Talk:Black body

Good Article nomination has failed

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

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

"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)

Radiation Emitted by a Human

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

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

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

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

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

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

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

$e$ ≈ 1!

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

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

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

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

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

Replace WMAP image with FIRAS spectrum?

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

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

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

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

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

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

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. [[6]] or [[7]]) is about 0.45 at 500 nm and still about 0.37 at 1000 nm, so the drop in emissivity can not explain such low emission at 1000 nm and higher frequencies.

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

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

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

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

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

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

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

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

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

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" [16], §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)

precise / perfect

Re [17]. 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" [18]) 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)

Temperature of Earth

I did the computation in the formula several times and I kept getting 246.78 which is ~2 K lower than the 248.573 shown on the page. I may have done something wrong or I did not understand the formula properly. My calculation was done as follows: X=[sqrt(1-Albedo)]*Rsun Y=Sqrt(X/2D) Tearth = Tsun*Y

So? Who is correct? BTW a result in this computation should not have six signicant digits as the inputs are 3 or 4 digits.75.185.95.241 (talk) 16:34, 17 April 2009 (UTC)

The numbers have been changed several times, e.g. here. It would be best to find source that does it out, and use their numbers, so that it's verifiable in a reliable source and we can see when it deviates from what the source says without rechecking the calculation. Dicklyon (talk) 21:44, 17 April 2009 (UTC)

75.185.95.241 your puzzlement about the Temperature of the Earth is not really surprising because the various formulas you can find for the same calculation generally include a term $\alpha \$ corresponding to the albedo. There are different opinions about what figure should be used for $\alpha \$, this article has used various values over the years, currently it uses $\alpha \$ = 0.367  ; Detailed_explanation Greenhouse effect uses $\alpha \$ =0.3 and gets another temperature. There are links to be found here[[19]] with even more figures.

Proper examination of the matter shows that Kirchoff's law of thermal radiation applies to planets, i.e. they are heated and cooled by radiation only, thus the temperature of a planet is independent of the albedo. This is true of all spherical bodies, a gold plated ball bearing would have the same temperature as the Earth, as would a glass marble. This truth is not accepted in greenhouse effect science which needs $\alpha =0.3$ to sustain its thesis.--Damorbel (talk) 17:11, 18 April 2009 (UTC)

That's incorrect, and a very bogus arguement by those who don't understand and want to deny greenhouse theory. The truth is that the absorptivity and reflectivity, though equal to each other, vary as a function of wavelength. Albedo is defined in the terms of the average reflectance and absorption of the incident solar spectrum, so it can be reduce to single number that determines how much energy the planet gets in from the sum. The emission, however, is more complicated, as the spectrum depends on temperature and the net emissivity averaged over the blackbody spectrum is therefore not constant. Greenhouse gasses are those gasses that are largely transparent to visible and near-IR solar radiation, but rather reflective (or absorptive and re-radiative) to the longer-wave IR of the planet's blackbody radition. This complexity is typical sidestepped by the definition of an effective temperature, which has a relation to actual temperature that is complicated by the greenhouse effect. What's being calculated here is an effective temperature, per its definition. Dicklyon (talk) 18:05, 18 April 2009 (UTC)

Incorrect? So a filter can generate or destroy energy? Before you go any further I suggest you read The Theory of Heat Radiation by Max Planck this nonsense that spectral effects, e.g. molecular resonance effects can create or destroy energy is becoming absurd. The (imaginary) blackbody effect supposedly posessed by the Earth means that energy disappears, arriving at one temperature and leaving at another without the possibility of disposing or aquiring energy in another way i.e. it is thermodynamically a closed system.

When you have finished with Planck, you can read Einstein on the matter particularly "Emission and Absorption of Radiation in Quantum Theory," 17 July 1916 available here: [20].

Can you not see that a body that absorbs only a fraction of the radiation incident on it is not going to be able to emit that absorbed energy with 100% efficiency? --Damorbel (talk) 21:13, 18 April 2009 (UTC)

Yes, I can see that it won't emit at 100% efficiency. The point is simply that the relevant spectral band are nearly disjoint, so the emission efficiency is not very related to the albedo. That's one reason the "effective" temperature is used; it ignores emission efficiency and tells you only temperature of blackbody would emit that amount of energy. If you understand the definitions, your confusion should be cleared up. It doesn't require understanding at the level of Einstein or Planck, but is consistent with them. Dicklyon (talk) 21:30, 18 April 2009 (UTC)

See details of wavelength-dependence and the greenhouse concept here. Dicklyon (talk) 21:39, 18 April 2009 (UTC)

You can only get away with not reading Planck and Einstein if you work it out for yourself, you need a thorough appreciation of the mechanism of absorption and emission. All matters concerning heat are by definition about atomic/molecular motions. The spectral image in your link is only about the transmission of EM radiation, it does not indicate if the curves, very, very far from a "black body in the infrared" assumption! Your spectral graph is about transmission only, it takes no account of emission. But GHGs emit radiation just as easily as they absorb it; in a gravitational field they always emit more than they absorb for the simple reason that their density decreases with the distance away from the source (the direction of maximum emission).

