some readers. Waleswatcher (talk) 21:28, 20 January 2012 (UTC)

This source says 500 degrees C also http://www.scientific.net/MSF.663-665.223 (see the Introduction section of the first page). Waleswatcher (talk) 21:35, 20 January 2012 (UTC)

OK, I forgot about scotopic vision, but I still question the idea that a blackbody viewed in the dark first looks grey. In the snippet of Partington that I was able to see, he mentions red, yellow, ... but no grey; can someone please quote for us what he says there? Even if one is well dark-adapted, I would think that the tail of warm bb spectrum would stimulate red cones before it stimulates rod; this is why pilots use red lights in the cockpit: to stimulate cones without stimulating rods (much). But if these guys report gray, we need to at least make it clear that this is only in scotopic vision, too dim to stimulate any cones and therefore below the level at which we can discriminate color, which is rather different from saying the color is gray. Also, the red will be first visible (i.e. at lowest temperature) only if there's no competing light and one is reasonably dark adapted; if you see a red glow in daylight, that has to be a lot hotter. Let's be more clear about such things. Dicklyon (talk) 22:50, 20 January 2012 (UTC)

As for "by emitting perfectly thermal radiation", the obvious reading is that perfectly modifies emitting, while it seems that you intend "perfectly thermal", which is itself a rather undefined concept here. I'm going to take out "perfectly" until we can find something better. Dicklyon (talk) 22:56, 20 January 2012 (UTC)

I agree that "emitting perfectly" is meaningless, there. Instead, a meaningful (and important) concept was there before the last edits: "a black body is the best possible emitter of thermal radiation", i.e. it emits more than any other body at the same temperature.--87.7.187.113 (talk) 08:12, 21 January 2012 (UTC)

And, is it relevant, here, scotopic vision or, anyway, how our eyes work? The important point here is that moderately hot objects are (objectively, spectroscopically) red and very hot objects are bluish.--87.7.187.113 (talk) 08:21, 21 January 2012 (UTC)

I agree. The gray/scotopic appearance is a distraction from the fact that the physical or colorimetric color is red. Dicklyon (talk) 17:43, 21 January 2012 (UTC)

Another important concept has disappeared from the lead (and from the article), in these days: the phrase about the ultraviolet catastrophe has been substituted by a trivial phrase which says that if a body does not radiate according the blackbody theory, it is not black. So, the following phrase, which tells the importance of the black body theory for the foundations of quantun theory, has lost its significance.--87.7.187.113 (talk) 08:43, 21 January 2012 (UTC)

• For Dicklyon, on page 466, Partington 1949 writes:

     According to Draper ¹¹ a solid body begins to emit red light (appearing to the eye as a faint "ghostly" grey) ¹² at 525°; the brightness or intensity rises rapidly with temperature and in addition to an increase of total radiation the distribution in the spectrum (on which the colour depends) changes with rise in temperature, the intensity of the short waves (violet) increasing more rapidly than that of the long waves (red). [ I have above detailed Partington's references ¹¹ and ¹².]

Draper 1878 writes:

....Thus Sir Isaac Newton fixed the temperature at which bodies become self-luminous at 635°; Sir Humphrey Davy at 812°; Mr Wedgwood at 947°; and Mr. Daniel at 980°. As respects the nature of the light emitted, there are similar contradictions.... [(635 – 32) x 5 / 9 = 335. I have not succeeded in finding where Newton says this. Perhaps some other editor will kindly enlighten us with that information?]

....The platinum and the voltaic battery were placed in a dark room, the temperature of which was 60° Fahr.; and after I had remained therein a sufficient length of time to enable my eyes to become sensible to feeble impressions of light, I caused the current to pass, gradually increasing its force until the platinum was visible. In several repetitions of this experiment it was uniformly found that ... the temperature of incandescence was 977° Fahr. [(977 – 32) x 5 / 9 = 525, and 525 + 273 = 798.]

....It is to be understood, of course, that this is in a very dark room.

