Temperature That ClicksEpisode 5 / Negative temperature is hotter than infinity

"Temperature bottoms out at 0 K and has no ceiling" ── both halves are wrong. Make β the quantity and the whole range lines up on one axis

Negative temperature is hotter than infinity In a system whose energy has a ceiling, \(T<0\) is realisable.
And it is not "cold" ── it is hotter than any positive temperature,
in the sense that on contact it is always the one that gives energy away.

Tools you'll need: Episode 2's \(\beta=d\ln\Omega/dE\), Episode 3's two-level system, logarithms The heart of this episode: the real quantity is 1/T, not T

Temperature has a floor (absolute zero) and no ceiling ── that is what we are taught. Neither half is quite right. As Episode 2 showed, the real quantity is not \(T\) but \(\beta=1/k_BT\): "gain one unit of energy, and by what factor does the count of ways multiply?" For ordinary matter more energy always means more ways, so \(\beta>0\). But in a system whose energy has a ceiling, the story changes ── near the ceiling, taking more energy makes the count of ways shrink. Then \(\beta<0\), i.e. \(T<0\). And that is not "colder than absolute zero" ── it is hotter than infinite temperature. It was made in nuclear spins in 1951 and in the motional degrees of freedom of cold atoms in 2013.

01Ordinary systems have \(\beta>0\) because they have no ceiling

A gas molecule can move arbitrarily fast. Its energy has no upper bound. So the more energy you add, the more combinations of velocities are possible ── the count of ways \(\Omega\) keeps growing.

Recall the definition \(\beta=d\ln\Omega/dE\) from Episode 2: if \(\Omega\) keeps growing, \(\beta\) is always positive. That is why "ordinary matter always has positive temperature." It is not a law of physics but merely a consequence of energy having no ceiling.

02Put a ceiling on it and you go over a hill

Now consider a system with a ceiling. The simplest is the two-level system from Episode 3: each particle is either down (energy 0) or up (energy \(\Delta E\)). With \(N\) of them the total energy runs from 0 to \(N\Delta E\) ── a genuine ceiling.

Count the ways and you get a hill

The number of states with \(n\) of the \(N\) particles up is \(\Omega=\binom{N}{n}\), with energy \(E=n\Delta E\).

Statefraction up, pways Ωβ = dlnΩ/dE
all down (lowest energy)01 (one way)\(+\infty\)
half and half0.5maximum0
all up (highest energy)11 (one way)\(-\infty\)

The count of ways is hill-shaped. The summit is at half and half, and past it adding energy makes the count fall. On the right of the summit \(d\ln\Omega/dE<0\), i.e. \(\beta<0\).
Solving explicitly, the relation to the upper-level occupancy \(p\) is

$$\beta=\frac{1}{\Delta E}\ln\frac{1-p}{p}$$
The heart of this episode ── β is continuous, T jumps

Raise \(p\) smoothly from 0 to 1 and:

$$\beta:\ +\infty\ \longrightarrow\ 0\ \longrightarrow\ -\infty \qquad\text{(one road, continuous)}$$ $$T:\ 0^{+}\ \longrightarrow\ +\infty\ \ \Big|\ \ -\infty\ \longrightarrow\ 0^{-} \qquad\text{(jumps in the middle)}$$

Only \(T\) jumps; \(\beta\) walks straight through without incident. So the eerie point called "infinite temperature" is an apparent singularity created by writing a reciprocal. This is why Episode 2 insisted the real quantity is \(\beta\).

03Why "hotter than infinity"

Which of two things is hotter is decided by which way heat flows when you put them in contact. As Episode 2 showed, heat flows toward the larger \(\beta\) (the side whose count of ways explodes on receiving energy is the "cold" one).

So line them up:

cold ←βT→ hot
absolute zero\(+\infty\)\(0^+\)the coldest there is
room temperature40 /eV300 K
the centre of the Sunsmall positive\(10^7\) K
infinite temperature0\(\pm\infty\)still only halfway
negative temperaturesmall negative\(-10^7\) Khotter still
complete inversion (all up)\(-\infty\)\(0^-\)the hottest there is

They line up on one ordering: the smaller \(\beta\), the hotter. A negative-temperature system has \(\beta<0\), so against any positive-temperature system (\(\beta>0\)) it is always the one that gives energy away ── hence "hot." And \(T=0^-\) (complete inversion) is the maximum temperature.

