# Spontaneous conversion of heat into work at negative temperatures

Consider a heavy macroscopic object moving in a gas. Friction causes its kinetic energy to be converted into heat. Thermodynamically, there is (effectively) no entropy associated with the kinetic energy because all the energy is concentrated in a single degree of freedom. Therfore, if an amount $J$ of energy is converted from kinetic energy into heat, the total entropy change is $J/T$, so we can see that this is a spontaneous process.

But now consider an object moving relative to a gas with negative temperature. Such a thing has been created in the laboratory, so this is not just idle theoretical speculation. If an amount $J$ of kinetic energy gets converted into heat, the total entropy change is still $J/T$, but now this is negative. This seems to mean that the opposite process - conversion of heat into kinetic energy, accelerating the object - would be spontaneous.

This generalises to all other processes that convert work into heat. For example, performing the Joule heating experiment with a negative-temperature gas should cause the paddle to turn, and negative-temperature gas flowing through a pipe should experience an accelerating force rather than a decelerating one. Just as superfluids have zero viscosity, it seems that negative-temperature fluids must have negative viscosity.

I realise that this does not lead to perpetual motion. As heat is converted into work the inverse temperature ($1/T$) will increase until it reaches zero. But what does look odd is that in some ways the arrow of time appears to be reversed.

I realise that experimentally we're very far from being able to produce the macroscopic quantities of negative-temperature fluids that would be required in order to observe these things. But is it possible in principle? And if it is, would we actually see the phenomena I described, or is there some fundamental reason why they wouldn't happen after all? And has such a connection between negative temperatures and the arrow of time been discussed or debated in the literature?

• I stumbled on this video some time ago describing about the paper you are mentioning. youtube.com/watch?v=yTeBUpR17Rw&noredirect=1 Maybe, it might be of some use to you. – user1966726 Oct 18 '13 at 9:40
• Thanks! I stumbled upon the same video the other day as well. After thinking about it some more, I think the spontaneous conversion of heat into work is basically the reason why mechanical systems with negative temperature are so unstable. – Nathaniel Oct 18 '13 at 11:00

## 2 Answers

What you describe is not possible in principle.

The temperature of a macroscopic system is defined by 1/T = dS/dE .
But S = k.Log W where W is the number of microscopic states so that we have :
1/T = k/W . dW/dE

Now the processes defining the temperature for usual systems are translation (gases and liquids) and vibration (solids) and for both dW/dE > 0 what explains why the temperature in classical thermodynamics is always positive.
However there is also the magnetic dipole energy and in this case when one applies a magnetic field (dE>0) the dipoles align with the field and dW<0 what means that the "magnetic" temperature is negative.

The above equations are not sufficient to define a temperature of a macroscopic system, energy equipartition is necessary. The latter is given in thermal equilibrium.
The reason why energy equipartition is necessary is that if it was not the case, the different degrees of freedom would have different 1/W . dW/dE thus different temperatures and there would not exist a unique temperature for the macroscopic system.
This is a well know phenomenon for low density gases where the statistics is no more Maxwell Boltzmann and for which one has to define 2 different temperatures - vibrational and translational. The system in this case has no more a well defined temperature - its behaviour must be studied by looking at detailed local interactions.

In a macroscopic system at a very low temperature in equilibrium we would have : translational temperature = vibrational temperature ~ 0. If such a system had only 2 (or N) possible spin states, then applying a magnetic field would bring the system out of equilibrium and for a (very) brief time we would have translational temperature = vibrational temperature ~ 0 and spin temperature < 0.
Even if a unique temperature is no more defined, one could say that the system as a whole has a kind of "non equilibrium negative" temperature.

Now if you move your macroscopic solid (assumed to be at T ~0) in such a system, you would immediately increase the number of microscopic translation and vibration states by collisions ("friction") what would increase the translational and vibrational temperatures. These degrees of freedom would in turn interact with the spin and increase the number of spin states e.g increase the spin temperature.

