Talk:Black body

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Contents 1 Headline text 2 Good Article nomination has failed 3 "Blackbody" vs. "Black body" 4 Blackbody vs emission spectrum 5 Sig figs in lava caption? 6 Graph inverted? 7 Planck's equation? 8 Earth Surface temperature 9 Mechanism? 10 First: equation or interpretation? 11 Temperature of the sun calculation 11.1 disputed 12 How to phrase the english 13 Visible Color 14 Some questions about Black Bodies 15 Black bodies and black holes 16 Factor Pi Wrong? 17 Minor Edit 18 Modified Blackbody 19 Radiation Emitted by a Human 19.1 e ≈ 1! 20 Replace WMAP image with FIRAS spectrum? 21 Spectrum 22 Visible light from room temp. objects? 23 Something is wrong with "the spectrum of an incandescent bulb in a typical flashlight"?! 24 Vandalism or cutesy physicist term?

[edit] Headline text

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Black body was a good article candidate, but did not meet the good article criteria at the time. Once the objections listed below are addressed, the article can be renominated. You may also seek a review of the decision if you feel there was a mistake. Date of review: 21/07/2006

[edit] Good Article nomination has failed

The Good article nomination for Black body has failed, for the following reason:

(This article needs a lot more sources. The entire Explanation section needs citation. I'm instead nominating this article for Unreferenced Good Article status) Talkstosocks 03:44, 22 July 2006 (UTC)

[edit] "Blackbody" vs. "Black body"

It looks strange to me to see "blackbody" used as a noun. In the texts I've seen, you normally say "black body" as a noun, and either "black-body" or "blackbody" as a compound adjective (e.g. for "black-body radiation").

I would vote to use "black body" anywhere we use it in noun form, and then hyphenate the adjective form for consistency. —Steven G. Johnson 22:54, Mar 29, 2004 (UTC).

• My vote is with you. - Omegatron
• The footnote on this isn't really relevant; hyphenating compound adjectives is a standard English language convention. -- Jrstewart 07:45, 20 August 2005 (UTC)

Black body says a color of an object is black.

Black-body or blackbody says a phonomena of an object is going to emmise or obsorb energy.-As my knowing of from hints of Blatt's book.--GyBlop 13:47, 27 February 2006 (UTC)

[edit] Blackbody vs emission spectrum

I don't understand how one generates a black-body radiation from a hole in a cavity. I learned that any matter generates an emission spectrum with clear bands, depending on its atomic composition. How is this converted by the cavity to a uniform spectrum that follows Planck law ? Are there also bands in the black-body radiation ?

If a substance only absorbs energy at certain wavelengths, which will happen if it is a dilute gas for example, then it will also only emit radiation at those wavelengths. That is, it is a grey body and not an ideal black body. (See also Kirchhoff's law.) —Steven G. Johnson 19:03, Aug 28, 2004 (UTC)

In all of my literature (such as Siegel and Howell), the term "Grey body" only applies to a body in which it's emissivity is not a function of wavelength, so if something shows distinct peaks it is *not* a grey body. Your description seems to imply that "grey body" only means that it's not black. Kaszeta 19:36, 28 Aug 2004 (UTC)

You're right, I'm over-using the "grey body" term. —Steven G. Johnson 19:56, Aug 29, 2004 (UTC)

Could some one help ? Pcarbonn 17:17, 10 Jul 2004 (UTC)

I'll take a stab at this, and hopefully someone will correct me if i'm wrong.

Note that a substance's emission and adsorption bands occur at the same frequencies. Whether the substance is emmitting more energy than is it adsorbing is just a matter of how much energy it has to emit versus how much radiation there is to adsorb.

Black-body (and grey-body) spectra are properties at thermal equilibrium — in this state, the substance by definition must be absorbing the same amount of energy as it is emitting —Steven G. Johnson 19:03, Aug 28, 2004 (UTC)

I disagree, but perhaps i am fundamentally mistaken. As i understand it, a substance's spectrum is an effect of how strongly it interacts with radiation at different frequencies, but the black body spectrum is a statement about the equilibrium statistics of radiation in the cavity. A small enough hole into an otherwise closed cavity is 'black' because any light falling on the hole from outside will bounce around the cavity for long enough to be adsorbed — the chance of it getting back out of the hole is sufficiently small. So the hole into the cavity is able to adsorb all light that falls on it. If a non-black body is in thermal equilibrium with its environment (including radiation field) then a passive measurement of the radiation from the body will be unable to distinguish it from a black body. All the gaps in its emission spectrum are filled in by radiation from the environment. Some frequencies are adsorbed and re-emitted, others simply reflected or scattered. An emission spectrum is not a thermodynamic equilibrium phenomenon. —Thomas w 16:02, 29 Aug 2004 (UTC)

Yes and no. Yes, the black body spectrum is derived from equilibrium statistics of the photon gas, and in this sense if you put an object in an large isothermal cavity the photon statistics in the vacuum surrounding the object will reach the same distribution regardless of the substance of the object. On the other hand, it is precisely from this situation that Kirchhoff's law is derived, showing that the amount of radiation being emitted by the body (as opposed to an infinitesimal hole in the cavity) is equal to its absorption. Yes, any time you observe thermal emission you are doing so in a system that is not in equilibrium (the observer/ambient environment is at a different temperature than the emitter), but black-body-like analyses assume that things are sufficiently static that equilibrium descriptions apply locally. Yes, it's true that in non-equilibrium conditions a substance may be absorbing more energy than it is emitting, or vice versa, but the emissivity is still closely related to absorptivity by Kirchoff's law (assuming that local equilibrium applies). —Steven G. Johnson 19:56, Aug 29, 2004 (UTC)

No part of an emission spectrum is completely black. While simple quantum transitions will dominate the spectrum, higher order (many-step) transitions, thermal doppler broadening of transitions and other effects (Heisenburg uncertainty relations?) will allow all substances to interact with all wavelengths of radiation to some extent. The effect of a cavity is that radiation is trapped in it for long enough to come into equilibrium with the substance forming the cavity at all wavelengths, not just those for which is has transitions that interact strongly with the radiation field.

The black-body spectrum depends only on the temperature of the cavity, and is independant of the substance the cavity is formed from.

This implies that your (Pcarbonn's) 11 July 2004 edit is incorrect on this subject.

