In Planck distribution of energy density for black body he solved the uv catastrophe by association of descrete energy to each mode, and by the equipartition theoreme from statistical mechanics we find the mean energy of each mode. What is the meaning of temperature in this case in the equipartition theorem? What do we mean by the temperature associated to the modes of the EM field?
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Mauricio
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This might help: How exactly does applying the Equipartition Theorem to radiation leads to UV catastrophe? – Mauricio Jan 21 '22 at 13:31
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Planck didn't "solve the uv catastrophe". The problem of the Rayleigh-Jeans method, implying infinite heat capacity of equilibrium EM radiation, later named as "UV catastrophe" by Ehrenfests, was known back then in 1900-1901 but the term didn't exist yet. This problem wasn't even primary Planck's motivation for his studies of blackbody radiation. He was interested in models of equilibrium EM radiation that could explain its thermodynamic properties, including observed black body spectra. He wasn't too concerned by the problem of the Rayleigh-Jeans method. – Ján Lalinský Jan 21 '22 at 18:20
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In these derivations of spectral function of equilibrium radiation, temperature $T$ is thermodynamic temperature of any material reservoir that is in thermodynamic equilibrium with the radiation.
This temperature $T$ implies values of some characteristics of the EM radiation, like radiation intensity at a given frequency or this intensity fluctuation statistical properties, so this temperature can in theory be measured directly from the radiation itself, without touching the reservoir with a thermometer. In this sense, the radiation itself "has" temperature $T$.
Ján Lalinský
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