When I woke this morning, the coloured book-covers beside my bed looked grey in the feeble light. I have no reason to dispute the statement of generally reliable source Partington 1949 based on his reliable sources, and if I did, I would be guilty of some form of own research or synthesis if I tried to post it in the Wikipedia, unless I showed more reliable sources than Partington.Chjoaygame (talk) 10:08, 21 January 2012 (UTC)

• For Waleswatcher, the source you cite http://www.scientific.net/MSF.663-665.223 reads: "Its visible glow changes from a dull red at about 500°C to bright yellowish and then to bright white around 5000°C. ... The apparent glow of the blackbody cavity was dim at 600°C, ...". The experiments were performed with a blackbody furnace with a range of temperature control between 600 – 1400°C. I see this as reason to be vague about the temperature in the lead.

Strictly speaking, the light from an incandescent body distinctly seen against a dark background is light that is not emitted in strict thermodynamic equilibrium, which requires the environment to be at the same temperature as the emitting body. It is perhaps also light that is emitted from a body not in local thermodynamic equilibrium such as would accurately guarantee the holding of Kirchhoff's and Planck's laws of thermal emission; local thermodynamic equilibrium usually requires spatio-temporal continuity of all thermodynamic intensive variables.

The lead is supposed to be a summary, not a detailed account. The temperature of first emission is certainly at several hundred degrees, but the precise figure probably depends on the kind of material. Draper deliberately avoided a surface that could be oxidized in the heat. He tested a platinum strip, which he chose because it had a non-corroded surface. It would have been a good reflector and therefore a poor emitter, at least under conditions to which Kirchhoff's and Planck's law of thermal emission apply; while powders yield different results. In the lead, I would prefer to leave a vague statement of the temperature than to get into difficult-to-verify details about different materials and surfaces. If some keen editor likes to put into the body of the article some such details, then we would be justified perhaps in summarizing some of that detail into the lead.Chjoaygame (talk) 10:21, 21 January 2012 (UTC)Chjoaygame (talk) 11:31, 21 January 2012 (UTC)

Chjoaygame, it looks like every single source you dug up agrees pretty much on 500C (not sure why you added 273 at one point up there). The only outlier is what one source says about Newton, which is first of all slightly unreliable, and in any case, who knows how accurately Newton could measure temperature. I think that can be safely discounted. Non-ideal materials will almost always emit less light, and in any case the lead is about a black body.

Dicklyon, our interpretation of the origin of the grey color is probably right, but it's just our interpretation. It can't go in the article without a cite.

I would be perfectly happy leaving out the whole scotopic/gray stuff. But if we talk about observations "understood to be in a very dark room" then we would need to say something to make it clear that we're no longer talking about color in the colorimetric sense. Dicklyon (talk) 17:43, 21 January 2012 (UTC)

87.7.187.113, the problem with "a black body is the best possible emitter of thermal radiation" is that it's not clear to the reader what is meant by "best". Emitting more radiation doesn't make something a better thermal emitter, because one characteristic of thermal radiation is the total intensity. Not only that, under some (non-equilibrium) circumstances black bodies emit less total radiation than other materials.

If we say perfectly thermal radiation (or ideal or exact or precise) the reader can click on thermal radiation, learn what it is, and understand the meaning of the phrase. But best possible emitter isn't clear and can't be understood like that. I think the whole sentence could be reworded to make it more clear that the spectrum that's emitted is of exactly thermal form, but I'm trying to make minimal changes. Anyway, some such adjective is needed, because that's one of the defining characteristics of an ideal BB. Waleswatcher (talk) 12:46, 21 January 2012 (UTC)

Reply to Waleswatcher about 273. I just wanted to facilitate comparison between the Centigrade and Kelvin temperatures which are cited in different parts of the article.

