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How does electromagnetic radiation get absorbed by an object (like a black body) in the classical regime?

In the classical picture, electromagnetic radiation is produced by the movement of charges, which produce a change in the electromagnetic field in space, which propagates as a wave. But, this change in the electromagnetic field can't be absorbed. If an electron wiggles, it's disturbance will permeate all space.

The only way radiation seems to be getting absorbed is by the following process:

The charges in the body absorbing the radiation start wiggling when they feel the radiation, and their motion in turn produces a second radiation. The original and the newly produced radiation interfere by superposition, and the resulting wave is weaker, i.e., attenuation.

So far so good, but then how do distinguish between the absorbed and emitted radiation from a blackbody?

P.S. This is the classical regime, and hence there are no photons.

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    $\begingroup$ Black body radiation can not be explained with a classical regime. It's a fundamental macroscopic effect of the quantum world. Can you force a round peg into a square hole? Sure, but why do that if we have the right pegs for the right holes, already? $\endgroup$ – CuriousOne Jul 6 '15 at 17:53
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In the classical picture, an incoming wave excites a (damped) oscillation in an atom. We imagine that the oscillation is of bound electron(s). The oscillating electron(s) then re-radiates electromagnetic waves, but importantly, these waves are emitted in a continuum of directions, following the spatial distribution emitted by an oscillating electric dipole.

What this means is that if you have a beam of radiation incident upon a set of atoms from some particular direction, then power is taken out of that beam and scattered into different directions. If considering the specific intensity of the original radiation and the radiation beyond (on the other side of) the set of atoms, this has the net appearance/effect of absorption.

I am not sure what you are asking w.r.t. distinguishing incident (absorbed) radiation and emitted blackbody radiation. By definition blackbody radiation is isotropic and has a spectrum that depends only on its temperature; this could be completely different from the spectrum of any incident radiation, which may also not be isotropic.

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  • $\begingroup$ So basically, the original radiation wave (as seen on the other side of the set of atoms) + the re-radiated wave = emitted radiation? $\endgroup$ – Sidd Jul 7 '15 at 9:52
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In a classical way excited particles emit the received energy in discrete quanta, later named photons. As a good description of photon emission one can take the process of stimulated emission.

The imagination, that the emission of energy from an excited electron is distributed all other the space is wrong. The best way to predict processes inside the nucleus sees to be a permanent existing EM field and disturbances in it. But for processes between electrons in two macroscopic bodies it is better to work with photons (stimulated emission/laser, black body radiation, photoelectric effect, CCD device of a camera, double slit experiment with single photons).

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    $\begingroup$ The question explicitly asks for the (outdated, of course) classical description, i.e. no photons. $\endgroup$ – ACuriousMind Jul 6 '15 at 15:25

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