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
• The footnote on this isn't really relevant; hyphenating compound adjectives is a standard English language convention. -- Jrstewart 07:45, 20 August 2005 (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 analyses assume that things are sufficiently static that equilibrium descriptions apply locally. Yes, it's true that in non-equilibrium conditions a substance may be absorbing more energy than it is emitting, or vice versa, but the emissivity is still closely related to absorptivity by Kirchoff's law (assuming that local equilibrium applies). —Steven G. Johnson 19:56, Aug 29, 2004 (UTC)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

—Thomas w

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

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

Mechanism?

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

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

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

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

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

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

disputed

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

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

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

If these are true, then the rate at which the sun radiates into all space is $(\sigma T_{s}^{4})(4\pi R_{s}^{2})$, 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)

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:

σT⁴(4πd²/πr²) + σu⁴(.125*4π(30 kly)²/πr²) = σt⁴(4πd²/πR² + 4π(30 kly)²/πR²(30 kly/d))

T⁴(d²/r²) + u⁴(.125(30 kly)²/r²) = t⁴d(d + 30 kly)/R²

lysdexia 11:44, 5 November 2005 (UTC)