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Of course I know what black-body radiation is, like everyone else who has taken a thermal or statistical physics course. But it was recently pointed out to me that one thing that is rarely taught (including to me) is what the mechanism for the radiation is.

At the particle level, how does black-body radiation arise (which radiative process is this)? In other words, where are the photons coming from?


To perhaps clarify a bit further, borrowing a comment from Kevin Driscoll on an answer:

The point of the question is that there are materials whose emission spectra are quite well described by a black-body spectrum. And yet we know that at the atomic level photon emission are caused by some quantum mechanical transitions. So, how is it that such an underlying quantum mechanical description gives rise to a black-body spectrum at everyday temperatures? What is the mechanics that causes light to be absorbed/emitted at all frequencies rather than at the discrete set we might expect from the electronic transitions in isolated atoms?

With a caveat from me:

That's a good re-statement, though I'd caution that the description need not a priori be quantum mechanical in nature - there is radiation in classical electrodynamics.

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marked as duplicate by John Rennie, ACuriousMind, Kyle Kanos, Qmechanic Sep 18 '15 at 14:46

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  • $\begingroup$ I enjoy this type of questions searching for the fundamentals underlying accepted phenomena. As a personal note: when comparing in my mind an atom (as a system that emits/absorbs EM quanta) with an equivalent classical system, I see that the later has no way of liberating kinetic energy w/o loosing components, like the emission of photons. Apparently matter can "cool down" and rearrange charges at these scales in a way that is not explained by QED, which accepts that charges emit EM quanta, but tries not to derive it from other more fundamental assumptions. $\endgroup$ – rmhleo Sep 17 '15 at 22:18
  • $\begingroup$ A good question that I have asked myself in the past and never satisfactorily answered. It seems to me that calculating specifically how the thermodynamic limit is approached by a real model (in this case of absorption/emission processes) is something that no one finds interesting enough to pursue. $\endgroup$ – Kevin Driscoll Sep 17 '15 at 22:51
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    $\begingroup$ I was under the impression that the blackbody radiation is not produced only by the electron transitions in atomic orbitals, but by interactions in the crystal structure which has significantly more degrees of freedom. The thermal oscillations of many atoms in the lattice can absorb and emit photons with any energy. Is that reasoning flawed? $\endgroup$ – mpv Sep 18 '15 at 9:31
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    $\begingroup$ possible duplicate of What are the various physical mechanisms for energy transfer to the photon during blackbody emission? $\endgroup$ – John Rennie Sep 18 '15 at 10:20
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    $\begingroup$ @RobJeffries Crystal was just an example for condensed matter. Plasma is still an electromagnetically interacting medium. Ions are colliding and scattering, free electrons can absorb or emit any photon via normal electromagnetic interaction. In plasma you are still not limited to a few electron transitions. $\endgroup$ – mpv Sep 18 '15 at 15:24
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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).

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  • $\begingroup$ The information need not be encoded in the radiation, but surely the photons came from somewhere. Are you saying that any radiative process at all will do it? Surely there must be one (or one class, or some but not others) process at work here. For instance, I'm pretty sure that photons from a source with a BB spectrum are not coming from stimulated emission. $\endgroup$ – Kyle Oman Sep 17 '15 at 21:14
  • $\begingroup$ @KyleOman Of course stimulated emission plays a role in bb radiation, how would you prevent stimulated emission! In a bb emitter (or anything in thermodynamic eqm. the ratio of emission to absorption (including stimulated emission as negative absorption) equals the Planck function. $\endgroup$ – Rob Jeffries Sep 17 '15 at 21:35
  • $\begingroup$ Stimulated emission to me implies line features in the spectrum, perhaps I'm wrong about this though. But more importantly in the context of my question, either the photons are being produced by some emission mechanism(s) that, because of the temperature of the BB, results in the famous spectrum, or the temperature is being imprinted into the radiation by some scattering process(es). Which emission mechanism(s) or scattering process(es)? $\endgroup$ – Kyle Oman Sep 17 '15 at 21:42
  • $\begingroup$ @KyleOman you are way off. BB radiation represents a detailed balance between all emission and absorption processes. Scattering just increases the effective optical depth and makes it more likely that an object would absorb everything incident upon it. Further edit. $\endgroup$ – Rob Jeffries Sep 17 '15 at 21:48
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    $\begingroup$ The point of the question is that there are materials whose emission spectra are quite well described by a blackbody spectrum. And yet we know that at the atomic level photon emission are caused by some quantum mechanical transitions. So, how is it that such an underlying quantum mechanical description gives rise to a blackbody spectrum at everyday temperatures? What is the mechanics that causes light to be absorbed/emitted at all frequencies rather than at the discrete set we might expect from the electronic transitions in isolated atoms? $\endgroup$ – Kevin Driscoll Sep 17 '15 at 22:58

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