I previously asked whether Planck's curve actually reaches zero, to which some of you answered that it only approaches it, which raises the question: doesn't this mean that a photon with extremely high frequency will have massive energy? I know that the intensity (number of photons) of such high-frequency photons is close to zero, but even if we have only one photon with high frequency the ultraviolet catastrophe would still be unresolved. So what is it I am missing?
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4 Answers
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The Planckian approaches zero fast enough: there can be some very high energy photons in a thermal distribution, but the probability of their presence drops fast; crucially, it drops faster than the energy rises. Qualitatively, the $\sim e^{-h \nu / k_B T}$ factor in the distribution can be multiplied by any power of $\nu$ and still yield a finite integral. Therefore, the energy carried by photons that have, say, frequencies higher than $N$ times $k_B T / h$ drops to zero as $N$ grows large.
You could find a $1 \mathrm{J}$ photon in the Sun's blackbody (where the average photon energy is of the order $10^{-19} \mathrm{J}$) but it is astronomically unlikely that you will, so on average the energy contribution by such photons is very small.
I've tried to give an intuitive answer; if you want a more formal one try to integrate the Planckian yourself to find the energy carried in a specific frequency band!
answered May 11, 2021 at 8:52
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17$\begingroup$ Caveat: the calculus skills required to "integrate the Planckian yourself" are not insignificant. $\endgroup$ May 11, 2021 at 20:41
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$\begingroup$ I agree! If OP wants to go down that route perhaps they'd benefit by taking the high-energy limit initially so that the calculation is easier, since they are interested in that contribution anyway... $\endgroup$ May 11, 2021 at 22:32
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$\begingroup$ @JacopoTissino you know how the OP said intensity is "number of photons"? How? I would have asked OP myself but his bio says it was last seen a year ago whereas u were seen 1 week ago. $\endgroup$– LawrenceNov 5, 2022 at 4:06
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Planck's Law relates the spectral radiance of the body $ B ( \nu, T ) $ to the frequency of the photon by $$B ( \nu, T ) = \frac{2 h \nu^2}{c^2} \frac{1}{e^{ h \nu/ k_B T} -1} . $$
It replaced the Rayliegh-Jeans Law (first put forward by Lord Rayleigh but the derivation was subsequently improved by Sir James Jeans) which gave $$B ( \nu, T ) = \frac{2 \nu^2 k_B T}{c^2} $$ which can be written as $$ \frac{2 \nu^2}{c^2} \times k_B T$$ and interpreted as $$ \# \mbox{ available states} \times \mbox{ energy of each state}. $$
Similarly Planck's Law can be written as $$ \frac{2 \nu^2}{c^2} \times h \nu \times \frac{1}{e^{h \nu/k_B T} - 1}$$ which is now interpreted as $$ \# \mbox{ available states} \times \mbox{ energy of each state} \times \mbox{ probability of state being occupied}.$$
Rayliegh-Jeans uses equipartition of energy to assume that each available state has the same probability of being occupied (namely unity) while Planck has the factor $$\frac{1}{e^{h \nu/k_B T} - 1}$$ which is the result for boson's (that photons are an example of).
It follows that there are two main differences between Rayliegh-Jeans and Planck: (i) the energy of each allowed state changes from $k_B T$ to $h \nu$, (ii) though the density of states increases with $\nu$, the higher the frequencies are less likely to be occupied.
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You are missing the Stefan-Boltzmannn law, which states that the total energy radiated by a perfect black body is $$P = \sigma T^4\,.$$ This also means that the total energy emitted is converging.
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The Planck curve is actually about probabilities.
You can always cut a very small interval of time and pretend that photons emitted in that interval don't fit the curve (and they will indeed not fit - they are a discrete events and the curve is pretty much analtytical).
On the other hand, if you wait long enough in order to get some very high energy photon, you will (highly likely) build a great deal of statistics over the photons near the maximum, so you won't get UV catastrophe.
