Talk:Black body/Archive 1

"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 phenomena of an object is going to emit or absorb energy.-As my knowing of from hints of Blatt's book.--GyBlop 13:47, 27 February 2006 (UTC)

In industry, "black body" is a noun that refers to a physical object; the black body. Alternatively, "blackbody", refers to the radiation emitted/absorberd/reflected/transmitted by a black body hence, blackbody radiation, a compound adjective. So, previous assessment was correct, at least in industry.The Lamb of God (talk) 04:18, 28 January 2009 (UTC)

I have a different theory, that doesn't depend on speaking for so-called "industry": there are two consistent approaches used; either noun "black body" and adjective "black-body", or "blackbody" for both. I find lots of books both ways, but none that crossover and use "blackbody" with one of the other forms. Dicklyon (talk) 04:53, 28 January 2009 (UTC)

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 analysis assumes 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 (Heisenberg 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 independent 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 sometimes 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 throughout 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)

I am sorry for taking so long to join the discussion. There remains great confusion about blackbody radiation in part because of the erroneous concept of universality. A historical review of the subject reveals that much of the problem occurred when Kirchhoff overextended the findings of Stewart. For a full discussion of thermal emission in cavities including Kirchhoff's missteps see: Pierre-Marie Robitaille. A Critical Analysis of Universality and Kirchhoff's Law: A Return to Stewart's Law of Thermal Emission, Progress in Physics, 2008, v. 3, 16-35; Pierre-Marie Robitaille. Blackbody Radiation and the Carbon Particle, Progress in Physics, 2008, v. 3, 36-55).JPMLR (talk) 03:27, 7 June 2008 (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 disappear. Also note that sharp lines are characteristic of low pressure gases. 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)

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)

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 File:Blackbody spectral radiance.gif--Maddog Battie 16:01, 4 August 2005 (UTC)

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.

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 emissivity , 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 insulating properties of the atmosphere can do is change the diurnal temperature extremes , not the mean . Is this correct ? -- Bob Armstrong

The discussion in this section of the article is decidedly oversimplified and has a false precision. To see this, try the same calculation for the planet Venus using the formula in the article. You will get about 340K instead of the actual value, which is over twice as high. The agreement for Earth is largely coincidental, especially to the precision of a few percent. While the climate change community may not be right about everything, they are right about this. Drphysics 16:04, 4 April 2007 (UTC)

Albedo (reflectivity) and emissitivty are indeed identical, but only for a given wavelength. Since there is very little overlap between the wavelengths of the energy emitted by the Sun and by the Earth a and e are independent, and indeed very different. The atmosphere is close to transparent to visible light while a strong absorber to IR, so we can have a greenhouse effect that heats the Earth. -- user:Thomas Palm —Preceding undated comment was added at 06:48, 24 July 2008 (UTC)

This section has been totally screwed up by Thomas Palm's inappropriate introduction of ( 1 - albedo ) as a correction. The problem is that he fails to make the compensating correction for the earth's emissivity in the equation for the earth's radiation to the rest of the celestial sphere. The discussion should have been left as simply the derivation of the black body temperature. In fact, with proper application of Kirchhoff's law emissivity and absorptivity being equal drop out of the equilibrium equation. Drphysics comment that the close agreement of the ( black body ) calculated temperature and the earth's temperature is coincidence is wrong. The relationship holds for all the inner planets except Venus. In fact what the calculation shows is that the temperature of Venus ( which incidentally has the highest albedo of all the planets and therefore should have the lowest temperature by these arguments ) MUST be generated internally and cannot be explained by the radiant energy it receives from the sun. This section is now useless and should either be removed or reverted to a discussion of the pure black body derivation. Bob Armstrong (talk) 02:45, 7 December 2008 (UTC)

It's actually not as screwed up as you think. Albdeo is not constant across wavelengths, so can't be applied to both the absorption and emission equally. In fact, since it's defined as the fraction of incident light reflected, it applies specifically to absorption. Emission depends on various things like greenhouse gases, and is generally handled by a different mechanism, namely the definition of effective temperature so that you don't have to know the details. This is not an actual temperature, but rather the temperature of a blackbody that would emit the same amount. It's done this way in reliable sources (e.g. this book that comes up with the same 255 K effective temperature). Dicklyon (talk) 08:19, 7 December 2008 (UTC)

