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Nugatory

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WannabeNewton

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Thanks a lot, I appreciate your help

What I understand is that classical equations assume that blackbodies emit radiation at at all frequencies and the radiation intensity is infinite at very high frequencies, but Planck assumed that light must be quantized to explain the actual distribution of the blackbody radiation which reaches a Max intensity at a certain frequency then it approaches zero at higher frequencies, I still have a problem understanding how Planck's assumption explained it.

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This is an Einstein's idea, used for photoefect (a simplified version only).

The black body radiation is just a sum over all possible wavelengths in the box.

There is a simple boundary condition for the standing waves, which gives the quantisation, not any energy corpuscles - the photons, are necessary, nor possible.

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Nugatory

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Informally... If light can only be emitted in quantized packets, then if there's not enough energy concentrated in one area of the object to create a packet of light at a given frequency then nothing at that frequency will be emitted. The higher the frequency of the light the more energy is needed, and the hotter the object the more energy is likely to be concentrated in one area. Combine these two results and you conclude (after some fairly huge oversimplifying of stuff discussed in that other thread) that for any temperature there are frequencies high enough that they're unlikely to be emitted; and that the hotter the object the higher the frequencies that will be emitted can be.Planck assumed that light must be quantized to explain the actual distribution of the blackbody radiation which reaches a Max intensity at a certain frequency then it approaches zero at higher frequencies, I still have a problem understanding how Planck's assumption explained it.

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Jano L.

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It should be said that quantum theory of thermal EM radiation (I believe due to Debye) leads to a divergent spectral function too:...A quantized theory of the electromagnetic field is sufficient in removing the divergence.

$$

\rho(\nu) = \frac{8\pi \nu^2}{c^3}\left(\frac{h\nu}{e^{h\nu / (k_B T)}-1} + \frac{h\nu}{2} \right)

$$

The divergent term ##h\nu/2## in the braces is the lowest energy eigenvalue of a quantum harmonic oscillator. This due to the presence of the so-called zero-point fluctuations of the field. The corresponding contribution to the spectral function diverges even more rapidly than the Rayleigh - Jeans function. It is usually removed by hand. The rest then gives the Planck function and resembles measurements very well.

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Jano L.

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Textbooks do not explain how Planck did it - as a matter of fact he derived his spectral function in a different way from the way that is most often taught today. For his approach, read his papers... I still have a problem understanding how Planck's assumption explained it.

http://www.ffn.ub.es/luisnavarro/nuevo_maletin/Planck (1900), Improvement of Wien's.pdf

http://www.ffn.ub.es/luisnavarro/nuevo_maletin/Planck (1900), Distribution Law.pdf

and for more details, his book

M. Planck, The theory of heat radiation, P. BLAKISTON'S SON & Co. 1914

https://archive.org/details/theheatradiation00planrich

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Jano L.

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The boundary condition of standing waves is not that important to the derivation of the Rayleigh-Jeans or Planck formula. One can do the same derivation for any cuboid in vacuum - the walls do not need to be reflective at all.Planck never assume any photons.

...There is a simple boundary condition for the standing waves, which gives the quantisation, not any energy corpuscles - the photons, are necessary, nor possible.

Quantization does not refer to discrete indexing of modes, but to discrete values of energy of each mode. The latter does not follow from the boundary condition of standing waves.

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The observed quantisation is always just a consequence of the boundary conditions,The boundary condition of standing waves is not that important to the derivation of the Rayleigh-Jeans or Planck formula. One can do the same derivation for any cuboid in vacuum - the walls do not need to be reflective at all.

and this applies to the atom, the Hall effect, flux quantisation in the superconductivity, tunneling effect, etc.; all the so-called quantum effects without exception.

Quantity of energy can be anything - there is no fundamental restriction,Quantization does not refer to discrete indexing of modes, but to discrete values of energy of each mode. The latter does not follow from the boundary condition of standing waves.

which shows just a black body.

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Jano L.

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Let's keep our focus on the equilibrium heat radiation. You said that the boundary condition of standing waves leads to quantization. Please explain why do you think that.The observed quantisation is always just a consequence of the boundary conditions,

and this applies to the atom, the Hall effect, flux quantisation in the superconductivity, tunneling effect, etc.; all the so-called quantum effects without exception.

In the common derivation of the equilibrium spectrum, energy of mode is supposed to have preferred values - multiples of ##h\nu##. So there is restriction in the calculation - values like ##1.3h\nu## are not allowed.Quantity of energy can be anything - there is no fundamental restriction,

which shows just a black body.

