Talk:Black body

"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

Blackbody vs emission spectrum

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

• A really simple cavity (http://physics.ucsd.edu/was-sdphul/labs/demos/modern/blackbody.html)
• Using cavities (http://www.omega.com/literature/transactions/volume1/calibrate2.html) for calibrating pyrometers. The graph at the start is hard to read but seems to back me up.
• Our question answered (http://www.newton.dep.anl.gov/askasci/phy99/phy99525.htm) by Ask a Scientist.
• A PDF file (http://orca.phys.uvic.ca/~tatum/stellatm/atm2.pdf) 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)