2
$\begingroup$

Planck's law for black body radiation was derived from a conductive cube with standing EM waves, which is a key component of the derivation.

Why then should the solution hold true to any other object, namely non-conductive objects with arbitrary geometries (it evidently does, but still, why)?

$\endgroup$

3 Answers 3

3
$\begingroup$

Planck's law isn't true for all objects, since emissivity can be wavelength-dependent (e.g. allowing for more radiation at longer wavelengths in passive radiative cooling coatings), or by allowing near-field couplings, metamaterials, or surface patterning allowing coherent emission. While one might quibble Planck's law is right when one controls for the emissivity, it is clear that one can create cases where it fails in useful ways. These cases are also fairly unusual.

The basic assumption is that the radiation is incoherent and in the far-field mode, plus that the energy forms a continuum. This makes sense for macroscopic objects, but for small particles there are deviations. However, most objects have a huge number of degrees of freedom oscillating incoherently and are typically separated by many wavelengths. Meanwhile, statistical mechanics all but guarantees that the emitting/absorbing state will be close to a maximum entropy statistical distribution, which means that the radiation field will also be close to its maximum entropy distribution... and the maximal entropy distribution of a radiation field for given energy is the Planckian distribution.

Hence most of the details of the emitting object are irrelevant: unless they are particular enough to get a very specific radiation field you will end up close to blackbody radiation just because it is the distribution of photons that is most likely.

$\endgroup$
2
$\begingroup$

Why then should the solution hold true to any other object, namely non-conductive objects with arbitrary geometries (it evidently does, but still, why)?

It shouldn't, and indeed it doesn't. Real bodies like rocks or human bodies do not emit radiation with spectrum of black body. The difference is quantified in terms of frequency-dependent and temperature-dependent emissivity $\epsilon$, so that intensity of emitted radiation of the real body is its emissivity times the Planck spectrum: $$ I_{body}(\omega, T) = \epsilon_{body}(\omega, T) I_{Planck}(\omega, T). $$

Emissivity different from 1 at any frequency range makes the emission spectrum different from the Planck one. For example, metals have quite low emissivity at infrared frequencies and lower frequencies than that.

The meaning of the Planck spectrum is not that all bodies radiate it, but that it is the spectrum of radiation in state of thermodynamic equilibrium. It is a limiting hypothetical state, because our world isn't in thermodynamic equilibrium, but there are some cases close to it.

The standard hypothetical way to achieve such radiation is to put a real body in a perfectly reflecting cavity, for example cube where walls are perfect conductors, so that no radiation can escape.

The real body's emitted radiation does not have the Planck spectrum, but it is reflected by the walls back to the body where it interacts with it. If the body stays inside the cavity for a long enough time, total radiation inside the cavity will approach equilibrium radiation and its spectrum will approach the Planck spectrum. So it is the reflected radiation from the walls interacting with the body that contributes substantially to making the equilibrium radiation: without that reflected radiation, total radiation couldn't achieve thermodynamic equilibrium and the Planck spectrum.

In reality there are no perfectly reflecting cavities, but there are other ways to get close to equilibrium, for example when the cavity walls are made of a dense heat insulating material (walls of a clay oven). Although the walls aren't perfectly reflecting, if they have the same temperature as the body inside, the radiation in the cavity will approach the equilibrium radiation.

$\endgroup$
0
$\begingroup$

IMO the elaborate and interesting answer by @AndersSandberg does not address the basics of the question.

The cubic (or parallelepipedic) cavity is only a simple way to compute the mode density (or states density in QM) $\rho(\nu)$ for the EM field. When it's size $L \rightarrow \infty$ ( ie is large enough with respect to the typicall involved wavelengths $\lambda=c/\nu$), the spatial density $\rho(\nu)/L^3$ becomes independant of the size $L$ and of the shape of the box. Hence the result does not rely on the shape of the box and provide volumic mode density proportional to $\nu^2$.

The key assumption is the "blackbody" one :

(1) there is some matter (surely not conducting walls) that is at thermodynamic equilibrium and exchange energy with the EM field to enable its thermalisation (2) this exchange involves, at each frequency, an absorptivity and an emissivity and these two are equal.

With this assumption, and Planck's hypothesis $E=h\nu$, the Boktzmam distribution of the energy in matter translates in Bose distribution for the radiation, with an energy per mode $h\nu/(\exp(-h\nu/kT)-1)$

The product of the two privides Planck's law for the volumic energy density per frequency.

$\endgroup$

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.