I took a statistical mechanics course last semester, and we learned that temperature is defined as:

$latex T = \frac{\partial E}{\partial S}&s=3$

where $latex E$ is energy and $latex S$ is entropy, or a measure of how many possible arrangements are available given a certain total energy. A positive temperature implies that a small positive change in energy causes a small positive change in entropy. In other words, adding a little bit more energy to a system causes there to be more ways to get that energy.

We almost always observe positive temperatures because normally, there are a huge number of possible energy levels for a system (infinite, actually). Adding a bit of energy allows the system to tap into previously inaccessible energy levels. This increases the number of possible ways we can rearrange the atoms in the energy levels to obtain the total energy. Thus, the entropy increases.

So then to have a negative temperature, we need to have a system where adding a bit of energy *decreases* the entropy. Luckily, there are some systems we can build that have very few energy levels. Let’s take, for example, a system of $latex N$ particles with only two energy levels: a ground state (zero energy) and an excited state with energy $latex \epsilon$.

Say we want the system to have an overall energy of 0. There is only one possible way to arrange the particles to get a total energy of 0, and that’s to put them all in the ground state. In other words, there’s only one available microstate for the system.

Next, we bump up the energy just a bit, to $latex \epsilon$. There’s only one way to get this total energy, and that’s to have one particle in the excited state with energy $latex \epsilon$ and the other $latex N – 1$ particles without any energy. But wait – we can choose any of the particles to be the one in the excited state. Thus, we have $latex N$ possible arrangements to get $latex E = \epsilon$.

You see that by giving a positive change to the energy, we have increased the entropy from zero to $latex \epsilon$. But this isn’t all. If we keep on adding little bits of energy $latex \epsilon$, we will reach a point where the entropy levels off. From some mathy arguments, we can discover that the system has maximum entropy when $latex E = N\epsilon/2$, or when we can put half of the particles in the ground state and half in the excited state. On second thought, let’s go into the mathy argument, since it’s not that bad.

For $latex E = \epsilon$, we need to pick one particle out of $latex N$ particles to be the one in the excited state. There are then $latex N$ number of possible ways to choose one particle.

For $latex E = 2\epsilon$, we need to pick two particles out of $latex N$ particles to be in the excited state. We could naively say that there are $latex N$ possible ways to pick the first particle and then $latex N-1$ ways to pick the second. However, since we don’t care about the order we choose the particles, two choices that could have been (particle #1) + (particle #2) and (particle #2) + (particle #1) are the same. That is, we overcount by two. Thus, there are $latex \frac{1}{2}N(N-1)$ possible ways to choose two particles.

And so on, and so forth. It turns out that mathematicians had studied this “choosing k out of N” thing way back in the 1600s, and they named it the binomial coefficient, or written out, $latex N \choose k$. The lumpy picture above is $latex 10 \choose k$ plotted as a function of $latex k$.

But back to negative temperature. Here, the energy essentially dictates how many particles we choose, and the entropy is proportional to the number of ways we can choose the particles, or the binomial coefficient plotted above: $latex N(k) \approx S(E)$. Thus, we see that entropy maxes out smack dab in the middle where $latex E = N\epsilon/2$, and the curve goes downhill past this point. This means that the slope $latex \frac{dS}{dE}$ goes below zero – adding a small bit of energy causes the entropy to *decrease*.

Voila. Negative temperature.

P.S.: The article and Wikipedia both mention that negative temperature is actually *hotter* than infinite temperature. We can see this from the binomial coefficient. The slope of the entropy curve goes to zero when entropy is maximized: $latex \frac{dS}{dE}|_{\mathrm{max} \, S} = 0$. Temperature is the reciprocal of this, so we see that at the peak, temperature is infinity. Below this point, temperature is positive since the slope is positive. Above this point, temperature becomes negative. Thus, we see that negative temperatures are actually *higher* than positive temperatures.

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I came across a slightly more advanced variation of this problem in a mechanics course earlier last semester, and for some reason I just couldn’t figure out how to find the kinetic energy. I found it instructive to go through a quick calculation…

The system is shown above – a simple rod swinging around a pivot that goes through one end. Assuming the usual things (the pivot at exactly the end of the rod, for example), there are two ways of finding the kinetic energy.

**1. Shifting the Moment of Inertia**

This is the classic way to approach this problem. Rotational kinetic energy is defined in terms of the moment of inertia about the pivot:

$latex T = \frac{1}{2}I_{\mathrm{pivot}}\dot{\theta}^2$

so we just use the parallel axis theorem to get the moment of inertia of a rod about its end,

$latex I_{\mathrm{pivot}} = I_{\mathrm{CM}} + M\left(\frac{\ell}{2}\right)^2$

where $latex \ell / 2$ is just the distance between the center of mass and the pivot, and $latex I_{\mathrm{CM}} = \frac{1}{12}M\ell^2$ can be easily found on Wikipedia, etc. or calculated. In fact, $latex I_{\mathrm{pivot}}$ is also readily available, but whatever.

Going through the calculation gives us

$latex I_{\mathrm{pivot}} = \frac{1}{3}M\ell^2$

which yields

$latex T = \frac{1}{6}M\ell^2\dot{\theta}^2$

**2. Center of Mass Velocity**

If you can’t use the parallel axis theorem (rare), the other method would be to decompose the kinetic energy into two parts: translational kinetic energy of the center of mass plus rotational kinetic energy of the rod about the center of mass. This is the same process used when calculating the kinetic energy of a spinning and translating rod (common problem in basic mechanics), for example. Thus, we write

$latex T = \frac{1}{2}I_{\mathrm{CM}}\dot{\theta}^2+\frac{1}{2}Mv_{\mathrm{CM}}^2$

where $latex I_{\mathrm{CM}} = \frac{1}{12}M\ell^2$ is the same as above, and the velocity of the center of mass is just

$latex v_{\mathrm{CM}} = \frac{1}{2}\ell\dot{\theta}$

or in other words, the angular velocity of the rod $latex \dot{\theta}$ times the distance from the pivot to the center of mass. As expected, this gives the same result

$latex T = \frac{1}{6}M\ell^2\dot{\theta}^2$

To be honest, this inaugural post was written partly because I wanted to play with LaTeX/Inkscape and partly to get something onto my blog so it wouldn’t look this shabby anymore.

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