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The statistical definition of temperature is the following:

$$\frac{1}{k_BT} = \frac{d\ln(\Omega)}{dE}$$

where $k_B$ is the Boltzmann constant, $T$ is the temperature, $\Omega$ is the possible number of states, and $E$ is the energy. Rearranging for T we have the following:

$$\begin{split} T & = \frac{1}{k_B} \frac{dE}{d\ln(\Omega)} \\[0.5em] T & \propto \frac{dE}{d\ln(\Omega)} \\[0.5em] \end{split}$$

The differential on the right handside: $\frac{dE}{d\ln(\Omega)}$ implies that if there was a negative change in energy with an increase in microstates, or a decrease in energy with an increase in microstates, the temperature of the system would be negative.

Can this occur, and if so what does it mean physically for the system?

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  • $\begingroup$ Comments removed; this is a friendly reminder that comments are not for short answers. A nice discussion of negative absolute temperatures is in the thermodynamics textbook by Schroeder. $\endgroup$
    – rob ♦
    Oct 22, 2021 at 16:18
  • $\begingroup$ @Rob are comments not useful for incomplete answers? I can answer "can this occur", but not "what does it mean". $\endgroup$
    – g s
    Oct 22, 2021 at 16:20
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    $\begingroup$ physics.stackexchange.com/questions/48615/… $\endgroup$
    – Señor O
    Oct 22, 2021 at 16:21
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    $\begingroup$ @gs If you'd like to partially answer the question, do so in an answer; there's no length limit on answers. Comments are for improving the question. Search Physics Meta for discussions about this policy. $\endgroup$
    – rob ♦
    Oct 22, 2021 at 16:22
  • $\begingroup$ This has been addressed many times on this site. Just search for negative temperatures... $\endgroup$
    – Yvan Velenik
    Oct 22, 2021 at 18:24

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A partial answer:

Yes, negative (Kelvin) temperature is possible and meaningful if you limit the number of high energy states. Consider the action of a laser, in which at first we add energy and increase entropy (positive temperature), but when the medium gets hot enough, we add more energy and decrease entropy (negative temperature). The negative temperature has more internal energy (it is hotter) than the positive temperature.

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  • $\begingroup$ Interesting, so once you get to a certain high energy, entropy decreases with increasing energy? $\endgroup$
    – Connor
    Oct 22, 2021 at 16:37
  • $\begingroup$ Yes, if you limit the number of high energy states. Pardon my very crude understanding, but: for the example of a laser, the more energy you pump in, the higher the proportion of atoms that are in a state associated with stimulated emission of the resonant frequency. There are fewer states with quality "stimulated emission of $\nu$" than generic high energy states, hence $d\Omega /dE <0$ $\endgroup$
    – g s
    Oct 22, 2021 at 20:35

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