Molecules absorb/emit EM energy because they have an electric and/or magnetic dipole; there is no other EM mechanism. Because of the dipole moment EM waves interact with them by producing a mechanical force that accelerating them, the same force that makes an electric motor turn - no dipole, no absorption/emission! Gas molecules have mechanical resonances that give the absorption an irregular spectrum. A black body is a hypothetical concept comprising an array of resonating dipoles, it takes no account of the need for real matter, with its kinetic exchange of energy, to be present. Temperature is a purely kinetic concept, it is only the dipole moment that links it to radiation. --Damorbel (talk) 10:20, 19 April 2009 (UTC)

Are there any sources that will help me understand your interpretation of greenhouse gasses; that is scientists who have worked it out and explained it? Are you saying that by invoking Einstein and Planck you can confidently prove that all planetary scientists are wrong? Interesting. I'd like to read more about that... Dicklyon (talk) 15:54, 19 April 2009 (UTC)

The ones I gave are the classical ones but the whole matter was started by G. Kirchhoff in 1862 with his paper "On the Relation Between the Emissive and the Absorptive Powersf Bodies for Heat and Light"; available in English in the book "The Laws of Radiation and Absorption" by D B Brace, I found a PDF for free on line but have since bought a copy through Amazon.

Kirchhoff's work was done 35 years before J J Thomson identified the electron as a particle, and even longer before Planck used it as a basis for his quantum theory and Einstein for his work on quantum mechanics. Like Newton's law of gravitation these works may be modified by new insights but never overturned completely.

If I read you correctly, you agree that the ratio of emissivity to absorptance is unity throughout the spectrum then you are nearly there; the key to understanding is that absorbed incoming radiation heats the planet whatever its wavelength, it just so happens that the Sun is hotter than the Earth. The Earth radiates heat to deep space because it is hotter than deep space, it doesn't matter how it is heated. All that Kirchhoff noted was that it is exactly the same mechanism that causes the Earth to radiate as lets it absorb, the cooling and heating occur via the same means. Kirchhoff didn't know about electrons and molecular dipoles but he was right about the ratio of absorption and emission. I may be repeating myself but a planet with an albedo >0 can never behave like a black body, a requirement for the Greenhouse Effect

I'm sorry that all those scientists who argue for the existance of a Greenhouse Effect do not seem to know about the radiation laws I am referring to but I can't help that, they never pursue the arguments fully but make various invalid assumptions, such as the Earth emitting as a black body, which give the wrong answer. --Damorbel (talk) 18:45, 19 April 2009 (UTC)

I do understand all that you say about emissivity equals absorptivity at every wavelength. All the planetary scientists agree with that, too. But they don't misinterpret it as you do; the greenhouse effect is precisely because the planet cannot radiate as a blackbody; if it could, its temperature would be equal to the effective temperature, by definition thereof. Dicklyon (talk) 19:33, 19 April 2009 (UTC)

Is it your argument that GHGs are what stops the planet radiating as a black body, i.e. they are responsible for the albedo? I am thinking that the reqirement for a black body is that it absorbs all incident radiation, i.e. it doesn't reflect any. --Damorbel (talk) 20:49, 19 April 2009 (UTC)

Yes and no; , since they absorb in the band where a 255K or so emitter would emit; but no they don't affect the albedo much, since very little of the sun's energy is in that band. Under no condition should you think of the earth as a blackbody; there are many things that prevent it being one; it's still a useful comparative model though. Dicklyon (talk) 02:09, 20 April 2009 (UTC)

Dicklyon - "GHGs reduce the earth's ability to radiate"? Not remotely true, H2O, CO2 etc. radiate powerfully in IR, just look here (Work your way to Meteosat 9 (full disk - WV), this shows just how much H2O (WV = water vapour) radiates. There is a wealth of images on this site.

"Under no condition should you think of the earth as a blackbody" - that is what I have been banging on about. How can it be "a useful comparative model" when it predicts (for the Greenhouse effect) an equilibrium temperature of −18 °C ? This −18 °C (255K) is fully part and parcel of the CO2/H2O effect that is allegedly saving us from being frozen to death.

