By Douglas B. Staple

2013-03-27 15:10:21 8 Comments

By conservation of energy, the solid is left in a lower energy state following emission of a photon. Clearly absorption and emission balance at thermal equilibrium, however, thermodynamic equilibrium is a statement of the mean behaviour of the system, not a statement that the internal energy is constant on arbitrarily short timescales. The energy has to come from somewhere during emission, and go somewhere during absorption.

Energy in a solid can be stored as kinetic and potential energy of electrons and nuclei, either individually or in collective modes such as phonons and plasmons. In thermal equilibrium energy will be stored more or less in various forms depending on the temperature and material. However, even if most of the thermal energy in a particular solid at temperature $T$ is stored in the form of phonons, it could be that phonons primarily interact with light indirectly via electrons, e.g. a phonon excites an electron in a phonon-electron interaction, which can interact with light via the EM field.

Given that light is an EM field, it makes sense to me that it is emitted and absorbed by charged particles. The electron-photon interaction is probably dominant for visible and ultraviolet light, given that metals are opaque, while semiconductors and insulators are transparent to (visible and UV) light with energy lower than their bandgap. However, once you get into energies in the IR and below, or X-rays and above, other mechanisms apparently take over. For example, on the high-energy end of the spectrum I've heard that gamma rays can interact directly with nuclear degrees of freedom, which is reasonable considering that gamma rays are emitted during a lot of nuclear reactions.

A review of absorption spectroscopy might give clues to important light-matter interactions over a broad range of wavelengths. Whether all of the these processes are involved in blackbody emission is a somewhat different question.

What physical processes mediate energy transfer during blackbody emission, and in which energy ranges are the various processes dominant?


@DilithiumMatrix 2013-04-02 16:54:37

This is a fantastic question, and a topic I was very confused about when I first took a class on Radiative Processes. The ultimate answer, as hinted at by @LubošMotl, is anything---if you start with a 'white-noise' of radiation (i.e. equal amounts of every frequency), it will equilibrate with the medium/material into a black-body distribution because of its thermal properties (see: Kirchhoff's Law, and the Einstein Coefficients). This is just like if you gave each molecule in a gas the same energy, they would settle to a Boltzmann Distribution.

In practice (and hopefully a more satisfying answer) is that it's generally a combination of line-emission and Bremsstrahlung, with Bremsstrahlung1 dominating at high temperatures ( $T \gtrsim 10^6 -10^7 K$ ). Lines are produced at myriads of frequencies depending on the substance of interest, and the thermodynamic properties (e.g. temperature). For everyday objects, I think the emission is primarily from molecular-vibrational lines. Individual lines are spread out by numerous thermodynamic broadening effects to cover larger portions of the spectrum. Finally, as per Kirchhoff's law, equilibrated objects can only emit up to the black-body spectrum. In practice, you'll still see emission/absorption lines imprinted, and additional sources of radiation.

Lets look at a breakdown of the relevant transitions as a function of energy level:
radio: nuclear magnetic energy levels (also cyclotron emission in the presence of moderate magnetic fields).
microwave: rotational energy levels
infrared: vibrational energy levels (molecules)
visible: electronic (especially outer electron transitions)
ultraviolet: electronic (especially outer/valence electron ejection/combination)
x-ray: electronic (inner electron transitions)
gamma-ray: nuclear transitions

1: Bremsstrahlung (German for 'braking radiation') is radiation produced by the acceleration of charged particles---most often electrons. This can happen between any combination of bound (in atoms) or unbound (free or in plasma) charges.

@Luboš Motl 2013-04-03 12:42:27

Good, relevant specifications, +1.

@Christoph 2013-04-04 20:49:10

it's Bremsstrahlung, not Brehmsstrahlung - people with the surname Brehm apparently are more likely to go into biology or medicine instead of physics ;)

@user10851 2013-04-05 01:37:39

Bremsstrahlung - clearly the answer of an astrophysicist :)

@Rob Jeffries 2015-03-01 22:28:33

I've never heard of free-free and bound-free transitions of hydrogen cations (the dominant continuum opacity in the visible part of the spectrum in the Sun) referred to as bremmstrahlung, however you spell it.

@Stephen J. Crothers 2016-02-10 15:01:40

In the following paper Professor Pierre-Marie Robitaille has argued that thermal emission is due to vibrations of nuclei within the lattice of a material, and hence also of a blackbody:

Robitaille, P.M. On the validity of Kirchhoff’s Law of thermal emission. IEEE Trans. Plasma Sci., 2003, v. 31, no. 6, 1263–1267.

@Rob Jeffries 2016-02-10 15:08:04

Thermal emission is not the same thing as blackbody emission.

