I understand that the thermal equilibrium between two bodies means that the two bodies attain the Same temperature. Therefore,there is no flow of a thermal energy between them. However, I don't know how i should understand the meaning of it when it comes to radiation and matter. Does it mean the photon and the matter in the question attain the same temperature? But actually before even Physicists knew about photons, How did they understand the meaning of that? What should I understand from Planck when he said in his paper and i quote "the thermal equilibrium of N identical resonators in the same stationary radiation field" What is the meaning of a stationary radiation field?
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$\begingroup$ Half your question $\endgroup$– BrianCommented Aug 10, 2022 at 11:15
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1$\begingroup$ See this related question and the discussion in Relativistic Radiation Hydrodynamics for a technical treatment of the thermodynamics and hydrodynamics of a gas of photons: such a gas needs the presence of some matter to "equilibrate", and its thermo/hydro-dynamics is possible only in some specific regimes. See also How do photons reach thermal equilibrium with the walls of the blackbody cavity?. $\endgroup$– QuilloCommented Sep 22, 2023 at 13:38
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2 Answers
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There are many questions here. I will answer the one in the title.
We imagine a system consisting of a chunk of matter surrounded with a bath of photons. The atoms forming the skin of the chunk of matter are being bombarded with photons; with each collision, energy is being shared back and forth between the vibrational modes of the atoms in the solid and the photon gas surrounding it.
After a while, this back-and-forth sharing of collision energy causes the system to settle into a state where on average the amount of energy leaving the chunk of matter and entering the photo gas is equal to the amount of energy leaving the photon gas and entering the chunk of matter.
This is the definition of thermodynamic equilibrium, and means that after a while the chunk of matter and the photons possess the same temperature.
answered Aug 10, 2022 at 5:43
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$\begingroup$ What does even temperature of a photon sense? Can you suggest a practical thermometer which can measure it? $\endgroup$– user326901Commented Aug 10, 2022 at 13:21
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$\begingroup$ @An_Elephant Temperature is a bulk property, so whether is it meaningful to talk about the temperature of a single photon is debateable. However, you can certainly look at the temperature of black body radiation: npl.co.uk/products-services/temperature/blackbody-radiation $\endgroup$– richardbCommented Aug 10, 2022 at 15:55
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A "Stationary radiation field" means that the intensity and spectrum of the radiation is not changing with time. For the matter and radiation to also be in thermal equilibrium, requires that the radiation has reached an equilibrium state with the matter that both absorbs and emits it. For every photon that is absorbed, another is emitted (at every frequency).
When this state is reached, the radiation field and the matter can be characterized by the same temperature. For the radiation field, it is the temperature of the blackbody Planck function that wholly determines its spectrum. For the matter, it is the temperature that determines the speed distribution and energy level occupancy of its constituents.
Note that it isn't merely that the energy being absorbed by the matter is the same as the energy being received by the matter. It is that this is true in any frequency window.