It gets worse than that, the formula calculating this prediction is given in this article $T_{E}=T_{S}{\sqrt {\frac {{\sqrt {1-\alpha }}R_{S}}{2D}}}$ making the equilibrium temperature very strongly dependent on albedo, here it puts $\alpha =0.367$, giving a temperature of $T_{E}=248.573\ \mathrm {K}$, very far from the 255K of the Greenhouse effect predictions. This alleged dependence of the planetary temperature on albedo can easily be shown to be nonsense, try putting albedo = 85% (like Venus), does anyone really believe that 2.5 times the cloud cover will drop the equilibrium temperature to 173K? --Damorbel (talk) 15:47, 21 April 2009 (UTC)

You're very confused; the 255 K is not an equilibrium temp with greenhouse effect; it's WITHOUT any greenhouse effect; it's a good thing we have some greenhouse gasses or the planet would be very cold, as you seem to understand. For venus, the "effective temperature" considering the albedo would be very much lower than the actual temperature, since Venus has very strong greenhouse effect. Dicklyon (talk) 16:45, 21 April 2009 (UTC)

I made no calculation including a greenhouse effect. I couldn't, I wouldn't know how to. All I say is the Earth's temperature is independent of the albedo because Kirchhoff's law of thermal radiation requires the emissivity/absorptivity ratio of planets to equal one . Didn't you agree with that? All I did was cite a formula used by the greenhouse faction, a formula that makes the Earth's temperature dependent on the albedo, a formula that does not have the ratio as one and therefore gives the wrong answer. --Damorbel (talk) 18:19, 21 April 2009 (UTC)

I agreed only that aborptivity and emissivity are equal to each other at every wavelength; but weighted over the very different spectra of insolation and emission, they are essentially unrelated, as the plots in the books I referenced illustrate. Dicklyon (talk) 20:54, 21 April 2009 (UTC)

The dominant fact remains that introducing the term ${\sqrt {1-\alpha }}$ (which has nothing to do with absorption and emission) into the temperature equation makes it thermally unbalanced and therefore invalid. Equally invalid is the concept that wavelength dependent effects can determine the totality of emission and absorption. The totality of thermal emission and absorption depends only on the exchange of EM energy and thermal (kinetic) energy by means of charge. Albedo arises from scattering/reflection processes which exclude energy exchange (mirrors are neither heated nor cooled by reflection). Reflection/scattering is redirection of energy without absorption, this distinction is never made in Greenhouse effect discussions, it is a shame that it renders the whole hypothesis invalid, it is the physics that is mistaken, not the mathematics. --Damorbel (talk) 08:36, 22 April 2009 (UTC)

The factor introduced to account for wavelength-weighted absorptivity due to all those mechanisms is just $1-\alpha$; you could call it anything different if you don't like to use albedo. In the effective temperature solution, it shows up as a fourth root (which someone decided to write as a square root of a square root. Since effective temperature is defined as ignoring emissivity, no corresponding factor for that is found to cancel it. If you want to know the temperature would be if average emmissivity equaled average aborptivity (each average over the corresponding relevant spectrum), then you can leave that out and you'll have an albedo-independent formula. But since we can measure and affect the albdeo and the greenhouse effect pretty much independently, that's not how it's usually done. Your "dominant fact" is nothing. Dicklyon (talk) 15:08, 22 April 2009 (UTC)

The expression ${\sqrt {1-\alpha }}$ is how the promoters of the Greenhouse effect introduce the albedo into the calculation of the planetary temperature. The idea is that the reflected light, the light we see when looking at Earth from far away, is to be subtracted from the incoming solar radiation when calculating the temperature. Now if $\alpha \,\!$ (the albedo), is 0.3 (30%) then the reflected part is the 30% and the absorbed part is (1-0.3)= 0.7 or 70%. Kirchhoff's law of thermal radiation states that the absorption coefficient equals the emission coefficient so there is (on average) a single temperature. If the Greenhouse hypothesis was correct and the Earth emmitted as a black body there would need to be be a real temperature and a second (lower) blackbody temperature (because a black body is a more effective radiator). Alternatively the planetary temperature would drop continuously because it was emitting more effectively than it was absorbing,--Damorbel (talk) 20:37, 22 April 2009 (UTC)