The spectrum depends on the substance because it depends on the emissivity (and thus, the absorptivity by Kirchhoff's law) of the substance. For a realistic material, you thus have a grey-body spectrum instead of a black-body spectrum. —Steven G. Johnson 19:03, Aug 28, 2004 (UTC)

Yes if the substance is not in thermal equilibrium with the radiation field, as for a hot body in a colder environment, but no for radiation in an enclosed cavity. Its colour is determined only by the equilibrium distribution of energy among the field modes, a function of the temperature of the system. The fact that some field modes come into equilibrium with the substance much faster than others is irrelevant. Rates get washed out as you take things into thermal equilibrium. —Thomas w 16:02, 29 Aug 2004 (UTC)

Well, that depends on what you mean by "emission" of the body. Kirchhoff's law is derived precisely under the assumption of thermal equilibrium (and, in particular, detailed balance), and shows that that body's emissivity equals (1 − reflectivity), or "absorptivity". (Of course, in equilibrium per se it is difficult to distinguish the emission of the body, since it is surrounded by a photon "gas" that does follow the black-body formula. I think this is what you mean, and Kirchhoff's law is sometims stated this way, but this is not the same thing as stating that the photon statistics within the body or leaving its surface, follow the black body law.) Another is to directly look at the derivation of the black-body formula, which assumes that the photons form a noninteracting "ideal gas"; as Landau &amp Lifshitz write (Statistical Physics: Part 1): "If the radiation is not in a vacuum but in a material medium, the condition for an ideal photon gas requires also that the interaction between radiation and matter should be small. This condition is satisfied in gases througout the radiation spectrum except for frequencies in the neighborhood of absorption lines of the material, but at high densities of matter it may be violated except at high temperatures. ... It should be remembered that at least a small amount of matter must be present if thermal equilibrium is to be reached in the radiation, since the interactions between the photons themselves may be regarded as completely absent." In a related vein, there was a recent Phys. Rev. Letter (Bekenstein, PRL 72 (16), 1994) that directly derives the statistics of photon quanta for an absorbing (ideal grey-body) material and shows that they are consistent with Kirchhoff's law (depending only on the absorptivity and the temperature). —Steven G. Johnson 19:56, Aug 29, 2004 (UTC)

Incidentally, the accuracy and precise applicability of Kirchhoff's law and black/grey-body formulas etcetera when applied to experimental non-equilibrium thermal emission (i.e. not objects within an isothermal enclosure) has apparently been much debated. See e.g. Pierre-Marie Robitaille, "On the validity of Kirchhoff's law of thermal emission, IEEE Trans. on Plasma Science 31 (6), 1263-1267 (2003) for a recent paper on the topic that reviews some of the literature (this author also takes a particular position in the controversy; I'm not sure how well accepted or well justified this position is without a more careful review). —Steven G. Johnson 19:56, Aug 29, 2004 (UTC)

Thank you all for your responses. Unfortunately, I cannot understand all of them (mainly because I do not know Kirchhoff's law). I guess I could further study this. Also, at some point, we should update the article to clarify this issue.

Just to clarify my concern, my question concerned the following paragraph:

In the laboratory, the closest thing to a black body radiation is the radiation from a small hole in a cavity : it 'absorbs' little energy from the outside if the hole is small, and it 'radiates' all the energy from the inside which is black. However, the spectrum (...) of its radiation will not be continuous, and only rays will appear whose wavelengths depend on the material in the cavity (see Emission spectrum). (...)

If this statement is wrong, please correct it ASAP in the article. I would also invite you to describe how one generates a black body radiation in the laboratory, and how its spectrum is measured with adequate precision. In particular, it would be useful to describe the photon field surrounding the cavity in the laboratory (very small energy ? in equilibrium ? with what ? ...), and the spectral resolution of the measuring equipment. Once we have that cleared, I believe that it will be much easier to discuss why black-body radiations in the laboratory have spectral rays, or not.

(actually, I would expect the measuring instrument to be also sensitive to some specific frequencies only, if it is made of ordinary matter. But I could be wrong again on this one: the human eye seems to respond to a wide range of frequencies: where is the trick ?)

Above, someone cites thermal doppler broadening of transitions as a way to broaden the bands. Because thermal velocity of atoms is so small compared to the speed of light, I would think that this effect would not be sufficient to remove the spectral rays (unless they are very very close to each other). Am I wrong ? Also, my (limited) understanding of the Heisenberg uncertainty principle makes me doubt that this could be another way to broaden the spectral rays (if it were, then how could we observe rays in some circumstances ?).

It is not that spectral features are completely smeared out by these effects, but rather that they imply that even substances with sharp lines interact with all wavelengths at least a tiny bit. It is only when radiation is trapped in a large cavity within the material for long enough to establish local equilibrium that all spectral features of the radiation dissapear. Also note that sharp lines are charicteristic of low pressure gasses. Solids are often much messier. —Thomas w

At the end, if we can say when spectral rays are observed and when they are not, we should probably update the emission spectrum article. (currently, it seems to say that they are always observed). Pcarbonn 20:02, 30 Aug 2004 (UTC)

I'm sorry i don't have time to respond in detail, but here are some of the better links that googling 'blackbody cavity' has furnished me with:

• A really simple cavity
• Using cavities for calibrating pyrometers. The graph at the start is hard to read but seems to back me up.
• Our question answered by Ask a Scientist.
• A PDF file that includes a nice explanation of how cavity radiation varying with substance would lead to perpetual motion.

—Thomas w

The picture of the colours of blackbody radiation looks like photoshop/gimp's blackbody gradient. I'm not sure if those are the true colors of the radition, so I'm going to upload a new image, using the information from http://www.vendian.org/mncharity/dir3/blackbody/UnstableURLs/bbr_color.html Zeimusu 01:37, 2005 Jan 13 (UTC)

[edit] Sig figs in lava caption?

The caption to the lava picture says: lava flows at about 1,000 to 1,200 °C (1,273 to 1,473 kelvins).

It's not apparent how many sig figs are meant for 1000 to 1200, but I can't believe it's 4 (particularly given the 'about' preceding). Perhaps the kelvin conversion should be correspondingly reduced from 4?

I know why this bugs you but "1,000 to 1,200 °C (1,300 to 1,500 kelvins)" would look wrong as well as the difference should 273. Maybe it would be better to drop the kelvins completely?--Maddog Battie 17:10, 4 August 2005 (UTC)

[edit] Graph inverted?

Aren't black-body curves normally shown with the X-axis representing frequency rather than wave-length? This will also show the ultraviolet catastrophe more clearly. CS Miller 20:19, July 11, 2005 (UTC)

I believe wave-length is more common.--Maddog Battie 16:01, 4 August 2005 (UTC)

Also someone might to add this image if its thought useful Image:Blackbody spectral radiance.gif--Maddog Battie 16:01, 4 August 2005 (UTC)

[edit] Planck's equation?