"every single source you dug up." Is this pejorative? It looks so. I was not digging for sources on that question; I was just dealing with other questions. Are you implying that there is something wrong with digging up sources? You feel comfortable to discount Newton as unreliable without even checking what he wrote.Chjoaygame (talk) 14:01, 21 January 2012 (UTC)

It wasn't intended to be pejorative at all - I apologize if it came across that way. I was impressed actually by the work you put into finding all of that; thanks for doing it. It seems that all the sources agree on a number that's pretty close to 500C, with the possible exception of Newton. I really don't think Newton should be regarded as reliable on this, particularly in the absence of the direct quote from him. Techniques for measuring these quantities have improved vastly since Newton's time. Waleswatcher (talk) 14:58, 21 January 2012 (UTC)

Isaac Newton and reliable sourcing

The second, partly controversial, edition of Ernst Mach's Die Principien der Wärmlehrere was published in 1900, and considered important enough to be translated by McCormack, T.J. into English in 1904 and further workers till 1942. I am looking at a 1986 English edition, edited by McGuinness, B., published by D. Reidel, Dordrecht, Holland, ISBN 90–277–2206–4. Einstein respected the work of Mach. Mach on pages 62–63 of the English considered the temperature studies of Newton. In those days, the term 'degree of heat' was used to refer to temperature. Newton used a linseed-oil thermometer. Ice just beginning to freeze, and the greatest degree of heat received by the thermometer from the contact of a man's body, were the reference temperatures, 0 and 1. Newton used iteratively his law of cooling in a flow of air to estimate higher temperatures. His high-temperature reference was red-hot iron. Taking the body temperature at 37 °C, Mach used Newton's published data to put temperatures to various phenomena. One might criticize the estimate 37 °C as likely to be a little too high, but let us accept it. Then Mach found Newton's temperature of boiling water to be 104 °C, and his temperature for melting lead to be 296 °C (current value 326 °C). Mach reports Newton's temperature of red heat as 600 °C; apparently that referred to an ordinary kitchen fire visible in daylight, I think. Newton's 1701 original was in Latin, in Phil. Trans. Roy. Soc. 270: 824–829, and translated into English in 1809 at Phil. Trans. Roy. Soc. (abridged) 4: 572-575. These are reproduced by ed. Cohen, I.B. (1958) Isaac Newton's Papers & Letters on Natural Philosophy and related documents, Cambridge University Press, pages 259–268. A little OR (naughty, naughty!) finds that Newton reported viewing in various conditions: at night in the dark, in the twilight immediately before sunrise, and in clear daylight. Looking in the dark at coal cooling from having been burning (not the metal observed by Draper, nor the fire reported by Mach above), Newton saw the faintest glow, by my calculation (114 x 37 / 12 = 351.5), using Newton's instructions, I think in agreement with Mach's procedure, at about 351 °C. Newton makes no comment about greyness or redness of the glow. Partington (1949) cites reports that some oxides glow at 210 °C. Newton's words were "Calor quo corpora ignita defervendo penitus desinunt in tenebris nocturnis lucere, & vicissim incalescendo incipiunt in iisdem tenebris lucere sed lucere tenuissima quae sentiri vix possit."

cavity radiation

I made a much more significant change to the paragraph that mentioned the cavity. The old version included a sentence "The failure of this theory in terms purely of radiation indicates that the spectrum of cavity radiation in thermodynamic equilibrium is due to the properties of the materials of the cavity walls or content, by which those materials obey Kirchhoff's law of thermal radiation." That's badly wrong. While it's true that if the walls emit and absorb radiation in quantized packets that could resolve the problem (and I think that's what Planck originally thought), that's not necessary (so "indicates" is very misleading) and the actual solution as currently understood is not that - it's that the radiation modes themselves are quantized, which has nothing to do with the material of the walls. Waleswatcher (talk) 13:04, 21 January 2012 (UTC)

Thank you for this comment, Waleswatcher, helpfully telling me that "That's badly wrong."

For me 'indicates' means 'points to', but does not mean 'implies', which I would have written if I had meant it. I think your reading of my word "indicates" is your own.