What makes you think the Earth as a whole is in thermal equilibrium? The amount of solar radiation incident on the Earth changes radically between winter and summer, so it is a driven oscillator even on short time scales. And there was increasing amount of solar energy trapped in stored oil over hundreds of millions of years, which we are now releasing in a (geologically) short time, and lots of other things driving the system in a nonequilibrium fashion. For that matter, all of biological evolution represents a nonequilibrium situation. Of course, we will eventually reach equilibrium at the heat death of the universe, but until then arguments about the planetary climate based entirely on thermal equilibrium seem bogus. In practice, we only ever have approximate thermal equilibrium locally, on smale scales of space and time (compared to the scale of the planet and geological time). And claiming that thousands of climate scientists have never made detailed studies of the radiative forcings in the system is absurd (and contradicted by mountains of publications...see the IPCC report for a summary). —Steven G. Johnson (talk) 16:54, 22 April 2009 (UTC)

Steven G. Johnson I am perhaps mistaken but I suppose it is me you are referring to. Let me say straight away, you are entirely correct in pointing out heat sources and sinks that cause deviation from thermal equilibrium. Perhaps you should take account of day/night variations also. The biological and fossil fuel effects you mention can best be calculated separately and, in so far as the will cause deviations in the average temperature, one must know the actual emissivity (absorptivity still equals emissivity) of the planet to find out the rate of cooling/heating. This is in stark contrast to the unity ratio that applies when calculating the temperature due to the Sun's radiation, as I said before, the unity ratio comes from the fact that radiative emission/absorption is by means of the same atomic/molecular charge, there is not one set of molecules/atoms emitting and another set absorbing.

I know I refer to "thermal equilibrium" which, strictly speaking, means there is no heat flow but in fact there is a great deal of heat flowing from the Sun. Steady state is a more accurate description. --Damorbel (talk) 17:28, 22 April 2009 (UTC)

"there is not one set of molecules/atoms emitting and another set absorbing" you say. But to a large extent, that's false. The involvement of differnet molecoles in the atmosphere and solids and liquids depends on their particular transitions; the transitions that do most of the absorbing of solar radiation are those matching the visible and near-IR and near-UV wavelengths where most of the solar energy is. The molecules involved in radiation are those that better match the approximate black-body spectrum of the planet. To a large extent, they are not the same molecules, though of course there's a lot of overlap, especially in the solids and liquids where the transitions are very spread out. Dicklyon (talk) 18:00, 22 April 2009 (UTC)

"The involvement of differnet .... where most of the solar energy is." Again yes! But the concept of a single temperature for the whole planet is also quite wrong but not misleading. The concept that the albedo influences the average planetary temperature is not only incorrect but misleading, it is dangerously wrong since it is causing money to be wasted on costly projects to save us from something that isn't going to happen, namely planetary temperature change due to CO2 in the atmosphere!--Damorbel (talk) 18:20, 22 April 2009 (UTC)

How can you assume that the Earth is in a steady state? This is not true on either short time scales (day/night, winter/summer) or on longer time scales (there is clear evidence of many kinds of long-term temperature variations). —Steven G. Johnson (talk) 22:15, 22 April 2009 (UTC)

The "assumption" is not expected to represent anything real; it's a calculational device, like "effective temperature" is; the thing that's easy to solve for is what would be the one temperature, which if constant, and if the earth radiated like a blackbody, would lead to radiation that matches the energy absorbed from the sun. It's not mysterious, but also not intended to be realistic. It's just what it is. Dicklyon (talk) 00:02, 23 April 2009 (UTC)

Since we agree that this assumption is unrealistic, and hence the results it produces are misleading, why is there a huge section of the article on black bodies devoted to this misuse of equilibrium assumptions? —Steven G. Johnson (talk) 15:35, 23 April 2009 (UTC)

The results aren't misleading if sensibly interpreted; we have this section because it is a common approach in reliable sources. The "effective temperature" gives you an idea what the temperature would be if the planet radiated as a blackbody. The difference relative to the real average temperature is mostly due to greenhouse gasses in the atmosphere. The method is one step toward understanding that. Dicklyon (talk) 16:04, 23 April 2009 (UTC)

Steven, "steady state" is not an assumption, it's an observation. Historically the solar system has a beginning, a life and an end, effectively its state is changing all the time. In any given period, much shorter than the total life, there is an average or "steady" state, during this period there will almost certainly be deviations from this "steady" state. But deviations observed, large or small, will probably have an identifiable cause, thus the benefit of identifying a steady state comes from discovering a cause of deviations, be it a little CO2 in the atmosphere or a supernova event in the Sun. --Damorbel (talk) 07:44, 23 April 2009 (UTC)

