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Why does Planck's law for black body radiation have that bell-like shape?

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I'm trying to understand Planck's law for the black body radiation, and it states that a black body at a certain temperature will have a maximum intensity for the emission at a certain wavelength, and the intensity will drop steeply for shorter wavelengths. Contrarily, the classic theory expected an exponential increase. I'm trying to understand the reason behind that law, and I guess it might have to do with the vibration of the atoms of the black body and the energy that they can emit in the form of photons. Could you explain in qualitative terms what's the reason? quantum-mechanics thermal-radiation wavelength share|improve this question edited Mar 21 at 14:28 asked Mar 21 at 9:10 clabacchio 1226

2   Note that a function which must be smooth, normalizable and everywhere non-negative is going to have one or more peaks ... – dmckee♦ Mar 21 at 13:29 @dmckee hehe that's clever but sadly doesn't help me in the physical side :) – clabacchio Mar 21 at 13:57 No, there isn't any physics in it. Still, the argument is very general and can be used as a handwavy motivation for lots of different systems having maxima in various parameters. It's just that you then have to understand the physics that sets any particular maximum. – dmckee♦ Mar 21 at 14:12 @dmckee true, and indeed understanding why the maximum exists is the "easy" part. The harder (to me) is what are the concurrent phenomena that create that curve. – clabacchio Mar 21 at 14:27 1   @clabacchio This can be understood better if you think of it as a resonance phenomenon. The various porcesses in the hot body, which are well known, do not occur with equal probability/rate. Given the temperature of the hot body, some processes are more probable than orhers, and there is a particular process that dominates the emission spectrum. This is what you get from the basic equation λ max T=c  , where c is a known constant. – JKL Mar 21 at 16:44

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The Planck distribution has a more general interpretation: It gives the statistical distribution of non-conserved bosons (e.g. photons, phonons, etc.). I.e., it is the Bose-Einstein distribution without a chemical potential. With this in mind, note that, in general, in thermal equilibrium without particle-number conservation, the number of particles n(E)  occupying states with energy E  is proportional to a Boltzmann factor. To be precise: n(E)=g(E)e −βE  Z    Here g(E)  is the number of states with energy E  , β=1 kT    where k  is the Boltzmann constant, and Z  is the partition function (i.e. a normalization factor). The classical result for n(E)  or equivalently n(λ)  diverges despite the exponential decrease of the Boltzmann factor because g(E)  grows unrealistically when the quantization of energy levels is not accounted for. This is the so-called ultraviolet catastrophe. When the energy of e.g. photons is assumed to be quantized so that E=hν  the degeneracy g(E)  does not outstrip the Boltzmann factor e −βE   and n(E)⟶0  as E⟶∞  , as it should. This result is of course due to Planck, hence the name of the distribution. It is straightforward to work this out explicitly for photons in a closed box or with periodic boundary conditions (e.g. see Thermal Physics by Kittel). I hope this was not too technical. To summarize, the fundamental problem in the classical theory is that the number of accessible states at high energies (short wavelengths) is unrealistically large because the energy levels of a "classical photon" are not quantized. Without this quantization, the divergence of n(E)  (equivalently, of n(λ)  ) would imply that the energy density of a box of photons is infinite at thermal equilibrium. This is of course nonsensical. share|improve this answer answered Mar 21 at 9:43 Joshua Barr 4447

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Joshua has beaten me to an answer, but I'll still post this since it's written at a simpler level. The reason you get a maximum because there are two effects that oppose each other. The number of modes per unit frequency rises as frequency squared, so as long as the energy of the modes is well below kT the energy is proportional to frequency squared. This is why the black body spectrum initially rises approximately as frequency squared. However the probability that a mode is excited falls exponentially as soon as the energy of the mode is greater than kT, so as the frequency goes to infinity the emitted radiation falls to zero. The net result of the two effects is that the emission first rises then falls again, and that's why there is a maximum in the middle. share|improve this answer edited Mar 21 at 10:07 answered Mar 21 at 9:53 John Rennie 39.5k11964

This (your response) is probably a more direct answer to his question. I.e., the increase in the degeneracy initially dominates and is then eventually overcome by the exponential suppression. This is a really generic behavior. – Joshua Barr Mar 21 at 10:05 If I can ask, what's the reason for the squared increase of the modes with less than kT energy? – clabacchio Mar 21 at 10:28 @clabacchio Put simply: In this case, there are more ways a particle can exist with a larger energy. To see this qualitatively, try considering a (quantum mechanical) particle in a 3D box. The quantitative result is different for photons than a massive particle, but the qualitative physics is the same. Note the dimensionality is crucial here. More information here. – Joshua Barr Mar 21 at 10:59