The experimental evidence is that matter is necessary to bring about a Planck distribution. This does not judge whether the modes themselves are quantized. In another place an editor has claimed that the radiation itself will reach a Planck distribution without the presence of matter, but has not offered a reliable source for this. He apparently thinks that his superior understanding of quantum electrodynamics trumps the ordinary Wikipedia need for reliable sourcing. Our sources here say, for example, that "The treatment of Kirchhoff's laws is nowhere more explicit and readable than in Planck ... (1913/1914) [Goody & Yung 1989, page 64]" and Planck insists on the need for matter. If there really were no need for the material, one would be wondering why such a long article on the black body. Your edit seems to presume that your understanding of "the actual solution as currently understood" does not need sourcing. This seems at odds with your new zeal for reliable sourcing. By the way, your unsourced opinion, that Planck originally thought that the walls emit and absorb radiation in quantized packets, is contrary to recognized and sourced opinion about what Planck originally thought. And it is not the occupation numbers of the modes that are quantized, but the modes themselves, I think. I hope this does not lead to the article on the black body becoming an article on quantum optics or quantum electrodynamics.Chjoaygame (talk) 14:01, 21 January 2012 (UTC)

It is a misuse of the word 'phenomenon' to write as if the ultraviolet catastrophe were are phenomenon when it is only a theoretical hypothesis that is known not to be realized in nature.Chjoaygame (talk) 14:06, 21 January 2012 (UTC)

Telling "you"? I didn't know you wrote that. Regarding the physics, I'd be happy to provide sources for the fact that occupation numbers of electromagnetic modes are quantized, and that's the resolution of the UV catastrophe. Any textbook on quantum field theory or "modern physics" will suffice, and there's a standard treatment of equilibrium cavity radiation that can be found in any text on stat mech. The material of the walls plays no role whatsoever in those calculations, but they (the calculations) produce the correct Planck spectrum. I'll wait and see if you really insist on that, but if you do I'll provide citations. As for Planck's beliefs, I'm not sure what you're referring to - I didn't put anything in the article about that, so there's no need for sources.

By the way, as for occupation numbers versus modes themselves - the quantity that is continuous in classical physics but discrete in QM is in fact the occupation numbers (if you like, the amplitude of the mode), NOT the wavelength. The wavelength is quantized even classically by the finite size of the cavity. That's a very common misunderstanding, so we should ensure the article makes it clear. Waleswatcher (talk) 15:16, 21 January 2012 (UTC)

Here's one reference that's readable in full online http://farside.ph.utexas.edu/teaching/sm1/statmech.pdf , section 8.10 p 181. He derives the classical UV catastrophe for cavity radiation and then shows how QM resolves it by quantizing the mode occupation numbers (see also sec. 8.6, p. 175). Note that the material of the walls plays absolutely no role - in fact he says explicitly that the boundary conditions aren't relevant so long as the cavity is larger than the wavelengths of interest. Any other text on stat mech has a similar derivation. Waleswatcher (talk) 15:59, 21 January 2012 (UTC)

Thank you Waleswatcher. I read on page 182 of Fitzpatrick: "Consider an enclosure whose walls are maintained at fixed temperature T." Why would he say that if the walls were irrelevant? There is a simple answer on page 184: "Well, as usual, quantum mechanics comes to our rescue." Then he refers us back to page 176, where he tells us about his equation (8.39): "This is known as the Planck distribution, after the German physicist Max Planck who first proposed it in 1900 on purely empirical grounds." Peviously, on page 175 he has told us "In fact, photons enclosed in a container of volume V, maintained at temperature T, can readily be absorbed or emitted by the walls." Also, Fitzpatrick has his history wrong. Planck was far from working on "purely empirical grounds".

You write that Fitzpatrick "says explicitly that the boundary conditions aren't relevant so long as the cavity is larger than the wavelengths of interest." I suppose you are there referring to Fitzpatrick's sentence: "As long as the smallest of these lengths, L, say, is much greater than the longest wavelength of interest in the problem, λ = 2π/k, then these assumptions should not significantly affect the nature of the radiation inside the enclosure." Since he has said that the photons can be absorbed or emitted by the walls, and that the walls are maintained at fixed temperature T, it is not reasonable to interpret this sentence of Fitzpatrick as you have just interpreted it. I don't see Fitzpatrick having explicitly used the words "boundary conditions" there.