In other words, it's not in a steady state, because there are numerous causes of unsteady behavior. Hence arguments against greenhouse effects based on "equilibrium" considerations are bogus (and, for that matter, circular: if you assume it's in steady state, then by assumption there can be no heating). —Steven G. Johnson (talk) 15:35, 23 April 2009 (UTC)

The notion of an equilibrium or steady-state solution is useful, and not bogus, even in the context of systems that are not quite in equilibrium and not quite in steady state. But you're right that many of the arguments against these analysis are bogus. Those who want to argue that greenhouse gasses are not the reason why the planet's average temperature is higher than the calculated equilibrium effective temperature need to show other reasons why the net emissivity is as low as it is; Damorble argues that it's lower because it should be equal to 1 minus the albedo, just like the absorptivity is; but his argument is bogus because it appeals to a misinterpretation of the fundamental physics by ignoring the wavelength dependence; it also ignores all that's known about planetary atmospheres; he pretends that his amateur analysis is better than the long-evolved understanding of the planetary science community. This is a typical argument by people whose real objection is political or religious, rather than scientific. Please note that I'm not among those who think that the scientific point of view ("SPOV") needs to trump other points of view in wikipedia; but so far I haven't seen any other points of view of this question in reliable sources; if there's a religious or political organization that has published an analysis the differs, in a reliable source, I'd have no issue with mentioning and citing that alternative POV. Dicklyon (talk) 17:35, 23 April 2009 (UTC)

Dicklyon, you never made it clear what is your source for Kirchhoff's laws, since he explicitly excludes wavelength sensitivity your argument that it has a wavelength sensitivity does not carry a lot of weight. What papers have you read on the matter, I would like a reference. I have read many of the publications of what you call "the planetary science community" and following an old tradition, I am pointing out an inconsistency. You make remarks that I am an amateur and use "bogus arguments", I think you intend these to be deprecating attacks on the quality of my contributions. If this is so, may I ask you to withdraw them? Steven may consider withdrawing his similar remarks. Thanking you in advance. --Damorbel (talk) 06:09, 24 April 2009 (UTC)

No, his argument is bogus because he's mis-applying Kirchhoff's law to a system that is not at thermal equilibrium, not even locally on the relevant lengthscale of the ground/atmosphere system. —G. Johnson (talk) 21:44, 23 April 2009 (UTC)

Steven, what is your source for Kirchhoff's law of thermal radiation? This is a very nice article but it has its limitations. Have you read the original papers? It was Kirchhoff who introduced the concept of a black body, you should have read what he has to say on the matter before describing anyting as "bogus". Have you read Max Planck on the matter, or Albert Einstein. These sources are a suitable introduction. If you have read them you will be able to say precisely where I misunderstand them. Until you have read them, I suggest you will have difficulty in explaining why the equation $T_{E}=T_{S}{\sqrt {\frac {{\sqrt {1-\alpha }}R_{S}}{2D}}}$ quoted from this article, is mistaken and that the term ${\sqrt {1-\alpha }}$ is the cause of the error. --Damorbel (talk) 06:09, 24 April 2009 (UTC)

Damorbel, you are correct in observing that my remarks represent "deprecating attacks on the quality of [your] contributions". That's because your "contributions" are stupid and pig-headed. I'm sure you're a great guy, fun to have a beer with and all, but what you keep saying is nonsense; your pretense that your analysis is "pointing out an inconsistency" in the planetary science is pure hogwash. I hope you don't misinterpret me and think I would take any of this back. Dicklyon (talk) 06:32, 24 April 2009 (UTC)

Stupid? Pig headed? That is how you see my objection to the equation $T_{E}=T_{S}{\sqrt {\frac {{\sqrt {1-\alpha }}R_{S}}{2D}}}$ ? I have given good sources to show it is invalid, you don't appear to have read them and you respond with abuse! Is this your idea of good wiki practice? --Damorbel (talk) 07:59, 24 April 2009 (UTC)

No, I admit that I continued to respond long after it was useful. Dicklyon (talk) 15:20, 24 April 2009 (UTC)

This article is a prime example of why Wikipedia cannot be trusted as a definitive source for , particularly , politically sensitive physics . The modification of the correct black body equation by adding an albedo term to the earth's absorptivity while blatantly failing to add the corresponding term for its emissivity is inexcusable . I'm glad to see Damorbel asserting exactly the same basic classical physical arguments I have made both here and on the Stefan-Boltzmann page over the last couple of years . That Wikipedia allows the absurd assertions that the albedo of a uniformly colored sphere will effect its equilibrium temperature , or that that doesn't matter because the earth is always just chasing its equilibrium , is a black mark against Wikipedia . I have implemented the correct classical physics at http://cosy.com/Science/TemperatureOfGrayBalls.htm . This simple implementation provides quantitative , experimentally falsifiable albeit sensible values . As pointed out above , leaving out the emissivity term for the earth leads to absurd results . Wikipedia needs to get a real physicist to clean up these pages despite however much they pray for the "greenhouse effect" to be real . Right now , this page is an embarrassment . Bob Armstrong (talk) 22:15, 25 April 2009 (UTC)