@Stephen J. Crothers 2016-02-11 06:21:47

@Rob Jeffries - You are incorrect. Blackbody emission is thermal emission. Planck's equation describes the thermal spectra of a blackbody, but nothing else. His book, in which he derives his equation, is titled 'The Theory of Heat Radiation'. Recall that heat = thermal.

@Stephen J. Crothers 2016-02-11 06:34:29

@Rob Jeffries - Your mechanism (H- dominant) for the the continuous spectrum of the photosphere is incorrect, although standard solar theory. Gases do not emit a continuous spectrum; only condensed matter can do that. Arguments for pressure broadening of gas spectra, and optically thick gases, do not produce a continuous spectrum at all. That the photosphere produces a continuous spectrum attests to its condensed matter state. This is confirmed by observations of eruption of solar flares, which produce concentric transverse waves in the photosphere.

@Rob Jeffries 2016-02-11 07:27:50

@StephenJCrowthers I'll address your first point (your second is absolute fantasy). I said that thermal emission is not blackbody emission, not the other way round. Please compare the spectrum of thermal bremsstrahlung with the Planck function.

@Stephen J. Crothers 2016-02-11 15:08:14

@Rob Jeffries - Blackbody emission is a form of thermal emission. Planck called it heat radiation, as in the title of his book: 'The Theory of Heat Radiation'. Kirchhoff's Law is called Kirchhoff's Law of Thermal Emission, or Kirchhoff's Law of Thermal Radiation. All materials at thermal equilibrium emit thermal (i.e. heat) radiation characteristic at a given temperature. Blackbodies emit what is called black radiation. Planck's equation describes thermal spectra of blackbodies.

@Stephen J. Crothers 2016-02-11 15:10:25

@Rob Jeffries - Unfortunately you are incorrect on the Sun. Gases do not emit continuous spectra. Solar scientists like to cling to a very old gas model, but the physical evidence is that the Sun is condensed matter (liquid metallic hydrogen) as explained here: Robitaille P.-M., Forty Lines of Evidence for Condensed Matter — The Sun on Trial: Liquid Metallic Hydrogen as a Solar Building Block, Progress in Physics, v.4, pp.90-142, 2013

@Stephen J. Crothers 2016-02-11 15:13:37

@Rob Jeffries - Gases can neither produce nor maintain transverse waves. Solar flares produce concentric transverse waves in the photosphere, as solar observatory photographs attest. Thus the photosphere is not a gas - it is condensed matter.

@Rob Jeffries 2016-02-11 17:18:15

Super, spot on. Good luck with that. Just out of interest - what makes you think Robitaille's ravings are correct and every other astrophysicist in the world is wrong? Or have you not read anything else? And you obviously still don't get it. Blackbody radiation is thermal radiation, but not all thermal radiation is blackbody radiation.

@Stephen J. Crothers 2016-02-12 16:00:41

@Rob Jeffries - no, I understood you, and I know that not all thermal emission is black. That is the very point. Not all thermal emission is black, and so Planck's equation does not describe all thermal emission, only blackbody emission. It is a scientific fact that gases do not emit continuous spectra and also a scientific fact that gases cannot produce or sustain transverse waves. Concentric transverse waves are produced in the photosphere by eruption of a solar flare. This has been observed and photographed. I didn't make these facts; Nature did. Why are you so hostile?

@Rob Jeffries 2015-09-18 15:06:40

That information is not contained within the bb radiation - all that can be gleaned is an emitting area and a temperature.

In practice the radiation can have arisen from any process where it is feasible for a photon at that frequency to be produced.

Of course to actually be a blackbody emitter there must also be a 100% chance that a photon at that frequency incident on the object is absorbed. This condition ensures that there are relevant radiative processes that are capable of emitting at that frequency too, since there are straightforward proportionalities (for instance) between the Einstein coefficients for absorption and both stimulated and spontaneous emission (the same is true of continuum processes too).

To perhaps over-elaborate, if you postulated a hypothetical object that is incapable of emitting light at some frequencies (e.g. a two-level atom with an Einstein spontaneous emission A coefficient approximating a delta function in frequency), you might never be able to make it thick enough to absorb at those frequencies and it couldn't be a blackbody. However, even for such a system there is a tiny chance of absorption at all frequencies, due to natural or doppler broadening. If you did make the material optically thick at all frequencies (ie physically very, very thick) then its output would still approximate a blackbody.

Therefore, if you wanted to answer probabilistically, then I would say that the most likely relevant emission process will be the inverse of whatever absorption process makes the blackbody object optically thick at that frequency.

