9
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ To be honest, I don't know enough statisical mechanics to give you answer, but in the Wikipedia article you link to it says " Confined point vortices are a system with bounded phase space as their canonical momenta are not independent degrees of freedom from their canonical position coordinates. Bounded phase space is the essential property that allows for negative temperatures, and such temperatures can occur in both classical and quantum systems." How this applies in general, I cannot say. $\endgroup$ Oct 1, 2020 at 19:15
  • $\begingroup$ I guess the next thing to do would be to express the entropy in terms of the canonical momenta and the position and apply the constraints and see what that does to S (i.e. is it concave or convex) and to $\partial S/\partial E$. $\endgroup$ Oct 1, 2020 at 19:18
  • $\begingroup$ Ironic to find this on the "hot question" list. $\endgroup$ Oct 2, 2020 at 18:10

2 Answers 2

20
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ Thank you professor for clarifying my professor's equilibrium remarks. $\endgroup$
    – Thormund
    Oct 2, 2020 at 2:43
  • $\begingroup$ Or better, to say that no isolated negative temperature system can exist in the real world. But you could also, to be fair, say this is just a special case of that no isolated systems, period, can exist in the real world, as there will always be some coupling to an outside system. The seeming difference between the two is a result of that the systems that we intuit as "seeming" isolated, like a hot gas in a box, don't have bounded energy levels. But $\endgroup$ Oct 3, 2020 at 2:58
  • $\begingroup$ there are no perfect boxes either, so in effect all temperatures are metastable equilibria, to the long-run temperature of the entire Universe which, in at least our current theoretical edifice, is 0 K. $\endgroup$ Oct 3, 2020 at 2:58
  • $\begingroup$ "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." In relativity, if a system is confined to a particular region with a fixed volume, then there is a bound to its energy; one it reaches that bound, it will be a black hole with its event horizon reaching the edges of the region. $\endgroup$ Oct 3, 2020 at 4:09
  • $\begingroup$ @Acccumulation Yes: nicely put. I decided not to go to GR in my answer but I am happy to see it here in the comments. $\endgroup$ Oct 3, 2020 at 8:09
10
$\begingroup$

You're pretty much right; in the case of spins, it's the fact that there's an upper bound on the system's energy that causes negative temperature, which is strongly related to the fact that there's a finite number of states.

With something like a gas, increasing energy always provides access to an increasingly large set of phase space because the area of a sphere in momentum space is proportional to the square of the momentum (area of sphere is $4 \pi r^2$), and the momentum scales with square root of energy. So in that case, the number of available microstates increases unbounded with energy.

With spins in a magnetic field, the lowest energy configuration is all spins aligned with the field and the highest energy configuration is all spins anti-aligned with the field. That's the key: there is a highest energy configuration, so adding more energy doesn't get you more configurations, and in fact if you start with half spins aligned and half spins anti-aligned, adding energy reduces the number of available states and so the temperature is negative.

$\endgroup$
3
  • $\begingroup$ Thank you for reinforcing my intuition $\endgroup$
    – Thormund
    Oct 2, 2020 at 2:45
  • $\begingroup$ Area of sphere is actually $4\pi r^2$ $\endgroup$
    – 1__
    Mar 6, 2023 at 19:02
  • $\begingroup$ @Lemoine thanks. Fixed. $\endgroup$
    – DanielSank
    Mar 6, 2023 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.