Blackbodies emit a continuous spectrum of radiation, whereas a cavity with reflective walls at thermal equilibrium contains a discrete spectrum.

According to Kirchoff, "smoothing out" the spectrum of the cavity radiation by looking at the average number of frequencies permitted by the cavity in a small frequency interval from $f$ to $f+df$, which leads to the Rayleigh-Jeans law which was later refined and became Planck's law, should give the blackbody radiation spectrum.

Kirchoff's argument was that when the cavity is at thermal equilibrium at some temperature, for which another blackbody at this temperature must have been placed in it for some time before being removed, allowing a certain frequency band to escape from this cavity into another cavity with opaque, say, perfectly absorbing walls at the same temperature should not lead to a change in temperature of the walls of the cavity with opaque walls for this would violate the 2nd law of thermodynamics.

This sounds very convincing but I cannot help but contemplate the following:

If said frequency band that can pass through the filter is constrained to, say, the lowest frequency that is permitted by the cavity and we make this frequency band ever tighter around precisely this lowest permitted frequency. The amount of radiation that the walls of the opaque cavity can radiate into the cavity with reflective walls grows ever smaller since the opaque walls emit a continuous spectrum. The radiation the goes from the reflective cavity into the opaque cavity, however, will not change because our frequency band is constrained to this frequency which the cavity permits discretely. The walls in the opaque cavity would absorb more than they can radiate away and so their temperature rises and the 2nd law of thermodynamics would be violated.

Please share your insights as to why this is an incorrect analysis.

  • $\begingroup$ I think you may have a magic filter that allows only long wavelength light to go from the reflective cavity to the opaque cavity, while allowing radiation of any wavelength to go from the opaque cavity to the reflective cavity. I may have misunderstood, but is this part of your setup? Such a thing is similar in spirit to Maxwell's demon. $\endgroup$
    – Andrew
    Nov 15, 2020 at 21:16
  • $\begingroup$ No, the filter blocks the same frequencies in both directions. The point is that because the spectrum of the cavity is discrete, the frequency band that can pass through the filter being constrained more and more tightly to this specific frequency will not affect the amount of radiation going from the reflective cavity to the opaque cavity whereas the blackbody emits a truly continuous spectrum and so this shrinking of the frequency band does affect how much the opaque cavity can radiate into the reflective cavity. This would cause the opaque walls to heat up. $\endgroup$
    – Quanta
    Nov 15, 2020 at 21:25
  • $\begingroup$ If I have understood well, I think the black body and the cavity will only be able to exchange the single frequency (or limited range of frequencies) allowed by the filter. The "other" black body frequencies will be damped in the reflective cavity due to the boundary conditions. At this frequency, there must be detailed balance between (a) radiation passing from the reflective to the opaque cavity, (b) opaque->reflective radiation, (c) radiation absorbed by the black body, and (d) radiation emitted by the black body. At other frequencies, there is only balance between c and d. $\endgroup$
    – Andrew
    Nov 15, 2020 at 21:33
  • $\begingroup$ I can see that such a balance is needed to prevent the 2nd law of thermodynamics from being violated. However my analysis shows that, using the fact that the blackbody spectrum is continuous whereas the cavity spectrum is discrete, we can restrict the amount of energy going from the opaque to the reflective cavity without restricting the energy flowing from the reflective to the opaque cavity. $\endgroup$
    – Quanta
    Nov 15, 2020 at 21:40
  • $\begingroup$ Seems fishy to me. If the cavity spectrum is discrete due to reflecting boundary conditions, then (a) only certain discrete frequencies can go from reflective to opaque (because there aren't "in between" frequencies in the reflective cavity) and (b) only discrete frequencies can go from opaque to reflective, because "in between" frequencies will be damped out in the reflective cavity. I think the 2nd law is fine :) $\endgroup$
    – Andrew
    Nov 15, 2020 at 21:43

3 Answers 3


Planck’s law states that emission of light is

$$ I_\nu(\hbar\omega) = g(\hbar\omega) f(\hbar\omega, T) $$

Let’s say that $I_\nu$ is the energy emitted from the surface of the blackbody per unit area per into per unit solid angle per unit time.