The equation reads:

I(v)= .....

where I(v)dv is ....

Which one is correct?

Both statements are correct. The first statement gives the expression for I(ν). The second statement explains what I(ν) means, but it does so by explaining what I(ν)dν means, because that is easier to understand. Its a standard way of explaining the meaning of an intensive physical quantity.

[edit] Earth Surface temperature

The derivation given seems totally logical to me . But the climate change community promulgates the relationship , essentially

  ( ( 1 - a ) * S % 4 ) = ( e * StephanBoltzmann * T ^ 4 ) 

( simplified from Meltdown : Predictable Distortion of Global Warming , Patrick J. Michael )

Where albedo , a , and emisivity , e , are independent so the surface temperature is a function of the ratio between them . My intuition is that they cannot be independent , and that's seems to be Kirchhoff's point . My intuition says that all changes in the insulative properties of the atmosphere can do is change the diurnal temperature extremes , not the mean . Is this correct ? -- Bob Armstrong

[edit] Mechanism?

I've taken a course on heat transfer, so I'm not totally in the dark, as it were, but I realize I don't have an intuition for this process. What is the mechanism by which an energetic atom releases a photon? They can't result from electrons jumping among energy levels because then we would see emission bands. I guess the intuition I have for black body radiation is that the little atoms are shaking around so the electric field of the electrons and protons is changing which results in a changing magnetic field and hence light. That seems a bit sketchy to me; in particular, the net charge of most atoms is zero, so how could a vibrating atom produce a photon? As a correlary, consider the impossible case of a single atom vibrating regularly; what would its emission spectrum look like? —BenFrantzDale 20:13, 13 October 2005 (UTC)

Hi - I think the answer I gave you on the Maxwell-Boltzmann page was not as helpful as it could have been. The bottom line is that in almost all practical cases, the photons and the atoms will interact to produce a black body spectrum. You are also right that as long as the atoms behave as line radiators, there will be problems generating black body photons with frequencies between the line frequencies. I think the answer is that you must have a continuum of electron energies - the atoms cannot behave strictly as line emitters. My experience has been with low temperature plasmas, and in these cases the densities of the atoms get so high that the energy levels are broadened until, when the black body limit occurs, they essentially form a continuum. This usually holds only over a certain frequency (i.e. energy) range. Outside that range, the plasma does not behave as a black body. In the case of a black body cavity, at a low enough temperature, the walls are emitting molecular infrared radiation which is easily broadened by mechanical vibration of the molecules, just as you thought. Also, if the walls are metallic, this implies a continuum of electron energy levels.

Check out the article Atomic line spectra for the mathematics. Instead of discrete levels, you could have a continuum of energy levels that were nevertheless Maxwell-Boltzmann distributed. The photons would have a black body distribution and the principle of detailed balancing describes the energy flow between the two at equilibrium. PAR 00:05, 14 October 2005 (UTC):

Thanks for the answer. I'm still a bit confused (which may be out of the scope of discussion for this article; I think my confusion may get into particle–wave duality). I feel like I have an understanding of atomic spectral lines; that makes sense to me. I also feel like I understand antennas; my understaning of antennas is more wave-like whereas my intuition of atomic spectral lines is of particles. From your answer above, it sounds like blackbody radiation is best explained in terms of particles. Is that correct? Then the continuum of wavelenghts results from a continuum of possible electron transitions? It strikes me as odd, though, that that continuum—the blackbody spectrum—is the same across most materials. Thanks. —BenFrantzDale 03:02, 14 October 2005 (UTC)

As i understand it, the central theorem, which predates quantum mechanics, it that at any given temperature and frequency, the ratio of emissivity to absorption has to be the same for any substance. If that weren't true, you could use filters to create a perpetual motion machine. If absorption is 1 (black) you get the black body spectrum. Here is another way to see why distinct electron energy levels don't put bumps in the spectrum coming from a cavity: Even for a line emitter, the probability of emission at any frequency is never zero. In a cavity with a small hole, a photon is likely to bounce of the walls many times before it escapes. A photon emitted (with high probability) at the frequency of an emission line has an comparably high probability of being absorbed in the next wall collision. It all evens out. What the discrete energy levels do do is keep thermal energy from from leaking into higher and higher energy levels, producing the ultraviolet catastrophe. --agr 10:28, 14 October 2005 (UTC)

There is no such thing as a theorem in physics - or anywhere else, outside mathematics and formal logic. Kirchhoff's law, just like anything else in physics, is therefore a theory, not a theorem 84.149.208.229 20:36, 30 March 2007 (UTC)

Yes, I forgot about that aspect of it, and that is the real answer. In the plasma example, if you have a volume of gas, all at the same temperature, that is much thicker than any photon's path to escape it, it will radiate as a black body. For those frequencies at the atomic line frequency, the photon's path is very short, because high emission/volume means high absorption. When you look at the gas, you are only looking at the radiation coming from at or near the surface. At frequencies between the lines, the emission is low so the absorption is low. Low absorption means the path of the photon is very large, and when you look at the gas, you see radiation both at the surface and deep into the gas. The fact that the emission per volume is low is exactly counteracted by the fact that the optical depth is large, and what you see is intensity that perfectly matches the intensity from the "on line" radiation. By "perfectly matches", I mean its in the same ratio as you would expect for a black body at that temperature!

We need to write this up and include it in the article. The thing that is missing is the detailed balancing. If they were not exactly matched, you could in principle set up a perpetual motion machine. What would that look like?

With regard to thinking it is strange that it's the same for all materials, you should look at the Maxwell distribution for massive particles. Is it strange that it is the same for any kind of particle? If not, then why should photons not equilibrate in the same way? Also since the black body spectrum fundamentally needs the radiation to be quantized in packets of energy in order to be derived, the particle viewpoint is indispensable. PAR 02:40, 15 October 2005 (UTC)

Interesting. That makes sense, I guess. I am still curious what the theoretical spectrum would be for a mass for which all atoms have the same energy. If absorption is involved, then it probably gets messy; I was initially thinking that the black body spectra would be a convolution of the Maxwell-Boltzmann distribution with the per-atom spectrum, but if absorption is inovlved then I guess it will be messier. —BenFrantzDale 22:48, 16 October 2005 (UTC)

If all the atoms had the same energy (and it wasn't the ground state) then you would have a laser. Thats how lasers work, a light source "pumps" a lot of atoms to the same energy, and then spontaneous emission begins the radiation output (A21 in the atomic spectra article), which stimulates the other atoms to emit in phase (B21 in the atomic spectra article). Usually there is absorption, because not all of the atoms get pumped, but if all the atoms were pumped, there would be no absorption (n1=0 in the atomic spectra article), at least not in front of the beam.