The person who should be embarrassed is you. Physicists are well aware that emissivity or absorptivity will not affect the equilibrium temperature if it's the same at all wavelengths; they're also well aware that's it varies strongly with wavelength and that albedo is just a weighted version, reduced to one number based on the solar spectrum, and that it's pretty much unrelated to a similarly weighted version using the blackbody emission spectrum. They define an "effective temperature" not to be absurd, not to calculate a meaningless number, but to calculate a baseline against which to compare the actual temperature that depends on the emissivity. They all know the earth is warmer than the "effective temperature" because it does not radiate like a blackbody. They've also studied the reasons for that in great detail, and have found the gasses in the atmosphere that absorb those radiated wavelenghts and re-radiate much of it back to the earth are part of the reason. They call these greenhouse gasses. Nothing here is political or religious until you get into people like you who are confused at the difference between greenhouse gasses and anthropomorphic greenhouse gasses, and have political or religious reasons to want to deny the latter and get confused and also deny the former. Yes, I'd be very embarrassed. Dicklyon (talk) 22:25, 25 April 2009 (UTC)

Your linked page includes the following bizarre statement: "The 'Black Body' Wikipedia page was the first place I saw the derivation of the Earth's temperature from the Sun's based on this equation." Bizarre, why? Well, glad you asked. Two things: first of all the wikipedia page does NOT calculate the Earth's temperature using this equation; it calculates only the "effective temperature", which is very clearly stated not to be the actual temperature; it references what the term means, very clearly. Second, it is bizarre that you would write a page like this, pretending to be some kind of a physics or planetary science expert, never having read any of the books about this kind of calculation. I first saw this in college over 35 years ago; greenhouse effect and understanding of planetary atmospheres and such long predates the concern over anthroporphic global warming that you seem to be so allergic to. Dicklyon (talk) 22:32, 25 April 2009 (UTC)

Just noticed your funny footnote that includes "I have seen it pointed out that Kirchhoff's equating of absorption and emission only applies at equilibrium. However absorption and emission spectra are generally rather constant for a substance over a given physical state." This gives me a clue what Damorbel was going on about and what some of the discussion above about equilibrium was being driven by. It's true that the spctrally averaged emissivity and absorptivity would be equal in an equilibrium situation, since the radiation being emitted and absorbed would be of the same black-body temperature. This is nowhere near true in the case of the radiation from the sun impinging on the earth. As several sources that I linked above show, the situation with the sun around 6000K and earth around 300K is so far from equilibrium that the spectra are almost totally disjoint. All your examples about grey spheres will have the same problem; the actually steady-state (not equilibrium) temperature of your balls in sunlight will depend not at all on how dark or light they look, but rather on how that albedo compares to their long-wave IR emissity. Maybe you need to do that experiment instead of the gedanken one you describe as "If it were the case, one would expect a ball coated with Magnesium Oxide with an albedo of about 0.9 to come to an equilibrium temperature of about -120c in a vacuum bottle sitting in room temperature surroundings," which simply displays your total failure to comprehend. Dicklyon (talk) 22:44, 25 April 2009 (UTC)

With respect to Venus you also have the absurd statement "The most heating this greenhouse theory could predict is raising the temperature back to the black body temperature. No 'runaway' effect could raise the temperature beyond that." What you mean here by "the black body temperature" is not even clear; I suppuse you mean the effective temperature that would have been computed when ignoring albedo? In any case, it's nonsense; the actual temperature can be any amount above the effective temperature; there's nothing that constrains the net emissivity to be greater than net absorptivity. Venus is in fact a good example of where greenhouse theory works out well to explain what would otherwise be uninterpretable, or interpretable only as an unknown internal heat source as you say "Confirming that Venus is much hotter than any simply radiantly heated object in its orbit could be . There must be some other internal source of heat." Your "confirmation" would be better called a "hallucination" or "divine inspiration" or "bullshit" or something like that. Your admission that "I do not understand how James Hansen could possibly have claimed that Venus's mean temperature could be due to heat trapping of any sort" begins to make sense... Dicklyon (talk) 23:00, 25 April 2009 (UTC)