The temperature scale was really a ring Take \(\beta\) as your axis and the whole range of temperature is a single line, \(-\infty<\beta<+\infty\). Rewrite it as \(T=1/k_B\beta\) and at \(\beta=0\) the value of \(T\) leaps to \(\pm\infty\), with the two ends \(T=0^{\pm}\) reappearing separately.
Intuitively, the temperature scale is closer to a ring than a line ── \(0^+ \to +\infty \to -\infty \to 0^-\) goes all the way round. Strictly it is an arc with both ends open, since \(0^+\) and \(0^-\) are not joined. All of the awkwardness comes from having written a reciprocal.

04Try it ── climb the entropy hill, then go over

On the left of the figure is the entropy \(S\) of the two-level system plotted against energy \(E\). It is hill-shaped. The slope of the tangent to this curve is exactly \(1/T=\partial S/\partial E\) ── Episode 2's definition itself.

Raise the upper-level occupancy \(p\) with the slider. The tangent tips over smoothly: rising (\(T>0\)) → horizontal (\(T=\infty\)) → falling (\(T<0\)). Nowhere is there a discontinuity. The right-hand panel marks the same state on both a \(\beta\) axis and a \(T\) axis ── you can see the \(\beta\) side pass through quietly while only the \(T\) side leaps from one end to the other.

Figure: left = the entropy S(E) of a two-level system (hill-shaped) with the tangent at the current state; its slope is 1/T. Right = the same state marked on a β axis (top) and a T axis (bottom). Raise p and β moves continuously while T jumps across ±∞
entropy S(E) tangent (slope = 1/T) current state

05It has actually been made

YearSystemMethod
1951nuclear spins (LiF crystal)
Purcell and Pound
Align the spins in a magnetic field, then reverse the field suddenly. The spins cannot follow, and find themselves biased toward the high-energy side ── a population inversion. The spin system alone holds \(T<0\) for several minutes
2013motional degrees of freedom of cold atoms
Braun et al.
An optical lattice puts a ceiling on the kinetic energy of potassium atoms (band structure), and the sign of the interaction is flipped. What was new is that negative temperature was achieved in motional degrees of freedom themselves
routinelylaser population inversionMore population upstairs than downstairs = formally \(T<0\). But it is not in thermal equilibrium, so whether to call it a "temperature" is a matter of taste (see the honest line below)

06So why does it not happen around us?

The answer goes back to section 01 ── everyday systems have no ceiling on kinetic energy. Molecules can always go faster, so \(\Omega\) keeps growing and \(\beta\) stays positive. To make a negative temperature you need to be able to treat a ceilinged degree of freedom on its own, cut off from the rest.

The nuclear-spin experiment worked because the spin system took minutes to exchange heat with the lattice (= the degree of freedom with no ceiling). For those minutes, the spins alone could carry a negative temperature as an independent "system." Put the other way round, negative temperature is always a quasi-equilibrium state ── wait long enough and heat always leaks into the unbounded degrees of freedom and the temperature returns to positive.

◇ ◇ ◇
The honest line ── the debate over "negative temperature"

Established: that with \(1/T=\partial S/\partial E\) as the definition, \(T<0\) is definable in a system whose energy has a ceiling and whose entropy is hill-shaped in \(E\); that such a system always releases energy on contact with a positive-temperature system (= hotter than any positive temperature); the realisations in nuclear spins (Purcell–Pound 1951) and in the motional degrees of freedom of cold atoms (Braun et al. 2013); and that using \(\beta=1/k_BT\) makes the entire range of temperature a single continuous axis ── all standard physics.

What is debated: (1) Negative temperature holds only in quasi-equilibrium. Wait for relaxation with unbounded degrees of freedom (lattice vibrations, say) and it always returns to positive; the premise is that there is a time window in which the subsystem alone can be treated as equilibrated. (2) The answer depends on which entropy you adopt. The body uses the usual (Boltzmann) entropy \(S=k_B\ln\Omega\); adopt the Gibbs volume entropy instead and the temperature is never negative ── a dispute broke out around 2014 over whether negative temperature exists at all, and it has not fully settled. The majority position is the one taken here (negative temperature is physically meaningful), but do note that the claim is definition-dependent. (3) A laser's population inversion can formally be written as \(T<0\), but being far from thermal equilibrium, some are cautious about calling it a temperature. (4) The figure idealises \(N\) independent two-level systems and ignores interactions.