Regarding the blackbody spectrum, as long as the photons are in thermal equilibrium with each other at all energies, you don't need to inquire into what caused it, any more than you need to inquire into what particular kinds of particles and collisions produced a Maxwell distribution of massive particles. The equilibrium distributions are the least messy of all. The difference between the two distributions lies in their "statistics" and their mass. Photons are massless bosons, massive particles are, I don't know, "Boltzons" or something. Massive particles are really bosons or fermions, but at high enough temperatures (i.e. room temperature), they both develop a Maxwell distribution. Check out the gas in a box article - this shows how all these distributions are related. PAR 00:17, 17 October 2005 (UTC)

[edit] First: equation or interpretation?

It's not clear in the article if Planck first discovered the black body spectrum equation and THEN he interpreted the result as quantized energy or if was the other way around. Actually, it does look a little bit as if it was the other way Around. But wasn't the actual order: the equation, which he achieved mainly because of an interpolation of other formulas known back then, and then the interpretation? -- Henrique 21 October 2005

[edit] Temperature of the sun calculation

I just wanted to point out that this calculation doesn't work out correctly. I know the average temperature of the Earth is indeed what it gives, but for some reason this calculation does not work. It gives a value of 5958 K instead of the 5770 K it says. I've tried using another method, calculating the Earth's surface temperature from the solar constant, earth-sun distance and sun radius and get a value of 278 K instead of 287 K...which I know is wrong. I'm just wondering what this is attributed to, I'm sure it's something simple.

207.195.69.58 08:08, 27 October 2005 (UTC) Rob Hewitt - 3rd year Engineering Physics

I inserted a reference for the derivation of the relationship between the surface temperature of a planet and its star. It's a common derivation that can be found in many introductory astronomy books. Planetary Science by George Cole and Michael Woolfson is just one example. References are very important though, and I should have placed one into the article sooner. JabberWok 17:58, 7 November 2005 (UTC)

Yes, good idea. I still think the fact that the radiation is lost to empty space is an important assumption to state, because the idea of two black (or grey) bodies in an enclosure is often used to illustrate many principles, especially the idea that the absorption coefficient of a grey body is equal to its emission coefficient. Anyone used to these kinds of arguments will ask "why aren't the two at the same temperature if they are in equilibrium?" which of course they are not. And the reason they are not is because of the loss of their radiation to empty space. PAR 20:08, 7 November 2005 (UTC)

[edit] disputed

The derivation doesn't make sense. The Earth's power is missing the areal fraction that the Sun's already has. If it were included, the distance parameter is cancelled out and both bodies at thermal equilibrium would be at temperatural equilibrium. Earth's areal fraction as 1 is consistent with all of its radiation power being sent back to the Sun. The two bodies then only have different temperatures because their effective areas are permanently different. Meseems that the equation finds a solar temperature near measured is a coincidence. To have a remission fraction of 1, Earth would either need to be in a space warp or have variable emissivity that would mock perfect remission: It would be a black body toward Sun and a white body everywhere else. The great albedo of Earth's surface and air and their solar losses before Sun's radiation hits our ground would conspire to coincide with the imbalanced blackbody equivalence, I think. So the derivation at least needs an explanation that its methodology is invalid or incomplete, and needs to be expanded to consider how Earth actively vents its heat. lysdexia 00:02, 3 November 2005 (UTC)

I think that the derivation is ok if you assume that:

1. The sun and the earth are both spheres.
2. The sun and the earth both radiate as homogeneous black bodies, each at their own temperature.
3. The sun is unaffected by the radiation from the earth.
4. The earth absorbs all the solar energy that it intercepts from the sun.
5. The rate at which the earth radiates energy is equal to the rate at which it absorbs solar energy.

If these are true, then the rate at which the sun radiates into all space is , i.e. the rate per area times the solar surface area. The earth catches a fraction of this radiation. That is then equated to the rate at which the earth radiates: and you get the result . Can you say at what point in this chain of reasoning you disagree?

As you say, there is no equilibrium here, but you can still have this disequilibrium when both the sun and the earth are behaving as perfect black bodies. This is because almost all of the sun's energy is being lost to empty space. If we enclosed the sun and the earth in a mirrored box, then the box would be filled with 5600 degree photons, and the earth would warm up to 5600 degrees, and we would have equilibrium. PAR 01:39, 3 November 2005 (UTC)

I already know the assumptions! #3 is most invalid. And radiation is only and fundamentally a consequence of Coulomb's law, an energetic transaction between electric charges, so that the sun loses energy to "empty space" is nonsense. Radiation from the sun is an interaction between its excited charges and all charges in the universe; in other words, matter must be present in space for the sun to radiate. Moreover, if a radiator is all that exists in space, the radiation power formula is wrong because there are no energy sinks; either the body's effective emissivity is 0, or its temperature is multivalued such that all of its radiation is regenerated into itself. The equation is missing a third expression, that of the radiation from the background. And I was thrown back by the setup because the equation was missing a negative sign to show whether the Earth was only a radiator or a regenerator for the Sun. If the equality had the sum of power regenerated to the Sun and radiated into the Universe, I think that it would get a more accurate solution for the solar temperature. lysdexia 06:17, 3 November 2005 (UTC)

I think that perhaps radiation is a little more complex then this. Firstly, electromagnetic radiation is just that: electromagnetic. Coulomb's laws, which only describes the electric field due to stationary charges cannot possibly be sufficient. Secondly, radiation can very much exist in vacuo without interacting with anything (though perhaps I am misunderstanding the point you are trying to make here). Threepounds 05:49, 9 November 2005 (UTC)

Ok, thanks, that clarifies things a lot for me. I have edited the assumptions to conform to the above list, and I think the section is now correct as it stands, but I would not want to remove the dispute tag until there was a consensus. PAR 11:56, 3 November 2005 (UTC)

Is Assumption #3 even worth stating? If the Earth emits as a black-body, the power it emits is:

Which seems like a lot, until you calculate the power the sun emits:

So by several orders of magnitude the radiation from Earth is irrelevant on solar-system scales. JabberWok 01:54, 4 November 2005 (UTC)

