1
$\begingroup$

Ultra violet catastrophe happens when we integrate the Rayleigh Jeans spectral density over all frequencies to calculate the total energy, which yields:

$$ \int_0^{+\infty} \frac{8\pi \nu^2}{c^3} k_B T \,\text{d}\nu \rightarrow +\infty.$$

This divergence caused problems as it was not observed in experiments, and was avoided when Bose-Einstein statistics for photon modes was used, so as to yield a finite result.

Apologies if this is a very dumb question, but would it not make sense that, for infinitely many frequencies, the energy tends to infinity? So it just the case that they did not observe this experimentally, or was there a deeper theoretical motivation as to why the UV catastrophe caused problems?

$\endgroup$
2
  • 1
    $\begingroup$ Pretty sure it was just that it wasn't observed. $\endgroup$
    – Kyle Kanos
    Feb 2, 2017 at 0:10
  • 1
    $\begingroup$ "UV catastrophe" never caused any problems, its historical role is a late anecdote, see Where did Rayleigh derive the ultraviolet catastrophe? (short answer is nowhere). It was a convenient way for textbook authors to "motivate" the quantization of light after the fact, and so it spread. Lorentz showed in 1908 that Planck's quanta were incompatible with classical electrodynamics, Ehrenfest circulated the nickname "UV catastrophe" only in 1911. $\endgroup$
    – Conifold
    Feb 2, 2017 at 0:53

2 Answers 2

8
$\begingroup$

My understanding is that the conventional explanation is somewhat oversimplified. If there were some hypothetical system that obeyed the Rayleigh-Jeans formula, then it would contain infinite energy if it were in thermal equilibrium (each mode would have an energy on the order of $k T$, and this would sum up to infinity). That doesn't mean that such a system would ever actually have infinite energy, because it would need to get that energy from somewhere in the first place. The ultraviolet catastrophe instead implies that such systems would never reach thermal equilibrium. They might keep absorbing energy endlessly, for example, which isn't observed experimentally.

$\endgroup$
1
  • $\begingroup$ It occurs to me that my last answer doesn't really answer your question. From an early 20th century perspective, I would guess the main argument was that it would destroy the developing theory of thermodynamics. If the theory predicts there are real physical systems that never thermalize, then nothing in the universe connected to that object would every thermalize either. In a modern perspective, we understand that real physical theories should be renormalizable, meaning the existence of infinitely many high-energy modes shouldn't contribute infinities to physically measurable quantities. $\endgroup$
    – user34722
    Feb 4, 2017 at 1:00
1
$\begingroup$

How could any experiment even logically measure infinite energy, in agreement with the classical theory? If a blackbody at arbitrary temperature radiated infinite energy, it would incinerate everything. It would be like an atomic bomb, but ... well ... infinitely more powerful. And yet there are (very good) blackbodies in the real world, and we're still here.

See also my question Ultraviolet catastrophe in a classical world.

$\endgroup$
2
  • $\begingroup$ So what's the physical meaning as to why some frequencies have a lower spectral energy density than others? $\endgroup$
    – SuperCiocia
    Feb 5, 2017 at 0:58
  • 1
    $\begingroup$ @SuperCiocia Classically, the energy of an EM wave is determined solely by its amplitude, so all frequencies of light have equal energy and are equally populated under the Boltzmann thermal energy distribution. Quantum mechanically, the energy of a photon of frequency $\nu$ is $E = h \nu$, where $h$ is Planck's constant. So the higher the frequency, the more energetic the photon, and the less it contributes statistically under the Boltzmann distribution. $\endgroup$
    – tparker
    Feb 5, 2017 at 4:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.