Below absolute zero

Is it possible to bring a temperature below absolute zero? And if so, what does, say, -1K look like? Surprisingly, negative temperatures are in fact possible, but only in some circumstances. To understand how, we need to get a better idea of what temperature actually means from a statistical thermodynamics point of view.

Suppose you have a system of $N$ particles$^{\dagger}$, each of which can have an energy level from a discrete list

$$
0 < E_1 < E_2 < E_3 < ...
$$

Now suppose that there are $N_1$ particles in state $E_1$, $N_2$ in $E_2$ and so on. Using combinatorics$^{\dagger_2}$ we can see that the number of ways this particular configuration can be achieved is given by

$$
\Omega = \frac{N!}{N_1!N_2!N_3!...}
$$

The next question is: if we know the total energy of the system $E_{total}$ can we work out the distribution of particles across the energy states? Well, we know the most likely distribution is that which can be achieved in the greatest number of ways. I.e. we're looking for $N_1, N_2,...$ that satisfies the following equations while maximizing the entropy $S = k_Bln(\Omega)$

 

$$
\begin{align}
\sum_k{N_k} - N &= 0 \\
\sum_k{N_kE_k} - E_{total}&= 0 \\
\end{align}
$$
 

Note that maximizing the entropy $S$ is the same as maximizing $\Omega$, but it makes later calculations easier as it allows us to apply Stirlings approximation  $ln(N!) \approx N\ ln(N) - N$.

So we need to maximize a function of the $N_k$ subject to two constraints.  This is a job for Lagrangian Multipliers.  To recap, this method says that finding a stationary point of $f(\overrightarrow{x})$ subject to $g(\overrightarrow{x}) = 0$ and $h(\overrightarrow{x}) = 0$ is equivalent to finding the $\overrightarrow{x}, \alpha, \beta$ that satisfy $g(\overrightarrow{x}) = 0$, $h(\overrightarrow{x}) = 0$, and $\partial L/\partial x_k = 0$ for all $k$, where $L(\overrightarrow{x},\alpha,\beta) = f(\overrightarrow{x}) - \alpha g(\overrightarrow{x}) - \beta h(\overrightarrow{x})$ ... a system of $n+2$ equations in $n+2$ variables$^{\dagger_3}$.

In our case, applying Stirling's approximation repeatedly to $L$, differentiating and simplifying, we get

$$
\frac{\partial L}{\partial N_k} = -ln(N_k) -\alpha -\beta E_k
$$
Setting the LHS to zero gives us
$$
N_k = e^{-\alpha}e^{-\beta E_k}
$$

This, sort of, gives us the distribution we were looking for.  The only problem is that we don't know what $\alpha$ and $\beta$ represent yet.  We can write $\alpha$ in terms of $\beta$ just by summing the above formula and rearranging:

$$
e^{-\alpha} = \frac{N}{\sum_i{e^{-\beta E_i}}}
$$

The next step is to show that, in fact, $\beta$ is just $1/k_BT$ where $T$ is the temperature.  To show this, note that we can rearrange the formula $dS=dQ/T$ to get

$$
T = \frac{dQ}{dS} = \frac{dQ}{d\beta} / \frac{dS}{d\beta}
$$

Now, it's long and tedious, but ultimately straightforward to differentiate the formulas for heat $Q = \sum_i{N_iE_i}$ and entropy $S = k_B\ ln(\Omega)$ to get 

$$
\begin{align}
\frac{dQ}{d\beta} &= -N\ var(E) \\
\frac{dS}{d\beta} &= -k_B\ \beta\ N\ var(E) \\
\end{align}
$$

 

And dividing one by the other gives

$$
T = \frac{1}{k_B\beta}
$$

Putting this all together we finally get a formula for $N_k$ in terms of just $N$ and $T$

$$
N_k =\frac{N}{\sum_i{e^{-\frac{E_i}{k_BT}}}} e^{-\frac{E_k}{k_BT}}
$$

Now of course if the system has an infinite number of energy levels $E_k$ it must follow that only positive temperatures $T$ are possible.  With a negative temperature the numbers of particles in each successive level is higher than in the last and the total energy $Q$ in the system would be infinite!  However, if there are only a finite number of energy levels there is nothing wrong with inserting a negative $T$ in the formula above.  In fact the negative temperatures correspond to systems with more internal energy than the positive temperatures do.

 

 

FOOTNOTES

• ($\dagger$) In fact it would be more correct to say $N$ degrees of freedom where the components of kinetic energy in the $x$,$y$, and $z$ directions all count as separate degrees of freedom, as do components of rotational energy, as well as a particles internal excitation state. However if movement is not possible then degrees of freedom and number of particles might be the same thing.
• ($\dagger_2$) To see why imagine an initial placement where $N_1$ particles have energy $E_1$ and so on. Then note you can permute those $N$ particles $N!$ ways without changing the number in each energy level. However, every genuinely unique assignment of particles to energy levels would be duplicated $N_1!N_2!...$ times since permuting particles within the same energy state would result in the same assignment.
• ($\dagger_3$) This is easy to prove (hint: consider an arbitrary coordinate system of the constrained space $\xi_i = \xi_i(\overrightarrow{x})$ and apply the chain rule).