Earth is still missing Sun's areal fraction. The only way that one can get a blackbody's temperature for Sun from Earth is to consider at least the mean galactic temperature as a function of distance or area subtended from Earth. The equation left with two expressions is meaningless because it assumes that Earth and other nonsolar bodies have a zero-temperature sink everywhere over them. lysdexia 14:57, 4 November 2005 (UTC)

Can you write down (here on the discussion page) what you think is the correct mathematical description? PAR 05:09, 5 November 2005 (UTC)

a guess:

σT⁴4πR²(πr²/4πd²) + σu⁴.125*4π(30 kly)²(πr²/.125*4π(30 kly)²) = σt⁴4πr²(πR²/4πd² + (4π(30 kly)² - πR²(30 kly/d))/4π(30 kly)²)

T⁴R²r²/d² + u⁴r² = t⁴r²(R²/d² + (4 - R²/d30 kly))

lysdexia 13:05, 5 November 2005 (UTC)

Thanks, could you give an explanation of the variables? PAR 15:03, 5 November 2005 (UTC)

I still see no reason to list this section as disputed. It was ment to show an example application of black body laws, as well as give a rough order of magnitude calculation. And it does this. JabberWok 17:05, 5 November 2005 (UTC)

meant

The variables are for three bodies. Guess which. lysdexia 18:01, 5 November 2005 (UTC)

[edit] How to phrase the english

There is disagrement (mainly with the user Lysdexia) over the sentences like the following:

• "The Sun emit that power..."
• "This is the power from the Sun that the Earth absorb:"
• "Even though the earth only absorb as a circular area πR2, it emit equally..."

The sentences should read "The Sun emits..." "The earth absorbs..." If this user insists that words like "emit" stay without the 's', then the phrasing of the sentence needs to change. For example it could be come "If the Sun were to emit that power..."

But as the sentences stand, they need an s. What do other people think? JabberWok 16:59, 5 November 2005 (UTC)

This is a copy of my response to Lysdexia's claim that his version is correct grammar (from User talk:Lysdexia). --best, kevin ···Kzollman | Talk··· 17:12, 5 November 2005 (UTC)

Perhaps I should be clearer. We are disputing this statement. Therefore, repeating it without justification does not help to resolve the conflict. For instance, I might reply by saying "It isn't grammatic." Where would that leave us? Can we agree that the verb forms "emit", "absorb" and "depend" are all used with plural subjects? For instance "Those people emit a foul order", "The cars absorb a lot of light", and "We depend on the kindness of others". If we can agree on this, then you must be claiming that "The Sun" and "The Earth" are plural. Is this what you are claiming? If so, I think it is in need of justification. --best, kevin ···Kzollman | Talk··· 16:35, 5 November 2005 (UTC)

I agree 100% with the above. To claims that the Sun are a plural entity and that the Earth are also a plural entities require justifications. ;) --chris.lawson 17:27, 5 November 2005 (UTC)

I never claimed that they were plural. If you didn't understand my edit summary, then you have no say on grammar. And does "lysdexia" sound like a he, illiterate? The sentences don't need introductory clauses every time that my conjugation is called out! They were implied from the section's introduction. Few people understand how to handle English words because teachers and students are dolts. All verbs in English dictionaries, believe it or not, are given in the "unconjugated" subjunctive mood, not the infinitive. (Infinitives end in -an.) As are clausal verbs. And I don't need an "if" or "whether" or "that" to begin a sentence, because the reader should know that the scenario to prove the blackbody outcomes wasn't real. lysdexia 18:01, 5 November 2005 (UTC)

Uh...what? All that is a red herring: the relevant discussion here is about subject-verb agreement. Subjects and verbs must agree in number. "The Sun" describes a singular object, which must take a singular verb. "The Earth" similarly describes a singular object. To use the plural form of a verb with either of these phrases is, simply, wrong. Also, I'd like to take this opportunity to remind you that Wikipedia has a policy of no personal attacks. Calling someone "illiterate" in an insulting manner, as you just did, is in violation of that policy.--chris.lawson 18:49, 5 November 2005 (UTC)

I am not an expert on English grammar, but I may be able to shed some light...I believe that Lysdexia is asserting that the sentence is (or should be) in the subjunctive. As an example, we might begin a sentence: "Should the Sun emit..." I'm guessing this example sounds right, even though it seems to use a plural verb with a singular noun -- it actually uses the subjunctive, as indicated by the word should.

Again, I am not a grammar expert. I don't know if Lysdexia is technically correct here. However, I have had significant exposure to written proofs and derivations, and Lysdexia's usage is not common for proofs or derivations. I suggest we conform to comman usage. -Rholton 19:57, 5 November 2005 (UTC)

But the sentences in question do not use subjunctive, did not use subjunctive when Lysdexia was accusing others of vandalism, and as far as I can tell, have NEVER used subjunctive. Again, a red herring and specious argument at best.--chris.lawson 21:54, 5 November 2005 (UTC)

Sorry to jump in here, but infinitives do not end in -an. Lysdexia is probably trolling. (Would that she were not!) Adam Bishop 20:05, 5 November 2005 (UTC)

I agree - this is one big troll and we are the fish. PAR 22:27, 5 November 2005 (UTC)

So calling me a troll isn't a personal attack but calling others illiterates is? The former's a malapropism and the latter's a truth. Whether or not it's taken as an attack is the reader's choice. Because Wikipedia forbids personal attacks, I suggest them not be. Rholton, the "proof" is based on flaky premises; the conclusion is assumed thence; they cannot be worden as a statement of truth, so they must take the subjunctive. And, yes, the verbs are in the subjunctive! The vandalism comment was about wrecking the meaning in my edit, not about what had been there. Adam, English infinitives do end in -an. You're a liar. Ye have been using the prospective mood the whole time. English's plural conjugation is -[e]th. I could use that instead, but the nouns I used were singular. English no longer conjugates by lot: Only the third person singular has an indicative mood; mostly the rest are subjunctive—that is, unless one reckons -h as an indicative. That seems safe. If only the writers knew what they were doing. lysdexia 02:32, 6 November 2005 (UTC)

I'm sure everyone can see that this is completely nonsensical and that you are contradicting yourself within your own paragraph. Stick to doing whatever else it is you do, and stop with the grammar lessons. Adam Bishop 16:14, 6 November 2005 (UTC)

If you could show it, you would; but you can't, so you're wrong. lysdexia 21:59, 6 November 2005 (UTC)

[edit] Visible Color

Looking around the web, I can't find a good spectrum picture. The one currently on this page, Image:Blackbody-colours-vertical.png, claims to be correct in hue and saturation, but the brightness is adjusted. That is useful, but it would be nice to have a spectrum image one could look at to accurately say "red-hot means xK". Other spectra I've seen have looked qualitatively more realistic, but don't seep to be calibrated. For example, one claims that candle light is dark red when clearly it isn't (I think).

