Like this video shows, blackbody enclosures held at the same temperature and having the same dimensions, albeit made of different materials, show, as expected, the same color of the pinhole, despite their different overall color.

If instead we have two blackbody enclosures of the same material, held at the same temperature, but having different dimensions, will the color of the pinholes differ?

I argue that they would be because the energies (not the energy densities $\rho$ in Planck’s, respectively, Wien’s laws) will be $\rho v_{small} \ne \rho v_{large}$, where $v$ with the subscript is volume. Besides, according to wave theory, frequency $\nu$ depends on the dimensions of the cavity $\nu = \frac{c}{2L}n$, where $L$ is the length of a enclosure, $c$ is the velocity of light and $n$ is the number of modes.

Do you agree with that? Is there an experiment demonstrating the above?


3 Answers 3


As you mentioned, photons in the box have quantised momenta based on the fact that they are standing waves, with an integer number of half wavelengths.

$\lambda = \frac{n}{2L}$

$p_n = \frac{h}{\lambda} = \frac{2hL}{n}$

There are photons of all allowed momenta in the box. Which momenta are allowed is determined by the expression for $p_n$ above, where n is all possible integers. So for a macroscopic system where L is much, much larger than $\lambda$, we can think of $\lambda$ as being a function of a continuous variable where, rather than being an integer, n can take on all any value (basically, we can ignore the step-like nature of $\lambda$).

For a photon gas (a type of Bose gas with zero chemical potential), the mean energy can be found using: $\overline{E} = \sum_i E_i \overline{N}_i$

Where $\overline{N}_i$ is the average number of particles which occupy a state with energy $E_i$. I'm not sure how much you know about statistical mechanics but for a photon gas $\overline{N}_i$ is:

$\overline{N}_i = \frac{1}{e^{E/kT}-1}$

We also know that for a photon:

$E_i= pc$

And so because we have concluded that we can approximate the allowed momenta as a continuous function of n this sum of $E_i N_i$ becomes an integral, which is a lot easier to solve! The result is:

$\overline{E} = \frac{\pi ^2 (kT)^4 V}{15(\hbar c)^3}$

Wow! The mean energy of the photon gas depends on $T^4$ and the volume... the important thing here is that this is the mean energy of the photon gas as a whole. If we divide by V we get the energy density, which only depends on $T^4$. This is an important result: the energy density in a black body cavity is only dependent on the temperature. We could take that energy per unit volume, which is carried by many photons with a broad spectrum of energies, and think of it as simply the energy of a single photon of frequency f:

$f = \frac{E_{density}}{h}$

Therefore our box at temperature T would consist of a number of photons, all with energy $E_{density}$. This energy is independent of volume and depends only on temperature, but the number of these photons in the box is dependent on the volume. It is the frequency of these hypothetical photons which determines the colour of light we see emitted from the box and it is independent of the volume. So a larger box (or larger opening) would simply allow more energy per unit time to escape, so the intensity of the radiation would be greater, but the colour of the light would be unaffected.

This photon with energy $E_density$ is purely hypothetical, but it is a useful pedagogical tool. The full blackbody spectrum, as devised by Plack, is given by:

$I = 2kT\frac{f^2}{c^2} \frac{hf/kT}{e^{hf/kt}-1}$

Where I is intensity. This is independent of volume.

I hope this makes sense.


The pinholes would have the same colour if the cavities are at the same temperature.

Imagine two different cavities, joined at their respective pinholes but isolated from the rest of the universe.

These two cavities would eventually arrive at thermal equilibrium at the same temperature with no net flow of energy through the join. That is as expected because the blackbody radiation field is isotropic.

We could also do this experiment with the same cavities, but this time insert a filter between the pinholes that only allows through a narrow range of wavelengths. The equilibrium must still be reached (it would take longer). We could choose any wavelength range for the filter and get the same result. This tells us that the flux at any particular wavelength - i.e. the blackbody radiation spectrum - is universal at a given temperature. It cannot depend on the materials or dimensions of the cavities.

The energy density in the two cavities will be the same, but the integrated energy content will not be. The flux of energy emerging from a pinhole is proportional to the energy density.

Treating the Planck function as a continuous spectrum implicitly assumes that $L$ is large enough that the photon energies can be summed over using an integral rather than a discrete summation, which requires that the separation between the different energy states is small compared with the average energy. i.e. Approximately $$ \frac{hc}{2L} \ll k_B T\ ,$$ $$ L \gg \frac{hc}{2k_B T}\ .$$ Which is a bit like saying $L$ must be much larger than a typical wavelength in the system. I don't think it would be appropriate to treat the spectrum as continuous if this were not true.

Experimental lab-based blackbody cavities could be of any (reasonable) size and emit a similar spectrum at the same temperature.


Black body radiation (BBR) results from the equilibriated thermal radiation that is emitted by the charges in a material as the material is heated. It is therefore a function of only temperature and is independent of material or geometry of the body, as also mentioned in the video you linked. Hence your question is answered in the negative.

Yes the total energy in the differently sized bodies would be different. It seems you then linked that to Planck's law to claim that color of the escaping photons would be different too.

This isn't true since, the total energy isn't different because there are photons of different frequencies but because there is more volume and therefore more photons. The spectrum is in fact same.

frequency ν depends on the dimensions of the cavity $ν=\frac{c}{2Ln}$

Yes that is the frequency of the $n^{th}$ stationary mode in the direction of $L$ of an EM wave but,as you may find in the initial steps of any derivation of BBR, the density of states and thereby the spectral energy density is independent of $L$. Hence geometry doesn't matter. This aspect results solely from wave theory itself, and has little to do with BBR per se.

  • $\begingroup$ But now you've added that bit about stars. Stars are not blackbody cavities and neither are their photospheres. $\endgroup$
    – ProfRob
    Jun 13, 2021 at 12:05
  • $\begingroup$ @ProfRob spectral lines from photosphere absorption not whitstanding, isn't their temp calculated assuming BBR? $\endgroup$
    – lineage
    Jun 13, 2021 at 12:06
  • $\begingroup$ Let's not start on that. Stars are not blackbodies, since their photospheres are not in thermal equilibrium. The spectrum can be approximated by an effective temperature but in detail it is quite dissimilar to the Planck function. Your general point that one can find approximations to blackbodies with a similar spectrum but a large range of sizes is sufficient, but this would be best just done in the lab. $\endgroup$
    – ProfRob
    Jun 13, 2021 at 12:08
  • $\begingroup$ I am not an expert in astrophysics but the deviation from the BBR fit ..is it significant enough for the current context ? Anyways I take your point and will drop the example. Feel free to add an experiment you may know in place of it, if you have the time. $\endgroup$
    – lineage
    Jun 13, 2021 at 12:12

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