What is the origin of blackbody radiation?

Question

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?

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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.

Answer 1

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 the inverse of any process where it is feasible for a photon at that frequency to be absorbed.

To actually be a blackbody emitter there must be a 100% chance that a photon (at any frequency, strictly speaking) 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), and that the object reaches detailed balance at the microscopic level if it is in thermodynamic equilibrium.

To perhaps over-elaborate, if you postulated a hypothetical object that is incapable of emitting or absorbing light at some frequencies (e.g. a two-level atom with an Einstein spontaneous emission A coefficient approximating a delta function in frequency), you would never be able to make it physically thick enough to absorb or emit at those frequencies and it couldn't be a blackbody. However, in any real system there is always 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 could still approximate a blackbody.

Therefore, if you wanted to answer probabilistically then the most likely relevant emission process(es) are the inverse of whatever absorption process(es) make the blackbody object opaque to radiation 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 (and a few molecular) transitions, but they are only important at discrete frequencies. Continuum opacity is provided by free-free and bound-free absorption due to metal atoms, ions and free electrons (contributed by potassium and sodium mainly) but actually the dominant opacity source in the Sun's photosphere is photo-detachment of electrons from H$^{-}$ ions and hence the dominant emission process across the visible continuum is free-bound emission when H$^{-}$ ions form.