No , you've now had multiple people point out the absurdity of your neglect of Kirchhoff . Your second sentence in your reply is simply wrong . albedo has nothing to do with solar or any other particular spectrum . It is simply the assumption of a flat spectrum . All the stuff about "effective temperature" is irrelevant obfuscation . It refers to the apparent temperature of radiant bodies observed from the outside . If you agree that the earth does not radiate as a black body , how can you possibly justify leaving the parameter out of the equation ? Forget any gasses or detailed structure . This is about the radiative equilibrium of balls depending on their temperatures and their spectra . It doesn't matter how the spectra are produced , all that matters is what the spectra of the bodies are . Until this section of the page got perverted by the unjustifiable addition of an absorptivity parameter to the left side of the equation while leaving the emissivity of the right side as a black body , the equation was correct . With your distortion , as has now been pointed out by others also , a white ball will be colder than a black ball in exactly the same radiant environment . Show me any experimental verification of that effect . You can't because it's absurd . It doesn't matter what either of us have read or not read , tho I am quite happy for anyone to judge me by my http://CoSy.com . What matters is LOGIC and experiment . Energy is conservative . Therefore the average over a cycle is equivalent to an equilibrium .

I know precisely how to extend my implementation of the Stefan-Boltzmann/Kirchhoff to spectra and expect to do it as I find time this year coming out with quantitative , experimentally testable numbers . So far as I can tell , you don't have a clue how to extend your analysis to spectra ( and your "greenhouse" gasses ) . In any case , the quantitative extension of your analysis to spectra is nowhere to be found on Wikipedia . Your statement that "with the sun around 6000K and earth around 300K is so far from equilibrium" shows such a lack of comprehension that it should by itself disqualify you from having control of any of these pages . It's your equation that predicts the absurdity of white balls being cold because they are white , not the correct equation .

The meaning of "black body temperature" in my paper is what is calculated by the Stefan-Boltzmann/Kirchhoff equations I implement . I'm talking about real temperatures that can be measured by real thermometers or determined by measuring radiant flux . The fact that you do not even understand that Venus is radiating on the order of 16 times the amount of energy that an object in its orbit is receiving from the sun shows your total incompetence . Actually in Hansen's defense ( only on this issue , not his general behavior ) I recently saw some comment that he did talk about some internal heat source in Venus in his original work .

I will repeat , this page is an embarrassment to Wikipedia . I again challenge you to present any experimental verification that albedo of a radiantly heated ball affects its mean temperature . I further challenge you to present the next steps in calculating the effects on temperature of differences in spectra , as with various gasses , of radiantly heated balls . My implementation has already been replicated by another person and has been translated into a more accessible array language which anyone can download and experiment with or extend .

This nonsense has been allowed to fester in Wikipedia long enough . I appeal to the masters of Wikipedia to enlist some real thermodynamicists to clean up these pages with experimentally verified and rational physics . Politics and religion should have nothing to do with it . Bob Armstrong (talk) 00:00, 27 April 2009 (UTC)

Albedo can be defined either as a wavelenght-dependent thing or as a weighted average, as in "broadband albedo is the reflectance in multiple bands integrated over the total solar spectrum." This depends on the solar spectrum. What definition do you prefer? Or do you ignore the definition, as you do with "effective temperature"? Dicklyon (talk) 01:57, 27 April 2009 (UTC)

Dicklyon, Albedo is a reflection/scattering process; as such it can introduce delays in the lightwave propagation producing spectral (colour) effects by interference, as well as by selective refraction. You should also realise that an important component of albedo comes from the incident wave encountering a change in density, another factor mostly independent of temperature and certainly having nothing to do with absorption and emission. Closely related is diffraction (see Diffraction grating). These (spectral) effects are quite independent of temperature. The magnitude of thermal absorption and emission depend only on the interaction of an EM and charge or a dipole, I have probably pointed out before that this occurs at a dimension far below optical wavelengths so it is not wavelength sensitive. The wavelength properties of the body confuguration govern the spectra of emission and absorption but they cannot alter the total energy.