In my observation, a set of lecture notes outlining a single semester intermediate level course intended for upper division undergraduates is not likely to be a reliable source for the proposition that the material of the walls is irrelevant, and this seems to be no exception.

As for your opinion about Planck's beliefs, I didn't say you had put anything about them in the article. I just noted that your opinion was unsourced and contrary to recognized and sourced opinion.

I am not about to insist on anything. I have learnt that such insistence often provokes further responses that do not advance matters.Chjoaygame (talk) 16:57, 21 January 2012 (UTC)

It's not that the walls are totally irrelevant - they do affect the spectrum, substantially so at long wavelength compared to the size of the cavity. But at short wavelength they really are irrelevant, and the material they are made of appears nowhere in that derivation, because it too is irrelevant. All you need to know are the dimensions of the cavity and the boundary conditions, and even those factors drop out at short wavelength. The temperature of course is relevant, but again, that doesn't necessarily have to do with the walls. For instance, you can do the calculation with periodic BCs for the modes, so that there are no walls at all. The result at short wavelength (or in the limit that the size of the cavity goes to infinity) is exactly the same.

• You are here repeating your assertions, but not offering argument in support of them. The reason, why the material of the wall appears nowhere in the derivation except in its announced assumptions, is that the derivation is slipshod and hardly physical. As you agree below, the quality of the source is poor.Chjoaygame (talk) 18:56, 21 January 2012 (UTC)

I don't disagree with you about the quality of that source, it was just the first thing google brought up. Waleswatcher (talk) 17:07, 21 January 2012 (UTC)

• Why did you put up a source that you got from Google like that?Chjoaygame (talk) 18:56, 21 January 2012 (UTC)

Here's a quote from "Modern Physics" by Tipler et al, p. 139: "In that year [1905] Einstein applied the same ideas [quantization] to the photoelectric effect and suggested that, rather than being a mysterious property of the oscillators in cavity walls and blackbody radiation, quantization is a fundamental characteristic of light energy."

• Tipler has his history wrong. Einstein's idea of quantizing the light itself was far from Planck's thinking and was considered so at the time. Nowadays it is common enough for careless writers like Tipler to make elementary mistakes like that.Chjoaygame (talk) 18:56, 21 January 2012 (UTC)

Here's a quote from http://www.itp.uni-hannover.de/~zawischa/ITP/blackbody.html: "Radiation emerging from a small opening in a cavity depends only on the temperature and not at all on the material of the cavity's walls." Waleswatcher (talk) 17:14, 21 January 2012 (UTC)

• Again a careless writer. The material of the walls must be opaque to all relevant wavelengths and must not be perfectly reflective for any of them. That's where the physics lies, and your source is really missing much of the physics.Chjoaygame (talk) 18:56, 21 January 2012 (UTC)

One more quote, from "Statistical Mechancs" by Huang (a standard text), p.278: "...making a cavity in any material...heating the material to a given temperature...If the cavity is sufficiently large, the thermodynamic properties of the radiation in the cavity should be independent of the nature of the wall. Accordingly we can impose on the radiation field any boundary condition that is convenient." Then he goes on and derives the Planck spectrum using quantization of occupation number. Waleswatcher (talk) 17:21, 21 January 2012 (UTC)

• Another careless writer, again missing much of the physics.Chjoaygame (talk) 18:56, 21 January 2012 (UTC)

Just to be clear, you think ALL those references (which include standard texts used in modern physics and statistical mechanics courses around the world) are wrong, while you are right? That's fine - you're entitled to your opinion - but you're not entitled to put your opinion into wikipedia.