Neither is emission. Emission is the flip side of absorption and both are wavelength dependent. It cannot simply be left out of the equations. For the flat spectrum of a gray body, it cancels the albedo. For a colored body the correction to the black body temperature will depend on the correlation of the spectrum of the body with the spectrum of the source versus the spectrum of the sink. "Effective temperature" so far as I can see has no relevance to anything other than correcting the temperature estimates of stars. The fact that while NASA's estimates of the temperatures of the inner planets is not what would be desired, the fact that only Venus is significantly different than its black body calculation must be considered. ( And the temperature of Venus, the same on the sides both facing and away from the sun, and greater than even the calculation for a disk, black facing the sun and white away, with far the highest albedo of all the planets, thus radiating far more energy than it can be receiving from the sun, cannot be explained by any greenhouse effect.64.93.126.163 (talk) 18:38, 7 December 2008 (UTC)

If you can provide a good source that discusses these observations, we can add something about that to the article. My point was simply that the approach currently in the article has support in reliable sources, and is self-consistent within the definitions given. It's not the only approach, and is not an accurate approach to estimating planatary temperature, and does not pretend to be. Dicklyon (talk) 20:03, 7 December 2008 (UTC)

Red : observed ; Green : calculated Black Body

It's easy to replicate the planetary temperatures versus black body calculation and I believe better estimates are available now than when I made the ( crude ) graph for Mercury , Venus , Earth , and Mars I've inserted here . The temperature of Venus is so high it radiates about 16 times the energy it can possibly be absorbing from the sun .

I really wish I had a reference to a more complete simple analysis of the temperature physics of radiantly heated colored balls but I have not . I even more strongly wish I could find simple laboratory experiments settling these issues unambiguously. I find it astounding that I can't .Bob Armstrong (talk) 01:19, 10 December 2008 (UTC)

Temperature relation between a planet and its star

This derivation seems OK to me. The only possible criticism is the name 'black body temperature', but terminology in physics is often misleading (greenhouse effect is another example) and if that name is the conventional one it should be kept. The reason for that slight reservation is that the emissivity (1-alpha) of the planet is taken to have one value (e.g. 0.633 for the Earth) in the visible, and another i.e. 1 for the infra-red. This model represents the fact that most solids are nearly black in the infra-red while being coloured in the visible. This is a much better model than using a true black body approximation i.e taking (1-alpha)= 1 or a gray-body model (1-alpha) = const.< 1 across the entire spectrum. See for example Bob Armstrong's remarks above, who explores a rather similar 'gray-body model' and shows that it can lead to unphysical results.

To refine the approximation you have to go to radiation transfer theory which would treat each infra-red wavelength separately. Bodies which are gray right across the spectrum (visible to infra-red) are a 'red herring' (please excuse the term) Deconvoluter (talk) 11:25, 7 September 2009 (UTC)

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 correlation, 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 understanding 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 wavelengths 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)

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

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)

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 canceled out and both bodies at thermal equilibrium would be at temperature 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 $(\sigma T_{s}^{4})(4\pi R_{s}^{2})$, i.e. the rate per area times the solar surface area. The earth catches a fraction $(\pi R_{E}^{2})/(4\pi D^{2})$ of this radiation. That is then equated to the rate at which the earth radiates: $(\sigma T_{E}^{4})(4\pi R_{E}^{2})$ and you get the result $(T_{E}/T_{S})^{4}=(R_{s}/2D)^{2}\,$. 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:
$P_{Eemt}=\left(\sigma T_{E}^{4}\right)\left(4\pi R_{E}^{2}\right)\,$

$=2\times 10^{17}W\,$

Which seems like a lot, until you calculate the power the sun emits:
$P_{Semt}=\left(\sigma T_{S}^{4}\right)\left(4\pi R_{S}^{2}\right)\,$

$=4\times 10^{26}W\,$

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)

I see that I am joining this discussion long after the controversy has raged and quieted. However, this section is still quite misleading and still raises questions (cf., the related question in discussion about Earth surface temperature). The derivation is highly oversimplified and neglects a number of significant effects. There is no reason to imagine that this calculation would give the right answer to within a few percent, and it is only through sheer luck and a bit of fudging that the answer is that close. If the same calculation is repeated for Venus, the error is a factor of two or so. This makes the concluding statement of the section wholly unjustified: "This is within three percent of the standard measure of 5780 kelvins which makes the formula valid for most scientific and engineering applications." Drphysics 16:23, 4 April 2007 (UTC)