From [1], an uncited book is quoted as saying

Assuming there is little light other than that emitted by the glowing charge in the furnace, you can judge a dull red glow to be from about 950°F (783K) to 1000°F (811K). Thereafter, as the temperature climbs, the red glow will brighten noticeably at about each 100 degree increment until it changes to orange at about 1600°F (1144K). The orange glow brightens through about 1900°F (1311K) where it begins to show a yellow tone. It will be quite yellow at about 2100°F (1422K), and it will show white at about 2400°F (1589K). It will be dazzling white at about 2600°F (1700K).

As I see it, an accurate spectrum would have colors sampled from an image of lava.

The current spectrum also perpetuates the misconception that things glow "blue hot" when the blue in most flames is due, I believe, to CO₂'s emission band. —BenFrantzDale 01:20, 28 November 2005 (UTC)

There is the article on the Planckian locus which gives the path of a black body through color space. That picture is, I think, calibrated. Also, at very high temperatures, the black body spectrum does have a blue tint to it, but no, flames are not at that temperature. PAR 01:28, 28 November 2005 (UTC)

That isn't quite what I had in mind. I followed that description to draw this image:

Assuming that description is correct, this spectrum shows the emission for 0K–1700K (one kelvin per pixel horizontally). Unless that calibration is inacruate, this should be a good stepping-off point for a WYSIWYG spectrum. —BenFrantzDale 01:49, 28 November 2005 (UTC)

I should add, that page describes this as "foundary colors", which is to say this is the colors you perceve hot metal as being in an otherwise dimly-lit environment with no reference whitepoint. Still, it seems like a useful reference given it answers the question of "what does red-hot mean?" —BenFrantzDale 01:57, 28 November 2005 (UTC)

Ok, you have included intensity in the above spectrum too. If you just did chromaticity, so that the colors were all displayed at the same intensity, it would go to pure red on the left, and if you extended it up to 10K degrees or more it would go towards white, then blue. PAR 02:15, 28 November 2005 (UTC)

[edit] Some questions about Black Bodies

1.I don't know why a double atom molecule has 2 freedom of rotation? Can anyone support me any pictures? Saying thanks first.

A double atom is like two point particles a fixed distance apart. If you had just two point particles, each would have 3 degrees of freedom (x1, y1, z1, x2, y2, z2), but when you require them to be a fixed distance d apart, that means that you have

and one degree of freedom is lost. So a double atom molecule has 5 degrees of freedom. You could say that they were x1, y1, z1, x2, y2 with z2 solvable from above, or you could say that they were the coordinates of the center of mass of the molecule (x, y, z) and the two angles θ and φ needed to describe the direction of the axis of the molecule. In that case, the θ and φ are the rotational degrees of freedom. Basically, there are only two because you only need two angles to specify a particular direction in space.

2.When estimating the T of the sun,why can we take to represent as ? My professor talked it the day before yesterday that it's reasonable enough.--HydrogenSu 20:13, 3 February 2006 (UTC)

If you have a function which is symmetric, then its mean is the same as its maximum. The black body curve is not perfectly symmetric, but its close enough for rough estimates.

HydrogenSu, you are doing it again - Ask these questions at the Wikipedia:Reference desk, and send me a note, and I will try to answer them. PAR 19:06, 5 February 2006 (UTC)

[edit] Black bodies and black holes

I'm an Italian student, I've just learned about Kirchoff's Law and I have a question. Black holes are virtually black bodies, because they can absorb every sort of radiation: but we also know they can't emit no radiation, acting in reality as if they were at a temperature of 0 K. So, do black holes violate Kirchoff's Law, and even the Second Principle of Thermodynamics? That looks pretty impossible, but I can't find any solution.

I am no expert, but I believe the answer lies with the application of Quantum Mechanics to understanding Black Holes. Stephen Hawking (who is an expert) has written exensively on this topic, and has advanced a Quantum theory of black holes that shows that black holes do in fact emit radiation, called Hawking radiation, and that the emission spectrum is precisely what you would expect from a black body. Although it may seem impossible for a black hole to emit radiation, Hawking has suggested that the source of the radiation is the creation of pairs of virtual particles and anti-particles immediately on either side of the event horizon. One particle, inside the event horizon, falls into the black hole. The anti-particle, just outside the event horizon, and having equal but opposite linear momentum, escapes from the black hole. Hawking's calculations show that the resulting spectrum exactly matches the functional form of black body radiation at a non-zero temperature, indicating that the temperature of a black hole is not 0 Kelvin. Over time, this process actually causes the black hole to "evaporate" at an ever increasing rate, until eventually the black hole disappears completely in a burst of high-energy gamma rays. In fact, some physicists and astronomers believe that this process may provide an explanation for gamma ray bursts, although there are other competing explanations as well. -- Metacomet 00:31, 11 February 2006 (UTC)

Hawking also showed that there is a connection between the thermodynamic concept of entropy and the surface area of the black hole's event horizon (see black hole thermodynamics and black hole entropy). In fact, as matter and energy fall into a black hole, the radius of the event horizon increases, which thereby increases the surface area of the sphere contained within the event horizon. Hawking has shown that the increase in the entropy associated with the increased surface area will be equal to or larger than the entropy associated with the infalling matter and energy, so that the black hole does in fact meet the requirements of the Second Law of Thermodynamics. The Second Law is, once again, on very firm ground. -- Metacomet 00:49, 11 February 2006 (UTC)

[edit] Factor Pi Wrong?

According to http://en.wikipedia.org/wiki/Planck's_law_of_black_body_radiation there shouldn't actually be a factor pi in the Planck law for the intensity.

Thomas

Right - I removed it. It was added by anonymous 129.16.117.172 and not caught. PAR 20:38, 16 April 2006 (UTC)

[edit] Minor Edit

I made a claim in the sun/planet temperature section.

 This is within three percent of the standard measure of 5780 kelvins which makes the formula valid for most
 scientific and engineering applications.

I don't feel that this will cause any objections however, in the next section the numbers do not work out quite as well, but are still within 10%. This signifies the formula's validity in my opinion as most engineering calculations have at least a 20% factor of safety. Anyone have any problems???