Out of curiosity, do you regard it as merely a coincidence that one can derive the Planck spectrum using periodic boundary conditions, where there are no walls at all? Waleswatcher (talk) 19:13, 21 January 2012 (UTC)

It's not just my opinion that those writers are careless. It's obvious in what they write, at least, supposing you are citing them accurately. Are you suggesting that a black body cavity radiation source can have glass walls or perfectly reflective walls? If so, why do you or other physicists take any interest in the ideal black body? Our reliable sources deny that a black body cavity radiation source can have non-opaque walls or perfectly reflecting walls, and I just go with them; it's not just my opinion. It seems that if enough writers are careless, you will accept their carelessness against our reliable sources. I note that you do not actually address the contents of our reliable sources. I note that you have not attempted to reply to my pointing out that your sources make elementary mistakes when they say that the nature of the wall is irrelevant. If the wall is really irrelevant, why does your cited source Fitzpatrick say so much about it and its interaction with the photons in the announcement of his discussion?

The "derivations" to which you refer are not physical derivations; they are just deductions from mathematical postulates. Those postulates are indeed derived properly by physical arguments, but they are being quoted without mentioning their proper physical derivations. The postulates are mathematical shells of which you effectively ignore the physical basis and empirical meaning. The idea of the black body is a physical one.

To derive by physical argument the Planck distribution without material is the challenge. You haven't even begun to cite a reliable source that can do it. I suggest you think about your "periodic" case. It has to refer to an infinite space if it has no walls. Are you seriously suggesting that experiments have tested infinite spaces with none but pure electromagnetic radiative content?Chjoaygame (talk) 21:00, 21 January 2012 (UTC)

I don't really know what you're arguing about at this point. If you think that the material the cavity walls are made of is relevant for black body radiation from the cavity, you're wrong and in explicit contradiction to multiple extremely reliable sources (like standard textbooks on statistical mechanics). I haven't seen you present any source that goes against that. If you're not disputing it, or you are but you don't think it should go into the article, then we're done discussing. If you want to propose alternate language to what's there now, please go ahead and we can discuss it.

As for glass, if the "cavity" is transparent to some frequencies then it obviously won't serve as a cavity for those frequencies. The sole requirement is that it be optically thick (i.e. opaque) to the relevant frequencies. As for periodic BCs, no, it's not infinite volume, it's finite precisely because of the BCs. Of course no one has done such an experiment, how could they? The point is that the derivation works exactly the same there (and I have no idea what you mean by "physical derivation", all derivations are mathematical by definition) and predicts exactly the same spectrum that's predicted for all cavities, regardless of their composition. Waleswatcher (talk) 21:17, 21 January 2012 (UTC)

Waleswatcher writes: "I have no idea what you mean by 'physical derivation'." Chjoaygame replies: Yes, that is the problem that concerns me here.

Waleswatcher goes on: "all derivations are mathematical by definition." Chjoaygame replies: It is a problem that you see it like that.

Your 'derivation' that 'makes no reference to the walls' uses a postulated mathematical formula for photons as bosons. The photon number is "not conserved". It sounds magically clever. No mention of walls. No filthy physics; just pure lovely mathematics!

But really it has fully hidden the physics from you, so that you cannot see that it makes a very essential reference to the walls.

While the number of photons in the cavity can vary in a way not open to particles of ponderable matter such as molecules, the law of conservation of energy still applies. When the photon number (number of photons in the cavity) varies in accord with the Bose-Einstein statistics, the energy of the photons comes from or goes to somewhere. You are looking at an 'open' system, from which photons can 'perish' or be annihilated, or into which photons can be 'created'. Not quite the same as entry and exit of molecules, which show conservation of matter, but still effectively the same from the viewpoint of energy conservation. The creation and perishing of the photons is just by interaction with the material of the walls or cavity material content, though that is hidden, or at least left implicit, not made explicit, in the mathematical formalism. So the mathematical formalism just hides the physics from you. Since you only look at the mathematical formalism, you don't see the 'derivation' as any more than that. From the mathematical formalism, you can't tell whether the walls come into it or not.