I agree. The physics is fine given the assumptions and the derivation is pretty standard, I think, but it needs a little revamping. I'm a PhD student in astrophysics at the University of Chicago studying the CMB and have seen this before. First, the "assumptions" section preceding the derivation should mention that the greenhouse effect is neglected as is internal energy generation. Second, the sentence beginning "In other words, the temperature of the Earth only depends on..." is too disconnected from the simplified assumptions going into the calculation and people will read it as an iron-clad fact. The three percent claim then sounds like a vindication of the assumptions, and juxtaposed next to the endorsement for use in science in engineering really does imply that a three percent accuracy is to be expected. This is an order of magnitude calculation and that should be reiterated - good to within a factor of a few. And I very much disagree with the assertion that it is "valid for most scientific and engineering applications." The scientists and engineers who are interested in the surface temperature of other planets are planetary astrophysicists, astro-biologists, and space agency engineers, for whom this would be dangerously inadequate. (If I were designing a probe to go to Venus, I'd sure like to know about a factor of two difference in surface temperature.) I don't disagree that this is a cool derivation and that the result is useful as a scaling relation (i.e. if the earth were four times as distant from the sun, the surface temperature would be half as large) but as it is the text is quite misleading. Moreover, I'm actually a bit surprised to even see this calculation here. The result is ultimately much more an aspect of planetary science than of the physics of blackbodies. You see this all the time in planetary science textbooks but never in thermo/radiative processes books. Of course, I can't seem to find an article in which I think it would fit better. If one existed, I would suggest moving it after these issues are addressed. AstroNerd2000 12:16, 23 August 2007 (UTC)

How to phrase the english

There is disagreement (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 common 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 worded 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)

Sup, guys... Let's clear this up. Lysdexia clearly has a confusion between the Modern English language and Anglo Saxon (Old English). While it is honourable to attempt to maintain all of the old verb forms, most of them were lost in late Old English. The -an ending for infinitives was the actual infinitive ending, but increasingly used was what is called the inflected infinitive, involving the familiar "to". For instance, I am under the impression that "to bear/carry" was "beran" but also "to berienne". The use of -[e]th as a plural conjugation is absurd. In Old English, the plural for all persons was similar ("-ath"), but even before Middle English, that had changed to the more Germanic "-en". By Modern English, we no longer had any endings for that. Lysdexia thinks he knows a bit about English historical linguistics, and wishes to impress that upon the community. Lysdexia, thank you, we acknowledge your "brilliance". Now that we have gotten past his absurd "show-off" of his broken knowledge regarding the historical grammar of the English language, we can move forward and use actual Modern English usage.--97.115.3.171 (talk) 17:59, 15 June 2008 (UTC) (Jon Sterling)

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 un-cited 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 perceive 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 chromaticism, 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)

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

$d^{2}=(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}+(z_{2}-z_{1})^{2}\,$

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 ${\bar {\lambda }}$ to represent as ${\mathcal {\lambda }}_{max}$? 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)

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 extensively 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)

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)

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)

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)

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 [3] 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)

Please fix this asap. The anisotropy map is counterproductive to the point being made. Dwt2004 16:48, 18 July 2007 (UTC)

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)

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)

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

File:Incandescent flashlight spectrum.gif

The image under discussion.

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. [[4]] or [[5]]) 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. [[6]][[7]] [[8]] 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. [9][10] 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.[11] 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 [12][13] 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)

This discussion is all very interesting -- but we really need some true, complete power spectrum plots for ordinary 1-100W 120v incandescent light bulbs! This should not be so hard to find?-69.87.203.221 00:16, 26 May 2007 (UTC)

There is no standard 100 W bulb, or any wattage bulb, hence it is meaningless to ask for the true spectrum of such a generic device. For instance, the temperature of all 100 W bulbs may not be the same or the filament might not be quite the same material with the same surface properties. The bulb's spectrum will change as it ages. The best you can do is to say that light bulbs are about 2500K blackbodies. In that vein, the graph of a 2500K or 3000K blackbody, as the one in this talk section, is about as good as it gets. The only thing missing is the emissivity of tungsten, which has some spectral shape that tends to attenuate the longer wavelengths somewhat. Also, depending on the type of glass used in the envelope, there will be further attenuation of the longer waves. For an emissivity curve, see tungsten emissivity. In any case, the spectrum of light bulbs is off topic for this article, except perhaps to mention in passing that incandescent lamps are approximately blackbodies at about 2500K. Drphysics 02:14, 21 June 2007 (UTC)

Vandalism or cutesy physicist term?

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

A bit of pedagogy

I feel that there is a (crucial) missing link between these two sentences:

"he found a mathematical formula fitting the experimental data in a satisfactory way. To find a physical interpretation for this formula, Planck had then to assume that the energy of the oscillators in the cavity was quantized"

Why did Planck have to assume quantization? This seems like the key to understanding his findings. This is probably obvious to anyone who has knowledge of these matters, but looking at the graph baffles me... Discrete temperatures, discrete energy levels, quantization? And what is (arb.)? a unit of energy? perhaps this should be stated. Anyway, I just think that a more complete explanation would benefit physics newbies. —The preceding unsigned comment was added by 24.203.133.6 (talk) 04:32, 11 April 2007 (UTC).

