The problem:

The standard derivation of the Rayleigh-Jeans law, in a cubic reflective cavity with a small blackbody in it, would have you believe that the cavity has an energy density of $kT$ for every frequency that creates a perfect standing wave.

This doesn't make any sense since this would mean that the blackbody is radiating at a finite power per unit area for each of the specific frequencies associated with each standing wave:

(In case you are unfamiliar with the derivation: http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/radpow.html#c1. This idea is actually pretty intuitive since absorptivity equals emissivity at thermal equilibrium.)

Since we can change the volume of the box this means that this is the case for every frequency...

So the blackbody is radiating a finite amount of power for every specific frequency and therefore radiates an infinite amount of energy for every frequency interval and will form a black hole as soon as it reaches a temperature of higher than absolute zero...

This is obviously nonsense.

It originates from the assumption that all the radiation in the box is contained in the perfect standing waves. I don't know why anyone would imagine that the blackbody will always be a good boy and only emit radiation that creates the perfect standing waves in the box or even know which frequencies these should be. In reality there is no mechanism that ensures that only and purely the perfect radiation persists in the cavity. Frequencies that are very close to the "perfect" ones will, as they form plane waves that bounce around the cavity to come back to their starting point, not overlap 100% perfectly with themselves but this does NOT mean that you can just wave your hand and POOF the associated energy density is zero! I do agree that frequencies that are too far from any "perfectly fitting" frequencies interfere so randomly and therefore destructively that they cannot build up and so can be ignored.

Why it is done this way:

Since the "almost perfect yet imperfect" frequencies don't form proper standing waves and are constantly changing as the EM radiation is bouncing around and interfering with itself in ever changing ways, you can wave bye bye to your breaking the field up in Fourier modes. This is a big loss because these modes were the perfect candidates for electromagnetic degrees of freedom of the system since there are countably many of them.

This forces us to make the additional assumption that the real degrees of freedom are the frequency bands around the perfect frequencies.

Perhaps this is not as straightforward given the definition of degrees of freedom as devised for a system of gas particles where things are much neater and clearer but this does not warrant all sorts of mental acrobatics and mathemagical gymnastics about why only perfect radiation should persists because it leads to big problems and is very confusing.

Please share your insights.


1 Answer 1


Yes the Rayleigh-Jeans law gives an infinite energy density. This is the ultra-violet catastrophe. The solution requires quantum mechanics and photons.

The business with "perfect standing waves" however is much simpler. The set of such waves form a complete orthonormal basis for the solutions and so any E&M wave bouncing round the cavity can be written as an (infinite) sum of such waves and so we may as well only consider the standing waves.

  • $\begingroup$ The problem I mention above makes the whole derivation of the Rayleigh-Jeans law come to a screeching halt long before you can even consider the secondary problem of the UV catastrophe. The UV catastrophe comes from an equation whose very derivation is invalid for the reasons I mentioned. The reason why I'm not talking about Planck's law is that when discussing it people tend to already assume the Rayleigh-Jeans law, the derivation of which I intend to discuss. The same argument makes the derivation of Planck's law unphysical. I will add this to the post, thank you. $\endgroup$
    – Quanta
    Commented Nov 23, 2020 at 16:18
  • $\begingroup$ About the second point you just added: imagine a spherical wave originating from the center of a cubical cavity over the time interval between it starting to radiate outwards and it first making contact with the walls. This function cannot be broken up into these "perfect" standing waves. If this wave is of a wavelength that cannot form a perfect standing wave it can never be broken up like this because it will never "settle" like any of the standing waves in your orthonormal basis since it doesn't overlap properly with itself. $\endgroup$
    – Quanta
    Commented Nov 23, 2020 at 16:24
  • $\begingroup$ Also, just arguing that the Rayleigh-Jeans derivation is correct doesn't really constitute an answer to my question since it doesn't adress the logical inconsitency with the infinite emission I adressed. $\endgroup$
    – Quanta
    Commented Nov 23, 2020 at 16:26
  • $\begingroup$ Your statementnabout the spherical wave not being able to be "broken up" is incorrect. Some quite famous people (Danel Bernoulli and D'Alembert) thought as you do, and asserted that Fourier was plainly wrong about any wave being able to be decomposed into standing waves--- but 200 years of mathematical progress and practical engineering experience have shown that Fourier was correct: Whatever wave you create in a cavity can be written as a linear superposition of standing waves. This is because the Laplace operator with cavity E&M boundary conditions is a self-adjoint opertor. $\endgroup$
    – mike stone
    Commented Nov 23, 2020 at 17:58
  • $\begingroup$ Can you send me the formula for how to write a function in terms of standing waves? $\endgroup$
    – Quanta
    Commented Nov 23, 2020 at 19:15

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