Why a Negative Absolute Temperature Is Hotter than Any Positive One — Epoche C1
In 1951 Edward Purcell and Robert Pound took a crystal of lithium fluoride out of a strong magnet and reversed the weak field at its new location faster than the nuclear spins inside the crystal could turn to follow. For the next several minutes the spins did not absorb radio-frequency power at their resonant frequency; they emitted it. Purcell and Pound described that state with a phrase that has been provoking objections ever since: the spin system, they said, was at a negative absolute temperature. The phrase looks like a category error. The third law of thermodynamics holds that absolute zero cannot be reached in finitely many steps, so a temperature below zero seems either impossible or an empty formalism. The confusion is real but sits in the wrong place: not in the physics, which follows in two lines from the definition of temperature, but in the habit of treating $T$, rather than its reciprocal, as the natural variable of thermodynamics. Correct that habit and a negative temperature turns out not to be colder than absolute zero at all. It is hotter than every positive temperature there is. The definition that does all the work Because the whole argument is a consequence of how temperature is defined, the definition has to be set out first, together with the reason it is the right one. In statistical mechanics the entropy of a system is fixed by counting: $S = k_\mathrm{B}\ln\Omega$, where $\Omega$ is the number of microstates — complete specifications of every microscopic degree of freedom — compatible with the system's total energy $E$, and $k_\mathrm{B}$ is Boltzmann's constant. Temperature is then not an extra ingredient. The fundamental relation $dE = T\,dS$ at fixed volume and particle number rearranges to $$\frac{1}{T} \;=\; \left(\frac{\partial S}{\partial E}\right)_{N,V},$$ so that $1/T$ is the entropy a system gains per unit of energy added to it. That definition earns its place by one short argument, needed again later. Put two systems in contact inside an isolating wall, so that they exchange energy but the total $E = E_1 + E_2$ is fixed. Entropy is additive, so $S_\mathrm{tot}(E_1) = S_1(E_1) + S_2(E - E_1)$, and equilibrium is the division that maximises it. Setting $dS_\mathrm{tot}/dE_1 = 0$ gives $\partial S_1/\partial E_1 = \partial S_2/\partial E_2$, that is, $1/T_1 = 1/T_2$. Temperature is the quantity that equalises across a diathermal wall — but what the calculation equalises is the slope $\partial S/\partial E$. The reciprocal is our addition. For ordinary matter the slope is positive and the distinction never arises. Adding energy to a gas or a solid always opens up more microstates, because a molecule can always move faster and kinetic energy has no ceiling; $\Omega$ grows without bound, $S$ rises with $E$, and $T > 0$ follows. The assumption doing the work is almost never stated: that the energy spectrum is unbounded above. Everything below turns on removing it. A ladder with a top rung The simplest system with a ceiling is a set of two-level spins, and working it through completely shows where the negative slope comes from. Take $N$ nuclear spins in a magnetic field, each with two available energies, $0$ and $\varepsilon$, where $\varepsilon$ is the Zeeman splitting — the energy difference between a spin aligned with the field and one aligned against it, proportional to the field strength. The total energy is $E = n\varepsilon$ if $n$ spins are excited, and it runs from $0$ to a hard maximum $N\varepsilon$. The ladder has a top rung. The microstate count is a binomial coefficient, because a microstate is settled by saying which $n$ of the $N$ spins are the excited ones: $$\Omega(n) \;=\; \binom{N}{n} \;=\; \frac{N!}{n!\,(N-n)!}.$$ Its maximum is at half filling, and one line shows why. The ratio of successive counts is $\Omega(n+1)/\Omega(n) = (N-n)/(n+1)$, which exceeds $1$ precisely when $n < (N-1)/2$. So the count climbs while fewer than half the spins are excited and falls thereafter, reaching $\Omega = 1$ at $n = N$: there is exactly one way for every spin to be excited, just as there is exactly one way for none to be. To get the slope, write $x = n/N$ for the excited fraction and use Stirling's approximation $\ln N! = N\ln N - N$ up to terms of order $\ln N$, negligible for the $10^{20}$-odd spins in a crystal. Then $$\ln\Omega = N\ln N - n\ln n - (N-n)\ln(N-n) = -N\left[x\ln x + (1-x)\ln(1-x)\right],$$ the $N\ln N$ terms cancelling when $n$ and $N-n$ are written as $xN$ and $(1-x)N$. Since $E = xN\varepsilon$, differentiating $S = k_\mathrm{B}\ln\Omega$ with respect to $E$ by the chain rule gives $$\frac{1}{T} \;=\; \frac{\partial S}{\partial E} \;=\; \frac{1}{N\varepsilon}\,\frac{\partial S}{\partial x} \;=\; \frac{k_\mathrm{B}}{\varepsilon}\,\ln\!\left(\frac{1-x}{x}\right).$$ Read off the three cases. For $x < 1/2$ the logarithm is positive and $T > 0$. At $x = 1/2$ it vanishes, so $1/T = 0$: entropy is at its maximum, $S = Nk_\mathrm{B}\ln 2$, which for a mole of spins is $R\ln 2 \approx 8.314 \times 0.693 \approx 5.76\,\mathrm{J\,K^{-1}\,mol^{-1}}$. For $x > 1/2$ the logarithm is negative and so is $T$. The same equation, inverted, delivers the population statement that is usually quoted alongside this one as though it were independent evidence. Exponentiating gives $x/(1-x) = e^{-\varepsilon/k_\mathrm{B}T}$, and since $x/(1-x) = n/(N-n)$ is the ratio of upper to lower occupation, this is the Boltzmann distribution for a two-level system, derived rather than assumed. For $T > 0$ the upper level is the less occupied; as $T \to \infty$ the ratio tends to $1$; for $T < 0$ the exponent changes sign and the upper level is the more occupied. That condition has a familiar name in another part of physics: population inversion, the state a laser or maser medium must be driven into before it can amplify rather than absorb. A negative-temperature spin system and an inverted laser medium are the same thing described in two vocabularies. What Purcell and Po