Planck had to quantize the oscillators to avoid the 'ultraviolet catastrophe' whereby the energy spectrum goes to infinity as the frequency increases (wavelength gets shorter), as shown in the "classical theory" curve in the spectrum. This would make the total energy unbounded, which is unacceptable. This was a long-standing problem of classical physics that was solved by quantization. Planck said the energy levels of the oscillators in the wall of the cavity had quantized energies. This was an ad hoc fix whose significance and relation to a coherent theory of matter and radiation was only understood long after Planck proposed it. It is not easy to see why the quantization affects the spectrum in this way. If I can think of a quick explanation, I'll post it.

The vertical scale on the blackbody spectrum is labeled 'arb.' meaning "arbitrary." The absolute units of the scale can be many different things. The purpose of the graph was to illustrate how the shape of the spectrum depends on temperature. The relative values are correct, as is the location of the peak wavelength. Numerical values can always be computed using the equations in the article. Drphysics 01:06, 21 June 2007 (UTC)

Problematic section on CMB

The content of the recently added section entitled "Cosmic microwave background radiation and Black body radiation of 2.7K" is nigh-on impenetrable. I don't think this is because of the subject's complexity, but because of the phrasing and grammar. I suggest that it either needs to be rewritten into sentences that parse properly in english, or removed altogether— as it stands now, that section does not help anyone's understanding.--cjllw ʘ TALK 01:44, 11 July 2007 (UTC)

I moved the paragraph here:

A nearly perfect black-body spectrum is exhibited by , see pict. with very small positive differences above. By Planck law given Hawking radiation is perfect black-body radiation emitted by black holes). This limit was already indicated by "Einstein, Albert Ueber einen die Erzeugung und Verwandlung des Lichtes betreffenden heuristischen Gesichtspunkt 1905" [14], §1": He showed by a simple Line integral of a constantly equal growing potential of equally distributed charges until an infinite border of an universe that this provokes the physics: Black body radiation defines the limit for all photons and its redshifts; Planck law gives the Hawking radiation and limits always the visible radius of any space by that lowest possible radiation of 2.7 K meaning that this minimum of a Cosmic microwave background radiation limits not only a Big bang.

The line of reasoning seems to be: introduce CMB, which was already mentioned elsewhere in the article. Then discuss why Hawking radiation from a black hole is also black body radiation, and that it still follows a Planck curve after a (gravitational) redshift. From this it is concluded/suggested that CMB results from black holes (maybe we're inside a black hole) and that this is an alternative to the big-bang theory. In this case it is original research and/or unverifiable due to lack of references. Han-Kwang 10:40, 11 July 2007 (UTC)

Correctly removed, IMO. Good work. --Michael C. Price ^talk 10:51, 11 July 2007 (UTC)

New Image

I made a new image if someone want to upload and place it there (i don't know how and it needs to be in SVG format) and a 3rd party needs to decide if it's better. I don't like the others because they have "arbitrary" units or units that are more obscure. There are several advantage to the simple units of "% of total watts per 100 nm" the graph is in. By this method, steradians and m^2 don't even need to be specified, and one spectrum is directly comparable to the others (area under all curves is the same since it's the total watts emitted). The units are completely valid. Also, it's nice to have an idea about typical filaments compared to the sun eventhough the glass and non-ideal nature of the filaments affect it as mentioned. But the sun nor anything else in the world has an ideal spectrum, so i think the utility of the comparison is outwieghs any complaints of lack of precision. I included two curves for incandescent to emphasize that temp is variable. Here's the PNG file:

black body curves —Preceding unsigned comment added by Zawy1 (talk • contribs) 20:58, 19 March 2008 (UTC)

Thats good, I agree the "arbitrary intensity" is not a good name for the vertical axis. But you have it normalized to an arbitrary wavelength interval: 100 nm, and also the relative intensities for different temperatures is lost. How about the same set of curves, with the vertical axis being "normalized intensity". Each curve is labelled with a numerical value equal to the integrated intensity for that curve. That way, the area under each curve is unity, and the relative intensity information is still there, and there are no arbitrary values introduced. PAR (talk) 19:31, 29 March 2008 (UTC)

°K

The graph in CIE 1931 color space incorrectly says T_C(°K). The Kelvin scale is not divided into degrees, rather, the units are just kelvins. BethelRunner (talk) 03:22, 7 April 2008 (UTC)

It would have been correct until about 1967 when the SI changed the notation; so, should we bother working on it? Dicklyon (talk) 03:28, 7 April 2008 (UTC)

Don't ask me; I'm just pointing it out. BethelRunner (talk) 02:53, 8 April 2008 (UTC)

Reorganization: incandescence, black body radiation, black body

(Original title: Move to black body radiation and/or merge with incandescence?)

