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Using the thermodynamic definition of temperature,

$$\frac{1}{T}=\frac{\partial S}{\partial U}\bigg|_{V,N}$$

negative temperatures are possible, in systems where the entropy decreases when energy is added. A few examples have been synthesized in the laboratory, such as population inversions in lasing systems and (in terms of motional degrees of freedom) by adiabatically manipulating a collection of cold atoms using a Feshbach resonance (see Braun, S.; Ronzheimer, J. P.; Schreiber, M.; Hodgman, S. S.; Rom, T.; Bloch, I.; Schneider, U. (2013). "Negative Absolute Temperature for Motional Degrees of Freedom". Science. 339 (6115): 52–55.).

The spectral radiance of blackbody radiation for a system at temperature $T$ is described by Planck's law:

$$B(\nu,T)=\frac{2h\nu^3}{c^2}\frac{1}{e{\frac{h\nu}{kT}}-1}$$

and the total luminosity, for a body of area $A$ and emissivity $\epsilon$, is described by the Stefan-Boltzmann equation:

$$L=\sigma A \epsilon T^4$$

It is clear that the spectral radiance is negative when you plug in negative temperatures, while the total luminosity is still positive. Do these equations still hold for negative-temperature systems? If so, how do we make sense of their results? If not, what do we replace them with?

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  • $\begingroup$ Best in mind the derivation of your formula for $B$ uses a geometric series obtained from a partition function $Z$ that won't converge with $T<0$ if $Z$ contains arbitrarily large frequencies. $\endgroup$
    – J.G.
    Commented Sep 23, 2018 at 21:15

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'Negative' temperature is a useful shorthand for describing certain kinds of systems where allowed energy levels are bounded from above. The notion of negative temperature is not the usual notion of temperature. Among other things, it does not describe systems in equilibrium.

Further, black-body radiation only exists for systems that are coupled to the electromagnetic field. That is, systems that can exchange heat via photons. All such systems that I can think of in real life do not have upper bounds on energy states, usually because thermal radiation has no upper bounds on energy. It may be possible to prepare a system that does couple to the EM field and has 'negative' temperature, but it would be a fairly exotic system and you'd have to calculate a special formula for its thermal radiation anyway.

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    $\begingroup$ Aren't population inversions in lasing systems necessarily coupled to the electromagnetic field? $\endgroup$
    – probably_someone
    Commented Sep 23, 2018 at 21:33
  • $\begingroup$ Edited for clarity. In those kinds of systems, the energy of the photons are not considered as part of the thermal energy of the system. $\endgroup$
    – A Nejati
    Commented Sep 23, 2018 at 21:37
  • $\begingroup$ Also, if this is not "the usual notion of temperature," then what is? As far as I knew, the thermodynamic notion was the most rigorous definition. $\endgroup$
    – probably_someone
    Commented Sep 23, 2018 at 21:39
  • $\begingroup$ Conventionally, the relation $\partial U = T \partial S$ only describes temperature for systems at equilibrium. $\endgroup$
    – A Nejati
    Commented Sep 23, 2018 at 21:41

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