So for example, the visible (almost) blackbody radiation from the Sun's photosphere you obviously have all the optical atomic and ionic (a few molecular) transitions, but also free-free and free-bound emission corresponding to the opacity contributed by ions (mainly H$^{-}$, the dominant opacity source in the photosphere). For different temperatures and different materials with different compositions, the dominant radiative processes will also be different - e.g. recombination radiation with atoms/ions at temperatures above $10^{4}$ K, molecular transitions at temperatures of hundreds of K.

@Marty Green 2013-04-05 14:10:28

In quantum mechanics you have a charge distribution, and if you track that charge distribution over time, then it would classically result in radiation. The question is: does the radiation calculated in this "semi-classical" way, calculating the quantum mechanical charge density and then applying Mawell's Equations...does this give you the correct radiation?

I do the comparison for the simplest possible case in this pair of blog articles on the s-p transition in Hydrogen, first doing the Copenhagen calculation with spontaneous emission and then doing it semi-classically by treating the hydrogen atoms as tiny antennas. I get the same answer both times.

@DilithiumMatrix 2013-04-05 14:18:46

Responses should be restricted to those that actually answer the question.

@Helder Velez 2013-04-03 12:03:49

from There Are No Pea-Shooters for Photons (pdf) By Marty Green

3. THE BLACK-BODY SPECTRUM. The ultraviolet catastrophe inherent in the Rayleigh-Jeans formula is an inevitable consequence of the equipartition theorem in classical mechanics. It is interesting, however, to think through the actual mechanism in detail. Why exactly to all frequencies of the radiation field get the same share of energy? The equipartition theorem is especially easy to understand for the case rigid diatomic molecules, where the energy is shared equally between the five modes: three translational and two rotational. If the average translation velocity of a molecule is 500 m/sec, then the average tangential velocity of a spinning molecule, taken about its center of mass, is also 500 m/sec. That is how equipartition works for mechanical energy. The question then becomes: how is this mechanical energy converted to radiant electromagnetic energy? *The simplest way is to allow the molecules to have a dipole moment*. Species like O2 and N2 will of course be electrically balanced (that’s why light passes through them so easily) but pretty much any molecule composed of two different atoms will have some dipole moment. When it it given rotational motion, it becomes an antenna. And as an antenna, it radiates. What is the frequency of the radiation? It is simply the rotational frequency of the molecule: in other words, the tangential velocity divided by the radius. The problem occurs if we let the radius become very small. The smaller the interatomic distance, the higher the frequency radiated by the spinning molecule. In theory there is no limit to how small the molecule might be, and how high the resulting frequency. There is however a well-known example to show that molecules do not in fact spin with arbitrarily high-speed. I am referring to the anomalous specific heat of hydrogen (and other light molecules) at very low temperatures. It is sometimes said that the rotational motions are “frozen out”. The interesting thing is that we can identify a mechanism which causes this: it derives from the de Broglie notion of matter waves. In order for the rotational motion to be driven independently of the translational motion, we rely on a clean hit between two molecules. This only works if the molecules are made of hard little billiard balls. What happens if the molecules are moving so slowly that their de Broglie wavelength becomes comparable to the interatomic spacing? When the incoming atoms are that big, you don’t get a clean strike which sets the target molecule spinning. You can’t help but strike both atoms at once, which imparts translational energy only. You can no longer drive the rotations, and that’s why the specific heat goes down. The law of specific heats breaks down at low temperatures because the equipartition theorem does not take into account the wave nature of matter. Without the equipartition theorem, there is no black body catastrophe.

@anna v 2013-04-03 05:44:28

Lets try this:

enter image description here

It is a plot that shows the peak temperature ( one could also find the average temperature) versus wavelength.

As others have pointed out a number of processes exist in a solid body, all of them of electromagnetic nature which will contribute to the wavelength plot.

Here is a table with frequencies:

enter image description here

Combining the information of the two figures, one can guess at the dominant processes involved in a black body radiating.

In the red curve, which is room temperatures, one sees as dominant electron volt transitions. These are the collective continua spectra coming from the vibrating molecules in the solid each molecule in the Van Der Waals field of all the rest. Since, as others have noted, molecules have electric dipoles, magnetic moments, there will be transitions in the temporary quantum mechanical solutions for each molecule, but the effect will be a continuum because the spectrum is composed of an incoherent addition of order 10^23 molecules. Even when spectral lines are excited in the molecules and the relaxation releases a photon, this photon can interact in a continuum with Compton etc scattering which will destroy most coherence and spectral lines, due to the huge number of molecules involved. As the temperatures go higher the process continues to be incoherent, just the energies involved larger.

