# What are the various physical mechanisms for energy transfer to the photon during blackbody emission?

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?

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I added QM and QFT tags because I think that might be where the answer lies. –  Douglas B. Staple Mar 27 '13 at 15:58
btw, gamma rays are on the very high energy part of the electromagnetic spectrum and their thermal emission happens in supenovae, whereas infrared are the lowest energy part of the spectrum. –  anna v Apr 2 '13 at 19:06
Sorry, this post is not to give an answer but rather to ask a naive question. Why is there this focus on electrons to emit light? Is it really not important that nucleii are charged as well? If I imagine a phonon as an excitation of a chain of charged beads in a neutralizing background, the resulting antena should radiate EM waves that have the frequency of the phonon...more or less, shouldn't it? –  gatsu Apr 5 '13 at 10:40
Yes, you are right, the lattice vibrations in a solid radiate just like an antenna. I think it's true that all of the thermal radiation properties of matter can be analyzed by finding the time-varying charge density (which is in principle calculable using quantum mechanics) and then just applying Maxwell's Equations to calculate the resulting radiation. –  Marty Green Apr 5 '13 at 12:13
@Marty I'm not sure if what you say is true or not, but if so then that's a part of the answer I'm looking for. –  Douglas B. Staple Apr 5 '13 at 13:06

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.

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Good, relevant specifications, +1. –  Luboš Motl Apr 3 '13 at 12:42
it's Bremsstrahlung, not Brehmsstrahlung - people with the surname Brehm apparently are more likely to go into biology or medicine instead of physics ;) –  Christoph Apr 4 '13 at 20:49
Oh boy, my bad! Haha, thanks @Christoph –  DilithiumMatrix Apr 4 '13 at 22:12
Bremsstrahlung - clearly the answer of an astrophysicist :) –  Chris White Apr 5 '13 at 1:37
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. –  Rob Jeffries Mar 1 at 22:28

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.

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Responses should be restricted to those that actually answer the question. –  DilithiumMatrix Apr 5 '13 at 14:18

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.

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Lets try this:

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:

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.

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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.

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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. –  Douglas B. Staple Apr 2 '13 at 17:00
@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. –  anna v Apr 2 '13 at 17:50
@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.) –  Nathaniel Apr 3 '13 at 2:58
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. –  Luboš Motl Apr 3 '13 at 12:40
@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 Apr 3 '13 at 13:25

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.

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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. –  Douglas B. Staple Mar 27 '13 at 15:54
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 Mar 28 '13 at 6:54
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. –  Luboš Motl Mar 28 '13 at 6:58
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. –  boyfarrell Mar 29 '13 at 16:24
@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. –  Douglas B. Staple Apr 1 '13 at 18:42