# Is it Possible for a System to have Negative Absolute Temperature?

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?

• 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.
– rob
Oct 22, 2021 at 16:18
• @Rob are comments not useful for incomplete answers? I can answer "can this occur", but not "what does it mean".
– g s
Oct 22, 2021 at 16:20
• physics.stackexchange.com/questions/48615/… Oct 22, 2021 at 16:21
– rob
Oct 22, 2021 at 16:22
• This has been addressed many times on this site. Just search for negative temperatures... Oct 22, 2021 at 18:24

• 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 $d\Omega /dE <0$