Exercises (solvable with this episode's ideas)
  1. Why can't you make a negative temperature in an ordinary gas?
    See the answer
    Because kinetic energy has no ceiling. Adding energy always increases the count of ways \(\Omega\), so \(\beta=d\ln\Omega/dE\) stays positive. Negative temperature requires a ceiling past which \(\Omega\) starts to fall.
  2. What does "negative temperature is hotter than infinity" mean?
    See the answer
    Heat flows toward the larger \(\beta\). Since \(\beta<0\) is smaller than any \(\beta>0\), a negative-temperature system is always the one that hands energy over. Hotter versus colder is defined by which side gives on contact, so in that sense it is the hottest.
  3. Why does \(T\) jump at \(\pm\infty\) while \(\beta\) does not?
    See the answer
    Because the real quantity is \(\beta\) and \(T=1/k_B\beta\) is its reciprocal. Zero is an unremarkable waypoint for \(\beta\), but taking a reciprocal makes it diverge. The singularity comes from the notation, not from the physics.
  4. Why doesn't a negative temperature last?
    See the answer
    Because the ceilinged degrees of freedom (spins) and the unbounded ones (lattice vibrations) eventually exchange heat. The moment it leaks, the whole system heads back to positive temperature. So negative temperature is a quasi-equilibrium phenomenon, requiring a separation of relaxation times.

Episode 5 summaryThe temperature scale was never a straight line

In a system whose energy has a ceiling, the count of ways \(\Omega\) is hill-shaped as a function of \(E\). Past the summit (half and half) \(d\ln\Omega/dE<0\), i.e. \(\beta<0\) and \(T<0\). For a two-level system \(\beta=\frac{1}{\Delta E}\ln\frac{1-p}{p}\), so you get there just by raising the occupancy \(p\) from 0 to 1.

And a negative temperature is not "cold" but hotter than any positive temperature ── heat flows toward the larger \(\beta\), so a \(\beta<0\) system is always the giver. Ordered from cold: \(0^+\to+\infty\ |\ -\infty\to0^-\), and only \(T\) jumps; \(\beta\) walks quietly through. Episode 2's insistence that "the real quantity is \(\beta\)" pays off here.

Realised in nuclear spins (1951) and cold atoms (2013). But it requires being able to isolate the ceilinged degrees of freedom, and waiting always returns it to positive temperature ── negative temperature is always a quasi-equilibrium phenomenon.

This document is Episode 5 of the "Temperature That Clicks" series, a reading piece for physics-loving high-schoolers and undergraduates. The statistical-mechanical definition of temperature \(1/T=\partial S/\partial E\); that in a system with a ceiling on energy (a two-level system) the entropy is hill-shaped in \(E\) and \(T<0\) on the high-energy side; that a negative-temperature system always releases energy to a positive-temperature one; the two-level relation \(\beta=\Delta E^{-1}\ln[(1-p)/p]\); and the realisations in nuclear spins (Purcell–Pound 1951) and in the motional degrees of freedom of cold atoms in an optical lattice (Braun et al. 2013) ── all established physics. That negative temperature holds only in quasi-equilibrium and relaxes back to positive temperature through unbounded degrees of freedom, that whether negative temperature is admissible depends on the choice of entropy (Boltzmann versus the Gibbs volume entropy) with a dispute around 2014 that has not fully settled, that opinions differ on calling a laser's population inversion a "temperature," and that the figure idealises independent two-level systems ── all spelled out in the body's "honest line." ── To print, use your browser's "Print" and "Save as PDF" (in the print version the slider and answers are frozen and hidden). Adjacent episodes: Episode 4, The second law only "almost" holds / Episode 6, Temperature is a period in imaginary time / Contents.

Print / make a PDF: ⌘+P (Ctrl+P on Windows). On screen, raising the upper-level occupancy tips the tangent from rising to horizontal to falling, and only T leaps across ±∞. "See the answer" opens each solution.