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I don't understand why quantization makes a peak on the blackbody radiation curve (so there is no UV catastrophe) and the relationship between that peak and quantization concept.

When the blackbody is heated, it starts to glow. All atoms start to vibrate. Total heating energy must be divided and equally shared by all atoms of blackbody. Right?

But what happens next?

Do some atoms get much more energy to radiate at higher frequency and waste the energy? What happens at the atomic level, so quantization can explain it? What does classical physics claim, so it can not explain the radiation curve?

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See physics.stackexchange.com/q/57561/28512 –  alanf Jul 10 '14 at 10:20
Maybe this question relates as well. –  NikolajK Jul 10 '14 at 10:50

3 Answers 3

This link gives a clear account of the difference between the classical and the need for the quantum mechanical formulation .

Blackbody radiation" or "cavity radiation" refers to an object or system which absorbs all radiation incident upon it and re-radiates energy which is characteristic of this radiating system only, not dependent upon the type of radiation which is incident upon it. The radiated energy can be considered to be produced by standing wave or resonant modes of the cavity which is radiating.

black body

The amount of radiation emitted in a given frequency range should be proportional to the number of modes in that range. The best of classical physics suggested that all modes had an equal chance of being produced, and that the number of modes went up proportional to the square of the frequency. spectrum

But the predicted continual increase in radiated energy with frequency (dubbed the "ultraviolet catastrophe") did not happen. Nature knew better.

Continuing perusing the link you will see the justification of the classical formula and the fact that the data does not follow it but follows the quantum induced formula.

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The blackbody spectrum has very little to do with the properties of radiation.

If there were no such thing as electromagnetic radiation, the heat in a solid body would distribute itself among all the vibrational modes of the atoms. Classically, all modes would get the same amount of energy. But in quantum mechanics, depending on the temperature, the higher-frequency modes don't always get their full share. This is the well-known low-temperature deviation from the law of Dulong and Petit.

Quantum Mechanics explains this anomaly, and it has nothing to do with the existence of electromagnetic radiation. It's just what you get when you solve the Schroedinger equation for coupled harmonic oscillators.

If you now let those vibrating atoms have an electric charge, they will radiate. You can calculate how much they radiate by using Maxwell's equations, and if you do, you will get the right answer: Plank's Law.

You don't need quantization to explain the black-body spectrum.

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Funny. The black body spectrum IS electromagnetic radiation. Those oscillators ARE charged. What you are saying is a parallel projection of similar solutions for non existent matter. Mathematics that is. –  anna v Aug 27 '14 at 3:29
Yes, it is funny what you can learn when you apply the mathematics to a hypothetical case. I'm glad you agree. –  Marty Green Aug 27 '14 at 14:54

Well, I would say that you are confused about the quantization idea. For quantization to be taken into account, physics of particles should be examined. For a simple example, particle in a box problem given in the beginning of quantum mechanics courses. So, quantization does not mean only atom, it can be examined for many situations in the real life.

Another thing I should mention that blackbody radiation curves are specific to material temperature. For a specific temperature, you have less amount of low energy radiation, then, there is a peak at some energy level where we can say radiation in this energy range is highly dense, then, decreasing to the higher energies. For example, you can easily find the concept multi-color blackbody through internet. It is used to identify optically thick (means close to be a blackbody) accretion disk, let's say around black holes. Changing temperature of accretion disk from innermost regions to outermost regions produces a superposition of different temperature blackbody radiation.

So, after some brief introduction to clarify some basic concepts, I will explain why quantization concept is included for a blackbody. The process in a black body is continous fully absorption of incident photon and emition at some other energy. (spectral density or black body curve with your saying) Since we are talking about photons in a finite closed space (sure we are inside the black body), we can define blackbody as a quantized system and photons should obey "planck distribution". (quite an easy concept you can find it through internet) And the important point, emitted photons from blackbody can not violate planck distribution that's why you have actually a peak in spectral density.

There should be also some intiution obtained from density of modes I believe. This concept favors independency of shape, volume etc. but dimension of the system is quite effective for the result.

And a final note: if there is no such distribution for a blackbody it would be really hard to keep its temperature quite long durations, it will just become a normal body.

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"Some statements are true, some statements are false, and some statements are so far off they're not even wrong". Sure applies here. –  WhatRoughBeast Jul 11 '14 at 3:14

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