If the blackbody is emitting into free empty space we know that the density of photon states is,

$$ g(\hbar\omega) = \frac{2\pi}{c^2h^3} \left(\hbar\omega\right)^2 $$

And because photons are bosons $f(\hbar\omega, T)$ is the Bose-Einstein distribution.

So two blackbody cavities radiating into free space will eventually come into equilibrium with each other by exchanging blackbody radiation over all wavelengths,

cavities in free space

If you now place a filter between the cavities which only passes energy at a single photon frequency.

cavities with filter

This does not change anything fundamental because they can still exchange energy and will eventually reach the same temperature.

It’s just like modifying the density of states with a delta function,

$$ I_\nu(\hbar\omega) = \delta(\hbar\omega_f)g(\hbar\omega) f(\hbar\omega, T) $$

  • $\begingroup$ I do not know if this is standard but naming a function “h” in a blackbody radiation context should be avoided, especially if you consider that this phenomenon is the origin of Planck constant h. $\endgroup$
    – Mauricio
    Nov 17, 2020 at 19:37
  • 1
    $\begingroup$ It's quite clear it's a function of energy $h(\hbar\omega)$ so I doubt that can be confused with a constant! Especially when $\hbar$ is used everywhere. However, to keep my only upvoter happy I have changed it to a $g$! The reason when doing calculations with both electronic and optical density of states I normally use $g(E)$ and $h(\hbar\omega)$ respectively. $\endgroup$
    – boyfarrell
    Nov 17, 2020 at 19:42

First, some experimental facts from observation. If you look thru a small hole into a cavity at temperature T, you will see the continuous Planck spectrum as given by his formula. 1) You will not see any variations due to modes of the cavity. 2) The emissivity of the cavity walls also makes no difference. The material on the walls (soot. silver, copper, wood, or cotton candy) makes no difference to the Planck spectrum seen. How can this be?

  1. The frequency density of resonant modes of the cavity were used in the derivation of the Planck spectrum. But no cavity has zero resistivity walls. For walls with some resistivity, standing waves of any frequency can meet the boundary conditions. The tails of these "non-resonant" waves extend into the walls, and quickly dissipate their energy in the walls' resistance. Lasting perhaps for only one oscillation, these waves have a very low Q compared to the resonant modes. Let's now consider the standing wave to be the sum of the reflections of a plane wave bouncing back and forth between opposite walls. For non-resonant frequencies, this photon perhaps makes 1 trip across the cavity before being absorbed. For a resonant frequency, the photon makes 1000 trips across the cavity before being absorbed. However, because emissivity=absorptivity, the thermally excited oscillators in the walls are emitting 1000 times more photons/sec for the non-resonant versus resonant frequency! If we now look at a volume in the middle of the cavity, we see the same average density of photons from 1000 photons x 1 trip or 1 photon x 1000 trips. The modes of the cavity make no difference in the photon number density at different frequencies.

  2. We could again use the bounce back and forth argument with Kirchoff's emissivity=absorptivity, but let's do his thermodynamic argument. Two cavities, made of different materials, are connected by a small hole with a filter that passes frequency $\nu \pm \Delta \nu$. The two cavities have both come to the temperature T. The power passing the filter in both directions must be the same or we could use the unbalanced flow of energy to do work. This would violate the second law of thermodynamics by getting useable work from two heat baths at the same temperature. Kirchoff concluded there was a universal spectrum, independent of cavity material, that emerged from all cavities, though it was left to Planck to deduce the actual function. You can also conclude that the cavities can't show any peaks due to modes. Otherwise, the filter could be set to pass a peak from the first cavity that the second cavity did not have. Power would again pass from a cavity at temperature T and warm a cavity of the same temperature T, violating the second law.

