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

Question

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?

Answer 1

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.