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UltrafastPED

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Do you mean Debye's model for specific heats? This is a semi-classical model (~1912). SeeIt should be said that quantum theory of thermal EM radiation (I believe due to Debye) leads to a divergent spectral function too:

$$

\rho(\nu) = \frac{8\pi \nu^2}{c^3}\left(\frac{h\nu}{e^{h\nu / (k_B T)}-1} + \frac{h\nu}{2} \right)

$$

The divergent term ##h\nu/2## in the braces is the lowest energy eigenvalue of a quantum harmonic oscillator. This due to the presence of the so-called zero-point fluctuations of the field. The corresponding contribution to the spectral function diverges even more rapidly than the Rayleigh - Jeans function. It is usually removed by hand. The rest then gives the Planck function and resembles measurements very well.

http://en.wikipedia.org/wiki/Debye_model

If you can quote the equation properly, can you please also provide your source? Then we can look for ourselves.

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Jano L.

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No, I mean straightforward application of quantum theory of harmonic oscillator to EM modes. See 19.2 inDo you mean Debye's model for specific heats?

L. Ballentine, Quantum mechanics - a modern development

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This is standard... check the calculations in QM, for example the hydrogen atom, and you should find the cause - explanation of the quantisation.You said that the boundary condition of standing waves leads to quantization. Please explain why do you think that.

with respect to an oscillator with fundamental frequency denoted as ##\nu## ...In the common derivation of the equilibrium spectrum, energy of mode is supposed to have preferred values - multiples of ##h\nu##. So there is restriction in the calculation - values like ##1.3h\nu## are not allowed.

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Jano L.

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Since you mentioned box, standing waves and boundary condition in relation to equilibrium radiation, I assumed you were talking about conditions imposed to EM waves inside a cavity: zero electric field along the walls. These boundary conditions do not imply any quantization of energy of EM modes.The black body radiation is just a sum over all possible wavelengths in the box.

There is a simple boundary condition for the standing waves, which gives the quantisation

...

This is standard... check the calculations in QM, for example the hydrogen atom, and you should find the cause - explanation of the quantisation.

But your second sentence above makes the impression that you're talking about boundary conditions imposed to Schroedinger's ##\psi## function instead. If you meant the latter, I agree that boundary conditions are an important part of the derivation of the discrete values ##(1/2)h\nu, (1+1/2)h\nu, (2+1/2)h\nu,...##

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Correct. After all there is no quantisation visible in the black body radiation - it's continuous function.These boundary conditions do not imply any quantization of energy of EM modes.

The derivation uses explicitly the condition for the standing waves.

This is just the standing waves condition.If you meant the latter, I agree that boundary conditions are an important part of the derivation of the discrete values ##(1/2)h\nu, (1+1/2)h\nu, (2+1/2)h\nu,...##

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Is it related to Boltzmann distribution?Informally... If light can only be emitted in quantized packets, then if there's not enough energy concentrated in one area of the object to create a packet of light at a given frequency then nothing at that frequency will be emitted. The higher the frequency of the light the more energy is needed, and the hotter the object the more energy is likely to be concentrated in one area. Combine these two results and you conclude (after some fairly huge oversimplifying of stuff discussed in that other thread) that for any temperature there are frequencies high enough that they're unlikely to be emitted; and that the hotter the object the higher the frequencies that will be emitted can be.

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I think you need to relearn quantum mechanics. Your understanding of quantization is lacking. The boundary conditions gives a discrete set of oscillation modes true, but that's not what quantization is. Quantization is the fact that each mode will be occupied by an integer number of particles. In classical physics each mode may have any amount of energy.This is standard... check the calculations in QM, for example the hydrogen atom, and you should find the cause - explanation of the quantisation.

with respect to an oscillator with fundamental frequency denoted as ##\nu## ...

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Jano L.

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This is probably just a misunderstanding. Apparently, parkner meant boundary conditions for the ##\psi## function of harmonic oscillator. These indeed play role in obtaining preferred values of energy ##(n+\frac{1}{2})h\nu##.I think you need to relearn quantum mechanics. Your understanding of quantization is lacking. The boundary conditions gives a discrete set of oscillation modes true, but that's not what quantization is. Quantization is the fact that each mode will be occupied by an integer number of particles. In classical physics each mode may have any amount of energy.

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Informally... If light can only be emitted in quantized packets, then if there's not enough energy concentrated in one area of the object to create a packet of light at a given frequency then nothing at that frequency will be emitted. The higher the frequency of the light the more energy is needed, and the hotter the object the more energy is likely to be concentrated in one area. Combine these two results and you conclude (after some fairly huge oversimplifying of stuff discussed in that other thread) that for any temperature there are frequencies high enough that they're unlikely to be emitted; and that the hotter the object the higher the frequencies that will be emitted can be.

Is that related to Boltzmann distribution ?

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