I became rather confused when looking at black body radiation and incandescence a few minutes ago, when trying to fix the dab page for black body radiation. First, there is really no ambiguity there - we know what they're looking for, and they're almost certainly looking for the information in this article. Also, this article's lede paragraph starts talking about "black bodies" the object, but ends up talking mostly about "black body radiation". Oddly, incandescence also manages to be mostly about black body radiation or duplicating information from here. Meanwhile, it would feel like the hierarchy should go more like "incandescence" > "black body radiation" > "black body".

I think the short-term solution would to make black body radiation the main article, with straight redirect from black body and maybe a merge from incandescence. Optimally, I'd personally really rather see incandescence be the primary article, but that would need a major overhaul and rethinking of the articles' organization. —AySz88\^-^ 07:17, 22 May 2008 (UTC)

But is incandescence really the same thing as BBR? The sun is incandescent, but it is not a black body. --Michael C. Price ^talk 10:27, 22 May 2008 (UTC)

Not really - but note that this article (black body) is the one that talks about stars and the sun, not incandescence. That's why I think this article (eventually) should really be moved and turned into incandescence. —AySz88\^-^ 16:59, 22 May 2008 (UTC)

Stars have spectral lines -- they are therefore not black bodies. --Michael C. Price ^talk 08:30, 23 May 2008 (UTC)

Indeed...? Thus, the above information should definitely not be in black body - probably Sun and incandescence and maybe even black body radiation, but not black body. —AySz88\^-^ 06:35, 25 May 2008 (UTC)

Yes, material should be moved between the two articles, which should be kept distinct. Stars are incandescent and are NOT black bodies. --Michael C. Price ^talk 19:26, 25 May 2008 (UTC)

The article seems clear on that point already, where it says "In astronomy, objects such as stars are frequently regarded as black bodies, though this is often a poor approximation." And it goes on to give an approximate but very useful application of the black-body radiation laws to estimate the temperature of a planent around a star; this doesn't make sense without the black-body radiation law, and is often treated in that context in physics books. Dicklyon (talk) 20:06, 25 May 2008 (UTC)

Oppose merge – the relationship between the concept of black body and incandescence is already discussed in several of these articles. The concepts are related, yet distinct. One article on both would be a big mess. The BBR concept is useful in analyzing the radiation of stars, because they behave approximately as incandescent black bodies, and of planets, even though they are not incandescent and even though they are very close to being blackbodies; the discussion in this article seems fine as it is, but some tuning can always improve it. Dicklyon (talk) 06:04, 23 May 2008 (UTC)

Oppose merge, they are distinct concepts. Some cleanup would be nice, though. - Eldereft ~(s)talk~ 23:52, 24 May 2008 (UTC)

(Re both above)

Uh, first, this isn't a vote... Also, the merge template is a little misleading; this isn't as much a merge proposal as a proposal to reorganize (but it was the closest thing that fit the method what I was thinking of, detailed below).

To clarify, I know and agree that incandescence, black body radiation, and black body are different and should be different articles. The problem, I think, is that nearly all the content for all three concepts are basically already mixed in black body! So I think this is already a mess for anyone looking for information on incandescence or black body radiation (why would they think to look at the black body article?).

I probably should have said this was a split instead of a merge, since the end result I was proposing is closer to a split. I'll change that now. But considering that all the information is here already, I think the quick way to do this (while having all three articles usable in the interim) is to redirect all three terms to a merged article in incandescence, then use standard summary style to reorganize that article and split it into the appropriate children articles.

We can bypass the whole merge/split thing by just overwriting black body radiation and incandescence with content from this article - though, this breaks page history and might cause GFDL/copyright confusion. I don't really care as much exactly which path is taken, but there should at least be this sort of reorganization. —AySz88\^-^ 06:35, 25 May 2008 (UTC)

So you need to more clearly identify what you want to split out, and to where. I think the black body radiation page contents is fairly sensible as it is, and it's not so big that a split is needed. Give an outline of how you think this should end up, so we can consider your proposal. Dicklyon (talk) 17:02, 25 May 2008 (UTC)

Huh? I think you mean this page, but this page is supposed to be about "black body" not "black body radiation". The black body radiation page is currently a disambiguation page and has no contents. Both black body and incandescence claim to cover "black body radiation".

My proposal is pretty much just to implement summary style on the topic of incandescence. There are many many ways to draw the line between incandescence and black body radiation, which is why I didn't really want to specify any one method. I can identify two main ways to draw the split:

• Have technical material in black body radiation and nontechnical in incandescence. This is somewhat like how it is now, except that incandescence less broad than black body, which is bad. Also, what should happen to technical material that semantically falls outside the realm of black body radiation?
• Restrict black body radiation to only black bodies and have all the information about grey bodies, humans, planets and stars, etc. placed in incandescence. Personally, I like this way better. In detail:
  • incandescence should mostly be about the general glowing-red-hot phenomenon (and related), with a summary of black body radiation showing how its concepts can be applied to various things. I think the majority of non-black-body things currently in this article, like the human radiation section and temperature of planet and star section, should be in incandescence.
  • black body radiation should contain the history of the topic, and the details and derivations of the mathematics and laws, focusing upon black bodies and perhaps mentioning its extension to grey bodies.
  • black body (currently this article) should just be a description of what a black body is and why it's important. I'm not totally sure that a separate article about "black bodies" should even exist - is there enough to say to warrant a distinct article from black body radiation?

