# Where is quantization used in deriving Planck's law?

There are several explanations for how Planck used quantization to explain blackbody radiation correctly without the ultraviolet catastrophe. I will follow this explanation.

For a cavity, the mode density per volume is $$8\pi\nu^2/c^3$$. Classically, the probability of occupation is equal for all modes and each mode has an energy $$kT$$. Thus, one obtains the Rayleigh-Jeans formula for blackbody radiation

$$I = \frac{8\pi\nu^2}{c^3}kT \, .$$

This blows up for large $$\nu$$. Planck's law instead says that the probability of occupation of a mode is $$\frac{1}{e^{h\nu/kT} - 1}$$ and the energy associated with that mode is $$h\nu$$. Therefore, we have $$I = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT} - 1} \, .$$ This equation behaves correctly for large $$\nu$$. Where exactly did one need quantization?

For instance, I understand that the Boltzmann distribution (which works for continuous distributions) has probability of occupation that goes as $$p(E) = e^{-E/kT}$$. Let energy be proportional to frequency (but not quantized) and this also averts the ultraviolet catastrophe and qualitatively produces the same shape as Planck's law.

EDIT

Nope, the above statement is wrong. I realized that if you actually do the integration correctly, you get the law below which still blows up for large $$\mu$$ $$I = \frac{8\pi\nu^2}{c^3}kT \, .$$ END OF EDIT

So where does the requirement of quantization exactly come in?

This question is closely related to this one and this one but is not a duplicate since those questions were asked at a more basic level.

From the classical Boltzmann theory you have the probability that the mode has energy $$E$$ $$p(E) = A e^{-E/kT} \tag{1}$$ (with an $$A$$ so that the total probability is $$\int_0^\infty p(E)\mathrm{d}E = 1$$).
From this distribution you get the average energy $$\bar{E} = \int_0^\infty p(E) E \mathrm{d}E = kT \tag{2}$$ This leads to the Raleigh-Jeans spectrum, but unfortunately it does not reproduce the experimentally measured black-body radiation.

Planck found that instead of equation (2) the average energy of a mode needs to be $$\bar{E} = \frac{h\nu}{e^{h\nu/kT}-1} \tag{3}$$ in order to correctly reproduce the experimentally measured black-body radiation. By curve-fitting he also found the numerical value of $$h = 6.6 \cdot 10^{-34} \text{Js}$$.

There was no physical explanation for equation (3) available until this time. But Planck saw (you may call it mathematical intuition) that $$p_n = A e^{-nh\nu/kT}, \text{with } n = 0, 1, 2, \dots \infty \tag{4}$$ (with an $$A$$ so that the total probability is $$\sum_{n=0}^\infty p_n = 1$$) would give the correct average energy (3). $$\bar{E} = \sum_{n=0}^\infty p_n nh\nu = \frac{h\nu}{e^{h\nu/kT}-1} \tag{5}$$ You can find the details of proving equation (5) from equation (4) for example in The Derivation of the Planck formula (pages 9-10).

Now, in equations (4) and (5) $$n$$ and $$E_n = nh\nu$$ obviously have discrete values, in contrast to $$E$$ in equations (1) and (2) having continuous values. The physical interpretation of this is that the mode doesn't have arbitrary energies $$E$$, but only discrete energies $$E_n = nh\nu$$. Or saying it more vividly: Light of frequency $$\nu$$ consists of particles, each having energy $$h\nu$$.

• Thank you. So to clarify, the equation $I = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}}$, which is obtained without quantization and only the use of Maxwell Boltzmann statistics also avoids the ultraviolet catastrophe and gives qualitatively the right shape (bell shape). It's just that quantitative experimental agreement required the use of Bose Einstein statistics. Is this right? – user1936752 Jan 13 '19 at 15:02
• @user1936752 I think you forgot a $-1$ in your comment But yes, $I = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}-1}$ was found empirically fitting the experimental data very well. And then a theory needed to be found to explain this. – Thomas Fritsch Jan 13 '19 at 15:09
• Sorry, maybe I have a misunderstanding but I meant it without the -1. If we only have Boltzmann statistics, $p(\nu) = 1/e^{h\nu/kT}$ (Note that we just equate energy with frequency and call the proportionality constant $h$. There is no quanitzation). Then you get a possible radiation law that says $I = \frac{8\pi\nu^2}{c^3}\frac{h\nu}{e^{h\nu/kT}}$. This would be a natural guess since it avoids the ultraviolet catastrophe but still keeps everything classical. My question is whether this theory is disproved only with detailed experimental data thereby requiring Planck's version with the -1? – user1936752 Jan 13 '19 at 15:15
• I think you user1936752 are conflating the Rayleigh-Jeans law (long wavelength "I") with Wien's law (short wavelength, your "comment I without -1"). Both were known to Planck and his formula/law that interpolates both short and long wavelength limits needed an explanation as described above. – hyportnex Jan 13 '19 at 15:28
• @user1936752 Now I seem I cannot follow you. From classical Maxwell-Boltzmann (without quantization) I get $I = \frac{8\pi\nu^2}{c^3}kT$ (without any $h$ entering the game). – Thomas Fritsch Jan 13 '19 at 15:29