Negative temperatures and negative absolute pressures are both possible in physical systems. Negative temperatures arise in (for example) populations of two-state systems, which have a maximum amount of internal energy they can contain; Negative pressure indicates a system in tension and is rare in fluids but common in solids.

My question is, if I start with a system with positive temperature and positive pressure, and pump energy into it until its temperature becomes negative, will its pressure become negative as well, or will it remain positive? Given a statistical ensemble with negative temperature it should be possible to define the pressure in the usual way, from $p/T = \partial S/ \partial V$, and my question is really about whether I should expect this quantity to be negative or positive.

A naïve argument that it would become negative is as follows: let's suppose that my system has the equation of state of an ideal gas ($pV=NRT$) but that it has a heat capacity that varies with $U$ in such a way as to allow negative temperatures. The only reason I'm making such a weird assumption is that I don't know what the actual equation of state should be - I'm not saying I think negative-temperature systems behave like gases. But if we make this assumption anyway then we can see that if $T$ becomes negative then $p$ must also be negative in order for the equation of state to hold, since $N$, $R$ and $V$ all remain positive.

Obviously no real system is likely to behave as described above. To reiterate, I'm not suggesting that a population of spins in a magnetic field would obey the gas equation. However, I would like to know whether real systems will typically behave in a qualitatively similar way, with the pressure and temperature both changing sign at the same time (and doing so by approaching positive infinity and then switching to negative infinity) or whether the temperature can change from positive to negative while the pressure remains finite and positive.


3 Answers 3


Negative temparature makes sense in statistical mechanics only if the associated Hamiltonian is bounded from above; otherwise the trace in the definition of the partition function does not exist.

In particular, a gas cannot have negative temperature since its Hamiltonian contains a kinetic energy term, which is not bounded from above. (Edit: Since the momentum operator is unbounded, the kinetic energy is unbounded, hence a Hamiltonian involving a kinetic term is unbounded. Thus the trace of $e^{-\beta H}$ is undefined for negative $\beta$. Thus the temperature must be positive. This is due to a purely mathematical property of the spectrum, independent of which energies can be realized in practice.)

Thus the question about its pressure at negative temperatures (and the argument with the ideal gas law) is moot.

Negative temperature is observed in certain spin systems living inside a crystal, and in systems of vortices in 2-dimensional plasmas. (This temperature is not the temperature of the crystal or plasma, but of the subsystem.) See, e.g., http://www.physics.umd.edu/courses/Phys404/Anlage_Spring11/Ramsey-1956-Thermodynamics%20and%20S.pdf

In these systems, the concept of pressure doesn't make much sense; at least it seems not to be used in this context. Formally, the pressure is of course defined. In a simple example, see Example 9.2.5 in my book
it changes sign with the temperature.

  • $\begingroup$ Please see the added emphasis in the question; I'm very well aware of what negative temperatures are. In theory, pressure should be definable for any statistical ensemble through $p/T = \partial S / \partial V$ - I just don't have any intuition about how that would behave for a spin system. $\endgroup$
    – N. Virgo
    Commented Oct 3, 2012 at 18:43
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    $\begingroup$ @Nathaniel: I don't know the behavior for a spin sytem, but I added a reference to a simple example where $P/T\ge 0$ for all temperatures. $\endgroup$ Commented Oct 4, 2012 at 11:26
  • $\begingroup$ Perfect, thanks - I think it's very likely that real systems will generally behave similarly to your two-state example. $\endgroup$
    – N. Virgo
    Commented Oct 4, 2012 at 11:36
  • $\begingroup$ Arnold, but physical systems are ultimately bound by the Hagedorn temperature, doesn't that imply that their Hamiltonians are also bounded? $\endgroup$
    – user56771
    Commented Oct 4, 2012 at 12:01
  • $\begingroup$ @user56771: No. See the edit to my answer. $\endgroup$ Commented Oct 4, 2012 at 13:51

A full answer to this question can be found in the supplementary material of this recent paper. The proof is very simple: $\partial S / \partial V$ has to be positive because if it weren't, the system could increase its entropy by spontaneously decreasing its volume, and therefore it wouldn't be stable. Since $\partial S / \partial V=p/T$, it follows that for a system to be in equilibrium at a negative temperature, it must also have a negative pressure. This argument requires the reasonable assumption that the system is contained within a rigid box.

Interestingly, the paper linked above is about the experimental realisation of a negative-temperature system using motional degrees of freedom, which in some ways does behave like a gas - so the weird assumptions in my original question weren't so far off the mark after all.


Negative temperature is a qualitative description of certain statistical ensembles. Those ensembles however are not thermal equilibrium states and not accessible by heating up a system. You cannot apply equilibrium thermodynamics to them either, so any gas law will not hold. So the answer is no, negative pressure and negative temperature are unrelated and no, you cannot heat up a system to negative temperatures and no, the ideal gas law does not apply.

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    $\begingroup$ They are in equilibrium; outside of equilibrium, the concept of temperature makes no sense. (Even in hydromechanics, local equilibrium is assumed!) $\endgroup$ Commented Oct 3, 2012 at 9:27
  • $\begingroup$ @ArnoldNeumaier, they are in an equilibrium, but not a thermal equilibrium. They have to rely on an external source of coherent energy, and as soon as you turn that off they equalize into a state real thermal equilibrium state which always has positive temperature. A real thermal equilibrium holds this equilibrium after being isolated. Just the mathematics is similar enough to real thermal equilibria that you can treat them the same way, but physically they're quite different. Specifically you cannot achieve negative temperatures by thermalization with a heat bath, or "heating up" $\endgroup$
    – A.O.Tell
    Commented Oct 3, 2012 at 12:56
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    $\begingroup$ @A.O.Tell the need to do that only arises from the difficulty of isolating the system. A perfectly isolated system with -ve temperature would be stable in the sense that its temperature would stay -ve. They're only unstable in the sense that they like to equilibriate with their surroundings, in much the same way as any other very very hot thing would. Also, please see the added emphasis in the question - I was very explicitly not suggesting the gas law would hold, but rather was using it as a way to approach the question. $\endgroup$
    – N. Virgo
    Commented Oct 3, 2012 at 18:48
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    $\begingroup$ @A.O.Tell: Wjhat is thre formal difference between equilibrium and thermal equilibrium? $\endgroup$ Commented Oct 4, 2012 at 11:29

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