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The question says it all. I believe a hypothetical perfect reflector is what's referred to as a "white body", but I might be wrong. From what I understand such a hypothetical perfect reflector should still emit thermal radiation based on its own temperature (as long as it's above 0 K, of course), just not absorb any thermal radiation from the environment. Is that not the case? The reason I ask is because I can't find a clear answer to it anywhere, and I only find texts and graphics suggesting that such a body wouldn't emit any thermal radiation at all, which does not sound right.

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  • $\begingroup$ My guess is that if it emits radiation, then it will quickly reach T=0, it should not be difficult to calculate how long this would take. $\endgroup$ Commented Mar 9 at 20:20
  • $\begingroup$ I'd hazard a guess that this is the reason why perfect reflectors don't exist. If the material admits different energy levels and transitions between them, then it can emit and absorb corresponding thermal radiation. But then it's not a perfect reflector. $\endgroup$ Commented Mar 9 at 21:54

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An idealized white body has a reflectivity $r$ of 1 and thus an emissivity $\varepsilon$ of 0 at all frequencies (the First Law, conservation of energy, states that reflection, transmission, and absorption must balance; the Second Law, entropy maximization or a prohibition from destroying entropy, states that a body at equilibrium in a thermal bath must absorb and emit equally, so absorptivity $a$ must equal emissivity $\varepsilon$).

So this white body (at temperature $T$) emits radiative flux $\varepsilon\sigma T^4$ (Stefan–Boltzmann constant $\sigma$), which is zero. In the absence of other heat transfer mechanisms (conduction, convection, latent heat, and so on), its temperature remains constant regardless of the conditions and surroundings, including impinging radiation. It shares this characteristic with the idealized transparent body.

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  • $\begingroup$ How can that be? It is said of thermal radiation: "Thermal radiation is generated when heat from the movement of charges in the material (electrons and protons in common forms of matter) is converted to electromagnetic radiation. All matter with a temperature greater than absolute zero emits thermal radiation." I don't see why this wouldn't apply to a perfectly reflective body, which would still have internal movement corresponding to its temperature. I can understand why a perfect reflector doesn't absorb thermal radiation, but not why it wouldn't emit it. $\endgroup$
    – Outis Nemo
    Commented Mar 9 at 21:11
  • $\begingroup$ It's an idealization. Real objects emit radiation, for the reason you give. (A good question would be: What is the essential nature of a low-emissivity material or surface? But this is a separate question.) $\endgroup$ Commented Mar 9 at 21:27
  • $\begingroup$ I just don't understand how it would be true even in the idealized case. Why would a perfectly reflecting body not emit thermal radiation based on its own internal temperature? As I mentioned above I can easily understand that it wouldn't absorb any thermal radiation, and thus not re-emit that, but I can't understand why it wouldn't itself emit some thermal radiation based on its own temperature, even in the idealized case. $\endgroup$
    – Outis Nemo
    Commented Mar 9 at 21:37
  • $\begingroup$ First of all, do you agree that the idealized white body has an absorptivity of 0? After all, it can’t both reflect all incoming radiation and also absorb some. $\endgroup$ Commented Mar 9 at 22:10
  • $\begingroup$ Then, do you agree that if a body absorbs no radiation, it can’t emit any? Otherwise, you could put it next to a gray body at the same temperature and it would cool off while heating the gray body, in violation of the Second Law. $\endgroup$ Commented Mar 9 at 22:13

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