The question of effective temperature is relevant. Effective temperature is the temperature of a body deduced from its radiation "signature", i.e. by measuring the output of EM energy. Take two cases, a gold plated sphere which reflects 99% of incident radiation i.e. it has an albedo of 0.99 and its absorbtivity is 0.01; take a charcoal sphere that reflects 1% (albedo 0.01) and absorbs 99% of incident radiation, put them in the same orbit as Earth and let them rotate so that they get to an even temperature. Now, if the nominal temperature is 283K, what is the effective temperature of the gold plated sphere and what is its thermal temperature? I think you will find that, being such a poor emitter, the gold sphere will be almost invisible in the infrared but is thermal temperature will be just the same as the charcoal. The charcoal will of course be very much brighter in the infrared, whereas the gold, with its higher albedo will be much brighter, if you are able to see it (c.f. Venus).--Damorbel (talk) 15:03, 27 April 2009 (UTC)

As to the balls you have, the equations correctly predict that the white one will have a much lower "effective temperature" than the black one, since the white one will absorb less energy from the sun, and therefore in steady state will emit less energy, and since the "effective temperature" is simply a measure of that emitted energy (the temperature of a black body that would emit the same amount of energy as a white sphere is going to be much lower than the temperature of the white sphere itself); if you don't accept that definition, and choose to keep erroneously confusing "effective temperature" with temperature, then of course you're going to have this conflict that you seems to torment you. Dicklyon (talk) 02:02, 27 April 2009 (UTC)

Your challenge to "present any experimental verification that albedo of a radiantly heated ball affects its mean temperature" is irrelevant, since the equations here don't say anything about mean temperature, only about effective temperature. Yet different colors and textures of ball do have different steady-state temperatures in the sun, depending on how the average absorptivity in the solar spectrum (1 – albedo) compares to the average emissivity in the band relevant to the actual radiation at a much lower temperature than the sun. That's why on your page showing thermometers in balls in the sun, the shiny one has a rather different temperature than the others; whether the black on or the white one comes out warmer depends on the emissivity of those surfaces at longer wavelengths than you can see. Dicklyon (talk) 02:11, 27 April 2009 (UTC)

Dicklyon, the Greenhouse effect, because of its thesis of "blackbody in the infrared", claims that, for the Earth, without the radiation effects of H2O,CO2 etc. would have a temperature of 254K. To justify this they use the albedo in their calculations, but the albedo is not a radiation effect of CO2 H2O etc., it is a reflection/scattering effect of the entire planet, an effect not related to thermal absorption/emission.--Damorbel (talk) 15:31, 27 April 2009 (UTC)

Don't worry, I don't have control of any of these pages. If you'd like to help, I suggest you try to understand reliable sources first; here is a good book that's very clear on the definitions and why the average emissivity and absorptivity are different so that it is "possible and convenient to treat the solar and terrestrial fluxes independently." And here's a book that should help clear up your delusions about Venus. Dicklyon (talk) 02:11, 27 April 2009 (UTC)

Your book contains the mistake I am talking about, I have seen many better argued but still incorrect publications, their better argument doesn't make them correct. I can find you books that say the Sun rotates round the Earth, that heat is phlogiston, even Sadi Carnot believed it was a substance called "caloric". For years "the vast majority of scientists" believed light travelled through a substance called Aether, they wrote books about it; the greenhouse effect is much the same, the "theology" of global warming says CO2 must be the cause,even if thermodynamics says it cannot be!--Damorbel (talk) 15:31, 27 April 2009 (UTC)

I've followed this whole discussion, and I pretty much have one point to make. If the greenhouse effect is a load of crap, why, pray tell, are there greenhouses? Dicklyon's physics apply equally well to both greenhouses and earth. The other two guys' versions are, from what I can tell, things made by people who've read something but didn't quite understand it (but who think they did). Headbomb {^ταλκ_{κοντριβς} – WP Physics} 06:32, 27 April 2009 (UTC)

The temperature of the Earth with no atmosphere.

I wrote a small comment on this at 11.25 AM on 7 sep 09 and placed it higher up on this page. I was trying to be polite. But I have now scrolled down and have had a glimpse of a nightmare i.e of what Wikipedia might have become.

As I implied earlier you can make any approximation you like in a model, but not all models are equal. Some of them are totally unrealistic.  The emissivity (1-alpha)  of solids in the infra-red is usually about 1 so it is sensible to ignore the albedo in that frequency range. In the visible range you can set (1-alpha) =1 but if you do , you would have to predict that photographs of the Earth from outer space would not look mainly bright blue but would be completely undetectable unless you had an infra-red camera. 

It is rather like using a steam hammer to crack a nut, but if you are unsure a good reference for Kirchoff's law is Landau and Lifshitz, Statistical Physics ,1958, p.176. As stated by others earlier on, you cannot combine the Sun and Earth into a single equilibrium system without raising the Earth to about 6,000 degs.C. The same applies when you add an atmosphere which is not at the same temperature as the ground. In effect the emissivity for a layer of CO2 or H2O is highly variable across the IR range. That does not mean that Kirchoff's law has broken down. Deconvoluter (talk) 16:52, 7 September 2009 (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? --[[::User:Angelo|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)