There are also several ways to do this. One way to get there (the method I think would work out best) is:

1. Temporarily merge incandescence, black body, and black body radiation into a single article at incandescence
2. Incrementally reorganize incandescence with a large subsection to become the proto-black body radiation
3. Apply normal summary style to respawn black body radiation and perhaps black body if needed

The main advantage is that this takes advantage of how Wikipedia naturally functions (insofar as summary style is natural), and it allows users to still at least can locate all the information at basically every intermediate stage. The disadvantage is that there's temporarily a monolithic article at incandescence, but it's already mostly organized (almost everything is in this article already).

Another way is to:

1. Cut/paste most of the contents from black body (this article) to black body radiation
2. Cut/paste contents which are broader than black body radiation into incandescence
3. Clean up the articles' structures

Though this second procedure is probably more intuitive, it needs a lot of up-front effort, and I don't think this can be broken into many smaller steps with good stopping points. If such stopping points are not necessary, then I suppose this could also work.

The first thing that I think we should agree on, though, is that the current state isn't really acceptable. Among other problems, I can't see how someone looking for information on "black body radiation" could have located it when the disambiguation page didn't have the "the main article is black body" comment. People expect black body to be an article covering black bodies, not black body radiation, so they'd go to incandescence (which also claims to be about "black body radiation", in bold no less, which I'll fix right now) and end up disappointed. —AySz88\^-^ 08:51, 26 May 2008 (UTC)

Oppose split – after reading the above clarified plan, I oppose it. I think the stuff about planets and stars and humans should remain here, since it is an analysis based on approximating them as blackbodies, and since planets and humans are not hot enough to be described by what is usually meant as "incandescent". Dicklyon (talk) 15:21, 26 May 2008 (UTC)

This is still not a vote; I'm not proposing any specific plan, but trying to form one which is agreeable. Do you at least agree that there exists a problem with the three articles' layout as they are currently? If you do, how would you rather fix the problem? —AySz88\^-^ 20:32, 26 May 2008 (UTC)

I understand it's not a vote; yet it is conventional to make it clear which editors are articulating which side of an issue. I explained my reason above, didn't I? As to whether I agree there's a problem, I'm still not sure I see it, other than that black-body radiation is a bit of a mess; I'd just make that a redirect to black body, and put the various other links in here. Incandescence can stay a small article as it is, qualitative, with no black-body theory in it. If there are parts of this article that would be better in incandescence, they can be moved (but what parts are you thinking of? as I mentioned above, the planet/star stuff can't be attacked without black-body equations, and humans are not incandescent). Dicklyon (talk) 21:00, 26 May 2008 (UTC)

I notice that incandescence was indeed a bit of a confused unsourced mess, with too much black-body concept mixed in. So I worked on cleaning it up, and started to source it. I don't think it needs to affect this article, though. Dicklyon (talk) 21:33, 26 May 2008 (UTC)

Sorry for the delay in replying. (I thought I posted a reply but apparently Firefox 3 must have crashed before I submitted it; my bad for multitasking with 200 tabs in a pre-release product...)

I don't think there should be "sides" so soon (at least, nothing so black and white as "support" and "oppose"), since the suggestion can be still changed however people wish.

I now see why a strict hierarchy might not be desired, since the connection is a bit looser than I originally surmised (while black body radiation is a subset of incandescence in the sense that only black bodies are considered in the former, it is a superset in the sense that only visible wavelengths are meant by the latter).

But I don't think we should be afraid of discussing black body radiation concepts inside incandescence, if that's the best or most insightful way to understand incandescence (and I think it probably is). I think the lava example, color temperature diagrams, etc. should also be more emphasized in incandescence instead of here (they're both more visible-light and useful for non-black-body understanding), though duplication wouldn't hurt.

I noticed that you amended the lede of incandescence to restrict its definition to visible light. While I now see that most (dictionary) definitions describe incandescence as visible light only, I don't think this is the best way to explain it. Due to its intrinsic connection with black-body radiation, it might just be better to be up front that the phenomenon simultaneously occurs in non-visible wavelengths at the same time as visible ones (using as an example the "red"-hot filaments emitting mostly in infrared, as already described in the article).

