What are Negative Absolute Temperatures

It’s a famous result in thermodynamics that you can’t reach absolute zero no matter how hard you try. Also, the definition of absolute zero makes it obvious that it’s meaningless to talk about colder temperatures. So, really, what are negative absolute temperatures? Just continue reading because here I’m going to make an attempt in answering that question.

Temperature

At first, it’s good to have a reminding discussion about the concept of temperature itself. In thermodynamics, temperature is that quantity that is the same between two systems in thermal equilibrium i.e. two systems that aren’t isolated from each other but don’t exchange heat. But this definition won’t be useful here because of its thermodynamical nature. Thermodynamics treats macroscopic systems regardless of their microscopic structure (that’s why you don’t need to change thermodynamics because of QM) but here we’ll have to work with the microscopic point of view. So we should consider other definitions of temperature. One definition that we’ll find useful, is through the equation $\frac 1 T \equiv \frac{d S}{dE}$ where S is the system’s entropy function and E is its internal energy. Also, it’ll be sometimes useful to consider the association of temperature with the average energy of the constituents of the system distributed among their degrees of freedom.

 The Boltzmann factor

Now let’s say we want to talk about the statistical mechanics of a system of particles. From QM, we know that each particle can only be in discrete energy levels and can have no energy in between. The relative number of particles occupying the energy levels $\varepsilon_1$ and $\varepsilon_2$, is given by the Boltzmann factor $\frac{n_2}{n_1}= \exp\left(-\frac{\varepsilon_2-\varepsilon_1}{kT}\right)$. In systems where there is an infinite number of energy levels (we always take it for granted that there is a ground state which bounds the energy from below, so the “energy ladder” actually continues to infinity only upwards), as particles gain more energy, they climb up the ladder and theoretically, this continues as long as you keep injecting energy to the system. There can be no equilibrium state in which the number of particles in a given energy state, is smaller than that of an energy level above it. Combining this fact with the form of Boltzmann factor given above, we understand that in such systems the temperature can only be positive.

 A simple model

But what about a system with a finite number of energy levels? For simplicity, I’m going to consider a system with two energy levels $\varepsilon_1$ and $\varepsilon_2$ and a constant number of particles n.  So we have $n=n_1+n_2$ and $\varepsilon=n_1\varepsilon_1+n_2\varepsilon_2$. Note that if we assume the energy to stay constant too, there will be no dynamics because the total energy and the number of particles will uniquely determine the state of the system. So the system is assumed only to have a constant number of particles but it can exchange energy with its surroundings. Now using the definition of entropy ($S=k \ln \Omega$) and noting that $\Omega=\frac{n!}{n_1!n_2!}$, with a little bit of algebra along with Stirling’s approximation, we can find that

$S=k \left[ n \ln n-\frac{\varepsilon-n\varepsilon_1}{\varepsilon_2-\varepsilon_1}\ln{(\frac{\varepsilon-n\varepsilon_1}{\varepsilon_2-\varepsilon_1})}+\frac{\varepsilon-n\varepsilon_2}{\varepsilon_2-\varepsilon_1} \ln{(-\frac{\varepsilon-n\varepsilon_2}{\varepsilon_2-\varepsilon_1})} \right]$.

Unlike the usual entropy functions where the increase in energy will always result in an increase of entropy, this function has a maximum at $\varepsilon=\frac n 2 (\varepsilon_1+\varepsilon_2)$ which shows that regardless of the amount of internal energy of the system, the dynamics will lead it to a point where the two energy levels have an equal number of particles in them. Now considering the Boltzmann factor, we find that this means the equilibrium temperature of this system is $T=\pm \infty$ (the reason for $\pm$ will become clear in a moment). Now let’s see what happens when we consider the temperature of this system. Taking the derivative of the above function w.r.t. energy, we find the temperature of the system to be $T= \frac{\varepsilon_2-\varepsilon_1}{k} \left[\ln\left(- \frac{\varepsilon-n\varepsilon_2}{\varepsilon-n\varepsilon_1} \right) \right]^{-1}$. This function tells us that the temperature is positive for $\varepsilon<\frac n 2 (\varepsilon_1+\varepsilon_2 )$ and negative for $\varepsilon>\frac n 2 (\varepsilon_1+\varepsilon_2 )$. But because the equilibrium energy is $\varepsilon=\frac n 2 (\varepsilon_1+\varepsilon_2)$, the dynamics of the system will lead it from positive temperatures to $T=+\infty$ and from negative temperatures to $T=-\infty$. Note that the system has no internal dynamics and it’s only the interaction with the environment that is leading it to equilibrium. But of course, it’s only a property of this simple model. You can do the same calculations for a three-level system which will be more interesting.

Negative absolute temperatures

From the simple model discussed above, it’s easy to see that negative temperature states are actually states with higher energy than positive temperature states. Also, the Boltzmann factor shows us that in such states, there are more particles in higher energy levels than the lower ones. What actually happens is that as the particles gain energy and climb up the ladder, at some point, they reach the highest energy level and there is no higher energy to climb to. So any energy added further just takes more particles from lower levels to higher levels. At some point, the number of particles in different levels may become equal which means the temperature is infinity. Note that there is no difference between plus or minus infinity and we should consider these two to be identical. If energy is again injected, the temperature rises from $-\infty$ and goes towards 0 from the negative side but again 0 itself can’t be reached. So if we want to have an absolute temperature axis ordered from coldest to hottest, we should have something like the below:

$0^+ , 1 , 2 , … , 273.15 , … , + \infty , -\infty , … , – 273.15 , … , -2 , -1 , 0^-$

 

From the above information, it seems more natural to use $\beta \equiv \ – \frac{1}{kT}$ as temperature. Because it takes both infinities to a single point($\beta=0$) and also sorts out different temperatures in the order of increasing energy. Another advantage is that this way because $0^+$ goes to $-\infty$ and $0^-$ goes to $+\infty$, you can get the two zeroes as far as possible from each other which is the right way of thinking about them. Also this way, it’s easier to think that you can’t reach these temperatures.

 Experimental realization

Now that we have some good theoretical knowledge about negative absolute temperatures, it’s time to talk about the experimental side. How can we have systems with a finite number of energy levels in the lab? Experimentalists use a trick to acquire such systems. The point is, although the total energy of any system forms a ladder that is only bounded from below, different degrees of freedom have different energy ladders. There are some degrees of freedom that actually have upper bounds too, like spin degrees of freedom which will have different energies in the presence of a magnetic field. Now what experimentalists do, is that they prepare the situation for the system somehow that the energy exchange between the doubly-bounded degree of freedom and other degrees of freedom becomes so much slower than the dynamics of the doubly-bounded degree of freedom itself. Also, the dynamics of other degrees of freedom become slower too. So in this way, they can think of the dynamics of the system being completely due to the doubly-bounded degree of freedom.

After this discussion, you may now expect me to talk about the modifications that we need to make to thermodynamics(especially the 2nd law) because of the existence of negative absolute temperatures. Of course, you can find some sources which actually do that. But here I don’t want to do this because as far as I know, physicists still can’t place such systems in a heat engine along with more conventional systems and so I prefer not to speculate about this issue.

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