The ideal black-body radiation curve (unlike the quantized emission seen from atomic spectra), is continuous over all frequencies. Many objects approximate ideal blackbodies and have radiation curves very similar in shape and continuity to that of an ideal black-body (often minus some emission and absorption lines from the atoms in an object, such as radiation curves seen from stars). I am wondering what exactly gives rise to a basically continuous black-body radiation curve in real objects? Since atomic energy states are quantized, it seems real life black-body curves would have some degree of measurable quantization to them (or perhaps the degree of quantization is so small the radiation curves look continuous).
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asked Jul 18, 2013 at 18:31
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3 Answers
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perhaps the degree of quantization is so small the radiation curves look continuous
Yes, this is the reason. The correspondence principle says that quantum mechanics has to become classical in the appropriate limit. One way to obtain an appropriate limit is with large numbers of particles. As you increase the number of particles in a material many-body system, you get more and more ways of putting together combinations of states for your material object. The density of states of the object grows very quickly (roughly exponentially) with the number of particles. Therefore the number of possible transitions between states also grows very rapidly.
The number of particles in a tungsten lightbulb filament is something like Avogadro's number. The exponential of Avogadro's number is really, really big.
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2$\begingroup$ Right. and one should remind oneself that even in simple potential problems the energy levels for large n (towards infinity) are really dense. google.gr/… . The complexity of real matter makes the continuum observed inevitable. $\endgroup$– anna vCommented Jul 18, 2013 at 19:35
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$\begingroup$ So, an object with a very small number of particles would not be able to emit a continuous radiation curve like the blackbody curve? And this would be why you need atoms in a gasesous state to observe their spectra, so that the atoms are separated and each atom's energy levels are not affected by others. $\endgroup$ Commented Jul 18, 2013 at 21:13
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1$\begingroup$ @Joshua: Yes, that's correct. $\endgroup$– user4552Commented Jul 18, 2013 at 21:16
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$\begingroup$ @Joshua Except that the best blackbodies in nature are gases - stars, the CMB. However even in gases you do not expect discrete, delta-function absorption lines (see my answer below). $\endgroup$– ProfRobCommented Sep 18, 2015 at 15:29
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$\begingroup$ Btw, 1 mol of tungsten is about 200 gram, which is enough to produce a couple of thousands of lightbulbs. $\endgroup$ Commented Apr 16, 2018 at 11:02
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This is the second time in only a few days that I've cited Luboš Motl's excellent answer to What are the various physical mechanisms for energy transfer to the photon during blackbody emission?. As Luboš points out, the precise microscopic mechanisms of the radiation are unimportant because the statistical properties ensure that it follows Planck's law.
To get the characteristic black body curve you just need enough ways to generate EM radiation. Typically thermal vibrations in whatever material you're looking at result in accelerated electrons and oscillating dipoles within the material, and both generate the electromagnetic waves. This isn't a resonant process, so you don't get sharp lines but just a continuum of frequencies.
answered Jul 18, 2013 at 18:43
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If the absorptivity of a medium really was discrete, then there would be no way it could emit blackbody radiation. The defining characteristic of a blackbody is that it absorbs light of all frequencies that are incident upon it (and that it is in thermal equilibrium). There is a close relationship (a direct proportionality) between the Einstein absorption and emission coefficients for atomic, ionic and molecular processes which ensures this.
So whilst you can imagine hypothetical materials with discrete absorption spectra caused by "delta function" spectral lines, you cannot also hypothesise that these would emit blackbody radiation - they would not.
In practice the absorption coefficients of real materials are not delta functions at fixed frequencies. Electronic transitions have finite widths - there is natural broadening, doppler broadening, pressure broadening. Real materials also have continuous absorption coefficients caused by photoionisation, free-free absorption, inelastic scattering etc. These effects cause the absorption coefficient to be non-zero at practically all frequencies. In those circumstances, to get a continuum blackbody we simply need to arrange to have enough material present that it is optically thick (that is, it has an optical depth much larger than unity) at all relevant frequencies. If that is so, and the material is in thermal equilibrium (energy levels populated according to Boltzmann factors etc.) then it will emit what is close-to-blackbody radiation.