Notyouravgjoe 20:36, 21APR06 (UTC)

[edit] Modified Blackbody

This harks back to an earlier discussion (blackbody vs. emission), but I was wondering if someone could add information about modified blackbodies? I'm not clear on how they are modified, I assume it depends on the object. However, at least for dust (interstellar dust), I think they are modified because their absorption and therefore emission wavelengths are limited by the size of the dust particle (e.g. dust particles < 1μm won't emit at wavelengths greater than a micron). Thus far, googling has only yielded qualitative results, e.g. [2], so I hope some expert can add a little. --Keflavich 04:13, 27 April 2006 (UTC)

[edit] Radiation Emitted by a Human

In the summary of my 02:55, 9 May 2006 (UTC), I misunderstood the previous version of JabberWok, where the surface temperature was reduced by clothing. If the surface is the surface of the clothes, then it is true. However, I still think that it is too complicated to explain here and we should not change temperatures to something guessed. Poszwa

Yes I was talking about the surface temperature of the clothes so that they'll cause you to emit less radiation to the environment overall. I don't like the current value of 164 Watts as that bring the minimum amount of food one would have to eat up to that incredible amount of 3,400 calories. Lets find a more reasonable way to estimate how much electromagnetic energy a person emits in a day. I guess I could walk down the hall to our supply room and put a thermometer on my shirt if you don't like my guessed value.JabberWok 21:01, 9 May 2006 (UTC)

It's 160 Watts because you would start to tremble increasing your heat production at the moment. I know it is hard to do it all day long and this number looks strange but with those assumptions it was right. I didn't want to guess any "more corrent" numbers and I didn't find any numbers to cite. Poszwa 00:29, 10 May 2006 (UTC)

Ok, so I know Wikipedia isn't the place for original research, but just out of curiosity I had to see for my self....So I went and grabbed a Fluke 51 thermometer out of the supply closet to find my average surface temperature. My exposed skin - face and arms - have an average temp of about 30.5 C, and my clothes - tshirt and pants - have an average temp of 25.8 C. And this is in a room with a temp of 20.2 C.

So, can we just go with human surface temp of about 28 C (301 K) and a room temp of about 20 C (293 K)? This results in a person being something like 95 Watts. (We could round and just say 100 Watts.)

And emitting 100 watts for 24 hours turns out to be...2,065 calories of energy! JabberWok 22:40, 9 May 2006 (UTC)

Your measured temperatures are OK but people normally don't radiate 100 Watts because part of the heat is transfered by other means and according to the link presented at the end of the article [3], it is a few tens of Watts. Showing an example that suggests that all of the energy consumption is radiated can be misleading. If we put 26-27 C and 20C ambient, we would get 71-83 W, which is probably more reasonable. Or we can use the example from the above-mentioned link, where they assume 34C skin and 23C ambient (but then the radiation is 130 W). Anyway, if the temperature is changed, the peak wavelength calculated below should also be changed. Poszwa 00:29, 10 May 2006 (UTC)

[edit] e ≈ 1!

The emissivity of human skin is apparently .98 [4], so it is a close approximations of a black body. Does the .02 less of 1 constitute that radiation which is reflected off of a skin sample that I then see as its color? --HantaVirus 15:55, 27 July 2006 (UTC)

The analysis in this section of the article explicitly neglects convective heat losses, attributing essentially all the metabolic heat loss to radiation. While the calculation using an 8 degree temperature difference gives heat loss roughly equal to resting metabolic rate (RMR), there is plenty of wiggle room in the assumed values. Most body surface area is likely to be covered by clothing or hair in an 20 C environment for a resting individual to be comfortable. Hence, a number closer to 6 degrees is probably justified. Maybe it is even lower. Perhaps the emissivity of clothing is significantly less than 1.

On the other hand, the convective coefficient can be estimated to be on the order of 10 W/m^2/K. One such measurement is for a flat, vertical plate that gives 10 (http://www.picotech.com/experiments/heat_transfer_coefficient/heat.html). Another gives a range of 5 to 35 for tubes in air at atmospheric pressure (http://www.cheresources.com/uexchangers.shtml). For comparison, the equivalent coefficient for radiation at about 20 C is about 6 W/m^2/K, found by taking the derivative of the Stefan-Boltzmann equation and evaluating at 300 K. By this comparison, convection dominates over radiation. Even if the convective coefficient is at the lower end of the range, convection and radiation are comparable. The link cited above from Hyperphysics (http://hyperphysics.phy-astr.gsu.edu/Hbase/thermo/coobod.html#c1) does not even attempt to estimate convection. The energy balance shown in the graphic is inconsistent, especially if convection turns out to be comparable to radiation.

The calculation in the article is far too conclusive. A more reasonable statement would be that radiation is significant, and probably comparable to convection. A full-blown calculation of the convective transfer rate would be required to say more.

Please forgive errors in form or style. This is my first comment here. Drphysics 19:32, 12 March 2007 (UTC)

[edit] Replace WMAP image with FIRAS spectrum?

What do folks think about replacing the WMAP anisotropies image with the actual cosmic blackbody spectrum measured by FIRAS on the COBE satellite? See Image:Firas_spectrum.jpg . The caption of the current figure, It is the most perfect blackbody emission known and corresponds to a temperature of 2.725 kelvins with an emission peak of of 160.4 GHz, is not well illustrated by the plot of the anisotropies. HEL 14:05, 8 October 2006 (UTC)

You have my vote. I kind of like the image with the error bars (e.g., http://www.astro.livjm.ac.uk/courses/phys134/cosmo.html, towards the bottom of the page), as it emphasizes just how good the fit is. That's a nitpick though. Tomrlutong 13:40, 22 January 2007 (UTC)

I have a copy of the plot as a gif (I think I got it from a NASA/COBE website) with the T = 2.725 kelvins legend on it. But I've never uploaded images to Wikipedia before! Is it kosher to upload something from NASA like this? HEL 22:31, 8 February 2007 (UTC)

Absolutely kosher. Go to [5] and fill in the blanks. Copy the stuff out of the information box and put in:

• Description=xxxxxxxx
• Source=NASA
• Date=2007-02-09 (i.e. upload date)
• Author=User:HEL
• Permission=PD-USGov-NASA

and for licensing select "original work of NASA - Public domain" PAR 04:01, 9 February 2007 (UTC)

[edit] Spectrum

The first spectrum picture doesn't look too accurate, especially the blue line. Can someone create a new version with actual mathematical functions to get it right?

Also, I'm curious what the spectrum would look like on a log-log plot, since it reminds me of a bandpass filter. — Omegatron 18:07, 13 October 2006 (UTC)

[edit] Visible light from room temp. objects?

