[Physics] the fundamental reason for existence of negative temperature in a given specific system

statistical mechanicstemperature

Question: Fundamentally, is the existence of negative temperatures a consequence of (a) the violation of entropy postulates, (b) inequilibrium, or (c) finite number of configurations?


Context: In my statistical mechanics class, we first began by claiming the existence of a function $S$, called entropy that contains all information of a (isolated) system (equivalently, the partition function as we move from microcanonical to canonical systems). We postulate several properties of the entropy function:

  • Entropy is concave,
  • $\frac{\partial S}{\partial E} > 0$,
  • $S$ is positively homogenous of degree 1, i.e.: Entropy is an extensive quantity, as exemplified by $S\left(\lambda E, \lambda X_1, \dots, \lambda X_m \right) = \lambda S\left(E, X_1, \dots, X_m \right),$ where $X_i$ are extensive parameters (thermodynamic quantities).

Then, if the system is in equilibrium, we can define the temperature of the system by $$\frac{1}{T} = \frac{\partial S}{\partial E},$$ where it is implicit that $X_i$ is held constant.

Now, considering the simplest model that yields negative temperatures: $N$ noninteracting two-level particles of fixed positions. It is easy to derive that the entropy $S$ as a function of energy $E$ is a parabola that decreases for $E > \frac{1}{2}\left( E_\text{max} – E_\text{min} \right)$, as seen in the graph here. My first thought was the violation of $\frac{\partial S}{\partial E} > 0$ (and hence the entropy postulate) is a consequence of the finite number of configurations, is the fundamental reason to the existence of negative temperature in this system. However, my tutor has repeatedly spoke of the violation of the entropy postulates as being the fundamental reason (is there circular logic here?), and my lecturer instead stating that negative temperatures are result of systems that are not in equilibrium.

Am I misunderstanding their points?


Remark 1: The finite number of configurations in a thermodynamic system is also mentioned in this wikipedia article here. The following sentence is succinct in describing the thought I had.

Thermodynamic systems with unbounded phase space cannot achieve negative temperatures: adding heat always increases their entropy. The possibility of a decrease in entropy as energy increases requires the system to "saturate" in entropy.

Remark 2: In the course of reading various posts on StackEx regarding negative temperatures, I had stumbled onto this, but it is somewhat beyond me, and unsure if it is relevant here.

Best Answer

Negative temperature is mainly to do with (c): a finite number of configurations. It is not a violation of entropy postulates or equilibrium, but I will qualify these statements a little in the following.

The heart of this is not to get 'thrown' by the idea of negative temperature. Just follow the ideas and see where they lead. There are two crucial ideas: first the definition of what we choose to call "temperature" $T$. It is defined by $$ \frac{1}{T} = \left( \frac{\partial S}{\partial U} \right)_{V} $$ where $U$ is internal energy and I put $V$ for the thing being held constant, but more generally it is all the various extensive parameters that appear in the fundamental relation for the system.

The next thing we need is a statement about stability. It is that in order for the system to be stable against small thermal fluctuations the entropy has to have a concave character as a function of $U$: $$ \frac{\partial^2 S}{\partial U^2} < 0 $$

One of the important points here is that we can satisfy the stability condition for either sign of the slope, and therefore for either sign of $T$. So a system having negative $T$ can satisfy the stability condition and therefore it can be in internal equilibrium. The negative temperature state is a thermal equilibrium state and that is the reason why we are allowed to use the word "temperature" to describe it.

Now we need to ask: but does it ever happen that there are equilibrium states in which the entropy goes down as the internal energy goes up? The answer can be yes when there is an upper bound to the energies that the system can reach. When this happens, as we add more and more energy to the system, we eventually squeeze it into a smaller and smaller set of possible states, so its entropy is decreasing. The classic example is a set of spins in a magnetic field.

And now I will qualify the above a little, as I said I would.

The thing is that no system really has an upper bound to its energy, because every system can have some form of kinetic energy, and this has no upper bound. When we treat spins in a magnetic field, for example, we should not forget that those spins are present on some particles, and those particles can move. The purely magnetic treatment ignores this degree of freedom, but the experimental realities do not. So in practice a spin system at negative spin temperature will begin to leak energy to its own vibrational degree of freedom (whose temperature is always positive, and you should note that the heat flow direction is from the thing at negative temperature to the thing at positive temperature, because this increases the entropy of both). This will eventually bring about the true equilibrium of both spin and vibration, and this will be a positive temperature. So your professor who said negative temperature was a non-equilibrium case was half right. The negative temperature is a metastable equilibrium, one whose lifetime gets longer as the coupling from the negative temperature aspect to other aspects of the system goes down.

This also bears on the issue of the entropy being concave. If the entropy has a region of negative slope at some energy then this negative slope will bring $S$ down as a function of $U$. But if in fact the system can access higher $U$ (via vibrational degrees of freedom, for example) then the $S(U)$ function must turn up again, not crossing zero, and this means it will have a region where it is convex ($\partial^2 S/\partial U^2 > 0$). That region will not be a stable equilibrium region. In practice a system having behaviours such as this in its entropy function will undergo a first order phase transition. It may be that something like this was in the mind of anyone who said they thought an entropy postulate was not being satisfied.

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