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If a rectangular pan has a constant and uniform temperature $T$ first, then put it in a vacuum. Considering the effect of thermal radiation, the temperature distribution of the rectangular blackbody will change. My question is, will the temperature of the corner be higher than the edge?

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First consider the pan as composed of small similar pieces. All of these small pieces radiate to the vacuum and thus loose energy and reduce the temperature. They all do that at the same rate. Then we see that as these pieces are in contact/vicinity of each other, they will capture some of the radiation emitted by their neighbors and they would also transmit energy through conduction, so they will not cool off at the same rate.

We can see that the particles at the corners, and then those at the edges have greater exposure to the vacuum and thus dissipate energy more efficiently. Even without doing a more detail calculation, it is clear that the corners would be cooler.

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Yes, I agree with the result you get. I believe a more accurate explaination is that thermal radiation is proportional to the expectation value of photon number(not the same rate everywhere). At the corner, the electricomagnetic mode's electric field is larger(The same reason with point discharge) and thus with larger expectation value of photon number. In this sense, the radiation from the corner is larger as there are more photons couple to the electron to emitt heat. – Xiao-Qi Sun Feb 3 '13 at 15:51

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