The article has this paragraph:

Interestingly, this means that every object around you is emitting electromagnetic waves with wavelengths of all values. Every object in the universe has heat, even the emptiness of space, and when the particles that make up an object vibrate on a microscopic level they radiate electromagnetic waves. These wavelengths are predominantly infrared (heat), but there is also a minute amount of visible light like red, yellow, green and blue. So, right now, you and everything around you is emitting visible light. The reason this light cannot be seen is that it has a very low intensity.

Which I think has some problems. The basic claim, that all objects emit electromagnetic waves at all wavelengths, kind of misses the whole 'ultraviolet catastrophie' point, and neglects quantitization of light. More specificaly, the article claims that "you and everything around you is emitting visible light," which I don't think is true. I calculate that a 300K object emits about 10^-34 watts at 400nm. Since the energy of a 400nm photon is 2.65x10^-19J, this means that a room temp object will almost never emit visible photons.

Could someone check my calculations before I edit the article? Tomrlutong 13:33, 22 January 2007 (UTC)

I think you absolutely right but the calculations are wrong. By my calculations, the blackbody intensity at λ=550 nm and T=300 kelvin is 3.195 x 10^-23 watts/m²/sr/m. If we multiply by π we get rid of the steradian term (sr) and have the total power per unit area per unit wavelength radiated by a black body at 550 nm, which would be 1.004 x 10^-22 watts/m²/m. If we say the luminosity function is about 100 nm wide, then multiplying by that (100 x 10^-9m) we get 1.004 x 10^-29 watts/m² which is the visible power emitted per square meter by a black body at room temperature. A 550 nm photon has an energy of hc/λ=3.612 x 10^-19 joules so that means that there are 2.779 x 10^-11 visible photons emitted per second per square meter. That means a black body with a surface area of one square meter, at room temperature, would emit about one visible photon every thousand years. If you don't fix this soon, I will, and thanks for pointing that out. PAR 02:57, 30 January 2007 (UTC)

[edit] Something is wrong with "the spectrum of an incandescent bulb in a typical flashlight"?!

As said in the article, the filament temperature appears to be about 4600 kelvins due to a peak emittance of around 630 nanometers. This temperature cannot be true since non-halogen tungsten lamps have filament temperature less then 3000K and low power lamps (~40W or less) have even lower temperature. For a flashlight bulb, a realistic temperature would be about 2500K.

However, more serious is the “impossible” shape of the spectrum: from spectrum it appears that the bulb in the visible range (380-780 nm) emits almost 80% of all emitted energy (area under the curve), i.e. it is a very efficient light source, emitting less then 20% in the infrared range. This, of course, cannot be true since an incandescent bulb is known to be very inefficient with less then 3-10% of energy in the visible range. I have constructed two Planck curves for 2500 and 3000K, and it can be seen that even for 3000K, maximum should be around 1000 nm, and the most of the spectra is in the invisible IR part of the spectra, without any maximum in the visible part.

So my question is: how to explain such a huge discrepancy between theory and the measurement?

It happened that I had access to a CCD spectrometer in the 200-1100 nm range, so I had measured spectrum of a flashlight bulb, and I have got a very similar spectrum, with a distinct maximum at about 650 nm, and almost no radiation in the 1000+ nm range, so it seems it is not an error in measuring.

But how this can be possible? I have considered that tungsten is not a perfect black body but a grey body, however spectral emissivity of tungsten (e.g. [[6]] or [[7]]) is about 0.45 at 500 nm and still about 0.37 at 1000 nm, so the drop in emissivity can not explain such low emission at 1000 nm and higher frequencies.

Does anybody have some explanation for this? (I.Niko 22:22, January 29 2007 (UTC))

I think you are probably right, a flashlight filament should emit as a grey body at about 3000 kelvin or less. I think the problem is that the radiation had to pass through glass which absorbed the infrared? What is the transmission vs wavelength function of the different types of glass the radiation had to pass through before it was measured? PAR 03:06, 30 January 2007 (UTC)

Yes, I thought about it, but I knew that glass is rather transparent in the near infrared. However, I have checked transmission curves for some types of glasses (e.g. [[8]][[9]] [[10]] and as it can be seen, all this types of glasses have almost flat 90% transmittance up to 2000 nm. So it appears that this should not be the source of so low emission in the IR. Unless for flashlight bulbs they use some weird type of glass that absorbs almost entire IR part of the spectra, but this doesn’t seam likely, since this would mean much more heating of glass envelope, which would be harmful for flashlight; common since would tell to use glass with good transmission for IR, to avoid overheating… (I.Niko 12:22, February 1 2007)

Ok, I see that the IR transmission is high and flat in those examples. I just always had a vague notion that the greenhouse effect was that visible light passed through the glass in the greenhouse, was absorbed inside and re-radiated as infrared, which could not escape through the glass, because it was opaque to infrared. I don't know if this was near IR, far, IR or what. Anyway I can think of three scenarios:

1. The flashlight curve is in error.
2. The flashlight curve is correct, but for a tungsten-halogen bulb, which, I believe operates at a higher temperature than a simple tungsten bulb.
3. The glass does absorb the infrared, and the references you have given are for the wrong kind of glass.

I will look into the second possibility. Can you think of any more? PAR 16:56, 1 February 2007 (UTC)

The tungsten curve is decidedly wrong and the reason is the responsivity of the sensor. CCDs are silicon, which typically cuts off rather sharply below one micrometer. [11][12] The spectrometer itself may also not be flat because of the grating blaze or any number of other reasons. The spectral calibration of such a device is nontrivial. You can be quite certain that the temperature of the filament is nowhere near 4600K since the melting point of tungsten is about 3700K.[13] This figure and its erroneous caption should be removed! Drphysics 16:45, 15 March 2007 (UTC)

I agree. This figure was apparently self-generated by a Wikipedia editor, and it appears that he/she did not properly calibrate for the spectral response of the spectrometer and detector. This is a fatal flaw; the image needs to be removed as soon as possible. (Aside: Plots of CCD spectral response available online [14][15] show similar response for a thinned, back-illuminated detector, with a peak in the vicinity of 700 nm.) --Srleffler 01:03, 16 March 2007 (UTC)

I have proposed deletion of the image. The discussion is located at Wikipedia:Images and media for deletion--Srleffler 01:30, 16 March 2007 (UTC)

[edit] Vandalism or cutesy physicist term?

"Pooped?" Colorful shortening of "put out"? It goes back thru many versions, I see. - robgood