Quantum gas goes below absolute zero
I don't really like the misleading headline, although it seems as though the actual paper does it as well (<a href="http://arxiv.org/abs/1211.0545" rel="nofollow">http://arxiv.org/abs/1211.0545</a>).  Temperature is conventionally defined as the average kinetic energy of a collection of particles.<p>However, in statistical mechanics, you derive thermodynamic relations from the principles of quantum mechanics.  It turns out there is a term β which for most purposes seems to match up with 1/kT, where k is the Boltzmann constant and T is classical temperature.  According to this article, there are some occasions where "conventional" T and quantum mechanical T do not match up.<p>The most famous relationship in statistical mechanics is:<p>S = k ln Ω<p>S is entropy.  Ω is the partition function for the microcanonical ensemble, and it corresponds to the number of quantum states available to a system. (In English: Ω is the number of different configurations that a system of particles can possible have.)<p>Another relation that can be derived is:<p>1/kT = ∂ln(Ω)/ ∂E<p>Substituting the first equation into the second, we get<p>1/T = ∂S/∂E<p>So what would a negative temperature imply?  That increasing energy leads to decreasing entropy.  My guess is that these researchers have created some system that does exactly that.