Because of the large number of interactions entering the black body radiation phenomenon, statistical methods have to be used as Lubos has answered.

@Nathaniel 2013-04-02 15:43:01

I'm not sure if it will completely answer your question, but you might well be interested in this paper (Smerlak, 2011 Eur. J. Phys. 32 1143. "A blackbody is not a blackbox."; arXiv version in case that link ever dies). It looks into black body radiation from a slightly different perspective than usual. Some of the best approximations to black bodies in nature are large volumes of gas, such as stars and planetary atmospheres. This pedagogical paper derives the black body spectrum by thinking about this more natural scenario, rather than the usual more artificial concept of a cavity with a small aperture.

What it all boils down to is the matter part of the system (the whole thing, not just a single atom of it) transitioning between different energy levels. For this to happen there has to be an interaction with the electromagnetic field. If the matter part of the system has a continuous spectrum of energy levels, and the matter and the radiation are in equilibrium, the result is that the radiation field has a Planck spectrum.

I get the feeling you're looking for something more specific than that - you want to know exactly why it is that a particular system of matter has a continuous spectrum of energy levels, and exactly what form its interactions with the radiation field take. I don't know the answer to that (I'd like to) but I thought this perspective might be useful nevertheless.

@Douglas B. Staple 2013-04-02 17:00:20

Excellent! I just read the paper you linked to, which was hugely helpful. You're right that I'm asking for something else, but this helps to understand the general problem. Thank you very much.

@anna v 2013-04-02 17:50:00

@DouglasB.Staple you could give him the bounty :) Nathaniel your last paragraph makes a mystery out of nothing. Each atom/molecule is a dipole or multipole and in a sense is a small antenna because it moves in the collective electric field of the other dipoles. The energies are continuous as there are a large number of different geometries creating different potentials. It loses kinetic energy by radiating and lowers the average kinetic energy lowering the temperature, no? black bodies slowly cool.

@Nathaniel 2013-04-03 02:58:22

@annav yeah, that's pretty much the level of explanation I'd be happy with. But as the other answers have pointed out, the precise nature of the interaction can take many forms - electric dipole radiation, magnetic dipole radiation, spectral lines, etc. In one sense these are all the same thing, but I think what Douglas wants to know is the details of which of these processes happens most in practice. (The answer will depend on the black body of course, with the mechanism for a plasma being quite different from that of a gas, and a solid cavity being different again.)

@Luboš Motl 2013-04-03 12:40:25

Sorry, @Nathaniel, but the paper you linked to is complete garbage. It tries to deny Kirchhoff's laws and allows "black bodies" not to absorb all the radiation. But black bodies are defined as (idealized) bodies that absorb all the radiation (that's what the adjective "black" really means: they don't reflect light: emission is irrelevant) and the absorption and emission are linked for each body and each frequency. All these errors by the author have many implications, e.g. in the preposterous claim that rarefied gases are black bodies. They're surely not.

@Nathaniel 2013-04-03 13:25:37

@LubošMotl I think you misunderstood it somehow. The main argument is just that if some matter and some radiation are in equilibrium at a given temperature, then the radiation must always have the same spectrum (the Planck one), regardless of the nature of the matter. This is analogous to the way a system's equilibrium distribution is independent of the nature of the heat bath it's in equilibrium with, and it follows from elementary principles of statistical mechanics. It's also, fundamentally, the basis of Planck's argument.

@Nathaniel 2013-04-03 13:26:13

The paper doesn't say rerefied gases are black bodies, it says that in large enough quantities they become good approximations to black bodies. Empirically this is absolutely true. It's why stars emit good approximations to black body spectra.

@Nathaniel 2013-04-07 03:37:05

@LubošMotl to elucidate a bit more, in Planck's argument the walls of the cavity will not necessarily have perfect absorbtivity - the point of the cavity is to allow the radiation to come into equilibrium with the matter in spite of this. Large enough quantities of gas have much the same effect, because scattering keeps the radiation in contact with the gas for long enough for it to approach equilibrium, even though in small quantities the absorptivity is far from 1.

@Rob Jeffries 2015-09-18 15:11:35

@Nathaniel it just boils down to whether they are optically thick or not. So long as photons at all frequencies are absorbed by the object/material then that object/material will behave like a blackbody if it is in thermal equilibrium.

@Luboš Motl 2013-03-27 15:33:30

It's exactly the point of thermodynamics – and statistical physics – that one doesn't have to know the microscopic origin of similar processes if he is only interested in thermodynamic and/or statistical properties.

