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I have read the Wikipedia article on how a cavity can represent a black body radiator but it has left me with 3 main questions.

  1. It says that the radiation inside the cavity is in thermal equilibrium with the walls of the cavity implying that the photons have a temperature but I don't understand how this can be given that they have no mass.

  2. It states that it is the actual hole in the box and not the cavity that is the black body and that a black body is a perfect absorber and emitter. Surely then if the only radiation entering the cavity comes from a single beam of light passing through the hole then all of this radiation must come out of the hole an not from the walls of the box which is very unlikely.

  3. It states that at a constant temperature this spectra will be produced but what decides the temperature, if it is an external source would this not alter the equilibrium inside the cavity. If the fixed temperature is a result of the source of radiation entering through the hole then isn't the equilibrium being formed between the source and the cavity walls rather than the centre if the cavity and the walls.

This probably doesn't make a lot of sense but I would appreciate any answers :)

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  • $\begingroup$ Related: physics.stackexchange.com/questions/170579/…. As far as a photon having a temperature is concerned, you could equally well treat it as a wave, with an energy related to amplitude. $\endgroup$ – user179430 Dec 24 '17 at 0:51
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  1. It says that the radiation inside the cavity is in thermal equilibrium with the walls of the cavity implying that the photons have a temperature but I don't understand how this can be given that they have no mass.

Photons, while massless, nonetheless carry energy. The energy is transferred to matter when the photons interact with matter. You feel this energy transfer when sunlight warms your body. A photon's energy is proportional to its wavelength. Because photons do carry energy and because that energy varies with wavelength, a collection of a large number of captive photons with different wavelengths have every characteristic of "temperature".

  1. It states that it is the actual hole in the box and not the cavity that is the black body and that a black body is a perfect absorber and emitter. Surely then if the only radiation entering the cavity comes from a single beam of light passing through the hole then all of this radiation must come out of the hole an not from the walls of the box which is very unlikely.

For one thing, it's a wikipedia article. Do not take wikipedia articles as being well-written or being correct. Take them with a grain of salt instead. Moreover, the portion of the article on cavity radiation is only three paragraphs long.

That said, the article does not say the hole is the black body. Both the cavity and the small hole are essential parts of what makes the cavity radiation be very close to ideal black body radiation. Take either one away and you don't have those close to ideal conditions.

Finally, an ideal black body (a perfect absorber and a perfect emitter) is just that -- an ideal. In other words, there is no such thing as an ideal black body. Cavity radiation however comes very close to this ideal.

  1. It states that at a constant temperature this spectra will be produced but what decides the temperature, if it is an external source would this not alter the equilibrium inside the cavity. If the fixed temperature is a result of the source of radiation entering through the hole then isn't the equilibrium being formed between the source and the cavity walls rather than the centre if the cavity and the walls.

Some external heat source keeps the cavity walls at a constant temperature.

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  • $\begingroup$ Thank you very much this answer was extremely helpful, only thing is still don't get is if the walls are being heated by an external heat source, wont some of this heat be transferred through the wall of the cavity and be radiated into the space inside, forming waves that do not come from the radiation entering through the hole. $\endgroup$ – cal Dec 26 '17 at 19:56

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