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

Question

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?

Answer 1

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

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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.

Answer 2

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.

Answer 3

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.

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.