# Can negative absolute temperature be achieved?

Thermodynamics does not allow the attainment of the absolute zero of temperature. Is then the term "negative temperature" a misnomer?

## 2 Answers

Your question seems to be predicated on the idea that negative absolute temperature is somehow supposed to be "colder" than absolute zero, and you are right that that would be non-sensical.

But actually, in a very precise sense, negative absolute temperatures are hotter than all positive temperatures, see also this question. This is simply a result of the statistical definition of temperature $T$, which is $$\frac{1}{T} = \frac{\partial S}{\partial E},$$ where $S$ is the entropy of the system and $E$ its energy content, so $\beta := \frac{1}{T}$ is actually the more natural quantity to think about in the physical formalism. See also this excellent answer by DanielSank for how the $\frac{1}{T}$ appears naturally as a Lagrange multiplier in thermodynamics.

From the above, we see that temperature is negative for system whose entropy decreases as the energy rises. Such systems are unusual, but they are not forbidden. Absolute zero, $T\to 0$, corresponds to $\beta \to \infty$. As the system gets hotter, $\beta$ decreases. $\beta\to 0$ looks weird in terms of temperature, since it corresponds to $T\to \infty$, but since it is $\beta$ that is of primary physical importance, it is not actually forbidden for it to cross zero and become negative. In terms of temperature, a system crossing the $\beta = 0$ point would have to be described as heating up all the way to "infinite positive temperature", then flipping the sign of the temperature and starting to go towards $T= 0$ again from $-\infty$.

• Would you mind someone suggesting an edit reminding of the difference between statistical temperature and kinetic temperature, and how that affects commonplace concepts of hot and cold? – can-ned_food May 21 '18 at 18:04

To expand on ACuriousMind's post, consider as a toy model a system with $2$ energy levels, say $0$ and $E>0$. These have respective probabilities $(1+e^{-\beta E})^{-1},\,e^{-\beta E}(1+e^{-\beta E})^{-1}$, so the mean energy is $E(1+e^{\beta E})^{-1}$. Clearly, this is $0$ at $\beta=+\infty$ ($T=0^+$), $E/2$ at $\beta=0$ ($T=\infty$) and $E$ at $\beta=-\infty$ ($T=0^-$), so a large negative $T$ is especially hot as measured by mean energy.

It turns out $2$-level systems cannot reach $\beta < 0$ while in thermal equilibrium. Lasers work by maintaining negative $T$ in systems with at least $3$ levels (well, most do). Thus the highest energy level is the most occupied. This is an example of population inversion.

• I like your answer except your argument is based on Boltzmann distribution which is valid only when the system is in equilibrium. – mithusengupta123 May 22 '18 at 17:20
• @mithusengupta123 If you have a preferred definition of temperature that works in non-equilibrium dynamics, you can work out which microstates have negative temperatures. As lasers prove, negative temperatures are certainly possible at least in equilibrium. – J.G. May 22 '18 at 17:25
• Could it be that you did an odd number of sign mistakes here? For example, in the probabilities, it should be $(1+e^{-\beta E})^{-1}$ and $e^{-\beta E} (1+e^{-\beta E})^{-1}$, respectively. Only then your statements make sense. – Merlin1896 May 22 '18 at 20:34

## protected by Qmechanic♦May 21 '18 at 19:03

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