Evidence of photon existence and Blackbody Radiation

  • #1
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How does the blackbody radiation prove the existence of photons or quanta as Planck described it, I've understood how the photoelectric effect proves the existence of photons, but the blackbody radiation seems quite vague to me. I would like a basic explanation for this, thanks in advance.
 

Answers and Replies

  • #3
WannabeNewton
Science Advisor
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A classical theory of the electromagnetic field would lead to an unphysical divergence in the power spectrum of the blackbody radiation. A quantized theory of the electromagnetic field is sufficient in removing the divergence. This does not mean it proves the existence of photon of field quantization; what it does, as with any other aspect of quantum theory or established physical theories in general, is corroborate theory with experiment.
 
  • #4
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Give this thread a read https://www.physicsforums.com/showthread.php?t=758363 and come back with any further questions.

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.
 
  • #5
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Planck never assume any photons.
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.
 
  • #6
Nugatory
Mentor
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5,698
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.
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.
 
  • #7
Jano L.
Gold Member
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...A quantized theory of the electromagnetic field is sufficient in removing the divergence.
It 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.
 
  • #8
Jano L.
Gold Member
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... I still have a problem understanding how Planck's assumption explained it.
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

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
 
  • #9
Jano L.
Gold Member
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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.
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.

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.
 
  • #10
15
0
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.
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.

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.
Quantity of energy can be anything - there is no fundamental restriction,
which shows just a black body.
 
  • #11
Jano L.
Gold Member
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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.
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.

Quantity of energy can be anything - there is no fundamental restriction,
which shows just a black body.
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.
 
  • #12
UltrafastPED
Science Advisor
Gold Member
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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:
$$
\rho(\nu) = \frac{8\pi \nu^2}{c^3}\left(\frac{h\nu}{e^{h\nu / (k_B T)}-1} + \frac{h\nu}{2} \right)
$$
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.
Do you mean Debye's model for specific heats? This is a semi-classical model (~1912). See
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.
 
  • #13
Jano L.
Gold Member
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Do you mean Debye's model for specific heats?
No, I mean straightforward application of quantum theory of harmonic oscillator to EM modes. See 19.2 in

L. Ballentine, Quantum mechanics - a modern development
 
  • #14
15
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You said that the boundary condition of standing waves leads to quantization. Please explain why do you think that.
This is standard... check the calculations in QM, for example the hydrogen atom, and you should find the cause - explanation of the quantisation.

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.
with respect to an oscillator with fundamental frequency denoted as ##\nu## ...
 
  • #15
Jano L.
Gold Member
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74
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.
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.

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,...##
 
  • #16
15
0
These boundary conditions do not imply any quantization of energy of EM modes.
Correct. After all there is no quantisation visible in the black body radiation - it's continuous function.
The derivation uses explicitly the condition for the standing waves.

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,...##
This is just the standing waves condition.
 
  • #17
439
13
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 it related to Boltzmann distribution?
 
  • #18
1,948
200
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## ...
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.
 
  • #19
Jano L.
Gold Member
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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 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##.
 
  • #20
439
13
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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