The black body radiation arises from all conceivable interactions between the electromagnetic field and the "black body" – from the electric dipole radiation, magnetic dipole radiation, and so on, and so on. But the virtue of thermodynamics and/or statistical physics is that even though this situation may look messy, the statistical/thermal properties of the resulting radiation may be exactly predicted as long as we know the temperature of the black body.

So ultimately all the emission boils down to the interaction terms in electromagnetism, $$ S = \int d^4x j^\mu A_\mu $$ but statistical physics or thermodynamics don't have to study any particular collection of many such interactions one-by-one because, as one may show using the thermodynamic or statistical methods, the resulting thermal properties and statistical distributions for the photons are completely universal.

When there are phonons at a nonzero temperature, they're also distributed in a black-body-like distribution similar to photons' and they interact with everyone else using all the allowed interactinos. But one doesn't have to assume any phonons in order to get the right distribution of photons. The photons will have a black body spectrum even in the vicinity of materials that contain almost no phonons. Whatever the degrees of freedom are, the photons near the heated source will behave as the black body radiation. The only necessary condition is the existence of some interactions that are able to transfer energy from the black body to the electromagnetic field. When the black body has a temperature, everything else follows and the electromagnetic field will ultimately reach the equilibrium with the black body i.e. it will contain the right black body radiation.

You should view the emission of black body radiation as an analogous process to the normal heat exchange between two bodies. At some temperature, they vibrate in various ways. Each of them may vibrate using different types of vibrations and rotations, one of them may be gas with freely moving molecules, the other one may be a solid with lots of harmonic oscillators. But when there's a sufficient interaction between these two bodies, the energy gets transferred from one to the other, the thermal equilibrium is reached, and the other body will exhibit the features we expect from a particular temperature of the body of this kind, regardless of the type of the other body it has interacted with and regardless of the microscopic interactions that were used in the heat transfer.

@Douglas B. Staple 2013-03-27 15:54:35

What you wrote explains why and how the underlying physical mechanisms are swept under the rug in the statistical physics and quantum-mechanical derivation of the blackbody spectrum. This is interesting, however it doesn't answer my question. I want to know if electronic transitions are the dominant interaction, and, if not, under which conditions given processes are dominant. For example, gamma rays can directly excite the nuclear degrees of freedom: this is the idea behind monoenergetic gamma rays for isotope identification. +1 for identifying the source of the problem.

@MarkWayne 2013-03-28 02:58:41

Douglas - Motl did answer the question, which was, "Exactly which fundamental physical process mediates the energy transfer in blackbody emission?" The answer is: "The black body radiation arises from all conceivable interactions between the electromagnetic field and the "black body" – from the electric dipole radiation, magnetic dipole radiation, and so on, and so on." Now you've amended the question to, "I want to know if electronic transitions are the dominant interaction," which is different. And I suspect that the answer depends on the material.

@Luboš Motl 2013-03-28 06:54:50

Dear @DouglasB.Staple, "sweeping under the rug" is a loaded expression because it carries a negative emotional connotation. In statistical physics and thermodynamics, this "sweeping under the rug" is a huge virtue. It's a defining property of these two disciplines in physics - the positive way to describe them is that they're able to determine certain macroscopic properties of objects and processes without the need to study the microscopic details. It's a great possibitility which helps physicists and engineers, too.

@Luboš Motl 2013-03-28 06:58:17

Otherwise "electronic transitions are the dominant interaction" isn't really a meaningful sentence because an "electronic transition" isn't an "interaction", it's a process. The interaction behind the process is e.g. the interaction between the electromagnetic field and an electric dipole moment of the atom - which may be reduced to $\phi \rho$ interaction terms. Yes, these interactions and transitions are key for the black body radiation at temperatures characteristic for atomic transitions - hundreds or thousands of K. At millions of kelvins, transitions in the nuclei would become the key.

@boyfarrell 2013-03-29 16:24:16

I think about it something like this. The material has a temperature, which means that the atoms are vibrating (phonons). The phonons are coupled with the electron system and so the phonons and electron reach an equilibrium. Electrons are charged particles so they can loose (and gain) energy by optical transitions which is results in black body radiation. So I think that electron excitation and relaxation are the source of black body radiation. But how they couple will the lattice is also important. A nice question.

@Douglas B. Staple 2013-04-01 18:42:11

@MarkWayne Saying "all conceivable interactions" avoids the question. Of course if you include all possible interactions, then somewhere in there you will get the process(es) responsible. Usually, when making a statement that X causes Y, it is implicit that X is the dominant process, although there may be others. When water falls from the sky we call it rain and say it's caused by water vapour condensing in the atmosphere; typically one doesn't mention, e.g., that some of the water might come from meteors breaking up in the atmosphere.

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