Finally after a very short time at equilibrium you would have again translational temperature = vibrational temperature = spin temperature > 0. And of course nothing special happened with the arrow of time.

• Right, but in the paper I linked (nature.com/news/quantum-gas-goes-below-absolute-zero-1.12146) it's not just the spin degrees of freedom that have a negative temperature, it's the motion degrees of freedom as well. – Nathaniel Jun 20 '13 at 8:45
• It is not a paper but a news article which makes several incorrect statements. dW/dE can only be negative if W presents an extremum. W of translation and vibration is monotonically increasing with energy e.g when E increases, W increases and W has not an extremum. Therefore for these processes dW/dE > 0 for any E. Only spin states present an extremum for W - f.ex for a 2 states system, W is maximum when half the spins is up and half is down. QED. – Stan Won Jun 24 '13 at 8:24
• Sorry, the actual paper is here sciencemag.org/content/339/6115/52 . I repeat: in this system, it is the motional degrees of freedom that have a negative temperature, not the spin ones. This is a Science paper, it's not flaky stuff. – Nathaniel Jun 24 '13 at 8:34
• The paper you mention is behind paywall so I can't say if your interpretation is or is not flaky stuff. What is clear and well known is that T may only be <0 if dW/dE has an extremum. Free translation has no extremum so dW/dE > 0 AND T>0. For explanation why, just read my answer or a paper explaining what negative temperature means in dynamics. My answer was : your experiment is impossible. – Stan Won Jun 27 '13 at 11:19
• It is publised in Science magazine, one of the two most prestigious journals that exist. The abstract states "Here, we prepared a negative temperature state for motional degrees of freedom." From these two facts, you can deduce that they prepared a negative temperature state for motional degrees of freedom. – Nathaniel Jun 27 '13 at 12:30

If you are still around, Nathaniel, you may be interesting in this followup paper, and the summary in the same issue. In essence, the authors argue that to have a maximum energy, which is needed for negative-temperature population inversion, one must necessarily be in the microcanonical ensemble. But in this ensemble, the only definition of entropy that is thermodynamically consistent is not the notion that Braun et al use, but a different one that is identical in the thermodynamic limit but is always monotonically increasing with energy, so negative temperature is not possible.

I am still digesting this argument myself, so I will leave it to you and any other contributors to decide its merits. But at the very least I think it is reasonable to say that while there is no controversy over what the Braun experiment did, some care should be taken in extrapolating its implications in the way that you are.

• Thanks. I've seen this paper and skimmed it, though not digested it fully. But I think they have a rather different idea of what entropy is from me. For me the only entropy is -$\sum p_i \log p_i$, with everything else being a special case or an approximation. But I think they come from a more old-school perspective where that formula is seen as the special case, since they never even mention it in their work. – Nathaniel Feb 16 '14 at 0:20
• I could be wrong but I think their argument is that (according to them) to get a negative temperature you need a Boltzmann distribution ($p_i\propto e^{-\beta E_i}$) with negative $\beta$, and (according to them) there's no reason to expect a system to be in a Boltzmann distribution unless it's in contact with a heat bath at the same temperature, which would have to be negative in this case... – Nathaniel Feb 16 '14 at 0:40
• ... But even if I accepted those premises, I would ask them to consider a population inversion in a large system, and then consider the statistics of one small part of it. The conclusion that it's in a negative-$\beta$ Boltzmann distribution is unavoidable. – Nathaniel Feb 16 '14 at 0:40
• My natural inclination would also be to define entropy in that way, or actually as the Von Neumann entropy, $S=tr(\rho ln(\rho))$. They do address this directly in the supplementary information. – Rococo Feb 16 '14 at 5:56
• A brief quote: "Clearly, entropic information measures themselves are a matter of convention , and there exists a large number of diﬀerent entropies ...Although it may seem desirable to unify information theory and thermodynamic concepts for formal or aesthetic reasons, some reservation is in order  when such attempts cause mathematical inconsistencies and fail to produce reasonable results in the simplest analytically tractable cases." – Rococo Feb 16 '14 at 5:56