I personally think that it would be better to include "radiation" in this article's name, since it seems to focus upon black body radiation a lot more than it discusses black bodies of themselves. —AySz88\^-^ 03:57, 8 June 2008 (UTC)

It is not very wiki-like to propose redefining incandescent to include non-visible wavelengths, just because it makes a better story. Let's stick with the usual definition, which is emitting enough black body radiation to give off visible light. Don't conflate these concepts. Dicklyon (talk) 07:31, 8 June 2008 (UTC)

And don't confuse incandescent with visible black body: incandescent is any hot body emitting visible light: stars are not black bodies but they are incandescent. --Michael C. Price ^talk 08:15, 8 June 2008 (UTC)

I never said anything about redefining incandescent. But the connection needs to be made somewhere, and it makes more sense to do so in the incandescence article than the black body article. —AySz88\^-^ 02:16, 12 June 2008 (UTC)

Confusing title

Would anyone mind changing the title to Black body (astronomy) as the name is ambiguous, it could refer to the bodies of black people. This proposal has nothing to do with the above proposal. Thanks, SqueakBox 21:39, 26 May 2008 (UTC)

The convention for titles is that they should usually be the shortest term usually used for a topic. Since this one is widely used in physics, astronomy, photography, lighting, etc., we should probably leave it; or possible swap it over to black-body radiation (I don't think we need both, and either name will do, since the topic is one and the terms are inseparable). Dicklyon (talk) 22:57, 26 May 2008 (UTC)

Perhaps Black body (science)--Michael C. Price ^talk 14:56, 7 June 2008 (UTC).

There is no need to move the article. I don't think that "black body" would be the ideal title for an article about the physiology of black people, or simply black physiology if you prefer. But it is the ideal title for this subject, being well-defined terminology, while "black body" in reference to physiology is a colloquialism. If such an article exists, a hatnote may be worth inserting. Ham Pastrami (talk) 12:48, 10 July 2008 (UTC)

Temperature relation between a planet and its star

This section is all very jolly but its completely WP:OR. I'm not particularly objecting, mind, just pointing that out. Meanwhile, its also wrong: the earth's albedo is about 30% so the idea that it absorbs all the suns energy is badly out; 14 oC is the average *sfc* temperature but that includes the greenhouse effect; the average radiative temperature is much lower William M. Connolley (talk) 22:30, 23 July 2008 (UTC)

I corrected the math and changed it to a calculation of the temperature of the Earth rather than the sun, which seems to make more sense to me, but then realized that the same calculation already exist at the page for effective temperature. Thomas Palm (talk) 07:22, 24 July 2008 (UTC)

Has anyone tried this equation on another solar body to see if it works? I calculated Mercury based on the values found in Wikipedia. Since it essentially has not atmosphere we don't have to worry about the additional complication of the greenhouse effect.

Using this equation, i get 433K for Mercury's temperature. Unless i'm doing something wrong, it's too high by ~100K for Mercury's equator and more so for higher latitudes. What's the deal?

Also if it's going to use albedo, shouldn't it specify which type of albedo: bond, geometric, or whatever? Though in Mercury's case both these albedo values are so close it doesn't matter.JW Bjerk (talk) 20:38, 22 November 2009 (UTC)

Exactly which equation were you using and what values did you use. Q Science (talk) 16:34, 23 November 2009 (UTC)

The assertion in this article that the surface temperature of a planet is somehow dependent on its albedo is cannot be sustained, it would mean that the probabiliy of absorption and emission in a thermal radiation field were not equal and thus no zero order (steady) thermal state could ever exist for a planet that was not a heat source/sink. For a longer explanation see [[15]] --Damorbel (talk) 17:58, 23 November 2009 (UTC)

precise / perfect

Re [16]. I agree that "most perfect" is invalid, but the edit has changed the sense. Are we saying that the spectrum is the most precisely known, or are we saying that it is the spectrum known to be closest to a pure black-body curve? Either assert would ideally come with some justification William M. Connolley (talk) 19:59, 6 August 2008 (UTC)

Lead section is too long

Lead section must have 4 paragraphs max as stated in WP:LEAD. Can I add the longintro template while we address the issue? --M4gnum0n (talk) 15:54, 4 September 2008 (UTC)

Or just go ahead and fix it... Dicklyon (talk) 16:46, 4 September 2008 (UTC)

BlackBody Simulators

Depending on ones internet browser, the image of a blackbody simulator is infringing on the title of the next section. I do not know if anyone wants to look into this. I am simply going to make the image smaller and see if that helps. The Lamb of God (talk) 15:33, 9 October 2008 (UTC)

Text vs Images

I'm strongly in favor of well-illustrated articles but this time I have to say that there is too little text and too many images. I'm not able to see any way to repair this myself. When I first saw the article, there were so many uncleared floats that the images overlapped the text. I've inserted a couple of clears but that makes the article look ugly, with too much whitespace.