You infer, and argue to me, that this means that the walls are irrelevant. Indeed, they might as well be made of glass, since the mathematical formalism leaves them invisible to you! But we have agreed that they do matter insofar as they mustn't be made of glass. We also need to remember that if there is no ponderable matter inside the cavity, then the walls have another necessary property: they are not allowed to be perfectly reflective.Chjoaygame (talk) 07:22, 23 January 2012 (UTC)

Again - I don't really know what you're arguing about at this point. I disagree with some of what you said above, but you're perfectly entitled to your opinion. But if you want to put something into the article that says that the walls of the cavity are important you'll need to provide a source that explicitly says so, and we'll have to give due weight to the fact that very reliable sources say exactly the opposite. Waleswatcher (talk) 13:58, 23 January 2012 (UTC)

Dear Waleswatcher, the importance of the walls is best explained by Planck himself, in his 1914 translated edition. It is true that the details of the walls don't matter: that unimportance of detail is what put Kirchhoff on the map in this part of physics. But their presence for the function of transduction between wavelengths is essential. And to do their transducing to thermodynamic equilibrium they need opacity and imperfection of reflectivity. Books that say the nature of the walls doesn't matter are just lazily or carelessly half-citing Planck, forgetting that although the detailed nature doesn't matter the general characters of opacity and imperfect reflectivity are essential.

As I mentioned above, you are relying on quantum mechanical argument, essentially the Bose-Einstein statistics, for your 'derivation without walls or material content'. You cannot see the physics in that. It is simply hidden in the formalism. But it is an ingredient in the original generation of the formalism. To exclude this you will need to exclude Planck's own treatment of his law. It may seem to you that the purely mathematical statements of the Bose-Einstein statistics allow one to ignore the presence or absence of the walls or in the case or perfectly reflecting walls the presence or absence of Planck's famous "speck of carbon", matter inside the cavity. But if you don't see how the walls are built into the mathematical formalism, that's all it will be for you, a mathematical formalism, not physics. Perhaps it may help if you actually read Planck 1914. It's a classic, at [2]. It is recommended by other present-day reliable sources; their explicit recommendations are what led me to read it. Sure, many things it says are superseded by later thoughts of Planck and massively superseded by subsequent developments of course. Likely you won't want to read absolutely every detail of it; fair enough.

I would say that your physics teachers have cheated you by not leading you to understand this. Physics really is about physics, it's not just a mathematical game. As far as I can work out, your good work in dealing with Dicklyon about the other matter is based on your understanding of the physics, against his lack of interest in it. This case of the walls is like that. The classic literature about this is clear as day, in particular from Ehrenfest, who was known as more or less a walking encyclopaedia of physics.

Some modern sources that perhaps may persuade you to take this seriously:

• Loudon, R. (2000), The Quantum Theory of Light, third edition, Oxford University Press, ISBN 0–19–850177–3, page 9: "The essence of the quantum theory of the radiation field is thus the association of a quantum harmonic oscillator with each mode of the field." This means that the modes of the field are distinct from the quantum harmonic oscillators that are respectively associated with them. Those quantum harmonic oscillators are the abstract representations of the emission and absorption events that are the births and deaths of photons. They represent the transitions of electrons from one orbital to another. It is perhaps true that even Loudon doesn't really quite seem to understand this fully, when he more or less reproduces Planck's ladder of states of the oscillator. The oscillator is not a field mode. It is something that is to be associated with the field mode, as a specifically tuned probe. We never detect photons as such. We only detect their effects on electrons and molecules. Our eyes are very well nano-engineered photon detectors, that work by having suitable molecules at the ready to absorb any relevant incident photon, and then the molecules are deformed by the excitation, so as to pass this message to the body of the cell in which they sit, and the cell magnifies the effect electrically and passes it to the brain, where we become aware of it.

• Mandel, L., Wolf, E. (1995), Optical Coherence and quantum optics, Cambridge University Press, ISBN 0–521–41711–2, [3], page 229: "In this chapter we will apply the correlation theory of scalar wavefields developed in the previous chapter to study radiation from localized primary and secondary scalar sources of any state of coherence." This means that the light is considered in terms of the material molecules that were its source.