So, the resolution of the op's original question is that the black body spectrum in cavities is universal, and does not show modes. This is not to say cavity modes can't be excited by an antenna in the cavity broadcasting power from a sine wave generator. Also, the black body spectrum from a radiator, not in thermal equilibrium with its $4\pi$ surrounds, is modified by the emissivity of its material.


I think you argument and conclusion are correct. A cavity which radiates only at discrete frequencies won't behave as blackbody radiator and will put out more energy than it receives at those discrete frequencies.

It is the assumption that is wrong - the idea that perfectly reflecting cavity with equilibrium radiation radiates at discrete frequencies. Equilibrium radiation means all frequencies are possible, not just some discrete ones.

The idea there are only waves of discrete frequencies inside the cavity probably comes from the usual derivation on Rayleigh-Jeans or Planck's formula, where the field is expanded into Fourier series.

Fourier series has the property that in expressing the function of position $x$, only sine waves with integer multiples of a fundamental wave number $\frac{\pi}{L}$ are present, where $L$ is size of the region where we seek to express the function as Fourier series. This region is usually taken to be the whole inside of the cavity, but nothing prevents us from taking a bigger box with side length $2L$.

With twice as big integrating region dimensions, we get twice as dense wave numbers and twice as dense corresponding frequencies $\omega_{nlm} = \frac{\pi c\sqrt{n^2+l^2+m^2} }{2L}$. Instead of a fundamental frequency (lowest) at $\frac{\pi c\sqrt{3}}{L}$ we get fundamental frequency at $\frac{\pi c\sqrt{3}}{2L}$, which is lower. Behold, radiation at lower frequency appeared, just due to using a different integrating region!

It is clear that position of singularities and their strengths are an artifact of the particular finite integrating region in the Fourier series method. They are correct for the used region, but there are infinitely many other choices.

If we use Fourier integral expansion instead of the Fourier series expansion, there is no $L$ in the formulae, and no discreteness in Fourier amplitude $\tilde{E}_x(k,l,m)$ as function of continuous wavenumbers $k,l,m$. It all becomes unique continuous quantities.

So equilibrium radiation inside a perfectly reflecting cavity is not (except near the cavity walls) physically different from that of a bigger cavity or radiation of a cavity that has walls made of blackbody at the same temperature.

  • $\begingroup$ Don't you run into trouble with the boundary condition that the electric field equals zero at the boundary if you use different integration regions? $\endgroup$
    – Quanta
    Nov 19, 2020 at 22:02
  • $\begingroup$ Why would we run into trouble? Function that is zero at some locus of points is still integrable. $\endgroup$ Nov 19, 2020 at 22:53
  • $\begingroup$ Hmm I don't get it. "It is clear that position of singularities and their strengths are an artifact of the particular finite integrating region in the Fourier series method." Are you saying that different size boxes will permit different frequencies? I'd argue that's a physical phenomenon, not a mathematical artifact if that's at all what you meant. Also, that point about the integral expansion is indeed interesting. You're saying that all kinds of frequencies show up that don't show up in the series expansion? Well in that case there's were I can see an artifact at play. $\endgroup$
    – Quanta
    Nov 20, 2020 at 0:17
  • $\begingroup$ If that's true that would be worrying since then the thermodynamic degrees of freedom that appear in the series expansion case are also due to an artifact and would make the whole derivation al lot more questionable. Not sure if I'm making sense, imma play around with Fourier transforms more... $\endgroup$
    – Quanta
    Nov 20, 2020 at 0:22
  • $\begingroup$ The cavity has physical size $L$ but the region used for defining Fourier series or Fourier transform of the field inside does not have to be the same as the cavity. It can be bigger. The bigger it is, the denser the Fourier modes, for the same cavity. The discrete Fourier modes, their positions in k-space and their intensities are purely mathematical artifact of the chosen integration region. They do not affect character of the equilibrium radiation inside. Only sum of those intensities over some interval, or integral of Fourier transform over some interval, are fixed by physics. $\endgroup$ Nov 20, 2020 at 2:13

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