Perhaps I'll take another crack at this later but I really do think the right remedy is attention from an expert in the subject. — Xiong熊talk* 12:47, 15 October 2008 (UTC)

Cut out confusing sentences in lead

Cut:

"If a perfect black body at a certain temperature is surrounded by other objects in thermal equilibrium at the same temperature, it will on average emit exactly as much as it absorbs, at every wavelength. Since the absorption is easy to understand—every ray that hits the body is absorbed—the emission is just as easy to understand.

A black body at temperature T emits exactly the same wavelengths and intensities which would be present in an environment at equilibrium at temperature T, and which would be absorbed by the body. Since the radiation in such an environment has a spectrum that depends only on temperature, the temperature of the object is directly related to the wavelengths of the light that it emits."

The above is not relevant to grasping what a black body is, and mis-directs the reader's focus to the characteristics of the environment, of the incoming radiation.

If you disagree, what is the point being made in the above? Perhaps it can be said another way, or later in the article.ToolmakerSteve (talk) 06:11, 3 November 2008 (UTC)

Yes, I disagree. Blackbody radiation (once called "normal radiation" [17]) is what you get when everything is at the same temperature. A blackbody in such an environment, if it absorbs all radiation that falls on it must emit normal radiation. This is key to understanding the blackbody spectrum, and is often presented that way in sources. I'm not saying it can't be improved, but simply removing that part because you don't quite get it is not a step in the right direction. Dicklyon (talk) 05:01, 6 November 2008 (UTC)

black-body radiation in non-perfect black bodies

I found it confusing that black bodies absorb all incoming radiation, yet "blackbody radiation" is talked about for many objects that are obviously not (perfect) black bodies, such as the Earth. This is especially pertinent now, because understanding global warming requires understanding greenhouse effect requires understanding blackbody radiation -- so I ended up here, even though I am interested in the Earth, not in some theoretical black body. Therefore, I have edited the lead section to speak more naturally to a reader such as myself, coming in with this purpose in mind. My apologies if I have done any damage to the pure physics attitude of the original article. Please help correct what I have done so that it can serve both sets of readers! ToolmakerSteve (talk) 06:14, 3 November 2008 (UTC)

I took out your statement "The formula for black-body radiation approximates the emissions from objects, even if they are not strictly black bodies, e.g. they reflect some radiation. Useful examples are Earth..." Obviously, to some extent such approximations may be useful, but to a much larger extent, understanding the large departures from the blackbody condition is what's useful. The earth is not very close to being a blackbody, which is why it is so much warmer than the blackbody-based calculation would suggest. If you want to have another go at making a better connection to greenhouse effect, etc., it would be best to base it on sources. Dicklyon (talk) 05:06, 6 November 2008 (UTC)

Cloud of light?

I'm sorry, but is there really such a thing as a cloud of light? I'm trying to understand what this article is saying. Is this something that exists in quantum physics? --Angelo (talk · contribs) 03:55, 18 August 2009 (UTC)

Is there really such a thing? Yes, for example inside a box, the contents of the box might be nothing but light; when it's in equilibrium with the box (everything at the same temperature), the radiation will have a black-body spectrum. Another ojbect inside that "cloud of light" in the box will also some to equilibrium with it, all the same temperature, as it absorbs and emits light. And yes, it's a quantum thing. Dicklyon (talk) 05:30, 18 August 2009 (UTC)

Perhaps a "thermal bath of photons," or something of that sort, would be a more standard description. The same concept appears in classical physics (via equipartition theorem among electromagnetic modes of the box), so it's not technically a "quantum thing", but quantum mechanics is needed to prevent an ultraviolet catastrophe. — Steven G. Johnson (talk) 17:00, 18 August 2009 (UTC)

An explanation which allows more people to understand?

Is it possible for there to be more examples and comparisons to real world situations in this article? I'm an electronic engineer, love science, and I need to know some practical stuff about IR radiation for a thermometer I'm building, this means knowing some stuff about black body radiation. But in everyone's (fantastic) effort to get to a perfect technical answer you've forgotten that the purpose of these articles is to educate. One method of improving people's understanding is to provide examples and 'for instance's. I believe the article would lead to more people interested in physics if you did. JakeAy (talk) 08:50, 28 October 2009 (UTC)

Have you read this article Infrared thermometer, or this one Pyrometer, these too Bolometer, Microbolometer? They may help you define the difficulties you are experiencing with this article and thus help to improve it.--Damorbel (talk) 13:25, 29 October 2009 (UTC)

I have added a few of these to "see also". However, there are a lot more that could (should) be added. That's why I also added Category:Infrared Q Science (talk) 15:16, 30 October 2009 (UTC)