• Razavy, M. (2011), Heisenberg's Quantum Mechanics, World Scientific, Singapore, ISBN 978–981–4304–10–8. This tells about Heisenberg's original quantum mechanics, how he considered the physics in constructing his original mathematical formalism. It is about the quantum harmonic oscillators as transitions between molecular orbitals mentioned above for Loudon.Chjoaygame (talk) 17:32, 23 January 2012 (UTC)

If the walls or material content really didn't matter, then the idea of a black body would become utterly redundant and we might as well delete the whole article.Chjoaygame (talk) 18:05, 23 January 2012 (UTC)

It's not clear to me what the argument is here. The whole point of using a cavity with a small hole is to make the effect of the wall material go to zero (in the limit of small hole); in that sense, it doesn't matter; but something needs to be there. There's another limit that might bother you, which is the limit of perfectly reflecting walls; that just makes the required hole size smaller to get to an given precision of blackbody spectrum. What matters is that there's some material to enable energy move between different frequencies; a thin gas of hydrogen atoms with only discrete states is probably not enough, but any non-crystalline solid is certainly fine. Dicklyon (talk) 18:24, 23 January 2012 (UTC)

• Thank you for this response Dicklyon. I agree. Something needs to be there. Waleswatcher is of the view that nothing needs to be there. The perfectly reflecting walls just have the effect of making the something need to be some matter within the cavity, the "speck of carbon". I agree with your next sentence. In this you and I agree against Waleswatcher.Chjoaygame (talk) 20:20, 23 January 2012 (UTC)

Chjoaygame, thanks for the sources. I'll read Planck when I have time, I'm curious. But (somewhat ironically) Planck actually isn't the best source for this, because the modern understanding arises from relativistic quantum field theory, a topic that wasn't fully developed until the 1960s. I don't agree with you that modern authors are sloppy or just roughly quoting Planck. They are describing something else - a calculation that one can do using QFT to derive the Planck spectrum. That calculation doesn't refer in any way to the material of the walls, nor does it matter what boundary condition you use to model them (as long as there is one). For instance you could couple the radiation field to harmonic oscillators, or you can just put periodic (no walls at all, but finite volume) or Dirichlet (perfectly reflecting) boundary conditions. By the way there's no problem with perfectly reflecting BCs. If you put any BC and also assume the radiation is in thermal equilibrium, you will get the Planck spectrum. It's true that radiation in a perfectly reflecting cavity would take a long time to equilibrate because it wouldn't interact with the molecules in the walls, but it actually would equilibrate eventually if only because of photon-photon interactions (which are weak but non-zero), or even gravitational interactions if you really want to push it. The time scale for that is much longer than for equilibrating with a wall made of some normal material, but it happens nonetheless. The fact remains that the necessary ingredients for the Planck spectrum are any boundary condition and thermal equilibrium, full stop.

As I suggested before, if you want to continue this discussion, please put forward some specific language you want to insert or change in the article, and we can discuss it. I don't think it's particularly useful or appropriate to keep discussing physics on this talk page in the absence of some disputed language. Thanks! Waleswatcher (talk) 18:41, 23 January 2012 (UTC)

One comment on this: "If the walls or material content really didn't matter, then the idea of a black body would become utterly redundant and we might as well delete the whole article." It's exactly the opposite. What makes the black body spectrum so powerful and interesting is that it is very, very general - that it doesn't in fact depend on the material. The fact that you can observe almost exactly the same spectrum in the cosmic microwave background (which did not originate from any kind of cavity or "body" in the normal sense) as you do from an oven or the sun or a hot piece of platinum is truly amazing. Waleswatcher (talk) 18:48, 23 January 2012 (UTC)

• Thank you Waleswatcher. I was wondering why you had not long ago taken the QFT line. I am told that it will take several lifetimes of the universe to see the looked for equilibrium; most of the matter of the universe will turn to radiation in that story, and then we will see the Planck distribution achieved; I have asked for a research grant to observe the process, but the bureaucracy is taking its time to process my request. As for the phrase 'black body spectrum': why bother with the words 'black body'; surely the right term would be just 'electromagnetic radiative spectrum'? I have said enough. The next step is up to you.Chjoaygame (talk) 20:20, 23 January 2012 (UTC)